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Hodge decomposition and degeneration of the spectral sequence
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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 $H^*(X,mathbbC)$ has a filtration such that the associated graded space is canonically identified with the direct sum of $H^p,q(X)$.
2) $H^k(X,mathbbC)$ 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 $H^p,q(X,mathbbC)$ is equal to
$F^p(H^p+q( X,mathbbC))cap overlineF^q(H^p+q( X,mathbbC))$ 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 $mathbbC$. 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)?
ag.algebraic-geometry complex-geometry hodge-theory
asked 4 hours ago
Alexander Braverman
3,8441238
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up vote
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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 $H^*(X,mathbbC)$ has a filtration such that the associated graded space is canonically identified with the direct sum of $H^p,q(X)$.
2) $H^k(X,mathbbC)$ 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 $H^p,q(X,mathbbC)$ is equal to
$F^p(H^p+q( X,mathbbC))cap overlineF^q(H^p+q( X,mathbbC))$ 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 $mathbbC$. 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)?
ag.algebraic-geometry complex-geometry hodge-theory
asked 4 hours ago
Alexander Braverman
3,8441238
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.
– R. van Dobben de Bruyn
3 hours ago
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.
– Alexander Braverman
3 hours ago
add a comment |Â
up vote
11
down vote
favorite
up vote
11
down vote
favorite
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 $H^*(X,mathbbC)$ has a filtration such that the associated graded space is canonically identified with the direct sum of $H^p,q(X)$.
2) $H^k(X,mathbbC)$ 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 $H^p,q(X,mathbbC)$ is equal to
$F^p(H^p+q( X,mathbbC))cap overlineF^q(H^p+q( X,mathbbC))$ 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 $mathbbC$. 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)?
ag.algebraic-geometry complex-geometry hodge-theory
asked 4 hours ago
Alexander Braverman
3,8441238
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 $H^*(X,mathbbC)$ has a filtration such that the associated graded space is canonically identified with the direct sum of $H^p,q(X)$.
2) $H^k(X,mathbbC)$ 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 $H^p,q(X,mathbbC)$ is equal to
$F^p(H^p+q( X,mathbbC))cap overlineF^q(H^p+q( X,mathbbC))$ 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 $mathbbC$. 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)?
ag.algebraic-geometry complex-geometry hodge-theory
ag.algebraic-geometry complex-geometry hodge-theory
asked 4 hours ago
Alexander Braverman
3,8441238
asked 4 hours ago
Alexander Braverman
3,8441238
asked 4 hours ago
Alexander Braverman
3,8441238
asked 4 hours ago
Alexander Braverman
3,8441238
asked 4 hours ago
Alexander Braverman
3,8441238
3,8441238
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.
– R. van Dobben de Bruyn
3 hours ago
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.
– Alexander Braverman
3 hours ago
add a comment |Â
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.
– R. van Dobben de Bruyn
3 hours ago
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.
– Alexander Braverman
3 hours ago
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.
– R. van Dobben de Bruyn
3 hours ago
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.
– R. van Dobben de Bruyn
3 hours ago
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.
– Alexander Braverman
3 hours ago
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.
– Alexander Braverman
3 hours ago
add a comment |Â
1 Answer
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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.
Let $X$ be 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 $F^iOmega^bullet_X/mathbf C=Omega_X/mathbf C^geq i$. This induces a descending filtration $F^bulletH^n_dR(X)=F^bulletH^n(X,mathbf C)$. 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 $(H^n(X,mathbf Z), F^bulletH^n(X,mathbf C))$ is a pure Hodge structure of weight n. Moreover we have isomorphisms $F^pH^n(X,mathbf C) cap overlineF^n-pH^n(X,mathbf C)cong H^q(X,Omega^p_X/mathbf C)$.
Remark 1: General theory of pure Hodge structures implies that there is a canonical isomorphism $H^n(X,mathbf C)cong oplus_p+q=nH^q(X,Omega^p_X/mathbf C)$.
Remark 2: Although we have a canonical decomposition $H^n(X,mathbf C)cong oplus_p+q=nH^q(X,Omega^p_X/mathbf C)$, 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
$$
operatornameTr_f:mathbf Rf_*Omega^q_X/k to Omega^q_Y/k.
$$
(Here we use that $operatornamedimX =operatornamedimY$ because in general the Trace map is map from $:mathbf Rf_*Omega^q_X/k$ to $Omega^q-d_Y/k[-d]$ where $d=operatornamedimX -operatornamedimY$).
Now note that the formation of Trace map is compatible with restrictions to open subsets. Hence, the composition $Tr_fcirc f^*:Omega^q_Y/k to Omega^q_Y/k$ is the identity morphism on an open dense subset $Usubset Y$ (because $f$ is birational!). Since $Omega^q_Y/k$ is a locally free sheaf, we conclude that $Tr_fcirc f^*$ is also the identity morphism. Therefore, $H^p(Tr_fcirc f^*)=H^p(Tr_f)circ H^p(f^*)$ 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
$$
E_1^p,q=H^q(X,Omega^p_X/mathbf C) Rightarrow H^p+q_dR(X).
$$
$$
downarrowf^*
$$
$$
E_1'^p,q=H^q(X',Omega^p_X'/mathbf C) Rightarrow H^p+q_dR(X').
$$
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) $F^pH^n_dR(X)cap overlineF^n-p+1H^n_dR(X)=0$. We know that that his equality holds for $X'$ (b/c $X'$ is projective) and we know that Hodge filtration $F^bulletH^n_dR(-)$ is functorial in $X$. Hence, argument in the Step $1$ says that
$$
F^pH^n_dR(X)cap overlineF^n-p+1H^n_dR(X) subset F^pH^n_dR(X')cap overlineF^n-p+1H^n_dR(X')=0.
$$
II) We have a canonical isomorphism $F^pH^n_dR(X)/F^p+1H^n_dR(X)cong H^q(X,Omega^p_X/mathbf C)$. 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 $F^pH^n_dR(X)+overlineF^n-p+1H^n_dR(X)=H^n_dR(X)$.
Since $F^pH^n_dR(X)$ and $overlineF^n-p+1H^n_dR(X)$ are disjoint inside $H^n_dR(X)$, it suffices to prove the equality
$$
operatornamedim(F^pH^n_dR(X))+operatornamedim(overlineF^n-p+1H^n_dR(X))=operatornamedim(H^n_dR(X)).
$$
Ok, Consequence II from Step $2$ implies that
$$
operatornamedim(F^pH^n_dR(X))=sum_igeq ph^i,n-i,
$$
$$operatornamedim(overlineF^n-p+1H^n_dR(X) = sum_igeq n-p+1h^i,n-i,
$$
$$
operatornamedim(H^n_dR(X))=sum_igeq 0h^i,n-i text, where h^p,q= operatornamedimH^q(X,Omega^p_X/mathbf C).
$$
Since we know that $F^pH^n_dR(X)$ and $overlineF^n-p+1H^n_dR(X)$ are disjoint we conclude that
$$
sum_igeq ph^i,n-i +sum_igeq n-p+1h^i,n-i leq sum_igeq 0h^i,n-i.
$$
Subtract $sum_igeq n-p+1h^i,n-i$ to obtain an inequality
$$
sum_igeq ph^i,n-i leq sum_ileq n-ph^i,n-i (*)
$$
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 $d=operatornamedim X$. Then we can rewrite both hand sides of $(*)$ in a different way.
$$
sum_igeq ph^i,n-i=sum_igeq ph^d-i,d-n+i=sum_jleq d-p=(2d-n)-(d-n+p) h^j,(2d-n)-j,
$$
$$
sum_ileq n-ph^i,n-i=sum_ileq n-ph^d-i,d-n+i=sum_jgeq d-n+ph^j,2d-n-j.
$$
But now observe that inequality (*) (applied to $n'=2d-n,p'=N-n+p$) guarantees that
$$
sum_ileq n-ph^i,n-i=sum_jgeq d-n+ph^j,2d-n-j leq sum_jgeq d-n+ph^j,2d-n-j =sum_ileq n-ph^i,n-i.
$$
Therefore $sum_igeq ph^i,n-i=sum_ileq n-ph^i,n-i$! And we showed that this is equivalent to the equality $$
F^pH^n_dR(X)+overlineF^n-p+1H^n_dR(X)=H^n_dR(X).
$$
In other words we showed that $(H^n(X,mathbf Z), F^bulletH^n(X,mathbf C))$ is a pure Hodge structure of weight $d$.
answered 30 mins ago
gdb
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1 Answer
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active
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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.
Let $X$ be 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 $F^iOmega^bullet_X/mathbf C=Omega_X/mathbf C^geq i$. This induces a descending filtration $F^bulletH^n_dR(X)=F^bulletH^n(X,mathbf C)$. 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 $(H^n(X,mathbf Z), F^bulletH^n(X,mathbf C))$ is a pure Hodge structure of weight n. Moreover we have isomorphisms $F^pH^n(X,mathbf C) cap overlineF^n-pH^n(X,mathbf C)cong H^q(X,Omega^p_X/mathbf C)$.
Remark 1: General theory of pure Hodge structures implies that there is a canonical isomorphism $H^n(X,mathbf C)cong oplus_p+q=nH^q(X,Omega^p_X/mathbf C)$.
Remark 2: Although we have a canonical decomposition $H^n(X,mathbf C)cong oplus_p+q=nH^q(X,Omega^p_X/mathbf C)$, 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
$$
operatornameTr_f:mathbf Rf_*Omega^q_X/k to Omega^q_Y/k.
$$
(Here we use that $operatornamedimX =operatornamedimY$ because in general the Trace map is map from $:mathbf Rf_*Omega^q_X/k$ to $Omega^q-d_Y/k[-d]$ where $d=operatornamedimX -operatornamedimY$).
Now note that the formation of Trace map is compatible with restrictions to open subsets. Hence, the composition $Tr_fcirc f^*:Omega^q_Y/k to Omega^q_Y/k$ is the identity morphism on an open dense subset $Usubset Y$ (because $f$ is birational!). Since $Omega^q_Y/k$ is a locally free sheaf, we conclude that $Tr_fcirc f^*$ is also the identity morphism. Therefore, $H^p(Tr_fcirc f^*)=H^p(Tr_f)circ H^p(f^*)$ 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
$$
E_1^p,q=H^q(X,Omega^p_X/mathbf C) Rightarrow H^p+q_dR(X).
$$
$$
downarrowf^*
$$
$$
E_1'^p,q=H^q(X',Omega^p_X'/mathbf C) Rightarrow H^p+q_dR(X').
$$
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) $F^pH^n_dR(X)cap overlineF^n-p+1H^n_dR(X)=0$. We know that that his equality holds for $X'$ (b/c $X'$ is projective) and we know that Hodge filtration $F^bulletH^n_dR(-)$ is functorial in $X$. Hence, argument in the Step $1$ says that
$$
F^pH^n_dR(X)cap overlineF^n-p+1H^n_dR(X) subset F^pH^n_dR(X')cap overlineF^n-p+1H^n_dR(X')=0.
$$
II) We have a canonical isomorphism $F^pH^n_dR(X)/F^p+1H^n_dR(X)cong H^q(X,Omega^p_X/mathbf C)$. 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 $F^pH^n_dR(X)+overlineF^n-p+1H^n_dR(X)=H^n_dR(X)$.
Since $F^pH^n_dR(X)$ and $overlineF^n-p+1H^n_dR(X)$ are disjoint inside $H^n_dR(X)$, it suffices to prove the equality
$$
operatornamedim(F^pH^n_dR(X))+operatornamedim(overlineF^n-p+1H^n_dR(X))=operatornamedim(H^n_dR(X)).
$$
Ok, Consequence II from Step $2$ implies that
$$
operatornamedim(F^pH^n_dR(X))=sum_igeq ph^i,n-i,
$$
$$operatornamedim(overlineF^n-p+1H^n_dR(X) = sum_igeq n-p+1h^i,n-i,
$$
$$
operatornamedim(H^n_dR(X))=sum_igeq 0h^i,n-i text, where h^p,q= operatornamedimH^q(X,Omega^p_X/mathbf C).
$$
Since we know that $F^pH^n_dR(X)$ and $overlineF^n-p+1H^n_dR(X)$ are disjoint we conclude that
$$
sum_igeq ph^i,n-i +sum_igeq n-p+1h^i,n-i leq sum_igeq 0h^i,n-i.
$$
Subtract $sum_igeq n-p+1h^i,n-i$ to obtain an inequality
$$
sum_igeq ph^i,n-i leq sum_ileq n-ph^i,n-i (*)
$$
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 $d=operatornamedim X$. Then we can rewrite both hand sides of $(*)$ in a different way.
$$
sum_igeq ph^i,n-i=sum_igeq ph^d-i,d-n+i=sum_jleq d-p=(2d-n)-(d-n+p) h^j,(2d-n)-j,
$$
$$
sum_ileq n-ph^i,n-i=sum_ileq n-ph^d-i,d-n+i=sum_jgeq d-n+ph^j,2d-n-j.
$$
But now observe that inequality (*) (applied to $n'=2d-n,p'=N-n+p$) guarantees that
$$
sum_ileq n-ph^i,n-i=sum_jgeq d-n+ph^j,2d-n-j leq sum_jgeq d-n+ph^j,2d-n-j =sum_ileq n-ph^i,n-i.
$$
Therefore $sum_igeq ph^i,n-i=sum_ileq n-ph^i,n-i$! And we showed that this is equivalent to the equality $$
F^pH^n_dR(X)+overlineF^n-p+1H^n_dR(X)=H^n_dR(X).
$$
In other words we showed that $(H^n(X,mathbf Z), F^bulletH^n(X,mathbf C))$ is a pure Hodge structure of weight $d$.
answered 30 mins ago
gdb
733113
add a comment |Â
up vote
3
down vote
I am far from being an expert in this area, but I will try to present my understanding of this subject.
Let $X$ be 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 $F^iOmega^bullet_X/mathbf C=Omega_X/mathbf C^geq i$. This induces a descending filtration $F^bulletH^n_dR(X)=F^bulletH^n(X,mathbf C)$. 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 $(H^n(X,mathbf Z), F^bulletH^n(X,mathbf C))$ is a pure Hodge structure of weight n. Moreover we have isomorphisms $F^pH^n(X,mathbf C) cap overlineF^n-pH^n(X,mathbf C)cong H^q(X,Omega^p_X/mathbf C)$.
Remark 1: General theory of pure Hodge structures implies that there is a canonical isomorphism $H^n(X,mathbf C)cong oplus_p+q=nH^q(X,Omega^p_X/mathbf C)$.
Remark 2: Although we have a canonical decomposition $H^n(X,mathbf C)cong oplus_p+q=nH^q(X,Omega^p_X/mathbf C)$, 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
$$
operatornameTr_f:mathbf Rf_*Omega^q_X/k to Omega^q_Y/k.
$$
(Here we use that $operatornamedimX =operatornamedimY$ because in general the Trace map is map from $:mathbf Rf_*Omega^q_X/k$ to $Omega^q-d_Y/k[-d]$ where $d=operatornamedimX -operatornamedimY$).
Now note that the formation of Trace map is compatible with restrictions to open subsets. Hence, the composition $Tr_fcirc f^*:Omega^q_Y/k to Omega^q_Y/k$ is the identity morphism on an open dense subset $Usubset Y$ (because $f$ is birational!). Since $Omega^q_Y/k$ is a locally free sheaf, we conclude that $Tr_fcirc f^*$ is also the identity morphism. Therefore, $H^p(Tr_fcirc f^*)=H^p(Tr_f)circ H^p(f^*)$ 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
$$
E_1^p,q=H^q(X,Omega^p_X/mathbf C) Rightarrow H^p+q_dR(X).
$$
$$
downarrowf^*
$$
$$
E_1'^p,q=H^q(X',Omega^p_X'/mathbf C) Rightarrow H^p+q_dR(X').
$$
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) $F^pH^n_dR(X)cap overlineF^n-p+1H^n_dR(X)=0$. We know that that his equality holds for $X'$ (b/c $X'$ is projective) and we know that Hodge filtration $F^bulletH^n_dR(-)$ is functorial in $X$. Hence, argument in the Step $1$ says that
$$
F^pH^n_dR(X)cap overlineF^n-p+1H^n_dR(X) subset F^pH^n_dR(X')cap overlineF^n-p+1H^n_dR(X')=0.
$$
II) We have a canonical isomorphism $F^pH^n_dR(X)/F^p+1H^n_dR(X)cong H^q(X,Omega^p_X/mathbf C)$. 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 $F^pH^n_dR(X)+overlineF^n-p+1H^n_dR(X)=H^n_dR(X)$.
Since $F^pH^n_dR(X)$ and $overlineF^n-p+1H^n_dR(X)$ are disjoint inside $H^n_dR(X)$, it suffices to prove the equality
$$
operatornamedim(F^pH^n_dR(X))+operatornamedim(overlineF^n-p+1H^n_dR(X))=operatornamedim(H^n_dR(X)).
$$
Ok, Consequence II from Step $2$ implies that
$$
operatornamedim(F^pH^n_dR(X))=sum_igeq ph^i,n-i,
$$
$$operatornamedim(overlineF^n-p+1H^n_dR(X) = sum_igeq n-p+1h^i,n-i,
$$
$$
operatornamedim(H^n_dR(X))=sum_igeq 0h^i,n-i text, where h^p,q= operatornamedimH^q(X,Omega^p_X/mathbf C).
$$
Since we know that $F^pH^n_dR(X)$ and $overlineF^n-p+1H^n_dR(X)$ are disjoint we conclude that
$$
sum_igeq ph^i,n-i +sum_igeq n-p+1h^i,n-i leq sum_igeq 0h^i,n-i.
$$
Subtract $sum_igeq n-p+1h^i,n-i$ to obtain an inequality
$$
sum_igeq ph^i,n-i leq sum_ileq n-ph^i,n-i (*)
$$
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 $d=operatornamedim X$. Then we can rewrite both hand sides of $(*)$ in a different way.
$$
sum_igeq ph^i,n-i=sum_igeq ph^d-i,d-n+i=sum_jleq d-p=(2d-n)-(d-n+p) h^j,(2d-n)-j,
$$
$$
sum_ileq n-ph^i,n-i=sum_ileq n-ph^d-i,d-n+i=sum_jgeq d-n+ph^j,2d-n-j.
$$
But now observe that inequality (*) (applied to $n'=2d-n,p'=N-n+p$) guarantees that
$$
sum_ileq n-ph^i,n-i=sum_jgeq d-n+ph^j,2d-n-j leq sum_jgeq d-n+ph^j,2d-n-j =sum_ileq n-ph^i,n-i.
$$
Therefore $sum_igeq ph^i,n-i=sum_ileq n-ph^i,n-i$! And we showed that this is equivalent to the equality $$
F^pH^n_dR(X)+overlineF^n-p+1H^n_dR(X)=H^n_dR(X).
$$
In other words we showed that $(H^n(X,mathbf Z), F^bulletH^n(X,mathbf C))$ is a pure Hodge structure of weight $d$.
answered 30 mins ago
gdb
733113
add a comment |Â
up vote
3
down vote
up vote
3
down vote
I am far from being an expert in this area, but I will try to present my understanding of this subject.
Let $X$ be 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 $F^iOmega^bullet_X/mathbf C=Omega_X/mathbf C^geq i$. This induces a descending filtration $F^bulletH^n_dR(X)=F^bulletH^n(X,mathbf C)$. 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 $(H^n(X,mathbf Z), F^bulletH^n(X,mathbf C))$ is a pure Hodge structure of weight n. Moreover we have isomorphisms $F^pH^n(X,mathbf C) cap overlineF^n-pH^n(X,mathbf C)cong H^q(X,Omega^p_X/mathbf C)$.
Remark 1: General theory of pure Hodge structures implies that there is a canonical isomorphism $H^n(X,mathbf C)cong oplus_p+q=nH^q(X,Omega^p_X/mathbf C)$.
Remark 2: Although we have a canonical decomposition $H^n(X,mathbf C)cong oplus_p+q=nH^q(X,Omega^p_X/mathbf C)$, 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
$$
operatornameTr_f:mathbf Rf_*Omega^q_X/k to Omega^q_Y/k.
$$
(Here we use that $operatornamedimX =operatornamedimY$ because in general the Trace map is map from $:mathbf Rf_*Omega^q_X/k$ to $Omega^q-d_Y/k[-d]$ where $d=operatornamedimX -operatornamedimY$).
Now note that the formation of Trace map is compatible with restrictions to open subsets. Hence, the composition $Tr_fcirc f^*:Omega^q_Y/k to Omega^q_Y/k$ is the identity morphism on an open dense subset $Usubset Y$ (because $f$ is birational!). Since $Omega^q_Y/k$ is a locally free sheaf, we conclude that $Tr_fcirc f^*$ is also the identity morphism. Therefore, $H^p(Tr_fcirc f^*)=H^p(Tr_f)circ H^p(f^*)$ 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
$$
E_1^p,q=H^q(X,Omega^p_X/mathbf C) Rightarrow H^p+q_dR(X).
$$
$$
downarrowf^*
$$
$$
E_1'^p,q=H^q(X',Omega^p_X'/mathbf C) Rightarrow H^p+q_dR(X').
$$
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) $F^pH^n_dR(X)cap overlineF^n-p+1H^n_dR(X)=0$. We know that that his equality holds for $X'$ (b/c $X'$ is projective) and we know that Hodge filtration $F^bulletH^n_dR(-)$ is functorial in $X$. Hence, argument in the Step $1$ says that
$$
F^pH^n_dR(X)cap overlineF^n-p+1H^n_dR(X) subset F^pH^n_dR(X')cap overlineF^n-p+1H^n_dR(X')=0.
$$
II) We have a canonical isomorphism $F^pH^n_dR(X)/F^p+1H^n_dR(X)cong H^q(X,Omega^p_X/mathbf C)$. 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 $F^pH^n_dR(X)+overlineF^n-p+1H^n_dR(X)=H^n_dR(X)$.
Since $F^pH^n_dR(X)$ and $overlineF^n-p+1H^n_dR(X)$ are disjoint inside $H^n_dR(X)$, it suffices to prove the equality
$$
operatornamedim(F^pH^n_dR(X))+operatornamedim(overlineF^n-p+1H^n_dR(X))=operatornamedim(H^n_dR(X)).
$$
Ok, Consequence II from Step $2$ implies that
$$
operatornamedim(F^pH^n_dR(X))=sum_igeq ph^i,n-i,
$$
$$operatornamedim(overlineF^n-p+1H^n_dR(X) = sum_igeq n-p+1h^i,n-i,
$$
$$
operatornamedim(H^n_dR(X))=sum_igeq 0h^i,n-i text, where h^p,q= operatornamedimH^q(X,Omega^p_X/mathbf C).
$$
Since we know that $F^pH^n_dR(X)$ and $overlineF^n-p+1H^n_dR(X)$ are disjoint we conclude that
$$
sum_igeq ph^i,n-i +sum_igeq n-p+1h^i,n-i leq sum_igeq 0h^i,n-i.
$$
Subtract $sum_igeq n-p+1h^i,n-i$ to obtain an inequality
$$
sum_igeq ph^i,n-i leq sum_ileq n-ph^i,n-i (*)
$$
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 $d=operatornamedim X$. Then we can rewrite both hand sides of $(*)$ in a different way.
$$
sum_igeq ph^i,n-i=sum_igeq ph^d-i,d-n+i=sum_jleq d-p=(2d-n)-(d-n+p) h^j,(2d-n)-j,
$$
$$
sum_ileq n-ph^i,n-i=sum_ileq n-ph^d-i,d-n+i=sum_jgeq d-n+ph^j,2d-n-j.
$$
But now observe that inequality (*) (applied to $n'=2d-n,p'=N-n+p$) guarantees that
$$
sum_ileq n-ph^i,n-i=sum_jgeq d-n+ph^j,2d-n-j leq sum_jgeq d-n+ph^j,2d-n-j =sum_ileq n-ph^i,n-i.
$$
Therefore $sum_igeq ph^i,n-i=sum_ileq n-ph^i,n-i$! And we showed that this is equivalent to the equality $$
F^pH^n_dR(X)+overlineF^n-p+1H^n_dR(X)=H^n_dR(X).
$$
In other words we showed that $(H^n(X,mathbf Z), F^bulletH^n(X,mathbf C))$ is a pure Hodge structure of weight $d$.
answered 30 mins ago
gdb
733113
I am far from being an expert in this area, but I will try to present my understanding of this subject.
Let $X$ be 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 $F^iOmega^bullet_X/mathbf C=Omega_X/mathbf C^geq i$. This induces a descending filtration $F^bulletH^n_dR(X)=F^bulletH^n(X,mathbf C)$. 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 $(H^n(X,mathbf Z), F^bulletH^n(X,mathbf C))$ is a pure Hodge structure of weight n. Moreover we have isomorphisms $F^pH^n(X,mathbf C) cap overlineF^n-pH^n(X,mathbf C)cong H^q(X,Omega^p_X/mathbf C)$.
Remark 1: General theory of pure Hodge structures implies that there is a canonical isomorphism $H^n(X,mathbf C)cong oplus_p+q=nH^q(X,Omega^p_X/mathbf C)$.
Remark 2: Although we have a canonical decomposition $H^n(X,mathbf C)cong oplus_p+q=nH^q(X,Omega^p_X/mathbf C)$, 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
$$
operatornameTr_f:mathbf Rf_*Omega^q_X/k to Omega^q_Y/k.
$$
(Here we use that $operatornamedimX =operatornamedimY$ because in general the Trace map is map from $:mathbf Rf_*Omega^q_X/k$ to $Omega^q-d_Y/k[-d]$ where $d=operatornamedimX -operatornamedimY$).
Now note that the formation of Trace map is compatible with restrictions to open subsets. Hence, the composition $Tr_fcirc f^*:Omega^q_Y/k to Omega^q_Y/k$ is the identity morphism on an open dense subset $Usubset Y$ (because $f$ is birational!). Since $Omega^q_Y/k$ is a locally free sheaf, we conclude that $Tr_fcirc f^*$ is also the identity morphism. Therefore, $H^p(Tr_fcirc f^*)=H^p(Tr_f)circ H^p(f^*)$ 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
$$
E_1^p,q=H^q(X,Omega^p_X/mathbf C) Rightarrow H^p+q_dR(X).
$$
$$
downarrowf^*
$$
$$
E_1'^p,q=H^q(X',Omega^p_X'/mathbf C) Rightarrow H^p+q_dR(X').
$$
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) $F^pH^n_dR(X)cap overlineF^n-p+1H^n_dR(X)=0$. We know that that his equality holds for $X'$ (b/c $X'$ is projective) and we know that Hodge filtration $F^bulletH^n_dR(-)$ is functorial in $X$. Hence, argument in the Step $1$ says that
$$
F^pH^n_dR(X)cap overlineF^n-p+1H^n_dR(X) subset F^pH^n_dR(X')cap overlineF^n-p+1H^n_dR(X')=0.
$$
II) We have a canonical isomorphism $F^pH^n_dR(X)/F^p+1H^n_dR(X)cong H^q(X,Omega^p_X/mathbf C)$. 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 $F^pH^n_dR(X)+overlineF^n-p+1H^n_dR(X)=H^n_dR(X)$.
Since $F^pH^n_dR(X)$ and $overlineF^n-p+1H^n_dR(X)$ are disjoint inside $H^n_dR(X)$, it suffices to prove the equality
$$
operatornamedim(F^pH^n_dR(X))+operatornamedim(overlineF^n-p+1H^n_dR(X))=operatornamedim(H^n_dR(X)).
$$
Ok, Consequence II from Step $2$ implies that
$$
operatornamedim(F^pH^n_dR(X))=sum_igeq ph^i,n-i,
$$
$$operatornamedim(overlineF^n-p+1H^n_dR(X) = sum_igeq n-p+1h^i,n-i,
$$
$$
operatornamedim(H^n_dR(X))=sum_igeq 0h^i,n-i text, where h^p,q= operatornamedimH^q(X,Omega^p_X/mathbf C).
$$
Since we know that $F^pH^n_dR(X)$ and $overlineF^n-p+1H^n_dR(X)$ are disjoint we conclude that
$$
sum_igeq ph^i,n-i +sum_igeq n-p+1h^i,n-i leq sum_igeq 0h^i,n-i.
$$
Subtract $sum_igeq n-p+1h^i,n-i$ to obtain an inequality
$$
sum_igeq ph^i,n-i leq sum_ileq n-ph^i,n-i (*)
$$
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 $d=operatornamedim X$. Then we can rewrite both hand sides of $(*)$ in a different way.
$$
sum_igeq ph^i,n-i=sum_igeq ph^d-i,d-n+i=sum_jleq d-p=(2d-n)-(d-n+p) h^j,(2d-n)-j,
$$
$$
sum_ileq n-ph^i,n-i=sum_ileq n-ph^d-i,d-n+i=sum_jgeq d-n+ph^j,2d-n-j.
$$
But now observe that inequality (*) (applied to $n'=2d-n,p'=N-n+p$) guarantees that
$$
sum_ileq n-ph^i,n-i=sum_jgeq d-n+ph^j,2d-n-j leq sum_jgeq d-n+ph^j,2d-n-j =sum_ileq n-ph^i,n-i.
$$
Therefore $sum_igeq ph^i,n-i=sum_ileq n-ph^i,n-i$! And we showed that this is equivalent to the equality $$
F^pH^n_dR(X)+overlineF^n-p+1H^n_dR(X)=H^n_dR(X).
$$
In other words we showed that $(H^n(X,mathbf Z), F^bulletH^n(X,mathbf C))$ is a pure Hodge structure of weight $d$.
answered 30 mins ago
gdb
733113
answered 30 mins ago
gdb
733113
answered 30 mins ago
gdb
733113
answered 30 mins ago
gdb
733113
733113
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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.
– R. van Dobben de Bruyn
3 hours ago
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.
– Alexander Braverman
3 hours ago