# Cohomology of Grassmannian and equivariant cohomology

Let $G=U(n)$ and $EU(n)=V_{n}(\mathbb{C}^{\infty})$ the infinit Stiefel manifold. So i think that $BU(n)$ is the infinite Grassmannian $G_r(\mathbb{C}^{\infty})$. We have $H_{BU(n)}(pt)= \mathbb{C}[x_{1}, \cdots, x_{n}]$. But from equivariant theory we have $H_{U(n)}(pt) \simeq H_{T}(pt)^{W}$ (where $T$ is a maximal torus in $U(n)$ and $W$ the Weyl group in $U(n)$). But, because $W$ is the permutation group $S_{n}$ it results $H_{T}(pt)\simeq \mathbb{C}[c_{1}, \cdots, c_{n}]$ where the $c_{i}$ are the simmetric polynomial in the $x_{i}$'s. So che cohomology of $BU(n)$ is isomorphic to $\mathbb{C}[x_{1}, \cdots, x_{n}]$ or $\mathbb{C}[c_{1}, \cdots, c_{n}]$?

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A fair amount of confusion here. Let's reverse notation. The cohomology of BT^n is the polynomial algebra on n generators x_i, each of degree 2. Its invariant algebra under the Weyl group, which is the symmetric group S_n, is the polynomial algebra on n generators s_i of degree 2i, the elementary symmetric functions in the x_i. The cohomology of BU(n) is the polynomial algebra on the Chern classes c_i of degree 2i, and the c_i map to the s_i under the natural map. In fact, one can justify that as a definition of the c_i. –  Peter May Jan 20 '13 at 18:58
The main lesson to learn here, I think, is that when you present a cohomology ring you should present it as a graded ring, announcing what the degrees are of the generators. I use $x_i^{(2)}, c_i^{(2i)}$ as indication, the first time I introduce a graded generator. –  Allen Knutson Jan 20 '13 at 19:18
But the isomorfism is $H^{*}(G_r()\infty)\simeq \mathbb{C}[x_{1}^(2), \cdots, x_{n}^{2}]$? –  Oscar1778 Jan 20 '13 at 21:51
No, it's not. $H_{BT} = {\mathbb Z}[x_1^{(2)},\ldots]$, $H_{BU(n)} = {\mathbb Z}[\ldots,c_i^{(2i)},\ldots]$, the latter being the $W$-invariants in the former. –  Allen Knutson Jan 22 '13 at 17:43