# Is there a formula for the total Chern Class of the tangent space of a projectivized vector bundle?

Let $V\rightarrow M$ be a complex vector bundle (of rank $k$) over a complex manifold $M$ (you can assume $M$ is compact if that helps, but it may not be relevant to my question). Let $\pi:\mathbb{P}V \rightarrow M$ be the projectivization of $V$.

$\textbf{Question}:$ Is there a formula for $c(T\mathbb{P}V)$, the total Chern class of the Tangent space of $\mathbb{P}V$?

My naive guess would be that it should be $\pi^*(c(TM))(1+c_1(\gamma^*))^{k+1}$, where $\gamma \rightarrow \mathbb{P}V$ is the tautological line bundle over $\mathbb{P}V$. I think my guess is correct if $M$ was just a point, or more generally if $V$ was a trivial bundle. But I do not know if this is correct in general.

The specific case for which I need an answer is when $M:= \mathbb{P}^1 \times \mathbb{P}^1$ and $V:= \mathcal{O}(d_1) \oplus \mathcal{O}(d_2)$.

$\textbf{Added Later}:$ It has been pointed out my guess is wrong in general. The correct answer is $$\pi^*(c(TM))c(\pi^*V \otimes \gamma^*).$$

No, your formula is not correct. You have to take into account the Chern classes of $V$. The relative tangent bundle $T_{\mathbb{P}V/M}$ is given by the so-called Euler exact sequence $$0\rightarrow \mathscr{O}_{\mathbb{P}V}\rightarrow \pi ^*V\otimes \gamma^* \rightarrow T_{\mathbb{P}V/M}\rightarrow 0\ ,$$ while $$0\rightarrow T_{\mathbb{P}V/M}\rightarrow T_{\mathbb{P}V}\rightarrow \pi ^*T_M\rightarrow 0\ .$$Putting things together we find $c(T_{\mathbb{P}V})=c(\pi ^*V\otimes \gamma^* )\,\pi ^*c(T_M)$.

Then use the standard formula for $c(\pi ^*V\otimes \gamma^* )$.

For any smooth fiber bundle

$$F\hookrightarrow P \stackrel{\pi}{\to} M$$

we have a short exact sequence of vector bundles over $P$

$$0\to VTP\to TP \to \pi^* TM\to 0,$$

where $VTP$ denotes the vertical tangent bundle defined as the kernel of the differential of $\pi$. If the bundle is holomorphic then the above is a short exact sequence of complex vector bundles and we deduce

$$c(TP)= c(VTP)\cdot \pi^* c(TM).$$

The classical Euler exact sequence argument shows that when $P=\mathbb{P}(V)$ that $\newcommand{\bC}{\mathbb{C}}$

$$\gamma^*\otimes \pi^*V \cong \underline{\bC}\oplus VTP,$$

where $\underline{\bC}$ denotes the trivial line bundle. Hence

$$c(TP)= c(\gamma^*\otimes \pi^*V)\cdot \pi^* c(TP).$$

In Section I.3 of Fulton-Lang Riemann-Roch algebra you can find an explicit formula for $c_k(L\otimes E)$, $L$ line bundle and $E$ vector bundle of rank $m$. More precisely

$$c_k(L\otimes E)=\sum_{j=1}^k \binom{m-j}{k-j} c_j(E)c_1(L)^{k-j}.$$ Note. The original answer had an error that I have now corrected. (Hat tip to abx).