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What is the Künneth formula for complete varieties?

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0

$begingroup$

I'm reading a part of Mumford's Abelian Varieties, and in the Chapter The theorem of the cube: II he claims that some "Künneth formula" tells us that if $L_1$ is a line bundle on a product $X times Y_1$ such that $L_1 cong p_2^*(M_1)$ for some line bundle $M_1$ on $Y_1$, then $p_2,*(L_1) cong M_1$.

Here $X$ is a complete variety over an algebraically closed field $k$, and $Y_1$ is a scheme of finite type over $k$.

The Künneth formula I know relates (co-)homology on product spaces with the (co-)homology on the base spaces, for example as in the stacks-project, tag 0BEC. I don't see how the claim should follow from this.

Does the claim follow from this Künneth formula, or is something else going on here?

algebraic-geometry vector-bundles cross-product abelian-varieties line-bundles

share|cite|improve this question

asked Mar 21 at 12:21

red_trumpetred_trumpet

1,025319

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Certainly the projection formula is going on here. Perhaps to apply it properly, one needs to use Kunneth.
  $endgroup$
  – aginensky
  Mar 21 at 14:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @aginensky Yeah by the projection formula we get: $p_2,* p_2^* M_1 = p_2,*(mathcalO otimes p_2^*M_1) = p_2,*(mathcalO_Xtimes Y_1) otimes M_1$, so we reduce this to show that $p_2,*mathcalO_X times Y_1 cong mathcalO_Y_1$. I think I can show this using "Cohomology and Base Change" (Hartshorne, p.290+291), but I still don't see any connection to the Künneth formula.
  $endgroup$
  – red_trumpet
  Mar 21 at 14:58
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Perhaps in saying that the structure sheaf of the product is the tensor product of the pullbacks of the two structure sheaves.
  $endgroup$
  – aginensky
  Mar 21 at 14:59
  

  
  
  
  

add a comment | 

0

$begingroup$

I'm reading a part of Mumford's Abelian Varieties, and in the Chapter The theorem of the cube: II he claims that some "Künneth formula" tells us that if $L_1$ is a line bundle on a product $X times Y_1$ such that $L_1 cong p_2^*(M_1)$ for some line bundle $M_1$ on $Y_1$, then $p_2,*(L_1) cong M_1$.

Here $X$ is a complete variety over an algebraically closed field $k$, and $Y_1$ is a scheme of finite type over $k$.

The Künneth formula I know relates (co-)homology on product spaces with the (co-)homology on the base spaces, for example as in the stacks-project, tag 0BEC. I don't see how the claim should follow from this.

Does the claim follow from this Künneth formula, or is something else going on here?

algebraic-geometry vector-bundles cross-product abelian-varieties line-bundles

share|cite|improve this question

asked Mar 21 at 12:21

red_trumpetred_trumpet

1,025319

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Certainly the projection formula is going on here. Perhaps to apply it properly, one needs to use Kunneth.
  $endgroup$
  – aginensky
  Mar 21 at 14:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @aginensky Yeah by the projection formula we get: $p_2,* p_2^* M_1 = p_2,*(mathcalO otimes p_2^*M_1) = p_2,*(mathcalO_Xtimes Y_1) otimes M_1$, so we reduce this to show that $p_2,*mathcalO_X times Y_1 cong mathcalO_Y_1$. I think I can show this using "Cohomology and Base Change" (Hartshorne, p.290+291), but I still don't see any connection to the Künneth formula.
  $endgroup$
  – red_trumpet
  Mar 21 at 14:58
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Perhaps in saying that the structure sheaf of the product is the tensor product of the pullbacks of the two structure sheaves.
  $endgroup$
  – aginensky
  Mar 21 at 14:59
  

  
  
  
  

add a comment | 

$begingroup$

I'm reading a part of Mumford's Abelian Varieties, and in the Chapter The theorem of the cube: II he claims that some "Künneth formula" tells us that if $L_1$ is a line bundle on a product $X times Y_1$ such that $L_1 cong p_2^*(M_1)$ for some line bundle $M_1$ on $Y_1$, then $p_2,*(L_1) cong M_1$.

Here $X$ is a complete variety over an algebraically closed field $k$, and $Y_1$ is a scheme of finite type over $k$.

The Künneth formula I know relates (co-)homology on product spaces with the (co-)homology on the base spaces, for example as in the stacks-project, tag 0BEC. I don't see how the claim should follow from this.

Does the claim follow from this Künneth formula, or is something else going on here?

algebraic-geometry vector-bundles cross-product abelian-varieties line-bundles

share|cite|improve this question

asked Mar 21 at 12:21

red_trumpetred_trumpet

1,025319

$endgroup$

share|cite|improve this question

asked Mar 21 at 12:21

red_trumpetred_trumpet

1,025319

share|cite|improve this question

asked Mar 21 at 12:21

red_trumpetred_trumpet

1,025319

asked Mar 21 at 12:21

red_trumpetred_trumpet

1,025319

asked Mar 21 at 12:21

red_trumpetred_trumpet

1,025319

1 Answer
1

active

oldest

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1

$begingroup$

It's always hard to reconstruct someone else's train of thought, but here's one attempted explanation.

As aginensky says, the first step seems to be to use the projection formula. This shows that $(p_2)_* L_1 = M_1 otimes (p_2)_* O_Z$ where I have written $Z$ to denote the product (to save typing).

So now it is enough to show that $(p_2)_* O_Z = O_Y_1$. To do this, recall that $(p_2)_* O_Z$ is defined by

beginalign* U &mapsto H^0((p_2)^-1(U), O_(p_2)^-1(U))\
&= H^0(U times X, O_U times X)
endalign*
Now Künneth tells us that this is isomorphic to
beginalign* &H^0(U,O_U) otimes_k H^0(X,O_X) =H^0(U,O_U)endalign*
using the fact that $X$ is complete.

share|cite|improve this answer

edited Mar 22 at 9:34

answered Mar 21 at 15:36

Asal Beag DubhAsal Beag Dubh

68515

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Seems correct. Two notes: In the last equation, it should be $otimes$ instead of $times$, and afaik $U mapsto H^0(Utimes X, mathcalO)$ already is a sheaf, so no need to sheafifiy here.
  $endgroup$
  – red_trumpet
  Mar 21 at 20:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are right about both things.
  $endgroup$
  – Asal Beag Dubh
  Mar 22 at 9:37
  

  
  
  
  

add a comment | 

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1

$begingroup$

It's always hard to reconstruct someone else's train of thought, but here's one attempted explanation.

As aginensky says, the first step seems to be to use the projection formula. This shows that $(p_2)_* L_1 = M_1 otimes (p_2)_* O_Z$ where I have written $Z$ to denote the product (to save typing).

So now it is enough to show that $(p_2)_* O_Z = O_Y_1$. To do this, recall that $(p_2)_* O_Z$ is defined by

beginalign* U &mapsto H^0((p_2)^-1(U), O_(p_2)^-1(U))\
&= H^0(U times X, O_U times X)
endalign*
Now Künneth tells us that this is isomorphic to
beginalign* &H^0(U,O_U) otimes_k H^0(X,O_X) =H^0(U,O_U)endalign*
using the fact that $X$ is complete.

share|cite|improve this answer

edited Mar 22 at 9:34

answered Mar 21 at 15:36

Asal Beag DubhAsal Beag Dubh

68515

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Seems correct. Two notes: In the last equation, it should be $otimes$ instead of $times$, and afaik $U mapsto H^0(Utimes X, mathcalO)$ already is a sheaf, so no need to sheafifiy here.
  $endgroup$
  – red_trumpet
  Mar 21 at 20:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are right about both things.
  $endgroup$
  – Asal Beag Dubh
  Mar 22 at 9:37
  

  
  
  
  

add a comment | 

1

$begingroup$

It's always hard to reconstruct someone else's train of thought, but here's one attempted explanation.

As aginensky says, the first step seems to be to use the projection formula. This shows that $(p_2)_* L_1 = M_1 otimes (p_2)_* O_Z$ where I have written $Z$ to denote the product (to save typing).

So now it is enough to show that $(p_2)_* O_Z = O_Y_1$. To do this, recall that $(p_2)_* O_Z$ is defined by

beginalign* U &mapsto H^0((p_2)^-1(U), O_(p_2)^-1(U))\
&= H^0(U times X, O_U times X)
endalign*
Now Künneth tells us that this is isomorphic to
beginalign* &H^0(U,O_U) otimes_k H^0(X,O_X) =H^0(U,O_U)endalign*
using the fact that $X$ is complete.

share|cite|improve this answer

edited Mar 22 at 9:34

answered Mar 21 at 15:36

Asal Beag DubhAsal Beag Dubh

68515

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Seems correct. Two notes: In the last equation, it should be $otimes$ instead of $times$, and afaik $U mapsto H^0(Utimes X, mathcalO)$ already is a sheaf, so no need to sheafifiy here.
  $endgroup$
  – red_trumpet
  Mar 21 at 20:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are right about both things.
  $endgroup$
  – Asal Beag Dubh
  Mar 22 at 9:37
  

  
  
  
  

add a comment | 

$begingroup$

It's always hard to reconstruct someone else's train of thought, but here's one attempted explanation.

As aginensky says, the first step seems to be to use the projection formula. This shows that $(p_2)_* L_1 = M_1 otimes (p_2)_* O_Z$ where I have written $Z$ to denote the product (to save typing).

So now it is enough to show that $(p_2)_* O_Z = O_Y_1$. To do this, recall that $(p_2)_* O_Z$ is defined by

beginalign* U &mapsto H^0((p_2)^-1(U), O_(p_2)^-1(U))\
&= H^0(U times X, O_U times X)
endalign*
Now Künneth tells us that this is isomorphic to
beginalign* &H^0(U,O_U) otimes_k H^0(X,O_X) =H^0(U,O_U)endalign*
using the fact that $X$ is complete.

share|cite|improve this answer

edited Mar 22 at 9:34

answered Mar 21 at 15:36

Asal Beag DubhAsal Beag Dubh

68515

$endgroup$

share|cite|improve this answer

edited Mar 22 at 9:34

answered Mar 21 at 15:36

Asal Beag DubhAsal Beag Dubh

68515

answered Mar 21 at 15:36

Asal Beag DubhAsal Beag Dubh

68515

answered Mar 21 at 15:36

Asal Beag DubhAsal Beag Dubh

68515