# 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?

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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.

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It is worth noting that Behrend and Fantechi's paper is a good source for reading about all of this material. –  Simon Rose Oct 3 '12 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? –  Xiaobo Zhuang Oct 4 '12 at 4:53
By the way, I think the automorphisms/deformations/obstructions of a curve should come from $H^i(C,T_C)$? –  Xiaobo Zhuang Oct 4 '12 at 4:54
Ah, of course, silly me. I have corrected that mistake. –  Simon Rose Oct 4 '12 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. –  Simon Rose Oct 4 '12 at 20:31