TITLE: Integral cohomology of $SU(n)$ - looking for constants
QUESTION [20 upvotes]: I am interested in explicit generators of the cohomology $H^\bullet(SU(n),\mathbb{Z})$. Let $\omega = g^{-1} dg$ be the Maurer-Cartan form on $SU(n)$. The forms $\alpha_3,\alpha_5,\dots,\alpha_{2n-1}$, defined by
$$ \alpha_k := \text{Tr}(\omega^{k})$$
are bi-invariant and define classes in de Rham cohomology.
It is well-known that the cohomology algebra $H^\bullet(SU(n),\mathbb{R})$ is an exterior algebra in $\alpha_3,\dots,\alpha_{2n-1}$. More precisely, my question is thus the following: find the optimal constants $C_k$ such that $C_k \cdot \alpha_k$ is an integral class.
For $n=2$, I have found in the literature that $C_3$ is $(24\pi^2)^{-1}$. But I cannot find a reference for the higher dimensions.
I expect that $C_k = q_k \cdot \pi^{-(k+1)/2}$, with $q_k$ some rational number, because of the volume of the sphere $\mathbb{S}^{2k-1}$, but I have no precise proof.
It seems from the comments in Cohomology of the unitary group that an interpretation in terms of transgression can solve my problem but I am not very familiar with this.
Thanks!

REPLY [5 votes]: I got confused by some of the remarks in the other answers and so decided to work it out for myself. Hopefully, the following is helpful to people who stumble upon this question like me.
One can indeed use transgression as discussed in Jeremy Daniel's answer in order to settle this problem, since the generators of $H^*(BU; \mathbb{Z})\cong \mathbb{Z}[c_1,c_2,\dots]$ are well known and given by the Chern classes. The following properties of the transgression map $T$ are important for us:

The transgression of nontrivial cup products is $0$, and therefore we
have that $(-1)^{k-1} (k-1)! T(ch_k) = T(c_k)$ for the Chern character/ Chern
classes
Transgressing the Chern character $ch_k$ two times in the path loop
fibration gives the Chern character $ch_{k-1}$ in one degree lower under the identification of $\Omega^2 BU \sim BU$, since
the Chern character is compatible with Bott periodicity
Transgression of an integral class must always give an integral class

It is clear from this that the transgression $T(ch_k) = (-1)^{k-1}\left(\frac{i}{2 \pi}\right)^k \frac{(k-1)!}{(2k-1)!} \alpha_{2k-1}$ cannot be the image of an integral generator under the coefficient homomorphism $H^{2k-1}(U;\mathbb{Z})\to H^{2k-1}(U;\mathbb{R})$ in general, since transgressing again in the path loop fibration would then yield that $T(T(ch_k)) = ch_{k-1}$ is an integral cohomology class, which is wrong for $k>2$.
What is claimed in Matthias Wendt's answer is still true: If we are in the stable range, then on a smooth map $f\colon S^{2k-1}\to U(n)$ which generates the homotopy group $\pi_{2k-1}(U(n)) = \mathbb{Z}$, this form evaluates to $\int_{S^{2k-1}} f^*  T(ch_k)=1$. The crucial point here is though that the image of such a generator under the Hurewicz homomorphism $h\colon \pi_{2k-1}(U) \to H_{2k-1}(U;\mathbb{Z})$ is divisible by exactly $(k-1)!$. This can be deduced from the last corollary in chapter 8 in
Bott, Raoul. The space of loops on a Lie group. Michigan Math. J. 5 (1958), no. 1, 35--61. doi:10.1307/mmj/1028998010. https://projecteuclid.org/euclid.mmj/1028998010,
which gives the corresponding statement for BU. This means that $(k-1)! T(ch_k)$ is an integral generator, which is consistent with the fact that transgression is an integral isomorphism and therefore must map the Chern classes to integral generators. Therefore, the optimal constant that makes the forms $\alpha_{2k-1}$ into integral generators should be
$a_k = \left(\frac{i}{2 \pi}\right)^k \frac{((k-1)!)^2}{(2k-1)!}$.