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Let $X$ be a projective integral scheme over $\mathbb{C}$.

If $X$ is smooth, then $\mathrm{h}^1(X,\mathcal{O}_X)$ is the dimension of the Albanese variety of $X$. Probably, even if $X$ is normal, this is still true. I am wondering whether this is always true.

Does $\mathrm{h}^1(X,\mathcal{O}_X)$ equal the dimension of the Albanese variety of $X$?

The Albanese variety of a normal projective variety $X$ is the identity component $\mathrm{Pic}^0_{X}$ of the Picard scheme of $X$ over $\mathbb{C}$. A reference for this is Mochizuki's appendix to his paper I was not able to find the answer to my question in this paper. Indeed, the "construction" of the Albanese of $X$ is done by using an equivariant alteration of $X$, and I can not see how to relate the dimension of the resulting Albanese variety to $\mathrm{H}^1(X,\mathcal{O}_X)$.

A variant of my question can be formulated as follows. I expect both questions to have a positive answer, but I might be wrong.

Is $\mathrm{h}^1(X,\mathcal{O}_X)\neq 0$ if and only if $X$ admits a non-constant map to some abelian variety?

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    $\begingroup$ Are you assuming $X$ normal, or not? If not, what's the definition of the Albanese? $\endgroup$
    – Will Sawin
    Commented Jul 21, 2019 at 6:04
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    $\begingroup$ The answer to the first question depends on what you mean by Albanese. The second question has a negative answer: consider a nodal or cuspidal rational curve. $\endgroup$
    – naf
    Commented Jul 21, 2019 at 7:28
  • $\begingroup$ @WillSawin In Corollary A.11 of the linked paper it is claimed that every (pointed) variety $X$ has an Albanese morphism. This is defined as the data of a semi-abelian variety $G$ and a universal morphism $X\to G$. However, as $X$ is proper, the Albanese variety $G$ will be (in this case) an abelian variety. In the case that $X$ is proper over $\mathbb{C}$, this coincides with $Pic^0_{red} = Pic^0$. $\endgroup$
    – Harry
    Commented Jul 21, 2019 at 8:21
  • $\begingroup$ @ulrich Thank you for your comment. Don't know how I overlooked that.... The answer to both questions is positive if $X$ is normal, right? For the first question, by Albanese I mean of $X$ I mean the data of a semi-abelian variety $G$ and a universal map $X\to G$. But since $X$ is proper, $G$ will be an abelian variety. $\endgroup$
    – Harry
    Commented Jul 21, 2019 at 8:25
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    $\begingroup$ Yes, if $X$ is normal then both questions have a positive answer. It is not true though that the Albanese is $Pic^0$ (even if $X$ is smooth); it is the dual abelian variety. $\endgroup$
    – naf
    Commented Jul 21, 2019 at 12:23

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