Question about Chern Character in Hatcher's book

87 Views Asked by At

I have a question about an argument from Allen Hatcher's script Vector Bundles and K-Theory in Cor. 4.4 (see page 110). Here the excerpt:

enter image description here

We consider a vector bundle $E \to S^{2n}$, Then for Chern classes we know (by cosidering cohomology groups of $S^{2n}$) that $c_1(E) =... c_{n-1}(E)=0$.

Futhermore by definitionof Chern character we have $ch(E)= dim E + s_n(c_1,\ldots, c_n)/n!$

My question is why holds

$$s_n(c_1,\ldots, c_n)/n!=\pm nc_n(E)/n!$$? (*)

The author refers to a recursion formula from page 63:

$s_n= \sigma_1 s_{n-1} - \cdots +(-1)^{n-1}n\sigma_n$.

where $\sigma_k$ are the $k$-th symmetric polynomials.

What I don't understand is why $s_n(c_1,\ldots, c_n)/n!=\pm nc_n(E)/n!$ and not $s_n(c_1,..., c_n)/n!=c_n^n(E)/n!$?

Indeed, here the symmetric polynomials are considered in variebles $t_i:= c_i(E)$ therefore $\sigma_1= \sum c_i(E)= c_n(E)$ and $\sigma_k=0$ for $k >1$ since all summands of $\sigma_k$ the containa factor $c_j$ with $j \neq n$. But this contracicts (*). Where is the error in my reasonings?

Thank you.

2

There are 2 best solutions below

0
On BEST ANSWER

I think you are confused about what $s_n$ means here. The notation $s_n(c_1,\dots,c_n)$ does not mean we are substituting the $c_i$ for the variables $t_i$ in the symmetric polynomial $t_1^n+\dots+t_n^n$. Rather, $s_n$ is defined as the polynomial which, when inputted the elementary symmetric polynomials in $t_1,\dots,t_n$, outputs $t_1^n+\dots+t_n^n$. That is, $$s_n(\sigma_1(t_1,\dots,t_n),\dots,\sigma_n(t_1,\dots,t_n))=t_1^n+\dots+t_n^n.$$ In particular, this means that when we apply the formula $$s_n= \sigma_1 s_{n-1} - \dots +(-1)^{n-1}n\sigma_n$$ to $s_n(c_1,\dots,c_n)$, we are substituting $c_i$ for $\sigma_i$, not for $t_i$. Since $c_i=0$ for $0<i<n$, we get just $$s_n(c_1,\dots,c_n)=(-1)^{n-1}nc_n.$$

0
On

I always find it helps to think about these things in terms of the Splitting Principle:

For every complex vector bundle $E \to X$ of rank $n$ (where maybe $X$ has to be paracompact) there is a "splitting space" $S(X)$ and a continuous function $f\colon S(X) \to X$ such that $f^*(E) \cong \oplus_{i=1} ^n L_i$ where $L_i \to S(X)$ is a complex line bundle and $f^*\colon H^*(X) \to H^*(S(X))$ is injective.

If we let $x_i = c_1(L_i)$ then $$f^*(c_k(E)) = \sigma_k(x_1,\dots,x_n)$$

where $\sigma_k\in \mathbb{Z}[x_1,\dots,x_n]$ is the $k$-th elementary symmetric polynomial. In particular $\sigma_1 = x_1 + \dots + x_n$ doesn't correspond to $c_n(E)$, it's $c_1(E)$.

The polynomial $x_1^k + \dots + x_n^k\in \mathbb{Z}[x_1,\dots,x_n]$ is symmetric so The Theory implies it can be expressed as a polynomial in the elementary symmetric poynomials, i.e. there is an $s_k \in \mathbb{Z}[y_1,\dots, y_n]$ (called a "Netwon Polynomial") such that $s_k(\sigma_1, \dots, \sigma_n) = x_1^k + \dots + x_n^k$. Then if $c_1(E),\dots,c_{n-1}(E)$ vanish so will $\sigma_1, \dots \sigma_{n-1}\in H^*(S(X))$, so the recursion formula reduces to

$$ s_n = (-1)^n n \sigma_n $$