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user267839
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Let $X$ be a smooth variety over complex numbers $\mathbb{C}$, say a threefold for sake of better intuition. Is there any geometrical intuition behind the fact that the Clemens-Griffiths component of the intermediate Jacobian

$$J(X)= H^1(\Omega^2,X)/H^3(X, \mathbb{Z})$$

is a birational invariant. Recall the Clemens-Griffiths component is called the associated polarized torus $(J(X), \theta)$ with non-degenerated divisor $\theta$.

I took some time to look in the proof and understood the single steps but haven't still any geometric intuition how to think about it. For example we know that $H^1(\Omega^2,X)$ is a priori not a birational invariant, but quotient out $H^3(X, \mathbb{Z})$ seems somehow to play important role in "clearing" the obstuctions that could occure.

Thinking of blowing up as archetypical examples for birational transformations, is there any picture one can have in mind how $H^3(X, \mathbb{Z})$ "cleans" the defect that in general prevents $H^1(\Omega^2,X)$ from beeing birational invariant. Is there any "geometry" behind?

Let $X$ be a variety over complex numbers $\mathbb{C}$. Is there any geometrical intuition behind the fact that the Clemens-Griffiths component of the intermediate Jacobian

$$J(X)= H^1(\Omega^2,X)/H^3(X, \mathbb{Z})$$

is a birational invariant. Recall the Clemens-Griffiths component is called the associated polarized torus $(J(X), \theta)$ with non-degenerated divisor $\theta$.

I took some time to look in the proof and understood the single steps but haven't still any geometric intuition how to think about it. For example we know that $H^1(\Omega^2,X)$ is a priori not a birational invariant, but quotient out $H^3(X, \mathbb{Z})$ seems somehow to play important role in "clearing" the obstuctions that could occure.

Thinking of blowing up as archetypical examples for birational transformations, is there any picture one can have in mind how $H^3(X, \mathbb{Z})$ "cleans" the defect that in general prevents $H^1(\Omega^2,X)$ from beeing birational invariant. Is there any "geometry" behind?

Let $X$ be a smooth variety over complex numbers $\mathbb{C}$, say a threefold for sake of better intuition. Is there any geometrical intuition behind the fact that the Clemens-Griffiths component of the intermediate Jacobian

$$J(X)= H^1(\Omega^2,X)/H^3(X, \mathbb{Z})$$

is a birational invariant. Recall the Clemens-Griffiths component is called the associated polarized torus $(J(X), \theta)$ with non-degenerated divisor $\theta$.

I took some time to look in the proof and understood the single steps but haven't still any geometric intuition how to think about it. For example we know that $H^1(\Omega^2,X)$ is a priori not a birational invariant, but quotient out $H^3(X, \mathbb{Z})$ seems somehow to play important role in "clearing" the obstuctions that could occure.

Thinking of blowing up as archetypical examples for birational transformations, is there any picture one can have in mind how $H^3(X, \mathbb{Z})$ "cleans" the defect that in general prevents $H^1(\Omega^2,X)$ from beeing birational invariant. Is there any "geometry" behind?

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user267839
user267839
  • 6k
  • 2
  • 11
  • 42

Clemens-Griffiths component birational invariant

Let $X$ be a variety over complex numbers $\mathbb{C}$. Is there any geometrical intuition behind the fact that the Clemens-Griffiths component of the intermediate Jacobian

$$J(X)= H^1(\Omega^2,X)/H^3(X, \mathbb{Z})$$

is a birational invariant. Recall the Clemens-Griffiths component is called the associated polarized torus $(J(X), \theta)$ with non-degenerated divisor $\theta$.

I took some time to look in the proof and understood the single steps but haven't still any geometric intuition how to think about it. For example we know that $H^1(\Omega^2,X)$ is a priori not a birational invariant, but quotient out $H^3(X, \mathbb{Z})$ seems somehow to play important role in "clearing" the obstuctions that could occure.

Thinking of blowing up as archetypical examples for birational transformations, is there any picture one can have in mind how $H^3(X, \mathbb{Z})$ "cleans" the defect that in general prevents $H^1(\Omega^2,X)$ from beeing birational invariant. Is there any "geometry" behind?