The homomorphism induced by holomorphic map preserves Hodge decomposition
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Let $f:Xto Y$ be a holomorphic map between two compact Kähler manifolds. Then the induced map $f^*:H^1(Y,mathbb{C})to H^1(X,mathbb{C})$ preserves the Hodge decomposition. Is there a reference for a proof of this statement? Or Can you give a proof?
In particular, can you explain what exactly does “preserves Hodeg decomposition” mean?
complex-geometry kahler-manifolds hodge-theory
edited Dec 10 '18 at 14:04
Michael Albanese
63.2k1598303
asked Dec 10 '18 at 2:41
DannyDanny
1,083412
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add a comment |
$begingroup$
Let $f:Xto Y$ be a holomorphic map between two compact Kähler manifolds. Then the induced map $f^*:H^1(Y,mathbb{C})to H^1(X,mathbb{C})$ preserves the Hodge decomposition. Is there a reference for a proof of this statement? Or Can you give a proof?
In particular, can you explain what exactly does “preserves Hodeg decomposition” mean?
complex-geometry kahler-manifolds hodge-theory
edited Dec 10 '18 at 14:04
Michael Albanese
63.2k1598303
asked Dec 10 '18 at 2:41
DannyDanny
1,083412
$endgroup$
add a comment |
$begingroup$
Let $f:Xto Y$ be a holomorphic map between two compact Kähler manifolds. Then the induced map $f^*:H^1(Y,mathbb{C})to H^1(X,mathbb{C})$ preserves the Hodge decomposition. Is there a reference for a proof of this statement? Or Can you give a proof?
In particular, can you explain what exactly does “preserves Hodeg decomposition” mean?
complex-geometry kahler-manifolds hodge-theory
edited Dec 10 '18 at 14:04
Michael Albanese
63.2k1598303
asked Dec 10 '18 at 2:41
DannyDanny
1,083412
$endgroup$
Let $f:Xto Y$ be a holomorphic map between two compact Kähler manifolds. Then the induced map $f^*:H^1(Y,mathbb{C})to H^1(X,mathbb{C})$ preserves the Hodge decomposition. Is there a reference for a proof of this statement? Or Can you give a proof?
In particular, can you explain what exactly does “preserves Hodeg decomposition” mean?
complex-geometry kahler-manifolds hodge-theory
complex-geometry kahler-manifolds hodge-theory
edited Dec 10 '18 at 14:04
Michael Albanese
63.2k1598303
asked Dec 10 '18 at 2:41
DannyDanny
1,083412
edited Dec 10 '18 at 14:04
Michael Albanese
63.2k1598303
asked Dec 10 '18 at 2:41
DannyDanny
1,083412
edited Dec 10 '18 at 14:04
Michael Albanese
63.2k1598303
edited Dec 10 '18 at 14:04
Michael Albanese
63.2k1598303
edited Dec 10 '18 at 14:04
Michael Albanese
63.2k1598303
63.2k1598303
asked Dec 10 '18 at 2:41
DannyDanny
1,083412
asked Dec 10 '18 at 2:41
DannyDanny
1,083412
asked Dec 10 '18 at 2:41
DannyDanny
1,083412
1,083412
add a comment |
add a comment |
1 Answer
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The Hodge decomposition is the isomorphism $H^1(Y; mathbb{C}) cong H^{1,0}_{bar{partial}}(Y)oplus H^{0,1}_{bar{partial}}(Y)$. The statement just means that $f^* : H^{1,0}_{bar{partial}}(Y) to H^{1,0}_{bar{partial}}(X)$ and $f^* : H^{0,1}_{bar{partial}}(Y) to H^{0,1}_{bar{partial}}(X)$.
answered Dec 10 '18 at 14:04
Michael AlbaneseMichael Albanese
63.2k1598303
$endgroup$
$begingroup$
Is there a proof on some book?
$endgroup$
– Danny
Dec 10 '18 at 14:48
1
$begingroup$
@Danny: Here's the exercise you need to do. Show first that a holomorphic mapping pulls back $(p,q)$-forms to $(p,q)$-forms, and then that $barpartial(f^*omega) = f^*(partialbaromega)$ when $f$ is holomorphic.
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– Ted Shifrin
Dec 19 '18 at 18:27
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
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active
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$begingroup$
The Hodge decomposition is the isomorphism $H^1(Y; mathbb{C}) cong H^{1,0}_{bar{partial}}(Y)oplus H^{0,1}_{bar{partial}}(Y)$. The statement just means that $f^* : H^{1,0}_{bar{partial}}(Y) to H^{1,0}_{bar{partial}}(X)$ and $f^* : H^{0,1}_{bar{partial}}(Y) to H^{0,1}_{bar{partial}}(X)$.
answered Dec 10 '18 at 14:04
Michael AlbaneseMichael Albanese
63.2k1598303
$endgroup$
$begingroup$
Is there a proof on some book?
$endgroup$
– Danny
Dec 10 '18 at 14:48
1
$begingroup$
@Danny: Here's the exercise you need to do. Show first that a holomorphic mapping pulls back $(p,q)$-forms to $(p,q)$-forms, and then that $barpartial(f^*omega) = f^*(partialbaromega)$ when $f$ is holomorphic.
$endgroup$
– Ted Shifrin
Dec 19 '18 at 18:27
add a comment |
$begingroup$
The Hodge decomposition is the isomorphism $H^1(Y; mathbb{C}) cong H^{1,0}_{bar{partial}}(Y)oplus H^{0,1}_{bar{partial}}(Y)$. The statement just means that $f^* : H^{1,0}_{bar{partial}}(Y) to H^{1,0}_{bar{partial}}(X)$ and $f^* : H^{0,1}_{bar{partial}}(Y) to H^{0,1}_{bar{partial}}(X)$.
answered Dec 10 '18 at 14:04
Michael AlbaneseMichael Albanese
63.2k1598303
$endgroup$
$begingroup$
Is there a proof on some book?
$endgroup$
– Danny
Dec 10 '18 at 14:48
1
$begingroup$
@Danny: Here's the exercise you need to do. Show first that a holomorphic mapping pulls back $(p,q)$-forms to $(p,q)$-forms, and then that $barpartial(f^*omega) = f^*(partialbaromega)$ when $f$ is holomorphic.
$endgroup$
– Ted Shifrin
Dec 19 '18 at 18:27
add a comment |
$begingroup$
The Hodge decomposition is the isomorphism $H^1(Y; mathbb{C}) cong H^{1,0}_{bar{partial}}(Y)oplus H^{0,1}_{bar{partial}}(Y)$. The statement just means that $f^* : H^{1,0}_{bar{partial}}(Y) to H^{1,0}_{bar{partial}}(X)$ and $f^* : H^{0,1}_{bar{partial}}(Y) to H^{0,1}_{bar{partial}}(X)$.
answered Dec 10 '18 at 14:04
Michael AlbaneseMichael Albanese
63.2k1598303
$endgroup$
The Hodge decomposition is the isomorphism $H^1(Y; mathbb{C}) cong H^{1,0}_{bar{partial}}(Y)oplus H^{0,1}_{bar{partial}}(Y)$. The statement just means that $f^* : H^{1,0}_{bar{partial}}(Y) to H^{1,0}_{bar{partial}}(X)$ and $f^* : H^{0,1}_{bar{partial}}(Y) to H^{0,1}_{bar{partial}}(X)$.
answered Dec 10 '18 at 14:04
Michael AlbaneseMichael Albanese
63.2k1598303
answered Dec 10 '18 at 14:04
Michael AlbaneseMichael Albanese
63.2k1598303
answered Dec 10 '18 at 14:04
Michael AlbaneseMichael Albanese
63.2k1598303
answered Dec 10 '18 at 14:04
Michael AlbaneseMichael Albanese
63.2k1598303
63.2k1598303
$begingroup$
Is there a proof on some book?
$endgroup$
– Danny
Dec 10 '18 at 14:48
1
$begingroup$
@Danny: Here's the exercise you need to do. Show first that a holomorphic mapping pulls back $(p,q)$-forms to $(p,q)$-forms, and then that $barpartial(f^*omega) = f^*(partialbaromega)$ when $f$ is holomorphic.
$endgroup$
– Ted Shifrin
Dec 19 '18 at 18:27
add a comment |
$begingroup$
Is there a proof on some book?
$endgroup$
– Danny
Dec 10 '18 at 14:48
1
$begingroup$
@Danny: Here's the exercise you need to do. Show first that a holomorphic mapping pulls back $(p,q)$-forms to $(p,q)$-forms, and then that $barpartial(f^*omega) = f^*(partialbaromega)$ when $f$ is holomorphic.
$endgroup$
– Ted Shifrin
Dec 19 '18 at 18:27
$begingroup$
Is there a proof on some book?
$endgroup$
– Danny
Dec 10 '18 at 14:48
$begingroup$
Is there a proof on some book?
$endgroup$
– Danny
Dec 10 '18 at 14:48
1
1
$begingroup$
@Danny: Here's the exercise you need to do. Show first that a holomorphic mapping pulls back $(p,q)$-forms to $(p,q)$-forms, and then that $barpartial(f^*omega) = f^*(partialbaromega)$ when $f$ is holomorphic.
$endgroup$
– Ted Shifrin
Dec 19 '18 at 18:27
$begingroup$
@Danny: Here's the exercise you need to do. Show first that a holomorphic mapping pulls back $(p,q)$-forms to $(p,q)$-forms, and then that $barpartial(f^*omega) = f^*(partialbaromega)$ when $f$ is holomorphic.
$endgroup$
– Ted Shifrin
Dec 19 '18 at 18:27
add a comment |
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