# Virtual dimension of moduli of stable maps

I apologize if this question was already asked somwhere else on this website. Let us consider $f:C \to X$ a stable map. This is a point in the moduli space of stable maps. It seems intuitively to me that any holomorphic global section of $f^*(T_X)$ gives a deformation of the stable map. It is well-known, but not clear to me, why the moduli space of stable maps in a neighbourood of this point looks like a closed subvariety in $H^0(C, f^*T_X)$ cut out by $h^1(C, f^*T_X)$ equations. Where do this equations come from? Is there a concrete way to see them? (e.g. some sections of $f^*(T_X)$ give forbidden deformations, or the same deformation,...).

-
This is a consequence of the deformation theory of morphisms with smooth target: deformations lie in $H^0(f^{\ast}T_X)$ and obstructions lie in $H^1(f^{\ast}T_X)$. This is explicitly covered, for example, in Hartshorne's notes on deformation theory (go to the section on deformations of a morphism). – Mike Skirvin Sep 18 '11 at 14:44
-1 for the choice of nickname! (just kidding. But I think with such a nickname you'll be perceived as an administrator or something by the new users..) – Qfwfq Sep 18 '11 at 15:09
we have a map from deformation space $H^0$ to obstruction space $H^1$ such that its zero locus corresponds to actual deformations. In symplectic geometry, this map is given by Cauchy-Riemann equation. – Mohammad F. Tehrani Sep 18 '11 at 15:14

If we are looking at the space of stable maps, we need to allow $C$ to deform as well. Let $C$ be a curve with the set of marked points $P$. There is a following exact sequence: \begin{eqnarray*} 0 &\to& \text{Aut}(C,P) \to H^0(C, f^*(T_X)) \\\ \to \text{Def}(C,P,f) &\to& \text{Def}(C,P) \to H^1(C,f^*T_X)) \\\ \to \text{Ob}(C,P,f) &\to& 0 \end{eqnarray*} Thus the virtual dimension is \begin{eqnarray*} \text{virdim} &=& \dim \text{Def} (C,P,f) - \dim \text{Ob}(C,P,f) \\\ &=& \dim \text{Def}(C,P) - \dim \text{Aut}(C,P) +h^0(f^*T_X) - h^1(f^*T_X) \end{eqnarray*}
As to why we need to substract $h^1$ from the dimension. Let $(O,m)$ be the local ring of the map $f$ at the moduli space. If the space is smooth at $f$ then the dimension would be $h^0$. But this is not often the case. Generally $h^0$ would be the dimension of Zariski tangent space ( $= \dim_k m/m^2$). If $O$ is not regular, one way to measure its irregularity is by using infinitesimal lifting property: given a map $\phi: O \to A$, can we lift it to an infinitesimal extension $A':$ $0\to J \to A' \to A \to 0$. This is the same as extending a deformation of $f$ over $A$ to a higher order. Such a lift is obstructed by an element $\alpha \in H^1(f^*T_X)\otimes J$ (local to global).
Let $O = (R,m_P)/I$ such that $R$ is regular having same Zariski tangent space. Then $\dim O$ is roughly $\dim R - \text{rank}(I) = \dim R - \dim I/m_pI$. Now $I/mP_I$ is the canonical obstruction for $O$ and it can be embedded into $H^1(C,f^*T_X)$, thus the dimension of $O$ is roughly $\dim R - \dim I/m_pI \approx h^0 -h^1$