# Normalization of the generator of third cohomology of a compact Lie group

It is proven in "Loop groups" by Pressley and Segal (Prop. 4.4.5, p. 49) that the left invariant 3-form $$\sigma$$ on a simply-connected compact Lie group $$G$$ whose value at the identity is given by $$\sigma(\xi, \eta, \zeta) = \langle [\xi, \eta], \zeta \rangle$$ defines an integral cohomology class (which I interpret as the statement that its periods, i.e. integrals over $$3$$-cycles, are integers) if and only if the invariant bilinear pairing $$\langle -, - \rangle$$ is such that $$\langle h_{\alpha}, h_{\alpha} \rangle \in 2 \mathbb Z$$ for every coroot $$h_{\alpha}$$. The proof relies on on the assertion that this is true for $$G = \mathrm{SU}(2)$$. However I calculated that with this construction of $$\sigma$$ the result is $$\int_{\mathrm{SU}(2)} \sigma = 48 \pi^2.$$ Therefore I think the value at the identity should be modified to $$\sigma(\xi, \eta, \zeta) = \frac{1}{4 8 \pi^2} \langle [\xi , \eta], \zeta \rangle,$$ and then the argument in the book goes through.

The question is where is the error - is it in Segal, Pressley or maybe my calculations are wrong? For reference I include a sketch of my calculation below. Perhaps it might be of use to someone in the future.

Since I'm working with a matrix group, left-invariant Maurer-Cartan takes the form $$\omega = g^{-1} dg$$. I parametrize $$g = \begin{bmatrix} x & y \\ - \overline{y} & \overline{x} \end{bmatrix}$$ with $$x,y$$ - complex numberd satisfying $$|x|^2+|y|^2=1$$. It is convenient to write $$x = \mathrm{cos}(\theta) e^{i \phi}$$, $$y = \mathrm{sin}(\theta) e^{i \psi}$$. Differential form $$\sigma = \mathrm{tr} \left( \omega \wedge [ \omega \wedge \omega] \right)$$ satisfies the assumptions mentioned above. After a few lines of calculations I get that $$\sigma = 12 \sin(2 \theta) d \theta \wedge d \phi \wedge d \psi.$$ Two factors of $$2 \pi$$ come from integrating $$\phi$$ and $$\psi$$ over $$[0, 2 \pi]$$. Integration with respect to $$\theta$$ over $$\left[ 0, \frac{\pi}{2} \right]$$ gives $$1$$. Hence the final result.

• Btw, why not divide by just $\pi^2$? $48 \in \mathbb{Z}$ afterall. Are you sure that you don't have added an extra factor $\pi$ in the Killing form? – Warlock of Firetop Mountain Apr 1 at 9:22
• Yes, 48 is an integer, but the point is that I want to find the "smallest possible" normalization which gives me integers (unless I misunderstood the proof presented in Pressley, Segal). – Blazej Apr 1 at 13:44