6
$\begingroup$

I have some difficulties understanding a very precise point of Switzer's proof of the existence and unicity of chern classes, which is Theorem 16.2 in his book. Unfortunatly there are many notations that I have introduce to write 100% of what is needed for my question to be self-contained.

So here we are : the theorem is

16.2. Theorem. Suppose $h^*$ is a cohomology theory with products such that for each $n$ there are elements $x_n\in h^2(CP^n)$ satisfying

  1. $h^*(CP^n)=h^*(pt)[x_n]/x_n^{n+1}$.
  2. The inclusion $i:CP^n \to CP^{n+1}$ gives $i^*x_{n+1} = x_n$.

Then for each $U(n)$-bundle $\xi$ over a CW-complex $X$ there are uniquely defined elements $c_0(\xi), c_1(\xi),\cdots, c_n(\xi)$ with $c_i(\xi) \in h^{2i}(X)$ depending only on the isomorphism class of $\xi$ and satisfying

  1. if $\xi\to X$ is a bundle and $f:Y\to X$ a map, then $c_i(f^*\xi) =f^*c_i(\xi)$
  2. $c_0(\xi) = 1$
  3. If $\gamma \to CP^n$ is the Hopf $U(1)$-bundle over $CP^n$, then $c_1(\gamma) = x_n$
  4. If $\xi$ is a $U(m)$-bundle and $\eta$ is a $U(n)$-bundle, both over X, then... (the convolution formula).

To prove this, the first step is to introduce the projectivisation $P(\xi)$ of a $U(n)$-bundle $\xi$, and to show with Leray-Hirsch that $h^*(P(\xi))=h^*(X)[1, y, y^2, \cdots, y^{n-1}]$ as modules over $h^*(X)$, where $y$ is found by taking the pullback of a generator of $h^2(CP^\infty)$ through $P(\xi)\to CP^{\infty}$ the classifying map of the tautological line bundle over $P(\xi)$.

Then the classes $c_i(\xi)$ are defined to be the coefficients for the linear dependence formula $$y^n=(-1)^{n+1}c_n(\xi)\cdot 1+(-1)^n c_{n-1}(\xi)y+\cdots+c_1(\xi)y^{n-1}$$

My issue is the following : to show the third point of the theorem, Switzer says the following

To prove 3. we simply observe that $P(\gamma) = CP^n$ and $\lambda_\gamma = \gamma$, $y_\gamma = x_n$, so $c_1\gamma = x_n$.

where

  • $\lambda_\xi$ stands for the tautological line bundle over $P(\xi)$
  • $\gamma$ is the standard tautological bundle over $CP^\bullet$ as stated in the hypothesis of the theorem
  • by $y_\gamma$ I guess that it is meant "the $y$ such that there is this basis obtained by the Leray-Hirsch theorem"

I can't see how this has a sense. Indeed, in the case of $\gamma$, the bundle to which we apply Leray-Hirsch theorem is $CP^n\to CP^n$, and so the basis is just $(1)$ over $h^*(CP^n)$... So there is no $y$ in this case. My concrete question would be : why does Switzer say that $y_\gamma=x_n$ ? Also, with this in mind, the linear dependence formula in the case of $\gamma$ gives nothing so I can't see why this could tell how $c_1(\gamma)$ can be computed...

Thank you for your help.

$\endgroup$

1 Answer 1

2
$\begingroup$

In your definitions it says $y := f^* x_\infty$ where $x_\infty$ is the generator of $h^2(\mathbb{CP}^\infty)$ and $f:P(\xi) \to \mathbb{CP}^\infty$ is the classifying map of the tautological line bundle over $P(\xi)$.

If $\xi = \gamma$ is the tautological line bundle over $X= \mathbb{CP}^n$ you have that $f:\mathbb{CP}^n \to \mathbb{CP}^\infty$ is inclusion into the colimit. Now $y=f^*x_\infty = x_n$ by the two assumptions on the cohomology theory. The linear independence formula now says $y^1 = (-1)^2 c_1(\gamma)$. So $x_n$ is the first chern class of $\gamma$.

$\endgroup$

You must log in to answer this question.

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