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.



Deformations of a blowup
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 $$h^0(X, \, T\_X)=0, \quad h^1(X, \, T\_X) = h^1(S, \,T\_S)+2.$$ 


