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Let $X$ be a smooth curve over $\mathbb{C}$. I wish to compute the deformation of the following data $(E,F,x)$. $E$ and $F$ are locally free sheaves over $X$ and $x$ is a point on $X$. They satisfy:

1) $E\longrightarrow F$ is a injective map of sheaves over $X$.

2) $F/E$ has length 1 and is supported at $x$.

How to compute the first order deformation of $(E,F,x)$ and describe it as a subspace of $H^{1}(End(E))\oplus H^{1}(End(F)) \oplus T_{x}X?$

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2 Answers 2

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A morphism $F \to E$ can be thought of as a representation of the quiver $$ A_2 = \{ \bullet \to \bullet \} $$ in the category $Coh(X)$. The category $Rep(A_2,Coh(X))$ is abelian, so what you need is $Ext^1$ in this category. If $S_1$ and $S_2$ denote the simple modules of the first and the second vertices of the quiver, there is an exact sequence $$ 0 \to E \otimes S_2 \to C \to F \otimes S_1 \to 0,\qquad(*) $$ where $C = \{F \to E\}$. Using the fact that $Ext^\bullet(S_i,S_i) = \Bbbk$, $Ext^\bullet(S_1,S_2) = \Bbbk[-1]$, $Ext^\bullet(S_2,S_1) = 0$, one gets a long exact sequence $$ 0 \to Hom(C,C) \to Hom(E,E) \oplus Hom(F,F) \to Hom(F,E) \to Ext^1(C,C) \to Ext^1(E,E) \oplus Ext^1(F,F) \to Ext^1(F,E) \to \dots $$ And the space $Ext^1(C,C)$ in its middle is the tangent space to deformations.

EDIT. To get the long exact sequence, note first that applying to $(*)$ the functor $Hom(-,E\otimes S_2)$ one gets $$ 0 \to 0 \to Hom(C,E\otimes S_2) \to Hom(E,E) \to Hom(F,E) \to Ext^1(C,E\otimes S_2) \to Ext^1(E,E) \to Ext^1(F,E) \to \dots $$ and applying to $(*)$ the functor $Hom(-,F \otimes S_1)$ one gets $$ Ext^i(C,F \otimes S_1) \cong Ext^i(F,F). $$ Now, finally, applying to $(*)$ the functor $Hom(C,-)$ and using the above identifications, one gets the desired sequence.

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  • $\begingroup$ This is very nice! I just want to comment on how this answer is equivalent to my answer below (and I give up the claim that it is not classical homological algebra of course!), the map into the sum of ext's being forgetting the chosen homotopy and the $Hom(F,E)$ component the fiber of this forgetting map. The conditions on the image in the $Ext^1$-terms come from the commutativity constrain, and the quotient on the term on the right from the effect of re-choosing $d_{\tilde{F}}$ and $d_\tilde{E}$ within a homotopy class on the choice of homotopy. $\endgroup$
    – S. carmeli
    Commented Mar 5, 2018 at 21:01
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    $\begingroup$ You can also represent this as the sheaf hypercohomology of the complex $\mathcal{H}om(E,E) \oplus \mathcal{H}om(F,F) \to \mathcal{H}om (F,E)$. The cohomology of this complex in the category of sheaves can be computed explictly as, in degree zero, a length one modification of $\mathcal{H}om (E,E)$ and, in degree one, a rank one skyscraper sheaf at $x$. The contribution of this skyscraper sheaf to the cohomology should be the map to $T_x X$. $\endgroup$
    – Will Sawin
    Commented Mar 6, 2018 at 16:58
  • $\begingroup$ @Sasha: Could you explain where does this long exact sequence come from? $\endgroup$
    – La folia
    Commented Mar 7, 2018 at 0:07
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    $\begingroup$ @Lafolia: I added some details to the answer. $\endgroup$
    – Sasha
    Commented Mar 7, 2018 at 6:40
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First note that the deformation space for this problem don't really seat there, since you forgot that the map $F \to E$ is part of the data, and in principle can be deformed as well. Moreover, if $X$ is complete, then the condition on the cokernel of the map $F \to E$ is open (clearly having a bigger cokernel is a closed condition and on the other hand the deformation can not eliminate the cokernel completely because the first Chern class [or the degree of the determinant] of $F$ and $E$ differ by 1). The conclusion of this is that in the case where $X$ is a complete curve, we only have to deform the morphism $F \to E$, the cokernel will be deformed in a way which is determined by the deformation of this morphism, and it imposes no conditions on this deformation.

The question of deforming a morphism $F \to E$ in general is quite interesting, because it is an example of a deformation problem which at least seems a little beyond classical homological algebra. Recall that the relation between deformations of $E$ and elements in the cohomology come from the fact that a deformation $\tilde{E}$ gives an extension $E \stackrel{\cdot \epsilon}{\to} \tilde{E} \stackrel{mod (\epsilon)}{\to} E$ where $\epsilon$ is the generator of the dual numbers $\mathbb{C}[\epsilon]$. "Rotating this triangle by $120^\circ$", you get a connecting morphism in the derived category $E \to E[1]$.

If you try to deform a morphism $\phi: F \to E $, you end up with a commutative diagram in $D^b(Coh(C))$:

$$ \require{AMScd} \begin{CD} F @>\phi>> E \\ @Vd_\tilde{F}VV @Vd_\tilde{E}VV \\ F[1] @>\phi[1]>> E[1] \end{CD} $$

The elements $d_\tilde{F}$ and $d_{\tilde{E}}$ represent the classes in $H^1(End(F))$ and $H^1(End(E))$ classifying the deformations, and the condition that the diagram is commutative in $D^b(Coh(C))$ imply restrictions on those, namely they have to map to the same thing in $Ext^1(F,E)$.

But this is not the hole story! The deformation of the morphism itself provide extra data to the deformation. This data, if you think about it, is something like $\textbf{choosing a reason for the diagram to commute up to homotopy}$.

In other words, the morphism $\tilde{F} \to \tilde{E}$ provides us, after passing to various injective resolutions of the things, with a given homotopy $\phi[1] d_{\tilde{F}} \cong d_{\tilde{E}} \phi$, defined up to homotopy-of-homotopies (just adding boundaries in this case).

To conclude, the deformations of morphisms $F \to E$ can be classified (I think) by homotopy classes of commutative-up-to-chosen-homotopy squares

$$ \require{AMScd} \begin{CD} F @>\phi>> E \\ @Vd_\tilde{F}VV @Vd_\tilde{E}VV \\ F[1] @>\phi[1]>> E[1] \end{CD} $$

(I don't know how to draw higher coherences hin this interface, sorry) in the DG, or better stable $\infty$-category $D^b_\infty(Coh(C))$. To the extent of my knowledge, this can not be written as a classical cohomology of anything, so you really have to work DG-ly here, but I might be wrong. This is not so bad, though, and with some effort one can actually write down what it means in concrete terms I think.

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