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I computed the CW structure on U(n) using morse theory.I want to verify my answer.So I was wondering if someone here can supply the answer as I can't find any source by googling.Also I want to know if there is any reference for this.

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See Theorem (2.2) in arxiv.org/abs/dg-ga/9506004 –  Christian Nassau Jul 3 '12 at 13:51
    
I think the cell structure is worked out in Milnor and Stasheff's Characteristic classes and/or Steenrod's Intro to Fibre Bundles. –  Scott Carter Jul 3 '12 at 13:59
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This can be found in the book "Cohomology Operations" by Steenrod and Epstein, Chapter 4, where they also do the parallel cases of orthogonal and symplectic groups, as well as the generalizations to Stiefel manifolds. For the original source in the unitary case they cite the paper of I. Yokota, "On the cellular decompositions of unitary groups", J. Inst. Polytech., Osaka City Univ. 7 (1956), 39-49. –  Allen Hatcher Jul 3 '12 at 16:11
    
I'm not sure I understand this question. Surely different Morse functions lead to different CW structures? –  Mark Grant Jul 4 '12 at 19:13
    
Mark: I think usually one is interested in decompositions with the minimal number of cells. –  Claudio Gorodski Jul 4 '12 at 20:41

1 Answer 1

up vote 2 down vote accepted

A simply-connected compact Lie group $G$ has the same rational homotopy type (and rational cohomology ring) of as a product of odd-dimensional spheres $S^{2m_1+1}\times\cdots\times S^{2m_r+1}$ where the $m_i$ are invariants called exponents and $r$ is the rank of $G$. (The exponents are related to many algebraic invariants of $G$).

In your case $U(n)$ is diffeomorphic to $S^1 \times SU(n)$ and the exponents of $SU(n)$ are $1,\ldots, n-1$. For instance, in case $n=3$, $SU(3)$ has exponents $1$, $2$ and $U(3)\sim S^1 \times S^3 \times S^5$ has a cell decomposition with one cell in each dimension $0$, $1$, $3$, $4$, $5$, $6$, $8$, $9$.

Classical references are A. Borel http://www.ams.org/journals/bull/1955-61-05/S0002-9904-1955-09936-1/home.html and H. Samelson http://www.ams.org/journals/bull/1952-58-01/S0002-9904-1952-09544-6/home.html.

Edit: There is the maybe easier following argument. In general, for a (locally trivial) fiber bundle $\pi:E\to B$ with typical fiber $F$, where $B$ and $F$ are CW-complexes, a cell decomposition for the total space $E$ is the same as that of the direct product $B\times F$, since each cell $c$ of $B$ is contractible and thus the bundle above it is trivial: $\pi^{-1}(c)\cong c\times F$.

Now consider the (principal) fiber bundle $SU(n-1)\to SU(n)\to S^{2n-1}$ for $n\geq2$ and apply induction.

Edit 2: The minimal number of cells in a CW-decomposition equals the sum of the Betti numbers over $\mathbf Z_2$ (or any field), by the Morse inequalities. We need only to know that $G$ and $S^{2m_1+1}\times\cdots\times S^{2m_r+1}$ have the same $\mathbf Z_2$-Betti numbers, so the minimal number of cells in this case is $2^r$, where $r$ is the rank of $G$.

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Could you please give some references? –  Guntram Jul 3 '12 at 16:26
    
Your claim "$G$ has the same homotopy type as a product of odd-dimensional spheres" is only true rationally, isn't it? –  Christian Nassau Jul 3 '12 at 16:28
    
Christian: I've edited my post, thanks. –  Claudio Gorodski Jul 3 '12 at 17:21
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If I'm not mistaken, a rationall cell decomposition need not come from an honest CW decomposition. –  Fernando Muro Jul 3 '12 at 22:07

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