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 Answers 2


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 $$s_n(c_1,\dots,c_n)=(-1)^{n-1}nc_n.$$


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

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


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.