If $0\longrightarrow M\longrightarrow E\longrightarrow N\longrightarrow 0$ is a short exact sequence of vector bundles on a surface. Here $M$ and $N$ are line bundles, and so rank $ E$=2. Also, if $L$ is an ample line bundle, we have that $\mu_L(M)>\mu_L(E)>\mu_L(N)$.

My question is this : Why does $\textrm{Ext}^1(N,M)/\textrm{Hom}(N,M)$ make sense?

From the initial exact sequence, by applying the functor Hom$(N,\_)$, we get the sequence $0\longrightarrow \textrm{Hom}(N,M)\longrightarrow \textrm{Hom}(N,E)\longrightarrow \textrm{Hom}(N,N)\longrightarrow \textrm{Ext}^1(N,M)\longrightarrow..$.

Similarly in the case if we apply Hom$(\_,M)$, we get $0\longrightarrow \textrm{Hom}(N,M)\longrightarrow \textrm{Hom}(E,M)\longrightarrow \textrm{Hom}(M,M)\longrightarrow \textrm{Ext}^1(N,M)\longrightarrow..$.

So how do we get this quotient?

  • $\begingroup$ Where did you read that the quotient makes sense? $\endgroup$ – Mariano Suárez-Álvarez Apr 22 '15 at 14:41
  • $\begingroup$ I am reading a paper - Stability of rank-3 Lazarsfeld-Mukai bundles on K3 surfaces. It is mentioned in that. $\endgroup$ – gradstudent Apr 22 '15 at 14:43
  • $\begingroup$ There is no sensible map from Hom to Ext in a general situation. $\endgroup$ – Mariano Suárez-Álvarez Apr 22 '15 at 14:58
  • $\begingroup$ Is there a natural action of Hom on Ext? $\endgroup$ – gradstudent Apr 22 '15 at 15:03
  • 2
    $\begingroup$ There is an action of $\hom(M,M)$ and of $\hom(N,N)$, but not of $\hom(M,N)$ $\endgroup$ – Mariano Suárez-Álvarez Apr 22 '15 at 15:06

As stated in the comments, this does not answer the question at all, but rather one that arose there.

Suppose our category had enough projectives. Choose a projective resolution $d_{\bullet}\!: P_{\bullet} \to N$. Using projectivity of $P_0$, resp. projectivity of $P_1$ and $\text{im}(f_0 \circ d_0) \subseteq \ker(E \to N) = M$, we get lifts of the identity $$\require{AMScd} \begin{CD} P_2 @>d_1>> P_1 @>d_0>> P_0 @>>> N @>>> 0 \\ @VVV @VVf_1V @VVf_0V @VV\text{id}V \\ 0 @>>> M @>>> E @>>> N @>>> 0. \end{CD} $$ It follows that $E \cong M \amalg_{P_1} P_0$ is the pushout. On the other hand, consider the complex $$0 \to \text{Hom}(P_0,M) \xrightarrow{d_0^*} \text{Hom}(P_1,M) \xrightarrow{d_1^*} \text{Hom}(P_2,M) \to \ldots$$ By construction, $f_1 \in \ker(d_1^*)$. This defines the class of our extension in $\text{Ext}^1(N,M) = H^1(\text{Hom}(P_{\bullet},M))$. Accordingly, it splits iff $f_1 \in \text{im}(d_0^*)$, and in fact, we have a bijection between $(d_0^*)^{-1}f_1$ and splittings of our extension*. Then $\text{Hom}(N,M) \cong \ker(d_0^*)$ acts on this simply transitively by addition.

*The splittings $M \amalg_{P_1} P_0 \cong E \to M$ are precisely given by $(\text{id}_M,\varphi)$ with $\varphi \circ d_0 = f_1$, that is, such that $\varphi \in (d_0^*)^{-1}f_1 \subseteq \text{Hom}(P_0,M)$.

  • $\begingroup$ Of course I was being silly in the comments to the question. The argument does apply to coherent sheaves, just by dualizing everything. Sorry. $\endgroup$ – Thomas Poguntke Apr 28 '15 at 18:51

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