2.1 Overview

We will discuss here the Virasoro constraints for stable maps. The constraints are equivalent to a procedure for removing the descendents of the canonical class. The procedures may be interpreted as series of the reactions (similar to the reactions discussed in the context of the $\mathrm {GW}/{\mathrm {PT}}$ descendent correspondence in [Reference Oblomkov, Okounkov and Pandharipande18, Section 3]). Our goal is to write the Virasoro constraints for Gromov–Witten theory in a form which is as close as possible to the Virasoro constraints of Conjecture 3 for stable pairs.

2.2 Gromov–Witten constraints: original form

The Virasoro constraints in Gromov–Witten theory were first proposedFootnote ¹⁵ in [Reference Eguchi, Hori and Xiong3]. We recall here the original form following [Reference Pandharipande20]. In Section 2.3, a reformulation which is more suitable for the $\mathrm {GW}/{\mathrm {PT}}$ correspondence will be presented.

In the discussion below, we fix a basis of $H^{*}(X)$ ,

(2.1) $$ \begin{align} \gamma_{0},\dots,\gamma_{r}\, , \quad \gamma_{i}\in H^{p_{i},q_{i}}(X)\, , \end{align} $$

for which $\gamma _{0}=1$ , $\gamma _{1}=c_{1}$ and $\gamma _{r}=[\operatorname {\mathrm {\mathsf {p}}}]$ . We assumeFootnote ¹⁶ $c_{1}\ne 0$ . We also fix a dual basis

$$ \begin{align*}\gamma^{\vee}_{0},\dots,\gamma^{\vee}_{r}\, , \quad \int_{X}\gamma_{i} \gamma_{j}^{\vee}=\delta_{ij}.\end{align*} $$

Footnote ¹⁷ The standard method of describing of the Virasoro constraints uses the generating function for the Gromov–Witten invariants (see [Reference Pandharipande20, section 4]):

$$ \begin{align*}F^{X}=\sum_{g\ge 0}u^{2g-2}\sum_{\beta\in H_{2}(X,\mathsf Z)} q^{\beta}\sum_{n\ge 0}\sum_{\stackrel{a_{1},\dots,a_{n}}{k_{1},\dots,k_{n}}}t_{k_{1}}^{a_{1}}\dots t_{k_{1}}^{a_{1}}\dots t_{k_{n}}^{a_{n}}\, \big\langle \tau_{k_{1}}(\gamma_{1})\dots\tau_{k_{n}}(\gamma_{n}) \big\rangle_{g,\beta}^{X,\mathrm{Con}},\end{align*} $$

where $\big \langle , \big \rangle ^{X,\mathrm {Con}}_{g,\beta }$ is the standard integral over stable maps with connected domains (and stable contracted components of all genera are permitted).

The degree $\beta =0$ summand $F_{0}^{X}$ of $F^{X}$ does not require knowledge of curves in X. We further split the degree 0 summand into summands of genus $g\le 1$ and genus $g\ge 2$ :

$$ \begin{align*}F^{X}_{0}=F^{X}_{0,g\le 1}+F^{X}_{0,g\ge 2}\, .\end{align*} $$

The $g\leq 1$ summand takes the form

$$ \begin{align*}F_{0,g\leq 1}^{X}=u^{-2}\sum_{i,j,k}\bigg(\frac{t_{0}^{i}t_{0}^{j}t_{0}^{k}}{3!}+\frac{t_{0}^{i}t_{0}^{j}t_{1}^{k}t_{0}^{0}}{2!}\bigg)\int_{X}\gamma_{i}\gamma_{j} \gamma_{k}-\sum_{i}\bigg(\frac{t_{0}^{i}}{24}+\frac{t_{1}^{i}t_{0}^{0}}{24}\bigg)\int_{X}\gamma_{i} c_{2}+ \ldots,\end{align*} $$

where the dots stand for terms divisible by $(t_{0}^{0})^{2}$ . The $g\geq 2$ summand $F^{X}_{0,g \ge 2}$ is determined by the string and dilaton equations from the constant maps contributions of [Reference Faber and Pandharipande4, Theorem 4].

Let $\widetilde {F}^{X}$ be the summand of $F^{X}$ with $\beta \neq 0$ . We define

$$ \begin{align*}Z_{0,*}^{X}=\exp(F_{0,*}^{X})\, ,\quad \widetilde{Z}^{X}=\exp(\widetilde{F}).\end{align*} $$

The Gromov–Witten bracket $\big \langle , \big \rangle ^{X,\mathrm {GW}}_{g,\beta }$ introduced in Section 1.6 corresponds to the partition function

$$ \begin{align*}Z^{X}_{0,g\le 1}\cdot \widetilde{Z}^{X} = \sum_{g\ge \mathbb{Z}}u^{2g-2}\sum_{\beta\in H_{2}(X,\mathsf Z)} q^{\beta}\sum_{n\ge 0}\sum_{\stackrel{a_{1},\dots,a_{n}}{k_{1},\dots,k_{n}}}t_{k_{1}}^{a_{1}}\dots t_{k_{1}}^{a_{1}}\dots t_{k_{n}}^{a_{n}}\, \big\langle \tau_{k_{1}}(\gamma_{1})\dots\tau_{k_{n}}(\gamma_{n}) \big\rangle_{g,\beta}^{X,\mathrm{GW}}.\end{align*} $$

The full partition function

$$ \begin{align*}Z^{X}=\exp(F^{X})=Z^{X}_{0,g\le 1} \cdot Z^{X}_{0,g \ge 2}\cdot \widetilde{Z}^{X}\,\end{align*} $$

corresponds to the standard disconnected Gromov–Witten bracket $\big \langle , \big \rangle ^{X,\bullet }_{g,\beta }$ ,

$$ \begin{align*}Z^{X}=\sum_{g\ge 0}u^{2g-2}\sum_{\beta\in H_{2}(X,\mathsf Z)} q^{\beta}\sum_{n\ge 0}\sum_{\stackrel{a_{1},\dots,a_{n}}{k_{1},\dots,k_{n}}}t_{k_{1}}^{a_{1}}\dots t_{k_{1}}^{a_{1}}\dots t_{k_{n}}^{a_{n}}\, \big\langle \tau_{k_{1}}(\gamma_{1})\dots\tau_{k_{n}}(\gamma_{n}) \big\rangle_{g,\beta}^{X,\bullet}.\end{align*} $$

The Virasoro operators $\mathrm {L}_{k}$ , $k\in \mathbb {Z}_{\ge -1}$ are differential operators which satisfy the Witt algebra relations,

$$ \begin{align*}[\mathrm{L}_{k},\mathrm{L}_{\ell}]=(k-\ell) \mathrm{L}_{k+\ell}.\end{align*} $$

The Virasoro conjecture [Reference Eguchi, Hori and Xiong3] states that the operators annihilate the partition function

(2.2) $$ \begin{align} \mathrm{L}_{k} \, Z^{X}=0 \,. \end{align} $$

For threefolds X, the operators are defined by

$$ \begin{align*} \mathrm{L}_{k}=& \sum_{m=0}^{\infty}\sum_{i=0}^{k+1}\bigg( [p_{a}+m-1]_{i}^{k}(C^{i})^{b}_{a}\tilde{t}^{a}_{m}\partial_{b,m+k-i}\\ &+ \frac{u^{2}}{2}(-1)^{m+1}[-p_{a}+1-m]_{i}^{k}(C^{i})^{ab}\partial_{a,m}\partial_{b,k-m-i-1} \bigg)\\ & +\frac{u^{-2}}{2}(C^{k+1})_{ab}t_{0}^{a}t^{b}_{0}\\ &-\frac{\delta_{k}}{24}\int_{X}c_{1}c_{2}, \end{align*} $$

where the Einstein conventions for summing over repeated indices are followed,

$$ \begin{align*}\tilde{t}_{m}^{a}=t^{a}_{m}-\delta_{a0}\delta_{m1}\, ,\quad \partial_{a,m}=\partial/\partial t_{m}^{a},\end{align*} $$

and $[x]^{k}_{j}=e_{k+1-j}(x,x+1,\dots ,x+k)$ .Footnote ¹⁸ The tensors in the equation are defined in terms of the dual basis

$$ \begin{align*}(C^{i})^{a}_{b}=\int_{X}\gamma_{a}^{\vee} c_{1}^{i}\gamma_{b}\, ,\quad (C^{i} )_{ab}=\int_{X}\gamma_{a}c_{1}^{i}\gamma_{b}\, ,\quad (C^{i})^{ab}=\int_{X}\gamma_{a}^{\vee} c_{1}^{i}\gamma_{b}^{\vee}.\end{align*} $$

2.3 Gromov–Witten constraints: correspondence form

We rewrite here the Virasoro constraints of Section 2 in the form most natural for the $\mathrm {GW}/{\mathrm {PT}}$ descendent correspondence. Since all of our results are for toric varieties, we specialize our discussion here to the case where X is a nonsingular projective threefold with only $(p,p)$ -cohomology.

We start by defining derivations $\mathrm {R}^{j}_{k}$ and quadratic differentials $\mathrm {B}^{k}$ on $\mathbb {D}_{\mathrm {GW}}^{X}$ by fixing the action on the generators:

• • The action of the derivation $\mathrm {R}^{j}_{k}$ on $\tau _{i}(\gamma )$ for $k\geq -1$ , $0\leq j \leq 3$ and $\gamma \in H^{2d}(X)$ is

  $$ \begin{align*}\mathrm{R}^{j}_{k}(\tau_{i}(\gamma))=[i+d-1]^{k}_{j}\, \tau_{k+i-j}(\gamma\cdot c_{1}^{j}),\end{align*} $$
  where $[x]^{k}_{j}=e_{k+1-j}(x,x+1,\dots ,x+k)$ and all terms $\tau _{\ell <-2}(\theta )$ are set to 0. As a special case,
  $$ \begin{align*}\mathrm{R}^{j}_{-1}(\tau_{i}(\gamma)) = \delta_{j}\, \tau_{i-1}(\gamma)\, .\\[-14pt]\end{align*} $$
  We will use the notation $\mathrm {R}_{k} = \sum _{j=0}^{3} \mathrm {R}_{k}^{j}\, .$

• • The action of the quadratic differential $\mathrm {B}^{k}$ on $\tau _{0}(\gamma )\tau _{0}(\gamma ^{\prime })$ is

  $$ \begin{align*}\mathrm{B}^{k}(\tau_{0}(\gamma)\tau_{0}(\gamma^{\prime}))=\int_{X}\gamma\gamma^{\prime} c_{1}^{k}\, .\\[-14pt]\end{align*} $$
  On all other quadratics terms, $\mathrm {B}^{k}$ acts by $0$ .

The differential operators $\mathrm {L}^{\mathrm {GW}}_{k}$ , for $k\geq -1$ , are then defined by the formula

$$ \begin{align*}\mathrm{L}_{k}^{\mathrm{GW}}= \mathrm{R}_{k}+\frac{u^{-2}}2\mathrm{B}^{k+1}+\frac{(\imath u)^{2}}{2}\mathrm{T}_{k}-\frac{\delta_{k}}{24}\int_{X}c_{1}c_{2}\, ,\\[-14pt] \end{align*} $$

where $\mathrm {T}_{k} = \sum _{j=0}^{3} \mathrm {T}_{k}^{j}$ and

(2.3) $$ \begin{align} \mathrm{T}^{j}_{k}=\sum_{m=-1}^{k-j+2}(-1)^{m+1}[2-m-d_{L}]_{j}^{k}\, :\tau_{m-1}\tau_{-m+k-j}(c_{1}^{j}):\ ,\\[-14pt]\nonumber \end{align} $$

where $d_{L}$ is the degree of the left term in the co-productFootnote ¹⁹ (as in Section 1.2). In formula (2.3), the symbol $::$ stands for the normal ordering convention: All negative descendents $\tau _{<0}(\gamma )$ are on the left of the positive descendents.

A calculation then yields the Virasoro bracket and the following bracket with $\tau _{k}(\mathsf {p})$ :

(2.4) $$ \begin{align} [\mathrm{L}^{\mathrm{GW}}_{n},\mathrm{L}^{\mathrm{GW}}_{k}]=(n-k)\, \mathrm{L}^{\mathrm{GW}}_{n+k}\, , \ \ \ \ \ \ [\mathrm{L}^{\mathrm{GW}}_{n},(k+1)!\, \tau_{k}(\mathsf{p})]= (k+n+2)!\, \tau_{n+k}(\mathsf{p})\,. \end{align} $$

Theorem 2.1 [Reference Givental8, Reference Iritani10]

Let X be a nonsingular projective toric threefold, and let $\beta \in H_{2}(X,\mathbb {Z})$ . For all $k\geq -1$ and $D\in \mathbb {D}^{X}_{\mathrm {GW}}$ , we have

$$ \begin{align*}\Big\langle \mathrm{L}^{\mathrm{GW}}_{k}(D) \Big\rangle_{\beta}^{X,\bullet}=0\, .\\[-14pt]\end{align*} $$

Theorem 2.1, which is exactly equivalent to constraints (2.2) for toric threefolds, was proven by Givental in two steps:

1. (i) Using the virtual localization formula of [Reference Graber and Pandharipande9], the Gromov–Witten theory of X is expressed in terms of graphs sums with descendent integrals over the moduli spaces of curves $\overline {M}_{g,n}$ at the vertices.

2. (ii) The Virasoro constraints, conjectured by Witten [Reference Witten33] for $\overline {M}_{g,n}$ and proven in [Reference Kontsevich12], are then used to establish the Virasoro constraints for X.

A second proof of Theorem 2.1, via the Givental–Teleman classificationFootnote ²⁰ of semisimple CohFTs, was given in [Reference Teleman30]. For varieties with nonsemisimple Gromov–Witten theory, the Virasoro constraints are known in very few cases.Footnote ²¹

2.4 Gromov–Witten constraints: stationary form

We rewrite the Virasoro constraints in Gromov–Witten theory of Section 2.3 in a form which preserves the algebra of stationary descendents,

$$ \begin{align*}\mathbb{D}^{X+}_{\mathrm{GW}}\subset \mathbb{D}^{X}_{\mathrm{GW}} \, .\\[-14pt]\end{align*} $$

We fix a basis (2.1) of the cohomology of X which satisfies the following further conventions. Let

$$ \begin{align*}\gamma_{1},\dots,\gamma_{s}\in H^{2}(X)\\[-14pt]\end{align*} $$

be a basis with $\gamma _{1}=c_{1}$ . Let

$$ \begin{align*}\gamma_{2s},\dots,\gamma_{s+1}\in H^{4}(X)\\[-14pt]\end{align*} $$

be a dual basis with respect to the Poincaré pairing. Let

$$ \begin{align*}\gamma_{0}=1\in H^{0}(X)\, ,\ \ \ \gamma_{2s+1}=\operatorname{\mathrm{\mathsf{p}}}\in H^{6}(X)\\[-14pt]\end{align*} $$

span the rest of the cohomology.Footnote ²² The Künneth decomposition of the diagonal is

$$ \begin{align*}\Delta=\sum_{i=0}^{2s+1}\gamma_{i}\otimes\gamma_{2s+1-i}\, .\\[-14pt]\end{align*} $$

Consider the term $\mathrm {T}_{k}$ . The only place for descendents of $1$ to appear in the operator $\mathrm {L}^{\mathrm {GW}}_{k}$ is in $\mathrm {T}_{k}^{0}$ . As most of the terms of $\mathrm {T}^{0}_{k}$ vanish by definition, we find

(2.5) $$ \begin{align} \frac{1}{2}\mathrm{T}^{0}_{k}=(k+1)!\, :\tau_{0}(1)\tau_{k-1}(\operatorname{\mathrm{\mathsf{p}}}):\,.\\[-14pt]\nonumber \end{align} $$

We denote the rest of the term by $\mathrm {T}^{\prime }_{k}$ ,

$$ \begin{align*}\mathrm{T}_{k} = \mathrm{T}^{\prime}_{k} + \mathrm{T}_{k}^{0}\, .\\[-14pt]\end{align*} $$

Inside the bracket $\langle , \rangle ^{X,\bullet }_{\beta }$ , the insertion $\tau _{0}(1)$ can be removed by the string equation (1.7). We are therefore led to define the operator

$$ \begin{align*}\mathcal{L}_{k}^{\mathrm{GW}}=\frac{(\imath u)^{2}}2\mathrm{T}^{\prime}_{k}+\mathrm{R}_{k} +\frac{u^{-2}}2\mathrm{B}^{k+1} +(\imath u)^{2}(k+1)!\,\mathrm{R}_{-1}\tau_{k-1}(\operatorname{\mathrm{\mathsf{p}}})\, ,\quad\ \ \mathrm{T}^{\prime}_{k}=\sum_{j>0}\mathrm{T}^{j}_{k}\, ,\\[-14pt]\end{align*} $$

where $\mathrm {R}_{k}=\sum _{j=0}^{3} \mathrm {R}_{k}^{j}$ and $\mathrm {R}_{-1}$ is the differentiation defined on the generators by

$$ \begin{align*}\mathrm{R}_{-1}\tau_{k}(\gamma)=\tau_{k-1}(\gamma)\, .\\[-14pt]\end{align*} $$

Inside the bracket $\langle , \rangle ^{X,\bullet }_{\beta }$ , we haveFootnote ²³

(2.6) $$ \begin{align} \mathcal{L}_{k}^{\mathrm{GW}}\, \stackrel{\langle,\rangle}{=}\, \widetilde{\mathrm{L}}_{k}^{\mathrm{GW}}+(\imath u)^{2}(1-\delta_{k})(k+1)!\, \widetilde{\mathrm{L}}_{-1}^{\mathrm{GW}}\tau_{k-1}(\operatorname{\mathrm{\mathsf{p}}})\, ,\\[-14pt]\nonumber \end{align} $$

where we have modified the Virasoro operators to exclude the descendents of $1$ :

(2.7) $$ \begin{align} \widetilde{\mathrm{L}}_{k}^{\mathrm{GW}}\, =\, \mathrm{L}_{k}^{\mathrm{GW}} - \frac{(\imath u)^{2}}{2} \mathrm{T}^{0}_{k} \, =\, \frac{(\imath u)^{2}}{2}\mathrm{T}^{\prime}_{k} + \mathrm{R}_{k}+\frac{u^{-2}}2\mathrm{B}^{k+1}-\frac{\delta_{k}}{24}\int_{X}c_{1}c_{2}\,. \end{align} $$

Though the operators $\mathcal {L}_{k}^{\mathrm {GW}}$ no longer satisfy the Virasoro bracket, the operators $\mathcal {L}_{k}^{\mathrm {GW}}$ preserve the subalgebra $\mathbb {D}_{\mathrm {GW}}^{X+}\subset \mathbb {D}_{\mathrm {GW}}^{X}$ .

Proposition 5. Let X be a nonsingular projective toric threefold, and let $\beta \in H_{2}(X,\mathbb {Z})$ . For all $k\geq -1$ and $D\in \mathbb {D}^{X+}_{\mathrm {GW}\circ }$ , we have

$$ \begin{align*}\Big\langle \mathcal{L}^{\mathrm{GW}}_{k}(D) \Big\rangle_{\beta}^{X,\bullet}=0.\end{align*} $$

Proof. The case $k=0$ follows because

$$ \begin{align*}\mathcal{L}_{0}^{\mathrm{GW}}-\mathrm{L}_{0}^{\mathrm{GW}}=\mathrm{T}_{0}^{0}=2:\tau_{0}(1)\tau_{-1}(\operatorname{\mathrm{\mathsf{p}}}):\end{align*} $$

and $\Big \langle \mathrm {T}_{0}^{0}\dots \Big \rangle ^{X,\bullet }_{\beta }=0$ . For the other case the argument is below.

Using equations (2.6) and (2.7), we have

(2.8) $$ \begin{align} \Big\langle \mathcal{L}^{\mathrm{GW}}_{k}(D) \Big\rangle_{\beta}^{X,\bullet} &= \Big\langle \mathrm{L}_{k}^{\mathrm{GW}}(D) +(\imath u)^{2}(k+1)!\, \mathrm{L}_{-1}^{\mathrm{GW}}(\tau_{k-1}(\operatorname{\mathrm{\mathsf{p}}}) D) \Big \rangle^{X,\bullet}_{\beta}\nonumber\\ &\quad -\frac{(\imath u)^{2}}{2} \Big \langle \mathrm{T}^{0}_{k}(D) +(\imath u)^{2}(k+1)!\, \mathrm{T}^{0}_{-1} (\tau_{k-1}(\operatorname{\mathrm{\mathsf{p}}})D) \Big \rangle^{X,\bullet}_{\beta}\,. \end{align} $$

The first bracket on the right side of equation (2.8) vanishes by Theorem 2.1. We can write the second bracket on the right as

$$ \begin{align*} &\frac{(\imath u)^{2}}{2}\Big \langle \mathrm{T}^{0}_{k}(D) +(\imath u)^{2}(k+1)!\, \mathrm{T}^{0}_{-1} (\tau_{k-1}(\operatorname{\mathrm{\mathsf{p}}})D) \Big \rangle^{X,\bullet}_{\beta} \\ &\qquad\qquad\qquad\qquad ={(\imath u)^{2}}\Big \langle (k+1)!\, \tau_{0}(1) \tau_{k-1}(\operatorname{\mathrm{\mathsf{p}}}) D +(\imath u)^{2}(k+1)!\, \tau_{0}(1)\tau_{-2}(\operatorname{\mathrm{\mathsf{p}}}) \tau_{k-1}(\operatorname{\mathrm{\mathsf{p}}})D \Big \rangle^{X,\bullet}_{\beta} \end{align*} $$

using equation (2.5). The right side of the above equation, after applying the commutator (1.8), is

$$ \begin{align*}{(\imath u)^{2}}\Big \langle (k+1)!\, \tau_{0}(1) \tau_{k-1}(\operatorname{\mathrm{\mathsf{p}}}) D +(\imath u)^{2}(k+1)!\, \tau_{-2}(\operatorname{\mathrm{\mathsf{p}}}) \tau_{0}(1) \tau_{k-1}(\operatorname{\mathrm{\mathsf{p}}})D \Big \rangle^{X,\bullet}_{\beta},\end{align*} $$

which vanishes after applying equation (1.9).

In our study of the $\mathrm {GW}/{\mathrm {PT}}$ descendent correspondence, we are interested in the Gromov–Witten bracket $\big \langle , \big \rangle ^{X,\mathrm {GW}}_{g,\beta }$ of Section 1.6 instead of the standard disconnected bracket $\big \langle , \big \rangle ^{X,\bullet }_{g,\beta }$ . Therefore, the following result is important for our study.

Proposition 6. Let X be a nonsingular projective toric threefold, and let $\beta \in H_{2}(X,\mathbb {Z})$ . For all $k\geq -1$ and $D\in \mathbb {D}^{X+}_{\mathrm {GW}\circ }$ , we have

$$ \begin{align*}\Big\langle \mathcal{L}^{\mathrm{GW}}_{k}(D) \Big\rangle_{\beta}^{X,\mathrm{GW}}=0.\end{align*} $$

Proof. Since $\mathcal {L}^{\mathrm {GW}}_{k}$ preserves $\mathbb {D}^{X+}_{\mathrm {GW}}$ , we have

$$ \begin{align*}\mathcal{L}^{\mathrm{GW}}_{k}(D) \in \mathbb{D}^{X+}_{\mathrm{GW}}.\end{align*} $$

Since the Gromov–Witten invariants corresponding to collapsed connected components of genus at least 2 always vanish in the presence of stationary descendents,

$$ \begin{align*}\Big\langle \mathcal{L}^{\mathrm{GW}}_{k}(D) \Big\rangle_{\beta}^{X,\bullet} = Z^{X}_{0,g\geq 2}\Big{|}_{\{t^{i}_{k}=0\}} \cdot \Big\langle \mathcal{L}^{\mathrm{GW}}_{k}(D) \Big\rangle_{\beta}^{X,\mathrm{GW}} .\end{align*} $$

Since $\Big \langle \mathcal {L}^{\mathrm {GW}}_{k}(D) \Big \rangle _{\beta }^{X,\bullet }$ vanishes by Proposition 5 and

$$ \begin{align*}Z^{X}_{0,g\geq 2}\Big{|}_{\{t^{i}_{k}=0\}} = \exp\left(\sum_{g=2}^{\infty} (-1)^{g} u^{2g-2}\, \frac{\chi(X)}{2} \, \int_{\overline{M}_{g}}\lambda_{g-1}^{3}\right)\\[-9pt]\end{align*} $$

is invertible,Footnote ²⁴ $\Big \langle \mathcal {L}^{\mathrm {GW}}_{k}(D) \Big \rangle _{\beta }^{X,\mathrm {GW}}$ also vanishes.

3.1 Intertwining property

We have already defined the operators $\mathrm {L}_{k}^{\mathrm {PT}}$ and $\mathcal {L}^{\mathrm {PT}}_{k}$ on $\mathbb {D}_{\mathrm {PT}}^{X}$ in Sections 1.2 and 1.3:

$$ \begin{align*}\mathrm{L}_{k}^{\mathrm{PT}}= \mathrm{T}_{k}+\mathrm{R}_{k} \, , \ \ \ \mathcal{L}^{\mathrm{PT}}_{k}= \mathrm{L}_{k}^{\mathrm{PT}}+(k+1)!\, \mathrm{L}_{-1}^{\mathrm{PT}}\mathsf{ch}_{k+1}(\operatorname{\mathrm{\mathsf{p}}})\, ,\\[-9pt]\end{align*} $$

for $k\geq -1$ . We also have

(3.1) $$ \begin{align} [\mathrm{L}_{n}^{\mathrm{PT}},\mathrm{L}_{k}^{\mathrm{PT}}]\, \stackrel{\langle,\rangle}{=}\, (k-n)\mathrm{L}_{n+k}^{\mathrm{PT}}\, , \ \ \ [\mathrm{L}_{n}^{\mathrm{PT}},(k-1)!\, \mathsf{ch}_{k}(\operatorname{\mathrm{\mathsf{p}}})]\, \stackrel{\langle,\rangle}{=}\, (n+k)!\, \mathsf{ch}_{n+k}(\operatorname{\mathrm{\mathsf{p}}})\,. \\[-9pt]\nonumber\end{align} $$

Equations (3.1) are parallel to equations (2.4) in Gromov–Witten theory.

The main computation of the paper is the intertwining property which relates the Virasoro operators for the stable pairs and Gromov–Witten theories via the descendent correspondence. We separate the argument into two cases: $k\leq 0$ and $k \geq 1$ . Proposition 7 covers the $k\leq 0$ case. The $k\geq 1$ case treated in Theorem 3.1 is harder.

Proposition 7 is proven in Section 3.3 except for steps at the end of the proof which will be completed in the proof of Theorem 3.1 in Sections 4–6. The argument is an intricate calculation based on a strategy of filtration.

Proposition 7. For $k=-1,0$ and $D\in \mathbb {D}^{X\bigstar }_{{\mathrm {PT}}}$ , we have

$$ \begin{align*}\mathfrak{C}^{\bullet}\circ \mathrm{L}_{k}^{\mathrm{PT}}(D)\, = \, (\imath u)^{-k}\, \widetilde{\mathrm{L}}_{k}^{\mathrm{GW}}\circ \mathfrak{C}^{\bullet}(D)\, \\[-9pt]\end{align*} $$

after the restrictions $\tau _{-2}(\operatorname {\mathrm {\mathsf {p}}})=1$ and $\tau _{-1}(\operatorname {\mathrm {\mathsf {p}}})=0$ .

Theorem 3.1. For all $k\ge 1$ and $D\in \mathbb {D}^{X\bigstar }_{{\mathrm {PT}}}$ , we have

$$ \begin{align*}\mathfrak{C}^{\bullet}\circ \mathrm{L}_{k}^{\mathrm{PT}}(D)=(\imath u)^{-k}\, \widetilde{\mathrm{L}}_{k}^{\mathrm{GW}}\circ \mathfrak{C}^{\bullet}(D)\,\\[-9pt] \end{align*} $$

after the restrictions $\tau _{-2}(\operatorname {\mathrm {\mathsf {p}}})=1$ and $\tau _{-1}(\gamma )=0$ for $\gamma \in H^{>2}(X)$ .

The evaluations of the left sides of the equalities in Proposition 7 and Theorem 3.1 require a slight generalization of the formulas (1.14)–(1.16) which govern the descendent correspondence on $\mathbb {D}^{X\bigstar }_{\mathrm {PT}}$ . Additional rules are required for

(3.2) $$ \begin{align} \widetilde{\mathsf{ch}}_{0}(\gamma), \widetilde{\mathsf{ch}}_{1}(\gamma)\ {\text{for }\gamma \in H^{>0}(X)} \ \ {\text{and}}\ \ \ \widetilde{\mathsf{ch}}_{2}(\delta)\ {\text{for }\delta \in H^{2}(X)\text{.}}\end{align} $$

The required rules take a very simple form since $\mathrm {L}_{k}^{\mathrm {PT}}(D)$ is at most linearFootnote ²⁵ in the classes (3.2) over $\mathbb {D}^{X\bigstar }_{\mathrm {GW}}$ :

(3.3) $$ \begin{align} \mathfrak{C}^{\circ}(\widetilde{\mathsf{ch}}_{0}(\gamma)) = -\int_{X}\gamma\, , \ \ \ \ \ \mathfrak{C}^{\circ}(\widetilde{\mathsf{ch}}_{0}(\gamma)M)=0\, , \end{align} $$

$$ \begin{align*} \mathfrak{C}^{\circ}(\widetilde{\mathsf{ch}}_{1}(\gamma)) = 0 \, , \ \ \ \ \ \ \ \ \ \ \ \ \mathfrak{C}^{\circ}(\widetilde{\mathsf{ch}}_{1}(\gamma)M) = 0, \end{align*} $$

where $M\in \mathbb {D}^{X\bigstar }_{\mathrm {PT}}$ . For $\mathfrak {C}^{\circ }(\widetilde {\mathsf {ch}}_{2}({\delta })M)$ with $M\in \mathbb {D}^{X\bigstar }_{\mathrm {PT}}$ , formulas (1.14)–(1.16) apply unchanged. The above rules are compatible with the $\mathrm {GW}/{\mathrm {PT}}$ descendent correspondence and will be established in Section 9.

The restrictions $\tau _{-2}(\operatorname {\mathrm {\mathsf {p}}})=1$ and $\tau _{-1}(\operatorname {\mathrm {\mathsf {p}}})=0$ in Proposition 7 are well-defined since both $\mathfrak {C}^{\bullet }\circ \mathrm {L}_{k}^{\mathrm {PT}}(D)$ and $\widetilde {\mathrm {L}}_{k}^{\mathrm {GW}}\circ \mathfrak {C}^{\bullet }(D)$ , $k=0,-1$ will be seen to lie in the commutative algebra generated by $\tau _{-2}(\operatorname {\mathrm {\mathsf {p}}})$ , $\tau _{-1}(\gamma )$ and $\mathbb {D}_{\mathrm {GW}}^{X+}$ . The commutation with $\tau _{-2}(p)$ and $\tau _{-1}(\operatorname {\mathrm {\mathsf {p}}})$ follows from equation (1.9).

Similarly, the restrictions $\tau _{-2}(\operatorname {\mathrm {\mathsf {p}}})=1$ and $\tau _{-1}(\gamma )=0$ for $\gamma \in H^{>2}(X)$ in Theorem 3.1 are well-defined since both $\mathfrak {C}^{\bullet }\circ \mathrm {L}_{k}^{\mathrm {PT}}(D)$ and $\widetilde {\mathrm {L}}_{k}^{\mathrm {GW}}\circ \mathfrak {C}^{\bullet }(D)$ , $k>0$ will be seen to lie in the commutative algebra generated by $\tau _{-2}(\operatorname {\mathrm {\mathsf {p}}})$ , $\tau _{-1}(\gamma )$ and $\mathbb {D}_{\mathrm {GW}}^{X\bigstar }$ . The algebra $\mathbb {D}_{\mathrm {GW}}^{X\bigstar }$ is generated by the essential descendents

$$ \begin{align*}\big\{ \, \tau_{i}(\gamma)\, | \, (i\geq 0, \gamma\in H^{>0}(X,\mathbb{Q}))\ \text{or}\ (i=0, \gamma\in H^{>2}(X,\mathbb{Q}))\, \big\}.\end{align*} $$

Again, commutation follows from equation (1.9).

3.2 Conventions for $(-1)!\mathsf {ch}_{1}(c_{1})$

In order to complete the definitions of the left sides of Proposition 7 and Theorem 3.1, we must also include the term $(-1)! \mathsf {ch}_{1}(c_{1})$ in the descendent correspondence $\mathfrak {C}^{\bullet }$ since such terms occur in $\mathrm {L}_{k}^{\mathrm {PT}}$ .

$\bullet $ The first case is

$$ \begin{align*}\mathfrak{C}^{\circ}((-1)!\mathsf{ch}_{1}(c_{1}))=0.\end{align*} $$

$\bullet $ The nonvanishing bumping term is given by

(3.4) $$ \begin{align} &\mathfrak{C}^{\circ}\Big((-1)!\mathsf{ch}_{1}(c_{1})\widetilde{\mathsf{ch}}_{k_{1}+2}(\gamma)\Big) =-\frac{(\imath u)^{-1}}{k_{1}!}\Bigg(\mathfrak{a}_{k_{1}-1}(c_{1}\gamma) +(\imath u)^{-1}\mathfrak{a}_{k_{1}-2}(c_{1}\gamma\cdot c_{1})\nonumber \\&\qquad\qquad\qquad \qquad\qquad\qquad\qquad\qquad \qquad\qquad\quad +(\imath u)^{-1}k_{1} \sum_{|\mu|=k_{1}-3}\frac{\mathfrak{a}_{\mu_{1}}\mathfrak{a}_{\mu_{2}}}{\text{Aut}(\mu)}(c_{1}\gamma\cdot c_{1})\Bigg)\, , \end{align} $$

where $k_{1}\ge 2$ .

$\bullet $ The higher bumping term is

$$ \begin{align*} \mathfrak{C}^{\circ}((-1)! \mathsf{ch}_{1}(c_{1})\widetilde{\mathsf{ch}}_{k_{1}+2}(\gamma)\widetilde{\mathsf{ch}}_{k_{2}+2}(\gamma^{\prime}))=\frac{(\imath u)^{-2}(k_{1}+k_{2}-1)}{k_{1}!k_{2}!}\mathfrak{a}_{k_{1}+k_{2}-2}(c_{1}\gamma\gamma^{\prime}),\end{align*} $$

$k_{1},k_{2}\ge 0, k_{1}+k_{2}>1$ . There is also an exceptional higher bumping term

$$ \begin{align*} \mathfrak{C}^{\circ}((-1)! \mathsf{ch}_{1}(c_{1})\widetilde{\mathsf{ch}}_{2}(\gamma)\widetilde{\mathsf{ch}}_{3}(\gamma^{\prime})) =\tau_{-2}(c_{1}\gamma\gamma^{\prime}).\end{align*} $$

3.3 Proof of Proposition 7

The cases $k=-1,0$ are special in two ways:

1. (i) We must use the exceptional cases of the operator $\mathfrak {C}^{\circ }$ , in the analysis for $k=-1,0$ .

2. (ii) While the operator $\widetilde {\mathrm {L}}_{k}^{\mathrm {GW}}$ for $k=-1,0$ has quadratic part $\frac {u^{-2}}{2}B^{k+1}$ , $\widetilde {\mathrm {L}}_{k}^{\mathrm {GW}}$ is a first order operator acting on the stationary sector of descendent algebra for $k>0$ .

For these reasons, we treat the $k=-1,0$ cases separately here.

The restrictions in the statement of Proposition 7 allow us freely use

(3.5) $$ \begin{align} \mathsf{ch}_{0}(\mathsf{p})=-1\, , \end{align} $$

which is compatible with $\mathfrak {C}^{\bullet }$ . Similarly, we can use

(3.6) $$ \begin{align} \mathsf{ch}_{1}(\mathsf{p})=0\,. \end{align} $$

Let us write down the corresponding operators explicitly:

$$ \begin{align*}\mathrm{L}_{-1}^{{\mathrm{PT}}}=\mathrm{R}_{-1}-(-1)!\, \mathsf{\mathsf{ch}}_{1}(c_{1}),\quad \widetilde{\mathrm{L}}_{-1}^{\mathrm{GW}}=\mathrm{R}_{-1}+\frac{u^{-2}}{2} \mathrm{B}^{0}.\end{align*} $$

$$ \begin{align*}\mathrm{L}_{0}^{{\mathrm{PT}}}=\mathrm{R}_{0}-\widetilde{\mathsf{ch}}_{2}(c_{1})-\frac{1}{2} \mathsf{ch}_{1}\mathsf{ch}_{1}(c_{1}),\quad \widetilde{\mathrm{L}}_{0}^{\mathrm{GW}}=\mathrm{R}_{0}+\frac{u^{-2}}{2}\mathrm{B}^{1}-\tau_{0}(c_{1})-\frac{1}{24}\int_{X}c_{1}c_{2}.\end{align*} $$

We have used equation (3.5) for $\mathrm {L}_{-1}^{{\mathrm {PT}}}$ . For ${\mathrm {L}}_{0}^{{\mathrm {PT}}}$ , only the $d_{L}=d_{R}=2$ summand is nonzero by equation (3.6).

Step 1. We check the statement for $D=1$ .

The left side of the equality of Proposition 7 for $k=-1$ is

$$ \begin{align*}\mathfrak{C}^{\bullet}(\mathrm{L}_{-1}^{\mathrm{PT}}(D))=-\mathfrak{C}^{\bullet}((-1)!\mathsf{ch}_{1}(c_{1}))=0.\end{align*} $$

The right side of the equality,

$$ \begin{align*} \imath u\, \widetilde{\mathrm{L}}_{-1}^{\mathrm{GW}}(\mathfrak{C}^{\bullet}(1))= \imath u\, \widetilde{\mathrm{L}}_{-1}^{\mathrm{GW}}(1)=0, \end{align*} $$

matches. For $k=0$ , the left side for $D=1$ is

$$ \begin{align*}\mathfrak{C}^{\bullet}(\mathrm{L}_{0}^{\mathrm{PT}}(1))=-\mathfrak{C}^{\bullet}(\widetilde{\mathsf{ch}}_{2}(c_{1}))=-\mathfrak{a}_{1}(c_{1})=-\tau_{0}(c_{1})-\frac{1}{24}\int_{X} c_{1}c_{2}.\end{align*} $$

The right side,

$$ \begin{align*}\widetilde{\mathrm{L}}_{0}^{\mathrm{GW}}(\mathfrak{C}^{\bullet}(1))=\widetilde{\mathrm{L}}_{0}^{\mathrm{GW}}(1)=-\tau_{0}(c_{1})-\frac{1}{24}\int_{X} c_{1}c_{2},\end{align*} $$

matches.

Step 2. We check the statement for $D=\widetilde {\mathsf {ch}}_{k+2}(\gamma )$ with $k\geq 0$ .

We must expand both sides of the equality of Proposition 7 in terms of $\tau $ . The following formula will be used:

(3.7) $$ \begin{align} (\imath u)^{k} \mathfrak{C}^{\circ}(\widetilde{\mathsf{ch}}_{k+2}(\gamma))&= \tau_{k}(\gamma) +\left(\sum_{i=1}^{k}\frac{1}{i}\right)\tau_{k-1}(\gamma\cdot c_{1})+ \left(\sum_{1\le i<j\le k}\frac{1}{ij}\right)\tau_{k-2}(\gamma\cdot c_{1}^{2}) \nonumber\\ &\quad +\sum_{|\mu|=k-1}\frac{\mu_{1}!\mu_{2}!}{\text{Aut}(\mu) k!}\bigg(\tau_{\mu_{1}-1}\tau_{\mu_{2}-1}(\gamma\cdot c_{1})+ \bigg(\sum_{i=1}^{\mu_{1}-1}\frac{1}{i}\bigg)\tau_{\mu_{1}-2}(\gamma \cdot c_{1}^{2})\tau_{\mu_{2}-1}(\operatorname{\mathrm{\mathsf{p}}})\nonumber\\&\quad +\bigg(\sum_{i=1}^{\mu_{2}-1}\frac{1}{i}\bigg)\tau_{\mu_{1}-1}(\operatorname{\mathrm{\mathsf{p}}})\tau_{\mu_{2}-2}(\gamma\cdot c_{1}^{2})\bigg)\nonumber\\ &\quad +\sum_{|\mu|=k-2}\frac{\mu_{1}!\mu_{2}!}{\text{Aut}(\mu)k!}\tau_{\mu_{1}-1}\tau_{\mu_{2}-1}(\gamma\cdot c_{1}^{2})\nonumber\\ &\quad +\sum_{|\mu|=k-3}\frac{\mu_{1}!\mu_{2}!\mu_{3}!}{\text{Aut}(\mu) (k-1)!}\tau_{\mu_{1}-1}\tau_{\mu_{2}-1}\tau_{\mu_{3}-1}(\gamma\cdot c_{1}^{2})\,. \end{align} $$

We split the analysis of the difference for

(3.8) $$ \begin{align} \mathfrak{C}^{\bullet}\circ \mathrm{L}_{-1}^{\mathrm{PT}}(D)- \imath u\, \widetilde{\mathrm{L}}_{-1}^{\mathrm{GW}}\circ \mathfrak{C}^{\bullet}(D) \end{align} $$

in stages according to the $\tau $ degree of terms. The second term of the difference is simpler since

$$ \begin{align*}\imath u\, \widetilde{\mathrm{L}}_{-1}^{\mathrm{GW}}\circ \mathfrak{C}^{\bullet}(D) = \imath u\, \mathrm{R}_{-1}(\mathfrak{C}^{\circ}(\widetilde{\mathsf{ch}}_{k}(\gamma))),\,\end{align*} $$

and the latter is a easy modification of equation (3.7). The first term is more involved since there are two parts: the action of $\mathrm {R}_{-1}$ and the interaction with $(-1)!\mathsf {ch}_{1}(c_{1})$ .

$\bullet $ We first study the $\tau $ linear terms of $(\imath u)^{k-1}\mathfrak {C}^{\bullet }\circ \mathrm {L}_{-1}^{\mathrm {PT}}(D)$ :

$$ \begin{align*} &\bigg( \tau_{k-1}(\gamma)+\Big(\sum_{i=1}^{k-1}\frac{1}{i}\Big)\tau_{k-2}(\gamma\cdot c_{1}) +\Big( \sum_{1\le i<j\le k-1}\frac{1}{ij}\Big)\tau_{k-3}(\gamma\cdot c_{1}^{2})\bigg) \\ &\qquad +\bigg( \frac{(\imath u)^{k-2}}{k!}\mathfrak{a}_{k-1}(\gamma\cdot c_{1})+\frac{(\imath u)^{k-3}}{k!}\mathfrak{a}_{k-2}(\gamma\cdot c_{1}^{2})\bigg) \\ &\quad =\bigg( \tau_{k-1}(\gamma)+\Big(\sum_{i=1}^{k-1}\frac{1}{i}\Big)\tau_{k-2}(\gamma\cdot c_{1}) +\Big( \sum_{1\le i<j\le k-1}\frac{1}{ij}\Big)\tau_{k-3}(\gamma\cdot c_{1}^{2})\bigg) \\ &\qquad +\frac{1}{k}\bigg(\tau_{k-2}(\gamma\cdot c_{1}) +\Big(\sum_{i=1}^{k-2}\frac{1}{i}\Big)\tau_{k-3}(\gamma\cdot c_{1}^{2})\bigg)+ \frac{1}{k(k-1)}\tau_{k-3}(\gamma\cdot c_{1}^{2}). \end{align*} $$

We have used here bumping with $(-1)!\mathsf {ch}_{1}(c_{1})$ from equation (3.4) to obtain the expression in the second line and an inversionFootnote ²⁶ of equation (1.13) to justify the second equality. After collecting together the coefficients in front of the $\tau $ ’s in the last expression, we obtain $R_{-1}(\mathfrak {C}^{\circ }(\widetilde {\mathsf {ch}}_{k}(\gamma )))$ , exactly as expected.

$\bullet $ We study next the $\tau $ -quadratic term of equation (3.8). Consider first the terms that have a co-product $(\gamma \cdot c_{1})^{L}_{i}\otimes (\gamma \cdot c_{1})^{R}_{i}$ as argument. Bumping with $(-1)!\mathsf {ch}_{1}(c_{1})$ does not produce such terms—only the terms of the second line of equation (3.7) contribute to the terms of equation (3.8). These terms cancel exactly.

$\bullet $ The $\tau $ -quadratic terms of difference (3.8) with argument $(\gamma \cdot c_{1}^{2})^{L}_{i}\otimes (\gamma \cdot c_{1}^{2})^{R}_{i}$ are slightly more involved. The second term of the difference has terms

$$ \begin{align*} \sum_{|\mu|=k-2}\frac{\mu_{1}!\mu_{2}!}{\text{Aut}(\mu)(k-1)!}\bigg(\Big(\sum_{i=1}^{\mu_{1}-1}\frac1i\Big)\tau_{\mu_{1}-2}(\gamma\cdot c_{1}^{2})\tau_{\mu_{2}-1}(\operatorname{\mathrm{\mathsf{p}}}) +\Big(\sum_{i=1}^{\mu_{2}-1}\frac1i\Big)\tau_{\mu_{1}-1}(\operatorname{\mathrm{\mathsf{p}}})\tau_{\mu_{2}-2}(\gamma\cdot c_{1}^{2})\bigg)\\ +\sum_{|\mu|=k-3} \frac{\mu_{1}!\mu_{2}!}{\text{Aut}(\mu)(k-1)!}\tau_{\mu_{1}-1}\tau_{\mu_{2}-1}(\gamma\cdot c_{1}^{2}), \end{align*} $$

where the term on the second line is a result of bumping with $(-1)!\mathsf {ch}_{1}(c_{1})$ . After simplifying the last expression, we obtain the corresponding $\tau $ -quadratic term of $R_{-1}(\mathfrak {C}^{\circ }(\widetilde {\mathsf {ch}}_{k+2}(\gamma )))$ as expected.

$\bullet $ The last step is to analyze the $\tau $ -cubic terms of the difference (3.8). Since bumping with $\mathsf {ch}_{1}(c_{1})$ is trivial, the terms match exactly.

Similarly, we must analyze the difference

(3.9) $$ \begin{align} \mathfrak{C}^{\bullet}\circ \mathrm{L}_{0}^{\mathrm{PT}}(D)- \widetilde{\mathrm{L}}_{0}^{\mathrm{GW}}\circ \mathfrak{C}^{\bullet}(D).\end{align} $$

Since both $\mathrm {R}_{0}$ on the stable pairs side and $\mathrm {R}_{0}^{0}$ on the Gromov–Witten side scale the descendents by the complex cohomological degree, the difference (3.9) is equalFootnote ²⁷ to

(3.10) $$ \begin{align} -\mathfrak{C}^{\bullet}\left((\widetilde{\mathsf{ch}}_{2}+\mathsf{ch}_{1}^{2}/2)(c_{1})\cdot D\right)-\bigg(\mathrm{R}_{0}^{1}+\frac{u^{-2}}2\mathrm{B}^{1}-\tau_{0}(c_{1})-\frac{1}{24}\int_{X}c_{1}c_{2}\bigg)\circ\mathfrak{C}^{\bullet}(D).\end{align} $$

If $D=\widetilde {\mathsf {ch}}_{k+2}(\gamma )$ , then $\mathrm {B}^{1}\circ \mathfrak {C}^{\bullet }(D)=0$ . We have already proved that the difference vanishes for $D=1$ . Since

$$ \begin{align*}\mathfrak{C}^{\bullet}(\mathsf{ch}_{1}\mathsf{ch}_{1}(c_{1})\widetilde{\mathsf{ch}}_{k+2}(\gamma))=0,\end{align*} $$

the difference (3.10) is equal to

(3.11) $$ \begin{align} -\mathfrak{C}^{\circ}(\widetilde{\mathsf{ch}}_{2}(c_{1})\widetilde{\mathsf{ch}}_{k+2}(\gamma)) -\mathrm{R}^{1}_{0}(\mathfrak{C}^{\circ}(\widetilde{\mathsf{ch}}_{k+2}(\gamma))).\end{align} $$

Comparing formulas (1.14) and (1.15), we conclude that the latter difference vanishes.

Indeed, let us expand both terms of equation (3.11). First,

$$ \begin{align*} \mathfrak{C}^{\circ}(\widetilde{\mathsf{ch}}_{2}(c_{1})\widetilde{\mathsf{ch}}_{k+2}(\gamma))&=-\frac{(\imath u)^{-1}}{k!}\mathfrak{a}_{k}(\gamma\cdot c_{1})-\frac{(\imath u)^{-2}}{k!}\mathfrak{a}_{k-1}(\gamma\cdot c_{1}^{2}) -\frac{(\imath u)^{-2}}{(k-1)!}\sum_{|\mu|=k-2}\frac{\mathfrak{a}_{\mu_{1}}\mathfrak{a}_{\mu_{2}}}{\operatorname{\mathrm{Aut}}(\mu)}(\gamma \cdot c_{1}^{2})\\&= -(\imath u)^{-k}\Bigg(\tau_{k-1}(\gamma\cdot c_{1})+\left(\sum_{i=1}^{k-1}\frac{1}{i}\right)\tau_{k-2}(\gamma\cdot c_{1}^{2})\Bigg)-\frac{(\imath u)^{-k}}{k}\tau_{k-2}(\gamma \cdot c_{1}^{2})\\&\quad -\frac{(\imath u)^{-k+2}}{(k-1)!} \sum_{|\mu|=k-2}\frac{\mu_{1}!\mu_{2}!}{\operatorname{\mathrm{Aut}}(\mu)}\tau_{\mu_{1}-1}\tau_{\mu_{2}-1}(\gamma\cdot c_{1}^{2}). \end{align*} $$

On the other hand,

$$ \begin{align*} \mathfrak{C}^{\circ}(\widetilde{\mathsf{ch}}_{k+2}(\gamma))&=\frac{1}{(k+1)!}\mathfrak{a}_{k+1}(\gamma)+ \frac{(\imath u)^{-1}}{k!}\sum_{|\mu|=k-1}\frac{\mathfrak{a}_{\mu_{1}}\mathfrak{a}_{\mu_{2}}}{\operatorname{\mathrm{Aut}}(\mu)}(\gamma\cdot c_{1})+\dots\\ &= (\imath u)^{-k}\Bigg(\tau_{k}(\gamma)+\left(\sum_{i=1}^{k}\frac{1}{i}\right)\tau_{k-1}(\gamma \cdot c_{1})\Bigg)\\&\quad +\frac{(\imath u)^{-k+2}}{k!}\sum_{|\mu|=k-1}\frac{\mu_{1}!\mu_{2}!}{\operatorname{\mathrm{Aut}}(\mu)} \tau_{\mu_{1}-1}\tau_{\mu_{2}-1}(\gamma \cdot c_{1})+\dots, \end{align*} $$

where we have used dots to stand for the terms that are of complex cohomological degree $3$ . Since

$$ \begin{align*}\mathrm{R}_{0}^{1}(\tau_{k}(\gamma))=\tau_{k-1}(\gamma\cdot c_{1}),\end{align*} $$

all the omitted terms are annihilated by $\mathrm {R}_{0}^{1}$ . The remaining terms of the difference (3.11) cancel.

Step 3. We check the statement for $D=\widetilde {\mathsf {ch}}_{k_{1}+2}(\gamma _{1})\widetilde {\mathsf {ch}}_{k_{2}+2}(\gamma _{2})$ with $k_{i}\geq 0$ .

We start with the difference (3.8):

(3.12) $$ \begin{align} &\mathfrak{C}^{\circ}(\mathrm{R}_{-1}(\widetilde{\mathsf{ch}}_{k_{1}+2}(\gamma_{1})\widetilde{\mathsf{ch}}_{k_{2}+2}(\gamma_{2})))-\mathfrak{C}^{\circ}((-1)!\widetilde{\mathsf{ch}}_{1}(c_{1})\widetilde{\mathsf{ch}}_{k_{1}+2}(\gamma_{1})\widetilde{\mathsf{ch}}_{k_{2}+2}(\gamma_{2}))\nonumber\\ &\quad -(\imath u)\mathrm{R}_{-1}(\mathfrak{C}^{\circ}(\widetilde{\mathsf{ch}}_{k_{1}+2}(\gamma_{1})\widetilde{\mathsf{ch}}_{k_{2}+2}(\gamma_{2}))) -(\imath u)\frac{u^{-2}}{2}\mathrm{B}^{0}(\mathfrak{C}^{\circ}(\widetilde{\mathsf{ch}}_{k_{1}+2}(\gamma_{1})),\mathfrak{C}^{\circ}(\widetilde{\mathsf{ch}}_{k_{2}+2}(\gamma_{2})))). \end{align} $$

Vanishing of the last expression follows from Proposition 12 and Proposition 13.

The difference (3.9) as above is equivalent to equation (3.10). Since we have already shown the vanishing for $D=1$ and $D=\widetilde {\mathsf {ch}}_{k+2}(\gamma )$ , we need only to check the vanishing of

(3.13) $$ \begin{align} &-\mathfrak{C}^{\circ}(\widetilde{\mathsf{ch}}_{2}(c_{1})D)-\frac{1}{2}\mathfrak{C}^{\bullet}(\mathsf{ch}_{1}\mathsf{ch}_{1}(c_{1})D)-R_{0}^{1}(\mathfrak{C}^{\circ}(D))\nonumber\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad-\frac{u^{-2}}{2}\mathrm{B}^{1}(\mathfrak{C}^{\circ}(\widetilde{\mathsf{ch}}_{k_{1}+2}(\gamma_{1})),\mathfrak{C}^{\circ}(\widetilde{\mathsf{ch}}_{k_{2}+2}(\gamma_{2}))). \end{align} $$

The vanishing follows from Propositions 12 and 13.

Step 4. We check the statement for $D=\widetilde {\mathsf {ch}}_{k_{1}+2}(\gamma _{1})\widetilde {\mathsf {ch}}_{k_{2}+2}(\gamma _{2})\widetilde {\mathsf {ch}}_{k_{3}+2}(\gamma _{3})$ with $k_{i}\geq 0$ .

The result follows immediately from the triple bumping relation (6.8) which holds in complete generality. No special cases require extra attention.

3.4 Proof of Theorem 1.1

The vanishings

(3.14) $$ \begin{align} \langle \mathcal{L}_{-1}^{\mathrm{PT}}(D)\rangle^{X,{\mathrm{PT}}}_{\beta}=0\ \ \ \ \text{and} \ \ \ \ \langle \mathcal{L}_{0}^{\mathrm{PT}}(D)\rangle^{X,{\mathrm{PT}}}_{\beta}=0 \end{align} $$

are simple to prove for all $D\in \mathbb {D}_{{\mathrm {PT}}}^{X}$ . For

$$ \begin{align*}\mathcal{L}_{-1}^{\mathrm{PT}}= \mathrm{R}_{-1} + \mathrm{R}_{-1}\mathsf{ch}_{0}(\mathsf{p}),\end{align*} $$

the vanishing (3.14) is immediate from the definition of $\mathrm {R}_{-1}$ and equation (1.1). For

$$ \begin{align*}\mathcal{L}_{0} = \mathrm{R}_{0}-{\widetilde{\mathsf{ch}}}_{2}(c_{1}) -\frac{1}{2} \mathsf{ch}_{1}\mathsf{ch}_{1}(c_{1})+ \mathrm{R}_{-1} \mathsf{ch}_{1}(\mathsf{p})\,\end{align*} $$

the vanishing (3.14) follows from the definition of $\mathrm {R}_{0}$ , the virtual dimension constraints and the divisor equation:

$$ \begin{align*} \Big\langle {\mathsf{ch}_{2}(c_{1}) \, \mathsf{ch}}_{k_{1}}(\gamma_{1})\cdots \mathsf{ch}_{k_{m}}(\gamma_{m})\Big\rangle_{\beta}^{X,{\mathrm{PT}}} \, =\, \int_{\beta} c_{1} \cdot \Big\langle { \mathsf{ch}}_{k_{1}}(\gamma_{1})\cdots \mathsf{ch}_{k_{m}}(\gamma_{m})\Big\rangle_{\beta}^{X,{\mathrm{PT}}}. \end{align*} $$

We now assume $k\geq 1$ . Using the intertwining property of Theorem 3.1, the stationary $\mathrm {GW}/{\mathrm {PT}}$ correspondence of Theorem 1.4 and the Virasoro constraints in Gromov–Witten theory, we can prove the stationary Virasoro constraints for stable pairs in the toric case.

Let $D\in \mathbb {D}_{{\mathrm {PT}}}^{X+}$ , so D is a monomial in the operators

$$ \begin{align*}\big\{ \, \widetilde{\mathsf{ch}}_{i}(\gamma)\, | \, i\geq 0,\ \gamma\in H^{>0}(X,\mathbb{Q})\ \big\} .\end{align*} $$

The first step is to check by hand that the Virasoro constraints

(3.15) $$ \begin{align} \Big\langle\mathcal{L}^{\mathrm{PT}}_{k}(D)\Big\rangle^{X,{\mathrm{PT}}}_{\beta}=0\, \end{align} $$

of Theorem 1.1 are compatible all with insertions of the form

(3.16) $$ \begin{align} \widetilde{\mathsf{ch}}_{0}(\gamma), \widetilde{\mathsf{ch}}_{1}(\gamma)\ {\text{for }\gamma \in H^{>0}(X)}\quad{\text{and}}\quad\widetilde{\mathsf{ch}}_{2}(\delta)\ {\text{for }\delta \in H^{2}(X)\text{.}} \end{align} $$

If any of the operators (3.16) appear in D, the Virasoro constraints (3.15) are true if the Virasoro constraints are true for the monomial obtained by dividing D by the occurring operators (3.16). We can therefore reduce to the case where D is a monomial in the operators

$$ \begin{align*}\big\{ \, \widetilde{\mathsf{ch}}_{i}(\gamma)\, | \, (i\geq 3, \gamma\in H^{>0}(X,\mathbb{Q}))\ \textit{or}\ (i=2, \gamma\in H^{>2}(X,\mathbb{Q}))\, \big\} .\end{align*} $$

In other words, $D\in \mathbb {D}^{X\bigstar }_{\mathrm {PT}}$ .

The next step is to apply Theorem 1.4:

(3.17) $$ \begin{align} (-q)^{d_{\beta}/2}\, \langle \mathcal{L}_{k}^{\mathrm{PT}}(D)\rangle^{X,{\mathrm{PT}}}_{\beta} =(-\imath u)^{d_{\beta}}\, \langle\mathfrak{C}^{\bullet}(\mathcal{L}_{k}^{\mathrm{PT}}(D))\rangle^{X,\mathrm{GW}}_{\beta} \end{align} $$

for all $k\geq 1$ . By the construction of the correspondence [Reference Pandharipande and Pixton26], the descendents of the point class do not interact with other descendents:

(3.18) $$ \begin{align}\mathfrak{C}^{\bullet}(\widetilde{\mathsf{ch}}_{k+2}(\operatorname{\mathrm{\mathsf{p}}})D)=(\imath u)^{-k}\tau_{k}(\operatorname{\mathrm{\mathsf{p}}})\mathfrak{C}^{\bullet}(D)\, , \end{align} $$

for every $D\in \mathbb {D}_{{\mathrm {PT}}}^{X\bigstar }$ .

By combining equations (3.17) and (3.18) and the intertwining statement of Theorem 3.1, we see

$$ \begin{align*} \langle\mathfrak{C}^{\bullet}(\mathcal{L}_{k}^{\mathrm{PT}}(D))\rangle^{\mathrm{GW}}_{\beta} &= \langle\mathfrak{C}^{\bullet}(\mathrm{L}_{k}^{\mathrm{PT}}(D))\rangle^{\mathrm{GW}}_{\beta} + (k+1)! \, \langle\mathfrak{C}^{\bullet}(\mathrm{L}_{-1}^{\mathrm{PT}}(\mathsf{ch}_{k+1}(\operatorname{\mathrm{\mathsf{p}}}) D))\rangle^{\mathrm{GW}}_{\beta} \\ & = (\imath u)^{-k}\langle \widetilde{\mathrm{L}}_{k}^{\mathrm{GW}}(\mathfrak{C}^{\bullet}(D))\rangle_{\beta}^{\mathrm{GW}}+ (\imath u)^{2-k}(k+1)!\, \langle \widetilde{\mathrm{L}}_{-1}^{\mathrm{GW}}(\tau_{k-1}(\operatorname{\mathrm{\mathsf{p}}})\mathfrak{C}^{\bullet}(D))\rangle_{\beta}^{\mathrm{GW}}\\ &= (\imath u)^{-k}\langle\mathcal{L}_{k}^{\mathrm{GW}}(\mathfrak{C}^{\bullet}(D))\rangle^{\mathrm{GW}}_{\beta}\\ &= 0,\end{align*} $$

where the last equality is by Proposition 6 which may be applied since

$$ \begin{align*}\mathfrak{C}^{\bullet}(D) \in \mathbb{D}_{\mathrm{GW}}^{X+}\,\end{align*} $$

by Proposition 4. We conclude

$$ \begin{align*}\langle \mathcal{L}_{k}^{\mathrm{PT}}(D)\rangle^{X,{\mathrm{PT}}}_{\beta}=0\end{align*} $$

as required.

We could have also used the intertwining property of Proposition 7 to prove the stable pairs vanishings (3.14) for $D\in \mathbb {D}_{{\mathrm {PT}}}^{X+}$ , but some additional care must be taken since the insertions $\mathsf {ch}_{0}(\mathsf {p})$ and $\mathsf {ch}_{1}(\mathsf {p})$ which occur in the terms

$$ \begin{align*}(k+1)! \, \langle\mathfrak{C}^{\bullet}(\mathrm{L}_{-1}^{\mathrm{PT}}(\mathsf{ch}_{k+1}(\operatorname{\mathrm{\mathsf{p}}}) D))\rangle^{\mathrm{GW}}_{\beta}\end{align*} $$

for $k=-1$ and $0$ are not covered by Proposition 7. We leave the details to the reader.

4.1 Overview

After an explicit study of various terms of the stationary Gromov–Witten Virasoro constraints in Section 4.2, we prove Theorem 3.1 in the basic case $D=1$ in Section 4.3.

4.2 Leading term

We analyze here the stationary Virasoro constraints on the Gromov–Witten side defined in Section 2.4.

The leading term $\mathrm {T}^{1}_{k}$ of $\mathrm {T}^{\prime }_{k}$ is of the form

$$ \begin{align*}\frac{1}{2}\mathrm{T}^{1}_{k}=\frac{k!}{u^{2}}\, \tau_{k}(c_{1})+\frac{1}{2}\sum_{a+b=k-2} (-1)^{d^{L}-1}(a+d^{L}-1)!(b+d^{R}-1)!\tau_{a}\tau_{b}(c_{1}),\end{align*} $$

where $a,b\ge 0$ in the sum. By the following result, the term $\mathrm {T}^{\prime }_{k}$ simplifies if we use the modified descendents $\mathfrak {a}_{i}$ .

Proposition 8. For all $k\geq -1$ ,

$$ \begin{align*}\mathrm{T}_{k}^{\prime}=-(\imath u)^{k-2}\sum_{a+b=k+2}(-1)^{d^{L}d^{R}}(a+d^{L}-3)!(b+d^{R}-3)!\frac{\mathfrak{a}_{a-1}\mathfrak{a}_{b-1}(c_{1})}{(a-1)!(b-1)!},\end{align*} $$

where the sum over all $a,b\ge 0$ and we use convention $\mathfrak {a}_{0}=0,\, \mathfrak {a}_{-1}/(-1)!=\tau _{-2}$ .

Proof. Using formula (1.13), we expand $\mathrm {T}^{\prime }_{k}$ in terms of $\mathfrak {a}_{i}$ to show that the quadratic and cubic in $c_{1}$ terms cancel. In the computation, we compare the expressions

$$ \begin{align*}[-a]^{k}_{2}=(-1)^{a} a!(k-a)!\left(\sum_{i=1}^{k-a}\frac{1}i-\sum_{i=1}^{a}\frac{1}i\right),\quad a\ge 0\, ,\ k\ge a,\end{align*} $$

$$ \begin{align*}[-a]^{k}_{3}=(-1)^{a}a!(k-a)!\left(\sum_{1\le i<j\le k-a}\frac{1}{ij}+\sum_{1\le i<j\le a}\frac{1}{ij}-\left(\sum_{i=1}^{k-a}\frac{1}{i}\right) \left(\sum_{i=1}^{a}\frac{1}{i}\right)\right),\quad a\ge 0,k\ge a\end{align*} $$

with the coefficients in equation (1.13).

The transformation (1.13) simplifies if we use the following operators and shorthand notations for the sums

$$ \begin{align*}\tilde{\mathfrak{a}}_{k}=\frac{(\imath u)^{k-1}}{k!}\mathfrak{a}_{k}\, , \quad \chi_{l}^{k}=\sum_{j=1}^{k}\frac{1}{j^{l}}\, , \quad \chi_{1,1}^{k}=\sum_{1\le i<j \le k}\frac{1}{ij}.\end{align*} $$

In the formulas below, all operators $\tilde {\mathfrak {a}}_{0}$ are set to be 0. We apply transformation to $\mathrm {T}_{k}^{1}$ to obtain

(4.1) $$ \begin{align} &\hspace{-20pt}\sum_{m=-1}^{k+1}(-1)^{d_{L}-1}(m+d_{L}-2)!(k-m-d_{L}+2)!\ \nonumber\\ &\quad \times\Bigg(\tilde{\mathfrak{a}}_{m}\tilde{\mathfrak{a}}_{-m+k}(c_{1})- \left(\chi_{1}^{m} -\chi_{1}^{k-m-1}\right)\tilde{\mathfrak{a}}_{m}\tilde{\mathfrak{a}}_{-m+k-1}(c_{1}^{2}) \nonumber\\ &\quad +\Big(\chi_{1}^{m}\chi_{1}^{k-m-2}+\chi_{2}^{m}+\chi_{1,1}^{m}+\chi_{2}^{-m+k-2}+\chi_{1,1}^{-m+k-2}\Big)\tilde{\mathfrak{a}}_{m}\tilde{\mathfrak{a}}_{-m+k-2}(c_{1}^{3})\Bigg)\,. \end{align} $$

To write the transformation of $\mathrm {T}^{2}_{k}$ , we split the sum for $\mathrm {T}^{2}_{k}$ into two subsums, the first with $d_{L}=2$ and the second with $d_{L}=3$ :

$$ \begin{align*} &\sum_{m=-1}^{k}(-1)(m)!(k-m)!(\chi_{1}^{k-m}-\chi_{1}^{m}) \Big(\tilde{\mathfrak{a}}_{m}(c_{1}^{2})\tilde{\mathfrak{a}}_{-m+k-1}(\operatorname{\mathrm{\mathsf{p}}})-\chi_{1}^{m-1}\tilde{\mathfrak{a}}_{m-1}\tilde{\mathfrak{a}}_{-m-k-1}(c_{1}^{3})\Big)\\ &\qquad+(m+1)!(k-m-1)!(\chi_{1}^{k-m-1}-\chi_{1}^{m-1}) \Big(\tilde{\mathfrak{a}}_{m}(\operatorname{\mathrm{\mathsf{p}}})\tilde{\mathfrak{a}}_{-m+k-1}(c_{1}^{2})-\chi_{1}^{k-m-2}\tilde{\mathfrak{a}}_{m}\tilde{\mathfrak{a}}_{-m-k-2}(c_{1}^{2})\Big). \end{align*} $$

Finally, the transformation of $\mathrm {T}^{3}_{k}$ to $\mathfrak {a}$ variables is

$$ \begin{gather*} \sum_{m=-1}^{k}(m+1)!(k-m-1)!\Big(\chi_{1,1}^{m-1}+\chi_{1,1}^{k-m-1}-\chi_{1}^{m-1}\chi_{1}^{k-m-1}\Big) \tilde{\mathfrak{a}}_{m}\tilde{\mathfrak{a}}_{-m+k-2}(c_{1}^{3}). \end{gather*} $$

After summing the terms $\mathrm {T}_{k}^{j}$ for $j=1,2,3$ , we find that only the first term in equation (4.1) does not cancel.

4.3 Intertwining for $D=1$

For the most of computations in Section 4, we will require the simplest case of the stationary $\mathrm {GW}/{\mathrm {PT}}$ transformation $\mathfrak {C}^{\bullet }$ of Section 1.7,

(4.2) $$ \begin{align} \mathfrak{C}^{\circ}(\widetilde{\mathsf{ch}}_{k+2}(\gamma))&=\frac{1}{(k+1)!}\mathfrak{a}_{k+1}(\gamma) +\frac{(\imath u)^{-1}}{k!}\sum_{|\mu|=k-1}\frac{\mathfrak{a}_{\mu_{1}}\mathfrak{a}_{\mu_{2}}(\gamma\cdot c_{1})}{\operatorname{\mathrm{Aut}}(\mu)} \nonumber\\ &\quad +\frac{(\imath u)^{-2}}{k!}\sum_{|\mu|=k-2}\frac{\mathfrak{a}_{\mu_{1}}\mathfrak{a}_{\mu_{2}} (\gamma\cdot c_{1}^{2})}{\operatorname{\mathrm{Aut}}(\mu)}+ \frac{(\imath u)^{-2}}{(k-1)!}\sum_{|\mu|=k-3}\frac{\mathfrak{a}_{\mu_{1}}\mathfrak{a}_{\mu_{2}}\mathfrak{a}_{\mu_{3}}(\gamma\cdot c_{1}^{2})}{\operatorname{\mathrm{Aut}}(\mu)}\,. \end{align} $$

Our first result is the simplest case of Theorem 3.1.

Proposition 9. For all $k\geq 1$ , we have

$$ \begin{align*}\mathfrak{C}^{\bullet}(\mathrm{L}^{\mathrm{PT}}_{k}(1))= (\imath u)^{-k}\, \widetilde{\mathrm{L}}_{k}^{\mathrm{GW}}(1).\end{align*} $$

Proof. Since the operators $\mathrm {R}_{k}$ annihilate $1$ , we must prove

(4.3) $$ \begin{align} \mathfrak{C}^{\bullet}(\mathrm{T}_{k})= {(\imath u)^{-k}}\left(\frac{(\imath u)^{2}}{2}\mathrm{T}^{\prime}_{k}\right)\,. \end{align} $$

From Section 1.2, we have the following formula on the stable pairs side:

$$ \begin{align*}\mathrm{T}_{k}=-\frac{1}{2}\sum_{a+b=k+2}(-1)^{d^{L}d^{R}}(a+d^{L}-3)!(b+d^{R}-3)!\, \widetilde{\mathsf{ch}}_{a}\widetilde{\mathsf{ch}}_{b}(c_{1}).\end{align*} $$

On the Gromov–Witten side, we have

$$ \begin{align*}\mathrm{T}_{k}^{\prime}=-(\imath u)^{k-2}\sum_{a+b=k+2}(-1)^{d^{L}d^{R}}(a+d^{L}-3)!(b+d^{R}-3)!\frac{\mathfrak{a}_{a-1}\mathfrak{a}_{b-1}(c_{1})}{(a-1)!(b-1)!}\, \end{align*} $$

by Proposition 8. Using equation (4.2), the quadratic term in the $\mathfrak {a}$ -insertions of $\mathfrak {C}^{\bullet }(\mathrm {T}_{k})$ exactly matches the full right side of equation (4.3). We will prove the other terms of $\mathfrak {C}^{\bullet }(\mathrm {T}_{k})$ all vanish.

The stable pairs term $\mathrm {T}_{k}$ is the sum of three subsums:

(4.4) $$ \begin{align} &\frac{1}{2}\sum_{a+b=k+2}\bigg( (a-2)!b!\, \widetilde{\mathsf{ch}}_{a}(c_{1})\widetilde{\mathsf{ch}}_{b}(\operatorname{\mathrm{\mathsf{p}}}) +a!(b-2)!\, \widetilde{\mathsf{ch}}_{a}(\operatorname{\mathrm{\mathsf{p}}})\widetilde{\mathsf{ch}}_{b}(c_{1}) \nonumber\\ &\qquad\qquad\qquad\qquad\qquad\qquad-(a-1)!(b-1)! \sum_{s+1\le \bullet,\star\le 2s}\alpha_{\bullet \star}\, \widetilde{\mathsf{ch}}_{a}(\gamma_{\bullet}) \widetilde{\mathsf{ch}}_{b}(\gamma_{\star})\bigg)\, , \end{align} $$

where last term usesFootnote ²⁸

$$ \begin{align*}c_{1}\cdot\gamma_{2s+1-\bullet}=\sum_{\star} \alpha_{\bullet \star} \gamma_{\star} .\end{align*} $$

After applying $\mathfrak {C}^{\bullet }$ to equation (4.4) we obtain quadratic, cubic and quartic monomials in $\mathfrak {a}$ . We will show the cubic and quartic terms vanish.

We start with the analysis of the quartic term of $\mathfrak {C}^{\bullet }(\mathrm {T}_{k})$ . The first term (4.4) yields the quartic part:

$$ \begin{gather*} \frac{1}{2}\int_{X} c_{1}^{3}\cdot \sum_{a+b=k+2}\bigg((a-2)!b!\cdot\frac{\mathfrak{a}_{b-1}(\operatorname{\mathrm{\mathsf{p}}})}{(b-1)!}\cdot \frac{(\imath u)^{-2}}{(a-3)!}\sum_{|\mu|=a-5}\frac{\mathfrak{a}_{\mu_{1}}(\operatorname{\mathrm{\mathsf{p}}})\mathfrak{a}_{\mu_{2}}(\operatorname{\mathrm{\mathsf{p}}})\mathfrak{a}_{\mu_{3}}(\operatorname{\mathrm{\mathsf{p}}})}{\operatorname{\mathrm{Aut}}(\mu)} \\+(b-2)!a!\cdot\frac{\mathfrak{a}_{a-1}(\operatorname{\mathrm{\mathsf{p}}})}{(a-1)!}\cdot \frac{(\imath u)^{-2}}{(b-3)!}\sum_{|\mu|=b-5}\frac{\mathfrak{a}_{\mu_{1}}(\operatorname{\mathrm{\mathsf{p}}})\mathfrak{a}_{\mu_{2}}(\operatorname{\mathrm{\mathsf{p}}})\mathfrak{a}_{\mu_{3}}(\operatorname{\mathrm{\mathsf{p}}})}{\operatorname{\mathrm{Aut}}(\mu)}\bigg)\,. \end{gather*} $$

The last term of equation (4.4) yields the following quartic part (with the sum over the same range of a and b as above):

$$ \begin{align*}-\frac{1}{2}\int_X c_1^3\, \cdot\, (a-1)!(b-1)!\cdot\frac{(\imath u)^{-1}}{(a-2)!}\sum_{|\mu'|=a-3}\frac{\mathfrak{a}_{\mu'_1}(\operatorname{\mathrm{\mathsf{p}}})\mathfrak{a}_{\mu'_2}(\operatorname{\mathrm{\mathsf{p}}})}{\operatorname{\mathrm{Aut}}(\mu')}\cdot\frac{(\imath u)^{-1}}{(b-2)!} \sum_{|\mu''|=b-3}\frac{\mathfrak{a}_{\mu''_1}(\operatorname{\mathrm{\mathsf{p}}})\mathfrak{a}_{\mu''_2}(\operatorname{\mathrm{\mathsf{p}}})}{\operatorname{\mathrm{Aut}}(\mu'')},\end{align*} $$

where, in both formulas, we have used convention $|\mu |=\sum _{i}\mu _{i}$ .

These two quartic parts cancel each other. Indeed, let us analyze the factor in front of

$$ \begin{align*}\frac{1}{2(\imath u)^{2}}\int_{X} c_{1}^{3}\cdot \mathfrak{a}_{\lambda_{1}}(\operatorname{\mathrm{\mathsf{p}}})\mathfrak{a}_{\lambda_{2}}(\operatorname{\mathrm{\mathsf{p}}})\mathfrak{a}_{\lambda_{3}}(\operatorname{\mathrm{\mathsf{p}}})\mathfrak{a}_{\lambda_{4}}(\operatorname{\mathrm{\mathsf{p}}})\end{align*} $$

in both expressions. For simplicity, let us assume $|\operatorname {\mathrm {Aut}}(\lambda )|=1$ . Then, the factor in the first quartic part is a sum with four terms:

(4.5) $$ \begin{align} \sum_{i=1}^{4}\left(\lambda_{i}+1\right)\left(\sum_{j\ne i}(\lambda_{j}+1)\right)\,. \end{align} $$

The factor in the second formula is a sum with three terms:

(4.6) $$ \begin{align} -\sum (\lambda_{i_{1}}+\lambda_{i_{2}}+2)(\lambda_{j_{1}}+\lambda_{j_{2}}+2)\, , \end{align} $$

where the sum is over all splittings

$$ \begin{align*}\{1,2,3,4\}=\{i_{1},i_{2}\}\cup\{j_{1},j_{2}\}.\end{align*} $$

The factors (4.5) and (4.6) are sums of 12 monomials of $\lambda _{i}+1$ and are opposites of each other. The case when $|\operatorname {\mathrm {Aut}}(\lambda )|>1$ is analogous.

Finally, we analyze the cubic terms. Let us first analyze the cubic terms of the form $\mathfrak {a}_{i}(\operatorname {\mathrm {\mathsf {p}}})\mathfrak {a}_{j}(\operatorname {\mathrm {\mathsf {p}}})\mathfrak {a}_{l}(\operatorname {\mathrm {\mathsf {p}}})$ . Since

$$ \begin{align*}\mathsf{ch}_{k+2}(c_{1})\mathsf{ch}_{0}(\operatorname{\mathrm{\mathsf{p}}})= (-1)\mathsf{ch}_{k+2}(c_{1}),\end{align*} $$

the cubic part of the first term of equation (4.4) with $b=0$ is:

(4.7) $$ \begin{align} -k\int_{X}\frac{c_{1}^{3}}{2(\imath u)^{2}}\sum_{|\mu|=k-1}\frac{\mathfrak{a}_{\mu_{1}}(\operatorname{\mathrm{\mathsf{p}}})\mathfrak{a}_{\mu_{2}}(\operatorname{\mathrm{\mathsf{p}}})\mathfrak{a}_{\mu_{3}}(\operatorname{\mathrm{\mathsf{p}}})}{\operatorname{\mathrm{Aut}}(\mu)}\, .\end{align} $$

A similar cubic part is produced by the second term of equation (4.4) with $a=0$ .

The other cubic parts of the first term of equation (4.4) are

(4.8) $$ \begin{align} \int_{X}c_{1}^{3}\sum_{a+b=k+2}\frac{b}{2 (\imath u)^{2}}\mathfrak{a}_{b-1}(\operatorname{\mathrm{\mathsf{p}}}) \sum_{|\mu|=a-4}\frac{\mathfrak{a}_{\mu_{1}}(\operatorname{\mathrm{\mathsf{p}}})\mathfrak{a}_{\mu_{2}}(\operatorname{\mathrm{\mathsf{p}}})}{\operatorname{\mathrm{Aut}}(\mu)} +\frac{b}{2\imath u}\mathfrak{a}_{b-1}(\operatorname{\mathrm{\mathsf{p}}})\sum_{|\mu|=a-3}\frac{\mathfrak{a}_{\mu_{1}}\mathfrak{a}_{\mu_{2}}(c_{1}^{2})}{\operatorname{\mathrm{Aut}}(\mu)}\,. \end{align} $$

A similar term is yielded by the second term of equation (4.4).

If $\operatorname {\mathrm {Aut}}(\mu )=1$ , then the factor in front of monomial

$$ \begin{align*}\frac{1}{2(\imath u)^{2}}\mathfrak{a}_{\lambda_{1}}(\operatorname{\mathrm{\mathsf{p}}})\mathfrak{a}_{\lambda_{2}}(\operatorname{\mathrm{\mathsf{p}}})\mathfrak{a}_{\lambda_{3}}(\operatorname{\mathrm{\mathsf{p}}})\end{align*} $$

of equation (4.8) is the sum of three terms

$$ \begin{align*}(\lambda_{1}+1)+(\lambda_{2}+1)+(\lambda_{3}+1)\end{align*} $$

and, hence, cancels with corresponding monomial from equation (4.7).

The cubic part of the last term of equation (4.4) is

$$ \begin{align*} -\frac{(a-1)}{2 \imath u} \sum_{\bullet,\star}\alpha_{\bullet \star}\, \mathfrak{a}_{b-1}(\gamma_{\star})\sum_{|\mu|=a-3}\frac{\mathfrak{a}_{\mu_{1}}\mathfrak{a}_{\mu_{2}}(c_{1}\cdot\gamma_{\bullet})}{\operatorname{\mathrm{Aut}}(\mu)}\\ - \frac{(b-1)}{2 \imath u}\sum_{\bullet,\star}\alpha_{\bullet \star}\, \mathfrak{a}_{a-1}(\gamma_{\star})\sum_{|\mu|=b-3}\frac{\mathfrak{a}_{\mu_{1}}\mathfrak{a}_{\mu_{2}}(c_{1}\cdot\gamma_{\bullet})}{\operatorname{\mathrm{Aut}}(\mu)}, \end{align*} $$

over all $a,b\geq 0$ satisfying $a+b=k+2$ . The sum cancels with the last term of equation (4.8).

5.1 Overview

The main result of Section 5 is a proof of Theorem 3.1 for

(5.1) $$ \begin{align} D\in \mathbb{D}^{1}_{\mathrm{PT}} \cap \mathbb{D}^{X\bigstar}_{{\mathrm{PT}}\circ} \, , \end{align} $$

where D is a product of $\widetilde {\mathsf {ch}}_{k_{i}}(\gamma _{i})$ satisfying

$$ \begin{align*}\gamma_{i}\cdot \gamma_{j}=0\ \ \text{for}\ \ i\ne j.\end{align*} $$

We treat the singleton $D= \widetilde {\mathsf {ch}}_{k}({\mathsf {p}})$ in Proposition 10. An intricate computation is required for Proposition 11 which settles the cases $D= \widetilde {\mathsf {ch}}_{k}(\gamma )$ , where

$$ \begin{align*}\gamma\in H^{i}(X)\ \ \ {\text{for }i=2\text{ and 4.}}\end{align*} $$

Finally, in Section 5.3, the general case (5.1) is formally deduced from the singletons.

5.2 Intertwining shift operators

We first relate the operators $\mathrm{R}_{\mathrm{k}}$ appearing in the Virasoro constraints on the stable pairs and Gromov–Witten sides. Recall,

(5.2) $$ \begin{align} \widetilde{\mathsf{ch}}_{k}(\alpha)=\mathsf{ch}_{k}(\alpha)+\frac{1}{24}\mathsf{ch}_{k-2}(\alpha\cdot c_{2} )\,, \end{align} $$

so $\widetilde {\mathsf {ch}}_{k}(\operatorname {\mathrm {\mathsf {p}}})=\mathsf {ch}_{k}(\operatorname {\mathrm {\mathsf {p}}})$ .

Proposition 10. For all $k\geq 1$ and all $i\geq 2$ , we have

$$ \begin{align*}\mathfrak{C}^{\bullet}(\mathrm{R}_{k}( \mathsf{ch}_{i}(\operatorname{\mathrm{\mathsf{p}}})))=(\imath u )^{-k} \, \mathrm{R}_{k}(\mathfrak{C}^{\bullet}( \mathsf{ch}_{i}(\operatorname{\mathrm{\mathsf{p}}}))).\end{align*} $$

Proof. The left side of the equation is

$$ \begin{align*}\mathfrak{C}^{\bullet}(\mathrm{R}_{k}(\mathsf{ch}_{i}(\operatorname{\mathrm{\mathsf{p}}})))=\mathfrak{C}^{\bullet}\left(\frac{(i+k)!}{(i-1)!}\mathsf{ch}_{i+k}(\operatorname{\mathrm{\mathsf{p}}})\right)= \frac{(i+k)!}{(i-1)!}\frac{\mathfrak{a}_{i+k-1}(\operatorname{\mathrm{\mathsf{p}}})}{(i+k-1)!}=\frac{(i+k)}{(i-1)!} \, \mathfrak{a}_{i+k-1}(\operatorname{\mathrm{\mathsf{p}}}) ,\end{align*} $$

where we have used the definition of $\mathrm {R}_{k}$ for stable pairs and equation (1.14) for the correspondence.

The right side of the equation is

$$ \begin{align*}\mathrm{R}_{k}(\mathfrak{C}^{\bullet}(\mathsf{ch}_{i}(\operatorname{\mathrm{\mathsf{p}}})))&=\mathrm{R}_{k}\left(\frac{\mathfrak{a}_{i-1}(\operatorname{\mathrm{\mathsf{p}}})}{(i-1)!}\right)\\ &=\mathrm{R}_{k}\left(\frac{\tau_{i-2}(\operatorname{\mathrm{\mathsf{p}}})}{(\imath u)^{i-2}}\right)\\ &=\frac{(i+k)!}{(i-1)!}\frac{\tau_{i+k-2}(\operatorname{\mathrm{\mathsf{p}}})}{(\imath u)^{i-2}}\\ &= \frac{(i+k)}{(i-1)!}(\imath u)^{k}\, \mathfrak{a}_{i+k-1}(\operatorname{\mathrm{\mathsf{p}}}),\end{align*} $$

where we have used equation (1.14) for the correspondence, equation (1.13) and the definition of $\mathrm {R}_{k}$ for Gromov–Witten theory. The two sides match.

Proposition 11. For all $k\geq 1$ , $\widetilde {\mathsf {ch}}_{i}(\gamma )\in \mathbb {D}^{\bigstar X}_{\mathrm {PT}}$ , $\gamma \in H^{\ge 2}(X)$ , we have

$$ \begin{align*}\mathfrak{C}^{\bullet}(\mathrm{L}^{\mathrm{PT}}_{k}(\widetilde{\mathsf{ch}}_{i}(\gamma)))=(\imath u)^{-k}\, \widetilde{\mathrm{L}}^{\mathrm{GW}}_{k}(\mathfrak{C}^{\bullet}(\widetilde{\mathsf{ch}}_{i}(\gamma))).\end{align*} $$

Proof. We start with the easiest case and proceed to the hardest case.

$\underline {\mathbf {Case}\ \gamma \in H^{6}(X).}$ The case $\gamma =\operatorname {\mathrm {\mathsf {p}}}$ follows immediately from the previous results:

$$ \begin{align*}\mathfrak{C}^{\bullet}(\mathrm{L}^{\mathrm{PT}}_{k}(\mathsf{ch}_{i}(\operatorname{\mathrm{\mathsf{p}}})))&=\mathfrak{C}^{\bullet}( \mathrm{T}_{k}\, \mathsf{ch}_{i}(\operatorname{\mathrm{\mathsf{p}}}) + \mathrm{R}_{k}(\mathsf{ch}_{i}(\operatorname{\mathrm{\mathsf{p}}})))\\ &= \mathfrak{C}^{\bullet}(\mathrm{T}_{k})\mathfrak{C}^{\bullet}(\mathsf{ch}_{i}(\operatorname{\mathrm{\mathsf{p}}}))+ (\imath u)^{-k}\mathrm{R}_{k}(\mathfrak{C}^{\bullet}(\mathsf{ch}_{i}(\operatorname{\mathrm{\mathsf{p}}}))\\ &= (\imath u)^{-k}\tilde{\mathrm{L}}^{\mathrm{GW}}_{k}(\mathfrak{C}^{\bullet}(\mathsf{ch}_{i}(\operatorname{\mathrm{\mathsf{p}}}))). \end{align*} $$

The second equality follows from Proposition 10 and equation (3.18). The third equality uses equation (4.3).

$\underline {\mathbf {Case}\ \gamma \in H^{4}(X).}$ We compute the difference

(5.3) $$ \begin{align} (\imath u )^{k}\mathfrak{C}^{\bullet}(\mathrm{R}_{k}(\widetilde{\mathsf{ch}}_{i}(\gamma)))-\mathrm{R}_{k}(\mathfrak{C}^{\bullet}(\widetilde{\mathsf{ch}}_{i}(\gamma)))\,. \end{align} $$

Since $\gamma \cdot c_{2}=0$ , we have $\widetilde {\mathsf {ch}}_{k}(\gamma )=\mathsf {ch}_{k}(\gamma )$ by (5.2).

We start by expanding the first term of the difference:

$$ \begin{align*} \mathfrak{C}^{\circ}(\mathrm{R}_{k}(\mathsf{ch}_{i}(\gamma)))&=\mathfrak{C}^{\circ}\left(\frac{(i+k-1)!}{(i-2)!}\mathsf{ch}_{k+i}(\gamma)\right)\\ & =\frac{(i+k-1)!}{(i-2)!}\left(\frac{\mathfrak{a}_{i+k-1}(\gamma)}{(i+k-1)!} + \frac{(\imath u)^{-1}}{(i+k-2)!}\sum_{\mu_{1}+\mu_{2}=i+k-3}\frac{\mathfrak{a}_{\mu_{1}}\mathfrak{a}_{\mu_{2}}}{2}(\gamma \cdot c_{1})\right). \end{align*} $$

To proceed, we invert the correspondence (1.13):

(5.4) $$ \begin{align} \frac{(\imath u)^{k}\mathfrak{a}_{k+1}}{(k+1)!}(\gamma)=\tau_{k}(\gamma)+\left(\sum_{i=1}^{k}\frac{1}{i}\right)\tau_{k-1}(\gamma\cdot c_{1})+\left(\sum_{1\le i<j\le k}\frac{1}{ij}\right)\tau_{k-2}(\gamma\cdot c_{1}^{2})\,. \end{align} $$

We then obtain

(5.5) $$ \begin{align} &(\imath u)^{k} \mathfrak{C}^{\circ}(\mathrm{R}_{k}(\mathsf{ch}_{i}(\gamma)))=\frac{(i+k-1)!}{(i-2)!}\left(\frac{\tau_{i+k-2}(\gamma)}{(\imath u)^{i-2}}+\left(\sum_{j=1}^{i+k-2}\frac{1}{j}\right)\frac{\tau_{i+k-3}(\gamma \cdot c_{1})}{(\imath u)^{i-2}}\right.\nonumber\\ &\qquad\qquad\qquad\qquad\qquad\qquad\left.+ \frac{(\imath u)^{-i+4}}{(i+k-2)!}\sum_{\mu_{1}+\mu_{2}=i+k-3}\mu_{1}!\mu_{2}!\frac{\tau_{\mu_{1}-1}\tau_{\mu_{2}-1}}{2}(\gamma\cdot c_{1})\right). \end{align} $$

We write the second term of the difference as

(5.6) $$ \begin{align} \mathrm{R}_{k}(\mathfrak{C}^{\circ}(\mathsf{ch}_{i}(\gamma)))=\mathrm{R}_{k}\left(\frac{\mathfrak{a}_{i-1}(\gamma)}{(i-1)!}+\frac{(\imath u)^{-1}}{(i-2)!}\sum_{\mu_{1}+\mu_{2}=i-3}\frac{\mathfrak{a}_{\mu_{1}}\mathfrak{a}_{\mu_{2}}}{2}(\gamma \cdot c_{1})\right)\,. \end{align} $$

After applying the inversion (5.4), we have

$$ \begin{align*}\mathrm{R}_{k}\left(\frac{\tau_{i-2}(\gamma)}{(\imath u)^{i-2}}+\left(\sum_{j=1}^{i-2}\frac{1}{j}\right)\frac{\tau_{i-3}(\gamma \cdot c_{1})}{(\imath u)^{i-2}}+\frac{(\imath u)^{4-i}}{(i-2)!}\sum_{\mu_{1}+\mu_{2}=i-3}\mu_{1}!\mu_{2}! \frac{\tau_{\mu_{1}-1}\tau_{\mu_{2}-1}}{2}(\gamma \cdot c_{1})\right).\end{align*} $$

We expand the above expression fully to obtain

(5.7) $$ \begin{align} &\frac{(i+k-1)!\tau_{i+k-2}(\gamma)}{(\imath u)^{i-2}(i-2)!} \nonumber\\&+\frac{(i+k-1)!}{(\imath u)^{i-2}(i-2)!}\left(\sum_{j=i-1}^{k+i-1}\frac1j\right)\tau_{i+k-3}(\gamma \cdot c_{1}) +\frac{(i+k-1)!}{(\imath u)^{i-2}(i-2)!}\left(\sum_{j=1}^{i-2}\frac{1}{j}\right)\tau_{i-k+3}(\gamma \cdot c_{1})\nonumber\\ &+ \frac{(\imath u)^{-i+4}}{(i-2)!}\sum_{\mu_{1}+\mu_{2}=i-3}\left((\mu_{1}+k+1)!\mu_{2}!\frac{\tau_{\mu_{1}+k-1}\tau_{\mu_{2}-1}}{2}(\gamma \cdot c_{1}) +\mu_{1}!(\mu_{2}+k+1)!\frac{\tau_{\mu_{1}-1}\tau_{\mu_{2}+k-1}}{2}(\gamma \cdot c_{1})\right), \end{align} $$

where we have used formula

$$ \begin{align*}[i]^{k}_{1}=\frac{(i+k)!}{(i-1)!}\sum_{j=i}^{i+k}\frac1j\end{align*} $$

in the expansion of $\mathrm {R}_{k}(\tau _{i-2}(\gamma ))$ .

To complete our computation of the difference (5.3), we observe several cancellations. The first term of equation (5.5) cancels with first term of equation (5.7). The second term of equation (5.5) almost cancels with the sum of the second and third terms of equation (5.7); the only term that does not cancel is

(5.8) $$ \begin{align} -\frac{(i+k-2)!}{(\imath u)^{i-2}(i-2)!}\tau_{i+k-3}(\gamma\cdot c_{1}). \end{align} $$

Finally, we rewrite the last term of equation (5.5) as

$$ \begin{align*}\frac{(\imath u)^{-i+4}}{(i-2)!}\sum_{\mu_{1}+\mu_{2}=i+k-3}(\mu_{1}+1)!\mu_{2}!\frac{\tau_{\mu_{1}-1}\tau_{\mu_{2}-1}}{2}(\gamma \cdot c_{1})+\mu_{1}!(\mu_{2}+1)!\frac{\tau_{\mu_{1}-1}\tau_{\mu_{2}-1}}{2}(\gamma \cdot c_{1}).\end{align*} $$

Then, we see that the last term of equation (5.6) cancels with the last term of equation (5.5) if $\mu _{1}\ge k+1$ and $\mu _{2}\ge k+1$ . Thus, the difference (5.3) equals

(5.9) $$ \begin{align} &\frac{(\imath u)^{-i+4}}{(i-2)!}\left(\sum_{\mu_{1}+\mu_{2}=i+k-3,\, \mu_{1}\le k}(\mu_{1}+1)!\mu_{2}!\frac{\tau_{\mu_{1}-1}\tau_{\mu_{2}-1}}{2}(\gamma \cdot c_{1})\right.\nonumber\\&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\left.+ \sum_{\mu_{1}+\mu_{2}=i+k-3,\, \mu_{2}\le k}\mu_{1}!(\mu_{2}+1)!\frac{\tau_{\mu_{1}-1}\tau_{\mu_{2}-1}}{2}(\gamma \cdot c_{1})\right). \end{align} $$

We now include the $\mathrm {T}_{k}$ and $\mathrm {T}_{k}^{\prime }$ terms in the difference. We have

(5.10) $$ \begin{align} &(\imath u)^{k}\mathfrak{C}^{\bullet}(\mathrm{L}^{{\mathrm{PT}}}_{k}(\mathsf{ch}_{i}(\gamma)))-{\widetilde{\mathrm{L}}}^{\mathrm{GW}}_{k}(\mathfrak{C}^{\bullet}(\mathsf{ch}_{i}(\gamma)))\nonumber\\ &\quad =(\imath u )^{k}\mathfrak{C}^{\bullet}(\mathrm{R}_{k}(\mathsf{ch}_{i}(\gamma)))-\mathrm{R}_{k}(\mathfrak{C}^{\bullet}(\mathsf{ch}_{i}(\gamma))) +(\imath u )^{k}\mathfrak{C}^{\bullet}(\mathrm{T}_{k}(\mathsf{ch}_{i}(\gamma)))-\frac{(\imath u)^{2}}{2}\mathrm{T}'_{k}(\mathfrak{C}^{\bullet}(\mathsf{ch}_{i}(\gamma))). \end{align} $$

Using equation (4.3), the $\mathrm {T}_{k}$ and $\mathrm {T}_{k}^{\prime }$ terms in equation (5.10) simplify to

(5.11) $$ \begin{align} &\frac{(\imath u)^{k}}{2}\sum_{a+b=k+2}(a-2)!b! \mathfrak{C}^{\circ}\left(\frac{\widetilde{\mathsf{ch}}_{a}(c_{1})\mathsf{ch}_{i}(\gamma)}{(\imath u)^{b-2}}\right)\tau_{b-2}(\operatorname{\mathrm{\mathsf{p}}})\nonumber\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\frac{(\imath u)^{k}}{2}\sum_{a+b=k+2} a!(b-2)!\tau_{a-2}(\operatorname{\mathrm{\mathsf{p}}})\mathfrak{C}^{\circ}\left( \frac{\widetilde{\mathsf{ch}}_{b}(c_{1})\mathsf{ch}_{i}(\gamma)}{(\imath u)^{a-2}}\right)\,. \end{align} $$

To complete our proof, we require the bumping formula (1.15):

(5.12) $$ \begin{align} \mathfrak{C}^{\circ}(\widetilde{\mathsf{ch}}_{k_{1}+2}(c_{1})\widetilde{\mathsf{ch}}_{k_{2}+2}(\gamma))=-\frac{1}{k_{1}!k_{2}!} (\imath u)^{-1}\mathfrak{a}_{k_{1}+k_{2}}(c_{1}\gamma)\,. \end{align} $$

Since $\gamma \in H^{4}(X)$ , all the other terms of equation (1.15) vanish. We apply the bumping formula (5.11). In particular, the first term of equation (5.11),

$$ \begin{align*}(a-2)!b! \mathfrak{C}^{\circ}\left(\frac{\widetilde{\mathsf{ch}}_{a}(c_{1})\mathsf{ch}_{i}(\gamma)}{(\imath u)^{b-2}}\right)\tau_{b-2}(\operatorname{\mathrm{\mathsf{p}}})=- (\imath u)^{-a-b-i+6}\frac{(a+i-2)!b!}{(i-2)!}\tau_{a+i-3}(\gamma\cdot c_{1})\tau_{b-2}(\operatorname{\mathrm{\mathsf{p}}})\, \end{align*} $$

cancels with the first term of equation (5.9). Similarly, the second term of equation (5.11) cancels with the second term of equation (5.9).

Let us observe that the term of last expression with $a=1$ by the exceptional bumping (3.4) turns into the terms of equation (5.9) with $\mu _{1}=k$ or $\mu _{2}=k$ . Similarly, the term with $b=0$ cancels out with the term (5.8).

Also, the assumption $\widetilde {\mathsf {ch}}_{i}(\gamma )\in \mathbb {D}^{X\bigstar }_{\mathrm {PT}}$ implies that $i\ge 2$ ; thus, no negative factorials appear in the above computations.

$\underline {\mathbf {Case}\ \gamma \in H^{2}(X).}$ If $\gamma \in H^{2}(X)$ , the $\mathrm {T}_{k}$ and $\mathrm {T}_{k}^{\prime }$ terms of the formula (5.10) acquires extra summands:

(5.13) $$ \begin{align} &(\imath u)^{k}\mathfrak{C}^{\bullet}(\mathrm{L}^{\mathrm{PT}}_{k}(\widetilde{\mathsf{ch}}_{i}(\gamma)))-\widetilde{\mathrm{L}}_{k}^{\mathrm{GW}}(\mathfrak{C}^{\bullet}(\widetilde{\mathsf{ch}}_{i}(\gamma)))\nonumber\\&\quad =(\imath u )^{k}\mathfrak{C}^{\bullet}(\mathrm{R}_{k}(\widetilde{\mathsf{ch}}_{i}(\gamma)))-\mathrm{R}_{k}(\mathfrak{C}^{\bullet}(\widetilde{\mathsf{ch}}_{i}(\gamma)))\nonumber\\&\qquad + \frac{(\imath u)^{k}}{2}\bigg[\sum_{a+b=k+2}(a-2)!b! \mathfrak{C}^{\circ}\left(\frac{\widetilde{\mathsf{ch}}_{a}(c_{1})\mathsf{ch}_{i}(\gamma)}{(\imath u)^{b-2}}\right)\tau_{b-2}(\operatorname{\mathrm{\mathsf{p}}})+a!(b-2)!\tau_{a-2}(\operatorname{\mathrm{\mathsf{p}}})\mathfrak{C}^{\circ}\left( \frac{\widetilde{\mathsf{ch}}_{b}(c_{1})\mathsf{ch}_{i}(\gamma)}{(\imath u)^{a-2}}\right)\nonumber\\&\qquad - \sum_{a+b=k+2}(a-1)!(b-1)!\sum_{0<\bullet,\star <2s+1} \alpha_{\bullet \star}\left(\mathfrak{C}^{\circ}(\widetilde{\mathsf{ch}}_{a}(\gamma_{\bullet})\cdot \mathsf{ch}_{i}(\gamma))\mathfrak{C}^{\circ}(\widetilde{\mathsf{ch}}_{b}(\gamma_{\star}))\right.\nonumber\\&\qquad \left.+\mathfrak{C}^{\circ}(\widetilde{\mathsf{ch}}_{a}(\gamma_{\bullet}))\mathfrak{C}^{\circ}(\widetilde{\mathsf{ch}}_{b}(\gamma_{\star})\cdot\mathsf{ch}_{i}(\gamma))\right)\bigg]\, , \end{align} $$

where we have usedFootnote ²⁹ $c_{1}\cdot \gamma _{2s+1-\bullet }=\sum _{\star } \alpha _{\bullet \star }\gamma _{\star }$ . Nevertheless, the strategy used in the previous case can be pursued also for $\gamma \in H^{2}(X)$ . The computation, which is carried out below, is of course more complicated.

We will study the difference

(5.14) $$ \begin{align} (\imath u )^{k}\mathfrak{C}^{\bullet}(\mathrm{R}_{k}(\widetilde{\mathsf{ch}}_{i}(\gamma)))-\mathrm{R}_{k}(\mathfrak{C}^{\bullet}(\widetilde{\mathsf{ch}}_{i}(\gamma)))\, \\[-15pt]\nonumber\end{align} $$

with $\gamma \in H^{2}(X)$ . The expansion of the first term is

(5.15) $$ \begin{align} (\imath u)^{k}\mathfrak{C}^{\circ}(\mathrm{R}_{k}(\widetilde{\mathsf{ch}}_{i}(\gamma))) &=(\imath u)^{k}\frac{(k+i-2)!}{(i-3)!}\mathfrak{C}^{\circ}(\widetilde{\mathsf{ch}}_{i+k}(\gamma))\nonumber\\ &=(\imath u)^{k}\frac{(k+i-2)!}{(i-3)!}\left(\frac{\mathfrak{a}_{i+k-1}(\gamma)}{(i+k-1)!} +\frac{(\imath u)^{-1}}{(i+k-2)!}\sum_{|\mu|=i+k-3}\frac{\mathfrak{a}_{\mu_{1}}\mathfrak{a}_{\mu_{2}}}{\operatorname{\mathrm{Aut}}(\mu)}(\gamma \cdot c_{1}) \right.\nonumber\\ &\quad \left. +\frac{(\imath u)^{-2}}{(i+k-2)!}\sum_{|\mu|=i+k-4} \frac{\mathfrak{a}_{\mu_{1}}\mathfrak{a}_{\mu_{2}}}{\operatorname{\mathrm{Aut}}(\mu)}(\gamma \cdot c_{1}^{2})\right.\nonumber\\ &\quad \left.+\frac{(\imath u)^{-2}}{(i+k-3)!}\sum_{|\mu|=i+k-5}\frac{\mathfrak{a}_{\mu_{1}}\mathfrak{a}_{\mu_{2}}\mathfrak{a}_{\mu_{3}}}{\operatorname{\mathrm{Aut}}(\mu)}(\gamma \cdot c_{1}^{2})\right)\,.\\[-15pt]\nonumber \end{align} $$

The second term of the difference (5.14) is more involved since we must transform the descendents $\mathfrak {a}$ to the standard descendents $\tau $ before applying the shift operator $\mathrm {R}_{k}$ :

(5.16) $$ \begin{align} \mathrm{R}_{k}(\mathfrak{C}^{\circ}(\widetilde{\mathsf{ch}}_{i}(\gamma)))&=\$\imath u)^{-(i-2)}\mathrm{R}_{k}\left(\tau_{i-2}(\gamma)+\left(\sum_{j=1}^{i-2}\frac{1}{j}\right) \tau_{i-3}(\gamma \cdot c_{1})+\left(\sum_{1\le j<l\le i-2}\frac{1}{jl}\right)\tau_{i-4}(\gamma\cdot c_{1}^{2})\right)\nonumber\\ &\quad +(\imath u)^{-(i-5)}\mathrm{R}_{k}\left(\frac{(\imath u)^{-1}}{(i-2)!}\left(\sum_{|\mu|=i-3}\frac{\mu_{1}!\mu_{2}!}{\operatorname{\mathrm{Aut}}(\mu)} \left(\tau_{\mu_{1}-1}\tau_{\mu_{2}-1}(\gamma \cdot c_{1})+ \left[\left(\sum_{j=1}^{\mu_{1}}\frac{1}{j}\right)\tau_{\mu_{1}-2}\tau_{\mu_{2}-1}\right.\right.\right.\right.\nonumber\\ &\quad \!\!\left.\left.\left. +\left(\sum_{j=1}^{\mu_{2}-1}\frac{1}{j}\right)\tau_{\mu_{1}-1}\tau_{\mu_{2}-2}\right](\gamma\cdot c_{1}^{2})\right)\right)\nonumber\\&\quad \left.+\frac{1}{(i-3)!}\sum_{|\mu|=i-5}\frac{\mu_{1}!\mu_{2}!\mu_{3}!}{\operatorname{\mathrm{Aut}}(\mu)} \tau_{\mu_{1}-1}\tau_{\mu_{2}-1}\tau_{\mu_{3}-1}(\gamma\cdot c_{1}^{2})\right)\,. \end{align} $$

Notice the upper limits in the first harmonic sum is $\mu _{1}$ ; the terms with $j=\mu _{1}$ correspond to the third term of equation (1.14).

We will study the right-hand side of equation (5.13) using equations (5.15) and (5.16) in three steps corresponding to the $\tau $ -degree.

$\bullet $ Consider first the $\tau $ -linear terms. The $\tau $ -linear terms of equation (5.15) are

(5.17) $$ \begin{align} (\imath u)^{k} \frac{(i+k-2)!}{(i-3)!}\left(\frac{1}{(\imath u)^{i+k-2}}\left( \tau_{i+k-2}(\gamma)+\left(\sum_{j=1}^{i+k-2}\frac{1}{j}\right)\tau_{i+k-3}(c_{1}\cdot\gamma)\right.\right. \nonumber\\ \left.\left. +\left(\sum_{1\le j<l\le i+k-2}\frac{1}{jl}\right)\tau_{i+k-4}(c_{1}^{2}\cdot \gamma)\right)\right).\end{align} $$

The $\tau $ -linear terms of equation (5.16) are more complicated:

(5.18) $$ \begin{align} &(\imath u)^{-i+2}\frac{(i+k-2)!}{(i-3)!}\left(\tau_{i+k-2}(\gamma)+\left(\sum_{j=i-2}^{i+k-2}\frac{1}{j}\right)\tau_{i+k-3}(\gamma\cdot c_{1})\right.\nonumber\\ &\quad \left.+\left(\sum_{i-2\le j<l\le i+k-2}\frac{1}{jl}\right) \tau_{i+k-4}(\gamma\cdot c_{1}^{2})+\left(\sum_{j=1}^{i-2}\frac{1}{j}\right)\left[ \tau_{i+k-3}(\gamma \cdot c_{1})+\left(\sum_{j=i-2}^{i+k-2}\frac{1}{j}\right)\tau_{i+k-4}(\gamma \cdot c_{1}^{2})\right]\right.\nonumber\\ &\quad \left.+\left(\sum_{1\le j<l\le i-2}\frac{1}{jl}\right) \tau_{i+k-4}(\gamma\cdot c_{1}^{2})\right)\,. \end{align} $$

The $\tau _{i+k-2}(\gamma )$ terms of equations (5.17) and (5.18) match so cancel in the difference (5.14). The $\tau _{i+k-3}(\gamma \cdot c_{1})$ terms in equations (5.17) and (5.18) almost cancel: The difference is

(5.19) $$ \begin{align} (\imath u)^{-i+2}\frac{(i+k-2)!}{(i-2)!}\tau_{i+k-3}(\gamma\cdot c_{1})\,. \end{align} $$

For the $\tau _{i+k-4}(\gamma \cdot c_{1}^{2})$ terms, we split the prefactor in equation (5.18) as

$$ \begin{align*}\sum_{j=1}^{i-2}\frac{1}{j}= \frac{1}{i-2}+\sum_{j=1}^{i-3}\frac{1}{j}\,\end{align*} $$

and the last coefficient of equation (5.18) as

$$ \begin{align*}\sum_{1\le j<l\le i-2}\frac{1}{jl} = \sum_{1\le j<l\le i-3}\frac{1}{jl}+ \frac{1}{i-2}\sum_{1\le j\le i-3} \frac{1}{j}.\end{align*} $$

Then, we see the difference of the $\tau _{i+k-4}(\gamma \cdot c_{1}^{2})$ terms in equations (5.17) and (5.18) is

(5.20) $$ \begin{align} (\imath u)^{-i+2}\frac{(i+k-2)!}{(i-2)!}\left(\sum_{j=1}^{i+k-2}\frac{1}{j}\right) \tau_{i+k-4}(c_{1}^{2}\cdot\gamma)\,. \end{align} $$

On the right-hand side of equation (5.13), the $\tau $ -linear terms (5.19) and (5.20) of the difference (5.14) are canceled with

$$ \begin{align*}\frac{(\imath u)^{k}}{2}\bigg[k!0! \mathfrak{C}^{\circ}\left(\frac{\widetilde{\mathsf{ch}}_{k+2}(c_{1})\mathsf{ch}_{i}(\gamma)}{(\imath u)^{-2}}\right)\tau_{-2}(\operatorname{\mathrm{\mathsf{p}}})+0!k!\tau_{-2}(\operatorname{\mathrm{\mathsf{p}}})\mathfrak{C}^{\circ}\left( \frac{\widetilde{\mathsf{ch}}_{k+2}(c_{1})\mathsf{ch}_{i}(\gamma)}{(\imath u)^{-2}}\right)\bigg]\end{align*} $$

using $\tau _{-2}(\operatorname {\mathrm {\mathsf {p}}})=1$ . In fact, after applying equation (1.15), we find

$$ \begin{align*} &(\imath u)^{k}\frac{k!}{(\imath u)^{-2}}\mathfrak{C}^{\circ}(\widetilde{\mathsf{ch}}_{k+2}(c_{1})\widetilde{\mathsf{ch}}_{i}(\gamma)) \\ &\quad = - \frac{(\imath u)^{k+1}}{(i-2)!}\bigg(\mathfrak{a}_{k+i-2}(c_{1}\gamma) +(\imath u)^{-1} \mathfrak{a}(c_{1}^{2}\gamma)\bigg)+\dots \\ &\quad = -\frac{(\imath u)^{2-i}}{(i-2)!}\bigg((k+i-2)!\bigg(\tau_{k+i-3}(\gamma\cdot c_{1})+ (\sum_{i=1}^{k+i-3}\frac{1}{i})\tau_{k+i-4}(\gamma\cdot c_{1}^{2}) \bigg)\\&\qquad + (k+i-3)!\tau_{k+i-4}(\gamma\cdot c_{1}^{2}) \bigg)+\dots. \end{align*} $$

where the dots stand for the $\tau $ -quadratic terms. The second equality follows from the formula (5.4).

$\bullet $ Consider next the $\tau $ -quadratic terms. We start with the quadratic terms of complex cohomological degree $2$ . The corresponding terms from equation (5.15) are

(5.21) $$ \begin{align} (\imath u)^{k}\frac{(k+i-2)!}{(i-3)!}\sum_{|\mu|=i+k-3} \frac{(\imath u)^{-\mu_{1}-\mu_{2}+2}\mu_{1}!\mu_{2}!}{(\imath u)(i+k-2)!\operatorname{\mathrm{Aut}}(\mu)} \tau_{\mu_{1}-1}\tau_{\mu_{2}-1}(\gamma \cdot c_{1})\,. \end{align} $$

The computation of the corresponding terms in equation (5.16) are more involved since the action of the shift operator $\mathrm {R}_{k}$ depends on the complex cohomological degree of the descendent:

(5.22) $$ \begin{align} &(\imath u)^{-i+4}\frac{1}{(i-2)!}\sum_{|\mu|=i-3}\frac{\mu_{1}!\mu_{2}!} {\operatorname{\mathrm{Aut}}(\mu)}\left(\frac{(\mu_{1}+k)!}{(\mu_{1}-1)!}\tau_{\mu_{1}+k-1}(\gamma\cdot c_{1})\tau_{\mu_{2}-1}(\operatorname{\mathrm{\mathsf{p}}}) \right.\nonumber\\ &\qquad\qquad\qquad\qquad\qquad\qquad\quad\qquad \left. +\frac{(\mu_{2}+k+1)!}{(\mu_{2})!}\tau_{\mu_{1}-1}(\gamma\cdot c_{1})\tau_{\mu_{2}-1+k}(\operatorname{\mathrm{\mathsf{p}}}) \right)\,. \end{align} $$

The linear combination of the first term of equation (5.22) with $\mu _{1}+k-1=a$ and second term with $\mu _{1}-1=a$ is equal to the corresponding term of eqiatopm (5.21) with $\mu _{1}-1=a$ . Hence, these cancel in the difference. Similar cancellations happen with rest of the terms. The resulting difference of equations (5.21) and (5.22) is

(5.23) $$ \begin{align} \frac{(\imath u)^{-i+4}}{(i-3)!}\left(\sum_{|\mu|=i+k-3,\, \mu_{1}\le k}\mu_{1}!\mu_{2}!\frac{\tau_{\mu_{1}-1}\tau_{\mu_{2}-1}}{\operatorname{\mathrm{Aut}}(\mu)}(\gamma \cdot c_{1}) + \sum_{|\mu|=i+k-3,\, \mu_{2}\le k}\mu_{1}!\mu_{2}!\frac{\tau_{\mu_{1}-1}\tau_{\mu_{2}-1}}{\operatorname{\mathrm{Aut}}(\mu)}(\gamma \cdot c_{1})\right). \end{align} $$

We will cancel equation (5.23) with the $\tau $ -quadratic terms of complex cohomological degree 2 in the sum

(5.24) $$ \begin{align} \frac{(\imath u)^{k}}{2}\bigg[\sum_{a+b=k+2}(a-2)!b! \mathfrak{C}^{\circ}\left(\frac{\widetilde{\mathsf{ch}}_{a}(c_{1})\mathsf{ch}_{i}(\gamma)}{(\imath u)^{b-2}}\right)\tau_{b-2}(\operatorname{\mathrm{\mathsf{p}}})+a!(b-2)!\tau_{a-2}(\operatorname{\mathrm{\mathsf{p}}})\mathfrak{C}^{\circ}\left( \frac{\widetilde{\mathsf{ch}}_{b}(c_{1})\mathsf{ch}_{i}(\gamma)}{(\imath u)^{a-2}}\right)\nonumber\\- \sum_{a+b=k+2}(a-1)!(b-1)!\sum_{\bullet,\star} \alpha_{\bullet \star}\, \left(\mathfrak{C}^{\circ}(\widetilde{\mathsf{ch}}_{a}(\gamma_{\bullet})\cdot \mathsf{ch}_{i}(\gamma))\mathfrak{C}^{\circ}(\widetilde{\mathsf{ch}}_{b}(\gamma_{\star}))\right.\nonumber\\\left.+\mathfrak{C}^{\circ}(\widetilde{\mathsf{ch}}_{a}(\gamma_{\bullet}))\mathfrak{C}^{\circ}(\widetilde{\mathsf{ch}}_{b}(\gamma_{\star})\cdot\mathsf{ch}_{i}(\gamma))\right)\bigg]\,. \end{align} $$

More precisely, the first and second terms of the last expression yield

$$ \begin{align*}\frac{(\imath u)^{-i+4}}{2}\bigg[ b \frac{(a+i-4)!(b-1)!}{(i-2)!}\tau_{a+i-5}(\gamma\cdot c_{1})\tau_{b-2}(\operatorname{\mathrm{\mathsf{p}}} )+ a \frac{(a-1)!(b+i-4)!}{(i-2)!}\tau_{a-2}(\operatorname{\mathrm{\mathsf{p}}})\tau_{b+i-5}(\gamma\cdot c_{1}) \bigg],\end{align*} $$

and the last two terms yieldFootnote ³⁰

$$ \begin{align*} &-\frac{(\imath u)^{-i-4}}{2}\bigg[ (a-1)\frac{(a+i-4)!(b-1)!}{(i-2)!} \tau_{a+i-5}(\gamma_{\bullet}\cdot \gamma)\tau_{b-2}(\gamma_{\star})\\ &\qquad\qquad\qquad\qquad\quad\qquad\qquad\qquad\qquad +(b-1)\frac{(a-1)!(b+i-4)!}{(i-2)!} \tau_{a-2}(\gamma_{\bullet}\cdot \gamma)\tau_{b+i-5}(\gamma_{\star}) \bigg].\end{align*} $$

The cancellation then follows from

$$ \begin{align*}\sum_{\bullet,\star}\alpha_{\bullet\star}\, (\gamma_{\bullet}\cdot \gamma)\otimes \gamma_{\star}=\operatorname{\mathrm{\mathsf{p}}}\otimes (\gamma\cdot c_{1}) \ \ \ \text{and}\ \ \ \sum_{\bullet,\star}\alpha_{\bullet\star}\, \gamma_{\bullet}\otimes(\gamma_{\star}\cdot \gamma)=(\gamma\cdot c_{1})\otimes \operatorname{\mathrm{\mathsf{p}}} .\end{align*} $$

We have cancelled all $\tau $ -quadratic terms of complex cohomological degree 2 in equation (5.13).

Let us also observe that the terms of equation (5.24) with $a=1$ and with $b=1$ cancel out by exceptional bumping with equation (3.4) with the term of equation (5.23) with $\mu _{1}=k$ or $\mu _{2}=k$ .

A longer computation is needed to deal with $\tau $ -quadratic terms of complex cohomological degree 3. Since all such terms have $\gamma \cdot c_{1}^{2}$ as an argument, we drop the cohomology insertion from the notation. The corresponding terms from equation (5.15) are:

(5.25) $$ \begin{align} &(\imath u)^{k}\frac{(k+i-2)!}{(i-3)!}\sum_{|\mu|=i+k-4}\frac{(\imath u)^{-\mu_{1}-\mu_{2}}}{\operatorname{\mathrm{Aut}}(\mu)(i+k-2)!}\left(\mu_{1}!\mu_{2}!+(\mu_{1}+1)!\mu_{2}!\left(\sum_{j=1}^{\mu_{1}}\frac{1}{j}\right)\right.\nonumber\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \left. +\mu_{1}!(\mu_{2}+1)!\left(\sum_{j=1}^{\mu_{2}}\frac{1}{j}\right)\right)\tau_{\mu_{1}-1}\tau_{\mu_{2}-1}\,. \end{align} $$

The corresponding terms from equation (5.16) are:

(5.26) $$ \begin{align} &\frac{(\imath u)^{-i+4}}{(i-2)!}\sum_{|\mu|=i-3}\frac{\mu_{1}!\mu_{2}!}{\operatorname{\mathrm{Aut}}(\mu)} \left[\frac{(\mu_{1}+k)!}{(\mu_{1}-1)!}\left(\sum_{j=\mu_{1}}^{\mu_{1}+k}\frac{1}{j}\right) \tau_{\mu_{1}+k-2}\tau_{\mu_{2}-1}\right.\nonumber\\ &\quad \left.+\frac{(\mu_{2}+k)!}{(\mu_{2}-1)!}\left(\sum_{j=\mu_{2}}^{\mu_{2}+k}\frac{1}{j}\right)\tau_{\mu_{1}-1}\tau_{\mu_{2}+k-2}+\left(\sum_{j=1}^{\mu_{1}}\frac{1}{j}\right)\left(\frac{(\mu_{1}+k)!}{(\mu_{1}-1)!}\tau_{\mu_{1}+k-2}\tau_{\mu_{2}-1}\right.\right. \nonumber\\ &\quad \left.\left.+\frac{(\mu_{2}+k+1)!}{\mu_{2}!}\tau_{\mu_{1}-2}\tau_{\mu_{2}+k-1}\right)+ \left(\sum_{j=1}^{\mu_{2}-1}\frac{1}{j}\right)\left(\frac{(\mu_{2}+k)!}{(\mu_{2}-1)!}\tau_{\mu_{1}-1}\tau_{\mu_{2}+k-2}\right.\right.\nonumber\\ &\quad \left.\left.+\frac{(\mu_{1}+k+1)!}{\mu_{1}!}\tau_{\mu_{1}+k-1}\tau_{\mu_{2}-2}\right)\right]\,. \end{align} $$

The expression (5.26) is simplified by the following strategy. We number the six $\tau $ -quadratic terms by their order of occurrence in equation (5.26). The first term of equation (5.26) combines with the third term. The second term combines with the fifth term. We also split off the summands with $j=\mu _{1}+k$ and $j=\mu _{2}+k$ from the first and second terms, respectively, as well as the summand with $j=\mu _{1}$ from the third term. Then, equation (5.26) equals

(5.27) $$ \begin{align} &\frac{(\imath u)^{-i+4}}{(i-2)!}\left(\sum \mu_{1}\frac{(\mu_{1}+k-1)!\mu_{2}!}{\operatorname{\mathrm{Aut}}(\mu)}\tau_{\mu_{1}+k-2}\tau_{\mu_{2}-1}+ \mu_{2}\frac{\mu_{1}!(\mu_{2}+k-1)!}{\operatorname{\mathrm{Aut}}(\mu)}\tau_{\mu_{1}-1}\tau_{\mu_{2}+k-2}\right.\nonumber\\ &\quad \left.+ \mu_{1}\frac{(\mu_{1}+k)!\mu_{2}!}{\operatorname{\mathrm{Aut}}(\mu)}\left(\sum_{j=1}^{\mu_{1}+k-1}\frac{1}{j}\right)\tau_{\mu_{1}+k-2}\tau_{\mu_{2}-1} +\mu_{2}\frac{\mu_{1}!(\mu_{2}+k)!}{\operatorname{\mathrm{Aut}}(\mu)}\left(\sum_{j=1}^{\mu_{2}+k-1}\frac{1}{j}\right)\tau_{\mu_{1}-1}\tau_{\mu_{2}+k-2}\right.\nonumber\\ &\quad \left. +(\mu_{2}+k+1)\frac{\mu_{1}!(\mu_{2}+k)!}{\operatorname{\mathrm{Aut}}(\mu)} \left(\sum_{j=1}^{\mu_{1}-1}\frac{1}{j}\right)\tau_{\mu_{1}-2}\tau_{\mu_{2}+k-1}\right.\nonumber\\ &\quad \left.+ (\mu_{1}+k+1)\frac{(\mu_{1}+k)!\mu_{2}!}{\operatorname{\mathrm{Aut}}(\mu)} \left(\sum_{j=1}^{\mu_{2}-1}\frac{1}{j}\right)\tau_{\mu_{1}+k-1}\tau_{\mu_{2}-2}\right.\nonumber\\ &\quad \left. +\frac{(\mu_{1}+k)!\mu_{2}!}{\operatorname{\mathrm{Aut}}(\mu)}\tau_{\mu_{1}+k-2}\tau_{\mu_{2}-1}+ \frac{(\mu_{1}-1)!(\mu_{2}+k+1)!}{\operatorname{\mathrm{Aut}}(\mu)}\tau_{\mu_{1}-2}\tau_{\mu_{2}+k-1}\right), \end{align} $$

where the sum is over $\mu _{1}\geq \mu _{2}$ , $|\mu |=i-3$ .

Let us fix an integer a satisfying $a>k-2$ . We observe that the sum of the first term from the first line of equation (5.27) with $\mu _{1}=a+2-k$ and the second term in the last line with $\mu _{2}=a+1-k$ will cancel with the first term of equation (5.25) with $\mu _{1}=a+1$ . Also, the sum of the second term from the first line with $\mu _{2}=a+2-k$ and the first term of the last line with $\mu _{1}=a+1-k$ will cancel with the first term of equation (5.25) with $\mu _{2}=a+1$ .

Similarly, the sum of the first term for the second line of equation (5.27) with $\mu _{1}=a+2-k$ and the first term from the third line of equation (5.27) with $\mu _{1}=a+2$ cancels with the second term of equation (5.25) with $\mu _{1}=a+1$ . Finally, the sum of the second term from the second line of equation (5.27) with $\mu _{1}=a+1$ and the last term from the last line of equation (5.27) with $\mu _{1}=a+1-k$ cancels with the last term of equation (5.25) with $\mu _{1}=a+1$ .

After all of these cancellations, we are left with

(5.28) $$ \begin{align} \sum \frac{(\imath u)^{-i+4}\mu_{1}!\mu_{2}!}{\operatorname{\mathrm{Aut}}(\mu)(i-3)!}\left(1+(\mu_{1}+1)\left(\sum_{j=1}^{\mu_{1}}\frac{1}{j}\right) +(\mu_{2}+1)\left(\sum_{j=1}^{\mu_{2}}\frac{1}{j}\right)\right)\tau_{\mu_{1}-1}\tau_{\mu_{2}-1}\, , \end{align} $$

where $\sum $ is the sum of two subsums: The first is over $\mu _{1}+\mu _{2}=i+k-4$ , $\mu _{1}\le k$ , and the second is over $\mu _{1}+\mu _{2}=i+k-4$ , $\mu _{2}\le k$ .

In the difference (5.13), the expression (5.28) is canceled by the corresponding $\tau $ -quadratic terms of complex cohomological degree 3 of equation (5.24). More precisely, the first and second terms of equation (5.24) yield

$$ \begin{align*}\frac{(\imath u)^{-i+4}}{2}\bigg[ b \frac{(a+i-4)!(b-1)!}{(i-2)!}\tau_{a+i-6}(\gamma\cdot c^{2}_{1})\tau_{b-2}(\operatorname{\mathrm{\mathsf{p}}} )+ a \frac{(a-1)!(b+i-4)!}{(i-2)!}\tau_{a-2}(\operatorname{\mathrm{\mathsf{p}}})\tau_{b+i-6}(\gamma\cdot c_{1}^{2}) \bigg],\end{align*} $$

after we apply equation (1.15) to these terms and drop $\tau $ -cubic terms and the terms of cohomological degree other than $3$ . In particular, the factors in first and second terms are produced by the $\mathfrak {a}$ -linear term of equation (1.15) proportional to $c_{1}$ .

The last two terms of equation (5.24) yieldFootnote ³¹

$$ \begin{align*} &-\frac{(\imath u)^{-i+4}}{2}\bigg[ (a-1)\frac{(a+i-4)!(b-1)!}{(i-2)!} \tau_{a+i-6}(\gamma_{\bullet}\cdot \gamma)\tau_{b-2}(\gamma_{\star}\cdot c_{1})\\ &\quad +(b-1)\frac{(a-1)!(b+i-4)!}{(i-2)!} \tau_{a-2}(\gamma_{\bullet}\cdot \gamma)\tau_{b+i-6}(\gamma_{\star}\cdot c_{1}) \bigg],\end{align*} $$

after we apply only the parts of equations (1.14) and (1.15) that are not $c_{1}$ -proportional, then we use the $\mathfrak {a}$ to $\tau $ the transition formula (5.4) and drop the $\tau $ cubic terms and the terms of homological degree other than $3$ .

Together these two sums combine and cancel the first term of equation (5.28). To cancel the last two terms of equation (5.28), we follow the same pattern. We first apply $c_{1}^{0}$ -part of equation (1.15) to the first and second terms of equation (5.24) and then apply $c_{1}$ -part of the $\mathfrak {a}$ to $\tau $ transition formula (5.4). Next, we apply the $c_{1}^{0}$ -parts of equations (1.14) and (1.15) and the $c_{1}^{1}$ -part of equation (5.4) to the last two terms of equation (5.24). After dropping the $\tau $ -cubic terms and the terms of complex cohomological degree other than $3$ , we exactly cancel the remaining terms of equation (5.28).

$\bullet $ Consider finally the $\tau $ -cubic terms. The cohomological arguments of these terms are $c_{1}^{2}\cdot \gamma $ , so as in the previous computation, we drop the cohomology insertion from the notation.

After expanding the corresponding terms of equation (5.15), we obtain

(5.29) $$ \begin{align} \frac{(\imath u)^{-i}(i+k-2)}{(i-3)!}\sum_{|\mu|=i+k-5}\frac{\mu_{1}!\mu_{2}!\mu_{3}!}{\operatorname{\mathrm{Aut}}(\mu)}\tau_{\mu_{1}-1}\tau_{\mu_{2}-1}\tau_{\mu_{3}-1}\, .\end{align} $$

On the other hand, the corresponding terms from equation (5.16) are more complicated:

$$ \begin{align*} &\frac{(\imath u)^{-i}}{(i-3)!}\sum_{|\mu|=i-5}\frac{\mu_{1}!\mu_{2}!\mu_{3}!}{\operatorname{\mathrm{Aut}}(\mu)}\left(\frac{(\mu_{1}+k+1)!}{\mu_{1}!}\tau_{\mu_{1}+k-1}\tau_{\mu_{2}-1}\tau_{\mu_{3}-1}\right.\\&\quad \left.+\frac{(\mu_{2}+k+1)!}{\mu_{2}!}\tau_{\mu_{1}-1}\tau_{\mu_{2}+k-1}\tau_{\mu_{3}-1}+\frac{(\mu_{3}+k+1)!}{\mu_{3}!}\tau_{\mu_{1}-1}\tau_{\mu_{2}-1}\tau_{\mu_{3}+k-1}\right). \end{align*} $$

In equation (5.29), we have $i+k-2=\sum _{j=1}^{3} (\mu _{j}+1)$ . Therefore, the difference between the last two expressions is the sum of the monomials

(5.30) $$ \begin{align} \left(\sum_{j, \mu_{j}\le k-2}(\mu_{j}+2)\right)\frac{(\imath u)^{-i} \mu_{1}!\mu_{2}!\mu_{3}!}{(i-3)!\operatorname{\mathrm{Aut}}(\mu)}\tau_{\mu_{1}-1}\tau_{\mu_{2}-1}\tau_{\mu_{3}-1}\, .\end{align} $$

Let us restrict our attention to the case when i is bigger than k; the other cases are analogous. After applying the reaction from the last line of equation (1.15), we obtain a formula for the expressions in the second line of equation (5.13):

(5.31) $$ \begin{align} &(a-2)!b! \mathfrak{C}^{\circ}\left(\frac{\widetilde{\mathsf{ch}}_{a}(c_{1})\widetilde{\mathsf{ch}}_{i}(\gamma)}{(\imath u)^{b-2}}\right)\tau_{b-2}(\operatorname{\mathrm{\mathsf{p}}})=\tau\mbox{-quadratic terms }\nonumber\\ &\qquad\qquad +\frac{(\imath u)^{-k-i}b!}{(i-2)!}\left(\sum_{|\mu|=a+i-6}\max(\max(\mu_{1}+1,\mu_{2}+1),i-2)\frac{\mu_{1}!\mu_{2}!}{\operatorname{\mathrm{Aut}}(\mu)}\tau_{\mu_{1}-1}\tau_{\mu_{2}-1}\right)\tau_{b-2}\, , \end{align} $$

(5.32) $$ \begin{align} &a!(b-2)!\tau_{a-2}(\operatorname{\mathrm{\mathsf{p}}})\mathfrak{C}^{\circ}\left( \frac{\widetilde{\mathsf{ch}}_{b}(c_{1})\widetilde{\mathsf{ch}}_{i}(\gamma)}{(\imath u)^{a-2}}\right)=\tau\mbox{-quadratic terms } \nonumber\\ &\quad +\frac{(\imath u)^{-k-i}a!}{(i-2)!}\left(\sum_{|\mu|=b+i-6}\max(\max(\mu_{1}+1,\mu_{2}+1),i-2)\frac{\mu_{1}!\mu_{2}!}{\operatorname{\mathrm{Aut}}(\mu)}\tau_{\mu_{1}-1}\tau_{\mu_{2}-1}\right)\tau_{a-2}\,. \end{align} $$

The terms of equations (5.31) and (5.32) with $\max (\mu _{1}+1,\mu _{2}+1)\le i-2$ contribute the monomials

(5.33) $$ \begin{align} (\imath u)^{-i}b\cdot\frac{\mu_{1}!\mu_{2}!(b-1)!}{(i-3)!}\tau_{\mu_{1}-1}\tau_{\mu_{2}-1}\tau_{b-2}\, ,\ \quad (\imath u)^{-i}a\cdot\frac{\mu_{1}!\mu_{2}!(a-1)!}{(i-3)!}\tau_{\mu_{1}-1}\tau_{\mu_{2}-1}\tau_{a-2}\, .\end{align} $$

Note $a+b=k+2$ in equation (5.13). Since $\max (\mu _{1}+1,\mu _{2}+1)\le i-2$ and $|\mu |=a+i-2$ or $|\mu |=b+i-2$ , we imply that $\mu _{1}+1,\mu _{2}+1\ge k-1$ . Thus, the corresponding terms of equations (5.31) and (5.32) cancel with the monomials (5.30) such that there is only one j with $\mu _{j}\le k-2.$

The terms in equations (5.31) and (5.32) with $\max (\mu _{1}+1,\mu _{2}+1)> i-2$ yield terms

$$ \begin{align*}(\imath u)^{-i}b(\mu^{\prime}+1)\cdot\frac{\mu_{1}!\mu_{2}! (b-1)!}{(i-2)!}\tau_{\mu_{1}-1}\tau_{\mu_{2}-1}\tau_{b-2}\, ,\quad\ (\imath u)^{-i}a(\mu^{\prime}+1)\cdot\frac{\mu_{1}!\mu_{2}!(a-1)!}{(i-2)!}\tau_{\mu_{1}-1}\tau_{\mu_{2}-1}\tau_{a-2},\end{align*} $$

where $\mu ^{\prime }=\max (\mu _{1},\mu _{2})$ . Both of these terms are of the form

(5.34) $$ \begin{align} (\imath u)^{-i}(\mu_{1}+1)(\mu_{2}+1)\cdot \frac{\mu_{1}!\mu_{2}!\mu_{3}!}{(i-2)!}\tau_{\mu_{1}-1}\tau_{\mu_{2}-1}\tau_{\mu_{3}-1}\, , \end{align} $$

with $\mu _{1}+1>i-2$ and $|\mu |=i+k-4$ . Since we assumed that $i> k$ , we have $\mu _{1}+\mu _{2}<k-4$ in equation (5.34). The discussed terms therefore combine to yield the sum of monomials

(5.35) $$ \begin{align} (\imath u)^{-i}(\mu_{1}+\mu_{2}+2)(\mu_{3}+1) \cdot \frac{\mu_{1}!\mu_{2}!\mu_{3}!}{(i-2)!}\tau_{\mu_{1}-1}\tau_{\mu_{2}-1}\tau_{\mu_{3}- 1}\, , \end{align} $$

where $\mu _{3}+1>i-2$ and $\mu _{1},\mu _{2}\le k-2$ .

The terms (5.35) combine with the terms from the expansion of the last two lines of equation (5.13). Indeed, since $\gamma _{\bullet }$ , $\gamma _{\star }$ in the last two lines of equation (5.13) are of complex cohomological degree $2$ , the $\tau $ -terms result from use of the $c_{1}^{1}$ -part of equation (1.14) and of the $c_{1}^{0}$ -part of equation (1.15). The expansion of these terms is a sum of monomials

(5.36) $$ \begin{align} -(\imath u)^{-i}(b-1)(a-1)\frac{(a+i-4)!\mu_{1}!\mu_{2}!}{(i-2)!} \tau_{a+i-5}\tau_{\mu_{1}-1}\tau_{\mu_{2}-1}\, , \end{align} $$

where $|\mu |=b-3$ .

The combination of equation (5.36) with $a=\mu _{3}-i+4$ , $b=\mu _{1}+\mu _{2}+3$ and equation (5.35) matches equation (5.30), since, in equation (5.36), we have

$$ \begin{align*}(b-1)(a-1)=(\mu_{1}+\mu_{2}+2)(\mu_{3}-i+3)=(\mu_{1}+\mu_{2}+2)(\mu_{3}+1)-(\mu_{1}+\mu_{2}+2)(i-2).\end{align*} $$

We have cancelled all $\tau $ -cubic terms.

The assumption $\widetilde {\mathsf {ch}}_{i}(\gamma )\in \mathbb {D}_{\mathrm {PT}}^{X\bigstar }$ implies $i\ge 3$ . Therefore, in the above computations, we do not see negative factorials in denominators.

5.3 Proof of Theorem 3.1 for $\mathbb {D}^{1}_{\mathrm {PT}} \cap \mathbb {D}^{X\bigstar }_{{\mathrm {PT}}}$

Theorem 3.1, for all $D\in \mathbb {D}^{1}_{\mathrm {PT}} \cap \mathbb {D}^{X\bigstar }_{{\mathrm {PT}}}$ , is an immediate consequence of Proposition 11 for singletons by the following simple argument. Let

$$ \begin{align*}D = \prod_{i=1}^{m} \widetilde{\mathsf{ch}}_{k_{i}}(\gamma_{i}) \in \mathbb{D}^{1}_{\mathrm{PT}} \cap \mathbb{D}^{X\bigstar}_{{\mathrm{PT}}},\end{align*} $$

where $\gamma _{i}\gamma _{j}=0\in H^{*}(X)$ for all $i\neq j$ .

By definition, for $k\geq 1$ ,

$$ \begin{align*} \mathfrak{C}^{\bullet}\left(\mathrm{L}^{\mathrm{PT}}_{k}(D)\right) & = \mathfrak{C}^{\bullet}\Big( \mathrm{L}^{\mathrm{PT}}_{k}\big( \prod_{i=1}^{m} \widetilde{\mathsf{ch}}_{k_{i}}(\gamma_{i}) \big)\Big)\\ & = \mathfrak{C}^{\bullet}\Bigg( \mathrm{T}_{k} \prod_{i=1}^{m} \widetilde{\mathsf{ch}}_{k_{i}}(\gamma_{i}) + \sum_{j=1}^{m} \mathrm{R}_{k}(\widetilde{\mathsf{ch}}_{k_{j}}(\gamma_{j})) \prod_{i\neq j} \widetilde{\mathsf{ch}}_{k_{i}}(\gamma_{i}) \Bigg). \end{align*} $$

Since $\gamma _{i}\gamma _{j}=0$ for $i\neq j$ ,

$$ \begin{align*}\mathfrak{C}^{\bullet}\Big( \mathrm{T}_{k} \prod_{i=1}^{m} \widetilde{\mathsf{ch}}_{k_{i}}(\gamma_{i})\Big)= (-m+1)\mathfrak{C}^{\bullet} (\mathrm{T}_{k}) \prod_{i=1}^{m} \mathfrak{C}^{\bullet}(\widetilde{\mathsf{ch}}_{k_{i}}(\gamma_{i})) + \sum_{j=1}^{m} \mathfrak{C}^{\bullet}(\mathrm{T}_{k} \widetilde{\mathsf{ch}}_{k_{j}}(\gamma_{j})) \prod_{i\neq j} \mathfrak{C}^{\bullet}(\widetilde{\mathsf{ch}}_{k_{i}}(\gamma_{i})).\end{align*} $$

By Proposition 11,

$$ \begin{align*} (\imath u)^{-k}\, \widetilde{\mathrm{L}}^{\mathrm{GW}}_{k}(\mathfrak{C}^{\bullet}(\widetilde{\mathsf{ch}}_{i}(\gamma_{i}))) & = \mathfrak{C}^{\bullet}\left(\mathrm{L}^{\mathrm{PT}}_{k}(\widetilde{\mathsf{ch}}_{i}(\gamma_{i}))\right) \\ & = \mathfrak{C}^{\bullet}\left( \mathrm{T}_{k}\right) \mathfrak{C}^{\bullet}\big( \widetilde{\mathsf{ch}}_{i}(\gamma_{i})\big) + \mathfrak{C}^{\bullet}\big( \mathrm{T}_{k}\widetilde{\mathsf{ch}}_{i}(\gamma_{i})\big) + \mathfrak{C}^{\bullet}\left(\mathrm{R}_{k}\big(\widetilde{\mathsf{ch}}_{k_{i}}(\gamma_{i})\big)\right). \end{align*} $$

We conclude

$$ \begin{align*} &\mathfrak{C}^{\bullet}\left(\mathrm{L}^{\mathrm{PT}}_{k}(D)\right) \\ &\quad =\sum_{j=1}^{m} (\imath u)^{-k}\, \widetilde{\mathrm{L}}^{\mathrm{GW}}_{k}(\mathfrak{C}^{\bullet}(\widetilde{\mathsf{ch}}_{j}(\gamma_{j}))) \prod_{i\neq j} \mathfrak{C}^{\bullet}(\widetilde{\mathsf{ch}}_{k_{i}}(\gamma_{i}))- (m-1) \mathfrak{C}^{\bullet} (\mathrm{T}_{k}) \prod_{i=1}^{m} \mathfrak{C}^{\bullet}(\widetilde{\mathsf{ch}}_{k_{i}}(\gamma_{i})) . \end{align*} $$

On the other hand,

$$ \begin{align*} &(\imath u)^{-k}\, \widetilde{\mathrm{L}}^{\mathrm{GW}}_{k} \left( \mathfrak{C}^{\bullet}(D)\right) \\ &\quad =\sum_{j=1}^{m} (\imath u)^{-k}\, \widetilde{\mathrm{L}}^{\mathrm{GW}}_{k}(\mathfrak{C}^{\bullet}(\widetilde{\mathsf{ch}}_{j}(\gamma_{j}))) \prod_{i\neq j} \mathfrak{C}^{\bullet}(\widetilde{\mathsf{ch}}_{k_{i}}(\gamma_{i}))- (m-1)(\imath u)^{-k}\left( \frac{(\imath u)^{2}}{2}\right)\mathrm{T}_{k} \prod_{i=1}^{m} \mathfrak{C}^{\bullet}(\widetilde{\mathsf{ch}}_{k_{i}}(\gamma_{i})) . \end{align*} $$

The proof is completed by applying equation (4.3).

6.1 Overview

We complete here the proof of Theorem 3.1. Since noninteracting insertions have already been treated in Section 5, we must address the interacting cases. In the desired equation,

(6.1) $$ \begin{align} \mathfrak{C}^{\bullet}\circ \mathrm{L}_{k}^{\mathrm{PT}}(D)=(\imath u)^{-k}\, \widetilde{\mathrm{L}}_{k}^{\mathrm{GW}}\circ \mathfrak{C}^{\bullet}(D)\, , \end{align} $$

the stable pairs descendent insertions of $D \in \mathbb {D}^{X\bigstar }_{{\mathrm {PT}}}$ can interact with each other via the $\mathrm {GW}/{\mathrm {PT}}$ descendent correspondence on both sides of equation (6.1). In addition, the stable pairs descendents can also interact with constant term of the Virasoro constraints on the left side. We must control all of these interactions.

6.2 Interactions among two insertions

We start with results which control the interactions of two descendent insertions.

Proposition 12. Let $\gamma ^{\prime }\in H^{2}(X)$ , $\gamma ^{\prime \prime }\in H^{4}(X)$ , and let $i\ge 3$ , $j\ge 2$ . Then, for $k\ge -1$ , we have

$$ \begin{align*}(\imath u )^{k}\, \mathfrak{C}^{\circ}(\mathrm{R}_{k}(\widetilde{\mathsf{ch}}_{i}(\gamma^{\prime})\widetilde{\mathsf{ch}}_{j}(\gamma^{\prime\prime})))=\mathrm{R}_{k}(\mathfrak{C}^{\circ}(\widetilde{\mathsf{ch}}_{i}(\gamma^{\prime})\widetilde{\mathsf{ch}}_{j}(\gamma^{\prime\prime}))).\end{align*} $$

Proof. We first compute the left side of the equation. After applying the shifts, we obtain

$$ \begin{align*}\mathrm{R}_{k}(\widetilde{\mathsf{ch}}_{i}(\gamma^{\prime})\widetilde{\mathsf{ch}}_{j}(\gamma^{\prime\prime})) =\frac{(i+k-2)!}{(i-3)!}\widetilde{\mathsf{ch}}_{i+k}(\gamma^{\prime})\widetilde{\mathsf{ch}}_{j}(\gamma^{\prime\prime})+\frac{(j+k-1)!}{(j-2)!}\widetilde{\mathsf{ch}}_{i}(\gamma^{\prime})\widetilde{\mathsf{ch}}_{j+k}(\gamma^{\prime\prime}).\end{align*} $$

We apply the correspondence to the both terms:

$$ \begin{align*} \mathfrak{C}^{\circ}(\mathrm{R}_{k}(\widetilde{\mathsf{ch}}_{i}(\gamma^{\prime})\widetilde{\mathsf{ch}}_{j}(\gamma^{\prime\prime})))&=(\imath u)^{-1}\left(\frac{1}{(i-3)!(j-2)!}+\frac{(j+k-1)}{(i-2)!(j-2)!}\right) \mathfrak{a}_{i+j+k-4}(\gamma^{\prime}\gamma^{\prime\prime})\\ &= (\imath u)^{-i-j-k+4}\frac{(i+j+k-3)!}{(i-2)!(j-2)!} \tau_{i+j+k-5}(\gamma^{\prime}\gamma^{\prime\prime}).\end{align*} $$

The right side of the equation is

$$ \begin{align*}\mathrm{R}_{k}(\mathfrak{C}^{\circ}(\widetilde{\mathsf{ch}}_{i}(\gamma^{\prime})\widetilde{\mathsf{ch}}_{j}(\gamma^{\prime\prime}))) &=\mathrm{R}_{k}\bigg(\frac{(\imath u)^{-1}}{(i-2)!(j-2)!}\mathfrak{a}_{i+j-4}(\gamma^{\prime}\gamma^{\prime\prime})\bigg)\\ &= (\imath u)^{-i-j+4} \frac{(i+j-4)!}{(i-2)!(j-2)!)}\mathrm{R}_{k}(\tau_{i+j-5}(\gamma^{\prime}\gamma^{\prime\prime}))\\ &=(\imath u)^{-i-j+4}\frac{(i+j+k-3)!}{(i-2)!(j-2)!}\tau_{i+j+k-5}(\gamma^{\prime}\gamma^{\prime\prime}),\end{align*} $$

which matches the left side.

Proposition 13. Let $\gamma ^{\prime },\gamma ^{\prime \prime }\in H^{2}(X)$ , and let $i,j\ge 3$ . Then, for $k\ge -1$ , we have

(6.2) $$ \begin{align} &\hspace{-17pt}(\imath u )^{k}\, \mathfrak{C}^{\circ}(\mathrm{R}_{k}(\widetilde{\mathsf{ch}}_{i}(\gamma')\widetilde{\mathsf{ch}}_{j}(\gamma'')))-\mathrm{R}_{k}(\mathfrak{C}^{\circ}(\widetilde{\mathsf{ch}}_{i}(\gamma')\mathsf{ch}_{j}(\gamma'')))\nonumber\\ &\quad =\sum_{a+b=k+2}(a-2)!b!\, \mathfrak{C}^{\circ}(\widetilde{\mathsf{ch}}_{i}(\gamma')\cdot\widetilde{\mathsf{ch}}_{j}(\gamma'')\cdot\widetilde{\mathsf{ch}}_{a}(c_{1}))\,\mathfrak{C}^{\circ}(\mathsf{ch}_{b}(\operatorname{\mathrm{\mathsf{p}}}))\nonumber\\&\qquad + a!(b-2)!\, \mathfrak{C}^{\circ}(\widetilde{\mathsf{ch}}_{i}(\gamma')\cdot\widetilde{\mathsf{ch}}_{j}(\gamma'')\cdot\widetilde{\mathsf{ch}}_{b}(c_{1}))\, \mathfrak{C}^{\circ}(\mathsf{ch}_{a}(\operatorname{\mathrm{\mathsf{p}}}))\nonumber\\ &\qquad - \sum_{a+b=k+2}(a-1)!(b-1)!\, \sum_{ \bullet,\star}\alpha_{\bullet\star}\left(\mathfrak{C}^{\circ}(\widetilde{\mathsf{ch}}_{a}(\gamma_{\bullet})\cdot \widetilde{\mathsf{ch}}_{i}(\gamma'))\mathfrak{C}^{\circ}(\widetilde{\mathsf{ch}}_{b}(\gamma_{\star})\widetilde{\mathsf{ch}}_{j}(\gamma''))\right.\nonumber\\&\qquad \left. +\, \mathfrak{C}^{\circ}(\widetilde{\mathsf{ch}}_{a}(\gamma_{\bullet})\widetilde{\mathsf{ch}}_{j}(\gamma''))\mathfrak{C}^{\circ}(\widetilde{\mathsf{ch}}_{b}(\gamma_{\star})\cdot \widetilde{\mathsf{ch}}_{i}(\gamma'))\right)\,. \end{align} $$

Proof. We follow the same strategy as in the proof of Proposition 11. We first compute

$$ \begin{align*}\mathrm{R}_{k}(\widetilde{\mathsf{ch}}_{i}(\gamma^{\prime})\widetilde{\mathsf{ch}}_{j}(\gamma^{\prime\prime}))= \frac{(k+i-2)!}{(i-3)!}\widetilde{\mathsf{ch}}_{i+k}(\gamma^{\prime})\widetilde{\mathsf{ch}}_{j}(\gamma^{\prime\prime}) +\frac{(k+j-2)!}{(j-3)!}\widetilde{\mathsf{ch}}_{i}(\gamma^{\prime})\widetilde{\mathsf{ch}}_{j+k}(\gamma^{\prime\prime}).\end{align*} $$

After applying the correspondence, we obtain

(6.3) $$ \begin{align} \mathfrak{C}^{\circ}(\mathrm{R}_{k}(\widetilde{\mathsf{ch}}_{i}(\gamma')\widetilde{\mathsf{ch}}_{j}(\gamma'')))=-\frac{1}{(i-3)!(j-2)!}\left[\frac{\mathfrak{a}_{i+j+k-4}(\gamma' \gamma'')}{\imath u}+ \frac{\mathfrak{a}_{i+j+k-5}(\gamma'\gamma'' \cdot c_{1})}{(\imath u)^{2}}+\right.\nonumber\\\left.(\imath u)^{-2}\sum_{|\mu|=i+j+k-6}\frac{f(i+k,j;\mu_{1},\mu_{2})}{\operatorname{\mathrm{Aut}}(\mu)}\mathfrak{a}_{\mu_{1}}\mathfrak{a}_{\mu_{2}} (\gamma'\gamma''\cdot c_{1})\right]- \frac{1}{(i-2)!(j-3)!}\left[\frac{\mathfrak{a}_{i+j+k-4}(\gamma'\gamma'')}{\imath u}+\right.\nonumber\\\left. \frac{\mathfrak{a}_{i+j+k-5}(\gamma'\gamma''\cdot c_{1})}{(\imath u)^{2}}+(\imath u)^{-2}\sum_{|\mu|=i+j+k-6}\frac{f(i,j+k;\mu_{1},\mu_{2})}{\operatorname{\mathrm{Aut}}(\mu)}\mathfrak{a}_{\mu_{1}} \mathfrak{a}_{\mu_{2}}(\gamma'\gamma''\cdot c_{1})\right], \end{align} $$

where $f(i,j;\mu _{1},\mu _{2})=\max (\max (i-2,j-2),\max (\mu _{1}+1,\mu _{2}+1))$ .

The second term of the difference is easier:

(6.4) $$ \begin{align} \mathrm{R}_{k}(\mathfrak{C}^{\circ}(\widetilde{\mathsf{ch}}_{i}(\gamma')\widetilde{\mathsf{ch}}_{j}(\gamma'')))= -\frac{(\imath u)^{-i-j+4}}{(i-2)!(j-2)!} \mathrm{R}_{k}\bigg( (i+j-4)!\bigg(\tau_{i+j-5}(\gamma'\gamma'')\nonumber\\ +\left(\sum_{s=1}^{i+j-4}\frac1s\right) \tau_{i+j-6}(\gamma'\gamma''\cdot c_{1})\bigg)+(\imath u)^{-2}\sum_{|\mu|=i+j-6}\frac{f(i,j;\mu_{1},\mu_{2})}{\operatorname{\mathrm{Aut}}(\mu)}\mathfrak{a}_{\mu_{1}}\mathfrak{a}_{\mu_{2}}(\gamma'\gamma''\cdot c_{1})\bigg)\,. \end{align} $$

We now analyze the difference. The $\tau $ -linear terms of complex cohomological degree $2$ in $(\imath u )^{k}$ times equations (6.3) and (6.4) are matching sums of the monomials:

$$ \begin{align*} (\imath u)^{-i-j+4}\frac{(i+j+k-4)!}{(i-2)!(j-2)!}(i+j-4) \tau_{i+j+k-5}(\gamma^{\prime}\gamma^{\prime\prime}).\end{align*} $$

The $\tau $ -linear terms of cohomological degree $3$ almost match. To be precise, the corresponding terms in equation (6.3) are sums the monomials:

$$ \begin{align*} (\imath u)^{-i-j+4}\frac{(i+j+k-4)!}{(i-2)!(j-2)!}(i+j-4)\left(\sum_{s=1}^{i+j+k-4}\frac1s\right)\tau_{i+j+k-6}(\gamma^{\prime}\gamma^{\prime\prime}\cdot c_{1}).\end{align*} $$

Respectively, the corresponding terms in equation (6.4) are sums of the same monomials plus an extra term

$$ \begin{align*} (\imath u)^{-i-j+4}\frac{(i+j+k-4)!}{(i-2)!(j-2)!}\tau_{i+j+k-6}(\gamma^{\prime}\gamma^{\prime\prime}\cdot c_{1}).\end{align*} $$

This extra term gets canceled by the term from the second line of equation (6.2) with $b=0$ because of equation (3.5).

Therefore, the difference of $(\imath u )^{k}$ times equations (6.3) and (6.4) consists only of the $\tau $ -quadratic terms of complex cohomological degree $3$ . We omit cohomological classes since all the cohomological arguments are $\gamma ^{\prime }\gamma ^{\prime \prime } \cdot c_{1}$ . The corresponding part of equation (6.3) is

(6.5) $$ \begin{align} \frac{(\imath u)^{-i-j+2}}{(i-2)!(j-2)!}\sum_{|\mu|=i+j+k-6}\frac{\mu_{1}!\mu_{2}!}{\operatorname{\mathrm{Aut}}(\mu)} \left[(i-2)f(i+k,j;\mu)+(j-2)f(i,j+k;\mu)\right]\tau_{\mu_{1}-1}\tau_{\mu_{2}-1}\, , \end{align} $$

where we assume that f vanishes whenever one of the argument is negative.

We must compare equation (6.5) with the expansion of the last four lines of equation (6.2). The first two of the last four lines of (6.2) expand to

$$ \begin{align*} &-\frac{(\imath u)^{-i-j+2}}{(i-2)!(j-2)!}\sum_{a+b=k+2}(i+j+a-6)b\frac{(i+j+a-7)!(a-1)!}{2}\tau_{i+j+a-8}\tau_{b-2}\\&+(i+j+b-6)a\frac{(i+j+b-7)!(a-1)!}{2}\tau_{i+j+b-8}\tau_{a-2}. \end{align*} $$

The last two lines of the last four lines of equation (6.2) expand to

$$ \begin{align*} &\frac{(\imath u)^{-i-j+2}}{(i-2)!(j-2)!}\sum_{a+b=k+2} (a-1)(b-1)\bigg((a+i-4)!(b+j-4)!\tau_{a+i-5}\tau_{b+j-5}\\ &\quad +(a+j-4)!(b+i-4)!\tau_{a+j-5}\tau_{b+i-5}\bigg). \end{align*} $$

These last two expressions are the $\tau $ -cubic contribution to equation (6.2) which results from the bumping of $\widetilde {\mathsf {ch}}_{i}(\gamma ^{\prime })\widetilde {\mathsf {ch}}_{j}(\gamma ^{\prime \prime })$ with the constant term $\mathrm {T}_{k}$ . The corresponding coefficient in front of $\tau $ -cubic monomial is given by the formula (6.7) below.

To complete the proof, we must match the coefficients in front of the terms in sums above. That is, we need to compare two expressions below for all $\mu $ satisfying $|\mu |=i+j+k-6$ :

(6.6) $$ \begin{align} &\hspace{-12pt}(i-2)f(i+k,j;\mu)+(j-2)f(i,j+k;\mu)-(\mu_{1}+1)f(i,j;\mu_{1}-k,\mu_{2})\nonumber\\&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad- (\mu_{2}+1)f(i,j;\mu_{1},\mu_{2}-k)\, , \end{align} $$

(6.7) $$ \begin{align} &\hspace{-12pt}[\mu_{1}+1]_{\le k}(\mu_{2}+1)+(\mu_{1}+1)[\mu_{2}+1]_{\le k}-[\mu_{1}-i+3]_{\ge 0}[\mu_{2}-j+3]_{\ge 0}\nonumber\\&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad-[\mu_{1}-j+3]_{\ge 0}[\mu_{2}-i+3]_{\ge 0}\, , \end{align} $$

where $[a]_{\le b}$ and $[a]_{\ge b}$ are cut off functions which equal a if a satisfies inequalities $a\ge b$ and $a\le b$ , respectively (and are zero otherwise). The matching now is a long and routine check. We give some details.

We can always assume $\mu _{1}\ge \mu _{2}$ and $i\ge j$ . Let us further assume k is small and $\mu _{1}\ge i+k$ . If $\mu _{2}\ge k $ , then the function (6.6) equals

$$ \begin{align*}(i+j-4)(\mu_{1}+1)-(\mu_{1}+1)(\mu_{1}-k+1)-(\mu_{2}-1)(\mu_{1}+1)=0. \end{align*} $$

The assumed inequalities force all terms in equation (6.7) to vanish.

Next, we assume all but last inequality are true, that is $\mu _{2}<k$ . Then the expression (6.6) becomes

$$ \begin{align*}(i+j-4)(\mu_{1}+1)-(\mu_{1}+1)(\mu_{1}-k+1)=(\mu_{1}+1)(\mu_{2}+1).\end{align*} $$

On the other hand, in equation (6.7), only the second expression does not vanish – the second expression matches equation (6.6). Rest of the case can be treated analogously.

6.3 Interactions among three insertions

The last interaction to consider is among three descendent insertions. Because of the stationary assumption, there is only one case to control.

Proposition 14. Let $\gamma ^{\prime },\gamma ^{\prime \prime },\gamma ^{\prime \prime \prime } \in H^{2}(X)$ , and let $i_{1},i_{2},i_{3}\ge 3$ , Then, for $k\ge -1$ , we have

$$ \begin{align*} (\imath u )^{k}\, \mathfrak{C}^{\circ}\left(\mathrm{R}_{k}( \widetilde{\mathsf{ch}}_{i_{1}}(\gamma^{\prime}) \widetilde{\mathsf{ch}}_{i_{2}}(\gamma^{\prime\prime}) \widetilde{\mathsf{ch}}_{i_{3}}(\gamma^{\prime\prime\prime}) )\right)- \mathrm{R}_{k}\left( \mathfrak{C}^{\circ}\big( \widetilde{\mathsf{ch}}_{i_{1}}(\gamma^{\prime}) \widetilde{\mathsf{ch}}_{i_{2}}(\gamma^{\prime\prime}) \widetilde{\mathsf{ch}}_{i_{3}}(\gamma^{\prime\prime\prime}) \big) \right)=0. \end{align*} $$

For the proof, we will use the explicit correspondence formula (1.16) for the triple interaction

(6.8) $$ \begin{align} \mathfrak{C}^{\circ}(\widetilde{\mathsf{ch}}_{i_{1}}\widetilde{\mathsf{ch}}_{i_{2}}\widetilde{\mathsf{ch}}_{i_{3}})(\gamma)=\frac{(|i|-6)(\imath u)^{-2}}{(i_{1}-2)!(i_{2}-2)!(i_{3}-2)!}\mathfrak{a}_{|i|-7}(\gamma),\end{align} $$

where $|i|=i_{1}+i_{2}+i_{3}$ .

Proof of Proposition 14

We first compute the left side of the equation. To start,

$$ \begin{align*} \mathrm{R}_{k}(\widetilde{\mathsf{ch}}_{i_{1}}(\gamma^{\prime})\widetilde{\mathsf{ch}}_{i_{2}}(\gamma^{\prime\prime})\widetilde{\mathsf{ch}}_{i_{3}}(\gamma^{\prime\prime\prime})) &= \ \, \frac{(i_{1}+k-2)!}{(i_{1}-3)!}\widetilde{\mathsf{ch}}_{i_{1}+k}(\gamma^{\prime})\widetilde{\mathsf{ch}}_{i_{2}}(\gamma^{\prime\prime}) \widetilde{\mathsf{ch}}_{i_{3}}(\gamma^{\prime\prime\prime})\\ &\quad +\frac{(i_{2}+k-2)!}{(i_{2}-3)!}\widetilde{\mathsf{ch}}_{i_{1}}(\gamma^{\prime})\widetilde{\mathsf{ch}}_{i_{2}+k}(\gamma^{\prime\prime}) \widetilde{\mathsf{ch}}_{i_{3}}(\gamma^{\prime\prime\prime})\\ &\quad +\frac{(i_{3}+k-2)!}{(i_{3}-3)!}\widetilde{\mathsf{ch}}_{i_{1}}(\gamma^{\prime})\widetilde{\mathsf{ch}}_{i_{2}}(\gamma^{\prime\prime}) \widetilde{\mathsf{ch}}_{i_{3}+k}(\gamma^{\prime\prime\prime}). \end{align*} $$

After applying the triple bumping and the transition from $\mathfrak {a}$ descendents to $\tau $ descendents, we obtain

(6.9) $$ \begin{align} \mathfrak{C}^{\circ}(\mathrm{R}_{k}(\widetilde{\mathsf{ch}}_{i_{1}}(\gamma')\widetilde{\mathsf{ch}}_{i_{2}}(\gamma'') \widetilde{\mathsf{ch}}_{i_{3}}(\gamma''')))=(|i|+k-6)(\imath u)^{-2}\left(\frac{1}{(i_{1}-3)!(i_{2}-2)!(i_{3}-2)!}\right.\nonumber\\ \left.+\frac{1}{(i_{1}-2)!(i_{2}-3)!(i_{3}-2)!}+\frac{1}{(i_{1}-2)!(i_{2}-2)!(i_{3}-3)!}\right)\mathfrak{a}_{|i|+k-7}(\gamma'\gamma''\gamma''')\nonumber\\ = (\imath u)^{-|i|-k+6}(|i|-6)\frac{(|i|+k-6)!}{(i_{1}-2)!(i_{2}-2)!(i_{3}-2)!} \tau_{|i|+k-8}(\gamma'\gamma''\gamma''')\,. \end{align} $$

On the other hand, the right side of the equation equals

$$ \begin{align*} \mathrm{R}_{k}(\mathfrak{C}^{\circ}(\widetilde{\mathsf{ch}}_{i_{1}}(\gamma')\widetilde{\mathsf{ch}}_{i_{2}}(\gamma'') \widetilde{\mathsf{ch}}_{i_{3}}(\gamma''')))&= \frac{(\imath u)^{-2}(|i|-6)}{(i_{1}-2)!(i_{2}-2)!(i_{3}-2)!}\mathrm{R}_{k}(\mathfrak{a}_{|i|-7}(\gamma'\gamma''\gamma'''))\\ &= (\imath u)^{-|i|+6} (|i|-6)\frac{(|i|+k-6)!}{(i_{1}-2)!(i_{2}-2)!(i_{3}-2)!} \tau_{|i|+k-8}(\gamma'\gamma''\gamma'''),\end{align*} $$

which matches $(\imath u)^{k}$ times equation (6.9).

6.4 Proof of Theorem 3.1

Let $k\ge 1$ , and let $D\in \mathbb {D}^{X\bigstar }_{{\mathrm {PT}}}$ . To prove the equality

$$ \begin{align*}\mathfrak{C}^{\bullet}\circ \mathrm{L}_{k}^{\mathrm{PT}}(D)=(\imath u)^{-k}\, \widetilde{\mathrm{L}}_{k}^{\mathrm{GW}}\circ \mathfrak{C}^{\bullet}(D),\end{align*} $$

after the restrictions $\tau _{-2}(\operatorname {\mathrm {\mathsf {p}}})=1$ and $\tau _{-1}(\gamma )=0$ for $\gamma \in H^{>2}(X)$ , we will expand both sides. The noninteracting case was already proven in Section 5.3. Equality in the general case will use Propositions 9, 10, 11, 12, 13 and 14.

In the formulas below, we will use shorthand notation for the constant term of $\mathrm {L}_{k}^{{\mathrm {PT}}}$ :

$$ \begin{align*}\mathrm{T}_{k}=\sum_{j} \mathrm{T}_{k,j}^{L}\mathrm{T}_{k,j}^{R},\end{align*} $$

where L and R denote the left and right sides in equation (1.3).

For $D=\prod _{i=1}^{\ell } D_{i} \in \mathbb {D}^{X\bigstar }_{{\mathrm {PT}}}$ , we have

(6.10) $$ \begin{align} \mathfrak{C}^{\bullet}(\mathrm{L}_{k}^{{\mathrm{PT}}}(D))&=\mathfrak{C}^{\bullet}(\mathrm{T}_{k} D+\mathrm{R}_{k}(D))\\ \nonumber &= \sum_{P^{\prime}}\sum_{j}\prod_{S\in P^{\prime}} \mathfrak{C}^{\circ}(\mathrm{T}_{k,j}^{S}D^{S})+\sum_{P^{\prime\prime}}\sum_{t=1}^{\ell(P^{\prime\prime})}\mathfrak{C}^{\circ}(\mathrm{R}_{k}(D^{S_{t}}))\prod_{S\in P^{\prime\prime},\, S\ne S_{t}}\mathfrak{C}^{\circ}(D^{S})\,. \end{align} $$

The first sum is over partitions $P^{\prime }$ of $\{1,\dots ,\ell ,L,R\}$ and

$$ \begin{align*}D^{S}=\prod_{i\in S\cap \{1,\dots,\ell\}}D_{i}\, ,\quad \mathrm{T}_{k,j}^{S}=\prod_{\gamma\in S\cap \{L,R\}} \mathrm{T}_{k,j}^{\gamma}.\end{align*} $$

The second sum is over partitions $P^{\prime \prime }$ of $\{1,\dots ,\ell \}$ and $P^{\prime \prime }=\{S_{1},\dots ,S_{\ell (P^{\prime \prime })}\}$ .

We must compare equation (6.10) with $(\imath u)^{-k}$ times

(6.11) $$ \begin{align} \ \ \ \mathrm{L}_{k}^{\mathrm{GW}}(\mathfrak{C}^{\bullet}(D)) & = \mathrm{L}_{k}^{\mathrm{GW}}(\sum_{P}\prod_{S\in P}\mathfrak{C}^{\circ}(D^{S}))\\ \nonumber &= \sum_{P^{\prime}}\mathrm{T}_{k}\prod_{S\in P^{\prime}}\mathfrak{C}^{\circ}(D^{S})+\sum_{P^{\prime\prime}}\sum_{t=1}^{\ell(P^{\prime\prime})} \mathrm{R}_{k}(\mathfrak{C}^{\circ}(D^{S_{t}}))\prod_{S\in P^{\prime\prime}, \, S\ne S_{t}}\mathfrak{C}^{\circ}(D^{S})\,. \end{align} $$

where both sums run over partitions $P^{\prime },P^{\prime \prime }$ of $\{1,\dots , \ell \}$ .

Since we only work with the stationary descendents, we can assume that the parts of partitions in the formulas have at most three elements. We will match the terms of equation (6.10) and $(\imath u)^{-k}$ times equation (6.11) depending on the size of $S_{t}$ .

$\bullet $ If $|S_{t}|=3$ , then the terms in equations (6.10) and (6.11) with $P^{\prime \prime }=\tilde {P}\sqcup S_{t}$ are matched by Proposition 14.

$\bullet $ If $|S_{t}|=2$ with $S_{t}=\{p,q\}$ , then we use Propositions 12 and 13 to match the terms of equation (6.10) with $P^{\prime \prime }=\tilde {P}\sqcup S_{t}$ and with $P^{\prime }$ equal to

$$ \begin{align*}\tilde{P}\sqcup \{S_{t},L\}\sqcup \{R\}\, ,\quad \tilde{P}\sqcup \{S_{t},R\}\sqcup \{L\}\, ,\quad \tilde{P}\sqcup\{p,R\}\sqcup\{q,L\}\, ,\quad \tilde{P}\sqcup \{p,L\}\sqcup \{q,R\},\end{align*} $$

with the terms of equation (6.11) with $P^{\prime \prime }=\tilde {P}\sqcup S_{t}$ .

$\bullet $ If $|S_{t}|=1$ with $S_{t}=\{p\}$ , then we use Proposition 10 and Proposition 11 to identify the terms of equation (6.10) with $P^{\prime \prime }=\tilde {P}\sqcup S_{t}$ and with $P^{\prime }$ equal to

$$ \begin{align*}\tilde{P}\sqcup\{p,L\}\sqcup\{R\}\, ,\quad \tilde{P}\sqcup\{p,R\} \sqcup \{L\}\end{align*} $$

with the terms of equation (6.11) with $P^{\prime \prime }=\tilde {P}\sqcup S_{t}$ .

$\bullet $ The terms of equation (6.10) with $P^{\prime }=\{L\}\sqcup \{R\} \sqcup \tilde {P}$ are equal to the terms of equation (6.11) with $P^{\prime }=\tilde {P}$ by Proposition 9.

The above four cases match all the terms in equations (6.10) and (6.11).

8.1 Vertex operators

Our goal here is to review the results of [Reference Oblomkov, Okounkov and Pandharipande18] and to explain how Theorem 1.4 can be derived from [Reference Oblomkov, Okounkov and Pandharipande18]. The full derivation is postponed to Section 9.

To state the main result of [Reference Oblomkov, Okounkov and Pandharipande18], we require negative descendents $\left \{ \mathfrak {a}_{k}\right \}$ for $k\in \mathbb {Z}_{<0}$ which are defined to satisfy the Heisenberg relations with positive descendents

(8.1) $$ \begin{align} [\mathfrak{a}_{k}(\alpha),\mathfrak{a}_{m}(\gamma)]=k\delta_{k+m}\int_{X}\alpha\cup\gamma\,. \end{align} $$

The descendents $\left \{ \mathfrak {a}_{k}\right \} $ for ${k\in \mathbb {Z}\setminus \{0\}}$ generate the $H^{*}(X)$ -algebra $\mathsf {Heis}_{X}$ .

For curve class $\beta \in H_{2}(X)$ , there is a geometrically defined Gromov–Witten evaluation $\langle \cdot \rangle _{\beta }$ map on the algebra generated by the nonnegative descendents. We can extend the evaluation map to the whole algebra $\mathsf {Heis}_{X}$ by defining

$$ \begin{align*}\big\langle\mathfrak{a}_{k}(\gamma)\Phi\big\rangle^{X,\mathrm{GW}}_{\beta}=\left[\int_{X} \big(-c_{1}\delta_{k+1}+\delta_{k+2}iu\big)\cdot \gamma\right]\, \big\langle \Phi\big\rangle^{X,\mathrm{GW}}_{\beta}\, ,\quad k<0.\end{align*} $$

We assemble the operators $\mathfrak {a}_{k}$ in the following generating function:

(8.2) $$ \begin{align} \phi(z)=\sum_{n>0}\frac{\mathfrak{a}_{n}}{n} \left(\frac{\imath zc_{1}}{u}\right)^{-n}+\frac{1}{c_{1}}\sum_{n<0}\frac{\mathfrak{a}_{n}}{n} \left(\frac{\imath zc_{1}}{u}\right)^{-n}\,. \end{align} $$

The main objects of study in [Reference Oblomkov, Okounkov and Pandharipande18] are the vertex operators

(8.3) $$ \begin{align} \mathrm{H}^{\mathsf{GW}}( x)=\sum_{k=0}^{\infty} \mathrm{H}^{\mathsf {GW}}_{k} x^{k+1}=\mathrm{Res}_{w=\infty}\left(\frac{\sqrt{dydw}}{y-w} :e^{\theta\phi(y)-\theta\phi(w)}:\right)\, , \end{align} $$

where y, w and x satisfy the constraints

(8.4) $$ \begin{align} ye^{y}=we^{w} e^{-x/\theta}\, , \ \quad \theta^{-2}=-c_{2}(T_{X})\,. \end{align} $$

Here, $\mathrm {Res}_{w=\infty }$ denotes $\frac {1}{2\pi \imath }$ times the integral along a small loop around $ w=\infty. $

Normally ordered monomials

$$ \begin{align*}\mathfrak{a}_{i_{1}}\mathfrak{a}_{i_{2}}\dots\mathfrak{a}_{i_{k}},\quad i_{1}\le i_{2}\le\dots\le i_{k}\end{align*} $$

form a linear basis of $\mathsf {Heis}$ . Respectively, we use $:\cdot :$ for the normal ordering operation

$$ \begin{align*}:\prod_{j}\mathfrak{a}_{i_{j}}:\,\,\,= \mathfrak{a}_{i_{1}}\mathfrak{a}_{i_{2}}\dots\mathfrak{a}_{i_{k}},\quad i_{1}\le i_{2}\le\dots\le i_{k}.\end{align*} $$

Extended $H^{*}(X)$ -linearly to the whole algebra $\mathrm {Heis}_{X}$ .

Let us notice that the equation (8.4) as well as the vertex operator (8.3) have symmetry

$$ \begin{align*}y\mapsto w,\quad w\mapsto y,\quad \theta\mapsto -\theta,\quad x\mapsto x.\end{align*} $$

This symmetry implies that the only even powers of $\theta $ appear in the expansion of equation (8.3) (see Lemma 15 from [Reference Oblomkov, Okounkov and Pandharipande18] for more discussions and further properties of the vertex operator).

The operators $\mathrm {H}^{\mathsf {GW}}_{k}$ are mutually commutative. To obtain explicit formulas for $\mathrm {H}^{\mathsf {GW}}_{k}$ , we use the Lambert function to solve equation (8.4) and express y in terms of $x,w$ . The integral in the definition of $\mathrm {H}^{\mathsf {GW}}_{k}$ can be interpreted as an extraction of the coefficient of $w^{-1}$ . The descendent classes

$$ \begin{align*}\mathrm{H}^{\mathsf{GW}}_{k}(\gamma) \in \mathsf{Heis}_{X}\end{align*} $$

are then obtained using the Sweedler coproduct. We also use the Sweedler coproduct conventions in

(8.5) $$ \begin{align} \mathrm{H}^{\mathsf{GW}}_{\vec{k}}(\gamma)=\prod_{i=1}^{m}\mathrm{H}^{\mathsf{GW}}_{k_{i}}(\gamma)\, , \ \ \quad \vec{k}=(k_{1},\dots,k_{m})\,. \end{align} $$

In the Sweedler conventions [Reference Kassel11], we abbreviate notation for the intersection with the small diagonal $\Delta _{n}\subset X^{n}$ with the pull-back of a class $\gamma \in H^{*}(X)$ :

$$ \begin{align*} H^{*}(X^{n})\ni [\Delta_{n}]\cdot \gamma=\sum_{k} \gamma_{1}^{k}\otimes \dots\gamma_{n}^{k}= \gamma_{(1)}\otimes\dots\otimes\gamma_{(n)}.\end{align*} $$

Thus, the formula (8.5) expands as

$$ \begin{align*}\prod_{i=1}^{m} \mathrm{H}^{\mathsf{GW}}_{k_{i}}(\gamma)=\prod_{i=1}^{m} \mathrm{H}^{\mathsf{GW}}_{k_{i}}(\gamma_{(i)}).\end{align*} $$

8.2 Stable pairs

The stable pairs analogues of the operators $\mathrm {H}^{\mathsf {GW}}_{\vec {k}}(\gamma )$ are products of $\mathrm {H}^{{\mathrm {PT}}}_{k}(\gamma )$ defined as follows.

The classes $\mathrm {H}^{{\mathrm {PT}}}_{k}(\gamma )$ are linear combinations of descendents on the moduli spaces of stable pairs. Let

$$ \begin{align*}\mathrm{H}^{{\mathrm{PT}}}_{k}(\gamma)= \pi_{*}\left( \mathrm{H}^{{\mathrm{PT}}}_{k}\cdot \gamma\right)\, \in \bigoplus_{n\in \mathbb{Z}} H^{*}(P_{n}(X,\beta)),\end{align*} $$

where the classes $\mathrm {H}^{{\mathrm {PT}}}_{k}\in \bigoplus _{n\in \mathbb {Z}} H^{*}(X\times P_{n}(X,\beta ))$ are defined by

$$ \begin{align*} \mathrm{H}^{{\mathrm{PT}}}(x)&=\sum_{k=0}^{\infty} x^{k+1} \mathrm{H}_{k}^{\mathrm{PT}} \\ &=\mathcal{S}^{-1}\left(\frac{x}{\theta}\right) \sum_{k=0}^{\infty} x^{k} \text{ch}_{k}(\mathbb{F}-\mathcal{O}),\end{align*} $$

where

$$ \begin{align*}\theta^{-2}=-c_{2}(T_{X})\, ,\ \ \ \ \mathcal{S}(x)=\frac{e^{x/2}-e^{-x/2}}{x}.\end{align*} $$

In particular, we have

$$ \begin{align*}\mathrm{H}^{{\mathrm{PT}}}_{k}=\text{ch}_{k+1}(\mathbb{F})+\frac{c_{2}}{24}\text{ch}_{k-1}(\mathbb{F})+\frac{7c_{2}^{2}}{5760}\text{ch}_{k-3}(\mathbb{F}) + \ldots.\end{align*} $$

8.3 Equivariant correspondence

All the definitions and construction introduced in Section 8.1 have canonical lifts to the equivariant setting with respect to a group action on the variety X. We review here the equivariant $\mathrm {GW}/{\mathrm {PT}}$ descendent correspondence [Reference Pandharipande and Pixton26].

The most natural setting is the capped vertex formalism of [Reference Maulik, Oblomkov, Okounkov and Pandharipande16, Reference Pandharipande and Pixton26] which we review briefly here. Let the three-dimensional torus

$$ \begin{align*}\mathsf{T}=\mathbb{C}^{*} \times \mathbb{C}^{*} \times \mathbb{C}^{*}\end{align*} $$

act on $\mathbf {P}^{1}\times \mathbf {P}^{1}\times \mathbf {P}^{1}$ diagonally. The tangent weights of the $\mathsf {T}$ -action at the point

$$ \begin{align*}\mathsf{p}=0\times 0\times 0 \in \mathbf{P}^{1}\times\mathbf{P}^{1}\times\mathbf{P}^{1}\end{align*} $$

are $s_{1},s_{2},s_{3}$ . The $\mathsf {T}$ -equivariant cohomology ring of a point is

$$ \begin{align*}H_{\mathsf{T}}(\bullet)=\mathbb{C}[s_{1},s_{2},s_{3}].\end{align*} $$

We have the following factorization of the restriction of class $c_{1}c_{2}-c_{3}$ of X to $\mathsf {p}$ :

$$ \begin{align*}c_{1}c_{2}-c_{3}=(s_{1}+s_{2})(s_{1}+s_{3})(s_{2}+s_{3}),\end{align*} $$

where $c_{i}=c_{i}(T_{X})$ .

Let $U\subset \mathbf {P}^{1}\times \mathbf {P}^{1}\times \mathbf {P}^{1}$ be the $\mathsf {T}$ -equivariant threefold obtained by removing the three $\mathsf {T}$ -equivariant lines $L_{1},L_{2},L_{3}$ passing through the point $\infty \times \infty \times \infty $ . Let $D_{i}\subset U$ be the divisor with $i^{th}$ coordinate $\infty $ . For a triple of partitions $\mu _{1},\mu _{2},\mu _{3}$ , let

(8.6) $$ \begin{align} \Big\langle\prod_{i}\, \tau_{k_{i}}(\mathsf{p})\, \Big|\, \mu_{1},\mu_{2},\mu_{3}\, \Big\rangle^{\mathsf{GW},\mathsf{T}}_{U,D}\, ,\quad \Big \langle\, \prod_{i} \text{ch}_{k_{i}}(\mathsf{p})\, \Big|\, \mu_{1},\mu_{2},\mu_{3}\, \Big \rangle^{\mathsf{PT},\mathsf{T}}_{U,D} \end{align} $$

denote the generating series of the $\mathsf {T}$ -equivariant relative Gromov–Witten and stable pairs invariants of the pair

$$ \begin{align*}D=\cup_{i} D_{i}\subset U\end{align*} $$

with relative conditions $\mu _{i}$ along the divisor $D_{i}$ .

The stable maps spaces are always taken with no contracted connected components of genus great than or equal to 2. The series (8.6) are the capped descendent vertices following the conventions of [Reference Oblomkov, Okounkov and Pandharipande18].

Theorem 8.1 [Reference Oblomkov, Okounkov and Pandharipande18]

After the change of variables $-q=e^{iu}$ the following correspondence between the two-leg capped descendent vertices holds:

$$ \begin{align*}\Big\langle\, \prod_{i} \mathrm{H}^{\mathsf{GW}}_{k_{i}}(\mathsf{p})\, \Big|\, \mu_{1},\mu_{2},\emptyset\, \Big\rangle^{\mathsf{GW},\mathsf{T}}_{U,D}= q^{-|\mu_{1}|-|\mu_{2}|}\Big\langle\, \prod_{i} \mathrm{H}^{{\mathrm{PT}}}_{k_{i}}(\mathsf{p})\, \Big|\, \mu_{1},\mu_{2},\emptyset\, \Big \rangle^{{\mathrm{PT}},\mathsf{T}}_{U,D}\end{align*} $$

$\mod (s_{1}+s_{3})(s_{2}+s_{3})$ .

The result of Theorem 8.1 has two defects. Since the third partition is empty, the result only covers the two-leg case. Moreover, the equality of the correspondence is not proven exactly, but only mod $(s_{1}+s_{3})(s_{2}+s_{3})$ . For the one-leg vertex with partitions $(\mu _{1}, \emptyset ,\emptyset )$ , Theorem 8.1 can be restricted in two ways to obtain the equality of the correspondence

$$ \begin{align*}\text{mod} \ \ (s_{1}+s_{3})(s_{1}+s_{2})(s_{2}+s_{3}).\end{align*} $$

8.4 Nonequivariant limit

By following the arguments of [Reference Pandharipande and Pixton26], a nonequivariant $\mathrm {GW}/{\mathrm {PT}}$ descendent correspondence for stationary insertions is derived in [Reference Oblomkov, Okounkov and Pandharipande18]. For our statements, we will follow as closely as possible the notation of [Reference Oblomkov, Okounkov and Pandharipande18, Reference Pandharipande and Pixton26].

Let $\mathsf {Heis}^{c}$ be the Heisenberg algebra with generators $\mathfrak {a}_{k\in \mathbb {Z}\setminus \{0\}}$ , coefficients $\mathbb {C}[c_{1},c_{2}]$ and relations

$$ \begin{align*}[\mathfrak{a}_{k},\mathfrak{a}_{m}]=k\delta_{k+m}c_{1}c_{2}.\end{align*} $$

Let $\mathsf {Heis}^{c}_{+}\subset \mathsf {Heis}^{c}$ be the subalgebra generated by the elements $\mathfrak {a}_{k>0}$ , and define the $\mathbb {C}[c_{1},c_{2}]$ -linear map

(8.7) $$ \begin{align} \mathsf{Heis}^{c}\rightarrow \mathsf{Heis}^{c}_{+}\, , \ \ \ \Phi\mapsto \widehat{\Phi} \end{align} $$

by $\widehat {\mathfrak {a}}_{k}= \mathfrak {a}_{k}$ for $k>0$ and

(8.8) $$ \begin{align} \widehat{\mathfrak{a}_{k} \Phi}=(-c_{1}\delta_{k+1}+\delta_{k+2}iu)\widehat{\Phi}\,,\ \ \ \text{for}\ k<0\,. \end{align} $$

When restricted to the subalgebra $\mathsf {Heis}^{c}_{+}$ , the $\mathbb {C}[c_{1},c_{2}]$ -linear map (8.7) is an isomorphism.

For a nonsingular projective threefold X and classes $\gamma _{1}, \ldots ,\gamma _{l}\in H^{*}(X)$ , the hat operation make no difference inside the Gromov–Witten bracket,

(8.9) $$ \begin{align} \langle \mathrm{H}_{\vec{k}}^{\mathrm{GW}}(\gamma)\rangle^{X,\mathrm{GW}}_{\beta}=\langle \widehat{\mathrm{H}}_{\vec{k}}^{\mathrm{GW}}(\gamma)\rangle^{\mathrm{GW}}_{\beta} \end{align} $$

because the treatment of the negative descendents on the left side is compatible with the treatment of the negative descendents by the hat operation.

Let $\vec {k}=(k_{1},\ldots ,k_{l})$ be a vector of nonnegative integers. Following [Reference Pandharipande and Pixton26], we define the following element of $\mathsf {Heis}^{c}_{+}$ :

$$ \begin{align*}\widetilde{\mathrm{H}}_{\vec{k}}=\frac{1}{(c_{1}c_{2})^{l-1}} \sum_{\text{set partitions }P\text{ of }\{1,\dots,\,\text{l}\}}(-1)^{|P|-1}(|P|-1)!\prod_{S\in P}\widehat{\mathrm{H}}^{\mathsf{GW}}_{\vec{k}_{S}},\end{align*} $$

where $\mathrm {H}^{\mathsf {GW}}_{\vec {k}_{S}}=\prod _{i\in S} \mathrm {H}^{\mathsf {GW}}_{k_{i}}$ and the element $\mathrm {H}_{k}^{\mathrm {GW}}\in \mathsf {Heis}^{c}$ is a linear combination of monomials of $\mathfrak {a}_{i}$ ; the expression is given by equation (8.3).

For classes $\gamma _{1},\ldots ,\gamma _{l}\in H^{*}(X)$ and a vector $\vec {k}=(k_{1},\ldots ,k_{l})$ of nonnegative integers, we define

$$ \begin{align*}\overline{\mathrm{H}_{k_{1}}(\gamma_{1})\dots\mathrm{H}_{k_{l}}(\gamma_{l})}=\sum_{\text{set partitions }P\text{ of }\{1,\dots,\,\text{l}\}}\, \prod_{S\in P}\widetilde{\mathrm{H}}_{\vec{k}_{S}}(\gamma_{S}),\end{align*} $$

where $\gamma _{S}=\prod _{i\in S}\gamma _{i}$ .

Theorem 8.2 [Reference Oblomkov, Okounkov and Pandharipande18]

Let X be a nonsingular projective toric threefold, and let $\gamma _{i}\in H^{\geq 2}(X,\mathbb {C})$ . After the change of variables $-q=e^{iu}$ , we have

$$ \begin{align*}\Big \langle\overline{ \mathrm{H}_{k_{1}}(\gamma_{1})\dots \mathrm{H}_{k_{l}}(\gamma_{l})} \Big \rangle_{\beta}^{\mathsf{GW}}= q^{-d/2}\Big \langle \mathrm{H}^{{\mathrm{PT}}}_{k_{1}}(\gamma_{1})\dots \mathrm{H}^{{\mathrm{PT}}}_{k_{l}}(\gamma_{l}) \Big \rangle_{\beta}^{\mathsf{PT}} ,\end{align*} $$

where $d=\int _{\beta } c_{1}$ .

8.5 Examples for $X=\mathbb {P}^{3}$

The prefactor $\mathcal {S}^{-1}\left (\frac {x}{\theta }\right )$ in front of $\sum _{k=0}^{\infty } x^{k}\mathsf {ch}_{k}(\mathbb {F}-\mathcal {O}) $ in the formula for $\mathrm {H}^{{\mathrm {PT}}}(x)$ has an expansion which the following initial terms:

$$ \begin{align*}1+\frac{c_{2}}{24}x^{2}+\frac{7c_{2}^{2}}{5760}x^{4}+\dots.\end{align*} $$

Therefore, the nonequivariant limit of $\mathrm {H}_{k}^{{\mathrm {PT}}}(\gamma )$ is

$$ \begin{align*}\left(\text{ch}_{k+1}(\gamma)+\frac{1}{24}\text{ch}_{k-1}(\gamma\cdot c_{2})\right).\end{align*} $$

On the Gromov–Witten side of the correspondence, we have

$$ \begin{align*} \langle \mathrm{H}_{1}^{\mathrm{GW}}(\gamma)\Phi\rangle=\langle\mathfrak{a}_{1}(\gamma)\Phi\rangle\, ,\ \quad \langle \mathrm{H}_{2}^{\mathrm{GW}}(\gamma)\Phi\rangle=\frac12\langle \mathfrak{a}_{2}(\gamma)\Phi\rangle,\end{align*} $$

$$ \begin{align*} \langle \mathrm{H}_{3}^{\mathrm{GW}}(\gamma)\Phi\rangle=\frac16\langle \mathfrak{a}_{3}(\gamma)\Phi\rangle+\frac{1}{24u^{2}}\langle c_{1}^{2}c_{2}\cdot\Phi\rangle,\end{align*} $$

$$ \begin{align*}\langle \mathrm{H}_{4}^{\mathrm{GW}}(\gamma)\Phi\rangle= \frac1{24}\langle \mathfrak{a}_{4}(\gamma)\Phi\rangle-\frac{i}{12 u}\langle\mathfrak{a}_{1}^{2}(c_{1}\cdot\gamma)\Phi\rangle-\frac{5i}{144u^{3}}\langle c_{1}^{3}c_{2}\cdot\Phi\rangle,\end{align*} $$

$$ \begin{align*} \langle\mathrm{H}_{5}^{\mathrm{GW}}\Phi\rangle=\frac1{120}\langle\mathfrak{a}_{5}(\gamma)\Phi\rangle-\frac{i}{24u}\langle\mathfrak{a}_{1}\mathfrak{a}_{2}(c_{1}\cdot \gamma)\Phi\rangle -\frac{1}{48u^{2}}\langle\mathfrak{a}^{2}_{1}(c_{1}^{2}\cdot\gamma)\Phi\rangle\\ +\frac{1}{24u^{2}}\langle\mathfrak{a}_{1}(c_{1}^{2}c_{2}\cdot\gamma)\Phi\rangle-\frac{1}{64u^{4}}\langle c_{1}^{4}c_{2}\cdot \Phi\rangle.\end{align*} $$

The operators $\mathfrak {a}_{k}$ are expressed in terms of standard descendentsFootnote ³²

(8.10) $$ \begin{align} \mathfrak{a}_{1}&=\tau_{0}-\frac{c_{2}}{24}\, , \\ \nonumber iu\mathfrak{a}_{2}/2&=\tau_{1}+c_{1}\cdot\tau_{0}\, ,\\ \nonumber -u^{2}\mathfrak{a}_{3}/3&=2\tau_{2}+3c_{1}\cdot\tau_{1}+c_{1}^{2}\cdot\tau_{0}\, , \\ \nonumber -iu^{3}\mathfrak{a}_{4}/4&=6\tau_{3}+11c_{1}\cdot\tau_{2}+6c_{1}^{2}\tau_{1}+c_{1}^{3}\cdot\tau_{0} \, ,\\ \nonumber u^{4}\mathfrak{a}_{5}/5&=24\tau_{4}+50c_{1}\cdot\tau_{3}+35c_{1}^{2}\cdot\tau_{2}+10c_{1}^{3}\cdot\tau_{1}+c_{1}^{4}\cdot\tau_{0}\,. \end{align} $$

The descendent correspondence of Theorem 8.2 implies relations for stable pairs and Gromov–Witten invariants of $\mathbf {P}^{3}$ . For example, for $\beta $ of degree 1,

(8.11) $$ \begin{align} - i q^{-2}\langle \text{ch}_{5}(\mathsf{L})\rangle &= \left(\frac1{u^{3}}\langle\tau_{3}(\mathsf{L})\rangle+\frac{22}{3u^{3}}\langle\tau_{2}(\mathsf{p})\rangle-\frac1{3u}\langle\tau_{0}\tau_{0}(\mathsf{p})\rangle\right)\, , \\ \nonumber q^{-2}\left(\langle \text{ch}_{6}(\mathsf{H})\rangle+\frac{1}{4}\langle \text{ch}_{4}(\mathsf{p})\rangle\right)&=\left( \frac{1}{u^{4}}\langle\tau_{4}(\mathsf{H})\rangle+\frac{25}{3u^{4}}\langle\tau_{3}(\mathsf{L})\rangle+ \frac{70}{3u^{4}}\langle\tau_{2}(\mathsf{p})\rangle\right.\\ \nonumber & \ \ \left.-\frac{1}{3u^{2}}\langle\tau_{0}\tau_{1}(\mathsf{L})\rangle+\frac{5}{3u^{2}}\langle\tau_{0}\tau_{0}(\mathsf{p})\rangle\right). \end{align} $$

Here, $\mathsf {p}$ is the class of point, $\mathsf {L}$ is the class of line and $\mathsf {H}$ is the class of hyperplane.Footnote ³³

8.6 Residues

To complete our proof of Theorem 1.4, we will compute the residues (8.3). More precisely, we will prove the following result.

Proposition 15. For $k_{i}\in \mathbb {Z}_{\geq 0}$ and $\gamma _{i}\in H^{\ge 2}(X)$ such that $\widetilde {\mathsf {ch}}_{k_{i}+2}(\gamma _{i})\in \mathbb {D}^{X\bigstar }_{{\mathrm {PT}}}$ , we have

$$ \begin{align*} \widetilde{\mathrm{H}}_{k_{1}+1}(\gamma_{1})&=\mathfrak{C}^{\circ}(\widetilde{\mathsf{ch}}_{k_{1}+2}(\gamma_{1}))\, ,\\ \widetilde{\mathrm{H}}_{k_{1}+1,k_{2}+1}(\gamma_{1}\cdot\gamma_{2})&=\mathfrak{C}^{\circ}(\widetilde{\mathsf{ch}}_{k_{1}+2}(\gamma_{1})\widetilde{\mathsf{ch}}_{k_{2}+2}(\gamma_{2}))\, , \\ \widetilde{\mathrm{H}}_{k_{1}+1,k_{2}+1,k_{3}+1}(\gamma_{1}\cdot\gamma_{2}\cdot\gamma_{3})&=\mathfrak{C}^{\circ}(\widetilde{\mathsf{ch}}_{k_{1}+2}(\gamma_{1})\widetilde{\mathsf{ch}}_{k_{2}+2}(\gamma_{2})\widetilde{\mathsf{ch}}_{k_{3}+2}(\gamma_{3})),\end{align*} $$

where the right side is defined by equations 1.14–1.16.

9.1 Preliminary computations

Before starting the proof of the Proposition 15, we compute the expansion of the terms of the residue formula (8.3).

Consider first the constraint equation (8.4). Solutions of the equation are formal power series in the variable

$$ \begin{align*}r=1/\theta\, , \ \ \ \ \theta^{-2}=-c_{2}(T_{X}).\end{align*} $$

We can solve the constraint equation iteratively in powers of r. Indeed, modulo $r^{1}$ , the constraint equation implies $w=y$ , and we start the expansion by

$$ \begin{align*}w(x,y) =y+O(r).\end{align*} $$

To find the next term of r in the expansion of $w(x,y)$ , we substitute

$$ \begin{align*}w(x,y)=y+f_{1}(x,y)r\end{align*} $$

into equation (8.4) and expand the result of the substitution in powers of r. The coefficient of $r^{1}$ in the expansion gives a linear equation which determines $f_{1}$ . After iterating the above procedure three times, we obtain

(9.1) $$ \begin{align} w(x,y)=y-xr\frac{y}{y+1}+(xr)^{2}\frac{y}{2(y+1)^{3}}+(xr)^{3}\frac{2y-1}{6(y+1)^{5}}+O(r^{4})\,. \end{align} $$

To see the expansion of the residue (8.3) has positive powers of $t=c_{1}$ , we use a change of variables:

(9.2) $$ \begin{align}y=v/t\, .\end{align} $$

The residue with respect to w on the right side of equation (8.3) is converted to a residue with respect to y via equation (9.1). Using equation (9.2), we will compute the residue with respect to v.

In the new variables, we have

$$ \begin{align*}\sqrt{dwdy}=\left(1-\frac{xrt}{2(v+t)}-\frac{(xr)^{2}t^{3}(4v-t)}{8(v+t)^{4}}\right)\frac{dv}{t}+O(r^{3}).\end{align*} $$

After we normal order the elements of the Heisenberg algebra in the expression for the vertex operator $\mathrm {H}^{\mathrm {GW}}(x)$ , the negative Heisenberg operators end up next to the vacuum $\langle \, |$ inside the bracket $\langle \cdot \rangle ^{\mathrm {GW}}$ . Relation (8.8), which governs interaction with $\langle \,|$ , yields the following factor in the expression under the residue:

(9.3) $$ \begin{align} \mathrm{E}&=\exp\left(-\frac{t}{2u}\left(\frac{w(y)^{2}-y^{2}}{r}\right)-\frac{t}{u}\left(\frac{w(y)-y}{r}\right)\right)\\[4pt] \nonumber &=\exp\left(\frac{xv}{u}\right) \left(1-\frac{trx^{2}v}{2u(v+t)}+\frac{t^{2}r^{2}(3xv^{2}+3txv+4t^{2}u)}{24u(v+t)^{3}}\right)+O(r^{3})\,. \end{align} $$

The inverse of $y-w$ in equation (8.3) becomes the factor

(9.4) $$ \begin{align} \mathrm{D}=-\frac{r}{w(y)-y}=\frac{v+t}{v}\left(1+\frac{t^{2}rx}{2(v+t)^{2}}+\frac{t^{3}r^{2}x^{2}(4v+t)}{12(v+t)^{4}}\right)+O(r^{3})\,. \end{align} $$

The elements of the Heisenberg algebra that participate in the residue formula are packed into the vertex operator

$$ \begin{gather*} \mathrm{V}=\mathrm{V}_{+}\cdot\mathrm{V}_{-}\, ,\quad \mathrm{V}_{+}(x,y)=\exp\left(\frac{1}{r}\sum_{n>0}\frac{\mathfrak{a}_{n}}{n(\imath ut)^{n}}( y^{-n}- w(y)^{-n})\right)\, ,\\[4pt] \mathrm{V}_{-}(x,y)=\exp\left(\frac{1}{rt}\sum_{n<0}\frac{\mathfrak{a}_{n}}{n(\imath ut)^{n}}( y^{-n}- w(y)^{-n})\right)\,. \end{gather*} $$

Thus, we need to compute the difference of powers in the expression for the vertex operators. Using formula for $w(y)$ (9.1), we obtain:

(9.5) $$ \begin{align} \frac{( yt)^{-n}-( w(y)t)^{-n}}{t^{n}r}&=\frac{nxt}{v^{n}(v+t)}+ nx^{2}rt^{2}\frac{((n+1)v+nt)}{v^{n}(v+t)^{3}}\nonumber\\[4pt] &\quad +nx^{3}r^{2}t^{3}\frac{n((n+1)(n+2)v^{2}+(2n^{2}+3n-1)tz+n^{2}t^{2})}{6v^{n}(v+t)^{5}}+O(r^{3})\,. \end{align} $$

The above calculations yield the leading terms of all algebraic expressions occurring in formula (8.3) for the vertex operator $\mathrm {H}^{\mathrm {GW}}(x)$ . As we will see in Section 9.2, the knowledge of these leading terms almost immediately leads to the simplest case of the descendent correspondence (1.14). For the other two cases, equations (1.15) and (1.16), we must analyze the interaction of two and three vertex operators $\mathrm {H}^{\mathrm {GW}}(x)$ . We apply standard vertex operator techniques to complete the proof of Proposition 15 in Section 9.2.

9.2 Proof of Proposition 15

9.2.1 Case $\widetilde {\mathrm {H}}_{k_{1}+1}(\gamma _{1})$

We start with the proof of the formula for the self-reaction. We must analyze the r expansion of the residue

(9.6) $$ \begin{align}\widetilde{\mathrm{H}}(x) = \widehat{\mathrm{H}}^{\mathrm{GW}}(x)=\operatorname{\mathrm{Res}}_{v=\infty}\frac{1}{t}\mathrm{E}\cdot \mathrm{D}\cdot\mathrm{V}_{+}\, .\end{align} $$

More precisely, we must compute the coefficients of

$$ \begin{align*}r^{i}t^{j}\, ,\ \ \ i+j\le 2.\end{align*} $$

By the argument of [Reference Oblomkov, Okounkov and Pandharipande18, Section 3.2], the coefficient of $rt^{j}$ vanishes. From the computations of the v expansions (9.1), (9.3), (9.4) and (9.5), the terms in front of $r^{i}$ , $i>0$ are proportional to t. The expression under the residue sign becomes

$$ \begin{align*}\exp\left(\frac{xv}{u}\right)\left(\frac{v+t}{t}+x\Sigma+\frac{x^{2}t}{v+t}\Sigma^{2}+\frac{x^{3}t^{2}}{(v+t)^{2}}\Sigma^{3}\right)+O(t^{3})+tO(r^{2})\, ,\quad\Sigma=\sum_{n>0}\frac{\mathfrak{a}_{n}}{(\imath uv)^{n}}.\end{align*} $$

After applying the residue operation to the last expression, we obtain the terms of formula (1.14) in the coefficients of the x-expansion.

9.2.2 Case $\widetilde {\mathrm {H}}_{k_{1}+1,k_{2}+1}(\gamma _{1}\cdot \gamma _{2})$

We show next that the double interaction term yields formula (1.15). The new computation that is needed for understanding the interaction term is $\widehat {\mathrm {H}}_{k_{1},k_{2}}$ . It is convenient to assemble the expressions into a generating series $\widehat {\mathrm {H}}(x_{1},x_{2})$ .

To compute $\widehat {\mathrm {H}}(x_{1},x_{2})$ , we must move all negative Heisenberg operators in the product of the vertex operators $\mathrm {H}^{\mathrm {GW}}(x_{1})\mathrm {H}^{\mathrm {GW}}(x_{2})$ to the left, next to the vacuum $\langle \, |$ . We use the standard vertex operator commutation relation to perform this reshuffling:

(9.7) $$ \begin{align} \mathrm{V}_{+}(x_{1},y_{1})\mathrm{V}_{-}(x_{2},y_{2} )=\mathrm{B}(x_{1},y_{1},x_{2},y_{2}) \mathrm{V}_{-}(x_{2},y_{2})\mathrm{V}_{+}(x_{1},y_{1} )\, , \end{align} $$

$$ \begin{align*}\mathrm{B}=\frac{(w_{2}-y_{1})(y_{2}-w_{1})}{(y_{2}-y_{1})(w_{2}-w_{1})},\end{align*} $$

where $w_{i}=w(x_{i},y_{i})$ . Using the computations of Section 9.1, we derive the following expansion:

$$ \begin{align*}\mathrm{B}=1-\frac{r^{2}y_{1}y_{2}x_{1}x_{2}}{(y_{1}-y_{2})^{2}(y_{1}+1)(y_{2}+1)}+O(r^{3}).\end{align*} $$

The negative Heisenberg operators interact with the vacuum $\langle \, |$ . We obtain

$$ \begin{align*}\widehat{\mathrm{H}}(x_{1},x_{2})=\operatorname{\mathrm{Res}}_{y_{1}=\infty}(\operatorname{\mathrm{Res}}_{y_{2}=\infty}(\mathrm{V}_{+}^{(1)}\mathrm{V}_{+}^{(2)}\mathrm{D}^{(1)}\mathrm{D}^{(2)}\mathrm{E}^{(1)}\mathrm{E}^{(2)}\mathrm{B}^{(12)})),\end{align*} $$

where $\mathrm {V}_{+}^{(i)}=\mathrm {V}_{+}(x_{i},y_{i})$ , $\mathrm {D}^{(i)}=\mathrm {D}(x_{i},y_{i})$ , $\mathrm {E}^{(i)}=\mathrm {E}(x_{i},y_{i})$ .

From equation (9.6), we see

$$ \begin{align*}\widehat{\mathrm{H}}(x_{1})\widehat{\mathrm{H}}(x_{2})&=\bigg(\operatorname{\mathrm{Res}}_{y_{1}=\infty}\mathrm{E}^{(1)}\cdot \mathrm{D}^{(1)}\cdot\mathrm{V}_{+}^{(1)}\bigg)\bigg(\operatorname{\mathrm{Res}}_{y_{2}=\infty}\mathrm{E}^{(2)}\cdot \mathrm{D}^{(2)}\cdot\mathrm{V}_{+}^{(2)}\bigg)\\ &=\operatorname{\mathrm{Res}}_{y_{1}=\infty}(\operatorname{\mathrm{Res}}_{y_{2}=\infty}(\mathrm{V}_{+}^{(1)}\mathrm{V}_{+}^{(2)}\mathrm{D}^{(1)}\mathrm{D}^{(2)}\mathrm{E}^{(1)}\mathrm{E}^{(2)})),\end{align*} $$

where the second equality holds because $V_{+}^{(i)}$ commute. We conclude, after the change of variables, the generating function $\widetilde {\mathrm {H}}(x_{1},x_{2})$ for $\widetilde {\mathrm {H}}_{k_{1},k_{2}}$ is given by

$$ \begin{align*}\widetilde{\mathrm{H}}(x_{1},x_{2})=\frac{1}{r^{2}t}\left(\widehat{\mathrm{H}}(x_{1},x_{2})-\widehat{\mathrm{H}}(x_{1})\widehat{\mathrm{H}}(x_{2})\right)=\operatorname{\mathrm{Res}}(\mathrm{V}_{+}^{(1)}\mathrm{V}_{+}^{(2)}\mathrm{D}^{(1)}\mathrm{D}^{(2)}\mathrm{E}^{(1)}\mathrm{E}^{(2)}\widetilde{\mathrm{B}}^{(12)})/(r^{2}t^{3}),\end{align*} $$

where $\operatorname {\mathrm {Res}}=\operatorname {\mathrm {Res}}_{v_{1}=\infty }\operatorname {\mathrm {Res}}_{v_{2}=\infty }$ and $\widetilde {\mathrm {B}}^{(12)}=\mathrm {B}^{(12)}-1$ . By expanding the scalar factor

$$ \begin{align*}\mathrm{D}^{(1)}\mathrm{D}^{(2)}\mathrm{E}^{(1)}\mathrm{E}^{(2)}\widetilde{\mathrm{B}}^{(12)}/(r^{2}t^{3})\end{align*} $$

in the operator inside the residue operation, we obtain

(9.8) $$ \begin{align}&\frac{tv_{1}v_{2}x_{1}x_{2}}{(v_{1}-v_{2})^{2}(v_{1}+t)(v_{2}+t)}\exp\left(\frac{x_{1}v_{1}+x_{2}v_{2}}{u}\right)\left(\frac{v_{1}+t}{t}+x_{1}\Sigma^{(1)}+\frac{x_{1}^{2}t}{v_{2}+t}\Sigma^{(1)}\Sigma^{(1)}\right)\nonumber\\ &\qquad\qquad\qquad\qquad\qquad\qquad\left(\frac{v_{2}+t}{t}+x_{2}\Sigma^{(2)}+\frac{x_{2}^{2}t}{v_{2}+t}\Sigma^{(2)}\Sigma^{(2)}\right)+O(t^{2})+O(r^{2}).\end{align} $$

The residue of the coefficient in front of $t^{-1}$ in equation (9.8) vanishes. The coefficient in front of $t^{0}$ is

$$ \begin{align*}\exp\left(\frac{x_{1}v_{1}+x_{2}v_{2}}{u}\right)\frac{x_{1}x_{2}}{(v_{1}-v_{2})^{2}}\left(v_{2}(1+x_{1}\Sigma^{(1)})+v_{1}(1+x_{2}\Sigma^{(2)})\right).\end{align*} $$

After applying the $\operatorname {\mathrm {Res}}$ operation, we obtain

$$ \begin{align*}\operatorname{\mathrm{Res}}_{v_{1}=\infty}\operatorname{\mathrm{Res}}_{v_{2}=\infty}\exp\left(\frac{x_{1}v_{1}+x_{2}v_{2}}{u}\right)\frac{x_{1}^{2}x_{2}v_{2}}{(v_{1}-v_{2})^{2}}\Sigma^{(1)}.\end{align*} $$

The coefficient in front of $x_{1}^{k_{1}+2}x_{2}^{k_{2}+2}$ in the last expression matches with the $\mathfrak {a}$ -linear terms of the right side of equation (1.15) that are proportional to $c_{1}^{0}$ .

Finally, we compute the coefficient in front of $t^{1}$ in equation (9.8):

$$ \begin{align*}&\frac{x_{1}x_{2}}{(v_{1}-v_{2})^{2}}\exp\left(\frac{x_{1}v_{1}+x_{2}v_{2}}{u}\right)\bigg[x_{1}x_{2}\Sigma^{(1)}\Sigma^{(2)}+x_{1}^{2}v_{2}\Sigma^{(1)}\Sigma^{(1)} +x_{2}^{2}v_{1}\Sigma^{(2)}\Sigma^{(2)}\nonumber\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad+ \left(\frac1v_{1}+\frac1v_{2}\right)\left(v_{2}(1+x_{1}\Sigma^{(1)})+v_{1}(1+x_{2}\Sigma^{(2)})\right)\bigg].\end{align*} $$

The residue of the terms from the first line of the last expression form the generating function of the $\mathfrak {a}$ -quadratic terms of the right-hand side of equation (1.15). The residue of the terms from the second line of the last expression form the generating function of the $c_{1}$ -proportional $\mathfrak {a}$ -linear terms of the right side of equation (1.15).