Deformations of a blowup

Question

Let $S$ be a smooth projective surface over $\mathbb{C}$. (I guess this can be more general—higher dimension, other ground fields, non-projective, maybe even singular?—and I'dd like to hear that.) Let $s \in S$ be a point. Let $\beta \colon X \to S$ be the blowup of $s \in S$. Suppose that $H^{i}(S, T_{S})$ is known for $i \in \{0,1,2\}$, as well as $\mathrm{Def}(S)$.

If I am not mistaken, the exceptional divisor $E$, which is a $(-1)$-curve, is rigid, in the sense that every deformation of $X$ also has a $(-1)$-curve. Therefore $$\mathrm{Def}(X) \cong \mathrm{Def}(X,E) \cong \mathrm{Def}(S,s) \cong \mathrm{Def}(S) \times T_{S,s},$$ where the last isomorphism is not canonical. (Rather $T_{S,s}$ is the kernel of the forgetful map $\mathrm{Def}(S,s) \to \mathrm{Def}(S)$.)

This question is about the cohomological side of the picture, i.e. $H^{i}(X,T_{X})$ for $i \in \{0,1,2\}$. My intuition says that $H^{1}(X,T_{X})$ should also increase with dimension two, whereas the obstruction space $H^{2}(X,T_{X})$ should stay the same. I've tried to fiddle around with the spectral sequence $$ H^{p}(S, R^{q}\beta_{*}T_{X}) \Longrightarrow H^{p+q}(X, T_{X}) $$ but I could not really come to the desired conclusions.

For $H^{0}(X, T_{X})$ we get the term $H^{0}(S, \beta_{*}T_{X})$.
For $H^{1}(X, T_{X})$ we get the terms $H^{1}(S, \beta_{*}T_{X})$ and $H^{0}(S, R^{1}\beta_{*}T_{X})$. Now $R_{1}\beta_{*}T_{X}$ is a skyscraper sheaf supported on $s$, and if I'm not mistaken, and vague geometric intuition makes me think that it is the tangent space $T_{S,s}$.
Finally, $R^{2}\beta_{*}T_{X} = 0$, so for $H^{2}(X, T_{X})$ we get the terms $H^{2}(S, \beta_{*}T_{X})$ and $H^{1}(S, R^{1}\beta_{*}T_{X})$.
But maybe this isn't the right way to approach the question…

So the main question is:

What are the $H^{i}(X,T_{X})$ for $i \in \{0,1,2\}$?

I've not been able to find this via google, though I guess this is pretty basic knowledge in deformation theory. But I'm pretty new to this field, so please bear with me.

Answer 1

The answer is the following and can be found in Hartshorne's book Deformation Theory, see in particular Exercise 10.5 page 83.

We work over an algebraically closed field $k$. Then there is an exact sequence of sheaves $$0 \to \beta_*T_X \to T_S \to k_s \oplus k_s \to 0,$$ inducing an exact sequence in cohomology $$0 \to H^0(X, \, T_X) \to H^0(S, \, T_S) \to k \oplus k \to H^1(X, \, T_X) \to H^1(S, \, T_S) \to 0$$ and an isomorphism $H^2(X, \, T_X) \cong H^2(S, \, T_S)$.

We can interpret this as follows.

First of all, the obstructions to first-order deformations of $X$ are the same as the obstructions to first-order deformations of $S$.

Next, the term $k \oplus k$ corresponds to the deformations of the point $s$ inside $S$. Then if the group of infinitesimal automorphisms of $S$ (that is, the term $H^0(S, \, T_S))$ maps surjectively onto the deformations of $s$ into $S$, then $H^1(X, \, T_X) \cong H^1(S, \, T_S)$, i.e. the first-order deformations of $X$ are just given by first-order deformations of $S$. Otherwise, moving $s$ gives nontrivial deformations of $S$.

The first case happens for instance when $S$ is an abelian surface. Then $\textrm{Aut}(S)$ is transitive, hence the map $H^0(S, \, T_S) \to k \oplus k$ is an isomorphism and this implies $$h^0(X, \, T_X)=0, \quad h^1(X, \, T_X) = h^1(S, \, T_S)=4.$$

The second case happens for instance when $S$ is a surface of general type. Then it is possible to prove that $S$ has at most finitely many automorphisms, hence $h^0(S, \, T_S)=0$. This yields $$h^0(X, \, T_X)=0, \quad h^1(X, \, T_X) = h^1(S, \,T_S)+2.$$

Answer 2

Let $S$ be a surface and $Z=\{p_1,...,p_n\}\subset S$ be a reduced subscheme of dimension zero. Let $\epsilon:\widetilde{S}\rightarrow S$ be the blow-up of $S$ at $Z$. Consider the exact sequence $$0\mapsto \epsilon^{*}\Omega_{S}\rightarrow \Omega_{\widetilde{S}}\rightarrow i_{*}\Omega_{E/Z}\mapsto 0$$ where $i:E\hookrightarrow\widetilde{S}$ is the exceptional divisor. Note that $\mathcal{H}om(i_{*}\Omega_{E/Z},\mathcal{O}_{\widetilde{S}}) = 0$ and by Grothendieck duality $\mathcal{E}xt^{1}(i_{*}\Omega_{E/Z},\mathcal{O}_{\widetilde{S}})\cong i_{*}T_{E/Z}(E)$. So dualizing the above exact sequence we get $$0\mapsto T_{\widetilde{S}}\rightarrow \epsilon^{*}T_{S}\rightarrow i_{*}T_{E/Z}(E)\mapsto 0.$$ Since $R^{1}\epsilon_{*}T_{\widetilde{S}} = 0$ we have $$0\mapsto\epsilon_{*}T_{\widetilde{S}}\rightarrow T_{S}\rightarrow T_{S|Z}\mapsto 0.$$ Now, $R^{i}\epsilon_{*}T_{\widetilde{S}} = 0$ for any $i > 0$. So $H^{i}(\widetilde{S},T_{\widetilde{S}})\cong H^{i}(S,\epsilon_{*}T_{\widetilde{S}})$ for any $i\geq 0$ and we get the following exact sequence in cohomology $$ \begin{array}{l} 0\mapsto H^{0}(\widetilde{S},T_{\widetilde{S}})\rightarrow H^{0}(S,T_S)\rightarrow K^{2n}\rightarrow H^{1}(\widetilde{S},T_{\widetilde{S}})\rightarrow H^{1}(S,T_S)\rightarrow 0 \rightarrow \\ \rightarrow H^{2}(\widetilde{S},T_{\widetilde{S}})\rightarrow H^{2}(S,T_S)\rightarrow 0\\ \end{array} $$ Since the map between the tangent spaces $H^{1}(\widetilde{S},T_{\widetilde{S}})\rightarrow H^{1}(S,T_S)$ is surjective and the map between the obstruction spaces $H^{2}(S,\epsilon_{*}T_{\widetilde{S}})\rightarrow H^{2}(S,T_S)$ is injective the map $Def_{\widetilde{S}}\rightarrow Def_S$ is smooth of relative dimension $2n-\dim H^{0}(S,T_S) + \dim H^{0}(\widetilde{S},T_{\widetilde{S}})$. This means that the obstructions to deforming $\widetilde{S}$ are exactly the obstructions to deforming $S$. The vector space $K^{2n}$ parametrizes the deformations of $Z$ inside $S$ and the spaces $H^{0}(\widetilde{S},T_{\widetilde{S}})$, $H^{0}(S,T_S)$ parametrize the infinitesimal automorphisms of $\widetilde{S}$ and $S$ respectively. If the map $$H^{0}(S,T_S)\rightarrow K^{2n}$$ is surjective then $H^{1}(\widetilde{S},T_{\widetilde{S}})\cong H^{1}(S,T_S)$ and the deformations of $\widetilde{S}$ are induced by deformations of $S$. Otherwise the deformations of $Z$ inside $S$ induce non-trivial deformations of $\widetilde{S}$.

For instance, take $S = \mathbb{P}^{2}$. Then $H^{0}(S,T_S)\cong T_{Id}Aut(\mathbb{P}^{2})$ has dimension $8$. If $n\leq 4$ the map $$T_{Id}Aut(\mathbb{P}^{2})\rightarrow K^{2n}$$ is surjective and $H^{1}(\widetilde{S},T_{\widetilde{S}})\cong H^{1}(\mathbb{P}^{2},T_{\mathbb{P}^{2}})$. Indeed if $n\leq 4$ there is an automorphism mapping $Z$ to any other set of $n$ points in general position and moving $Z$ inside $\mathbb{P}^{2}$ just induces trivial deformations of $\widetilde{S}$. Furthermore, since $\mathbb{P}^{2}$ itself is rigid we have $H^{1}(\mathbb{P}^{2},T_{\mathbb{P}^{2}})=0$.

More generally, let $X$ be a smooth variety and $Z\subseteq X$ be a smooth subvariety. Then $$T^{1}Def_{(X,Z)} = H^{1}(X,T_{X}(-log (Z))) = H^{1}(Bl_{Z}X,T_{Bl_{Z}X}) = T^{1}Def_{Bl_{Z}X}.$$