# How do we get the quotient $Ext^1(N,M)/Hom(N,M)$?

If $0\longrightarrow M\longrightarrow E\longrightarrow N\longrightarrow 0$ is a short exact sequence of torsion free coherent sheaves on a surface. Here $M$ is a line bundle, $E$ a vector bundle of rank 2 and $N$ the quotient. The exact sequence corresponds to the Harder Narasimhan filtration of $E$ with respect to an ample $L$. So we have that $\mu_L(M)>\mu_L(E)>\mu_L(N)$.

In the paper I am looking at, they consider the Moduli stack of extensions of the above type, call this $\mathcal{E}$. There is a natural projection $p:\mathcal{E}\longrightarrow\mathcal{M}(v(M))\times\mathcal{M}(v(N))$. Here $\mathcal{M}(v(M))$ denotes the moduli space of coherent sheaves on $S$ with mukai vector $v$. They say that the fiber over a point $(M,N)$ is the quotient $[\textrm{Ext}^1(N,M)/\textrm{Hom}(N,M)]$?

How does the quotient make sense? What is the action of $Hom(N,M)$ on Ext$^1(N,M)$. They further say that action is trivial. But I am not able to understand what the action is at all.

So how do we get this quotient? Thanks in advance!

The action is trivial, as you wrote. Generally if $G/S$ acts trivially on $X/S$ and $G$ is abelian then any object in the quotient stack $[X/G]=B_X G$ has an automorphism group canonically isomorphic to $G$ (for $G$ non abelian, it could be a conjucagy form of $G$). Here it is easy to check that if $$0\to M\xrightarrow{i} E \xrightarrow{p} N \to 0$$ then $$u\mapsto \phi =id + i\circ u \circ p$$ defines an isomorphism between $Hom(N,M)$ and the automorphism group of your exact sequence (fixing the extremities, that is $\phi\circ i =i$ and $p\circ \phi =p$).
Given any $f\in \mathrm{Hom}(N, M)$, we can form an automorphism of the short exact sequence $0 \to M \xrightarrow{\alpha} E \xrightarrow{\beta} N\to 0$ in this way: Let $\sigma:E\to E$ be $\sigma = 1_E + \alpha\circ f\circ \beta$. Then $(1_M, \sigma, 1_N)$ is an automorphism of the exact sequence in question. It may be the action mentioned in that paper.