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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?

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    $\begingroup$ The geometry is that when you blow up a curve in a threefold, you modify the intermediate Jacobian by adding the Jacobian of the curve. This seems pretty geometric to me. $\endgroup$
    – abx
    Commented Jun 25, 2020 at 16:33
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    $\begingroup$ At the level of intuition --so a very sloppy comment (I assume that we are speaking about threefolds here). Suppose you take a (say) smooth threefold and you want to blow it up in a (say) smooth curve. $H^{2,1}(X)$ will not tell you much about the curve you are blowing up (just its genus). However, if you quotient out by $H^3(X, \mathbb{Z})$ then you want to recover the Jacobian of curve, which by Torelli determines the curve up to isomorphism (and smooth projective birational curves are isomorphic). Do not take it as a complete answer, but I found this picture helpful back in the days. $\endgroup$
    – Enrico
    Commented Jun 25, 2020 at 16:35
  • $\begingroup$ @abx: where can I find a proof of this fact you have quoted. $\endgroup$
    – user267839
    Commented Jun 25, 2020 at 17:21
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    $\begingroup$ In Clemens-Griffiths of course. $\endgroup$
    – abx
    Commented Jun 25, 2020 at 19:22
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    $\begingroup$ An excellent modern survey by Beauville: arxiv.org/pdf/1507.02476.pdf; see section 3 for the intermediate Jacobian. $\endgroup$
    – Evgeny Shinder
    Commented Jun 28, 2020 at 23:06

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