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?

  • 3
    $\begingroup$ 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. $\endgroup$ – Ted Shifrin May 28 '13 at 20:19
  • 1
    $\begingroup$ 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. $\endgroup$ – Sam Lisi May 29 '13 at 0:19

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, $x\neq 0$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.