I am looking for a proof for the following statement



> 
> A smooth projective variety $X$ has maximal Albanese dimension if and
>  only if the cotangent bundle of $X$ is generically generated by its
>  global sections, that is,$$ H^0(X,\Omega\_X^1)\otimes\mathcal O\_X\to
>  \mathcal \Omega\_X^1$$ is surjective at the generic point of $X$.
> 
> 
> 



Maximal Albanese variety
In characteristic $0$ that follows from Sard's theorem / generic smoothness. In characteristic $p$, I believe that this is false (I still need to check how double point singularities, $\text{Zero}(x\_1^2 + \dots + x\_{n-1}^2+x\_n^2 + x\_{n+1}^p)$, affect the cotangent sheaf). 


**Characteristic 0.**In characteristic $0$, the map of sheaves above is the pullback map, $$d(\text{alb}\_X)^\dagger: \text{alb}\_X^\*\Omega\_{\text{Alb}\_X/k} \to \Omega\_{X/k}.$$ Since the Albanese morphism from $X$ to its image, $\text{alb}\_X(X)$, is generically smooth, the dimension of the image equals the rank of $d\text{alb}\_X^\dagger$ at a generic point of $X$.


**Positive characteristic.** However, this seems to be false in positive characteristic. Here is an e-mail that I wrote a few years ago about positive characteristic counterexamples to a similar question. I added some TeX formatting to the e-mail.


"Dear Chris,


I saw Davesh today, and we talked more about the issue we were discussing on Friday regarding purely inseparable covers of Abelian varieties. I now believe that there are serious problems whenever the dimension is $> 1.$ Let $k$ be an algebraically closed field of characteristic $p > 0$. Let $A$ be an Abelian variety of dimension $g > 1$. For a "generic" element $f$ in $k(A)$, the corresponding rational transformation $f : A \dashrightarrow \mathbb{P}^1$ "regularizes" on a blowing up, $u : A' \to A$, such that $A'$ is smooth over $k$. Now consider the following Cartesian diagram, $$ \begin{array}{rrl} X & \xrightarrow{i} & A' \\ g~\downarrow & & \downarrow~f \\ \mathbb{P}^1 &\xrightarrow{F}& \mathbb{P}^1\end{array}$$ where the bottom horizontal $k$-morphism $\mathbb{P}^1 \to \mathbb{P}^1$ ~~is the~~ arises from the Frobenius morphism . The purely inseparable morphism to consider is $i$.


The issue is this: for $f$ generic, the "critical" points of $f$ are just finitely many ordinary double points (I am assuming that $p$ is different from $2$). If $\text{dim}(A) = 1$, then the points of $X$ above such critical points are non-normal. When you normalize, then $X$ is smooth, and your previous argument is valid, i.e., $X$ is an elliptic curve. However, if $\text{dim}(A) > 1$, then the points of $X$ above these finitely many critical points are, in fact, already normal. Thus, $X$ is normal yet singular. So $X$ cannot be an Abelian variety.


Best regards,


Jason" 


By construction, the morphism $i$ is finite. Thus, $X$ has maximal Albanese dimension. By construction $H^0(A',\Omega\_{A'/k})$ generically generates $\Omega\_f$, and thus the $i$-pullback also generically generates $\Omega\_g$. If the dimension of the Albanese of $X$ were strictly greater than $\text{dim}(A)$, then there would be more etale $\ell$-sheeted covers of $X$ than there are of $A$, for $\ell$ prime to the characteristic. Yet, since $i$ is purely inseparable, the pullback map on etale fundamental groups is an isomorphism. If the Albanese morphism of $X$ were different from $i$, it would be a generically finite morphism that factors $i$. Since $i$ has prime degree $p$, that is impossible. Thus, $i$ is the Albanese morphism.


Since $i$ is purely inseparable, the rank of $d(\text{alb}\_X)^\dagger$ at a general point is at most $\text{dim}(A)-1$. Finally, since $\Omega\_i$ is the pullback of $\Omega\_F$, i.e., the pullback of $\mathcal{O}\_{\mathbb{P}^1}(-2)$, there are no nonzero global sections of $\Omega\_i$. Thus, from the usual exact sequence, $$0 \to \text{Image}(d(\text{alb}\_X)^\dagger) \to \Omega\_{X/k} \to \Omega\_i \to 0,$$ it follows that the rank of the following sheaf homomorphism at a general point of $X$ equals $\text{dim}(A)-1$, $$H^0(X,\Omega\_{X/k})\otimes\_k \mathcal{O}\_X \to \Omega\_{X/k}.$$


