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Let $G$ be a semi simple Lie group (or real reductive), $\mathfrak{g}$ its lie algebra and $B$ its killing form. We can defined the 3-form $k$ by $$k(X,Y,Z)=B([X,Y],Z).$$ with $X,Y,Z\in \mathfrak{g}$. In fact, $k$ is nothing but the structure constants.

It is easy to prove that $k$ is a closed forme on $G$. For example, let $\nabla$ be the connection on $TG$ defined by $\nabla_XY=0$. Its torsion is $T(X,Y)=-[X,Y]$. Then $$d=e^i\wedge\nabla_{e_i}+i_{T}.$$ where $e_i$ is a base of $\mathfrak{g}$ and $e^i$ is the dual base. It is easy to verifier $e^i\wedge\nabla_{e_i}k=0$ and $i_{T}k=0$. So $dk=0$.

Can someone give some explanations of the 3 closed form $k$? For example,

  1. What is the topology mean of its cohomology class $[k]\in H^3(G)$.
  2. How about the case $[k]=0$ or $[k]\neq0$?
  3. If $G$ is a compact semi simple Lie group, can we say more about the form $k$?
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up vote 3 down vote accepted

Of course, this has been well-studied. Cartan himself pointed out this form. You can read about it in his little 1936 book La topologie des groupes de Lie. There are, of course, more modern references, such as Adams' Lectures on Lie groups.

To answer your questions:

  1. The form $k$ is always nonzero on Lie subgroups of type $SO(3)$ or $SU(2)$

  2. Of course, it can happen that $k=0$ (just take a torus) or $[k]=0$ (just take $G=SL(2,\mathbb{R})$). If the group contains a subgroup isomorphic to $SO(3)$ or $SU(2)$, though, it can't be cohomologous to zero.

  3. In the case that $G$ is compact and simple, then $[k]$ generates $H^3_{dR}(G)$. When $G$ is only semi-simple, this need not be true, as the dimension of $H^3_{dR}(G)$ can be bigger than $1$.

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Thanks for the answer, I like so much Cartan's book. And… tells the whole story. – shu May 29 '12 at 18:43

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