$H^1(X,O_X)$, holomorphic $1$-forms, and $b_1(X)/2$ for normal $X$.

Question

Suppose $X$ is a normal projective variety over $\mathbb C$. In the case $X$ is smooth according to Hodge theory $h^1(X,O(X))$ is the dimension of the space of holomorphic $1$-forms on $X$ and this number is equal as well to the half of the first Betti number $b_1(X)/2$ .

I would like to know what happen in the case when $X$ is singular and normal.

1) Is there some relation (equality or inequality) between $h^1(X,O(X))$ and $b_1(X)$? For example does $b_1(X)=0$ imply $h^1(X,O(X))=0$?

2) Suppose that $h^1(X,O(X))=n$ is it true that there is a canonical $n$-dimensional space of $1$-forms on $X$, holomorphic outside of its singularities? (if yes, can something be said about their behaviour at singularities?)

Is there some pedagogical reference treating these questions?

Answer 1

(Although I have pretty much "retired" from Mathoverflow, I will answer this, since the answer is nice but probably not all that well known.)

Theorem. If $X$ is complex normal projective variety, then it is still true that $b_1(X)=2h^1(\mathcal{O}_X)$.

Proof. Let $\pi:\tilde X\to X$ be a desingularization. Since $X$ is normal, the fibres of $\pi$ are connected. Therefore $\pi_*\mathbb{Z}$ (with analytic topology) is connected. It follows that $H^1(X,\mathbb{Z})\to H^1(\tilde X, \mathbb{Z})$ is injective since it can be identified with the edge map for Leray. Therefore the, a priori mixed, Hodge structure on $H^1(X)$ is pure of type $\lbrace (1,0),(0,1)\rbrace$. Consequently $$b_1(X)=2\dim [H^1(X)]^{(0,1)}=2 \dim im[H^1(X,\mathcal{O}_X)\to H^1(\tilde X, \mathcal{O}_{\tilde X}] = 2h^1(\mathcal{O}_X)$$

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Regarding your question 2, you can construct an Albanese map $X\to Alb(X)$ to the torus associated to the dual Hodge structure on $H_1(X)$. The space of $1$-forms on $Alb(X)$ will pullback to a space of the required dimension on $X$.

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To address your comments: a general reference for mixed Hodge structures is the book by Peters and Steenbrink (although this may be bit a heavy). And yes, the argument does work in the Moishezon case, and a bit more generally. I guess that I may as well admit the above argument was extracted from paper in Duke from 1990; this contains some more details and elaborations as well.