most recent 30 from http://mathoverflow.net 2013-06-19T09:39:59Z http://mathoverflow.net/feeds/user/30618 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/120631/cohomology-of-fine-grassmannian-manifold Oscar1778 2013-02-02T22:56:41Z 2013-02-02T22:56:41Z

<p>Let it consider the fibration $U(n) \rightarrow V_{n}(\mathbb{C}^{k}) \rightarrow Gr(n,k)$ where $U(n)$ is the unitary group, $V_{n}(\mathbb{C}^{k})$ the Stiefel manifold and $Gr(n,k)$ the Grassmannian manifold. Now I want to use a spectral sequence in order to calculate the coohomology of finite Grassmannian manifold. So how can I calculate $H^{s}(V_{n}(\mathbb{C}^{k}))$ and $H^{m}(U(n))$?</p>

http://mathoverflow.net/questions/118819/on-direct-limit-of-stiefel-mainfold Oscar1778 2013-01-13T15:44:47Z 2013-01-28T01:22:00Z

<p>I'd like to build a model for the space $EU(n)$: the total space of universal bundle $\pi:EU(n) \rightarrow BU(n)$. $\;$ $EU(n)$ must be a conctractible space on which $U(n)$ acts freely. So I consider Stiefel manifold $V_{n}(\mathbb{C}^{k})$ of $\;$ $n-$frame in $\mathbb{C}^{k}$ and the associated fiber bundle $$ V_{n-1}(\mathbb{C}^{k-1}) \rightarrow V_{n}(C^{k}) \rightarrow S^{2k-1} $$ The induced sequence in homotopy shows that all homotopy gruops vanish for $k$ large. Indeed, the natural action of $U(n)$ on $V_{n}(\mathbb{C}^{k})$ is free (the quotien is the Grassmannian manifold). So I can choose $$ EU(n)=\lim_{k \to \infty} V_{n}(\mathbb{C}^{k}) = V_{n}(\mathbb{C}^{\infty}) $$ the direct limit of $V_{n}(\mathbb{C}^{k})$. How could I prove that the action of $U(n)$ on $V_{n}(\mathbb{C}^{k})$ is still free? In other words, why is the limit of a free action free?</p>

http://mathoverflow.net/questions/119751/lie-groups-bundle Oscar1778 2013-01-24T12:20:53Z 2013-01-24T23:21:24Z

<p>Given compact Lie groups $H \subset K \subset G$, there is a fiber bundle $ \frac{K}{H} \rightarrow \frac{G}{H} \rightarrow \frac{G}{K}$. Do you have a simple proof of this?</p>

http://mathoverflow.net/questions/119417/cohomology-of-grassmannian-and-equivariant-cohomology Oscar1778 2013-01-20T18:31:23Z 2013-01-20T18:31:23Z

<p>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 $\mathbb{C}[c_{1}, \cdots, c_{n}]$?</p>

http://mathoverflow.net/questions/118662/cohomology-of-quotient-space/118809#118809 Oscar1778 2013-01-13T13:51:54Z 2013-01-13T13:51:54Z

<p>Have you some reference to study how Bruhat decomposition works? How can I use Lefschetz? However, do you know a complete reference in this topic?</p>

http://mathoverflow.net/questions/119751/lie-groups-bundle/119760#119760 Oscar1778 2013-01-24T15:33:33Z 2013-01-24T15:33:33Z

In my situation $G=U(n)$ and $K=N_{G}(T)$ is the normalizer of $T$ in $U(n)$ and $T$ is a maximal torus in $U(n)$ (i.e. the subgroup of diagonal matrix). So I ask you if $U(n) \rightarrow U(n)/N_{G}(T)$ is a principal $N_{G}(T)$-bundle (i.e. $N_{G}(T)$ acts freely on U(n)). Thaks.

http://mathoverflow.net/questions/119417/cohomology-of-grassmannian-and-equivariant-cohomology Oscar1778 2013-01-20T21:51:28Z 2013-01-20T21:51:28Z

But the isomorfism is $H^{*}(G_r()\infty)\simeq \mathbb{C}[x_{1}^(2), \cdots, x_{n}^{2}]$?