# Splitness of a short exact sequence on a curve

Let $C$ be a curve with genus $g > 1$. Consider the product $C \times C$, with natural projections $p_1$ and $p_2$ (from the first and second factor, respectively) to $C$.

Consider the following short exact sequence of sheaves on $C \times C$:

$$0 \rightarrow p_2^* K_C^{\otimes 2} \rightarrow p_2^* K_C^{\otimes 2} \otimes \mathcal{O}_{C \times C}(\Delta) \rightarrow K_C \rightarrow 0$$

where $K_C$ is the canonical bundle of $C$, $\Delta$ is the diagonal in $C \times C$, and the map $p_2^* K_C^{\otimes 2} \otimes \mathcal{O}_{C \times C}(\Delta) \rightarrow K_C$ is the restriction map to $\Delta$. The last entry $K_C$ is a line bundle on $\Delta$, which is a sheaf on $C \times C$ supported on $\Delta$.

Now consider the projection ${p_1}_*$ of this sequence. Is the resulting sequence on $C$ split?

• I think I understand, but maybe it would be good to say what the maps are and how $K_C$ can live on $C\times C$. – Hoot May 21 '16 at 18:02
• @Hoot Good point. Clarifications added. – Anonymous May 21 '16 at 18:38

The answer is still negative, though I feel the previous answer is incorrect. Let $L=p_1^*K_C, M=p_2^*K_C$ and then saying $0\to p_{1*}M^2\to p_{1*}M^2(\Delta)\to K_C\to 0$ (the last is surjective, since the corresponding $R^1$ is zero) splits is same as sayining after twisting by $K_C^{-1}$, the section $1$ can be lifted. This implies in particular $H^0(L^{-1}\otimes M^2(\Delta))\neq 0$. But $p_{2*}(L^{-1}\otimes \mathcal{O}(\Delta))=0$ since $g>1$ and thus the $H^0=0$ which proves that the sequence that you need is not split.

• When you wrote $p_{1*}p_2^*M^2$, did you mean $p_{1*}M^2$? – Anonymous May 23 '16 at 17:42
• Yes, that is what I meant. Will edit it. – Mohan May 23 '16 at 17:48
• Could you provide some more details? For example, when you said "section 1 can be lifted", why does that imply that $H^0(C \times C, L^{-1} \otimes M^2(\Delta)) \neq 0$? Also, what did you mean by ${p_2}_*(L^{-1} \otimes \mathcal{O}(\Delta)) = 0$? – Anonymous May 23 '16 at 17:57
• One has $p_{1*}M^2(\Delta)\otimes K_C^{-1}\to \mathcal{O}_C\to 0$, by twisting the above sequence by $K_C^{-1}$. If the first sequence splits, so does the second and then one has an inclusion $\mathcal{O}_C\subset p_{1*}M^2(\Delta)\otimes K_C^{-1}$, which means the latter has a non-zero section. By projection formula, this is same as saying $H^0(L^{-1}\otimes M^2(\Delta))\neq 0$. Looking at the other projection, we see that $H^0(K_C^2\otimes p_{2*}(L^{-1}\otimes\mathcal{O}(\Delta))\neq 0$. I do not understand your confusion about the last part. – Mohan May 23 '16 at 18:24
• Forgive me for being slow on this. So by projection formula we obtain $H^0(C, {p_1}_*M^2(\Delta) \otimes K_C^{-1}) = H^0(C, {p_1}_*(M^2(\Delta) \otimes p_1^*K_C^{-1})) \neq 0$. But how is this related to the conclusion? We are not considering the splitness of ${p_2}_*$ of the original sequence, so how do we draw a contradiction? – Anonymous May 23 '16 at 19:05

I came up with an alternative answer. Let $\mathcal{M}_{g,n}$ be the moduli space of Riemann surfaces of genus $g$ with $n$ marked points. Then our sequence

$$0 \rightarrow {p_1}_* p_2^* K_C^{\otimes 2} \rightarrow {p_1}_* (p_2^* K_C^{\otimes 2} \otimes \mathcal{O}_{C \times C}(\Delta)) \rightarrow K_C \rightarrow 0$$

is in fact the same as

$$0 \rightarrow H^0(C,K_C^2) \otimes \mathcal{O}_C \rightarrow T^*\mathcal{M}_{g,1}|_{C} \rightarrow K_C \rightarrow 0$$

which is dual to

$$0 \rightarrow T_C \rightarrow T\mathcal{M}_{g,1}|_{C} \rightarrow H^1(C,T_C) \otimes \mathcal{O}_C \rightarrow 0$$

This last sequence is maximally non-split.