Let $X$ be a curve of genus $g \geq 2$. Using Geometric Invariant Theory, we can construct a moduli space $\mathcal{M}(r,d)$ of vector bundles on $X$ of rank $r$ and degree $d$. The details of this construction are a bit over my head for now, however I would like to at least be able to prove that the dimension of this moduli space is $r^{2}(g-1)+1$.

I'm lacking understanding of one key fact. In Michael Thaddeus' paper (http://www.math.columbia.edu/~thaddeus/papers/odense.pdf) he mentions that the tangent space to $\mathcal{M}(r,d)$ at any stable bundle $E$ satisfies the following

$T_{E} \mathcal{M}(r,d) \simeq H^{1}(\rm{End}E)$

Can anyone help me understand this? I don't understand Thaddeus' argument. Given the above fact, it's trivial to apply Hirzebruch-Riemann-Roch and complete the derivation of the dimension of the moduli space.

  • 5
    $\begingroup$ This is not an answer. A bundle $E$ is described by a cocycle $\{g_{i,j}\}_{i,j\in I}$ and you are looking for first order deformations of it, namely bundles described by $\{\tilde{g}_{i,j}\}_{i,j\in I}$, where $\tilde{g}_{i,j}=g_{i,j}(1+\epsilon a_{i,j})$. Here $\epsilon$ is an infinitesimal number, i.e. $\epsilon^2=0$. The condition $\{\tilde{g}_{i,j}\}$ cocycle is equivalent to $\{a_{i,j}\}\in H^1(\mathcal{E}nd(E))$ and hence your thesis. Here $\mathcal{E}nd(E)$ is the endomorphism sheaf. This is explained in Mukai, Moduli of vector bundles on $K3$ surfacesand symplectic manifolds. $\endgroup$
    – User3773
    Mar 18, 2016 at 14:55
  • $\begingroup$ Thanks, I like this way of thinking about it. When you refer to a cocycle condition you mean that $\tilde{g}_{ij}\tilde{g}_{jk}\tilde{g}_{kl}=1$ correct? Assuming that the $g$ also satisfy the cocycle condition, and that $\epsilon^{2} =0$, I get the condition that $g_{jk}a_{jk}g_{kl} + a_{ij}g_{jk}g_{kl}+g_{jk}g_{kl}a_{kl}=0$. Is it this that implies somehow that $\{a_{ij}\} \in H^{1}( \rm{End}E)$? $\endgroup$
    – Benighted
    Mar 18, 2016 at 18:56
  • 3
    $\begingroup$ Yes, it is. To conclude you need to use the fact that $\{g_{ij}\}$ is a cocycle and that the $\{a_{ij}\}$ commutes with them. Moreover, notice that you eventually want to find a relation like $a_{ij}-a_{jk}+a_{ki}=0$, i.e. $\{a_{ij}\}$ is an additive cocycle with values in the sheaf of endomorphisms of $E$. As I said this is not a complete answer because there are details to check and important hidden facts, but once again I recall Mukai's paper for a detailed reference. $\endgroup$
    – User3773
    Mar 19, 2016 at 0:23
  • $\begingroup$ See people.math.harvard.edu/~gaitsgde/grad_2009/SeminarNotes/… This comes down to the fact that the dimension (=rank of the tangent complex) of $\text{Bun}_GX$ at a bundle $P$ is computed by considering the maps $\text{Spec}k[\varepsilon]/\varepsilon^2\to \text{Bun}_GX$ restricting to $\text{Spec}k\to\{P\}$. By definition (of $\text{Bun}_GX$'s functor of points) this is the same as a first-order deformation of $P$, which is given by an $H^1$ as in your question. $\endgroup$
    – Pulcinella
    Mar 16, 2021 at 15:36
  • $\begingroup$ @User3773 Your comment gives a very good opinion. Just one question, Mukai write in page 151 that: $g_{ij}^{-1}a_{ij}g_{ij}+a_{jk}-a_{ik}=0$ is equivalent to say $\{a_{ij}\}$ is a cocycle in $H^1(X,End(E))$. But the cocycle condition should be $a_{ij}+a_{jk}-a_{ik}=0$, how can these be equivalent? $\endgroup$
    – Z. Liu
    Nov 21, 2023 at 6:52

2 Answers 2


Here's the answer from a differential geometry perspective.

Given a unitary vector bundle $E \to M$, $M$ complex, a $\bar \partial$ operator $D: \Omega^0(E) \to \Omega^{0,1}(E)$ (where the latter is defined as sections of $\Lambda^{0,1}(M) \otimes E$) is a linear operator that satisfies the Leibniz rule on smooth forms; the space of them is called $\mathcal D$. The integrability condition says that there is a holomorphic structure on $E$ with $D$ as its $\bar \partial$ operator if and only if $D^2: \Omega^0(E) \to \Omega^{0,2}(E)$ is trivial (extending this via Leibniz to $(0,1)$-forms). (That is, the equation that $D$ come from a holomorphic structure is the flat connection equation.)

Now, the space of $\bar \partial$ operators is affine over the space $\Omega^{0,1}(\text{End}(E))$, so this is its tangent space at any operator. Also note that the group of unitary automorphisms $\mathcal U$ of $E$ acts on $\mathcal D$, such that if $D$ has $D^2 = 0$, so does $u(D)$.

What is the derivative of this action? The action sends $D \mapsto D - (Du)u^{-1}$, and noting that the Lie algebra of $\mathcal U$ is $\Omega^0(\text{End}(E))$, we see that the differential at $u \mapsto D - (Du)u^{-1}$ is $a \mapsto D a$.

Lastly let us compute the tangent space to the space of solutions $\mathcal S$ to $D^2 = 0$ in $\mathcal D$ (what we are interested in is the moduli space $\mathcal S/\mathcal G$.). This is the space of solutions of the linearization, which we compute at a solution $D$ as follows: if $d \in \Omega^{0,1}(\text{End}(E))$, then $(D+d)(D+d)\sigma = D(d\sigma) + d(D\sigma)+d^2$, and we combine $D(d\sigma) + d(D\sigma)$ into one operator, $(Dd)\sigma$. So the equation linearizes to $Dd = 0$.

Then the tangent space to the moduli space $\mathcal D/\mathcal S$, at least at an irreducible connection so that the group $\mathcal G$ acts freely, is $\text{ker}(D)/\text{im}(D)$ - the first cohomology group $H^1(\text{End}(E))$, provided we're using $D$ as our derivative operator. That is to say, it's the first cohomology of the holomorphic vector bundle $\text{End}(E)$, holomorphic structure coming from the structure induced by $D$, as desired.

Another way to understand this should come from the Narasimhan-Seshadri theorem, which I don't understand as well.


Warning: I have seen the construction of the moduli of (stable) vector bundles as a manifold via a symplectic quotient (in a course with notes available here), rather than as a scheme via GIT quotient, which I don't understand well. However, my understanding is that the two are analagous, and this is supported based on the fact that the argument Thaddeus appears to be making is the same as in the symplectic case. In particular, both for both types of quotient $X//G$, we first find some subset $Y \subset X$ where $G$ acts nicely enough to take a nice quotient, and $X//G:= Y/G$.

I think the argument in mind is that whenever you have a (sufficiently nice) group $G$ acting (sufficiently nicely) on a (sufficiently nice) scheme $X$ (so that the quotient makes sense as an orbit space), the projection map $T_x Y \rightarrow T_{[x]}(X//G)$ has kernel equal to the image of the infinitesimal action of $G$ on $Y$, $\mathfrak{g}\rightarrow T_x Y$ induced by the action of $G$ on $Y$. In Thaddeus' argument, he demonstrates an exact sequence:

$$\mathfrak{gl}(\chi,\mathbb{C})\rightarrow T_EY \rightarrow H^1(\operatorname{End}(E)) \rightarrow 0$$

After this, he observes that the first map is the infinitesimal action, so we have that $H^1(\operatorname{End}(E))$ is isomorphic to the cokernel of the infinitesimal action, which gives the result we wanted by the general theory mentioned above.


You must log in to answer this question.

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