Integral cohomology of $SU(n)$ - looking for constants

Question

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!

Answer 1

It seems the required coefficient is given in

• R. Bott and R. Seeley. Some remarks on the paper of Callias: "Axial anomalies and index theorems on open spaces". Comment. Math. Phys. 62 (1978), 235-245. link to paper on Project Euclid

The discussion on p.237 says that $$ \left(\frac{i}{2\pi}\right)^m\frac{(m-1)!}{(2m-1)!}\alpha_{2m-1} $$ corresponds, under the identification ${\rm H}^{2m-1}({\rm U}(n),\mathbb{Z})\cong {\rm Hom}({\rm H}_{2m-1}({\rm U}(n),\mathbb{Z}),\mathbb{Z})$, to a generator of the image of the Hurewicz map $\pi_{2m-1}({\rm U}(n))\to{\rm H}_{2m-1}({\rm U}(n),\mathbb{Z})$.

Answer 2

OK, I have understood how it works. This is classical material but it seems difficult to find an answer in the literature for the non-specialist; so let me explain.

We consider the universal bundle $E U(n) \rightarrow B U(n)$ with fiber $U(n)$. Since $E U(n)$ is contractible, the Serre spectral sequence of the fibration gives information on the cohomology of $U(n)$. I recall that $H^{\bullet}(BU(n),\mathbb{Z}) = \mathbb{Z}[c_1,c_2,\dots,c_n]$, where $c_p$ is of degree $2p$. What the spectral sequence says is that to each $c_p$ corresponds a class $x_p$ in $H^{2p-1}(U(n),\mathbb{Z})$ which transgresses $c_p$.

In order to answer my question, we need to understand how to express these $x_p$ in terms of $c_p$.

Let $I = I(\mathfrak{u}(n))^{U(n)}$ be the space of invariant polynomials on $\mathfrak{u}(n)$. The Chern-Weil isomorphism $W$ gives that $I \cong H^{2\bullet}(BU(n),\mathbb{R})$.

If $P$ is of degree $l$ in $I$, we can consider the cohomology class $W(P)$ and transgress it to get a cohomology class of degree $2l-1$ in $U(n)$; denoting by $TP$ the class obtained, it can be shown that $$ TP = \frac{(-1)^{l-1}}{2^{l-1} {2l-1 \choose l} } P(\omega \wedge [\omega,\omega]^{l-1})$$ (see equation 3.10 in Chern-Simons, Characteristic Forms and Geometric Invariants ; a minus one seems to be missing in the denominator).

The conclusion is that we have to know which invariant polynomials give integral classes on $H^{2\bullet}(BU(n),\mathbb{Z})$. But of course, these are just the polynomials used in order to compute the Chern classes of a vector bundle.

Since the polynomials that I consider in my question correspond to the Chern character, one has to relate the Chern character and Chern class in order to answer. Take $l=2$ for instance. We define the polynomial $ch_2(A) = \frac{-1}{8\pi^2}\text{Tr}(A^2)$. If $c_1$ and $c_2$ are the first and second Chern polynomials, one has $$ ch_2 = \frac{1}{2}(c_1^2 - 2c_2).$$ The transgression $Tc_1^2$ vanishes so that $Tch_2 = Tc_2$ is an integral class (and this is optimal for the constant).

This gives that $\frac{1}{48\pi^2} \text{Tr}(\omega \wedge [\omega,\omega]) = \frac{1}{24\pi^2} \text{Tr}(\omega^3)$ is an integral class, as was claimed.

Edit --- With the same method, one can compute the constants for $l=3$ but this is quite subtle. First, we define $$ ch_3 = \frac{1}{6}(c_1^3 - 3c_1 c_2 + 3c_3)$$ so that $ch_3(A) = \frac{1}{6}\Big(\frac{1}{2\pi i}\Big)^3 \text{Tr}(A^3)$. The transgression of $c_1^3$ is zero and the transgressions of $c_1 c_2$ and $c_3$ are of course integral. This shows that the transgression of $2ch_3$ is integral.

Computing, this gives $T(2ch_3) = \frac{1}{24 \pi^3} \frac{1}{10} \text{Tr}(\omega^3)$. (I forget the $i$-factor)

If one uses Matthias formula, he will find that $\frac{1}{480 \pi^3} \text{Tr}(\omega^3)$ is integral, which is better by a factor $2$.

In fact, the class $c_1 c_2 - c_3$ is always even. This is one of the so-called Schwarzenberger condition. This shows that in fact $T(ch_3)$ is integral and explains the $2$-factor.

This discussion shows that the optimal constant can be difficult to obtain, from both methods (since indeed, there is no guarantee that the Hurewicz map maps to a non-divisible class). Anyway, the picture is very nice :-).

Remark that in general the formula of Matthias exactly says that the transgression of $ch_l$ is integral, which is not obvious to me.

Answer 3

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 $a_k = \left(\frac{i}{2 \pi}\right)^k \frac{((k-1)!)^2}{(2k-1)!}$.