Let $X$ be a smooth scheme and $Z\subset X$ a smooth subscheme. Consider the blow-up
$$\pi:\widetilde{X}:=Bl\_{Z}X\rightarrow X$$
of $X$ along $Z$.


**What is the relation between the cohomology of the tangent bundle $T\_{\widetilde{X}}$ and the cohomology of the tangent bundle $T\_{X}$ of $X$ ?**



Cohomology of tangent bundles
There is an exact sequence
$$
0 \to \pi^\*\Omega\_X \to \Omega\_{\tilde X} \to i\_\*\Omega\_{E/Z} \to 0,
$$
where $E$ is the exceptional divisor and $i:E \to \tilde X$ is its embedding.
Dualizing it one gets
$$
0 \to T\_{\tilde X} \to \pi^\*T\_X \to i\_\*T\_{E/Z}(E) \to 0.\qquad(\*)
$$
On the other hand, since $E = P(N)$ is the projectivization of the normal bundle, one has the relative Euler sequence
$$
0 \to O\_E \to p^\*N(-E) \to T\_{E/Z} \to 0,
$$
where $p:E \to Z$ is the projection. It implies that $p\_\*T\_{E/Z}(E) = N$, hence pushing forward $(\*)$ one gets
$$
0 \to \pi\_\*T\_{\tilde X} \to T\_X \to j\_\*N \to 0,
$$
where $j:Z \to X$ is the embedding. Finally, writing down the cohomology long exact sequence for this you can compute $H^i(T\_{\tilde X}) = H^i(\pi\_\*T\_{\tilde X})$.


