# Cohomology of tangent bundles

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 $X$ ?

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 $H^i(T_{\tilde X}) = H^i(\pi_*T_{\tilde X})$.
• Thank you. The isomorphism $H^{i}(T_{\widetilde{X}})=H^{i}(\pi_{*}T_{\widetilde{X}})$ can be also seen as a consequence of the fact that $R^{i}\pi_{*}T_{\widetilde{X}} = 0$. From the last exact sequence you wrote to prove that $H^{i}(\pi_{*}T_{\widetilde{X}}) = H^{i}(T_{X})$ we need to have $H^{i}(j_{*}N) = 0$. Oct 25, 2013 at 17:25
• I never wrote that $H^i(\pi_*T_{\tilde X}) = H^i(T_X)$. Oct 25, 2013 at 17:39
• Yes, I know. For that equality one needs to have $H^{i}(j_{*}N) = 0$. Oct 25, 2013 at 17:47
• I am sorry but I did not understand how I get the second exact sequence from the first. It seems to me that dualizing the first exact sequence you wrote one has $$0\mapsto T_{E/Z}\rightarrow T_{\widetilde{X}}\rightarrow \pi^{*}T_{X}\mapsto 0$$ instead of $(\ast)$. Oct 25, 2013 at 18:55
• This is what derived functors do. One can check that for a divisorial embedding $i:E \to M$ and a locally free sheaf $F$ on $E$ one has $\mathcal{H}om(i_*F,O_M) = 0$, $\mathcal{E}xt^1(i_*F,O_M) = i_*F(E)$. This is an instance of the Grothendieck duality. Oct 25, 2013 at 19:28