\max(3,N)\,; \\ p=3, &d=2. \end{cases} \end{equation*} \end{theorem} Assuming these theorems, we can formulate our main result on the bi-Hamiltonian cohomology, from which Theorem~\ref{thm:mainstatement} follows: \begin{theorem} \label{thm:mainmain} The bi-Hamiltonian cohomology \(BH_d^p \) vanishes for\vspace*{-3pt} \begin{align*} \begin{cases} p < d & d \geq 2\,; \\ p > d+N & d \geq 2\,; \\ d \leq p \leq d+N & d > \max (3,N)\,; \\ p = 2 & d = 2, \end{cases} \end{align*} unless \((p,d) = (2,3)\), in which case it is isomorphic to \(\bigoplus_{i=1}^N C^\infty (u^i) \), the space of central invariants. \end{theorem} The regions of this theorem are visualized in Figure \ref{vanishingregions}. \begin{figure}[ht] \hspace*{-.5cm}\begin{tabular}{ccc} \includegraphics[scale=.6]{carlet-et-al_fig1}&\hspace*{-7mm}&\includegraphics[scale=.6]{carlet-et-al_fig2}\\ \smaller (a) The case \(N \geq 3\).&\hspace*{-7mm}&\smaller(b) The case \(N = 2\). \end{tabular} \caption{\smaller All bi-Hamiltonian cohomology groups are zero in region~\(A\), except for the black dot, which is given by the central invariants. All groups are unknown in region \(B\), except for the white dot, which vanishes.}\label{vanishingregions} \end{figure} \begin{proof} Using the isomorphism between \(BH_d^p \) and \(H_d^P (\hcF [\l ])\) in the required range, all the vanishing statement follow from the exact sequence~\eqref{eq:MainExact} as both the second and the fourth terms are zero. For $(p,d)=(2,3)$, the second term is zero, which implies that $H^2_3(\hcF[\lambda])\cong H^3_3(\hcA[\lambda])$, and $H^3_3(\hcA)\cong \bigoplus_{i=1}^N C^\infty (u^i)$ by Theorem \ref{resultp=d}. \end{proof} \begin{remark} Observe that the cohomology of \(\hcA [\l ] \) is still unknown on the subcomplexes $p=d+1, \dotsc, d + N$ for $d< N$, unless $(p,d)=(3,2)$ or unless \(N =1 \). The last case has been determined completely in \cite[Prop.\,4]{cps14-2}. The key to determining the cohomology completely would likely lie in an extension of the proof of Proposition \ref{prop32}, where one would have to study more carefully the transformation \(\th_i^0 \mto \bar{\th}_i^0 \). This transformation is trivial in the case \(N =1 \), so the subtlety does not occur there. \end{remark} We conclude this section with one more piece of notation that we use throughout the rest of the paper: for a multi-index \(I = \{ i_1, \dotsc, i_s\} \), we write \(f^I = \prod_{i \in I} f^i\), \(\theta_I^t = \theta_{i_1}^t \dotsb \theta_{i_s}^t \), etc. \section{The first vanishing theorem} \label{vanishing} In this section we give a proof of Theorem \ref{vanth}, based on the proof of \cite{cps15}. This section does not contain any new results, but has the main purpose of recalling some objects that will be used later.\par The presentation of the proof given here is improved over \cite{cps15}, mainly by focusing less on the intricacies of spectral sequences and more on the structure and decomposition of the spaces and differentials involved. This exposition is somewhat less detailed as a result and the reader is expected to be familiar with spectral sequence techniques for graded complexes; more details can be found in \cite{cps15}.\enlargethispage{.5\baselineskip}% \subsection{} \label{degu} Let $\deg_u$ be the degree on $\hcA$ defined by assigning\vspace*{-3pt} \begin{equation*} \deg_u u^{i,s} = 1, \quad s >0 \end{equation*} and zero on the other generators. The operator $D_\lambda$ splits in the sum of its homogeneous components\vspace*{-3pt} \begin{equation*} D_\lambda = \Delta_{-1} + \Delta_0 + \dots, \end{equation*} where $\deg_u \Delta_k= k$. To the degree $\deg_u + \deg_\theta$ we associate a decreasing filtration of $\hcA[\lambda]$. Let us denote by $\tensor*[^1]{E}{}$ the associated spectral sequence. The zero page $\tensor*[^1]{E}{_0}$ is simply given by $\hcA[\lambda]$ with differential $\Delta_{-1}$:\vspace*{-3pt} \begin{equation*} (\tensor*[^1]{E}{_0}, \tensor*[^1]{d}{_0}) = (\hcA[\lambda], \Delta_{-1}). \end{equation*} To find the first page $\tensor*[^1]{E}{_1}$, we have to compute the cohomology of this complex. \subsection{} Let us compute the cohomology of the complex $(\hcA[\lambda], \Delta_{-1})$. The differential can be written as \begin{equation*} \Delta_{-1} = \sum_i (-\lambda+u^i) f^i \hat{d}_i, \end{equation*} where $\hat{d}_i$ is the de\,Rham-like differential \begin{equation*} \hat{d}_i = \sum _{s \geq 1} \theta_i^{s+1} \frac{\partial }{\partial u^{i,s}}. \end{equation*} It is convenient to split $\hcA$ in a direct sum \begin{equation*} \hcA = \hcC \oplus \Bigl(\bigoplus_{i=1}^N \hcC_i^\textnormal{nt} \Bigr) \oplus \hcM. \end{equation*} Here \begin{equation*} \hcC=C^\infty(U)[\theta^0_1,\dots,\theta^0_N,\theta^1_1,\dots,\theta^1_N], \end{equation*} and \begin{equation*} \hcC_i = \hcC \llbracket \{ u^{i,s}, \theta_i^{s+1} \mid s\geq1 \} \rrbracket, \end{equation*} while $\hcC_i^\textnormal{nt}$ denotes the subspace of $\hcC_i$ spanned by nontrivial monomials, \ie all monomials that contain at least one of the variables $u^{i,s}$, $\theta_i^{s+1}$ for $s\geq 1$. By $\hcM$ we denote the subspace of $\hcA$ spanned by monomials which contain at least one of the mixed quadratic expressions \begin{equation*} u^{i,s} u^{j,t}, \quad u^{i,s} \theta_j^{t+1}, \quad \theta_i^{s+1} \theta_j^{t+1} \end{equation*} for some $s,t \geq1$ and $i\not= j$. \begin{lemma} The differential $\Delta_{-1}$ leaves invariant each direct summand in \begin{equation} \label{split1} \hcA[\lambda] = \hcC[\lambda] \oplus \Bigl(\bigoplus_{i=1}^N \hcC_i^\textnormal{nt}[\lambda] \Bigr) \oplus \hcM[\lambda], \end{equation} and in particular maps $\hcC[\lambda]$ to zero. \end{lemma} \begin{proof} It is easy to check that \begin{align*} &\hat{d}_i (\hcC) = 0, &\hat{d}_i (\hcM) \subseteq \hcM, \\ &\hat{d}_i (\hcC_i^\textnormal{nt}) \subseteq \hcC_i^\textnormal{nt}, &\hat{d}_i (\hcC_j^\textnormal{nt}) = 0 \quad i\not= j, \end{align*} from which the lemma follows immediately. \end{proof} The cohomology of $\hcA[\lambda]$ is therefore the direct sum of the cohomologies of the summands in the direct sum~\eqref{split1}, and in particular \begin{equation*} H(\hcC[\lambda], \Delta_{-1}) = \hcC[\lambda]. \end{equation*} Let us first observe that the cohomology of the de\,Rham complex $(\hcC_i, \hat{d}_i)$ is trivial in positive degree. \begin{lemma} \quad$H(\hcC_i, \hat{d}_i) = \hcC$. \end{lemma} \begin{proof} The proof is completely analogous to the standard proof of the Poincaré lemma. \end{proof} In particular we have that \begin{equation*} H(\hcC_i^\textnormal{nt}, \hat{d}_i) = 0, \end{equation*} therefore the kernel of $\hat{d}_i$ in $\hcC_i^\textnormal{nt}$ coincides with $\hat{d}_i(\hcC_i)$. \begin{lemma} \quad $\dpl H(\hcC_i^\textnormal{nt}[\lambda], \Delta_{-1}) = \frac{\hat{d}_i (\hcC_i)[\lambda]}{(-\lambda+u^i)\hat{d}_i (\hcC_i)[\lambda]}$. \end{lemma} \begin{proof} On $\hcC_i^\textnormal{nt}[\lambda]$ the differential $\Delta_{-1}$ is equal to $(-\lambda+u^i) f^i \hat{d}_i$. Its kernel coincides with the kernel of $\hat{d}_i$ on $\hcC_i^\textnormal{nt}[\lambda]$, which is $d_i(\hcC_i)[\lambda]$. Its image is $(-\lambda+u^i)\hat{d}_i (\hcC_i)[\lambda]$. \end{proof} Finally we prove that the complex $(\hcM[\lambda], \Delta_{-1})$ is acyclic. \begin{lemma} \quad $H(\hcM[\lambda], \Delta_{-1}) =0$. \end{lemma} \begin{proof} This lemma can be proved by induction on $N$. Denote, for convenience, the corresponding space and the differential by $\hcM[\lambda]_{(N)}$ and $\Delta_{-1,(N)}$. We also use in the proof the notation $\hcA_{(N)}$ and $\hcC_{(N)}$. The differential $\Delta_{-1}$ is naturally the sum of two commuting differentials, \begin{equation*} \Delta_{-1,(N)}=\Delta_{-1,(N-1)}+(-\lambda+u^N) f^N \hat d_N. \end{equation*} The cohomology of $(-\lambda+u^N) f^N \hat{d}_N$ on $\hcM[\lambda]_{(N)}$ is equal to the direct sum of two subcomplexes, $\hcC_{(N)}\otimes_{\hcC_{(N-1)}}\hcM[\lambda]_{(N-1)}$ and $$ \frac{\hat d_N(\hcC^\textnormal{nt}_N)\otimes_{\hcC_{(N)}} \left(\left(\bigoplus_{i=1}^{N-1} \hcC_i^\textnormal{nt}[\lambda] \right) \oplus\hcC_{(N)}\otimes_{\hcC_{(N-1)}}\hcM[\lambda]_{(N-1)}\right)}{(-\lambda+u^N)}. $$ On the first component the induced differential is equal to $\Delta_{-1,(N-1)}$, so we can~use the induction assumption. On the second component the induced differential is equal~to $$\left.\left(\Delta_{-1,(N-1)}\right)\right|_{\lambda=u^N},$$ so, up to rescaling by non-vanishing functions, it is a de\,Rham-like differential acting only on the second factor of the tensor product. This second factor can be identified with $\hcC_{(N)}\otimes_{\hcC_{(N-1)}} \hcA_{(N-1)}/\hcC_{(N-1)}$, so the possible non-trivial cohomology is quotiented out (\cf the standard proof of the Poincaré lemma). \end{proof} This completes the computation of the cohomology of the complex $(\hcA[\lambda],\Delta_{-1})$: {\let\MakePointrait\relax \begin{proposition} \begin{equation} \label{cohA} H(\hcA[\lambda], \Delta_{-1}) = \hcC[\lambda] \oplus \biggl( \bigoplus_{i=1}^N \frac{\hat{d}_i (\hcC_i)[\lambda]}{(-\lambda+u^i)\hat{d}_i (\hcC_i)[\lambda]} \biggr). \end{equation} \end{proposition} } \subsection{}\label{sec:1E1} The first page $\tensor*[^1]{E}{_1}$ of the first spectral sequence is given by the cohomology of the complex $H(\hcA[\lambda], \Delta_{-1})$ with the differential induced by the operator $\Delta_0$: \begin{equation*} (\tensor*[^1]{E}{_1}, \tensor*[^1]{d}{_1}) = (H(\hcA[\lambda], \Delta_{-1}), \Delta_0). \end{equation*} We recall the formula for the operator $\Delta_0$ in the appendix. To get the second page $\tensor*[^1]{E}{_2}$ of the first spectral sequence we have to compute the cohomology of this complex. Let $\deg_{\theta^1}$ be the degree on $\hcA$ defined by setting \begin{equation*} \deg_{\theta^1} \theta^1_i =1 \quad i=1, \dots,N \end{equation*} and zero on the other generators. The operator $\Delta_0$ splits in its homogeneous components \begin{equation*} \Delta_0 = \Delta_{0,1} + \Delta_{0,0}+\Delta_{0,-1}, \end{equation*} where $\deg_{\theta^1} \Delta_{0,k} =k$. To the degree $\deg_{\theta^1} - \deg_{\theta}$ we associate a decreasing filtration of $H(\hcA[\lambda], \Delta_{-1})$, and denote by $\tensor*[^2]{E}{}$ the associated spectral sequence. The zero page $\tensor*[^2]{E}{_0}$ is given by $H(\hcA[\lambda], \Delta_{-1})$ with the differential induced by $\Delta_{0,1}$: \begin{equation*} (\tensor*[^2]{E}{_0}, \tensor*[^2]{d}{_0}) = (H(\hcA[\lambda], \Delta_{-1}), \Delta_{0,1}). \end{equation*} The first page $\tensor*[^2]{E}{_1}$ is given by the cohomology of this complex. \subsection{} \label{D01} To obtain a simple expression for the action of $\Delta_{0,1}$ on the cohomology~\eqref{cohA}, it is convenient to perform a change of basis in $\hcA$. Let $\Psi$ be the invertible operator that rescales the generators of $\hcA$ as follows \begin{equation*} u^{i,s} \mto (f^i)^{\sfrac{s}2} u^{i,s}, \quad \theta_i^s \mto (f^i)^{\psfrac{s+1}2} \theta_i^s. \end{equation*} The operator $\Delta_{0,1}$ has a simpler form when conjugated with $\Psi$, and since $\Psi$ leaves invariant all the subspaces that we consider, such conjugation does not affect the computation of the cohomology. \begin{lemma} \label{lemmaD01} The operator $\Delta_{0,1}$ acts on the cohomology~\eqref{cohA} as $\Psi \tilde{\Delta}_{0,1} \Psi^{-1}$, where \begin{equation*} \tilde\Delta_{0,1} =\sum_i (-\lambda+u^i) \theta_i^1\, \frac{\partial }{\partial u^i} +\sum_{i,j} (-\lambda+u^j) (\gamma_{ij} \theta_j^1 - \gamma_{ji} \theta_i^1) \theta^0_j \,\frac{\partial }{\partial \theta_i^0} +\sum_i \theta_i^1 \cE_i \end{equation*} and leaves invariant each of the summands in Equation \eqref{cohA}. Here $\cE_i$ is the Euler operator that multiplies any monomial $m$ by its weight $w_i(m)$ defined by \begin{equation*} w_i(u^{i,s}) = \frac{s}2 +1, \quad w_i(\theta_i^{s-1}) = \frac{s}2-1 \quad s\geq1 \end{equation*} and zero on the other generators. \end{lemma} \begin{proof} Recall that $\Delta_{0,1}$ is the $\deg_u =0$ and $\deg_{\theta^1} = 1$ homogeneous component of the differential $D_\lambda$. An explicit expression can be found in~\cite{cps15}. By a straightforward computation, we have that $\Psi^{-1} \Delta_{0,1} \Psi$ is equal to $\tilde{\Delta}_{0,1}$ plus two extra terms \begin{align*} &-\sum_{i,j}\sum_{s\geq1} (-\lambda+u^i) \left(\sfrac{f^i}{f^j} \right)^{\psfrac{s+1}2} \left( (s+2) \gamma_{ji} \theta_i^1 +s \gamma_{ij}\theta_j^1 \right) u^{j,s}\, \frac{\partial }{\partial u^{i,s} } \\ &+\sum_{i,j}\sum_{s\geq2} (-\lambda+u^j) \left(\sfrac{f^i}{f^j} \right)^{\sfrac{s}2} \left( (1-s) \gamma_{ij} \theta_j^1-(1+s) \gamma_{ji} \theta_i^1 \right) \theta_j^s\, \frac{\partial }{\partial \theta_i^s}. \end{align*} The following formulas are useful in the computation of the conjugated operator: \begin{align*} \Psi^{-1} \frac{\partial }{\partial u^{i,s}} \Psi &= (f^i)^{\sfrac{s}2} \frac{\partial }{\partial u^{i,s}}, \quad \Psi^{-1} u^{i,s} \Psi = (f^i)^{-\sfrac{s}2} u^{i,s},\\ \Psi^{-1} \frac{\partial }{\partial \theta_i^s} \Psi &= (f^i)^{\psfrac{s+1}2} \frac{\partial }{\partial \theta_i^s}, \quad \Psi^{-1} \theta_i^s \Psi = (f^i)^{-\psfrac{s+1}2} \theta_i^s,\\ \Psi^{-1} \frac{\partial }{\partial u^i} \Psi&= \frac{\partial }{\partial u^i} + \sum_{j} \frac{\partial \log f^j}{\partial u^i} \sum_{s\geq0} \Bigl( \frac{s}2\, u^{j,s} \frac{\partial }{\partial u^{j,s}} + \frac{s+1}2\, \theta_j^s \frac{\partial }{\partial \theta_j^s} \Bigr). \end{align*} By construction the operator $\Delta_{0,1}$ induces a map on the cohomology~\eqref{cohA}, and so does the conjugated operator $\Psi^{-1} \Delta_{0,1} \Psi$. Let us make a few easy to check observations in order to simplify this operator: \begin{enumerate} \item $\tilde{\Delta}_{0,1}$ maps $\hcC[\lambda]$ to itself, while the two extra terms send it to zero; \item the two extra terms, when $j\not= i$, send $\hat{d}_i(\hcC_i)[\lambda]$ to $\hcM[\lambda]$ which is trivial in cohomology; \item both $\tilde{\Delta}_{0,1}$ and the extra terms for $j=i$ map $\hat{d}_i(\hcC_i)[\lambda]$ to $\hcC^{\textup{nt}}_i[\lambda]$, and, because they need to act on cohomology, they actually send it to $\hat{d}_i(\hcC_i)[\lambda]$; \item terms in $\hat{d}_i(\hcC_i)[\lambda]$ which are proportional to $\lambda - u^i$ actually vanish in cohomology, so we can set $\lambda$ equal to $u^i$; this in particular kills the $i=j$ part of the extra terms. \end{enumerate} The lemma is proved. \end{proof} Let us identify \begin{equation} \label{ide} \frac{\hat{d}_i (\hcC_i)[\lambda]}{(-\lambda+u^i)\hat{d}_i (\hcC_i)[\lambda]} \simeq \hat{d}_i (\hcC_i) \end{equation} by setting $\lambda$ equal to $u^i$. Let $\cD_i$ be the operator induced by $\Delta_{0,1}$ on $\hat{d}_i (\hcC_i)$ by this identification. Its explicit form is given in the next corollary. \begin{corollary} The operator $\cD_i$ on $\hat{d}_i (\hcC_i)$ is given by $\cD_i= \Psi \tilde{\cD}_i \Psi^{-1}$ with \begin{equation*} \tilde{\cD}_i =\sum_k \theta_k^1 \biggl[ (u^k - u^i) \biggl(\frac{\partial }{\partial u^k} + \sum_j \gamma_{jk} \theta_k^0 \frac{\partial }{\partial \theta_j^0} \biggr) + \sum_j (u^i - u^j) \gamma_{jk} \theta_j^0 \frac{\partial }{\partial \theta_k^0} +\cE_k \biggr]. \end{equation*} \end{corollary} The first page of the second spectral sequence is therefore given by the following direct sum \begin{equation} \label{2E1} \tensor*[^2]{E}{_1} \simeq H(\hcC[\lambda],\Delta_{0,1}) \oplus \biggl( \bigoplus_{i=1}^N H\bigl(\hat{d}_i(\hcC_i), \cD_i\bigr) \biggr). \end{equation} \subsection{} \label{cohC} A vanishing result for the cohomology of $\hcC[\lambda]$ is obtained by a simple degree counting argument. \begin{proposition} \label{vanC} The cohomology $H^p_d(\hcC[\lambda], \Delta_{0,1})$ vanishes for all $(p,d)$, unless \begin{equation*} d=0, \dots,N, \quad p=d, \dots, d+N. \end{equation*} \end{proposition} \begin{proof} The possible bi-degrees of the elements of $\hcC$ are precisely those excluded in the proposition. \end{proof} \subsection{} \label{cohdC} We have the following vanishing result for the cohomology of $(\hat{d}_i (\hcC_i), \cD_i)$. \begin{proposition} \label{vandC} The cohomology $H^p_d(\hat{d}_i(\hcC_i), \cD_i)$ vanishes for all $(p,d)$, unless \begin{equation*} d=2,\dots, N+2, \quad p=d, \dots, d+N-1. \end{equation*} \end{proposition} \begin{proof} To prove this result let us introduce a third spectral sequence. For fixed $i$, let $\deg_{\theta^1_i}$ be the degree that assigns degree one to $\theta_i^1$ and degree zero to the remaining generators. Consider the decreasing filtration associated to the degree $\deg_{\theta_i^1} - \deg_\theta$. Let $\tensor*[^3]{E}{}$ be the associated spectral sequence. Let $\cD_{i,1}$ be the homogeneous component of $\cD_i$ with $\deg_{\theta_i^1} =1$, \ie $\cD_{i,1} = \Psi \tilde{\cD}_{i,1} \Psi^{-1}$ with \begin{equation*} \tilde{\cD}_{i,1} = \theta_i^1 \biggl[ \sum_j (u^i - u^j) \gamma_{ji} \theta_j^0 \frac{\partial }{\partial \theta_i^0} +\cE_i \biggr]. \end{equation*} The zero page $\tensor*[^3]{E}{_0}$ is given by $\hat{d}_i(\hcC_i)$ with differential $\cD_{i,1}$: \begin{equation*} (\tensor*[^3]{E}{_0}, \tensor*[^3]{d}{_0}) = (\hat{d}_i(\hcC_i), \cD_{i,1}). \end{equation*} To prove the proposition it is sufficient to prove the vanishing of the cohomology of this complex in the same degrees, which we will do in the next lemma. \end{proof} \begin{lemma} The cohomology $H^p_d(\hat{d}_i(\hcC_i), \cD_{i,1})$ vanishes for all $(p,d)$, unless \begin{equation*} d=2,\dots, N+2, \quad p=d, \dots, d+N-1. \end{equation*} \end{lemma} \begin{proof} As before let us work with the operator $\tilde{\cD}_{i,1}$. Let $\m$ be a monomial in the variables $u^{i,s}$, $\theta_i^{s+1}$ for $s\geq1$. For $g\in\hcC$, we have \begin{equation*} \tilde{\cD}_{i,1}\bigl(g \hat{d}_i (\m)\bigr) = \theta_i^1 \biggl( \sum_j (u^i - u^j) \gamma_{ji} \theta_j^0 \frac{\partial }{\partial \theta_i^0}\, g + (w_i(g) + w_i(\m) -1) g \biggr) \hat{d}_i(\m), \end{equation*} where $w_i$ is the weight defined in Lemma \ref{lemmaD01}. Therefore $\tilde{\cD}_{i,1}$ leaves $\hcC \hat{d}_i(\m)$ invariant for each monomial $\m$. We will now prove that the cohomology of the subcomplex $\hcC \hat{d}_i(\m)$ vanishes for all monomials $\m$, except for the case $\m=u^{i,1}$, therefore the cohomology of $\hat{d}_i(\hcC_i)$ is just given by the cohomology of $\hcC \hat{d}_i(u^{i,1})$. Notice that $\hat{d}_i(\m)$ is nonzero only for $w_i(\m) \geq \sfrac32$, and the case $w_i(\m)=\sfrac32$ corresponds to $\m=u^{i,1}$ and $\hat{d}_i(\m) = \theta_i^2$. Let us split $\hcC = \hcC^i_0 \oplus \theta_i^0 \hcC^i_0$, where $\hcC^i_0$ is the subspace spanned by monomials that do not contain $\theta_i^0$. Given $g \in \hcC^i_0$ we have \begin{equation*} \tilde{\cD}_{i,1} \bigl(g \hat{d}_i(\m) \bigr)= \theta_i^1(w_i(\m)-1)g \hat{d}_i(\m). \end{equation*} Notice that the coefficient $w_i(\m)-1$ is non-vanishing, therefore $\tilde{\cD}_{i,1}$ is acyclic on the subcomplex $\hcC^i_0 \hat{d}_i(\m)$. For $g \in \theta_i^0 \hcC^i_0$, the differential $\tilde{\cD}_{i,1}$ maps $g \hat{d}_i(\m)$ to \hbox{$\theta_i^1(w_i(\m)-\sfrac32)g \hat{d}_i(\m) \in \theta^0_i \hcC^i_0 \hat{d}_i(\m)$} plus an element in $\hcC^i_0 \hat{d}_i(\m)$. It is well-known that when a complex $(C,d)$ contains an acyclic subcomplex $C'$, its cohomology is given by the cohomology of a subspace $C''$ complementary to $C'$ with differential given by the restriction and projection of $d$ to $C''$. In the present case this implies that the cohomology of $\hcC \hat{d}_i(\m)$ is equivalent to the cohomology of $\theta_i^1 \hcC^i_0$ with differential given by the operator of multiplication by the element $\theta_i^1(w_i(\m)-\sfrac32)$. Such complex is acyclic as long as $w_i(\m) \not= \sfrac32$. The only nontrivial case is when $\m = u^{i,1}$, and in such case the cohomology is given by \begin{equation*} \theta_i^0 \hcC^i_0 \hat{d}_i(u^{i,1}) =\hcC^i_0 \theta_i^0 \theta_i^2. \end{equation*} Counting the degrees of the possible elements in this space we obtain the vanishing result above. \end{proof} \subsection{} From the previous two propositions it follows that $\tensor*[^2]{E}{_1}$ is zero if the bi-degree $(p,d)$ is not in one of the two specified ranges, \ie in their union given in Theorem \ref{vanth}. Clearly the vanishing of $\tensor*[^2]{E}{_1}$ in certain degrees implies the vanishing of $\tensor*[^1]{E}{_2}$ and consequently of $H(\hcA[\lambda], D_\lambda)$ in the same degrees. This concludes the proof of Theorem \ref{vanth}. \section{The cohomology of \texorpdfstring{$(\hat{d}_i(\hcC_i), \cD_i)$}{dhat(Ci,Di)}} \label{subcomplexdCi} In this section we extend the vanishing result of Section \ref{cohdC} to a computation of the full cohomology of the complex $(\hat{d}_i(\hcC_i), \cD_i)$. First, we can represent the space $\hat{d}_i(\hcC_i)$ as a direct sum \begin{equation*} \hat{d}_i(\hcC_i) = \hcC_0^i \theta_i^2 \oplus \hcC^i_0 \theta_i^0 \theta_i^2 \oplus \hcC \otimes \hat{d}_i(V_i), \end{equation*} where, as before in Section \ref{cohdC}, we denote by $\hcC^i_0$ the subspace of $\hcC$ spanned by monomials that do not contain $\theta_i^0$. We denote by $V_i$ the space of polynomials in $u^{i,\geq 1},\theta_i^{\geq2}$ of standard degree $\geq2$. \begin{lemma} The differential $\cD_i$ leaves invariant the spaces $\hcC_0^i \theta_i^2$ and $\hcC \otimes \hat{d}_i(V_i)$, while \begin{equation*} \cD_i (\hcC_0^i \theta_i^0 \theta_i^2) \subset \hcC \theta_i^2 = \hcC_0^i \theta_i^2 \oplus \hcC^i_0 \theta_i^0 \theta_i^2. \end{equation*} \end{lemma} \begin{proof} As before we can equivalently work with $\tilde{\cD}_i$. The statement is a simple check, noticing that $[\tilde{\cD}_i, \hat{d}_i ]_+ = - \theta_i^1 \hat{d}_i$. \end{proof} As we know from Section \ref{cohdC} the cohomology is a subquotient of $\hcC^i_0 \theta_i^0 \theta^2_i$. Therefore the subcomplexes $\hcC_0^i \theta_i^2$ and $\hat{d}_i(V_i)$ are acyclic and the cohomology is given by \begin{equation*} H(\hat{d}_i(\hcC_i), \cD_i) = H(\hcC_0^i \theta_i^0\theta_i^2, \cD_i'), \end{equation*} where $\cD_i'$ is the restriction and projection of $\cD_i$ to $ \hcC_0^i \theta_i^0\theta_i^2$. Explicitly $\cD_i' = \Psi \tilde{\cD}_i' \Psi^{-1}$ is given by removing the terms in $\tilde{\cD}_i$ that decrease the degree in $\theta_i^0$, which gives \begin{equation*} \tilde{\cD}_i' =\sum_{k\not= i} \theta_k^1 \biggl[ (u^k - u^i) \biggl(\frac{\partial }{\partial u^k} + \sum_{j\not= i} \gamma_{jk} \theta_k^0 \frac{\partial }{\partial \theta_j^0} \biggr) + \sum_j (u^i - u^j) \gamma_{jk} \theta_j^0 \frac{\partial }{\partial \theta_k^0} +\cE_k \biggr]. \end{equation*} Notice that $\cE_i$ maps $\hcC^i_0 \theta_i^0 \theta^2_i$ to zero, since both $\theta_i^1$ and $\theta_i^0 \theta_i^2$ have degree $w_i$ equal to zero. We can now split $\hcC^i_0 \theta_i^0 \theta^2_i$ in the direct sum $\hcC^i_{0,1} \theta_i^0 \theta^2_i \oplus \hcC^i_{0,1} \theta_i^0 \theta_i^1 \theta^2_i$, where $\hcC^i_{0,1}$ is the subspace of $\hcC^i_0$ spanned by monomials that do not depend on $\theta_i^1$. Since $\tilde{\cD}_i'$ does not act on $ \theta_i^1 \theta^2_i$ or $\theta_i^0 \theta_i^1 \theta^2_i$, we can reduce our problem to computing the cohomology of the complex $(\hcC^i_{0,1}, \tilde{\cD}_i')$. Let us denote by $\hat{\delta}_k^i$ the coefficient of $\theta_k^1$ in $\tilde{\cD}_i'$, \ie \begin{equation*} \tilde{\cD}_i' = \sum_{k\not=i} \theta_k^1 \hat{\delta}_k^i. \end{equation*} \begin{lemma} The cohomology $H^p_d(\hcC^i_{0,1}, \tilde{\cD}_i')$ is nontrivial only in degrees $d=0$ and $p=0, \dots, N-1$. In degree $(d=0,p)$ it is isomorphic to $C^\infty (u^i) \otimes \bigwedge^p \R^{N-1}$ and is represented by an element \begin{equation*} F = \sum_{\substack{J \subseteq [n]\setminus \{ i \} \\ | J|=p} } F^J(u^1, \dots, u^N) \theta^0_J \in \bigcap_{k \not =i} \ker \hat{\delta}_k^i, \end{equation*} which depends on a single function of the variable $u^i$. \end{lemma} \begin{proof} We represent the space of coefficients $\hcC^i_{0,1}$ as a direct sum $\bigoplus_{\ell,t=0}^{n-1} K^{\ell,t}$, where an element of $K^{\ell,t}$ can be written down as $$ \sum_{\substack{I\subset [n]\setminus \{i\} \\ |I|=\ell} } f^I \theta^1_I \cdot \sum_{\substack{J\subset [n]\setminus \{i\} \\ |J|=t }} \theta^0_JF_{I,J}(u^1,\dots,u^n). $$ The action of $\mathcal{D}_i'$ can be described, in both cases, as a map $K^{\ell,t}\to K^{\ell+1,t}$ given by the following formula on the components of the corresponding vectors: $F_{I,J}\mto G_{S,T}$, where $$ G_{S,T} = \sum_{s\in S} \frac{\d}{\d u^{s}}F_{I\setminus\{s\},T} + (A_{s;t})^J_{T} F_{I\setminus\{s\},J}, $$ where the coefficients of the matrices $(A_{s;t})^J_T$ can easily be reconstructed from the formula for the operator $\tilde{\mathcal{D}}_i'$. So, this way we can describe each of the subcomplexes $K^{\sbullet,t}\theta^0_i\theta^2_i$, $K^{\sbullet,t}\theta^0_i\theta^1_i\theta^2_i$, $t=0,\dots,n-1$, as a tensor product of the de\,Rham complex of smooth functions in $n-1$ variable $u^k$, $k\not= i$, with a vector space whose basis is indexed by monomials of degree $t$ in $\theta^0_q$, $q\not=i$. The differential (the restriction of $\tilde{\mathcal{D}}_i'$ to this subcomplex) is equal to the de\,Rham differential $\sum_{p\not=i} \theta^1_p \sfrac{\d}{\d u^p}$ twisted by a linear map: \begin{equation} \label{eq:deformed-de-Rham} \sum_{p\not=i} \theta^1_p \cdot \Bigl(\frac{\d}{\d u^p} + A_{p;t}\Bigr). \end{equation} (the coefficients of $A_{p;t}$ depend on whether we consider the case of $K^{\sbullet,t}\theta^0_i\theta^2_i$ or $K^{\sbullet,t}\theta^0_i\theta^1_i\theta^2_i$, but the shape of the differential is the same in both cases). The cohomology of the differential~\eqref{eq:deformed-de-Rham} is isomorphic to the cohomology of the de\,Rham differential $\sum_{p\not=i} \theta^1_p\, \sfrac{\d}{\d u^p}$. It is represented by the differential forms of order~$0$, that is, it is non-trivial only for $\ell=0$, whose vector of coefficients $F_{\emptyset,J}$ solves the differential equations \begin{equation*} \frac{\partial F_{\emptyset,J}}{\partial u^p} + (A_{p;t})_J^T F_{\emptyset,T} = 0 \end{equation*} for $p\not=i$. The solution of this equation is uniquely determined by the restriction $F_{\emptyset,J}|_{u_p=0,\ p\not=i}$, that is, by a single function of $u^i$. So, finally, we obtain the statement of the lemma. \end{proof} Taking into account the action of $\Psi$ we obtain the cohomology the complex $(\hat{d}_i(\hcC_i), \mathcal{D}_i)$. \begin{proposition} \label{lem13} The cohomology of $(\hat{d}_i(\hcC_i), \mathcal{D}_i)$ is nontrivial only in degrees equal to $(p,d)=(2,2),\dotsc, (N+1,2)$ and $(p,d)=(3,3),\dots, (N+2,3)$. In the degrees $(2+t,2)$ and $(3+t,3)$ it is isomorphic to $C^\infty(u^i) \otimes \bigwedge^t\mathbb{R}^{N-1}$, $t=0, \dots, N-1$. More precisely, representatives of cohomology classes in degrees $(2+t,t)$ and $(3+t,3)$ are given respectively by elements of the form \begin{equation*} F \cdot (f^i)^{t/2 +2} \theta_i^0 \theta_i^2, \qquad G \cdot (f^i)^{t/2 +3} \theta_i^0 \theta_i^1 \theta_i^2 \end{equation*} for $F$, $G$ representatives of $H^t_0(\hcC^i_{0,1},\tilde{\cD}_i')$ as given in the previous lemma. \end{proposition} \section{The cohomology of \texorpdfstring{$(\hcC[\lambda], \Delta_{0,1})$}{C[l],D01} at \texorpdfstring{$p=d$}{p=d}} \label{subcomplexpd} In this section we extend the result of Section \ref{cohC} by computing the cohomology of the subcomplex of $(\hcC[\lambda], \Delta_{0,1})$ defined by setting $p=d$. From Proposition \ref{vanC} we already know that the complex \((\hcC [\l ], \Delta_{0,1}) \) is non-trivial only for \(d \in \{ 0, \dotsc, n\}\) and \(p \in \{ d, \dotsc, d+n\} \). As usual, as the differential is of bidegree \((p,d) = (1,1)\), it splits in subcomplexes of constant \(p-d \). Here we consider the case $p=d$. \begin{proposition} \label{pro14} For $p=d$ the cohomology of the complex $(\hcC[\l], \Delta_{0,1})$ is given by \begin{equation*} H_p^p (\hcC [\l ], \Delta_{0,1}) \simeq \begin{cases} \R [\l ] & p = 0,\\ \bigoplus_{i=1}^N C^\infty (u^i) \th_i^1 & p= 1,\\ 0 & \text{else.} \end{cases} \end{equation*} \end{proposition} \begin{proof} For $p=d$ the complex $\hcC[\lambda]$ is equal to \begin{equation*} C^\infty (U) [ \theta^1_1, \dots, \theta_N^1 ]. \end{equation*} Let us compute the cohomology of $\tilde{\Delta}_{0,1}$. Because there is no dependence on $\theta^0_k$ and the degree $w_k$ of $\theta_k^1$ is zero, the differential simplifies to \begin{equation*} \tilde{\Delta}_{0,1} = \sum_i \delta_i, \quad \delta_i = (-\lambda+ u^i) \theta_i^1 \frac{\partial }{\partial u^i}. \end{equation*} We will let \(J \subseteq \{ 1, \dotsc, N\} \) denote a multi-index and write $\theta_J^1$ for the lexicographically ordered product $\prod_{j\in J} \theta_j^1$. For each of the \(\th_1^1, \th_2^1, \dotsc, \th_N^1 \), we can define a degree \(\deg_{\th_i^1} - \deg_\th \), which again induces a decreasing filtration. The filtration associated to~\(\th_i^1\) has \(\delta_i \) as differential on the zeroth page of the spectral sequence. Considering all these filtrations, we get the following picture: \begin{equation*} \xymatrix@C=0pt{ & C^\infty (U) [\l ] \th_k^1 \ar[dd]^(0.3){\delta_i}|!{[dl];[dr]}\hole \ar[rr]^{\delta_j} && C^\infty(U)[\l ] \th_j^1 \th_k^1\ar[dd]^{\delta_i} \\ C^\infty (U)[\l ] \ar[dd]_{\delta_i} \ar[rr]_(0.3){\delta_j} \ar[ru]^{\delta_k} && C^\infty (U)[\l ] \th_j^1 \ar[dd]^(0.3){\delta_i} \ar[ru]_{\delta_k} & \\ & C^\infty (U)[\l ] \th_i^1 \th_k^1 \ar[rr]_(0.35){\delta_j}|!{[ur];[dr]}\hole && C^\infty (U)[\l ] \th_i^1 \th_j^1 \th_k^1 \\ C^\infty (U)[\l ] \th_i^1 \ar[rr]_{\delta_j} \ar[ru]_{\delta_k} && C^\infty (U)[\l ] \th_i^1 \th_j^1 \ar[ru]_{\delta_k}& } \end{equation*} So the complex can be visualized as an \(N\)-dimensional hypercube with a term in every corner. On the first page of the \(\th_1^1 \)-spectral sequence, the differential is \(\sum_{j \neq 1} \delta_j \), and we can use the \(\th_2^1\)-filtration to get another spectral sequence. This procedure can be repeated inductively. Consider an element in \(C^\infty (U)[\l ] \th_J^1\). Clearly it is in $\ker \delta_1$ if $J$ contains $1$ or if it does not depend on $u^1$: \begin{equation*} \ker \delta_1 = \bigoplus_{J \ni 1} C^\infty (U)[\lambda]\theta_J^1 \oplus \bigoplus_{J \not\ni 1} C^\infty (u^2, \dots, u^N)[\lambda] \theta_J^1, \end{equation*} where $C^\infty (u^2, \dots, u^N)$ denotes the functions in $C^\infty (U)$ which are constant in $u^1$. On the other hand, we clearly have \begin{equation*} \im \delta_1 = \bigoplus_{J \ni 1}(u^1-\l) C^\infty (U) [\l ] \th_J^1, \end{equation*} therefore the first page of the spectral sequence is \begin{equation*} H(\hcC[\lambda], \delta_1) = \bigoplus_{J \ni 1} \frac{C^\infty (U)[\lambda]}{(u^i-\lambda)C^\infty (U)[\lambda]} \theta_J^1\oplus \bigoplus_{J \not\ni 1} C^\infty (u^2, \dots, u^N)[\lambda] \theta_J^1. \end{equation*} As these arguments do not depend on the \(\th_i^1 \) for \(i \neq 1 \) in any way, on the first page of the spectral sequence we can use the \(\th_2^1 \) filtration and use the same arguments to find the first page of its spectral sequence. Completing the induction, we get the following result for the \(\tilde{\Delta}_{0,1}\)-cohomology on \(\hcC [\l ]\): \begin{equation*} \bigoplus_{ J \subseteq \{ 1, \dotsc, N\}} \frac{C^\infty (\{ u^j\}_{j \in J})[\l ]}{\sum_{j \in J} (u^j - \l)} \th_J^1, \end{equation*} where the sum in the denominator is an ideal sum. If \(|J| \geq 2\), this ideal sum contains the invertible element \(u^i - u^j = (u^i - \l) - (u^j -\l)\) for \(i, j \in J\), so the cohomology is zero. The cohomology of $\tilde{\Delta}_{0,1}$ is therefore nontrivial only in degree zero, where it equals $\R[\lambda]$, and in degree one, where it is given by the sum $\bigoplus_{i=1}^N C^\infty (u^i) \th_i^1$. To find the cohomology of $\Delta_{0,1}$ we need to take into account the action of the operator $\Psi$. Hence the cohomology of $\Delta_{0,1}$ in degree one is $\bigoplus_{i=1}^N C^\infty (u^i) f^i(u) \th_i^1$. The proposition is proved. \end{proof} \section{A vanishing result for \texorpdfstring{$\tensor*[^1]{E}{_2}$}{1E2} at \texorpdfstring{$(p,d)=(3,2)$}{(p,d)=(3,2)}} \label{subcomplexpdpu} We now go back to the first spectral sequence $\tensor*[^1]{E}{}$ associated with $\deg_u$ in Section \ref{degu} and prove a vanishing result for its second page. \begin{proposition} \label{prop32} The cohomology of the complex $(H(\hcA[\lambda], \Delta_{-1}), \Delta_0)$ vanishes in degree $(p,d)=(3,2)$. \end{proposition} \begin{proof} In Section \ref{sec:1E1} the vanishing result for $\tensor*[^1]{E}{_2}$ is proved by introducing a filtration in the degree $\deg_{\theta^1}$. In order to extend the vanishing to the case $(p,d)=(3,2)$, we split the differential $\Delta_0$ in a different way. Recall that the operator $\Delta_0$ is by definition the homogeneous component of $D_\lambda$ of degree $\deg_u$ equal to zero. It induces a differential on the first page $\tensor*[^1]{E}{_1}$ of the first spectral sequence, that is on the cohomology $H(\hcA[\lambda ], \Delta_{-1}) $ given by Equation \eqref{cohA}. From Lemma \ref{lem13} we know that the cohomology of this complex is vanishing for $\deg_u$ positive. We can therefore limit our attention to the subcomplex with $\deg_u$ equal to zero \begin{equation*} \tensor*[^1]{E}{_1^0} =\hcC[\lambda]\oplus \bigoplus_{i=1}^N \frac{\hcC \llbracket \theta^{\geq 2}_i\rrbracket^\textnormal{nt} [\lambda]}{(\lambda-u^{i})\hcC \llbracket\theta^{\geq 2}_i\rrbracket^\textnormal{nt} [\lambda]}, \end{equation*} where the superscript in $\hcC\llbracket \theta^{\geq 2}_i\rrbracket^\textnormal{nt}$ indicates that every monomial should include at least one $\theta_i^{\geq 2}$. Let us denote by $\deg_{\theta^0}$ the degree that counts the number of $\theta^0_j$, $j=1, \dots, N$, and split $\Delta_0$ it its homogeneous components \begin{equation*} \Delta_0 = \Delta_0^1 + \Delta_0^0, \end{equation*} where $\deg_{\theta^0} \Delta_0^k = k$. The decreasing filtration on $\tensor*[^1]{E}{_1^0}$ associated to the degree $\deg_\theta - \deg_{\theta^0}$ induces a spectral sequence $\tensor*[^4]{E}{}$, whose zero page is clearly $\tensor*[^1]{E}{_1^0}$, with differential $\tensor*[^4]{d}{_0} = \Delta_0^1$. The first page $\tensor*[^4]{E}{_1}$ is given by the cohomology of $(\tensor*[^1]{E}{_1^0},\Delta_0^1)$ which we now consider. The form of $\Delta_0^1$ can be easily derived from the explicit expression of $\Delta_0$, see the appendix. When acting on $\tensor*[^1]{E}{_1^0}$ it simplifies to the following operator, which for simplicity we still denote $\Delta_0^1$: \begin{equation*} \Delta_0^1 = \frac{1}{2} \sum_{i} \tilde\theta_i^0 \sum_{s\geq1} \theta^{s+1}_i \frac{\d}{\d \theta^s_i}, \end{equation*} with \begin{equation*} \tilde\theta_i^0:= f^i \theta_i^0 + \sum_{j\not=i} (u^i -u^j) \frac{f^j \d_j f^i}{f^i} \theta_j^0. \end{equation*} We consider now the spectral sequence on $\tensor*[^1]{E}{_1^0}$ induced by the degree $\deg_{\theta^{\geq2}}$, which assigns degree one to all $\theta_i^s$ with $s\geq2$. Let $\Delta_0^1 = \Delta_0^{1,0} + \Delta_0^{1,1}$, where \begin{equation*} \Delta_0^{1,0} = \frac{1}{2} \sum_{i} \tilde\theta_i^0 \sum_{s\geq2} \theta^{s+1}_i \frac{\d}{\d \theta^s_i}, \quad \Delta_0^{1,1} = \frac{1}{2} \sum_{i} \tilde\theta_i^0 \theta^{2}_i \frac{\d}{\d \theta^1_i}, \end{equation*} are of degree $\deg_{\theta^{\geq2}} \Delta_0^{1,k} = k$. We can rewrite our complex as \begin{equation*} \hcC[\lambda]\oplus \bigoplus_{i=1}^N \bigoplus_{k \geq1} \frac{\hcC\llbracket\theta^{\geq 2}_i \rrbracket^{(k)} [\lambda]}{(\lambda-u^{i})\hcC \llbracket \theta^{\geq 2}_i \rrbracket^{(k)} [\lambda]}, \end{equation*} where $\hcC \llbracket \theta^{\geq 2}_i\rrbracket^{(k)}$ denotes the homogeneous polynomials with $\deg_{\theta^{\geq2}}$ equal to $k$. Each of the summands is invariant under $\Delta_0^{1,0}$, so it forms a subcomplex whose cohomology we can compute independently. Notice that the differential vanishes on $\hcC[\lambda]$, while it acts like multiplication by $\tilde{\theta}^0_i$ on the $k=1$ subcomplex $$ \hcC \theta^2_i \to \hcC \theta^3_i \to \hcC \theta^4_i \to \cdots, $$ which is therefore acyclic except for the first term, where the cohomology is given by the kernel of the multiplication map, \ie the ideal of $\tilde\theta_i^0$ in $\hcC$ multiplied by $\theta^2_i$. The first page of the spectral sequence is therefore given by \begin{equation} \label{firstpage} \hcC[\lambda] \oplus \bigoplus_i \frac{\hcC \tilde{\theta}_i^0 \theta_i^2 [\lambda]}{(\lambda-u^i)\hcC \tilde{\theta}_i^0 \theta_i^2 [\lambda]} \oplus \bigoplus_{k\geq 2} \bigoplus_{i} H(\hcC \llbracket \theta^{\geq 2}_i\rrbracket^{(k)}, \Delta_0^{1,0}). \end{equation} While it is not difficult to compute the cohomology groups appearing in the third summand, it can be easily seen that they give no contribution to $\tensor*[^1]{E}{_2}$. Indeed, we know from Lemma \ref{lem13} that the cohomology with standard degree $d\geq4$ is a subquotient of $\hcC[\lambda]$, but the minimal degree of elements in the third summand above is $d = 5$. On this page the differential is induced by $\Delta_0^{1,1}$, which has $\deg_{\theta^{\geq 2}}$ equal to one. When acting on the second summand $\hcC \tilde{\theta}_i^0 \theta_i^2$ it vanishes, since it produces a mixed term $\theta^2_i \theta^2_j$ which cannot be in $\hcC \llbracket \theta^{\geq 2}_i\rrbracket^{(k)}$ for $k\geq 2$. Therefore the cohomology of the first two summands is determined by the kernel and the image of the map \begin{equation*} \Delta_0^{1,1}: \hcC[\lambda] \to \bigoplus_i \frac{\hcC \tilde{\theta}_i^0 \theta_i^2 [\lambda]}{(\lambda-u^i)\hcC \tilde{\theta}_i^0 \theta_i^2 [\lambda]}. \end{equation*} The image can be computed in the following way: first of all, it is clear that an element in the image is a linear combination of $\theta^2_i$, $i=1,\dots,N$, where the coefficient of each $\theta^2_i$ does not depend on $\theta^1_i$ and is in the ideal generated by $\tilde\theta_i^0$ in $\hcC$. Therefore the image is a subspace of\vspace*{-3pt} \begin{equation} \label{imageA011} \bigoplus_{i=1}^N \frac{ \hcC_1^i \tilde{\theta}_i^0 \theta_i^2 [\lambda]}{(\lambda-u^i)}, \end{equation} where $\hcC_1^i$ is the subspace of $\hcC$ generated by monomials that do not depend on $\theta_i^1$. Second, it is sufficient to consider the fact that the image of the ideal \hbox{$\prod_{j\not=i} (-\lambda+u^j) \hcC[\lambda]$} under $\Delta_{0}^{1,1}$ is\vspace*{-3pt} \begin{equation*} \frac{\hcC_1^i \tilde{\theta}_i^0 \theta_i^2 [\lambda]}{(\lambda-u^i)} \end{equation*} to conclude that the image of $\Delta_0^{1,1}$ is the whole space~\eqref{imageA011}. So, the cohomology of $\Delta_0^{1,1}$ on the second term in~\eqref{firstpage} is\vspace*{-3pt} \begin{equation*} \bigoplus_{i=1}^N \frac{ \hcC_1^i \tilde{\theta}_i^0 \theta_i^1 \theta_i^2 [\lambda]}{(\lambda-u^i)}. \end{equation*} In particular, we see that it cannot give any contribution to the cohomology of degree $(p,d)=(3,2)$. The second page of the spectral sequence associated to $\deg_{\theta^{\geq2}}$ is\vspace*{-3pt} \begin{equation} \label{cohA01} \ker \Delta_0^{1,1}|_{\hcC[\lambda]} \oplus \bigoplus_{i} \frac{ \hcC_1^i \tilde{\theta}_i^0 \theta_i^1 \theta_i^2 [\lambda]}{(\lambda-u^i)} \oplus \bigoplus_{k\geq 2} \bigoplus_{i} H\bigl(H(\hcC \llbracket \theta^{\geq 2}_i \rrbracket^{(k)}, \Delta_0^{1,0}), \Delta_0^{1,1}\bigr), \end{equation} where, as discussed before, the third summand does not give any contribution to $\tensor*[^1]{E}{_2}$, and can therefore be ignored here. Since $\Delta_0^1$ vanishes on this page, Equation \eqref{cohA01} gives the cohomology of $(\tensor*[^1]{E}{_1^0}, \Delta_0^1)$ which coincides with the first page $\tensor[^4]{E}{_1}$ of the spectral sequence $\tensor*[^4]{E}{}$. The differential $\,^4d_1$ on $\,^4E_1$ is the one induced by $\Delta_0^0$, the degree $\deg_{\theta^0}$ zero part of $\Delta_0$. The three summands in Equation \eqref{cohA01} are invariant under the action of the differential $\Delta_0^0$, which in particular vanishes on the second term. To see this, observe that since the standard degree of the second term is $d=3$ and that of the third term is $d\geq5$, there can be no terms mapped between these two spaces by $\Delta_0^0$, nor from the second space to itself. The third term cannot map to the first one, since $\Delta_0^0$ cannot remove more than one $\theta^{\geq2}$. The operator $\Delta_0^0$ has to increase the standard degree and the $\theta$-degree by one, while keeping $\deg_{\theta^0}$ unchanged. This can only be achieved on $\hcC[\lambda]$ by increasing $\deg_{\theta^1}$ by one, implying $\Delta_0^0 = \Delta_{0,1}$, which is given in Lemma \ref{lemmaD01}. Explicitly: \begin{multline*} \Delta_{0}^0 = (u^i - \l)f^i \th_i^1 \Bigl(\frac{\d}{\d u^i} - (\d_i \log f^k) \th_k^1\, \frac{\d}{\d \th_k^1} - \frac{1}{2} (\d_i \log f^k) \th_k^0\, \frac{\d}{\d \th_k^0} \Bigr) \\ - \frac{1}{2} (u^j-\l) \d_i f^j \th_j^1 \th_j^0\, \frac{\d}{\d \th_i^0} + \frac{1}{2} f^i \tilde{\theta_i^0} \th_i^1\, \frac{\d}{\d \th_i^0} + (u^i - \l) f^j\, \frac{\d_j f^i}{f^i}\, \th_j^0 \th_i^1\, \frac{\d}{\d \th_i^0}. \end{multline*} From this formula it is easy to see that $\Delta_0^0$ maps $\hcC[\lambda]$ to itself. Finally, from the formula for $\Delta_0$, we easily see that there are no terms that remove the dependence on~$\theta^2_i$ in the second summand in Equation \eqref{cohA01}, therefore such summand cannot map to the first. To get the second page $\,^4E_2$ we therefore need to compute the cohomology of the differential $\Delta_0^0$ on $\ker \Delta_0^{1,1}|_{\hcC[\lambda]} $. The discussion so far was for general bidegrees $(p,d)$. However to be able to say something more we need to restrict to the subcomplex $p=d+1$. We see that an element proportional to $\theta^1_i$ is in the kernel of $\Delta_0^{1,1}$ if and only if it is also proportional either to $(-\lambda+u^i)$ or to $\tilde\theta^0_i$. Therefore, it can be represented as a sum over all subsets $I\subset \{1,\dots,n\}$, $|I|=t$, of the elements of the form $$ \sum_{j=1}^n F^j (u,\lambda) \theta^0_j \cdot \prod_{i\in I} (-\lambda+u^i) \theta^1_i + \sum_{i\in I} G^i(u) \tilde \theta^0_i \theta^1_i \cdot \prod_{\substack{j\in I \\ j\not=i}} (-\lambda+u^j) \theta^1_j. $$ This representation naturally splits the kernel of $\Delta_0^{1,1}$ into two summands, let us call them $F$ and $G$. Observe that the splitting of the $p=d+1$ part of the kernel of $\Delta_0^{1,1}$ on $\hcC[\lambda]$ into the direct sum $F\oplus G$ defines a filtration for the operator $\Delta_0^0 = \Delta_{0,1}$. We can see this by using the base change \(\Psi \). First, define \begin{equation*} \bar{\th}_i^0 \coloneqq \Psi^{-1} \tilde{\th}_i^0 = \th_i^0 +2 (u^j - u^i) \gamma_{ji} \th_j^0. \end{equation*} From the formula above for $\Delta_0^0$ we have that we can write \(\Delta_0^0 = \Psi \bar{\Delta} \Psi^{-1} \), for \begin{align*} \bar{\Delta} &= (u^i - \l) \th_i^1 \frac{\d}{\d u^i} + (u^i - \l) \gamma_{ji} \th_i^1 \th_i^0 \frac{\d}{\d \th_j^0} - (u^i -\l) \gamma_{ji} \th_i^1 \th_j^0 \frac{\d}{\d \th_i^0} + \frac{1}{2} \bar{\th}_i^0 \th_i^1 \frac{\d}{\d \th_i^0}. \end{align*} The first three terms preserve \(F = \Psi^{-1}F\), while the last sends \(F \) to \(\bar{G} \coloneqq \Psi^{-1}G\). Moreover, the entire operator preserves \(\bar{G} \). Furthermore, the parts \(F \to F \) and \(\bar{G} \to \bar{G} \) form deformed de\,Rham differentials \(d + A \). Therefore, the only possible cohomology is in the lowest degree in $\theta^1_\sbullet$, which is zero for $F$ and $1$ for $G$. So, only nontrivial cohomology in the case $p=d+1$ is possible in the degree $(t+1,t)=(1,0)$ and $(t+1,t)=(2,1)$. This implies the the cohomology of degree $(3,2)$ is equal to zero. \end{proof} \begin{remark}\label{DeltapresG} Note that it is not directly clear from the definitions that $\bar{\Delta} \bar{G} \subset \bar{G}$. However, we know that \(\bar{\Delta} \) must preserve the kernel of $\Delta_0^{1,1}$ twisted by \(\Psi \), which is \(F \oplus \bar{G} \). Moreover, looking at the \(\lambda \)-degree, we see that for elements of \(\bar{G}\) it is one less \(\deg_{\th^1} \) while for elements of \(F\) it is at least \(\deg_{\th^1} \). As \(\deg_{\th^1} \bar{\Delta} =1 \), and none of its terms increase the \(\lambda\)-degree by more than \(1\), this proves that \(\bar{\Delta} \) cannot map \(\bar{G} \) outside of \(\bar{G} \). A more direct proof requires Ferapontov's flatness equations for $f^i$ \cite{Ferapontov01}. We give this calculation in the appendix. \end{remark} \begin{remark} In the proof, we restricted to \(p = d+1\). In order to extend the argument, one would have to show that the transformation \(\th_i^0 \mto \bar{\th}_i^0 \) is invertible. This would allow for a splitting similar to the splitting in \(F\) and \(G\) here. \end{remark} \section{Proofs of the main theorems} \label{cohomologyA} In this section we collect all results from the rest of the paper to compute the cohomology of the complex $(\hcA[\lambda], D_\lambda)$, proving Theorems \ref{resultp=d} and \ref{extravanishing}. \begin{proof}[Proof of Theorem \ref{resultp=d}] As observed in Section \ref{D01}, the first page $\tensor*[^2]{E}{_1}$ is given by the direct sum~\eqref{2E1}. From Lemma \ref{lem13} and Proposition \ref{pro14} we get \begin{equation*} (\tensor*[^2]{E}{_1})^p_p \cong \begin{cases} \R [\l ] & p = 0,\\ \bigoplus_{i=1}^N C^\infty (u^i) \th_i^1 & p = 1,\\ \bigoplus_{i=1}^N C^\infty (u^i) \th_i^0 \th_i^2 & p =2,\\ \bigoplus_{i=1}^N C^\infty (u^i) \th_i^0 \th_i^1 \th_i^2 & p = 3,\\ 0 & \text{else}. \end{cases} \end{equation*} On this first page, the differential $\tensor*[^2]{d}{_1}$ must lower the spectral sequence degree $\deg_{\theta^1} - \deg_{\theta}$ by one, in other words, since the differential must still be of bidegree $(1,1)$, it must leave the degree $\deg_{\theta^1}$ unchanged, which is impossible on this subcomplex. Hence, the differential \(\tensor*[^2]{d}{_1} \) is equal to zero, and \((\tensor*[^2]{E}{_2})_p^p \cong (\tensor*[^2]{E}{_1})_p^p \). On the second page, the differential $\tensor*[^2]{d}{_2}$ must lower the spectral sequence degree by two, \ie it must be of degree $\deg_{\theta^1}$ equal to $-1$. Therefore, on this subcomplex the differential can only be non-trivial between \(p =1\) and \(p =2\). Looking back at the formula for \(\Delta_0 \), one can easily identify the terms of degree $\deg_{\theta_1} = -1$, which give \begin{align*} \Delta_{0,-1}= \sum_{i} \frac{1}{2} \Big[ \sum_j(u^j-\l) \big(\d_i f^j \th_j^0 \th_j^2 +f^j \frac{\d_j f^i}{f^i} (\th_i^0 \th_j^2 - \th_j^0 \th_i^2) \big)+ f^i \th_i^0 \th_i^2 \Big] \frac{\d}{\d \th_i^1}. \end{align*} $\Delta_{0,-1}$ induces an operator on $H(\hcA[\lambda], \Delta_{-1})$. Since we are interested only in the differential at degree $p=1$, we need to consider just the action of such operator on $\hcC[\lambda]$, which is, taking into account the identification~\eqref{ide} \begin{equation*} \sum_i \frac12 \Big[ (u^i-u^j) \frac{f^j}{f^i} \partial_j f^i \theta^0_j \theta^2_i + f^i \theta^0_i \theta^2_i \Big] \frac{\d}{\d \th_i^1}. \end{equation*} The image of $\hcC[\lambda]$ through this operator is thus in $\bigoplus_i H(\hat{d}_i(\hcC_i), \cD_i)$, where the first term, being in $\hcC^i_{0,1} \theta_i^2$ vanishes. Hence, the only surviving term is \(\frac{1}{2} f^i \th_i^0 \th_i^2 \sfrac{\d}{\d \th_i^1}\), which gives an isomorphism \(\tensor*[^2]{d}{_2} : (\tensor*[^2]{E}{_2})_1^1 \to (\tensor*[^2]{E}{_2})_2^2 \). The differential is therefore zero on $(\tensor*[^2]{E}{_2})^p_p$ for $p\not=1$ and an isomorphism for $p=1$, so \((\tensor*[^2]{E}{_3})_p^p \) is zero unless $p=0$ or $p=3$, when it is equal to $(\tensor*[^2]{E}{_2})_p^p$. This spectral sequence has no other non-trivial differentials, so \((\tensor*[^2]{E}{_\infty})_p^p \) has the same form. As \(\tensor*[^2]{E}{} \implies\tensor*[^1]{E}{_2} \), we get that \((\tensor*[^1]{E}{_2})_p^p \) is of this form as well. Because all differentials must have \((p,d)\)-bidegree \((1,1)\), there can be no higher non-trivial differentials on this part of the first spectral sequence. Now, \(\tensor*[^1]{E}{} \implies H(\hcA [\l ], D_\l)\), yielding the result. \end{proof} \begin{proof}[Proof of Theorem \ref{extravanishing}] We take Theorem \ref{vanth} as a starting point. Then the extra vanishing at degrees $d=N, N+1$ follows from Lemma \ref{lem13}, and the vanishing at $(3,2)$ follows from Proposition \ref{prop32}. \end{proof} \renewcommand{\thesection}{\Alph{section}} \setcounter{section}{0} \section*{Appendix. Formula for and calculations with \texorpdfstring{$\Delta_0$}{D0}} \refstepcounter{section}\label{sec:D0} We recall from~\cite{cps15} the formula for the degree $\deg_u$ zero part of \hbox{the operator\,$D_\lambda$}.\vspace*{-3pt} \begin{align*} \Delta_0 & = (-\lambda+u^i)f^i\theta_i^{1}\frac{\d}{\d u^i}\\[-3pt] & + \! \! \sum_{\substack{ s=a+b \\ s, a \geq 1; b\geq 0 }} \! \! (-\lambda+u^i) \binom {s}{b} \partial_j f^i u^{j,a} \theta_i^{1+b} \frac{\d}{\d u^{i,s}} + \! \! \sum_{\substack{ s=a+b \\ s, a \geq 1; b\geq 0 }} \! \! \binom {s}{b} f^i u^{i,a} \theta_i^{1+b} \frac{\d}{\d u^{i,s}} \\[-3pt] & + \! \! \frac{1}{2}\sum_{\substack{ s=a+b \\ s \geq 1; a,b\geq 0 }} \! \! (-\lambda+u^i) \binom{s}{b} \d_jf^i u^{j,1+a} \theta_i^b \frac{\d}{\d u^{i,s}} +\frac{1}{2} \! \! \sum_{\substack{ s=a+b \\ s \geq 1; a,b\geq 0 }} \! \! \binom{s}{b} f^i u^{i,1+a} \theta_i^b \frac{\d}{\d u^{i,s}} \\[-3pt] & +\frac{1}{2} \! \! \sum_{\substack{ s=a+b \\ s \geq 1; a,b\geq 0 }} \! \! (-\lambda+u^i) \binom{s}{b} f^i\frac{\d_i f^j}{f^j} u^{j,1+a}\theta_j^b \frac{\d}{\d u^{i,s}} +\frac{1}{2} \! \! \sum_{\substack{ s=a+b \\ s \geq 1; a,b\geq 0 }} \! \! \binom{s}{b} f^i u^{i,1+a}\theta_i^b \frac{\d}{\d u^{i,s}} \\[-3pt] & -\frac{1}{2} \! \! \sum_{\substack{ s=a+b \\ s \geq 1; a,b\geq 0 }} \! \! (-\lambda+u^j) \binom{s}{b} f^j\frac{\d_j f^i}{f^i} u^{i,1+a}\theta_j^b \frac{\d}{\d u^{i,s}} -\frac{1}{2} \! \! \sum_{\substack{ s=a+b \\ s \geq 1; a,b\geq 0 }} \! \! \binom{s}{b} f^i u^{i,1+a}\theta_i^b \frac{\d}{\d u^{i,s}} \\[-3pt] & + \frac{1}{2} \sum_{\substack{ s=a+b \\ s,a,b\geq 0 }} (-\lambda+u^j)\binom{s}{b} \d_if^j \theta_j^a \theta_j^{1+b} \frac{\d}{\d \theta_i^s} + \frac{1}{2} \sum_{\substack{ s=a+b \\ s,a,b\geq 0 }} \binom{s}{b} f^i \theta_i^a \theta_i^{1+b} \frac{\d}{\d \theta_i^s} \\[-3pt] & + \frac{1}{2} \sum_{\substack{ s=a+b \\ s,a,b\geq 0 }} (-\lambda+u^j)\binom{s}{b} f^j\frac{\d_j f^i}{f^i} \theta_i^a \theta_j^{1+b} \frac{\d}{\d \theta_i^s} +\frac{1}{2} \sum_{\substack{ s=a+b \\ s,a,b\geq 0 }} \binom{s}{b} f^i \theta_i^a \theta_i^{1+b} \frac{\d}{\d \theta_i^s} \\[-3pt] & - \frac{1}{2} \sum_{\substack{ s=a+b \\ s,a,b\geq 0 }} (-\lambda+u^j)\binom{s}{b} f^j\frac{\d_j f^i}{f^i} \theta_j^a \theta_i^{1+b} \frac{\d}{\d \theta_i^s} - \frac{1}{2} \sum_{\substack{ s=a+b \\ s,a,b\geq 0 }} \binom{s}{b} f^i \theta_i^a \theta_i^{1+b} \frac{\d}{\d \theta_i^s}. \end{align*} The direct proof that \(\bar{\Delta} \bar{G} \subset \bar{G} \) in Proposition \ref{prop32} is given below. Recall that its validity is deduced more abstractly in Remark \ref{DeltapresG} as well. \begin{lemma} The operator \(\bar{\Delta} \) preserves \(\bar{G} \), where\vspace*{-3pt} \begin{align*} \bar{\Delta} &= (u^i - \l) \th_i^1 \frac{\d}{\d u^i} + (u^i - \l) \gamma_{ji} \th_i^1 \th_i^0 \frac{\d}{\d \th_j^0} - (u^i -\l) \gamma_{ji} \th_i^1 \th_j^0 \frac{\d}{\d \th_i^0} + \frac{1}{2} \bar{\th}_i^0 \th_i^1 \frac{\d}{\d \th_i^0}\\[-3pt] \tag*{and} \bar{G} &= \bigoplus_{i=1}^N C^\infty (U) \Big[ \big\{ (u^j -\l) \th_j^1\big\}_{j=1}^N \Big] \bar{\th}_i^0 \th_i^1. \end{align*} \end{lemma} \begin{proof} When calculating the action of \(\bar{\Delta} \) on an element of the form \(G(u) \bar{\th}^0_i \th^1_i \in \bar{G}\), we get the following (where \(i\) is a fixed index, and \(k\), \(l\), and \(m\) are summed over)\vspace*{-3pt} \begin{align*} \bar{\Delta}G(u) \bar{\th}^0_i \th^1_i &= \frac{\d}{\d u^k} \big(G (\th_i^0 + 2(u^l - u^i) \gamma_{li} \th_l^0)\big) \th_i^1 (u^k-\l)\th_k^1 \\ & \quad + G \gamma_{mk} \th_k^0 \frac{\d}{\d \th_m^0} \big(\th_i^0 + 2 (u^l - u^i)\gamma_{li} \th_l^0\big) \th_i^1 (u^k-\l)\th_k^1\\ &\quad - G \gamma_{lk} \th_l^0 \frac{\d}{\d \th_k^0} (\th_i^0 + 2(u^m - u^i) \gamma_{mi} \th_l^0) \th^1_i (u^k -\l) \th_k^1\\ & \quad +\frac{1}{2} G \frac{\d}{\d \th_k^0}\big(\th_i^0 + 2(u^l - u^i) \gamma_{li} \th_l^0 \big) \th_i^1 \bar{\th}_k^0 \th_k^1 \end{align*} \begin{align*} \phantom{\bar{\Delta}G(u) \bar{\th}^0_i \th^1_i}&= \frac{\d G}{\d u^k} \bar{\th}_i^0 \th_i^1 (u^k - \l) \th_k^1 + 2 G \gamma_{ki} \th_k^0 \th_i^1 (u^k - \l)\th_k^1\\ &\quad + 2 G (u^l - u^i) \d_k \gamma_{li} \th_l^0 \th_i^1(u^k-\l)\th_k^1 + G\gamma_{ik} \th_k^0 \th_i^1 (u^k -\l)\th_k^1 \\ &\quad + 2G (u^l-u^i) \gamma_{lk} \gamma_{li} \th_k^0 \th_i^1 (u^k -\l)\th_k^1 \\ &\quad- 2 G (u^k-u^i) \gamma_{lk} \gamma_{ki} \th_l^0 \th_i^1 (u^k-\l)\th_k^1 + G (u^k- u^i) \gamma_{ki} \th_i^1 \bar{\th}_k^0 \th_k^1 \end{align*} Using Equation \eqref{eq:dgammaijk} for the third term if \(i,k,l\) distinct, that part of the third term adds up to the sixth term. \begin{align*} \bar{\Delta}G(u) \bar{\th}^0_i \th^1_i &= \frac{\d G}{\d u^k} \bar{\th}_i^0 \th_i^1 (u^k - \l) \th_k^1 + 2 G \gamma_{ki} \th_k^0 \th_i^1 (u^k - \l)\th_k^1\\ &\quad + 2 G (u^k - u^i) \d_k \gamma_{ki} \th_k^0 \th_i^1(u^k-\l)\th_k^1 + G\gamma_{ik} \th_k^0 \th_i^1 (u^k -\l)\th_k^1 \\ &\quad + 2 G (u^l-u^i) \gamma_{lk} \gamma_{li} \th_k^0 \th_i^1 (u^k -\l)\th_k^1 + G (u^i-\l) \gamma_{ki} \bar{\th}_k^0 \th_i^1 \th_k^1\\ &\quad + 2 G (u^l-u^k) \gamma_{lk} \gamma_{ki} \th_l^0 \th_i^1 (u^k-\l)\th_k^1 - G \gamma_{ki} \bar{\th}_k^0 \th_i^1 (u^k- \l) \th_k^1 \end{align*} By the definition of \(\bar{\th}_k^0 \), the last two terms drop out against half of the second term. So we get \begin{align*} \bar{\Delta}G(u) \bar{\th}^0_i \th^1_i &= \frac{\d G}{\d u^k} \bar{\th}_i^0 \th_i^1 (u^k - \l) \th_k^1 + G (\gamma_{ik} + \gamma_{ki}) \th_k^0 \th_i^1 (u^k - \l)\th_k^1\\ &\quad + 2 G (u^k - u^i) \d_k \gamma_{ki} \th_k^0 \th_i^1(u^k-\l)\th_k^1 \\ &\quad + 2 G (u^l-u^i) \gamma_{lk} \gamma_{li} \th_k^0 \th_i^1 (u^k -\l)\th_k^1 + G (u^i-\l) \gamma_{ki} \bar{\th}_k^0 \th_i^1 \th_k^1 \end{align*} By Equation \eqref{udgamma}, we get \begin{align*} \bar{\Delta}G(u) \bar{\th}^0_i \th^1_i &= \frac{\d G}{\d u^k} \bar{\th}_i^0 \th_i^1 (u^k - \l) \th_k^1 - G u^i\d_i \gamma_{ik} \th_k^0 \th_i^1 (u^k - \l)\th_k^1\\ &\quad - 2 G u^i \d_k \gamma_{ki} \th_k^0 \th_i^1(u^k-\l)\th_k^1 \\ &\quad - 2 G u^i \gamma_{lk} \gamma_{li} \th_k^0 \th_i^1 (u^k -\l)\th_k^1 - G \gamma_{ki} (u^i-\l) \th_i^1 \bar{\th}_k^0 \th_k^1 \end{align*} Applying Equation \eqref{dgammaij} gives \begin{align*} \bar{\Delta}G(u) \bar{\th}^0_i \th^1_i &= \frac{\d G}{\d u^k} \bar{\th}_i^0 \th_i^1 (u^k - \l) \th_k^1 - G \gamma_{ki} (u^i-\l) \th_i^1 \bar{\th}_k^0 \th_k^1 \end{align*} Multiplying with a factor \(\prod_{j \in I} (u^j - \l) \th_j^1 \) does not change the calculation, so we can extend this calculation to all of \(\bar{G}\), showing that \(\bar{\Delta} \) does indeed preserve this space. \end{proof} \backmatter \bibliographystyle{jepalpha} \bibliography{carlet-et-al} \end{document}