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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.

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  • $\begingroup$ 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? $\endgroup$ – Warlock of Firetop Mountain Apr 1 at 9:22
  • $\begingroup$ 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). $\endgroup$ – Blazej Apr 1 at 13:44

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