# on a Deformation long exact sequence of moduli space of stable maps

I am reading the book "mirror symmetry" by Hori,Katz,Klemm,etc. And I want to understand the following Deformation long exact sequence

\begin{align} 0 & \to Aut(Σ, p_1, . . . , p_n, f)\to Aut(Σ, p_1, . . . , p_n) &\newline \to Def(f) &→ Def(Σ, p_1, . . . , p_n, f) → Def(Σ, p_1, . . . , p_n) &\newline \to Ob(f) &\to Ob(Σ, p_1, . . . , p_n, f) \to 0 \end{align}

it connects three deformation theory:
1. deformation of stable curves
2. deformation of maps(with fixed source)
3. deformation of stable maps(with possible changing source curves)

And my understanding goes as follows:
Let $\mathscr{X}=M_{g,n}$ be the moduli stack of algebraic curves(genus $g$, n-marked point), and let $\mathscr{Y}=M_{g,n}(X,\beta)$ be the moduli stack of stable maps. Then there is a natural "forgetful" morphism:
$\pi : \mathscr{Y} \to \mathscr{X}$
by forgeting the "map".

We have a distinguished triangle of cotangent complexes in the derived category $D^{-} (\mathscr O_{\mathcal{Y}})$:

$$\pi^* L_{\mathscr{X}}\to L_{\mathscr Y}\to L_{\mathscr{Y}/\mathscr{X}}\to \cdot$$

Now apply $R\mathscr{Hom}$, we have a long exact sequence:

\begin{align} \mathscr Ext ^0 (L_{\mathscr Y/\mathscr X },\mathcal O_{\mathscr Y }) & \to \mathscr Ext^0 (L_{\mathscr Y}, \mathcal O_{\mathscr Y} ) \to \mathscr Ext^0 (\pi^* L_{\mathscr X},\mathcal O_{\mathscr Y} )& \newline \to \mathscr Ext ^1 (L_{\mathscr Y/\mathscr X },\mathcal O_{\mathscr Y }) & \to \mathscr Ext^1 (L_{\mathscr Y}, \mathcal O_{\mathscr Y} ) \to \mathscr Ext^1 (\pi^* L_{\mathscr X},\mathcal O_{\mathscr Y} )& \newline \to \mathscr Ext ^2 (L_{\mathscr Y/\mathscr X },\mathcal O_{\mathscr Y }) & \to \mathscr Ext^2 (L_{\mathscr Y}, \mathcal O_{\mathscr Y} ) \to \mathscr Ext^2 (\pi^* L_{\mathscr X},\mathcal O_{\mathscr Y} ) \end{align}

My questions:
(1). is it an exact sequence of sheaves on $\mathscr Y$ with the first long exact sequence as its stalks?
(2). If (1) is true, then how to see $\mathscr Ext^i (\pi^* L_{\mathscr X},\mathcal O_{\mathscr Y} )$ (i=0,1,2) corresponds to Aut,Def,Ob of curves? And why the two ends of the exact seqence vanishes?

I don't believe that this is correct. The easiest way to see this is to look at your second question: The automorphisms/deformations/obstructions of a curve come from $H^i(C, T_C)$, i.e. they are the sheaves

$R^i p_*\omega_{U/\overline{\mathcal{M}_{g,n}}}^\vee$

where $p : U \to \overline{\mathcal{M}_{g,n}}$ is the universal family, and $\omega_{U/\overline{\mathcal{M}_{g,n}}}$ the relative dualizing sheaf. But these do not depend on $\overline{\mathcal{M}_{g,n}}(X, \beta)$ !

In the end, I think the issue is that you have the wrong exact sequence. What you want (to produce the relative obstruction theory) is the complex

$R^i p_*f^*T_X$

where the maps $p, f$ arise in the universal diagram

$\overline{\mathcal{M}_{g,n}}(X, \beta) \longleftarrow_p U \longrightarrow_f X$

It is not obvious to me that your sheaves should be the same as these ones.

• It is worth noting that Behrend and Fantechi's paper is a good source for reading about all of this material. Commented Oct 3, 2012 at 19:16
• Yes, of course you are right, In my mind, $L_{\mathscr Y/\mathscr X}$ provides the relative theory, $L_{\mathscr Y}$ provides the deformation theory of stable maps, and $\pi^* L_{\mathscr X}$ should provide some "sub"deformation theory? I want a way to formulate the deformation long exact sequence, but it seems that Behrend and Fantechi's paper does not give me the answer? Commented Oct 4, 2012 at 4:53
• By the way, I think the automorphisms/deformations/obstructions of a curve should come from $H^i(C,T_C)$? Commented Oct 4, 2012 at 4:54
• Ah, of course, silly me. I have corrected that mistake. Commented Oct 4, 2012 at 14:18
• I think that the idea is that a perfect obstruction theory is a morphism of complexes $\mathcal{E}^bullet \to L^\bullet$ which induces isomorphism/surjections in various degrees. Certainly, the cotangent complex is an example of such an obstruction theory, but it is not necessarily the right one to use. Commented Oct 4, 2012 at 20:31