Tangent space of the stack $\overline{\mathcal{M}}_{g,n}(X,\beta)$.

Question

Let $\overline{\mathcal{M}}_{g,n}(X,\beta)$ be the moduli stack of stable maps $f$ from genus $g$, $n$-marked curve $C$ to a variety $X$ i nth curve class $\beta \in H^2(X,\mathbb{Z})$. I would like to know why the hyperext group $$ Ext^1_C(f^*\Omega_X\rightarrow \Omega_C(\sum_{i=1}^n),\mathcal{O}_C) $$ represents tangent space of the moduli and the obstruction lies in
$$ Ext^2_C(f^*\Omega_X\rightarrow \Omega_C(\sum_{i=1}^n),\mathcal{O}_C)? $$ I am aware that the deformation-obstrcution of a map $f:C\rightarrow X$ with a fixed curve $C$ is governed by $H^i(C,f^*T_X)$ for $i=1,2$ and the automorphism-deformation of a $n$-marked points $(C;p_1,\dots,p_n)$ is governed by $Ext^i_C(\Omega_C(\sum_{i=1}^n),\mathcal{O}_C)$ for $i=0,1$. However, I don't know how to combine them into one package in the hyperext groups above.

I would appreciate it if someone could kindly explain how to obtain and understand the hyperext groups above.

Answer 1

The classical reference is Illusie's PhD thesis "complexe cotangent et deformations".There the result without the marked points is proven in a very, very general context using the cotangent complex; then one needs to know that for a smoooth variety, or for a nodal curve (or, again, in many more cases) the cotangent complex is isomorphic in the derived category to the cotangent sheaf.

Fixing the case of the marked points can easily be done by hand (they're smooth points after all): I can expand on this step if it helps. Or, one can go modern and use logarithmic geometry.

If you want to get a feeling for why such a result is true, a very short reference is a paper by Ziv Ran on deformations of morphisms: Algebraic Curves and Projective Geometry Lecture Notes in Mathematics Volume 1389, 1989, pp 246-253 Deformations of maps

Another approach would be to sit down and prove it yourself using as only references a few classical facts about deformation theory (say, what you learn from the first ten pages of Artin's Tata lectures). Whether this is a useful exercise or a waste of time depends on your personality. Again, details upon request.

Answer 2

Let $L$ be the complex $[ f^\ast \Omega_X \rightarrow \Omega_C(D) ]$, concentrated in degrees $[-1,0]$ on $C$.

One way to understand the obstructions is to understand the deformations for affine curves first. Observe first that there is a canonical equivalence of categories between the category of deformations of a pointed map $f : C \rightarrow X$, with $C$ not-necessarily proper, and the category of extensions

$G(C) = \mathrm{Ext}(L, \mathcal{O}_C)$.

The meaning of the category on the right is the category of extensions $E$ of $\Omega_C(D)$ by $\mathcal{O}_C$ together with a trivialization of the induced extension of $f^\ast \Omega_X$ by $\mathcal{O}_C$. One checks explicitly that the category of extensions of $\Omega_C(D)$ by $\mathcal{O}_C$ is canonically equivalent to the category of deformations of the pointed curve $C$ and that splitting the induced extension of $f^\ast \Omega_X$ is the same as extending the map $f : C \rightarrow X$ to a map from the deformed curve to $X$.

Note that $G$ is a commutative group stack (what is sometimes called a "Picard stack"). It is easy to check that $G$ is non-empty when $C$ is an affine curve (use the fact that deformations of nodal curves are unobstructed and the formal criterion of smoothness for $X$), which means that any obstructions to deforming a proper curve come from the failure of local deformations to glue. There is a standard way to compute these local obstructions---e.g., by Cech cohomology, noting that mutatis mutandis, Cech cohomology still works for group stacks---and this yield a class in $H^1(C, G)$. Cohomology of a commutative group stack is the same as cohomology of the associated complex, so we get an obstruction in

$H^1(C, L^\vee[1]) = H^2(C, L^\vee) = \mathrm{Ext}^2(L, \mathcal{O}_C)$.