Cohomology of Grassmannian 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?
 A: 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
  
  
*
  
*$f^\ast: H^\ast(X) \longrightarrow H^\ast(Y)$ is injective.
  
*$f^\ast E = L_1 \oplus \cdots \oplus L_n$ 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.
