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I am teaching a course on Hodge theory and I realised that I don't understand something basic.

Let first $X$ be a compact Kahler manifold. Let $H^{p,q}(X)=H^q(X,\Omega^p_X)$ where $\Omega^p_X$ is the sheaf of holomorphic $p$-forms. Let also $\Omega^*_X$ denote the holomorphic de Rham complex. Then there two statements:

1) The spectral sequence coming from the stupid filtration on $\Omega^*_X$ degenerates at $E_1$. This is equivalent to saying that $H^*(X,\mathbb{C})$ has a filtration such that the associated graded space is canonically identified with the direct sum of $H^{p,q}(X)$.

2) $H^k(X,\mathbb{C})$ is canonically isomorphic to the direct sum of $H^{p,q}(X)$ with $p+q=k$.

The 2nd statement implies the 1st (since for degeneration of the spectral sequence it is enough to check that the two spaces have the same dimension). I want to understand to what extent the 1st statement implies the 2nd. In fact, the theory of harmonic forms implies that $H^{p,q}(X,\mathbb{C})$ is equal to $F^p(H^{p+q}( X,\mathbb{C}))\cap \overline{F^q(H^{p+q}( X,\mathbb{C}))}$ where $F^*$ stands for the Hodge filtration (induced by the stupid filtration on the de Rham complex). My question is whether it is possible to prove this fact without using harmonic forms if we already know the degeneration of the spectral sequence? For example, assume that $X$ is a complete smooth algebraic variety over $\mathbb{C}$. Then Deligne and Illusie give an algebraic proof of the degeneration of the spectral sequence using reduction to characteristic $p$. Is it true that the Hodge decomposition holds in this case as well? How to prove this? (I know that if $X$ is not necessarily Kahler then it might happen that the spectral sequence degenerates but the Hodge decomposition fails, but these examples are not algebraic, so the question still makes sense). Basically, I want to understand whether it is possible to prove the existence of Hodge decomposition without ever using the Dolbeaut complex (or any explicit resolution of the holomorphic or algebraic de Rham complex)?

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  • $\begingroup$ My understanding is that the theory of variations of Hodge structures shows that we should not expect the Hodge decomposition to be too canonical (algebraically), otherwise we could do it in families. This suggests at least that there should be some non-algebraic input (but maybe throwing in the complex conjugation is already enough). I would be interested in a more complete and coherent answer. $\endgroup$ – R. van Dobben de Bruyn Sep 29 '18 at 19:20
  • $\begingroup$ Well, the problem is really with the proof, not with the statement - at least for Kahler manifolds you can construct the decomposition intersecting various terms of the Hodge filtration with their complex conjugates. I would like to see a proof of this statement which doesn't use harmonic forms. $\endgroup$ – Alexander Braverman Sep 29 '18 at 19:26
  • $\begingroup$ Take an elliptic curve $E$ with regular differential 1-form $\omega$. Then the statement of Hodge theory basically says that the class of $\omega$ is not a multiple of the class of $\bar\omega$. Equivalently, the integral $\int_E \omega\wedge\bar\omega$ is not zero. Is there any way to prove that this integral is not zero besides showing that (up to a simple factor) it is strictly positive because it is the volume of the fundamental domain? $\endgroup$ – Anton Mellit Jan 4 at 12:28
  • $\begingroup$ By the way, if you are satisfied with this argument for elliptic curves, then you can probably upgrade it to arbitrary projective varieties using hard Lefschetz. $\endgroup$ – Anton Mellit Jan 4 at 12:36
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I am far from being an expert in this area, but I will try to present my understanding of this subject.

First of all, this is true that Hodge decomposition holds for smooth proper varieties over $\mathbf C$. However, the standard proof (that is explained below) uses the projective case as a black box.

Secondly, degeneration of Hodge-to-de Rham spectral sequence does imply that there is a filtration on $H^{n}(X,\mathbf C)$ s.t. the associated graded pieces are isomorphic to $H^{p,q}(X)$. In particular, it means that there is an abstract isomorphism of $\mathbf C$ vector spaces $H^n(X,\mathbf C)\cong \oplus_{p+q=n}H^q(X,\Omega^p_{X/\mathbf C})$. But I don't know any way to pick a canonical one just assuming degeneration of the spectral sequence.

Moreover, there is an example of a smooth compact non-Kahler complex manifold X, s.t. the associated Hodge-to-de Rham spectral sequence degenerates but the the Hodge filtration on the de Rham cohomology $H^n_{dR}(X)$ doesn't define a pure Hodge structure. An explicit situation of such a manifold is a Hopf surface $X=(\mathbf C^2 - \{0,0\})/q^{\mathbf Z}$, where $q\in \mathbf C^*$ and $|q|<1$ (action is diagonal). Then one can prove that Hodge-to-de Rham spectral sequence degenerates for $X$ but $H^1(X,\mathcal O_X)=0\neq \mathbf C=H^0(X,\Omega^1_{X/\mathbf C})$. Hence the Hodge filtration doesn't define a pure Hodge structure on $H^1_{dR}(X)$ because of the failure of hodge symmetry. Although, it doesn't imply that there is no canonical isomorphism $H^n(X,\mathbf C)\cong \oplus_{p+q=n}H^q(X,\Omega^p_{X/\mathbf C})$, it shows that it's unlikely that the 2nd statement can be a consequence of the first.

All of that being said, I will explain how to construct pure Hodge structures on cohomology of any proper smooth variety.

Let us start with $X$ being a smooth algebraic variety over $\mathbf C$. Then one can define a descending filtration on the algebraic de Rham complex $\Omega^{\bullet}_{X/\mathbf C}$ as $F^i\Omega^{\bullet}_{X/\mathbf C}=\Omega_{X/\mathbf C}^{\geq i}$. This induces a descending filtration $F^{\bullet}H^n_{dR}(X)=F^{\bullet}H^n(X,\mathbf C)$. The main result is that this construction defines a pure Hodge structure under the properness assumption on X.

Theorem 1: Let $X$ be a smooth and proper variety over $\mathbf C$, then a pair $(H^n(X,\mathbf Z), F^{\bullet}H^n(X,\mathbf C))$ is a pure Hodge structure of weight n. Moreover we have isomorphisms $F^{p}H^n(X,\mathbf C) \cap \overline{F^{n-p}H^n(X,\mathbf C)}\cong H^{q}(X,\Omega^p_{X/\mathbf C})$.

Remark 1: General theory of pure Hodge structures implies that there is a canonical isomorphism $H^n(X,\mathbf C)\cong \oplus_{p+q=n}H^{q}(X,\Omega^p_{X/\mathbf C})$.

Remark 2: Although we have a canonical decomposition $H^n(X,\mathbf C)\cong \oplus_{p+q=n}H^{q}(X,\Omega^p_{X/\mathbf C})$, a priori there is no morphism $H^q(X,\Omega^p_{X/\mathbf C}) \to H^n(X,\mathbf C)$. I don't know any purely algebraic construction of these maps, which doesn't use the fact that Hodge filtration defines a pure Hodge structure as an input. I believe this is the crucial place where theory of harmonic forms comes into play. Once you have such morphisms, Theorem $1$ is basically equivalent to the degeneration of the Hodge-to-de Rham spectral sequence.

Remark 3: I think (I may be wrong) that there is no purely algebraic proof of Theorem $1$ as well. However, what I will show below is that if one takes Theorem 1 in the projective case as an input (this is the situation where $X^{an}$ is a Kahler manifold, so one can use theory of harmonic forms), then he/she can deduce the general form of Theorem $1$ in a purely algebraic way.

Before going to explain this proof, I need to mention a lemma which eventually be the key step to reduce a case of a proper variety to the case of a projective one.

Lemma 1: Let $f:X \to Y$ be a proper birational morphism between smooth connected varieties over a field $k$, then the induced morphism $H^p(f^*):H^p(Y,\Omega^q_{Y/k}) \to H^p(X,\Omega^q_{X/k})$ is injective for any $p$ and $q$.

Proof: Recall that $H^p(f^*)$ is induced by a morphism $$ \Omega^q_{Y/k} \to \mathbf Rf_*f^*\Omega^q_{Y/k} \to \mathbf Rf_*\Omega^q_{X/k}. $$ Compose it with the Trace map $$ \operatorname{Tr_f}:\mathbf Rf_*\Omega^q_{X/k} \to \Omega^q_{Y/k}. $$ (Here we use that $\operatorname{dim}X =\operatorname{dim}Y$ because in general the Trace map is map from $:\mathbf Rf_*\Omega^q_{X/k}$ to $\Omega^{q-d}_{Y/k}[-d]$ where $d=\operatorname{dim}X -\operatorname{dim}Y$).

Now note that the formation of Trace map is compatible with restrictions to open subsets. Hence, the composition $Tr_f\circ f^*:\Omega^q_{Y/k} \to \Omega^q_{Y/k}$ is the identity morphism on an open dense subset $U\subset Y$ (because $f$ is birational and Y is normal!). Since $\Omega^q_{Y/k}$ is a locally free sheaf, we conclude that $Tr_f\circ f^*$ is also the identity morphism. Therefore, $H^p(Tr_f\circ f^*)=H^p(Tr_f)\circ H^p(f^*)$ is the identity morphism as well. So $H^p(f^*)$ is injective as desired.

Now let's come back to the proof of Theorem 1 in the proper case.

Step 1: Chow lemma and Hironaka's resolution of singularities allow us to find a smooth and projective variety $X'$ together with a map $f:X' \to X$ s.t. $f$ is birational and proper. Then there is a canonical morphisms of Hodge-to-de Rham spectral sequences for $X$ and $X'$. Namely, let $$ E_1^{p,q}=H^q(X,\Omega^p_{X/\mathbf C}) \Rightarrow H^{p+q}_{dR}(X). $$ $$ \downarrow{f^*} $$ $$ E_1'^{p,q}=H^q(X',\Omega^p_{X'/\mathbf C}) \Rightarrow H^{p+q}_{dR}(X'). $$

Note that Lemma $1$ guarantees that $f^*$ is injective on the first page. Since we know that $E'$ degenerates on the first page ($d_1'^{p,q}=0$ for any $p,q$) we conclude that $d_1^{p,q}=0$ (because $f^*$ is compatible with differentials and injective). So $E_2^{p,q}=E_1^{p,q}$ and $E_2'^{p,q}=E_1'^{p,q}$, therefore $f^*$ is also injective on the second page. Then the same argument shows that $d_2^{p,q}=0$ for any $p,q$. Keep going to show that $E$ degenerates on the first page ($E_1^{p,q}=E_{\infty}^{p,q}$).

Step 2: Two consequences from the Step 1.

I) $F^pH^n_{dR}(X)\cap \overline{F^{n-p+1}H^n_{dR}(X)}=0$. We know that that his equality holds for $X'$ (b/c $X'$ is projective) and we know that Hodge filtration $F^{\bullet}H^n_{dR}(-)$ is functorial in $X$. Hence, argument in the Step $1$ says that $$ F^pH^n_{dR}(X)\cap \overline{F^{n-p+1}H^n_{dR}(X)} \subset F^pH^n_{dR}(X')\cap \overline{F^{n-p+1}H^n_{dR}(X')}=0. $$ II) We have a canonical isomorphism $F^pH^n_{dR}(X)/F^{p+1}H^n_{dR}(X)\cong H^{q}(X,\Omega^p_{X/\mathbf C})$. This is a direct consequence of the degeneration of the Hodge-to-de Rham spectral sequence.

Step 3: The last thing we need to check is that $F^pH^n_{dR}(X)+\overline{F^{n-p+1}H^n_{dR}(X)}=H^n_{dR}(X)$.

Since $F^pH^n_{dR}(X)$ and $\overline{F^{n-p+1}H^n_{dR}(X)}$ are disjoint inside $H^n_{dR}(X)$, it suffices to prove the equality $$ \operatorname{dim}(F^pH^n_{dR}(X))+\operatorname{dim}(\overline{F^{n-p+1}H^n_{dR}(X)})=\operatorname{dim}(H^n_{dR}(X)). $$ Ok, Consequence II from Step $2$ implies that $$ \operatorname{dim}(F^pH^n_{dR}(X))=\sum_{i\geq p}h^{i,n-i}, $$ $$\operatorname{dim}(\overline{F^{n-p+1}H^n_{dR}(X)} = \sum_{i\geq n-p+1}h^{i,n-i}, $$ $$ \operatorname{dim}(H^n_{dR}(X))=\sum_{i\geq 0}h^{i,n-i} \text{, where } h^{p,q}= \operatorname{dim}H^q(X,\Omega^p_{X/\mathbf C}). $$

Since we know that $F^pH^n_{dR}(X)$ and $\overline{F^{n-p+1}}H^n_{dR}(X)$ are disjoint we conclude that $$ \sum_{i\geq p}h^{i,n-i} +\sum_{i\geq n-p+1}h^{i,n-i} \leq \sum_{i\geq 0}h^{i,n-i}. $$ Subtract $\sum_{i\geq n-p+1}h^{i,n-i}$ to obtain an inequality $$ \sum_{i\geq p}h^{i,n-i} \leq \sum_{i\leq n-p}h^{i,n-i} (*) $$ Our goal is to prove that this is actually an equality. Here comes the miracle! We use an additional symmetry between hodge numbers of interchange left and right hand sides of this inequality to actually show that it is an equality! Namely, we apply Serre duality to see that $h^{p,q}=h^{d-p,d-q}$ where $d=\operatorname{dim} X$. Then we can rewrite both hand sides of $(*)$ in a different way. $$ \sum_{i\geq p}h^{i,n-i}=\sum_{i\geq p}h^{d-i,d-n+i}=\sum_{j\leq d-p=(2d-n)-(d-n+p)} h^{j,(2d-n)-j}, $$ $$ \sum_{i\leq n-p}h^{i,n-i}=\sum_{i\leq n-p}h^{d-i,d-n+i}=\sum_{j\geq d-n+p}h^{j,2d-n-j}. $$

But now observe that inequality (*) (applied to $n'=2d-n,p'=N-n+p$) guarantees that $$ \sum_{i\leq n-p}h^{i,n-i}=\sum_{j\geq d-n+p}h^{j,2d-n-j} \leq \sum_{j\geq d-n+p}h^{j,2d-n-j} =\sum_{i\leq n-p}h^{i,n-i}. $$

Therefore $\sum_{i\geq p}h^{i,n-i}=\sum_{i\leq n-p}h^{i,n-i}$! And we showed that this is equivalent to the equality $$ F^pH^n_{dR}(X)+\overline{F^{n-p+1}H^n_{dR}(X)}=H^n_{dR}(X). $$ In other words we showed that $(H^n(X,\mathbf Z), F^{\bullet}H^n(X,\mathbf C))$ is a pure Hodge structure of weight $d$.

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  • $\begingroup$ Thank you. The first part (what is written before the Theorem) is more or less what I wrote originally (in particular the Hopf surface is precisely the example that I meant when I wrote "If $X$ is not necessarily Kahler then it might happen that the spectral sequence degenerates but the Hodge decomposition fails). So, it is clear that the degeneration of the spectral sequence alone doesn't imply Hodge decomposition, but the question is whether it is possible to prove the latter in reasonable generality without using harmonic forms. I mentioned non-projective varieties just as an example. $\endgroup$ – Alexander Braverman Sep 30 '18 at 2:01
  • $\begingroup$ @AlexanderBraverman Sorry, I partially misunderstood your question. I thought that your main question is "to what extent the 1st statement implies the 2nd". I tried to say that it unlikely that there might be an algebro-geometric proof of the Hodge decomposition. One needs some inpute outside of algebraic geometry to split the Hodge filtration (like harmonic forms in complex geometry or Galois representation in the $p$-adic version). For example, for proper smooth schemes over a finite extension $K/\mathbf Q_p$ one has a canonical decomposition ... $\endgroup$ – gdb Sep 30 '18 at 5:52
  • $\begingroup$ ...canonical decomposition $H_{et}^n(X\otimes_{K} \bar{K}, \mathbf Q_p)\otimes \mathbf C_p \cong \oplus_{i+j=n}H^i(X,\Omega^j_{X/K})\otimes_{\mathbf Q_p}\mathbf C_p(-j)$ but there is no such canonical decomposition for schemes over $\mathbf C_p$. In this situation the action of Galois group helps to obtain a splitting. So, it is unlikely that one can prove the Hodge decomposition without any further input from analysis (or reduction to the $p$-adic case. But this will not prove hodge symmetry). $\endgroup$ – gdb Sep 30 '18 at 5:55
  • $\begingroup$ Also look at this question mathoverflow.net/questions/28265/…. It seems that rather deep (functional) analysis is required in any of these approaches. $\endgroup$ – gdb Sep 30 '18 at 6:06
  • $\begingroup$ I don't mind functional analysis. What I in some sense do mind is that the theory of harmonic forms requires not only a choice of the Kahler but also working with a particular (Dolbeault) resolution of the de Rham complex. My feeling is that in a "good" theory no particular resolution should appear. The statement which implies the resolution is very clear - it just says that the Hodge filtration can be split by means of its complex conjugate. It seems very unfortunate to me that you can't prove such a fact without working with a particular resolution. $\endgroup$ – Alexander Braverman Oct 1 '18 at 4:33

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