[Math] Chern class of tautological line bundle

characteristic-classesdifferential-geometryvector-bundles

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?

Best Answer

From purely algebraic arguments:

  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, $x\neq 0$.

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