I would like to prove (or find a counterexample to) the following statement:

Let $X$ be a complex analytic scheme and let $\pi: Y \to X$ be its universal cover. Let $F$ be a coherent sheaf on $X$ and suppose that $L\otimes F\cong F $ for all $L\in \operatorname{Pic}^0(X)$. Then there exists a sheaf $G$ on $Y$, essentially unique up to deck transformations, such that $F\cong \pi_*G$. Edit: In light of Jason's counterexample, I will additionally assume that $\pi_1(X)$ is Abelian.

Edit: I am willing to add the following strong assumptions on $X$ (although I'm not sure they are needed). Namely that $X$ is compact and that the normalizations of all irreducible components of $X$ are simply-connected.


(1) I'm also (maybe even more) interested in the algebraic setting and would happy to see the correct formulation of the above statement in the algebraic category.

(2) I stated the above in the generality in which I think it might be true, but the situation I am actually interested is far more specific, so if the above is false, I would be interested in additional hypothesis (on either $F$ or on $X$) such that the statement would be true.


A simple example is where $X$ is rational curve with one node (e.g. a nodal cubic curve). Then $Y$ is an infinite chain of $\mathbb{P}^1$s meeting in nodes. A torsion free, rank 1, degree 0 sheaf on $X$ is either a degree 0 line bundle (which does not push forward from $Y$ but fails the assumption) or is the push forward of the structure sheaf of one of the components of $Y$.

In general, I do not want to assume that $X$ is either reduced or irreducible.

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    $\begingroup$ What is $Pic^0$ for an arbitrary complex analytic scheme? $\endgroup$ – Mohan Jun 6 '18 at 22:20
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    $\begingroup$ I would say that $\operatorname{Pic}^0(X)$ is the group of line bundles $L$ with $c_1(L)=0$. $\endgroup$ – Jim Bryan Jun 6 '18 at 22:33
  • $\begingroup$ If $X=\mathbb{C}^*$, then Picard group is trivial. Take $F=\mathcal{O}$, which can not be the direct image of a sheaf from the universal cover, since no non-zero sheaf will have direct image coherent. $\endgroup$ – Mohan Jun 6 '18 at 22:41
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    $\begingroup$ I guess I could include a properness assumption on $X$, although in the case of $\mathbb{C}^*$ aren't there non-trivial line bundles in the analytic topology? There are certainly non-trivial flat line bundles. $\endgroup$ – Jim Bryan Jun 6 '18 at 23:09
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    $\begingroup$ Universal covers (even in proper case) are seldom finite, so I don't see how you will have a coherent sheaf as the direct image in general. $\endgroup$ – Mohan Jun 6 '18 at 23:37

Let $E$ be the elliptic curve. Let $E_1$ and $E_2$ be two different double covers of $E$, with $E = E_1 /x_1$ and $E=E_2/x_2$ for two-torsion points $x_1,x_2$.

Let $M$ be the minimal resolution of singularities of $ E_1 \times E_2 /\langle (a,b) \to (-a,-b), (a,b) \to (a+ x_1, x_2-b) \rangle $.

Then $M$ is an Enriques surface, with fundamental group $\mathbb Z/2$. We can embed $E$ into $M$ either horizontally or vertically. In the horizontal case the inverse image in the double cover is $E_1$, and in the verticle case the inverse image is $E_2$.

By combining these, we get an embedding of $E$ into $M \times M$, whose inverse image in the universal cover is $E$, mapping to $E$ by the multiplication by two map.

On the other hand the Picard group of $M \times M$ is $\mathbb Z/2 \times \mathbb Z/2$ (as long as it is defined using rational Chern classes, and not integral Chern classes, for which the statement is trivially false as $\mathcal O_X$ on $M$ is a counterxample). Pulled back to $E$, these line bundles are all the two-torsion vector bundles.

So the question becomes:

Is any coherent sheaf on $E$ which is isomorphic to its own tensor product with any two-torsion line bundle equal to a pushforward for $E$ under the multiplication-by-two map?

Then the answer is "no", because indecomposable vector bundles on $E$ are classified by their determinant, so every indecomposable rank two bundle has this property, but none is a pushforward under multiplication by two as that would force it to have rank a multiple of four.

However, it may still be true if the fundamental group is cyclic.

  • $\begingroup$ Thanks Will. Very clever counter example. I clearly still need to find the correct general statement to pose and prove which will apply in my particular example (which is a bit too baroque to pose as its own question). This example is helpful (shows that I should perhaps have some hypothesis about the support of my sheaf vis a vis the fundamental group of $X$). $\endgroup$ – Jim Bryan Jun 16 '18 at 15:30

That is false. I am taking a break from something else, so I will mostly refer to other MO answers. I gather from your example that you allow $X$ to be singular. Then Simpson proved that every finitely presented group $G$ is isomorphic to the fundamental group of a (usually singular) complex projective variety $X$, Theorem 12.1 of the following.

Simpson, Carlos
Local systems on proper algebraic V-manifolds.
Pure Appl. Math. Q. 7 (2011), no. 4,
Special Issue: In memory of Eckart Viehweg, 1675–1759.

I found this reference from Andy Putman's answer to this MO question.
Fundamental groups of compact Kähler manifolds

Let $G$ be a nontrivial finitely presented group whose Abelianization is trivial, e.g., the alternating group $\mathfrak{A}_n$ for $n\geq 5$. Then $\text{Pic}^0(X)$ is trivial. Thus the structure sheaf of $X$ has the invariance property. Yet the structure sheaf is not the pushforward of a coherent sheaf, because the rank of the structure sheaf, $1$, is not divisible by the order of $G$.

Let me anticipate the next question: "What if $E_\rho\otimes F$ is isomorphic to $F^{\oplus r}$ for every complex representation $\rho$ of $\pi_1(X)$ with finite dimension $r$ and with associated vector bundle $E_\rho$ on $X$?" However, the fantastic answers to the following question resolve this negatively; there are many finitely presented groups having no nontrivial representation, and then you can take $F$ to be the structure sheaf.
Finitely presented sub-groups of GL(n,C)

Edit. I just noticed the edit by the OP that allows (and insists) that $X$ be highly reducible. That makes things much simpler. Let $<g_1,\dots,g_m|p_1,\dots, p_n>$ be a presentation of $G$. For each generator $g_i$, let $X'_i \cong C_i\times \mathbb{P}^3$ be an irreducible component, where $(C_i,x_i)$ is a nodal plane cubic with a marked point $p_i$. Let $(C_0,y_1,\dots,y_m)$ be an $m$-marked curve of genus $0$, and let $X'_0$ be $C_0\times \mathbb{P}^3$.

Begin with $X'$, the reducible $4$-fold obtained by gluing each $X'_i,$ $i=1,\dots,m$, to $X'_0$ by identifying $\{x_i\}$ with $\{y_i\}$, inducing an identification of $\{x_i\}\times \mathbb{P}^3$ with $\{y_i\}\times \mathbb{P}^3$. The fundamental group of $X'$ is the free group on $m$-generators $<g_1,\dots,g_m>$.

Moreover, for every element $p$ of the fundamental group, the subgroup $<p>$ of the fundamental group is the image fundamental group under pushforward for an unramified, finite morphism of complex projective schemes, $$f_p:D_p\times \mathbb{P}^3\to X',$$ where $D_p$ is an $n$-gon of $\mathbb{P}^1$s that lifts to $\widetilde{X}'$ once we separate a single node. Choose a closed immersion, $$e_p:D_p\hookrightarrow \mathbb{P}^3.$$ The graph of $e_p$ in $D_p\times \mathbb{P}^3$ is a copy of $D_p$.

For each $j=1,\dots,n$, let $W_j$ denote the image under $f_{p_j}$ of the graph of $e_{p_j}$. This is an $n$-gon in $X'$ whose image fundamental group in $<g_1,\dots,g_m>$ equals $<p_j>$. If the embeddings $e_{j}$ are chosen sufficiently general, then these curves $W_j$ are pairwise disjoint. Let $X_j$ be a copy of $\mathbb{P}^3$, and glue this to $X'$ by identifying $W_j$ with time image of $e_{p_j}$ in $X_j$. Then the reduced, complex projective scheme $X$ obtained by gluing each $X_j$ to $X'$ has fundamental group $< g_1,\dots,g_m|p_1,\dots,p_n>$.

  • $\begingroup$ Thanks for your answer. Clearly I should have included some hypotheses on $\pi_1(X)$. In the case I am most interested in, I can take $\pi_1(X)$ to be $\mathbb{Z \times Z}$. $\endgroup$ – Jim Bryan Jun 7 '18 at 16:35
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    $\begingroup$ @JimBryan. I believe that should be true. Probably one of the experts on Abelian varieties can answer that better than me. $\endgroup$ – Jason Starr Jun 7 '18 at 18:17

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