Proof. Let $\mathfrak {he}\left (W_{i}\right )$ denote the number of hanging half-edges along the boundary of $\left (W_{i}\right )_{+}$ , and consider the triple

(2.2) $$ \begin{align} \left(\mathfrak{d}\left(W_{i}\right),\chi\left(W_{i}\right),-\mathfrak{he}\left(W_{i}\right)\right). \end{align} $$

For every i, $W_{i}$ is a compact subsurface of Z, and so the three quantities are well-defined integers. We claim that at every step in the growing process, the triple (2.2) strictly reduces with respect to the lexicographic order.

Indeed, assume we do not halt after i steps, and let $O_{1},\ldots ,O_{k}$ be the list of $4g$ -gons in $Z\setminus W_{i}$ which are added to $W_{i}$ in order to obtain $W_{i+1}$ . By the choice of $4g$ -gons, it is clear that $\mathfrak {d}\left (W_{i+1}\right )\le \mathfrak {d}\left (W_{i}\right )$ . If the inequality is strict, we are done. So assume $\mathfrak {d}\left (W_{i+1}\right )=\mathfrak {d}\left (W_{i}\right )$ . This means that $\partial \left (W_{i}\right )$ contains no long blocks nor long chains, so it is boundary reduced, and that the edges in the complements in Z of the half-blocks and half-chains at $\partial \left (W_{i}\right )$ all belong to $\partial \left (W_{i+1}\right )$ . In other words, let $p_{1},\ldots ,p_{m}$ be these complements in Z: So $p_{j}$ is either a half-block or a half-chain. The equality $\mathfrak {d}\left (W_{i+1}\right )=\mathfrak {d}\left (W_{i}\right )$ means that all the edges in $p_{1},\ldots ,p_{m}$ belong to $\partial W_{i+1}$ .

It is easy to see that in this case $\chi \left (W_{i+1}\right )\le \chi \left (W_{i}\right )$ : The number of new $4g$ -gons and vertices in $W_{i+1}$ at most balances the number of new edges. Let V denote the set of internal vertices in $p_{1},\ldots ,p_{m}$ (so not at their endpoints). We have strict inequality $\chi \left (W_{i+1}\right )<\chi \left (W_{i}\right )$ if and only if some $v\in V$ belongs to $W_{i}$ or to two different complements from $p_{1},\ldots ,p_{m}$ .

Now, assume that $\mathfrak {d}\left (W_{i+1}\right )=\mathfrak {d}\left (W_{i}\right )$ and $\chi \left (W_{i+1}\right )=\chi \left (W_{i}\right )$ . Then $W_{i}$ is boundary reduced and each of the complements $p_{1},\ldots ,p_{m}$ is a connected piece of $\partial W_{i+1}$ . If $O_{j}$ touches a half-block of $\partial W_{i}$ , its annexation adds a net of $\left (2g-1\right )\left (4g-2\right )-2=8g\left (g-1\right )$ hanging half-edges. Every $4g$ -gon along a half-chain of $\partial W_{i}$ also adds on average a net of $8g\left (g-1\right )$ hanging half-edges. So if we add at least one $4g$ -gon at the $\left (i+1\right )$ st step, $-\mathfrak {he}$ strictly decreases. So indeed the triple (2.2) strictly decreases lexicographically in every step.

Finally, there are at most finitely many steps in which $\mathfrak {d}\left (W_{i}\right )$ decreases because this is a nonnegative integer. So it is enough to show there cannot be infinitely many steps in which $\mathfrak {d}\left (W_{i}\right )$ is constant. If $\mathfrak {d}\left (W_{i+1}\right )=\mathfrak {d}\left (W_{i}\right )$ , then $W_{i}$ is boundary reduced. If $\chi \left (W_{i}\right )$ keeps decreasing, then eventually we hit the bound $\chi \left (W_{i}\right )<\chi _{0}$ and halt. If $\mathfrak {d}\left (W_{i}\right )$ and $\chi \left (W_{i}\right )$ are constant, then $\mathfrak {he}\left (W_{i}\right )$ increases constantly, but in every tiled surface W, $\mathfrak {he}\left (W\right )\le 4g\mathfrak {d}\left (W\right )$ , so there cannot be infinitely many steps of this type too. This proves the lemma.

Proof. Note that every boundary edge of $W_{0}$ is necessarily an h-image of a boundary edge of Y so that $\mathfrak {d}\left (W_{i}\right )\le \mathfrak {d}\left (W_{0}\right )\le \mathfrak {d}\left (Y\right )$ . In every step, we add at most $\frac {\mathfrak {d}\left (W_{i}\right )}{2g-1}\le \frac {\mathfrak {d}\left (Y\right )}{2g-1}$ $4g$ -gons. So it is enough to bound the number of steps performed in the growing process until it terminates. There are at most $\mathfrak {d}\left (Y\right )$ steps in which $\mathfrak {d}\left (W_{i}\right )$ strictly decreasesFootnote ¹¹ ( $\mathfrak {d}\left (W_{i+1}\right )<\mathfrak {d}\left (W_{i}\right )$ ), so there are at most $\mathfrak {d}\left (Y\right )+1$ possible values of $\mathfrak {d}\left (W_{i}\right )$ . In steps where $\mathfrak {d}\left (W_{i}\right )$ is unchanged, $W_{i}$ is boundary reduced, so by definition $\chi \left (W_{i}\right )\ge \chi _{0}$ (otherwise, the process terminates). Let $\pi _{0}\left (Y\right )$ denote the number of connected components of Y. For all i, $W_{i}$ is a subsurface of Z with at most $\pi _{0}\left (Y\right )$ connected components, and by the classification of surfaces, $\chi \left (W_{i}\right )\le 2\pi _{0}\left (Y\right )$ . There are at most $2\pi _{0}\left (Y\right )-\chi _{0}$ steps with $\mathfrak {d}\left (W_{i}\right )$ fixed and $\chi \left (W_{i}\right )$ strictly decreasing. Finally, when $\mathfrak {d}\left (W_{i}\right )$ is constant there are at most $2\pi _{0}\left (Y\right )-\chi _{0}+1$ possible values of $\chi \left (W_{i}\right )$ , and if $\mathfrak {d}\left (W_{i+1}\right )=\mathfrak {d}\left (W_{i}\right )$ and $\chi \left (W_{i+1}\right )=\chi \left (W_{i}\right )$ , then $\mathfrak {he}\left (W_{i+1}\right )\ge \mathfrak {he}\left (W_{i}\right )+8g\left (g-1\right )$ and $\mathfrak {he}\left (W_{i}\right )\le 4g\mathfrak {d}\left (W_{i}\right )\le 4g\mathfrak {d}\left (Y\right )$ , so there are at most $\frac {\mathfrak {d}\left (Y\right )}{2\left (g-1\right )}$ steps with the same value of $\mathfrak {d}\left (W_{i}\right )$ and $\chi \left (W_{i}\right )$ . Overall there are at most

(2.3) $$ \begin{align} \mathfrak{d}\left(Y\right)+\left(\mathfrak{d}\left(Y\right)+1\right)\left[\left(2\pi_{0}\left(Y\right)-\chi_{0}\right)+\left(2\pi_{0}\left(Y\right)-\chi_{0}+1\right)\cdot\frac{\mathfrak{d}\left(Y\right)}{2\left(g-1\right)}\right] \end{align} $$

steps in the growing process. Define $B\left (Y\right )$ to be $\frac {\mathfrak {d}\left (Y\right )}{2g-1}$ times equation (2.3).

Proof. By Lemma 2.11 and the halting conditions of the growing process, it is clear that every such morphism in $\mathcal {R}$ satisfies R1 and R2. Given a morphism $h\colon Y\to Z$ as in Definition 2.13, $h\left (Y\right )$ (also named $W_{0}$ ) is a quotient of Y and therefore the number of cells in $h\left (Y\right )$ is bounded. From Lemma 2.12, we now conclude that that there is a bound on the number of cells in any $W_{f}$ with $f\in \mathcal {R}$ . This shows that $\mathcal {R}$ is finite as there are finitely many tiled surfaces with given bounds on the number of cells and finitely many morphisms between two given compact tiled surfaces.

It remains to show that $\mathcal {R}$ is a resolution. By the way it was constructed, it is clear that every morphism $h\colon Y\to Z$ with $\partial Z=\emptyset $ decomposes as

(2.4) $$ \begin{align} Y\stackrel{f}{\to}W_{f}\hookrightarrow Z \end{align} $$

and that $f\in \mathcal {R}$ . To show uniqueness, assume that h decomposes in an additional way

(2.5) $$ \begin{align} Y\stackrel{\varphi}{\to}W_{\varphi}\hookrightarrow Z, \end{align} $$

where $W_{\varphi }$ is the result of the growing process for some $h'\colon Y\to Z'$ with $\partial Z'=\emptyset $ . We claim that equations (2.4) and (2.5) are precisely the same decompositions of h. Indeed, the growing process defined by $h'\colon Y\to Z'$ takes place entirely inside $W_{\varphi }$ and does not depend on the structure of $Z'\backslash W_{\varphi }$ : In the $\left (i+1\right )$ st step of the growing process, the decision whether or not to annex more $4g$ -gons and where depends only on the structure and boundary of $W_{i}$ . Consequently, the growing process defined by the morphism $h'\colon Y\to Z'$ has the exact same output, in terms of the resulting element we add to $\mathcal {R}$ , as the growing process defined by the composition $Y\stackrel {\varphi }{\to }W_{\varphi }\hookrightarrow Z$ . But because the growing process is deterministic, the latter is identical to the growing process defined by $h\colon Y\to Z$ .

Proof of Proposition 2.15.

Suppose that $f\colon \mathrm {Core}\left (J\right )\to W_{f}$ satisfies $\chi \left (W_{f}\right )\ge \chi _{0}$ . In particular, $W_{f}$ is strongly boundary reduced by R2. Let $j\in \mathrm {Core}\left (J\right )$ be a vertex and assume without loss of generality that $J=p_{*}\left (\pi _{1}\left (\mathrm {Core}\left (J\right ),j\right )\right )$ (the fact that $p_{*}\left (\pi _{1}\left (\mathrm {Core}\left (J\right ),j\right )\right )\le \pi _{1}\left (\Sigma _{g},o\right )=\Gamma _{g}$ is conjugate to J follows from [Reference Magee and PuderMP22a, Prop. 5.3 and Cor. 4.11]). Define $K\stackrel {\mathrm {def}}{=}\pi _{1}\left (W_{f},f\left (j\right )\right )$ . As $W_{f}$ is compact, K is finitely generated. Let $p_{K}\colon \left (K\backslash \widetilde {\Sigma _{g}},k\right )\to \Sigma _{g}$ be the pointed covering space with $\pi _{1}\left (K\backslash \widetilde {\Sigma _{g}},k\right )=K$ . By the unique lifting property from the theory of covering spaces [Reference HatcherHat05, Prop. 1.33 and 1.34], as $W_{f}$ is connected, there is a unique lift $\alpha $ of the restricted covering map $p_{W_{f}}\colon W_{f}\to \Sigma _{g}$ to $K\backslash \widetilde {\Sigma _{g}}$ , which maps $f\left (j\right )$ to k. We will show that $\alpha $ gives an isomorphism between $W_{f}$ and $\mathrm {Core}\left (K\right )$ .

First, we show that $\alpha \left (W_{f}\right )\subseteq \mathrm {Core}\left (K\right )$ . Recall that $W_{f}$ is the result of the growing process for some $h\colon \mathrm {Core}\left (J\right )\to Z$ with $\partial Z=\emptyset $ . Consider $W_{0}\stackrel {\mathrm {def}}{=} h\left (\mathrm {Core}\left (J\right )\right )\subseteq Z$ . Recall that $W_{f}=\mathsf {SBR}\left (W_{0}\hookrightarrow Z\right )$ . By the unique lifting property, $\alpha \circ f$ is the natural morphism $\mathrm {Core}\left (J\right )\to K\backslash \widetilde {\Sigma _{g}}$ , which, by [Reference Magee and PuderMP22a, Lem. 5.4], has an image contained in $\mathrm {Core}\left (K\right )$ . Hence, $\alpha \left (W_{0}\right )\subseteq \mathrm {Core}\left (K\right )$ . As $\mathrm {Core}\left (K\right )$ is strongly boundary reduced [Reference Magee and PuderMP22a, Prop. 5.3], we have that $\mathsf {SBR}\left (\alpha \left (W_{0}\right )\hookrightarrow K\backslash \widetilde {\Sigma _{g}}\right )$ is contained in $\mathrm {Core}\left (K\right )$ . By [Reference Magee and PuderMP22a, Lem. 4.7],

$$\begin{align*}\alpha\left(W_{f}\right)=\alpha\left(\mathsf{SBR}\left(W_{0}\hookrightarrow Z\right)\right)\subseteq\mathsf{SBR}\left(\alpha\left(W_{0}\right)\hookrightarrow K\backslash\widetilde{\Sigma_{g}}\right)\subseteq\mathrm{Core}\left(K\right). \end{align*}$$

Now, Z is a covering space of $\Sigma _{g}$ and we may assume it is connected (because $\mathrm {Core}\left (J\right )$ is). Thus, Z is identical to $L\backslash \widetilde {\Sigma _{g}}$ for some $L=\pi _{1}\left (Z,h\left (j\right )\right )$ . By property R2, $W_{f}$ is strongly boundary reduced, and so by [Reference Magee and PuderMP22a, Cor. 4.11] its embedding in Z is $\pi _{1}$ -injective. In other words, $K\le L$ and, therefore, there is a morphism $m\colon \left (K\backslash \widetilde {\Sigma _{g}},k\right )\to \left (Z,f\left (j\right )\right )$ . By the unique lifting property, the composition $m\circ \alpha \colon \left (W_{f},f\left (j\right )\right )\to \left (Z,f\left (j\right )\right )$ must be identical to the embedding $\left (W_{f},f\left (j\right )\right )\hookrightarrow \left (Z,f\left (j\right )\right )$ and therefore $\alpha $ is injective. So $\alpha \left (W_{f}\right )$ is a strongly boundary reduced subtiled surface of $\mathrm {Core}\left (K\right )$ with fundamental group K. By [Reference Magee and PuderMP22a, Lem. 5.7], it follows that $\alpha \left (W_{f}\right )\supseteq \mathrm {Core}\left (K\right )$ . We conclude that $\alpha \colon W_{f}\to \mathrm {Core}\left (K\right )$ is an isomorphism, and R3 is proven.

To prove R4, suppose that $K\in \mathfrak {MOG}\left (J\right )$ . Let $h\colon \mathrm {Core}\left (J\right )\to K\backslash \widetilde {\Sigma _{g}}$ be the natural morphism. By the definition of the resolution $\mathcal {R}_{J}$ , the morphism h factors as $\mathrm {Core}\left (J\right )\stackrel {f}{\to }W_{f}\stackrel {}{\hookrightarrow }K\backslash \widetilde {\Sigma _{g}}$ for some $f\in \mathcal {R}_{J}$ . Because $h\left (\mathrm {Core}\left (J\right )\right )\subseteq \mathrm {Core}\left (K\right )$ (by [Reference Magee and PuderMP22a, Lem. 5.4]) and because $\mathrm {Core}\left (K\right )$ is strongly boundary reduced, we have $W_{f}\subseteq \mathrm {Core}\left (K\right )$ .

Let C be a connected component of the difference between the thick version of $\mathrm {Core}\left (K\right )$ and the thick version of $W_{f}$ . As $\mathrm {Core}\left (K\right )$ is compact, $\overline {C}$ is compact. As $\mathrm {Core}\left (K\right )$ is connected, $\overline {C}$ must intersect $\partial W_{f}$ and in particular has at least one boundary component. Since $W_{f}$ is boundary reduced, $\overline {C}$ is not homeomorphic to a disc, and so $\chi \left (\overline {C}\right )\le 0$ . Now,

$$\begin{align*}\chi\left(K\right)=\chi\left(\mathrm{Core}\left(K\right)\right)=\chi\left(W_{f}\right)+\sum_{C}\chi\left(\overline{C}\right), \end{align*}$$

the sum being over all connected components as above. We conclude that $\chi \left (W_{f}\right )\ge \chi \left (K\right )= \chi _{\max }\left (J\right )\ge \chi _{0}$ . By R2, $W_{f}$ is strongly boundary reduced and by R3, $W_{f}=\mathrm {Core}\left (M\right )$ for some subgroup M. But then $M\in \mathfrak {MOG}\left (J\right )$ , $\chi \left (\mathrm {Core}\left (M\right )\right )=\chi \left (K\right )$ , and every connected component C as above satisfies $\chi \left (\overline {C}\right )=0$ . As $\overline {C}$ has at least one boundary component, it must be an annulus. But then $\mathrm {Core}\left (M\right )$ is a deformation retract of $\mathrm {Core}\left (K\right )$ , so $M=K$ up to conjugation and so $W_{f}=\mathrm {Core}\left (M\right )=\mathrm {Core}\left (K\right )$ .

Proof. Construct $\mathcal {R}_{\gamma }=\left \{ f\colon \mathrm {Core}\left (\left \langle \gamma \right \rangle \right )\to W_{f}\right \} $ as $\mathcal {R}\left (\mathrm {Core}\left (\left \langle \gamma \right \rangle \right ),0\right )$ from Definition 2.13. By Theorem 2.14 and Proposition 2.15, the elements in $\mathcal {R}_{\gamma }$ with $\chi \left (W_{f}\right )=0$ are precisely the core surfaces of the subgroups in $\mathfrak {MOG}\left (\left \langle \gamma \right \rangle \right )$ . So it only remains to show that $\mathfrak {MOG}\left (\left \langle \gamma \right \rangle \right )$ are precisely $\left \langle \gamma _{0}^{~m}\right \rangle $ with $m\mid q$ .

But f.g. subgroups $K\le \Gamma _{g}$ with $\chi \left (K\right )=0$ are necessarily cyclic. Assume $K=\left \langle \delta \right \rangle \in \mathfrak {MOG}\left (\left \langle \gamma \right \rangle \right )$ , so $\gamma \in \left \langle \delta \right \rangle $ , and we may assume that $\gamma $ is a positive power of $\delta $ (otherwise switch to $\delta ^{-1}$ ). Every finitely generated subgroup of $\Gamma $ of infinite index is free (e.g., [Reference ScottSco78]), and a subgroup of finite index of $\Gamma _{g}$ is isomorphic to $\Gamma _{h}$ for some $h\ge g$ and so cannot be generated by less then $2g$ elements. We conclude that the subgroup $\left \langle \delta ,\gamma _{0}\right \rangle \le \Gamma $ is free. Because there is a relation $\gamma _{0}^{~q}=\delta ^{k}$ for some $k\in \mathbb {N}$ , it must be a cyclic subgroup. By definition, $\gamma _{0}$ is not a proper power, and so $\delta $ must be a positive power of $\gamma _{0}$ , and hence $\delta =\gamma _{0}^{~m}$ for some $m\mid q$ .

Proof. Let $\mathcal {A}\subseteq \mathrm {End}(V^{\lambda })$ be the algebra generated by the $\mathcal {E}_{\mu ,R_{1},R_{2}}^{\lambda }$ (over all $\mu )$ and $\mathcal {A}_{\mu }\subseteq \mathcal {A}$ be the algebra generated by the $\mathcal {E}_{\mu ,R_{1},R_{2}}^{\lambda }$ with a fixed value of $\mu $ . Suppose that $Q\in \mathrm {Tab}(\lambda )$ . The formula for the action of $\mathcal {E}_{\mu ,R_{1},R_{2}}^{\lambda }$ on $v_{Q}$ is

(4.3) $$ \begin{align} \mathcal{E}_{\mu,R_{1},R_{2}}^{\lambda}\left(v_{Q}\right)=\mathbf{1}\left\{ \mu_{m}(Q)=\mu,\,Q\lvert_{>m}=R_{2}\right\} \frac{1}{\sqrt{d_{\mu}}}v_{Q\lvert_{\le m}\sqcup R_{1}}. \end{align} $$

It is clear that the $\mathcal {E}_{\mu ,R_{1},R_{2}}^{\lambda }$ are an orthonormal set of elements in $\mathrm {End}(V^{\lambda })$ . It follows from equation (4.3) that

$$\begin{align*}\mathcal{E}_{\mu_{1},R_{1},R_{2}}^{\lambda}\mathcal{E}_{\mu_{2},R_{3},R_{4}}^{\lambda}=\mathbf{1}\left\{ \mu_{1}=\mu_{2},R_{3}=R_{2}\right\} \frac{1}{\sqrt{d_{\mu_{1}}}}\mathcal{E}_{\mu_{1},R_{1},R_{4}}^{\lambda}, \end{align*}$$

so the $\mathcal {E}_{\mu ,R_{1},R_{2}}^{\lambda }$ are an orthonormal basis for the algebra $\mathcal {A}$ , and those with a fixed $\mu $ are an orthonormal basis for $A_{\mu }$ . Furthermore, we have

$$\begin{align*}\mathcal{A}=\bigoplus_{\mu\subset\lambda,\mu\vdash m}\mathcal{A}_{\mu}. \end{align*}$$

For $\mu \vdash m$ , let $p_{\mu }\in \mathbf {C}[S_{m}]$ be the central idempotent projection onto the $\mu $ -isotypic component of $\mathbf {C}[S_{m}]$ and let $P_{\mu }^{~\lambda }$ be the image of $p_{\mu }$ in $\mathrm {End}(V^{\lambda })$ . The element $P_{\mu }^{~\lambda }$ is the orthogonal projection onto $\left (V^{\lambda }\right )_{\mu }$ , and $P_{\mu }^{~\lambda }$ is in the center of $Z(\lambda ,m,n)$ . Hence, for every $z\in Z(\lambda ,m,n)$ , we can write

$$\begin{align*}z=\bigoplus_{\mu\vdash m,\,\mu\subset\lambda}z^{(\mu)}, \end{align*}$$

where $z^{(\mu )}\stackrel {\mathrm {def}}{=} P_{\mu }^{\lambda }zP_{\mu }^{\lambda }.$ Moreover, if we let $\mathcal {B}_{\mu }$ be the algebra generated by the image of $\mathbf {C}[S_{m}]$ in $\mathrm {End}(V_{\mu }^{\lambda })$ , each $z^{(\mu )}$ must be in the commutant $\mathcal {B}^{\prime }_{\mu }$ of $\mathcal {B}_{\mu }$ in $\mathrm {End}(V_{\mu }^{\lambda }$ ). On the other hand, if $z=\bigoplus _{\mu \vdash m,\,\mu \subset \lambda }z^{(\mu )}$ and each $z^{(\mu )}\in \mathcal {B}^{\prime }_{\mu }$ , then $z\in Z(\lambda ,m,n)$ . This shows that

(4.4) $$ \begin{align} Z(\lambda,m,n) & =\bigoplus_{\mu\vdash m,\,\mu\subset\lambda}\mathcal{B}^{\prime}_{\mu}. \end{align} $$

Since $V^{\mu }$ is an irreducible module for $\mathbf {C}[S_{m}]$ , the algebra generated by $\mathbf {C}[S_{m}]$ in $\mathrm {End}(V^{\mu })$ is the whole of $\mathrm {End}(V^{\mu })$ . Hence, under the isomorphism of Lemma 3.2, the algebra $\mathcal {B}_{\mu }$ is identified with $\mathrm {End}(V^{\mu })\otimes \mathbf {C}\mathrm {Id}_{V_{\lambda /\mu }}$ . By a classical theorem, due to Tomita [Reference TomitaTom67] in the generality of von Neumann algebras,Footnote ¹² the commutant of a tensor product is the tensor product of the two commutants. Therefore, still using the isomorphism of Lemma 3.2, we have

$$\begin{align*}\mathcal{B}^{\prime}_{\mu}\cong\mathbf{C}\mathrm{Id}_{V^{\mu}}\otimes\mathrm{End}(V_{\lambda/\mu}). \end{align*}$$

This space is the algebra $\mathcal {A}_{\mu }$ , so $\mathcal {B}^{\prime }_{\mu }=\mathcal {A}_{\mu }$ and

$$\begin{align*}Z(\lambda,m,n)=\bigoplus_{\mu\vdash m,\,\mu\subset\lambda}\mathcal{B}^{\prime}_{\mu}=\bigoplus_{\mu\vdash m,\,\mu\subset\lambda}\mathcal{A}_{\mu}=\mathcal{A} \end{align*}$$

as required.

Proof. The first inequality is given by Liebeck and Shalev in [Reference Liebeck and ShalevLS04, Lem. 2.1]. To bound $d_{\lambda }$ from above, note that the standard tableaux of shape $\lambda $ can be obtained by choosing which $\lambda _{1}$ elements of $[n]$ are in the first row (of which there are at most $\binom {n}{b_{\lambda }}$ choices) and choosing the remaining $b_{\lambda }$ numbers’ locations outside the first row, of which there are at most $b_{\lambda }!$ choices. Hence,

$$\begin{align*}d_{\lambda}\leq b_{\lambda}!\binom{n}{b_{\lambda}}\leq n^{b_{\lambda}}.\\[-41pt] \end{align*}$$

Proof. By assumption, $n\ge k+2b_{\lambda }\ge k+2b_{\nu }$ , so $n-b_{\lambda }\ge \frac {n}{2}$ and $n-k-b_{\nu }\ge \frac {n-k}{2}$ and Lemma 4.2 applies. The statement now follows from Lemma 4.2 together with the inequality $\binom {p}{q}\geq (\frac {p}{q})^{q}$ , holding for $p,q\in \mathbf {N}$ with $p\ge q\ge 1$ .

Proof. If $k=0$ the statement is trivial, so we may assume $k\ge 1$ . We prove equation (4.6) as a consequence of the following slightly stronger statement:

(S) If $\lambda _{1}+\nu _{1}\ge n+\ell _{\mathrm {cox}}(\sigma )$ , then for any $T\in \mathrm {Tab}(\lambda /\nu )$ , $A_{0}\subseteq \left [m+1,n\right ]$ of size $\lambda _{1}-\nu _{1}$ and any unit vector u in

$$\begin{align*}W_{A_{0}}\stackrel{\mathrm{def}}{=}\mathrm{span}\left(\left\{ w_{T'}\,\middle|\,T'\in\mathrm{Tab}(\lambda/\nu),\,\mathrm{top}(T')=A_{0}\right\} \right) \end{align*}$$

we have

(4.7) $$ \begin{align} |\langle\sigma w_{T},u\rangle|\leq\left(\frac{\ell_{\mathrm{cox}}(\sigma)}{\nu_{1}+\lambda_{1}-n}\right)^{d(\sigma\mathrm{top}(T),A_{0})}. \end{align} $$

The proposition follows from (S) by using the bound $\ell _{\mathrm {cox}}(\sigma )\leq k^{2}$ and setting $A_{0}=\mathrm {top}(T')$ , $u=w_{T'}$ .

Let $D\stackrel {\mathrm {def}}{=} d\left (\sigma \mathrm {top}(T),A_{0}\right )$ . We prove (S) by induction on $\ell \stackrel {\mathrm {def}}{=}\ell _{\mathrm {cox}}(\sigma )$ . The base case of the induction is $\ell =0$ . Then $\sigma =\mathrm {id}$ and

$$\begin{align*}|\langle\sigma w_{T},u\rangle|=|\langle w_{T},u\rangle|=0 \end{align*}$$

unless $\mathrm {top}(T)=A_{0}$ , meaning $D=0$ . On the other hand, if $D=0$ , then

$$\begin{align*}|\langle\sigma w_{T},u\rangle|\leq1=0^{0}=\left(\frac{\ell}{\nu_{1}+\lambda_{1}-n}\right)^{D} \end{align*}$$

as required.

For the inductive step, for $\ell \geq 1$ we write $\sigma =s_{j}\sigma '$ , where $\ell _{\mathrm {cox}}(\sigma ')=\ell -1$ and $j\in \left [m+1,n-1\right ]$ . Two scenarios can occur.

(i) Suppose $s_{j}A_{0}=A_{0}$ . In this case, by the definition of the action of the Coxeter generators in equation (3.1), $s_{j}u$ is a unit vector in $W_{A_{0}}$ . Also, by the invariance of the distance function under $s_{j}$ ,

$$\begin{align*}d(\sigma'\mathrm{top}(T),A_{0})=d(\sigma\mathrm{top}(T),s_{j}A_{0})=d(\sigma\mathrm{top}(T),A_{0})=D. \end{align*}$$

The inductive hypothesis then yields

$$\begin{align*}|\langle\sigma w_{T},u\rangle|=|\langle\sigma'w_{T},s_{j}u\rangle|\leq\left(\frac{\ell-1}{\nu_{1}+\lambda_{1}-n}\right)^{D}\leq\left(\frac{\ell}{\nu_{1}+\lambda_{1}-n}\right)^{D}, \end{align*}$$

as required.

(ii) Suppose otherwise that $s_{j}A_{0}\neq A_{0}$ . This means that exactly one of j and $j+1$ are in $A_{0}$ . We write

(4.8) $$ \begin{align} u=\sum_{T'\in\mathrm{Tab}(\lambda/\nu)\colon\mathrm{top}(T')=A_{0}}\beta_{T'}w_{T'}. \end{align} $$

For each $T'$ with $\mathrm {top}(T')=A_{0}$ , we have

$$\begin{align*}|\mathrm{ax}({\fbox{\(j+1\)}}_{T'},{\fbox{\(j\)}}_{T'})|\geq\nu_{1}+\lambda_{1}-n. \end{align*}$$

From equation (4.8) and the formula for the action of Coxeter generators (3.1), we can therefore write $s_{j}u=w_{1}+w_{2}$ , where $w_{1}\in W_{s_{j}A_{0}}$ and $w_{2}\in W_{A_{0}}$ are orthogonal vectors with $\|w_{1}\|\leq 1$ and $\|w_{2}\|\leq (\nu _{1}+\lambda _{1}-n)^{-1}$ . Hence,

$$\begin{align*}\left|\langle\sigma w_{T},u\rangle\right|=\left|\langle\sigma'w_{T},s_{j}u\rangle\right|\leq\left|\langle\sigma'w_{T},w_{1}\rangle\right|+\left|\langle\sigma'w_{T},w_{2}\rangle\right|. \end{align*}$$

Note that $d(\sigma '\mathrm {top}(T),s_{j}A_{0})=d\left (\sigma \mathrm {top}(T),A_{0}\right )=D$ , so by the inductive hypothesis

(4.9) $$ \begin{align} |\langle\sigma w_{T},u\rangle|\leq\left(\frac{\ell-1}{\nu_{1}+\lambda_{1}-n}\right)^{D}+\frac{1}{\nu_{1}+\lambda_{1}-n}\left(\frac{\ell-1}{\nu_{1}+\lambda_{1}-n}\right)^{d(\sigma'\mathrm{top}(T),A_{0})}. \end{align} $$

By the triangle inequality,

$$ \begin{align*} D-1=d(\sigma'\mathrm{top}(T),s_{j}A_{0})-d(s_{j}A_{0},A_{0})\leq d(\sigma'\mathrm{top}(T),A_{0}), \end{align*} $$

so using $\nu _{1}+\lambda _{1}\geq n+\ell $ we obtain from equation (4.9)

$$ \begin{align*} |\langle\sigma w_{T},u\rangle| & \leq\left(\frac{\ell-1}{\nu_{1}+\lambda_{1}-n}\right)^{D}+\frac{1}{\nu_{1}+\lambda_{1}-n}\left(\frac{\ell-1}{\nu_{1}+\lambda_{1}-n}\right)^{D-1}\\ & =\frac{(\ell-1)^{D-1}\ell}{(\nu_{1}+\lambda_{1}-n)^{D}}\leq\left(\frac{\ell}{\nu_{1}+\lambda_{1}-n}\right)^{D}, \end{align*} $$

as required.

Proof. Since $V^{\lambda (n)/\nu (n-k)}$ is finite-dimensional with dimension independent of n (as long as $n\ge \left |\lambda \right |$ ), it suffices to prove the result in the case that $\sigma \in S_{k}$ is a Coxeter generator $s_{i}$ with $i\in [k-1]$ . Interpreted as an element of $S^{\prime }_{k}\leq S_{n}$ via equation (4.10), $\sigma $ corresponds to the Coxeter generator $s_{i+n-k}\in S_{n}$ .