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?