Proof. Clearly $\tilde{\rho}(\tilde{K})\lhd \tilde{\rho}(\Gamma)$ since $\tilde{K}\lhd \Gamma$ and $\tilde{\rho}$ is a homomorphism. Suppose $A\in {\mathfrak{g}}$ and $f\in \mathcal{O}_{\mathbb{X}}$. Now, if $\tilde{\rho}( \gamma_1 \gamma_2)\in \tilde{\rho}(\tilde{K})$ with $ \gamma_1\in \ker(\rho)$ and $ \gamma_2\in \ker(\sigma)$, then

\begin{equation*}\tilde{\rho}( \gamma_1 \gamma_2)(A\otimes f)=\rho( \gamma_1 \gamma_2)\otimes \tilde{\sigma}( \gamma_1 \gamma_2)(A\otimes f)=\rho( \gamma_2)A\otimes \tilde{\sigma}( \gamma_1)f,\end{equation*}

where the homomorphism $\rho\otimes\tilde{\sigma}$ is defined in (2.1). Hence $({\mathfrak{g}}\otimes_{{\mathbb{C}}}\mathcal{O}_{\mathbb{X}})^{\tilde{\rho}(\tilde{K})}={\mathfrak{g}}^{\rho(\ker\sigma)}\otimes_{{\mathbb{C}}}\mathcal{O}_{\mathbb{X}}^{\tilde{\sigma}(\ker \rho)}.$ By employing Lemma 2, this proves the claim.

Proof. Define $\varphi: {\mathfrak{g}}\otimes_{{\mathbb{C}}}{\mathcal{O}_{\mathbb{X}}}\rightarrow {\mathfrak{g}}\otimes_{{\mathbb{C}}}{\mathcal{O}_{\mathbb{X}}}$ on simple tensors by $\varphi(A\otimes f)=\tau_1(A)\otimes \tilde{\tau}_2f$, where $A\in {\mathfrak{g}}$, $f\in {\mathcal{O}_{\mathbb{X}}}$ and $\tilde{\tau}_2f(z)=f(\tau_2^{-1}z)$ for $z\in \mathbb{X}$, and extend φ ${\mathbb{C}}$-linearly to the whole space. It is a straightforward verification that φ is Lie algebra isomorphism and that $\varphi\circ\tilde{\rho}(\gamma)=\tilde{\rho}'(\gamma)\circ \varphi$ for all $\gamma\in \Gamma$, that is, it intertwines $\tilde{\rho}$ and $\tilde{\rho}'$. This implies that $a\in {\mathfrak{g}}\otimes_{{\mathbb{C}}}{\mathcal{O}_{\mathbb{X}}}$ is invariant with respect to the action defined by $\tilde{\rho}$ if and only if $\varphi(a)$ is invariant with respect to the action defined by $\tilde{\rho}'$. Thus φ restricts to an isomorphism between $\mathfrak{A}$ and $\mathfrak{A}'.$

Proof. The group $\mathrm{Aut}(\mathfrak{sl}_2)$ is isomorphic to $\mathrm{PSL}_2({\mathbb{C}})$ by the adjoint representation. A finite subgroup of $\mathrm{PSL}_2({\mathbb{C}})$ leaves a Hermitian inner product invariant and is therefore conjugate to a subgroup of $\mathrm{PSU}_2$ which in turn is isomorphic to $\mathrm{SO}_3$. Therefore, the statement of the lemma is equivalent to the analogue statement for the Lie group $\mathrm{SO}_3$ instead of $\mathrm{Aut}(\mathfrak{sl}_2)$. Klein showed that the subgroups of $\mathrm{SO}_3$ are precisely the orientation preserving isometries of $\mathbb{R}^3$ that fix the regular pyramids, regular polygons, regular tetrahedrons, regular octahedrons and regular icosahedrons, centred at the origin [Reference Klein17]. This yields the groups listed in the statement, respectively. We may moreover assume that the vertices of the polyhedra have norm 1. If we then take two polyhedra of the same type, there is an element of $\mathrm{SO}_3$ that transforms one to the other. This group element conjugates the symmetry groups of the two polyhedra and proves the last statement of the lemma.

Proof. By Lemma 5, we know that the finite subgroups of $\mathrm{Aut}(\mathfrak{sl}_2)$ are given by

\begin{equation*}C_N,\quad D_N,\quad A_4,\quad S_4, \quad A_5.\end{equation*}

We will show that only the subgroups as given in the statement are simultaneously subgroups of $\mathrm{Aut}(T)$ as well.

Let T be a complex torus. It follows from [Reference Miranda25, Proposition III.1.11] that any automorphism σ of T is of the form $\sigma(z)=\epsilon z+\alpha$, where ϵ is a suitable root of unity (for which T has multiplication by ϵ) and $\alpha\in T$. This immediately gives a semi-direct product structure $\mathrm{Aut}(T)=\mathrm{Aut}_0(T) \ltimes t(T)$ where $\mathrm{Aut}_0(T)$ is the group of automorphisms fixing zero and t(T) is the group of translations of T. Now, take two automorphisms $\sigma,\sigma'$ defined by $\sigma(z)=\epsilon z+\alpha$ and $\sigma'(z)=\epsilon'z+\alpha'$. Then $[\sigma,\sigma'](z)=z-\alpha'-\epsilon'\alpha+\epsilon\alpha'+\alpha$, so that $[\sigma,\sigma']$ is indeed a translation. Hence $[\mathrm{Aut}(T),\mathrm{Aut}(T)]\subset t(T)$.

Now, let $r\in t(T)$ be any translation of T, say $r(z)=z+\alpha$. We will show that there are $s\in \mathrm{Aut}_0(T)$ and $r'\in \mathrm{Aut}(T)$ such that $r=[s,r']$. Suppose $s(z)=-z$ (any torus has multiplication by −1) and let $r'(z)=z-\frac{\alpha}{2}$. Then $sr'(z)=-z+\frac{\alpha}{2}$. Hence $[s,r'](z)=-(-z-\frac{\alpha}{2}-\frac{\alpha}{2})=z+\alpha=r(z)$. This proves the claim that $[\mathrm{Aut}(T),\mathrm{Aut}(T)]=t(T)$.

For the proof that C_N, D_N are subgroups of $\mathrm{Aut}(T)$, as well as A ₄ for a suitable torus T, we refer to Lemma 7.

Let us now argue that S ₄ and A ₅ are not subgroups of $\mathrm{Aut}(T)$ for any complex torus T. To see that S ₄ is not a subgroup of $\mathrm{Aut}(T)$, note that $[S_4,S_4]=A_4$, which is non-abelian, whereas for $\Gamma\subset \mathrm{Aut}(T)$, the commutator subgroup $[\Gamma,\Gamma]\subset t(T)$ is abelian.

Finally, suppose for a contradiction that $A_5\subset \mathrm{Aut}(T)$ for some complex torus T. Then $A_5\cong C_{\ell} \ltimes H$ for some normal subgroup H ≠ 1, since $\mathrm{Aut}(T)=\mathrm{Aut}_0(T) \ltimes t(T)$. However, A ₅ is simple and thus it cannot have a proper normal subgroup. This shows that the list above is all there is in the intersection of finite subgroups of $\mathrm{Aut}(\mathfrak{sl}_2)$ and $\mathrm{Aut}(T)$, for all complex tori T.

Proof. We leave it to the reader to verify that the subgroups in the statements satisfy the group relations and thus are indeed of the mentioned isomorphism class.

We start with the cyclic groups. If an element r of $\mathrm{Aut}(T)$ fixes an element p of T, then $t(p) r t(p)^{-1}\in \mathrm{Aut}_0(T)$ (where $t(p)(z)=z+p$) and $t(p) \langle r\rangle t(p)^{-1}$ is as described in 1a. If on the other hand r has no fixed points, then it is as described in 1b.

Now for the dihedral groups, consider first the abelian case $D_2=C_2\times C_2$. This group occurs precisely once in t(T), as in 2a, since both groups have precisely three elements of order 2.

Suppose now that $D_2\subset \mathrm{Aut}(T)$ has an element s which is not contained in t(T). Using a conjugation as we did to classify the cyclic groups, we may assume that $s\in\mathrm{Aut}_0(T)$, so that $s(z)=-z$. There must also be a non-trivial element of D ₂, say r, contained in t(T). Indeed, the elements of order 2 in $\mathrm{Aut}(T)\setminus t(T)$ are the maps $z\mapsto -z+b$. A product of two such maps is in t(T). Thus the group D ₂ is as described in 2b with N = 2.

For the remaining dihedral groups $D_N\subset\mathrm{Aut}(T)$ ($N\ge 3$), we notice that $[D_N,D_N]=\langle r^2\rangle\subset [\mathrm{Aut}(T),\mathrm{Aut}(T)]\subset t(T)$ implies that $r\in t(T)$. At least one of the order 2 elements $D_N\setminus\langle r \rangle$ must be outside of t(T) because D_N is non-abelian. Taking a conjugate of the group, we may again assume that such an element, say s, fixes 0. Thus, we arrive at the remaining groups described in 1b.

Finally, for the group A ₄, we argue as follows. The derived subgroup $[A_4,A_4]=\langle r_1,r_2\rangle\subset [\mathrm{Aut}(T),\mathrm{Aut}(T)]\subset t(T)$ is uniquely determined by the only three elements in t(T) of order 2. The element s must be outside of t(T) for A ₄ is non-abelian. Taking a conjugate of the group, we may assume that s fixes zero, which leaves two options: $s(z)=\mathrm{e}^{2\pi {i}/3}z$ and $s(z)=\mathrm{e}^{4\pi {i}/3}z$. Both options generate the same group, since there is an automorphism of A ₄ sending s to s ².

Proof. A point $p\in T$ is a ramification point of π if the multiplicity of π at p, denoted by $\mathrm{mult}_p(\pi)$, is at least 2. By [Reference Miranda25, Theorem III.3.4], $\mathrm{mult}_p(\pi)$ equals $|\Gamma_p|$, where $\Gamma_p=\{\gamma\in\Gamma: \gamma\cdot p=p\}$. It is clear there are no branch points if $\Gamma\subset t(T)$ since a translation has no fixed points. Let now $\Gamma=C_3\subset \mathrm{Aut}_0(T)$ where $T={\mathbb{C}}/{\mathbb{Z}}\oplus {\mathbb{Z}}\omega_6$. It is straightforward to check that $|\Gamma_p| \gt 1$ if and only if $p\in \{0,1/2,\omega_6/2,(1+\omega_6)/2\}$. The only points that are in the same Γ-orbit, are $1/2$ and $\omega_6/2$. Hence the total number of branch points of π after deleting $\Gamma\cdot \{0\}$ equals 2. The other cases follow by similar arguments.

Proof. We will only prove the claims for $\Gamma=C_3$; the rest follows in the same way. We know by Lemma 1 that $\mathcal{O}_{\mathbb{T}}=\mathbb{C}[\wp,\wp']$. Using (5.4) we see that ${\mathbb{C}}[\wp']\subset \mathcal{O}_{\mathbb{T}}^{\chi_0}$, ${\mathbb{C}}[\wp']\wp^2\subset \mathcal{O}_{\mathbb{T}}^{\chi_1}$ and ${\mathbb{C}}[\wp']\wp\subset \mathcal{O}_{\mathbb{T}}^{\chi_2}.$ It remains to show that $\mathbb{C}[\wp,\wp']$ is a subset of ${\mathbb{C}}[\wp']\oplus {\mathbb{C}}[\wp']\wp^2\oplus {\mathbb{C}}[\wp']\wp$. If $\wp^a\wp'^b$ is a monomial with $a\ge 3$, then we can substitute $\wp^3=\frac{1}{4}(\wp')^2+\frac{g_2}{4}\wp+\frac{g_3}{4}$, using (5.2), to obtain a polynomial in $\wp$ and $\wp'$ where all exponents of $\wp$ are less than a. Repeating this substitution finitely many times, we see that $\wp^a\wp'^b$ is an element of ${\mathbb{C}}[\wp']\oplus {\mathbb{C}}[\wp']\wp^2\oplus {\mathbb{C}}[\wp']\wp$. Hence $\mathbb{C}[\wp,\wp']$ is indeed a subset of ${\mathbb{C}}[\wp']\oplus {\mathbb{C}}[\wp']\wp^2\oplus {\mathbb{C}}[\wp']\wp$.

Proof. We know that $\mathcal{O}^{\tilde{\sigma}_\alpha(C_N)}_{\mathbb{T}}$ is the space of $\Lambda_{(\alpha)}$-periodic meromorphic functions which are holomorphic on $T\setminus \Lambda_{(\alpha)}$. With this perspective, $\mathcal{O}_{\mathbb{T}}^{\tilde{\sigma}_{\alpha}(C_N)}=\mathcal{O}_{\mathbb{T}/\sigma_{\alpha}(C_N)}$. Now use Lemma 1 with T taken to be the complex torus $T/\sigma_{\alpha}(C_N)$. Thus $\mathcal{O}_{\mathbb{T}}^{\tilde{\sigma}_{\alpha}(C_N)}=\mathbb{C}[\wp_{\Lambda_{(\alpha)}}, \wp_{\Lambda_{(\alpha)}}'].$

Proof. Let $v=\frac{\wp'}{\tilde{\wp}}$ and first take $N\geqslant 3$. Assume r is given by $r:z\mapsto z+\alpha$. To prove the first item, note that $(r^k\tilde{\wp}) = -2(k\alpha)+ ((k+1)\alpha)+((k-1)\alpha)$ and hence

\begin{equation*}(1/r^k\tilde{\wp})=-((k-1)\alpha)-((k+1)\alpha)+2(k\alpha).\end{equation*}

Using the divisor in (5.5), we see that $r^kv=r^k\left(\frac{\wp'}{\tilde{\wp}}\right)$ has order 1 poles at $(k-1)\alpha$, $k\alpha$ and $(k+1)\alpha$, which are distinct when $N\geqslant 3$. Therefore, $P_j=\sum_{k=0}^{N-1}\frac{r^k\wp'}{\omega_N^{kj}r^k\tilde{\wp}}$ has at most order 1 poles in $\mathcal{S}=\{0,\alpha,2\alpha,\dots,(N-1)\alpha\}$.

To prove the second statement, we observe that the divisor $(P_j)$ is invariant under r, since r acts by a non-zero scalar on P_j. Hence if P_j has a pole, then it has poles of order 1 everywhere in $\mathcal{S}$.

For the third item, we consider a number of expansions about z = 0. The summands of P_j that are responsible for the potential occurrence of pole at z = 0 are $r^{-1}v, v$ and rv. It follows from the definition of $\wp$ in (2.2) that about z = 0, we have

\begin{equation*}\frac{1}{\wp(z)-\wp(\alpha)}=z^2+O(z^4),\quad \wp'(z)=-\frac{2}{z^3}+O(z).\end{equation*}

Thus,

(5.7)\begin{align} v(z)=\frac{\wp'}{\tilde{\wp}}(z)= -\frac{2}{z}+O(z). \end{align}

For $N\geqslant 3$, we have $r\tilde{\wp}(z)=-\wp'(\alpha)z+O(z^2).$ Observe that $\wp'(\alpha)\neq 0$ as soon as $N\geqslant 3$, because zeros of $\wp'$ appear at half lattice points (i.e. 2-torsion points). We see $\frac{1}{r\tilde{\wp}}(z)= -\frac{1}{\wp'(\alpha)}\frac{1}{z}+O(1)$ and $r\tilde{\wp}'(z)=-\wp'(\alpha)+O(z)$ from which we infer

\begin{equation*}rv(z)=\frac{1}{z}+O(1).\end{equation*}

Similarly, one argues that $r^{-1}v(z)=\frac{1}{z}+O(1).$ Thus indeed

\begin{equation*}\mathrm{Res}_{z=0}(P_j)=\mathrm{Res}_{z=0}\left(\sum_{k=0}^{N-1}\omega_N^{-kj}r^kv\right)=-2+\omega_N^{-j}+\omega_N^{j}.\end{equation*}

If N = 2, then $P_{1}=\frac{\wp'}{\tilde{\wp}}-r\frac{\wp'}{\tilde{\wp}}$ and it is easy to see that it has order 1 poles in $\{0,\alpha\}$. The expansion in (5.7) still holds, thus $\frac{\wp'}{\tilde{\wp}}(z)=-\frac{2}{z}+O(z)$ and one can show that $r\frac{\wp'}{\tilde{\wp}}(z)=\frac{2}{z}+O(1)$ about z = 0.

Now consider P_j for $j\equiv 0\mod N$. Observe that it is invariant under r. Thus P ₀ descends to a function $\tilde{P}_0$ on the torus $\tilde{T}=T/\langle r\rangle$. If P ₀ has poles, they are of order 1 and if this is the case, then $\tilde{P}_0$ would have a simple pole on $\tilde{T}$. This is not possible due to the well-known fact that if a compact Riemann surface has a meromorphic function with a single simple pole, it must be isomorphic to the Riemann sphere, cf. [Reference Miranda25, Proposition II.4.11]. Hence P ₀ is constant.

Finally, $sP_{0}=-P_{0}$ from which it follows that $P_{0}=0$. This concludes the proof.

Proof. The space of meromorphic functions on a complex torus T is denoted by $\mathcal{M}(T)$. Let D be a divisor on T. The degree $\deg(D)$ of $D=\sum_{p\in T}n_p(p)$ is defined as $\deg(D)=\sum_{p\in T}n_p.$ Introduce the space of meromorphic functions on T with poles bounded by D, denoted by L(D):

\begin{equation*}L(D)=\{f\in \mathcal{M}(T): \mathrm{div}(f)\geqslant -D\}.\end{equation*}

Note that L(D) is a complex vector space. The Riemann–Roch theorem, specialised to genus 1 Riemann surfaces [Reference Miranda25, Proposition V.3.14], says that if D is a divisor on a complex torus T and $\deg(D) \gt 0$, then $\dim_{{\mathbb{C}}}L(D)=\deg(D)$. Let $D=\sum_{j=0}^{N-1}(j\alpha).$ We see that $\dim_{{\mathbb{C}}}L(D)=\deg(D)=N$. By construction, we have $P_j\in L(D)^{\chi_j}$, and hence $\dim_{{\mathbb{C}}}L(D)^{\chi_j}= 1$ if $j\not\equiv 0\mod N$. Therefore, $L(D)={\mathbb{C}}\langle 1,P_{1},\ldots, P_{N-1}\rangle.$ We proceed inductively. For each $k\in \{1,\ldots, N-1\}$, going from $L((m-1)D)^{\chi_k}$ to $L(mD)^{\chi_k}$, we add the one-dimensional subspace generated by a product of P_j’s such that $0\neq \prod_{j\in \{j_1,\ldots, j_m\}}P_j\in L(mD)^{\chi_k}$ to $L((m-1)D)^{\chi_k}$. In particular, for $k\not\equiv 0\mod N/2$ and letting $Q=(P_{-j}P_j)^{\lfloor \frac{m-1}{2}\rfloor}$ (where $\lfloor x\rfloor$ denotes the integer part of x), we claim that

\begin{equation*}L(mD)^{\chi_k}/L((m-1)D)^{\chi_k}=\begin{cases} {\mathbb{C}} P_kQ+L((m-1)D)^{\chi_k},\quad \text{if}\,\, m\,\, \text{odd}\\ {\mathbb{C}} P_{-k}P_{2k}Q+L((m-1)D)^{\chi_k},\quad \text{if}\,\, m\,\, \text{even}. \end{cases}\end{equation*}

Here we used that $N\geqslant 3$, so that there are enough integers k such that $P_{2k}\neq 0$. For $k\equiv 0\mod N/2$ and m even, we need to change $P_{-k}P_{2k}$ to $P_{-k-n}P_{2k+n}\in\mathcal{O}_{\mathbb{T}}^{\chi_k}$ for some suitable $n\in{\mathbb{Z}}$. This shows that $\mathcal{O}_{\mathbb{T}}=\bigcup_{j\in \mathbb{N}}L(jD)$ is generated as a ${\mathbb{C}}$-algebra by $P_{1},\ldots, P_{N-1}$ whenever $N\geqslant 3$.

To prove the second claim, notice that $r P_{-j}P_j=P_{-j}P_j$ and that it has poles of order 2 in ${\Lambda}_{(\alpha)}$ if and only if $j\not\equiv 0\mod N$. If z ₀ is a zero of P _−j, then $-z_0$ is a zero of P_j since $P_{-j}(z_0)=-P_j(-z_0)$ by Lemma 12. Hence the divisor of $P_{-j}P_j$ equals $(P_{-j}P_j)=-2(0)+(z_0)+(-z_0)$. Now recall that $\wp_{{\Lambda}_{(\alpha)}}$ has an order 2 pole at z = 0 and is even, hence $(\wp_{{\Lambda}_{(\alpha)}})=-2(0)+(w_0)+(-w_0)$ for some $w_0\in T'$. We see that indeed $P_{-j}P_j=c_1\wp_{{\Lambda}_{(\alpha)}}+c_2$ for some $c_1,c_2\in {\mathbb{C}}$ and $c_1\neq 0$ if and only if $j\not\equiv 0\mod N$. Therefore, ${\mathbb{C}}[\wp_{{\Lambda}_{(\alpha)}}]={\mathbb{C}}[P_{-j}P_j]$ if and only if $j\not\equiv 0\mod N$.

Proof. Let $j\in {\mathbb{Z}}$ and write $f=P_{2j}P_{-j}^2-P_{-2j}P_j^2$. Clearly $\gamma f=f$ for all $\gamma\in \langle s,r\rangle\cong D_N$ for any $N\in \mathbb{N}$. In particular, f is even and hence has at most order 2 poles, for any choice of j. By Proposition 2, we see that for all $k\not\equiv 0\mod N$ we have that $f=\lambda P_{-k}P_{k}+\mu$ for some $\lambda,\mu\in {\mathbb{C}}$.

Assume now that $N\geqslant 3$, $k=\pm j$ and $j\not \equiv 0 \mod N/2$. If µ would be zero and $z_0\in T$ is a zero of P_j, then $f(z_0)=P_{2j}(z_0)P_{-j}^2(z_0)=0.$ Thus, z ₀ would also be a zero of P _−j or $P_{2j}$. However, since the set of zeros of P_k is invariant under r, the set of zeros of P_j must then coincide with the zero set of P _−j or $P_{2j}$. Since the order and location of the poles of P_j, P _−j and $P_{2j}$ already coincide, this means that $(P_j)=(P_{-j})$ or $(P_j)=(P_{2j})$, that is, P_j is either a constant multiple of P _−j or $P_{2j}$. This is absurd, which therefore means that µ cannot be zero.

Proof. First recall that $G=C_{\ell}\subset \mathrm{Aut}_0(T)$ and that K consists of translations of T. By Lemma 11, we, therefore, have $\mathcal{O}_{\mathbb{T}}^K={\mathbb{C}}[\wp,\wp']$ for some $\wp$ associated to a suitable lattice. Observe that the statement is trivially true for $\ell=1$, hence we take $\ell\in \{2,3,4,6\}$.

Assume the normal form of $(\mathfrak{sl}_2\otimes_{{\mathbb{C}}}\mathcal{O}_{\mathbb{T}})^{K}$ to be

\begin{equation*}(\mathfrak{sl}_2\otimes_{{\mathbb{C}}}\mathcal{O}_{\mathbb{T}})^{K}={\mathbb{C}}\langle E,F,H\rangle \otimes_{{\mathbb{C}}}\mathcal{O}_{\mathbb{T}}^K={\mathbb{C}}\langle E,F,H\rangle \otimes_{{\mathbb{C}}}{\mathbb{C}}[\wp,\wp']\cong \mathfrak{sl}_2\otimes_{{\mathbb{C}}}\mathcal{O}_{\mathbb{T}/K},\end{equation*}

with $\mathcal{O}_{\mathbb{T}}^K$-linear brackets $[H,E]=2E$, $[H,F]=-2F$ and $[E,F]=H$. Now, since K is normal in Γ, $g\in G$ acts on ${\mathbb{C}}\langle E,F,H\rangle \otimes_{{\mathbb{C}}}\mathcal{O}_{\mathbb{T}}^K$ and $\{g\cdot E,g\cdot F,g\cdot H\}$ forms again a standard $\mathfrak{sl}_2$-triple, where the action is defined in (2.1). By assumption, $g\cdot H=H$. Furthermore, $[H,g\cdot E]=2g\cdot E$, so $g\cdot E=E\otimes k_1$ for some function $k_1\in \mathcal{O}_{\mathbb{T}}^K$. Similarly, $g\cdot F=F\otimes k_2$ for some $k_2\in \mathcal{O}_{\mathbb{T}}^K$. We have $[g\cdot E,g\cdot F]=H$ and this implies that $k_2=k_1^{-1}$. The only units in $\mathcal{O}_{\mathbb{T}}^K={\mathbb{C}}[\wp,\wp']$ are the non-zero constants. Therefore, $g\cdot E=\omega_{\ell}^jE$ and $g\cdot F=\omega_{\ell}^{-j}F$, for some $j\in \{1,\ldots, \ell-1\}$ such that $\gcd(j,\ell)=1$. Notice that this forces $j\in \{1,\ell-1\}$, due to the limited amount of choices for $\ell$.

We now decompose ${\mathbb{C}}\langle E,F,H\rangle \otimes_{{\mathbb{C}}}{\mathbb{C}}[\wp,\wp']$ into its $C_{\ell}$ isotypical components:

\begin{equation*}({\mathbb{C}}\langle E,F,H\rangle \otimes_{{\mathbb{C}}}{\mathbb{C}}[\wp,\wp'])^{G}={\mathbb{C}}\langle E\otimes p_1,F\otimes p_2,H\rangle\otimes_{{\mathbb{C}}} {\mathbb{C}}[J],\end{equation*}

where $p_1\in {\mathbb{C}}[\wp,\wp']^{\chi_{\ell-j}}$, $p_2\in {\mathbb{C}}[\wp,\wp']^{\chi_j}$ and J generates $\mathcal{O}_{\mathbb{T}}^{\Gamma}$, which follows from Lemma 10.

Its Lie structure is given by

\begin{equation*}[H,E\otimes p_1]=2E\otimes p_1,\quad [H,F\otimes p_2]=-2F\otimes p_2,\quad [E\otimes p_1,F\otimes p_2]=H\otimes p_1p_2,\end{equation*}

where $p_1p_2\in {\mathbb{C}}[J]$. We may assume that $g\cdot E=\omega_{\ell}E$ and $g\cdot F=\omega_{\ell}^{-1}F$ after applying an automorphism that interchanges the isotypical components if necessary (e.g. $\mathrm{Ad}\begin{pmatrix} 0 & 1\\ -1 &0 \end{pmatrix}$) and assume therefore that $p_1\in {\mathbb{C}}[\wp,\wp']^{\chi_{\ell-1}}$, $p_2\in {\mathbb{C}}[\wp,\wp']^{\chi_{1}}$.

Proof. First of all, we would like to stress that we can view $\Psi\in \mathrm{Aut}_{\mathcal{O}_{\mathbb{T}}}(\mathfrak{sl}_2\otimes_{{\mathbb{C}}}\mathcal{O}_{\mathbb{T}})$ as a holomorphic $\mathrm{Aut}(\mathfrak{sl}_2)$-valued function on $\mathbb{T}$, in which case we use the notation $\Psi(z)$. We shall use both interpretations interchangeably and let the argument of Ψ be either an element of $\mathbb{T}$ or an element of $\mathfrak{sl}_2\otimes_{{\mathbb{C}}}\mathcal{O}_{\mathbb{T}}$.

The first part claims that for $k\in K$, if we have $\Psi(k\cdot z)=\rho(k)\Psi(z)$, then there is a normal form for $\mathfrak{K}$. Let $\{x_1,x_2,x_3\}$ be a basis for $\mathfrak{sl}_2$. Then for $j=1,2,3,$ we have that $X_j(z):=\Psi(z)x_j$ is K-invariant:

\begin{equation*}k\cdot X_j(z)=k\cdot \Psi(z)x_j=\rho(k)\Psi(k^{-1}\cdot z)x_j=X_j(z).\end{equation*}

Since $\Psi\in \mathrm{Aut}_{\mathcal{O}_{\mathbb{T}}}(\mathfrak{sl}_2\otimes_{{\mathbb{C}}}\mathcal{O}_{\mathbb{T}})$ is an inner automorphism, we have

\begin{equation*}{\mathbb{C}}\langle x_1,x_2,x_3\rangle\otimes_{{\mathbb{C}}}\mathcal{O}_{\mathbb{T}}=\Psi({\mathbb{C}}\langle x_1,x_2,x_3\rangle\otimes_{{\mathbb{C}}}\mathcal{O}_{\mathbb{T}})={\mathbb{C}}\langle X_1,X_2,X_3\rangle\otimes_{{\mathbb{C}}}\mathcal{O}_{\mathbb{T}}.\end{equation*}

Taking the invariants, we obtain a normal form

\begin{equation*}\mathfrak{K}={\mathbb{C}}\langle X_1,X_2,X_3\rangle\otimes_{{\mathbb{C}}}\mathcal{O}_{\mathbb{T}}^K\cong \mathfrak{sl}_2\otimes_{{\mathbb{C}}}{\mathbb{C}}[\wp,\wp'],\end{equation*}

with the brackets inherited from $\mathfrak{sl}_2$ and where $\wp,\wp'$ generate $\mathcal{O}_{\mathbb{T}}^K$.

To prove the second item, we will argue as follows. Let $\gamma=gk\in \Gamma=G \ltimes K$ and $x\in \mathfrak{sl}_2$. By (6.1),

\begin{equation*}\gamma\cdot \Psi(z)x=\Psi(z)\tilde{\rho}(\gamma)x.\end{equation*}

Since G is cyclic, it leaves a CSA of $\mathfrak{sl}_2$ invariant. Hence there is a basis $\{X_1,X_2,X_3\}\subset \mathfrak{sl}_2$, with X ₁ a generator in the CSA, such that with respect to this basis, $\tilde{\rho}(\gamma)=\tilde{\rho}(gk)=\mathrm{diag}(1,\chi_1(g),\chi_{\ell-1}(g))$, where χ_j are the characters of G, g generates G and $k\in K$. Now, $\Psi(X_1\otimes 1)$ generates a CSA of $\mathfrak{K}$ and $g\cdot \Psi(X_1\otimes 1)=\Psi(X_1\otimes 1)$ for all $g\in G$ by (6.1). We can, therefore, apply Lemma 14 to arrive at the conclusion of the existence of a normal form for $\mathfrak{A}$ and the stated form of it.