2
$\begingroup$

Let $G$ be a complex semisimple group and $\mathcal{O} \subset G$ a conjugacy class, i.e. $\mathcal{O} = \{gag^{-1} : g \in G\}$ for some $a \in G$. Let $\Omega$ be the Cartan 3-form on $G$ defined by $$ \Omega(x^L, y^L, z^L) = \langle x, [y, z] \rangle,\quad (x,y,z \in \mathrm{Lie}(G)) $$ where $\langle,\rangle$ is the Killing form, and $x^L, y^L, z^L$ are the left-invariant vector fields generated by $x,y,z$.

Question: Is the pullback of $\Omega$ to $\mathcal{O}$ trivial in cohomology?

$\endgroup$
3
  • $\begingroup$ What do you mean by $x^L$? $\endgroup$
    – YCor
    Jul 29, 2022 at 20:25
  • 1
    $\begingroup$ The left-invariant vector field generated by $x$. $\endgroup$ Jul 29, 2022 at 20:33
  • 3
    $\begingroup$ Thanks. One particular case: if $G=\mathrm{SL}_2(\mathbf{R})$ then conjugacy classes are manifolds of dimension $\le 2$ so their cohomology vanish in degree 3. $\endgroup$
    – YCor
    Jul 29, 2022 at 20:53

1 Answer 1

4
$\begingroup$

Yes, it is exact, and there is in fact a canonical 2-form on each conjugacy class whose derivative is your $\Omega$. This was an important observation when studying D-branes in WZW models, see, e.g. https://arxiv.org/abs/hep-th/0008038 or https://arxiv.org/abs/hep-th/0205233.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.