Here $U(n)$ is the unitary group, consisting of all matrix $A \in M_n (\mathbb{C})$ such that $AA^*=I$

Problem How to calculate the integer cohomology group $H^*(U(n))$ of $U(n)$? What if $O(n)$ replace $U(n)$?

My primitive idea is that: as for $U(n)$, it is a Lie group and can acts transitively on the unit sphere with a isotropy group (or say stabilizer) $U(n-1)$. Hence, $U(n)/U(n-1) = S^{2n-1}$ and we have a fibre bundle $$ \pi : U(n) \to S^{2n-1}$$ with fibre $U(n-1)$. Then, we try to use spectral sequence and Leray's theorem to calculate.

Although it seems that we should calculate it by induction, I don't know how to continue. Moreover, someone could use any other method as long as it work.

  • $\begingroup$ Your method seems like the correct one. Have you tried working through the spectral sequence for some small n? $\endgroup$ – Robert Short May 31 '15 at 12:08

I'll expand on the hint given by Robert. $U(1)$ is just $S^1 \subset \mathbb{C}$, and its cohomology ring is the exterior algebra $\Lambda[c_1]$. Here, the subscript of the generator indicates its degree.

For $n = 2$, we have, as you described, a fibre bundle $U(1) \to U(2) \to S^3$. Thus there is a Serre spectral sequence with \begin{equation} E_2^{p,q} = H^p(S^3,\mathcal{H}^q U(1)) \cong H^p(S^3) \otimes H^q(U(1)) \Rightarrow H^{p+q}(U(2)). \end{equation}

We have $E_2^{p,q} \cong \Lambda[c_1,c_3]$. By lacunary reasons, this spectral sequence collapses on the second page, and so we deduce $H^*(U(2)) \cong \Lambda[c_1,c_3]$.

In general, the spectral sequence for the fiber bundle $U(n-1) \to U(n) \to S^{2n-1}$ always collapses on the second page, and you can use induction to prove the proposition $H^*(U(n)) \cong \Lambda[c_1,c_3,\ldots,c_{2n-1}]$.

I like Mimura and Toda's The topology of Lie groups as a reference for these types of calculations. If I recall, they also discuss how to obtain the above proposition using Morse theory (or maybe just a rational decomposition of $U(n)$).

You also asked about $O(n)$, which is trickier. First, $O(n)$ is not path-connected, but has two path components each homeomorphic to $SO(n)$, so we may as well consider $SO(n)$ instead. There is a section in Hatcher (3.D) that is dedicated to computing the mod 2 cohomology of $SO(n)$, using a cell decopmosition.

You can also compute the mod 2 cohomology of $O(n)$ and $SO(n)$ using spectral sequences, but the problem here is that the analogous spectral sequences need not collapse, so the analysis is more subtle. One can instead induct on Stiefel manifolds, which generalize $O(n)$. I learnt how to do this from May et al's notes on characteristic classes. In particular, see Theorem 2.4 for the cohomology groups of $SO(n)$, although the ring structure requires consideration of Steenrod operations. May also discusses cohomology away from the prime 2.

  • $\begingroup$ Thank you! I'm a little confused with your notation $Λ[c_1,c_3,…,c_{2n−1}]$? Could you tell me about it ? $\endgroup$ – Hang Jun 1 '15 at 11:25
  • $\begingroup$ By $\Lambda[c_1,c_3,\ldots,c_{2n-1}]$ I mean the exterior algebra on the generators $c_1, \ldots, c_{2n-1}$. Think of it as a polynomial algebra, but with skew-commutativity imposed instead. Also, we have $c_i \in H^i$, which gives the grading in cohomology. For example, $H^6(U(3))$ is freely generated by the element $c_1 c_5$ (and $c_3^2 = 0$). $\endgroup$ – JHF Jun 2 '15 at 1:16
  • $\begingroup$ Sorry, would you mind give some details about how to get $H^∗(U(2))$ . After I read Bott's book , I still cannot understand how you get $E^2=\Lambda[c_1,c_3]$ $\endgroup$ – Hang Jul 6 '15 at 14:55
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    $\begingroup$ Right, sorry about the delay! Anyway, we know that $U(2)$ is a $S^1$-bundle over $S^3$. The homology of $S^1$ is $\Lambda[c_1]$, and the homology of $S^3$ is $\Lambda[c_3]$. The $E^2$ page of the Serre spectral sequence is just $H^*(S^3) \otimes H^*(S^1)$ since the base $S^3$ is simply-connected. So $E^2 = \Lambda[c_3] \otimes \Lambda[c_1] \cong \Lambda[c_1,c_3]$. $\endgroup$ – JHF Jul 18 '15 at 5:02

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