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Motivation:

For a given rank 2 vector bundle we want to know how many theta-characteristic valued twisted endomorphisms it has.

Setting:

Let $C$ be a smooth algebraic curve over a field of characteristic $\neq 2$ and let $\vartheta$ be a theta characteristic on $C$ ($\vartheta^2 \cong K_C$), which does not have any nontrivial global sections. Furthermore we are given line bundles $L_1,L_2$ on $C$ and we consider an extension $[E] \in \operatorname{Ext}^1(L_2,L_1)$. We take this exact sequence, dualize it and tensor it with $\vartheta \otimes L_2$. We find the exact sequence:

$$0 \to \vartheta \to \vartheta \otimes E^\vee \to \vartheta \otimes L_1^{-1}\otimes L_2 \to 0$$

Looking at the long exact sequence in cohomology and using $H^0(C,\vartheta) = H^1(C,\vartheta) = 0$ we find: $H^0(C,\vartheta \otimes L_1^{-1}\otimes L_2) \cong H^0(C,\vartheta \otimes E^\vee \otimes L_2)$ (1). If we tensor our original short exact sequence by $\vartheta\otimes L_1^{-1}$ we get

$$0 \to \vartheta \to \vartheta \otimes L_1^{-1}\otimes E \to \vartheta \otimes L_1^{-1} \otimes L_2 \to 0.$$

From the long exact sequence we derive $H^0(C,\vartheta\otimes L_1^{-1}\otimes E) \cong H^0(C,\vartheta\otimes L_1^{-1}\otimes L_2)$ (2). Now we tensor the original short exact sequence by $\vartheta \otimes E^\vee$ and then look at the long exact sequence:

$$0 \to H^0(\vartheta\otimes E^\vee \otimes L_1) \to H^0(\vartheta \otimes E^\vee \otimes E) \to H^0(\vartheta \otimes E^\vee\otimes L_2)\xrightarrow{\delta}$$ $$ H^1(\vartheta\otimes E^\vee \otimes L_1) \to H^1(\vartheta \otimes E^\vee \otimes E) \to H^1(\vartheta \otimes E^\vee\otimes L_2) \to 0$$

Now to learn something about $H^0(\vartheta \otimes E^\vee \otimes E)$ one has to understand $\delta$. If we apply (1), (2) and Serre duality we get the following new map

$$\Delta: H^0(\vartheta \otimes L_1^{-1}\otimes L_2) \cong H^0(\vartheta \otimes E^\vee \otimes L_2) \xrightarrow{\delta} H^1(\vartheta \otimes E^\vee \otimes L_1) \cong H^0(\vartheta \otimes E \otimes L_1^\vee)^\vee \cong H^0(\vartheta \otimes L_1^{-1} \otimes L_2)^\vee$$.

The Questions:

Can one understand (or describe in a different way) this map $\Delta$ in terms of the extension. In other words what is the map $\operatorname{Ext}^1(L_2,L_1) \to \left( H^0(L_1^{-1}\otimes L_2)^\vee \right)^{\otimes 2}$? What is the rank of $\Delta$? Furthermore i suspect that $\Delta$ viewed as a bilinear form is antisymmetric, because the parity of $h^0(\operatorname{End}(E)\otimes \vartheta)$ is constant in families by a result of Mumford.

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1 Answer 1

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I claim that I can give a reasonably "geometric" formula.

Consider the surface $C \times C$ and the diagonal $\Delta$. The the line bundle $\vartheta \boxtimes \vartheta$ has vanishing cohomology in each degree by the Künneth formula, so the global sections of the line bundle $(\vartheta \boxtimes \vartheta) \otimes \mathcal O(\Delta)$ are the same as the global sections of the quotient, i.e. the restriction of $(\vartheta \boxtimes \vartheta) \otimes \mathcal O(\Delta)$ to $\Delta$, which is $\vartheta \otimes \vartheta \otimes K_C^{-1} = \mathcal O_C$. In particular there is a canonical section $f \in H^0( C \times C, (\vartheta \boxtimes \vartheta) \otimes \mathcal O(\Delta))$. Pulling $f$ back on the map switching the two copies of $C$ has the effect of negating the residue at $\Delta$, hence negating $f$, so $f$ is antisymmetric.

Now for $a, b $ in $H^0( C, \vartheta \otimes L_1^{-1} \otimes L_2)$, we can form the section $a \otimes b - b\otimes a$ in $$H^0(C \times C, ( \vartheta \otimes L_1^{-1} \otimes L_2) \boxtimes ( \vartheta \otimes L_1^{-1} \otimes L_2) ) = H^0( C, \vartheta \otimes L_1^{-1} \otimes L_2)^{\otimes 2}. $$ Restricted to the diagonal $\Delta$, this section becomes $ab-ba=0$, so $a\otimes b - b \otimes a$ in fact gives a section of

$$H^0(C \times C, \mathcal O(-\Delta) \otimes ( ( \vartheta \otimes L_1^{-1} \otimes L_2) \boxtimes ( \vartheta \otimes L_1^{-1} \otimes L_2) ) ).$$

Thus the product $f ( a\otimes b - b \otimes a)$ gives a section in $$ H^0(C\times C, (\vartheta \boxtimes \vartheta) \otimes ( \vartheta \otimes L_1^{-1} \otimes L_2) \boxtimes ( \vartheta \otimes L_1^{-1} \otimes L_2))$$

$$=H^0(C\times C, ( K_C \otimes L_1^{-1} \otimes L_2) \boxtimes ( K_C\otimes L_1^{-1} \otimes L_2)).$$ $$= H^0(C, K_C \times L_1^{-1} \otimes L_2)^{\otimes 2}. $$

Since $H^0(C, K_C \times L_1^{-1} \otimes L_2)$ is dual to $\operatorname{Ext}^1(L_2,L_1)$ by Serre duality, we can pair this twice with a class in $\operatorname{Ext}^1(L_2,L_1)$ to obtain a number.

This pairing is the bilinear form you're looking for, which is indeed antisymmetric. Whether that expression is better for you is not clear. This description may be used to calculate the rank of the bilinear form if the Ext class is supported at a divisor of small degree, but I think one could do that more directly (if the support of the Ext class is small relative to the degree of $L_2 \otimes L_1^{-1}$, the vector bundle splits as a direct sum of line bundles, and one can use this description to more easily calculate its global sections.)

I don't have time to write up the proof right now but I can if it would be helpful.

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  • $\begingroup$ Thanks a lot for the quick answer. A description in terms of a multiplication on some variety is indeed what i hoped for although i don't see the identification with the boundary map yet. Could you maybe give a hint where to start the proof? $\endgroup$ Commented May 21 at 8:04
  • $\begingroup$ Regarding the last paragraph: I think for my purposes i need to consider extensions where the resulting vector bundle is a general stable vector bundle. Nothing should split there (im unsure wheter you mean $E$ or $E\otimes E^\vee$. But i still would like to understand what you meant: What is the support of an ext class? Does it have something to do with $H^1(L_1^{-1}\otimes L_2) = H^0(K_C \otimes L_1\otimes L_2^{-1})^{\vee}$? $\endgroup$ Commented May 21 at 8:08
  • $\begingroup$ @clemens_nollau Yes, by the support of an ext class I mean the smallest (in degree) closed subscheme $Z$ of the curve such that the associated linear form factors through the restriction of $H^0(C, K_C \otimes L_1^{-1} \otimes L_2$ to that $H^0(Z, K_C \otimes L_1^{-1} \otimes L_2)$. Equivalently, this is a closed subscheme $Z$ such that the extension splits when pulled back to $\operatorname{Ext}^1(L_2(-Z),L_1)$. The support will not be small for a general extension so this perspective won't be so helpful to you. $\endgroup$
    – Will Sawin
    Commented May 21 at 10:12
  • $\begingroup$ @clemens_nollau Idea of the proof: First check that the bilinear form associated to $\delta$ comes from the natural bilinear form $(\vartheta \otimes E \otimes L_1^{-1}) \otimes (\vartheta \otimes E^\vee \otimes L_2) \to (K_C \otimes L_1^{-1} \otimes L_2)$. (The analogous statement is true for the connecting map of an arbitrary extension of vector bundles). Now $E^\vee$ is $E$ tensored with $L_1^{-1}\otimes L_2^{-1}$ so the two other maps you use to create $\Delta$ are actually dual to each other. This duality uses the determinant so the resulting bilinear form is symplectic. $\endgroup$
    – Will Sawin
    Commented May 21 at 10:20
  • $\begingroup$ @clemens_nollau Now fix $Z$ that the extension class factors through. We can choose $Z$ to be a union of isolated points. Express $E$ as the subbundle of $L_1(Z) \oplus L_2$ consisting of sections $(s_1,s_2)$ where for each section of $K_C \otimes L_1^{-1} $ over $Z$, multiplying by $s_2$ and taking the linear form associated to the Ext class is equivalent to multiplying by $s_1$ and summing the residue at each point of $z$. This requires one of the ways of explicitly expressing the Serre dualiy map by residues. $\endgroup$
    – Will Sawin
    Commented May 21 at 10:26

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