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Let $G_r$ the infinite complex Grassmannian manifold. We know that $H^{*}(G_r)=\mathbb{C}[x_{1}, \cdots, x_{n}]$ where $x_i$ are the Chern classes of tautological bundle. But $H^{*}(G_r)$ is also isomorphic to the ring $\mathbb{C}[c_{1}, \cdots, c_{n}]$ where $c_i$ are the symmetric polynomials in $y_i$ where $y_i$ are the variables in $\mathbb{C}[y_1, \cdots, y_n]$. How can I see Chern classes as symmetric polynomials?

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up vote 11 down vote accepted

Recall the splitting principle for complex vector bundles.

Theorem. (Complex Splitting Principle) For all rank $n$ complex vector bundles $p: E \longrightarrow X$, there exists a manifold $Y$ and a map $f: Y \longrightarrow X$ such that

  1. $f^\ast: H^\ast(X) \longrightarrow H^\ast(Y)$ is injective.

  2. $f^\ast E = L_1 \oplus \cdots \oplus L_n$ where the $L_i$'s are complex line bundles.

The splitting principle tells us that for purposes of computation, we may consider a complex vector bundle as a Whitney sum of complex line bundles.

Given such a splitting $f^\ast E = L_1 \oplus \dots \oplus L_n$, write $y_k = c_1(L_k)$. Then by the Whitney product formula and naturality of Chern classes, we have \begin{align} f^\ast c(E) & = c(f^\ast E) \\ & = c(L_1 \oplus \cdots \oplus L_n) \\ & = \prod_{k = 1}^n c(L_k) \\ & = \prod_{k = 1}^n (1 + c_1(L_k)) \\ & = \prod_{k = 1}^n (1 + y_k), \end{align} which upon expansion shows that $f^\ast c_k(E)$ is the $k^\text{th}$ elementary symmetric polynomial in the $y_i$'s.

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Very good answer! – ArthurStuart Jan 22 '13 at 0:03
Do you have an explicit example and a reference for a proof of Complex Splitting Principle? – ArthurStuart Jan 22 '13 at 10:09
It will almost certainly be in Milnor-Stasheff. Else Hatcher has a (partial) book on vector bundles on his website, and it will likely be in there. – Juan S Jan 22 '13 at 21:54
I didn't understood why $ \prod_{k = 1}^n c(L_k) = \prod_{k = 1}^n (1 + c_1(L_k)) $... – ArthurStuart Jan 23 '13 at 15:09
$c(L_k)$ is the total Chern class of a complex line bundle, so $$c(L_k) = 1 + c_1(L_k) + c_2(L_k) + \cdots = 1 + c_1(L_k)$$ as $c_i(L_k) = 0$ for $i \geq 2$ when $L_k$ is a complex line bundle. – Henry T. Horton Jan 23 '13 at 16:52

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