# Chern class of tautological line bundle

I'm studying characteristic classes from the Chern-Weil construction (via connection and curvature). I'm trying to compute some simple examples. Let $E$ be the tautological line bundle over projective space $P(\mathbf{C}^n)$. I want to show that the first Chern class of $E$ does not vanish. I suppose I could just introduce a connection on $E$ using local trivializations (is there a natural choice?), patch things together, compute the curvature and from this the first Chern class. However, that sounds a bit tedious. Are there more elegant ways to compute it?

• I would endorse computing some examples from scratch. If you're familiar with the mechanism of moving frames in differential geometry, this is a snap. You could do it with transition functions, yes. – Ted Shifrin May 28 '13 at 20:19
• After you have done the computation this way, I think it would be instructive to see how Milnor-Stasheff do it. I learned a lot from working through that section of their book. – Sam Lisi May 29 '13 at 0:19

1. If $$E \to X$$ is a complex vector bundle of complex dimension $$n$$, then the top Chern class $$c_n(E) \in H^{2n}(X; \mathbb R)$$ is equal to the Euler class $$e(E)$$.
2. Every $$n$$-dimensional complex manifold $$X$$ (such as $$\mathbb {CP}^n$$) is orientable. In algebraic terms, orientability can be seen as the top homology group $$H_{2n}(X; \mathbb R)$$ having rank one, and we can choose an orientation, i.e. a choice of generator $$[X] \in H_{2n}(X; \mathbb R)$$.
Now let $$\mathbb {CP}^n = X$$ and $$x$$ be the Euler class of the tautological line bundle over $$\mathbb {CP}^n$$. Then there is a nice argument on p22 Example 2.9.3 of these lecture notes on characteristic classes (from the algebraic topology point of view). Essentially, it boils down to showing that
$$\langle e(T \mathbb {CP}^n), [\mathbb {CP}^n] \rangle =(n+1)(-1)^n \langle x^n, [\mathbb {CP}^n] \rangle$$
where $$\langle,\rangle : H^*(\mathbb {CP}^n, \mathbb R) \otimes_{\mathbb R} H_*(\mathbb {CP}^n, \mathbb R) \to \mathbb R$$ is the canonical non-degenerate pairing between homology and cohomology and $$x^n$$ is the cup product of $$x$$ with itself $$n$$ times.
We know from standard theory that for an orientable manifold $$M$$, the pairing $$\langle e(TM), [M] \rangle = \chi(M)$$ -- the Euler characteristic. The homology groups of $$\mathbb {CP}^n$$ have rank one in every even grading that is $$\leq 2n$$ and zero otherwise, so $$\chi(M)= n+1$$. Hence, $$x\neq 0$$.