[Math] Tangent Space to Moduli Space of Vector Bundles on Curve

algebraic-geometrycomplex-geometry

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.

Best Answer

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.

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