2
$\begingroup$

This question is motivated by the question Tangent Space to Moduli Space of Vector Bundles on Curve which is about the computation of first-order deformations of a vector bundle, A.K.A, the tangent space at a rational point of the moduli space of vector bundles over a curve. It is well known that the first order deformations of a vector bundle $E$ are given by the elements of $H^1(C,End(E))$, but I don't quite understand the proof. By definition, we have a covering $\{U_{i}\}$ of $C$ where $E$ trivializes, and in addition, we have transition maps $$g_{ij}:U_{i}\cap U_{j}\rightarrow Gl(r)$$ satisfying the cocycle condition. And the big trick comes next where they said that the deformations of $E$ (vector bundles over $C_{\epsilon}$ being $\epsilon^2=0$ ) are given by transition functions of the form $g_{ij}(1+\epsilon a_{ij})$ with $a_{ij}\in \Gamma(U_{i}\cap U_{j}, \mathfrak{Gl}(r))$ being $\mathfrak{Gl}(r))$ the Lie algebra of $Gl(r)$. Can anybody explain it to me with patience this last part. I really want to understand it.

$\endgroup$
2
  • 3
    $\begingroup$ Intuitively, a deformation of the vector bundle would entail a 1-parameter family of vector bundles $E_t$ over $C$, such that $E_0 = E$. This would give rise to a family $g_{ij}(t) : U_{ij} \to GL_r$ of transition functions with $g_{ij}(0) = g_{ij}$. The deformation can be read off by taking derivative of the curve of transition functions at $t = 0$; $a_{ij} = g_{ij}'(0)$ is exactly that. This is a map from $U_{ij}$ which lands into the tangent space, i.e., Lie algebra $T_0 GL_r = \mathfrak{gl}_r$. The trick with $\varepsilon^2 = 0$ allows one to do this trick in the algebraic category. $\endgroup$ Jul 15, 2021 at 22:47
  • $\begingroup$ Thank you for your response @BalarkaSen but I would like a rigorous proof of this fact, in order to achieve a better understanding. $\endgroup$ Jul 16, 2021 at 6:17

1 Answer 1

0
$\begingroup$

Question: "Can anybody explain it to me with patience this last part. I really want to understand it."

Answer: If $C$ is a projective curve over a field $k$ and $E$ is a coherent $\mathcal{O}_C$-module, you may (following Serenesi "Deformations of algebraic schemes") define the Quot scheme $Q:=Quot^C_E$, parametrizing coherent quotients $f:E \rightarrow F$ of $E$. Let $K:=ker(f)$. The tangent space of $Q$ at $(F,f)$ is given by the following formula:

$$T_{[f]}(Quot^C_E):=H^0(C, Hom_{\mathcal{O}_C}(K,F)).$$

The obstruction space $Ob_{[f]}(Quot^C_E)$ is given by the following formula:

$$Ob_{[f]}(Quot^C_E) \cong Ext^1_{\mathcal{O}_C}(K,F).$$

If $E$ and $F$ are locally trivial and of finite rank it follows $K$ is locally trivial of finite rank and

$$Ob_{[f]}(Quot^C_E) \cong Ext^1_{\mathcal{O}_C}(K,F) \cong Ext^1_{\mathcal{O}_C}(\mathcal{O}_C, K^*\otimes F) \cong $$

$$H^1(C, K^* \otimes F).$$

Hence if $E,F$ are of finite rank and locally trivial you get the tangent space

$$T_{[f]}(Quot^C_E):=H^0(C, K^* \otimes F)$$

and the obstruction space

$$Ob_{[f]}(Quot^C_E) \cong H^1(C, K^*\otimes F).$$

As I recall it: There is a detailed proof in the above mentioned book but it takes several pages. The book gives an introdution to the Hilbert and Quot schemes in the "language of schemes". Maybe this helps - a book is more reliable than online notes.

Note: In your case you should try to construct your moduli space $M_C$ using the quot-scheme. You will need a locally trivial (or coherent) sheaf $F$ on $C$ such that your vector bundles $[E]\in M_C$ are quotients of $F$. Then the above formula(s) give the tangent and obstruction space. If $E=K$ is finite rank locally trivial, it follows $E^*\otimes E \cong End_{\mathcal{O}_C}(E)$ hence

$$Ob_{[f]}(Quot^C_F)\cong H^1(C, E^*\otimes E) \cong H^1(C, End_{\mathcal{O}_C}(E)).$$

Hence you must construct a bundle $F$ and a quotient $f: F \rightarrow E$ with $ker(f)=E$.

Example: One choice is $F:=E \oplus E$. There is a canonical surjective map onto $E$ with kernel $E$. Hence

$$M_C := Quot^C_{E\oplus E}.$$

Any coherent sheaf $F\in Ext^1_{\mathcal{O}_C}(E,E)$ can be used.

$\endgroup$
4
  • $\begingroup$ But this is not what I have asked. I want to know what happens with the transition functions. $\endgroup$ Jul 16, 2021 at 22:18
  • $\begingroup$ @SamanthaSmith - yes, the above construction is another type of construction that is more general. Still it may be useful to have several approaches - this is why I wrote the post- you wanted to achieve a "better understanding". $\endgroup$
    – hm2020
    Jul 17, 2021 at 11:35
  • $\begingroup$ Maybe I have expressed myself poorly @hm2020. I really appreciate your answer, and I understand it better due to the book of Sernesi, but I still don't understand where those transition functions arise. $\endgroup$ Jul 17, 2021 at 11:38
  • $\begingroup$ @SamanthaSmith - You may use the Quot scheme $Quot^C_{E\oplus E}$. Then your formula follows from the book of Sernesi. $\endgroup$
    – hm2020
    Jul 17, 2021 at 15:25

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .