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 $k$?



Topologic or geometric mean of the structure constants of a semi simple lie algebra
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 $1$.


