Proof The fact that $F^\delta _\bullet $ is an algebra filtration is proven in [Reference Enriquez and ZerbiniEZ], Prop 5.5(a). The inclusion $p^*{\mathcal {O}}(C)\subset F^\delta _1\mathcal O_{hol}(\tilde C)$ follows from the $n=0$ case of Prop 5.5(b) in [Reference Enriquez and ZerbiniEZ]. Statement (b) follows from [Reference Enriquez and ZerbiniEZ], Theorem C.

Proof Statements (a) and (c) are consequences of Theorem A(b) in [Reference Enriquez and ZerbiniEZ] with J equal to the element $J_\sigma $ defined in Definition 1.2 of [Reference Enriquez and ZerbiniEZ]. Statement (b) follows from [Reference Enriquez and ZerbiniEZ], Theorem A(a).

Proof The first statement follows from [Reference Enriquez and ZerbiniEZ], Proposition 5.10, and the second statement follows from its combination with Lemma 2.4(c).

Proof This is straightforward.

Proof The residue map induces a linear map $O_S/\partial (O_S)\to (\mathbb C^S)_0$ , where $(\mathbb C^S)_0\subset \mathbb C^S$ is the set of tuples whose sum is equal to $0$ .

If $f\in O_S$ is a representative of an element of the kernel of $O_S/\partial (O_S)\to (\mathbb C^S)_0$ , then the residue of $fdz$ at any $s\in S$ is zero, so that the function $z\mapsto \int _{z_0}^z f(u)du$ is single-valued around each $s\in S$ . It follows that there exist $a,b\in \mathbb C$ such that $z\mapsto \int _{z_0}^z f(u)du-az-bg_1(z)$ belongs to $O_S$ , and therefore $f\equiv a+bg^{\prime }_1=a-bE_2$ mod $\partial (O_S)$ . This implies that the kernel of $O_S/\partial (O_S)\to (\mathbb C^S)_0$ is spanned by $1$ and $E_2$ .

The map $O_S/\partial (O_S)\to (\mathbb C^S)_0$ is surjective, because a preimage of $(a_s)_s\in (\mathbb C^S)_0$ is $\sum _{s\in S}a_sT_{\tilde s}(g_1)$ . Therefore $O_S/\partial (O_S)$ fits in an exact sequence

$$ \begin{align*}0\to\mathrm{Span}_{\mathbb C}(1,E_2)\to O_S/\partial(O_S)\to (\mathbb C^S)_0\to 0. \end{align*} $$

A basis of $O_S/\partial (O_S)$ is therefore the union of a basis of $\mathrm {Span}_{\mathbb C}(1,E_2)$ and of the preimage of a basis of $(\mathbb C^S)_0$ . Since $(1,E_2)$ is linearly independent and since ${((T_{\tilde s}-id)(g_1))_{s\in S\smallsetminus \{0\}}}$ is such a preimage, the image of

(2.5.1) $$ \begin{align} 1,E_2,((T_{\tilde s}-id)(g_1))_{s\in S\smallsetminus\{0\}} \end{align} $$

in $O_S/\partial (O_S)$ forms a $\mathbb C$ -basis of $O_S/\partial (O_S)$ .

The linear isomorphism $-\cdot dz : O_S\to \Omega ({\mathcal {E}}_S)$ , $f\mapsto f\cdot dz$ is such that $(-\cdot dz) \circ \partial =d$ (equality of linear maps $O_S\to \Omega ({\mathcal {E}}_S)$ ). This linear isomorphism therefore induces a linear isomorphism $-\cdot dz : O_S/\partial (O_S)\to \Omega ({\mathcal {E}}_S)/d(O_S)$ . The statement follows from this and from the fact that the claimed family is the image of (2.5.1) by $-\cdot dz$ .

Proof This follows from Lemmas 2.5 and (2.3.3).

Proof (a) follows from the fact that for any $r\geq 0$ , one has $(\log (t))^r=O(t^{-\alpha })$ at the neighborhood of $0$ for any $\alpha>0$ . (b) is an immediate direct check. Let us prove (c). The fact that $\mathrm {ev}$ and $\mathrm {can}$ are algebra morphisms can be verified directly. Let us prove the injectivity of $\mathrm {can}$ (see also [BGF], Lemma 3.350). Assume that $(P,\phi )\in \mathbb C[X]\oplus \mathcal F$ and $\mathrm {can}(P,\phi )=0$ . If P is nonzero, let $d\geq 0$ be its degree and $a_dX^d$ be its dominant term. Then $\mathrm {can}(P,\phi )(t)\sim a_d(\log (t))^d$ at the neighborhood of $0$ , which contradicts $\mathrm {can}(P,\phi )=0$ , therefore $P=0$ . Then $\mathrm {can}(0,\phi )=\phi $ , which implies $\phi =0$ .

Proof This follows from the fact that for any $z\in \tilde {{\mathcal {E}}}_S$ and $t\in ]0,\delta [$ , the map $a\mapsto k_{\mathrm {sect}(t)}(a)(z)$ is an algebra morphism $\mathrm {Sh}(V)\to \mathbb C$ (see Definition 4.2(c)).

Proof Since $\omega _{1,0}$ has a pole of order $1$ at $0$ , for any $z\in \tilde {{\mathcal {E}}}_S$ , the map $t\mapsto \int _{\mathrm {sect}(t)}^z \omega _{1,0}+\mathrm {log}(t)$ is an analytic function; denoting by $G(z)$ its value at 0, it therefore belongs to $G(z)+\mathcal F$ . One then has $G(z) =G(\mathrm {sect}(\delta /2))+\int _{\mathrm {sect}(\delta /2)}^z \omega _{1,0}$ for any $z\in \tilde {{\mathcal {E}}}_S$ , which implies the holomorphicity statement.

Proof (a) By (3.1.1), it suffices to prove that for any $r\geq 0$ and $a\in [\underbrace {\left (\begin {smallmatrix}1\\0\end {smallmatrix}\right )|\ldots | \left (\begin {smallmatrix}1\\0\end {smallmatrix}\right )}_{r\text { factors}}|a^*]$ , with $a^*\in \mathrm {Sh}^*(V)$ , one has $k(a,z)\in \mathbb C[X]_{\leq r}\oplus \mathcal F$ . The element $k(a,z)\in \mathbb C^{]0,\delta [}$ is the map taking $t\in \,]0,\delta [$ to the first term of the sequence of equalities

$$ \begin{align*} &I_{\mathrm{sect}(t)}([\underbrace{\omega_{1,0}|\ldots|\omega_{1,0}}_{r\text{ factors}}|\kappa(a^*)])(z) =I_{\mathrm{sect}(t)}([\underbrace{dG|\ldots|dG}_{r\text{ factors}}|\kappa_*(a^*)])(z) \\ &=\frac{1}{r!}\,I_{\mathrm{sect}(t)}\big([(G-G(\mathrm{sect}(t)))^r \odot \kappa_*(a^*)]\big)(z) \\ & =\sum_{l=0}^r \frac{1}{l!(r-l)!}\,(-G(\mathrm{sect}(t))^l I_{\mathrm{sect}(t)}\big([G^{r-l}\odot \kappa_*(a^*)]\big)(z) \end{align*} $$

where $\odot $ is the linear map ${\mathcal {O}}_{hol}(\tilde {{\mathcal {E}}}_S)\otimes \mathrm {Sh}({\mathcal {O}}_{hol}(\tilde {{\mathcal {E}}}_S)) \to \mathrm {Sh}({\mathcal {O}}_{hol}(\tilde {{\mathcal {E}}}_S))$ given by $f\odot (a_1\otimes \cdots \otimes a_n):=(fa_1)\otimes \cdots \otimes a_n$ . One shows that for any $l\in [\![0,r]\!]$ , the map ${t\mapsto I_{\mathrm {sect}(t)}[G^{r-l}\odot \kappa _*(a^*)]}$ belongs to $\mathbb C+\mathcal F$ . It follows from Lemma 3.6 that for any $l\in [\![0,r]\!]$ , the function $t\mapsto (-G(\mathrm {sect}(t))^l$ belongs to $(t\mapsto (\mathrm {log}(t))^l)+\mathcal F$ . Since $((t\mapsto (\mathrm {log}(t))^l)+\mathcal F)\cdot (\mathbb C+\mathcal F)\subset \mathbb C(t\mapsto (\mathrm {log}(t))^l)+\mathcal F$ , $k(a,z)$ belongs to $\sum _{l=0}^r \mathbb C(t\mapsto (\mathrm {log}(t))^l)+\mathcal F$ , which is $\mathbb C[X]_{\leq r}+\mathcal F$ .

Statement (b) follows from (a), Lemma 3.5 and the injectivity of $\mathrm {can}$ (see Lemma 3.4(c)).

Proof (a) For each $z\in \tilde {{\mathcal {E}}}_S$ , the composition $\mathrm {Sh}(V)\stackrel {k(-,z)}{\to } \mathbb C[X]\oplus \mathcal F\stackrel {\mathrm {ev}}{\to }\mathbb C$ is an algebra morphism as it is a composition of such morphisms (see Lemma 3.8(b) and Lemma 3.4(c)). The map $k_{\overrightarrow 0} : \mathrm {Sh}(V)\to \mathbb C^{\tilde {{\mathcal {E}}}_S}$ is the composition $\mathrm {Sh}(V)\to \mathrm {Sh}(V)^{\tilde {{\mathcal {E}}}_S}\to \mathbb C^{\tilde {{\mathcal {E}}}_S}$ , where the first map is the diagonal morphism and the second map is the product $\prod _{z\in \tilde {{\mathcal {E}}}_S}(\mathrm {ev}\circ k(-,z))$ . Since both the diagonal morphism and $\mathrm {ev}\circ k(-,z)$ are algebra morphisms for each $z\in \tilde {{\mathcal {E}}}_S$ , $k_{\overrightarrow 0}$ is an algebra morphism, which proves (a).

(b) Both sides belong to $\mathbb C[X]\oplus \mathcal F$ , which is contained in $\mathbb C^{]0,\delta [}$ , so it suffices the prove that they coincide as functions on $]0,\delta [$ . For $t\in \,]0,\delta [$ , one has

$$ \begin{align*} k(a,z)(t) &=k_{\mathrm{sect}(t)}(a)(z)=k_{\mathrm{sect}(t)}(a^{(1)})(\mathrm{sect}(\delta/2)) k_{\mathrm{sect}(\delta/2)}(a^{(2)})(z) \\ & =k(a^{(1)},\mathrm{sect}(\delta/2))(t)\cdot k_{\mathrm{sect}(\delta/2)}(a^{(2)})(z) \end{align*} $$

where the first and last equalities follow from definitions, and the middle equality follows from [Reference Enriquez and ZerbiniEZ], Lemma 2.5.

(c) Applying $\mathrm {ev}$ to the identity of (b), for any $a\in \mathrm {Sh}(V)$ and any $z\in \tilde {{\mathcal {E}}}_S$ one finds

$$ \begin{align*}k_{\overrightarrow 0}(a)(z)=k_{\overrightarrow 0}(a^{(1)})(\mathrm{sect}(\delta/2)) k_{\mathrm{sect}(\delta/2)}(a^{(2)})(z) \end{align*} $$

(equality in $\mathbb C$ ). Since, for any $b\in \mathrm {Sh}(V)$ , the map $z\mapsto k_{\mathrm {sect}(\delta /2)}(a^{(2)})(z)$ belongs to ${\mathcal {O}}_{hol}(\tilde {{\mathcal {E}}}_S)\subset \mathbb C^{\tilde {{\mathcal {E}}}_S}$ , it follows that the map $z\mapsto k_{\overrightarrow 0}(a)(z)$ also belongs to ${\mathcal {O}}_{hol}(\tilde {{\mathcal {E}}}_S)$ .

(d) This statement follows from (a) and (c).

Proof Obvious.

Proof (a) The first statement follows from the specialization of Lemma 3.8(a) for $\mathrm {deg}(a)\leq 0$ . The second statement follows from the fact that for any pair $(z,t)$ , the map $a\mapsto k_{\mathrm {sect}(t)}(a)(z)$ is an algebra morphism $\mathrm {Sh}^*(V)\to \mathbb C$ , and from the first statement; the last statement follows from the second statement by composing with $\mathrm {ev}_0$ .

(b) Let $a\in \mathrm {Sh}^*(V)$ . Since $\Delta (\mathrm {Sh}^*(V))\subset \mathrm {Sh}^*(V)\otimes \mathrm {Sh}(V)$ , the equality from Lemma 3.10(b) takes place in $\mathbb C\oplus \mathcal F$ . Applying $\mathrm {ev}_0$ to it, one obtains that for every $z\in \tilde {{\mathcal {E}}}_S$

(equality in $\mathbb C$ ). The first part of (b) then follows from the fact that for any $a\in \mathrm {Sh}(V)$ , the map $z\mapsto k_{\mathrm {sect}(\delta /2)}(a^{(2)})(z)$ belongs to ${\mathcal {O}}_{hol}(\tilde {{\mathcal {E}}}_S)\subset \mathbb C^{\tilde {{\mathcal {E}}}_S}$ . The second part of (b) follows from its first part, combined with statement that the maps is an algebra morphism $\mathrm {Sh}^*(V)\to \mathbb C^{\tilde {{\mathcal {E}}}_S}$ , which follows from (a).

(c) This follows from (b) and from Lemma 3.13.

Proof One checks that the restrictions of $k_{\overrightarrow 0}$ and to $\mathrm {Sh}^*(V)$ coincide. Let us compute $k_{\overrightarrow 0}(\left (\begin {smallmatrix}1\\0\end {smallmatrix}\right ))$ . For $z\in \tilde {{\mathcal {E}}}_S$ , $k(\left (\begin {smallmatrix}1\\0\end {smallmatrix}\right ),z)$ is the function $]0,\delta [\, \ni t\mapsto k_{\mathrm {sect}(t)}(\left (\begin {smallmatrix}1\\0\end {smallmatrix}\right )) (z)=\int _{\mathrm {sect}(t)}^z \omega _{1,0}$ . By Lemma 3.6, this function belongs to $(t\mapsto -\mathrm {log}(t)+G(z))+\mathcal F=P(\mathrm {log}(t))+\mathcal F$ , where $P(X)=G(z)-X\in \mathbb C[X]$ . It follows that , hence that . By Lemma 3.13, this implies the statement.

Proof (a) Fix $a\in \mathrm {Sh}(V)$ and $z\in \tilde {{\mathcal {E}}}_S$ . By [Reference Enriquez and ZerbiniEZ], Lemma 2.5, one has the identity $k_{\mathrm {sect}(t)}(a)(z)=k_{\mathrm {sect}(t)}(a^{(1)})(z_0)\cdot k_{z_0}(a^{(2)})(z)$ for any $t\in \,]0,\delta [$ . One derives from this the equality (in $\mathbb C^{]0,\delta [}$ ) $(t\mapsto k_{\mathrm {sect}(t)}(a)(z))=(t\mapsto k_{\mathrm {sect}(t)}(a^{(1)})(z_0)) \cdot k_{z_0}(a^{(2)})(z)$ . Since both sides actually belong to the subalgebra $\mathbb C[X]\oplus \mathcal F\subset \mathbb C^{]0,\delta [}$ , then this may be viewed as an equality in $\mathbb C[X]\oplus \mathcal F$ . Applying to both sides the morphism $\mathrm {ev} : \mathbb C[X]\oplus \mathcal F\to \mathbb C$ , one derives the equality (in $\mathbb C$ ) $k_{\overrightarrow 0}(a)(z)=k_{\overrightarrow 0}(a^{(1)})(z_0)\cdot k_{z_0}(a^{(2)})(z)$ .

It follows that, for any $a\in \mathrm {Sh}(V)$ , one has $(z\mapsto k_{\overrightarrow 0}(a)(z))=(z\mapsto k_{\overrightarrow 0}(a^{(1)})(z_0)) \cdot k_{z_0}(a^{(2)})(z)$ (equality in $\mathbb C^{\tilde {{\mathcal {E}}}_S}$ ). Both sides actually belong to ${\mathcal {O}}_{hol}(\tilde {{\mathcal {E}}}_S)$ , and are respectively equal to the two sides of (3.2.1), which proves (a).

(b) Equation (3.2.1) implies the inclusion $k_{\overrightarrow 0}(\mathrm {Sh}(V)) \subset k_{z_0}(\mathrm {Sh}(V))$ , and also implies the relation $k_{z_0}(a)=k_{\overrightarrow 0}(S(a^{(1)}))(z_0) \cdot k_{\overrightarrow 0}(a^{(2)})$ for any $a \in \mathrm {Sh}(V)$ , where S is the antipode of $\mathrm {Sh}(V)$ , which implies the opposite inclusion $k_{z_0}(\mathrm {Sh}(V)) \subset k_{\overrightarrow 0}(\mathrm {Sh}(V))$ .

Proof The identity

(3.3.2) $$ \begin{align} d(k_{\mathrm{sect}(\delta/2)}([\,t\,| \left(\begin{smallmatrix}n\\a\end{smallmatrix}\right)])=k_{\mathrm{sect}(\delta/2)}(t)\cdot \omega_{n,a} \end{align} $$

for any t and $(n,a)$ as above follows from [Reference Enriquez and ZerbiniEZ], Equation (2.1.1). Let us define the map $\partial ^V : \mathrm {Sh}(V)\to \mathrm {Sh}(V)\otimes V$ by setting $\partial ^V(1)=0$ and $\partial ^V([t|v]):=t\,\otimes \,v$ for any $t\in \mathrm {Sh}(V)$ and $v\in V$ . For $t\in \mathrm {Sh}(V)$ , denote $\partial ^V(t)$ by $t^{[1]} \otimes t^{[2]}$ . Then (3.3.2) implies the identity

(3.3.3) $$ \begin{align} d(k_{\mathrm{sect}(\delta/2)}(t))=k_{\mathrm{sect}(\delta/2)}(t^{[1]})\cdot \omega(t^{[2]}), \end{align} $$

where $\omega : V\to \Omega _{hol}(\tilde {{\mathcal {E}}}_S)$ is the map given by $\left (\begin {smallmatrix}n\\a\end {smallmatrix}\right )\mapsto \omega _{n,a}$ .

Applying Lemma 3.16 with $z_0:=\mathrm {sect}(\delta /2)$ , one finds that, for any $t\in \mathrm {Sh}(V)$ ,

(3.3.4) $$ \begin{align} k_{\overrightarrow 0}(t)=k_{\overrightarrow 0}(t^{(1)})(\mathrm{sect}(\delta/2))\cdot k_{\mathrm{sect}(\delta/2)}(t^{(2)}) \end{align} $$

(equality in ${\mathcal {O}}_{hol}(\tilde {{\mathcal {E}}}_S)$ ), which gives rise to the first equality in

$$ \begin{align*} d(k_{\overrightarrow 0}(t)) &=k_{\overrightarrow 0}(t^{(1)})(\mathrm{sect}(\delta/2))\cdot d(k_{\mathrm{sect}(\delta/2)}(t^{(2)})) \\ &=k_{\overrightarrow 0}(t^{(1)})(\mathrm{sect}(\delta/2))\cdot k_{\mathrm{sect}(\delta/2)}((t^{(2)})^{[1]}) \cdot\omega((t^{(2)})^{[2]}) \\ & =k_{\overrightarrow 0}((t^{[1]})^{(1)})(\mathrm{sect}(\delta/2))\cdot k_{\mathrm{sect}(\delta/2)}((t^{[1]})^{(2)}) \cdot\omega(t^{[2]})\\& =k_{\overrightarrow 0}(t^{[1]}) \cdot \omega(t^{[2]}), \end{align*} $$

where the second equality follows from (3.3.3), the third equality follows from $(id\otimes \partial ^V)\circ \Delta =(\Delta \otimes id) \circ \partial ^V$ , and the fourth equality follows from (3.3.4). One derives $d(k_{\overrightarrow 0}(t))=k_{\overrightarrow 0}(t^{[1]}) \cdot \omega (t^{[2]})$ , from which the claimed identity follows.

Proof Let $\tilde s\in \tilde S$ . It follows from (2.4.7) that the collection of functions $(T_{\tilde s}g_n)_{n\geq 0}$ satisfies the relation $d(T_{\tilde s}g_n)=\sum _{k=0}^{n-1}(-1)^{n-k}\,T_{\tilde s}g_k\cdot T_{\tilde s}(E_{n-k+1})\cdot dz$ for any $n\geq 1$ . Let us prove the statement by induction on $n\geq 0$ . For $n=0$ , one has $T_{\tilde s}g_0=1\in F_0^\delta \mathcal {O}_{hol}(\tilde {{\mathcal {E}}}_S)$ . Assume that $T_{\tilde s}g_k\in F_k^\delta \mathcal {O}_{hol}(\tilde {{\mathcal {E}}}_S)$ for $k<n$ . Since $T_{\tilde s}E_{l}\in {\mathcal {O}}({\mathcal {E}}_S)$ for $l\geq 2$ , one has $T_{\tilde s}(E_l)\cdot dz \in \Omega ({\mathcal {E}}_S)$ for these values of l, hence $d(T_{\tilde s}g_n)=\sum _{k=0}^{n-1}(-1)^{n-k}T_{\tilde s}g_k \cdot T_{\tilde s}(E_{n-k+1})\cdot dz\in F_{n-1}^\delta \mathcal {O}_{hol}(\tilde {{\mathcal {E}}}_S)\cdot p^*\Omega ({\mathcal {E}}_S)$ , and therefore $T_{\tilde s}g_n\in F_n^\delta \mathcal {O}_{hol}(\tilde {{\mathcal {E}}}_S)$ .

Proof Let us prove the statement by induction on r. If $r=0$ , the statement is clear. If $r>0$ , assume the statement to be true until $r-1$ , so that $\tilde {\operatorname {\mathrm {\Gamma }}}\left (\begin {smallmatrix}n_1 &n_2 &\ldots &n_{r-1}\\a_1 &a_2 &\ldots &a_{r-1}\end {smallmatrix};-\right )$ belongs to $F^\delta _{r-1+n_1+\cdots +n_{r-1}}\mathcal {O}_{hol}(\tilde {{\mathcal {E}}}_S)$ . By Lemma 4.2(b), one has $T_{a_r}g_{n_r}\in F_{n_r}^\delta \mathcal {O}_{hol}(\tilde {{\mathcal {E}}}_S)$ . The facts that $F_\bullet ^\delta \mathcal {O}_{hol}(\tilde {{\mathcal {E}}}_S)$ is an algebra filtration and that $dz\in p^*\Omega ({\mathcal {E}}_S)$ then imply that the right-hand side of (3.3.1) belongs to $F^\delta _{r-1+n_1+\cdots +n_r}\mathcal {O}_{hol}(\tilde {{\mathcal {E}}}_S)\cdot p^*\Omega ({\mathcal {E}}_S)$ , which implies that $\tilde {\operatorname {\mathrm {\Gamma }}}\left (\begin {smallmatrix}n_1 &n_2 &\ldots &n_r\\a_1 &a_2 &\ldots &a_r\end {smallmatrix};-\right )$ belongs to $F^\delta _{r+n_1+\cdots +n_r}\mathcal {O}_{hol}(\tilde {{\mathcal {E}}}_S)$ . This concludes the proof by induction.

Proof One has $O_S={\mathcal {O}}({\mathcal {E}}_S)\subset F^\delta _1{\mathcal {O}}(\tilde {{\mathcal {E}}}_S)\subset F^\delta _\infty {\mathcal {O}}(\tilde {{\mathcal {E}}}_S)$ . Moreover, Lemma 4.2 with $\tilde s=0$ and $n=1$ implies $g_1\in F^\delta _1{\mathcal {O}}(\tilde {{\mathcal {E}}}_S) \subset F^\delta _\infty {\mathcal {O}}({\mathcal {E}}_S)$ . Together with Proposition 4.3 and the fact that $F^\delta _\infty {\mathcal {O}}({\mathcal {E}}_S)$ is an algebra, this implies that $\mathcal G\subset F^\delta _\infty {\mathcal {O}}({\mathcal {E}}_S)$ . By [Reference Enriquez and ZerbiniEZ], Theorem C, one has $F^\delta _\infty {\mathcal {O}}({\mathcal {E}}_S)=A_{{\mathcal {E}}_S}$ , which implies the result.

Proof Let us prove inductively on $d\geq 0$ that if $a_0,\ldots ,a_d\in O$ and $\sum _i a_ig_1^i\in {\mathcal {O}}_{hol}(\mathbb C)$ , then $a_0 \in \mathbb C$ and $a_1=\ldots =a_d=0$ . If $d=0$ , then $a_0 \in O \cap {\mathcal {O}}_{hol}(\mathbb C)=\mathbb C$ , which proves the statement for $d=0$ . Assume the statement at step $d-1$ and let us prove it at step d. Let $a_0,\ldots ,a_d \in O$ be such that $f:=\sum _i a_ig_1^i \in {\mathcal {O}}_{hol}(\mathbb C)$ . Then

$$ \begin{align*} \sum_{j=0}^{d-1}g_1^j\bigg(\sum_{i>j}a_i\binom{i}{j}(2\pi\mathrm i)^{i-j}\bigg)&=\sum_{(i,j)|j<i}a_i{i \choose j}g_1^j(2\pi\mathrm i)^{i-j}=\sum_i a_i\big((g_1+2\pi\mathrm i)^i-g_1^i\big)\\ &=(T_\tau-id)(f)\in {\mathcal{O}}_{hol}(\mathbb C), \end{align*} $$

which by the induction assumption implies that $\sum _{i>0}a_i(2\pi \mathrm i)^{i} \in \mathbb C$ and for any $j=1,\ldots ,d-1$ , $\sum _{i>j}a_i {i \choose j}(2\pi \mathrm i)^{i-j}=0$ . This implies $a_2=\ldots =a_d=0$ and $a_1 \in \mathbb C$ . Then $a_0+a_1g_1 \in {\mathcal {O}}_{hol}(\mathbb C)$ . Assume $a_1 \neq 0$ . Since $g_1$ has a simple pole at $0$ with residue $1$ , and since $a_0+a_1g_1 \in {\mathcal {O}}_{hol}(\mathbb C)$ , $a_0$ has a simple pole at $0$ with residue $-a_1$ , which contradicts $a_0 \in O$ . Therefore $a_1=0$ . Then $a_0\in {\mathcal {O}}_{hol}(\mathbb C)\cap O=\mathbb C$ , which implies the statement at step d.

Proof Let P belong to the kernel of this morphism and let d be its degree as a polynomial in X. If $P\neq 0$ , then there exist $f_0,\ldots ,f_d\in O_S$ with $f_d\neq 0$ , such that $P=\sum _{i=0}^d f_i \otimes X^i$ . Then $\sum _{i=0}^d f_ig_1^i=0$ . Applying $(T_\tau -id)^d$ to this equality, using the invariance of $f_i$ under $T_\tau $ and the relation $(T_\tau -id)^d(g_1^i)=\delta _{i,d}(2\pi \mathrm {i})^dd!$ for $i\in [\![0,d]\!]$ , one obtains $f_d=0$ , a contradiction.

Proof One has for any $n\geq 1$ the identity

$$ \begin{align*} b_n=\sum_{k\geq 1}\frac{1}{k!}\sum_{\substack{n_1+\cdots+n_k=n,\\(n_1,\ldots,n_k)\in\mathbb Z_{\geq1}^k}}a_{n_1}\cdots a_{n_k}, \end{align*} $$

which together with the algebra filtration properties implies (a). This identity implies that, for any $n\geq 1$ ,

$$ \begin{align*} \overline b_n=\sum_{k\geq 1}\frac{1}{k!}\sum_{\substack{n_1+\cdots+n_k=n,\\(n_1,\ldots,n_k)\in\mathbb Z_{\geq1}^k}}\overline a_{n_1}\cdots \overline a_{n_k}, \end{align*} $$

where $\overline a_i,\overline b_i$ are the classes of $a_i,b_i$ in $\mathrm {gr}_i(A)$ , from which one derives (b).

Proof Equation (2.4.5) implies the identity

$$ \begin{align*}1+\sum_{n\geq 1}g_n\,\alpha^{n}=\exp\Big((1\otimes X)\,\alpha-\sum_{r\geq2}\frac{(-1)^{r}}{r}\,((E_r-e_r)\otimes 1)\,\alpha^r\Big), \end{align*} $$

(equality in $(O\otimes \mathbb C[X])[[\alpha ]]$ ) where for $n\geq 1$ , $g_n$ is identified with its image in $O\otimes \mathbb C[X]$ . The first statement in (a) follows from this equality, combined with Lemma 4.7(a) and the relations $X\in F_1\mathbb C[X]$ and $E_r-e_r\in F_rO$ for any $r\geq 2$ , which imply $1\otimes X\in F_1(O\otimes \mathbb C[X])$ and $(E_r-e_r)\otimes 1\in F_r(O\otimes \mathbb C[X])$ for $r\geq 2$ . The second statement in (a) follows from Lemma 4.7(b), which implies that

$$ \begin{align*}1+\sum_{n\geq 1}\overline g_n &=\exp\Big(\overline{1\otimes X}-\sum_{r\geq2}\frac{(-1)^{r}}{r}\,\overline{(E_r-e_r)\otimes 1}\Big) \\ &=\exp\Big(X-\sum_{r\geq2}\frac{(-1)^{r}}{r}Y^r\Big)=e^{X-Y}(1+Y), \end{align*} $$

(equality in the subalgebra $\mathbb C[[X]]\oplus Y^2\mathbb C[[X,Y]]$ of $\mathbb C[[X,Y]]$ ), which in turn implies that $\overline g_n=(X-Y)^n/n!+Y(X-Y)^{n-1}/(n-1)!=(X-Y)^{n-1}(X+(n-1)Y)/n!$ .

(b) If $\sum _n\lambda _ng_n=0$ is a nontrivial linear dependence relation, let $i:=\mathrm {max}\{i|\lambda _i\neq 0\}$ . Then $0=\sum _n\lambda _ng_n\in F_i(O[g_1])$ , and the image of this element in $\mathrm {gr}_i(O[g_1])$ is $\lambda _i\overline {g_i}$ , which is nonzero, this yielding a contradiction.

(c) Let D be the endomorphism of $O\otimes \mathbb C[X]$ defined by $D(f\otimes P):=\partial f\otimes P-fE_2\otimes P'$ . Then D is a derivation of $O\otimes \mathbb C[X]$ , which is interwined with $\partial $ under the specialization of the isomorphism from Lemma 4.6 to $S=\{0\}$ .

The derivation $\partial $ of O is of filtration degree 1, i.e., $\partial (F_kO)\subset F_{k+1}O$ for any $k\geq 0$ . The associated graded endomorphism $\mathrm {gr}(\partial )$ is then a degree 1 derivation of $\mathbb C\oplus Y^2\mathbb C[Y]$ . The image of the derivation $Y^2\partial /\partial Y$ of $\mathbb C[Y]$ is contained in $Y^2\mathbb C[Y]$ , therefore $Y^2\partial /\partial Y$ induces a derivation of $\mathbb C\oplus Y^2\mathbb C[Y]$ of degree 1, which can be seen to coincide with $\mathrm {gr}(\partial )$ . One deduces from this that the derivation $\partial \otimes \mathrm {id}$ of $O \otimes \mathbb C[X]$ is of filtration degree 1, and that its associated graded is the derivation $Y^2 \partial /\partial Y$ of $\mathbb C[X] \oplus Y^2 \mathbb C[X,Y]$ .

The endomorphism $f\otimes P\mapsto -E_2f\otimes P'$ is a derivation of $O\otimes \mathbb C[X]$ , which is also of filtration degree 1, as $E_2$ has filtration degree 2 and $P\mapsto P'$ has filtration degree $-1$ . The associated graded of the endomorphism $f\mapsto E_2f$ of O (resp. $P\mapsto P'$ of $\mathbb C[X]$ ) is the endomorphism $f\mapsto Y^2f$ of $\mathbb C \oplus Y^2\mathbb C[Y]$ (resp. $\partial /\partial X$ of $\mathbb C[X]$ ), therefore the associated graded of $f\otimes P\mapsto -E_2f\otimes P'$ is the degree 1 derivation of $\mathbb C[X]\oplus Y^2\mathbb C[X,Y]$ given by $-Y^2\partial /\partial X$ .

It follows that D has filtration degree 1, and that its associated graded is the derivation of $\mathbb C[X] \oplus Y^2\mathbb C[X,Y]$ induced by the derivation $Y^2({\partial \over \partial Y}-{\partial \over \partial X})$ of $\mathbb C[X,Y]$ .

We now prove by induction on $k\geq 0$ the equality of subspaces of $O[g_1]$

(4.2.3) $$ \begin{align} F_{k+1}(O[g_1])=\partial(F_k(O[g_1]))+\mathrm{Span}_{\mathbb C}(1,g_1,\ldots,g_{k+1}). \end{align} $$

One has $F_0(O[g_1])=\mathbb C$ and $F_1(O[g_1])=\mathrm {Span}_{\mathbb C}(1,g_1)$ , which implies (4.2.3) for $k=0$ . Assume $k>0$ and $F_{k}(O[g_1])=\partial (F_{k-1}(O[g_1]))+\mathrm {Span}_{\mathbb C}(1,g_1,\ldots ,g_{k})$ , and let us show (4.2.3). The inclusion of the right-hand side in the left-hand side is obvious, so one has to show that this inclusion is an equality.

Recall that the space $\partial (F_k(O[g_1]))$ is a subspace of $F_{k+1}(O[g_1])$ . It follows from the commutativity of

and from the surjectivity of $F_k(O[g_1])\to \mathrm {gr}_k(O[g_1])$ that the image of this subspace in $\mathrm {gr}_{k+1}(O[g_1])$ is equal to the image of $\mathrm {gr}_k(\partial ) : \mathrm {gr}_k(O[g_1])\to \mathrm {gr}_{k+1}(O[g_1])$ , which is identified with the map

(4.2.4) $$ \begin{align} Y^2 \left({\partial\over\partial X}-{\partial\over\partial Y}\right) : \mathbb CX^k\oplus Y^2\cdot\mathbb C[X,Y]_{k-2} \to\mathbb CX^{k+1}\oplus Y^2\cdot\mathbb C[X,Y]_{k-1}. \end{align} $$

The image of (4.2.4) is obviously contained in $Y^2\cdot \mathbb C[X,Y]_{k-1}$ , which induces a linear map

(4.2.5) $$ \begin{align} Y^2\left({\partial\over\partial X}-{\partial\over\partial Y}\right) : \mathbb CX^k\oplus Y^2\cdot\mathbb C[X,Y]_{k-2} \to Y^2\cdot\mathbb C[X,Y]_{k-1}. \end{align} $$

The kernel of the linear endomorphism $Y^2({\partial \over \partial X}-{\partial \over \partial Y})$ of $\mathbb C[X,Y]$ is $\mathbb C[X+Y]$ , which implies that the kernel of the linear map $Y^2({\partial \over \partial X}-{\partial \over \partial Y}) : \mathbb C[X,Y]_k\to \mathbb C[X,Y]_{k+1}$ is the one-dimensional vector space $\mathbb C\cdot (X+Y)^k$ . The kernel of (4.2.5) is the intersection of this vector space with the source of (4.2.5), which is zero as the coefficient of $YX^{k-1}$ in $(X+Y)^k$ is nonzero; this implies that the map (4.2.5) is injective. The source and target of (4.2.5) both have dimension k; the equality of these dimensions, together with the injectivity of (4.2.5), implies that (4.2.5) is a linear isomorphism. All this implies that the image of (4.2.4) is equal to $Y^2\cdot \mathbb C[X,Y]_{k-1}$ , therefore that the image of $\partial (F_k(O[g_1]))$ by $F_{k+1}(O[g_1])\to \mathrm {gr}_{k+1}(O[g_1])\simeq \mathbb CX^{k+1}\oplus Y^2\cdot \mathbb C[X,Y]_{k-1}$ is $Y^2\cdot \mathbb C[X,Y]_{k-1}$ .

On the other hand, $\mathrm {Span}_{\mathbb C}(1,g_1,\ldots ,g_k)\subset F_k(O[g_1])$ , which implies that the image of $\mathrm {Span}_{\mathbb C}(1,g_1,\ldots ,g_{k+1})$ under $F_{k+1}(O[g_1])\to \mathrm {gr}_{k+1}(O[g_1])$ is $\mathbb C\,\overline g_{k+1}$ , whose image in $\mathbb CX^{k+1}\oplus Y^2\cdot \mathbb C[X,Y]_{k-1}$ is given by (a).

Then the image of $\partial (F_k(O[g_1]))+\mathrm {Span}_{\mathbb C}(1,g_1,\ldots ,g_{k+1})$ under $F_{k+1}(O[g_1])\to \mathrm {gr}_{k+1}(O[g_1])$ is the sum of the images of $\partial (F_k(O[g_1]))$ and $\mathrm {Span}_{\mathbb C}(1,g_1,\ldots ,g_{k+1})$ under this map, which is equal to the sum $Y^2\cdot \mathbb C[X,Y]_{k-1}+\mathbb C(X-Y)^{k}(X+kY)/(k+1)!$ , that can be simplified into $\mathbb C\cdot X^k\oplus Y^2\cdot \mathbb C[X,Y]_{k-1}\simeq \mathrm {gr}_{k+1}(O[g_1])$ .

It follows that the image of $\partial (F_k(O[g_1]))+\mathrm {Span}_{\mathbb C}(1,g_1,\ldots ,g_{k+1})$ under the projection $F_{k+1}(O[g_1])\to \mathrm {gr}_{k+1}(O[g_1])$ is equal to its target. The space $\partial (F_k(O[g_1]))+\mathrm {Span}_{\mathbb C}(1,g_1,\ldots ,g_{k+1})$ also contains $\partial (F_{k-1}(O[g_1]))+\mathrm {Span}_{\mathbb C}(1,g_1,\ldots ,g_{k})$ , which by the induction hypothesis is equal to $F_k(O[g_1])$ , and therefore also to the kernel of the projection $F_{k+1}(O[g_1])\to \mathrm {gr}_{k+1}(O[g_1])$ . All this implies the equality $\partial (F_k(O[g_1]))+\mathrm {Span}_{\mathbb C}(1,g_1,\ldots ,g_{k+1})=F_{k+1}(O[g_1])$ , which proves the induction step. This proves (4.2.3) for any $k\geq 0$ .

Let us now prove by induction on $k\geq 0$ the equality of subspaces of $F_{k+1}(O[g_1])$

(4.2.6) $$ \begin{align} \partial(F_k(O[g_1]))\cap \mathrm{Span}_{\mathbb C}(1,g_1,\ldots,g_{k+1})=\{0\}. \end{align} $$

One has $F_0(O[g_1])=\mathbb C$ hence $\partial (F_0(O[g_1]))=0$ , which implies (4.2.6) for $k=0$ . Assume $k>0$ and $\partial (F_{k-1}(O[g_1]))\cap \mathrm {Span}_{\mathbb C}(1,g_1,\ldots ,g_{k})=\{0\}$ , and let us show (4.2.6). Let $P\in F_k(O[g_1])$ , $(\lambda _i)_{i\in [0,k+1]}\in \mathbb C^{k+2}$ be such that $\partial (P)=\sum _i\lambda _i g_i$ . The image of this equality under the projection $F_{k+1}(O[g_1])\to \mathrm {gr}_{k+1}(O[g_1])$ is $\lambda _{k+1}\overline {g_{k+1}}=\mathrm {gr}_k(\partial )(\overline P)$ , where $\overline P$ is the image of P in $\mathrm {gr}_k(O[g_1])$ . The map $\mathrm {gr}_k(\partial ) : \mathrm {gr}_k(O[g_1])\to \mathrm {gr}_{k+1}(O[g_1])$ is injective (as this maps can be identified with (4.2.5) which has been proved to be injective), and its image does not contain $\overline {g_{k+1}}$ (as this image has been proved to be identified to $Y^2\mathbb C[X,Y]_{k-1}$ and in view of the identification of $\overline {g_{k+1}}$ in (a)). It follows that $\lambda _{k+1}=0$ and $\overline P=0$ , therefore in the equality $\partial (P)=\sum _i\lambda _i g_i$ the left-hand side belongs to $\partial (F_{k-1}(O[g_1]))$ and the right-hand side to $\mathrm {Span}_{\mathbb C}(1,g_1,\ldots ,g_{k})$ , which by the induction assumption implies $\partial (P)=0$ and $\lambda _0=\ldots =\lambda _k=0$ . This proves the induction step, therefore (4.2.6) holds for any $k\geq 0$ .

Then

(4.2.7) $$ \begin{align} \nonumber O[g_1] &=\sum_{k \geq 0}F_{k+1}(O[g_1]) =\sum_{k \geq 0}\partial(F_k(O[g_1]))+\mathrm{Span}_{\mathbb C}(1,g_1,\ldots,g_{k+1}) \\ &\nonumber =\sum_{k \geq 0}\partial(F_k(O[g_1])) +\sum_{k \geq 0}\mathrm{Span}_{\mathbb C}(1,g_1,\ldots,g_{k+1}) \\ &=\partial \left(\sum_{k \geq 0}F_k(O[g_1])\right)+ \mathrm{Span}_{\mathbb C}(g_i,i\geq 0) \nonumber\\ & =\partial(O[g_1]) +\mathrm{Span}_{\mathbb C}(g_i,i\geq0), \end{align} $$

where the second equality follows from (4.2.3).

For any $k,l \geq 0$ , one has

$$ \begin{align*}\partial(F_kO[g_1]) &\cap \mathrm{Span}_{\mathbb C}(1,g_1,\ldots,g_{l+1}) \subset \partial(F_{\mathrm{max}(k,l)}O[g_1]) \\ &\cap \mathrm{Span}_{\mathbb C}(1,g_1,\ldots,g_{\mathrm{max}(k,l)+1})=0, \end{align*} $$

where the last equality follows from (4.2.6). This implies

(4.2.8) $$ \begin{align} \forall k,l\geq0,\quad \partial(F_kO[g_1]) \cap \mathrm{Span}_{\mathbb C}(1,g_1,\ldots,g_{l+1})=0. \end{align} $$

Then

(4.2.9) $$ \begin{align} \nonumber\partial(O[g_1]) \cap \mathrm{Span}_{\mathbb C}(g_i,i\geq 0) &=(\cup_{k \geq 0}\partial(F_kO[g_1])) \cap (\cup_{l \geq 0}\mathrm{Span}_{\mathbb C}(1,g_1,\ldots,g_{l+1}) \\ & =\cup_{k,l \geq 0} \partial(F_kO[g_1]) \cap \mathrm{Span}_{\mathbb C}(1,g_1,\ldots,g_{l+1})=0, \end{align} $$

where the last equality follows from (4.2.8).

The statement then follows from the combination of (4.2.7) and (4.2.9).

Proof The relations $T_1g_1=g_1$ and $T_\tau g_1=g_1+2\pi \mathrm {i}$ , together with the commutativity of the various $T_a$ , $a\in \mathbb C$ , implies that for any $s\in S$ the function $(T_{\tilde s}-1)(g_1)$ is $\Lambda $ -invariant. It has simple poles at $0$ and s, which implies that it belongs to $O_S$ , which proves (a). One clearly has $T_s(O)\subset O_S$ for any $s\in S$ , so that (a) implies the inclusion $O_S\supset \sum _{s\in S}T_s(O)+\sum _{s\in S\smallsetminus \{pr(0)\}}\mathbb C\cdot (T_{\tilde s}-1)(g_1)$ . If now $f\in O_S$ , let $(f_s)_{s\in S}$ be the collection where $f_s\in \mathbb C((z-\tilde s))$ is the local expansion of f at $\tilde s$ . The map $O\to \mathbb C((z))\to \mathbb C((z))/z^{-1}\mathbb C[[z]]$ is surjective, so for any $s\in S$ one can find $o_s\in O$ whose image by this map coincides with the image of $f_s$ by the map $\mathbb C((z-\tilde s))\simeq \mathbb C((z))\to \mathbb C((z))/z^{-1}\mathbb C[[z]]$ . Then $g:=f-\sum _{s\in S}T_s(o_s)$ belongs to $O_S$ and has at most simple poles at S. Then $g-\sum _{s\in S\smallsetminus pr(0)}\mathrm {res}_s(g\cdot dz)\cdot (T_{\tilde s}-id)(g_1)$ belongs to $O_S$ and is regular on $S\smallsetminus pr(0)$ ; hence it belongs to O and has at most a simple pole at $pr(0)$ , which implies that it is constant. This proves the desired opposite inclusion.

Proof (a) The automorphism $T_\tau $ of ${\mathcal {O}}_{mer}(\mathbb C,\Lambda \cup (a+\Lambda ))$ restricts to automorphisms of the subalgebras $O[g_1]$ and $T_a(O[g_1])$ , therefore to an automorphism of the vector subspace $O[g_1]+T_a(O[g_1])$ .

The O-algebra morphism $O[X]\to O[g_1]$ induced by $X \mapsto g_1$ intertwines the linear endomorphism $T_\tau -id$ of $O[g_1]$ with the linear endomorphism $\mathrm {exp}(2\pi \mathrm {i}\partial _X)-1$ of $O[X]$ , where $\partial $ is the derivation of $O[X]$ given by $O\mapsto 0$ and $X\mapsto 1$ . On the other hand, $\mathrm {exp}(2\pi \mathrm {i}\partial _X)-1={\mathrm {exp}(2\pi \mathrm {i}\partial _X)-1\over \partial _X}\circ \partial _X$ , where ${\mathrm {exp}(2\pi \mathrm {i}\partial _X)-1\over \partial _X}$ is a linear automorphism of $O[X]$ as $\partial _X$ is locally nilpotent, and $\partial _X$ is surjective. It follows that the linear endomorphism $\mathrm {exp}(2\pi \mathrm {i}\partial _X)-1$ of $O[X]$ is surjective. It follows from this surjectivity, from the interwtining of this operator with $T_\tau -id$ and from the surjectivity of the map $O[X]\to O[g_1]$ that the linear endomorphism $T_\tau -id$ of $O[g_1]$ is surjective. This implies the surjectivity of the linear endomorphism $T_\tau -id$ of $T_a(O[g_1])$ , and therefore that of the linear endomorphism $T_\tau -id$ of $O[g_1]+T_a(O[g_1])$ .

Let now f be as in the (a). It follows from the surjectivity of the linear endomorphism $T_\tau -id$ of $O[g_1]+T_a(O[g_1])$ that there exists $\tilde f \in O[g_1]+T_a(O[g_1])$ , such that $(T_\tau -id)(\tilde f)=(T_\tau -id)(f)$ (equality in $O[g_1]+T_a(O[g_1])$ ). Then $f-\tilde f$ belongs to ${\mathcal {O}}_{mer}(\mathbb C,\Lambda \cup (a+\Lambda ))$ and is invariant both under $T_1$ and $T_\tau $ , which implies that it belongs to $O_{\{pr(0),pr(a)\}}$ , which by Lemma 4.9(b) for $S=\{pr(0),pr(a)\}$ is equal to $O+T_a(O)+\mathbb C(T_a-id)(g_1)$ , and is therefore contained in $O[g_1]+T_a(O[g_1])$ . The statement then follows from $\tilde f\in O[g_1]+T_a(O[g_1])$ and $f-\tilde f\in O[g_1]+T_a(O[g_1])$ .

(b) The proof is by induction on $n\geq 1$ . For $n=1$ , the statement follows from (a). Assume the statement at step n and let us prove it at step $n+1$ . Let $g\in {\mathcal {O}}_{mer}(\mathbb C,\Lambda \cup (a+\Lambda ))$ be such that $T_1(g)=g$ and $(T_\tau -id)^{n+1}(g) \in O[g_1]+T_a(O[g_1])$ . Then $f:=(T_\tau -id)(g)$ is such that $T_1(f)=f$ as $T_1$ commutes with $T_\tau -id$ , and $(T_\tau -id)^n(f) \in O[g_1]+T_a(O[g_1])$ . By the induction assumption, this implies $f\in O[g_1]+T_a(O[g_1])$ . Then one has $T_(g)=g$ and $(T_\tau -id)(g)\in O[g_1]+T_a(O[g_1])$ , which by (a) implies $g\in O[g_1]+T_a(O[g_1])$ . This proves the statement at step $n+1$ .

(c) This follows from (b) and from the invariance of $g_1^n\cdot T_a(g_1)$ under $T_1$ , which follows from that of $g_1$ and of $T_a(g_1)$ , and from $(T_\tau -id)^{n+2}(g_1^n\cdot T_a(g_1))=0$ .

Proof For $s\in S$ , one has $T_{\tilde s}(O)\subset O_S\subset O_S[g_1]$ , where the first inclusion follows from Lemma 4.9(b). Moreover, $T_{\tilde s}(g_1)=g_1+(T_{\tilde s}-id)(g_1)\in O_S[g_1]$ by Lemma 4.9(a). Then $T_{\tilde s}(O[g_1])$ is the subalgebra of ${\mathcal {O}}_{mer}(\mathbb C, pr^{-1}(S))$ generated by $T_{\tilde s}(O)$ and $T_{\tilde s}(g_1)$ , which are both contained in $O_S[g_1]$ , which implies the inclusion $T_{\tilde s}(O[g_1])\subset O_S[g_1]$ as $O_S[g_1]$ is an algebra. As $O_S[g_1]$ is stable under summation, this implies the inclusion $O_S[g_1]\supset \sum _{s\in S}T_{\tilde s}(O[g_1])$ .

In order to show the opposite inclusion, let us first show that $\sum _{s\in S}T_{\tilde s}(O[g_1])$ is a subalgebra of ${\mathcal {O}}_{mer}(\mathbb C, pr^{-1}(S))$ . For each $s\in S$ , the subspace $T_{\tilde s}(O[g_1])$ is a subalgebra of ${\mathcal {O}}_{mer}(\mathbb C, pr^{-1}(S))$ . So it is enough to prove that for any $s\neq t\in S$ , one has $T_{\tilde s}(O[g_1])\cdot T_{\tilde t}(O[g_1])\subset T_{\tilde s}(O[g_1])+T_{\tilde t}(O[g_1])$ . By translation invariance, it suffices to prove this for $t=0$ , in which case $\tilde t=0$ , i.e., to prove

(4.3.2) $$ \begin{align} O[g_1]\cdot T_a(O[g_1])\subset O[g_1]+T_a(O[g_1]) \end{align} $$

for $a\in \mathbb C\smallsetminus \Lambda $ .

One has $O\cdot T_a(O)\subset O_{\{pr(0),pr(a)\}}=O+T_a(O)+\mathbb C\cdot (T_a-id)(g_1)$ where the first inclusion follows from the fact that both O and $T_a(O)$ are contained in $O_{\{pr(0),pr(a)\}}$ and that the latter set is an algebra and the second inclusion follows from Lemma 4.9(b) for $S=\{pr(0),pr(a)\}$ . This implies $O\cdot T_a(O)\subset O[g_1]+T_a(O[g_1])$ . Now $O[g_1]\cdot T_a(O[g_1])$ is the sub- $\mathbb C[g_1,T_a(g_1)]$ -module of ${\mathcal {O}}_{mer}(\mathbb C, pr^{-1}(S))$ generated by $O\cdot T_a(O)$ , therefore in order to prove (4.3.2) it suffices to show that $O[g_1]+T_a(O[g_1])$ is a sub- $\mathbb C[g_1,T_a(g_1)]$ -module of ${\mathcal {O}}_{mer}(\mathbb C, pr^{-1}(S))$ , i.e., is stable under multiplication by $g_1$ and $T_a(g_1)$ . The inclusions $O[g_1]\cdot g_1\subset O[g_1]$ and $T_a(O[g_1])\cdot T_a(g_1)\subset T_a(O[g_1])$ are obvious, it therefore remains to prove

(4.3.3) $$ \begin{align} O[g_1]\cdot T_a(g_1)\subset O[g_1]+T_a(O[g_1]),\quad T_a(O[g_1])\cdot g_1\subset O[g_1]+T_a(O[g_1]). \end{align} $$

For $f\in O$ , one has $f\cdot T_a(g_1)=f(a)T_a(g_1)+(f-f(a))(T_a-id)(g_1)+(f-f(a))g_1$ . Then $f(a)T_a(g_1)\in \mathbb CT_a(g_1)$ , $(f-f(a))g_1\in O[g_1]$ and $(f-f(a))(T_a-id)(g_1)\in O$ as $f-f(a)$ belongs to O and vanishes at a, while $(T_a-id)(g_1)$ belongs to $O_{\{0,a\}}$ and has only a simple pole at a. It follows that

(4.3.4) $$ \begin{align} O\cdot T_a(g_1)\subset O[g_1]+\mathbb CT_a(g_1). \end{align} $$

Then one has $Og_1^n\cdot T_a(g_1)\subset O[g_1]g_1^n+\mathbb CT_a(g_1)g_1^n\subset O[g_1]+T_a(O[g_1])$ by virtue of (4.3.4) and (4.3.1). This proves the first inclusion in (4.3.3), the second inclusion is a consequence of it (applying $T_{-a}$ and replacing a by $-a$ ).

This ends the proof of the fact that $\sum _{s\in S}T_{\tilde s}(O[g_1])$ is a subalgebra of ${\mathcal {O}}_{mer}(\mathbb C,pr^{-1}(S))$ .

One has $O_S=\sum _{s\in S}T_s(O)+\sum _{s\in S\smallsetminus \{pr(0)\}}(T_{\tilde s}-id)(g_1)\subset \sum _{s\in S}T_{\tilde s}(O[g_1])$ , where the equality follows from Lemma 4.9(b) and the inclusion follows from $(T_{\tilde s}-id)(g_1)\in O[g_1]+T_{\tilde s}(O[g_1])$ for any $s\in S\smallsetminus \{pr(0)\}$ . One also has $g_1\in T_0(O[g_1])\subset \sum _{s\in S}T_{\tilde s}(O[g_1])$ . These two statements, together with the fact that $\sum _{s\in S}T_{\tilde s}(O[g_1])$ is an algebra, implies the inclusion $O_S[g_1]\subset \sum _{s\in S}T_{\tilde s}(O[g_1])$ .

Proof One has

$$ \begin{align*} O_S[g_1] &=\sum_{s\in S}T_{\tilde s}(O[g_1]) =\sum_{s\in S}T_{\tilde s}(\partial(O[g_1])+\mathrm{Span}_{\mathbb C}\{g_n\,|\,n\geq 0\}) \\ & =\sum_{s\in S}T_{\tilde s}(\partial(O[g_1]))+\sum_{s\in S}T_{\tilde s}(\mathrm{Span}_{\mathbb C}\{g_n\,|\,n\geq 0\}) \\ &=\partial(\sum_{s\in S}T_{\tilde s}(O[g_1]))+\sum_{s\in S}T_{\tilde s}(\mathrm{Span}_{\mathbb C}\{g_n\,|\,n\geq 0\}) \\ & =\partial(O_S[g_1])+\sum_{s\in S}T_{\tilde s}(\mathrm{Span}_{\mathbb C}\{g_n\,|\,n\geq 0\}), \end{align*} $$

where the first and last equalities follow from Lemma 4.11, the second equality follows from (4.2.2), the third equality follows from the linearity of $T_{\tilde s}$ , and the fourth equality follows from the commutativity of $\partial $ and $T_{\tilde s}$ .

Proof One has

$$ \begin{align*} O_S[g_1] &=\sum_{s \in S}T_{\tilde s}(O[g_1]) =\sum_{s\in S}T_{\tilde s}(\mathbb C1+\mathrm{Span}\{g_n|n>0\}+\partial(O[g_1])) \\ & =\mathbb C1+\sum_{s \in S}T_{\tilde s}(\mathrm{Span}\{g_n|n>0\}) +\partial(\sum_{s\in S} T_{\tilde s}(O[g_1])) \\ &=\mathbb C1+\sum_{s \in S}T_{\tilde s}(\mathrm{Span}\{g_n|n>0\})+\partial(O_S[g_1]) \end{align*} $$

where the second equality follows from Lemma 4.8(c), the third equality follows from the commutativity of $T_{\tilde s}$ with $\partial $ , the fourth equality follows from Lemma 4.11; this implies that the said sum map is surjective.

Let us prove its injectivity. Let $P \in O_S[g_1]$ and $S \ni s \mapsto t_s \in \mathrm {Span}\{g_n|n>0\}$ and $\lambda \in \mathbb C$ be such that

$$ \begin{align*}\lambda+\partial(P)+\sum_{s \in S}T_{\tilde s}(t_s)=0. \end{align*} $$

By Lemma 4.11, there exists a map $S \ni s \mapsto P_s \in O[g_1]$ such that $P=\sum _{s \in S}T_{\tilde s}(P_s)$ . Then

$$ \begin{align*}\lambda+\sum_{s \in S}T_{\tilde s}(\partial(P_s)+t_s)=0 \end{align*} $$

(equality in $O_S[g_1] \subset {\mathcal {O}}_{mer}(\mathbb C,p^{-1}(S))$ ).

For each $s \in \tilde S$ , the equality $T_{\tilde s}(\partial (P_s)+t_s)=-\lambda -\sum _{s'\in S\smallsetminus \{s\}}T_{\tilde s'}(\partial (P_s')+t_s')$ implies that the set of poles of $T_{\tilde s}(\partial (P_s)+t_s)$ is contained in $p^{-1}(s) \cap p^{-1}(S\smallsetminus \{s\})$ , which is empty; so $T_{\tilde s}(\partial (P_s)+t_s)$ is an element of $O[g_1]$ which is holomorphic on $\mathbb C$ , therefore is constant by Lemma 4.5. Therefore there exists a map $S \ni s \mapsto \lambda _s \in \mathbb C$ such that $T_{\tilde s}(\partial (P_s)+t_s)=\lambda _s$ for any $s \in S$ , and therefore

$$ \begin{align*}\lambda+\sum_{s \in S}\lambda_s=0. \end{align*} $$

Then $\partial (P_s)=T_{\tilde s}^{-1}(\lambda _s)-t_s=\lambda _s-t_s$ for any $s \in S$ . By Lemma 4.8(c), one has for any $s \in S$ , $t_s-\lambda _s=0$ and $\partial (P_s)=0$ . Since $t_s\in \mathrm {Span}\{g_n|n>0\}$ and by Lemma 4.8(b), one derives $t_s=\lambda _s=0$ . Therefore $\partial (P)=\sum _s \partial (P_s)=0$ , $\lambda =\sum _s \lambda _s=0$ , which implies the injectivity.

Proof This follows from the shuffle identity satisfied by the functions $\tilde \Gamma $ , which follows from the algebra morphism status of $k_{\overrightarrow 0}$ .

Proof Both $\mathrm {int}(T_{\tilde s}(g_n)\cdot \tilde {\operatorname {\mathrm {\Gamma }}}\left (\begin {smallmatrix}n_1 &n_2 &\ldots &n_{r}\\a_1 &a_2 & \ldots &a_{r}\end {smallmatrix};-\right ))$ and $\tilde {\operatorname {\mathrm {\Gamma }}}\left (\begin {smallmatrix}n_1 &n_2 &\ldots &n_{r}&n\\a_1 &a_2 & \ldots &a_{r}&\tilde s\end {smallmatrix};-\right )-\tilde {\operatorname {\mathrm {\Gamma }}}\left (\begin {smallmatrix}n_1 &n_2 &\ldots &n_{r}&n\\a_1 &a_2 & \ldots &a_{r}&\tilde s\end {smallmatrix};z_0\right )$ are elements of ${\mathcal {O}}_{hol}(\tilde {{\mathcal {E}}}_S)$ . By (3.3.1) their images by $d : {\mathcal {O}}_{hol}(\tilde {{\mathcal {E}}}_S)\to \Omega _{hol}(\tilde {{\mathcal {E}}}_S)$ coincide, and they both vanish at $z_0$ . It follows that

(4.4.1) $$ \begin{align} \mathrm{int}(T_{\tilde s}(g_n)\cdot \tilde{\operatorname{\mathrm{\Gamma}}}\left(\begin{smallmatrix}n_1 &n_2 &\ldots &n_{r}\\a_1 &a_2 & \ldots &a_{r}\end{smallmatrix};-\right))=\tilde{\operatorname{\mathrm{\Gamma}}}\left(\begin{smallmatrix}n_1 &n_2 &\ldots &n_{r}&n\\a_1 &a_2 & \ldots &a_{r}&\tilde s\end{smallmatrix};-\right)-\tilde{\operatorname{\mathrm{\Gamma}}}\left(\begin{smallmatrix}n_1 &n_2 &\ldots &n_{r}&n\\a_1 &a_2 & \ldots &a_{r}&\tilde s\end{smallmatrix};z_0\right). \end{align} $$

This identity implies the statement.

Proof By induction on r. For $r=0$ , and $f\in O_S[g_1]$ , one has $\mathrm {int}(\partial (f))=f-f(z_0)\in F_0(\mathcal G)$ . Assume the statement at step $r-1$ . For $f\in O_S[g_1]$ , $n_1,\ldots ,n_r\geq 0$ and $a_1,\ldots ,a_r\in \tilde S$ , integration by parts together with (3.3.1) yields

$$ \begin{align*} \mathrm{int}(\partial(f)\cdot \tilde{\operatorname{\mathrm{\Gamma}}}\left(\begin{smallmatrix}n_1 &n_2 &\ldots &n_{r}\\a_1 &a_2 & \ldots &a_{r}\end{smallmatrix};-\right)) & =f\cdot \tilde{\operatorname{\mathrm{\Gamma}}}\left(\begin{smallmatrix}n_1 &n_2 &\ldots &n_{r}\\a_1 &a_2 & \ldots &a_{r}\end{smallmatrix};-\right)-f(z_0)\cdot \tilde{\operatorname{\mathrm{\Gamma}}}\left(\begin{smallmatrix}n_1 &n_2 &\ldots &n_{r}\\a_1 &a_2 & \ldots &a_{r}\end{smallmatrix};z_0\right) \\ & \quad-\mathrm{int}(f\cdot T_{a_r}(g_{n_r})\cdot \tilde{\operatorname{\mathrm{\Gamma}}}\left(\begin{smallmatrix}n_1 &n_2 &\ldots &n_{r-1}\\a_1 &a_2 & \ldots &a_{r-1}\end{smallmatrix};-\right)). \end{align*} $$

Note that $f\cdot \tilde {\operatorname {\mathrm {\Gamma }}}\left (\begin {smallmatrix}n_1 &n_2 &\ldots &n_{r}\\a_1 &a_2 & \ldots &a_{r}\end {smallmatrix};-\right )\in F_r(\mathcal G)$ and $f(z_0)\cdot \tilde {\operatorname {\mathrm {\Gamma }}}\left (\begin {smallmatrix}n_1 &n_2 &\ldots &n_{r}\\a_1 &a_2 & \ldots &a_{r}\end {smallmatrix};z_0\right )\in \mathbb C\subset F_0(\mathcal G)$ . Moreover, $f\in O_S[g_1]$ and $T_{a_r}(g_{n_r})\in O_S[g_1]$ by Lemma 4.12, which implies that $f\cdot T_{a_r}(g_{n_r})\in O_S[g_1]$ . By Lemma 4.12, one may then decompose $f\cdot T_{a_r}(g_{n_r})$ as $\partial (g)+\sum _{(n,a)\in \mathbb Z_{\geq 0}\times \tilde S} \lambda _{n,a}T_a(g_n)$ , with $g\in O_S[g_1]$ and $\lambda _{n,a}\in \mathbb C$ . Then

$$ \begin{align*} &\mathrm{int}(f\cdot T_{a_r}(g_{n_r})\cdot \tilde{\operatorname{\mathrm{\Gamma}}}\left(\begin{smallmatrix}n_1 &n_2 &\ldots &n_{r-1}\\a_1 &a_2 & \ldots &a_{r-1}\end{smallmatrix};-\right)) \\ & =\mathrm{int}(\partial(g)\cdot \tilde{\operatorname{\mathrm{\Gamma}}}\left(\begin{smallmatrix}n_1 &n_2 &\ldots &n_{r-1}\\a_1 &a_2 & \ldots &a_{r-1}\end{smallmatrix};-\right))+\sum_{(n,a)\in\mathbb Z_{\geq0}\times\tilde S} \lambda_{n,a} \mathrm{int}(T_a (g_n)\cdot \tilde{\operatorname{\mathrm{\Gamma}}}\left(\begin{smallmatrix}n_1 &n_2 &\ldots &n_{r-1}\\a_1 &a_2 & \ldots &a_{r-1}\end{smallmatrix};-\right)). \end{align*} $$

By the induction assumption, $\mathrm {int}(\partial (g)\cdot \tilde {\operatorname {\mathrm {\Gamma }}}\left (\begin {smallmatrix}n_1 &n_2 &\ldots &n_{r-1}\\a_1 &a_2 & \ldots &a_{r-1}\end {smallmatrix};-\right ))$ belongs to $F_{r-1}(\mathcal G)$ . Moreover, for each $(n,a)$ , $ \mathrm {int}(T_a (g_n)\cdot \tilde {\operatorname {\mathrm {\Gamma }}}\left (\begin {smallmatrix}n_1 &n_2 &\ldots &n_{r-1}\\a_1 &a_2 & \ldots &a_{r-1}\end {smallmatrix};-\right ))\in F_r(\mathcal G)$ by Lemma 4.16. All this implies $\mathrm {int}(\partial (f)\cdot \tilde {\operatorname {\mathrm {\Gamma }}}\left (\begin {smallmatrix}n_1 &n_2 &\ldots &n_{r}\\a_1 &a_2 & \ldots &a_{r}\end {smallmatrix};-\right ))\in F_r(\mathcal G)$ , which concludes the inductive argument.

Proof For $r\geq 0$ , set

$$ \begin{align*}\mathrm{gr}_r(\mathcal G):= \sum_{\substack{n_1,\ldots ,n_{r}\geq 0,\\ a_1,\ldots ,a_{r}\in \tilde S}} O_S[g_1]\cdot \tilde{\operatorname{\mathrm{\Gamma}}}\left(\begin{smallmatrix}n_1 &n_2 &\ldots &n_{r}\\a_1 &a_2 & \ldots &a_{r}\end{smallmatrix};-\right). \end{align*} $$

One then has $F_r(\mathcal G)=\sum _{r'\in [\![0,r]\!]}\mathrm {gr}_{r'}(\mathcal G)$ . We will prove the inclusion $\mathrm {int}(F_r(\mathcal G))\subset F_{r+1}(\mathcal G)$ by induction on $r\geq 0$ . Let us first prove that $\mathrm {int}(F_0(\mathcal G))\subset F_1(\mathcal G)$ .

For $f\in O_S[g_1]$ , one has $\mathrm {int}(\partial (f))=f-f(z_0)\in O_S[g_1]$ , which implies that

(4.4.2) $$ \begin{align} \mathrm{int}(\partial(O_S[g_1]))\subset O_S[g_1]\in F_0(\mathcal G). \end{align} $$

It follows from (4.4.1) with $r=0$ that, for $s\in S$ and $n\geq 0$ , one has $\mathrm {int}(T_{\tilde s}(g_n))=\tilde {\operatorname {\mathrm {\Gamma }}}\left (\begin {smallmatrix}n\\\tilde s\end {smallmatrix};-\right )-\tilde {\operatorname {\mathrm {\Gamma }}}\left (\begin {smallmatrix}n\\\tilde s\end {smallmatrix};z_0\right )$ , which implies that

(4.4.3) $$ \begin{align} \mathrm{int}(T_{\tilde s}(g_n))\in F_1(\mathcal G). \end{align} $$

Then

$$ \begin{align*} & \mathrm{int}(F_0(\mathcal G)) =\mathrm{int}(O_S[g_1])=\mathrm{int}(\partial(O_S[g_1])+\sum_{s\in S}T_{\tilde s}(\mathrm{Span}_{\mathbb C}\{g_n\,|\,n\geq0\})) \\ & \quad =\mathrm{int}(\partial(O_S[g_1]))+\sum_{s\in S,n\geq0}\mathbb C\cdot \mathrm{int}(T_{\tilde s}(g_n)) \subset F_0(\mathcal G)+\sum_{s\in S,n\geq0}F_1(\mathcal G)=F_1(\mathcal G), \end{align*} $$

where the first equality follows from $F_0(\mathcal G)=O_S[g_1]$ , the second equality follows from Lemma 4.12 and the inclusion follows from (4.4.2) and (4.4.3). This proves the initial step of the induction.

Assume that $r\geq 1$ and that $\mathrm {int}(F_{r-1}(\mathcal G))\subset F_r(\mathcal G)$ . Let us prove that $\mathrm {int}(F_r(\mathcal G))\subset F_{r+1}(\mathcal G)$ . One has $F_{r}(\mathcal G)=F_{r-1}(\mathcal G) +\mathrm {gr}_r(\mathcal G)$ , so by the induction assumption it suffices to prove that $\mathrm {int}(\mathrm {gr}_{r}(\mathcal G))\subset F_{r+1}(\mathcal G)$ .