Let $X$ be a smooth projective variety over a field $k$ with char$k=0$ and $\mathcal{L}$ be a very ample line bundle on $X$. Let $\mathcal{F}$ be a coherent sheaf on $X$. It is well-know that if $\mathcal{F}\otimes \mathcal{L}\cong \mathcal{F}$, then $\text{dim}(\text{supp}\mathcal{F})=0$. Actually it can be proved by looking at the degree of the Hilbert polynomial associated with $\mathcal{F}$.

Now Let $X$ and $Y$ be two smooth projective varieties over a field $k$ with char$k=0$ and $\mathcal{L}$ be a very ample line bundle on $X$. Let $p: X\times Y\to X$ be the projection. Let $\mathcal{F}$ be a coherent sheaf on $X\times Y$ and we consider $p(\text{supp}\mathcal{F})$, the projection to $X$ of the support of $\mathcal{F}$. If $\mathcal{F}\otimes p^*\mathcal{L}\cong \mathcal{F}$, do we have $$\text{dim}(p(\text{supp}\mathcal{F}))=0?$$

By studying the Hilbert polynomial it can be proved that $\text{dim}(\text{supp}R^ip_*\mathcal{F})=0$. But this seems to be weaker than the result I expect.

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    $\begingroup$ Note that if $F$ has this property, then so does $F\otimes p_Y^* E$ for any sheaf $E$ on $Y$. In particular, you know that $dim(supp (p_*(F\otimes p_Y^*E)))=0$ for any $E$. Taking $E$ to be a high power of an ample line bundle gives you what you want. $\endgroup$ – t3suji Dec 4 '16 at 7:25
  • $\begingroup$ @t3suji I'm sorry why this gives what I want? Could you give some more details? $\endgroup$ – Zhaoting Wei Dec 4 '16 at 16:25
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    $\begingroup$ The claim is that for any (quasi-compact noetherian) scheme $S$ and any coherent sheaf $F$ on $S\times{\mathbb P}^n$, we have $p(supp(F))=supp(p_*(F(n)))$ for $n\gg 0$. Here $p$ is the projection onto $S$. $\endgroup$ – t3suji Dec 4 '16 at 22:16
  • $\begingroup$ @t3suji Is it a standard result in algebraic geometry? $\endgroup$ – Zhaoting Wei Dec 5 '16 at 2:41
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    $\begingroup$ I don't have a reference, but it is pretty straightforward: enough to consider the case $S=Spec(R)$, and then a coherent sheaf $F$ is given by a finitely generated graded $R[x_0,\dots,x_n]$-module $M=\bigoplus M_d$; for large enough $D$, $M_D$ generates $M_{\ge D}:=\bigoplus_{d\ge D} M_d$, so the annihilator ideal of $M_D$ in $R$ is contained in the annihilator ideal of $M_{\ge D}$. $\endgroup$ – t3suji Dec 5 '16 at 6:05

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