Let $\mathbb{C}^k\hookrightarrow E\to B$ be a complex vector bundle. Let $\mathbb{CP}^{k-1}\hookrightarrow\mathbb{P}(E)\to B$ be its projectivization. We can consider the tautological line bundle $L$ over $\mathbb{P}(E)$ which is the line bundle $$L= \{([x],V) \in \mathbb{P}(E)\times E | \ V \in [x] \} \to \mathbb{P}(E)$$ $$([x],V)\mapsto [x].$$ I would like to compute the first Chern class of this line bundle.

In the case when $E = B\times \mathbb{C}^k$ is the trivial vector bundle, then $\mathbb{P}(E) = B\times \mathbb{CP}^{k-1}$ and the tautological line bundle is $L = B\times \mathcal{O}(-1)$ where $\mathcal{O}(-1)\to \mathbb{CP}^{k-1}$ is the tautological line bundle. Therefore in this case we get that the first Chern class is given by $- P.D.([B\times \mathbb{CP}^{k-2}])$. But what can we say in general?

Motivation: study the normal bundle of $\mathbb{P}(E)$ inside $L$ in order to understand the blow-up along a submanifold.

  • $\begingroup$ See this MO question. $\endgroup$ Apr 10, 2019 at 14:20
  • $\begingroup$ I would be really surprised if there were an easy description of the firt chern class for general $E \to B$. I'm happy to be wrong though. $\endgroup$ Apr 10, 2019 at 17:09

1 Answer 1


I suspect in your post that $n = k-1$. In any case you're off by a sign. After all, it is the first Chern class of $\mathcal{O}(1)$ on $\mathbb{P}^n$ which is (the Poincare dual of) a hyperplane!

The general case is similar. If you think of the first Chern class as being Poincare dual to a generic non-zero section, then note that you can think of sections of $\mathcal{O}(1)$ on $\mathbb{P}(V)$ as cutting out a a family of hyperplanes over your base.

By the way, the cohomology of the projective bundle is given by

$$H^*(\mathbb{P}(V)) = H^*(B)[\zeta]/(\zeta^n + c_1\zeta^{n-1} + \ldots + c_n)$$

Here $\zeta = c_1(\mathcal{O}(1))$.

  • $\begingroup$ Well spotted! But, this does not answer the question of how can we express the first Chern class. $\endgroup$ Apr 10, 2019 at 16:11

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .