Third cohomology group of a simple Lie algebra

Question

How can I show that the third Lie cohomology (with trivial coefficients $\mathbb R$) of a simple Lie algebra is generated by the Maurer-Cartan form ?

I searched on internet and in some books but didn't found anything.

Starting from the answer I take any cocycle $\omega$ and the Killing form on the simple Lie algebra $\mathfrak g$ denoted by $\langle \cdot, \cdot \rangle$ I can construct $T_{a, b} \in \mathfrak g$ such that $\langle T_{a,b}, c \rangle = \omega (a,b,c)$. Not sure how to continue.

Any reference will be helful.

Bonus question : Who was the first to come up with such a result ?

Answer 1

The proof is in the paper on Lie algebra cohomology by Chevalley and Eilenberg, in $1948$ (and in Koszul's paper of $1950)$. Chevalley and Eilenberg proved that $H^3(\mathfrak{g},K)$ is nonzero for every nonzero semisimple Lie algebra $\mathfrak{g}$ and every field $K$ of characteristic zero. This works as follows:

Let $B(x,y)$ denote the Killing form of $\mathfrak{g}$. Define a $3$-cochain in $C^3(\mathfrak{g},K)$ by $$ g(x,y,z):=B(x,[y,z])=B([x,y],z). $$ In the second equation we use that $B$ is invariant. By Lemma $4.4.4$ of the thesis of Florian Kickinger, page $150$, every invariant $p$-cochain in $C^p(\mathfrak{g},K)$ of a semisimple Lie algebra over a field of characteristic zero is a $p$-cocycle in $Z^p(\mathfrak{g},K)$. Here $g$ is invariant, if $$ g([x_1,x],x_2,\ldots ,x_n)+\cdots +g(x_1,\cdots,x_{p-1},[x_p,x])=0 $$ for all $x,x_1,\ldots ,x_p\in \mathfrak{g}$. Now, for $p=3$ and the above $g$ we have \begin{align*} g([x_1,x],x_2,x_3) & + g(x_1,[x_2,x],x_3)+g(x_1,x_2,[x_3,x]) \\ & = B([[x_1,x],x_2], x_3)+B([x_1,[x_2,x]],x_3)+B([x_1,x_2],[x_3,x]) \\ & = B(([x_1,x_2],x],x_3)-B([x_1,x_2],[x,x_3]) \\ & = 0. \end{align*} Hence $g \in Z^3(\mathfrak{g},K)$. Recall that by definition we have \begin{align*} 0 & = (d_3g)(x,x_1,x_2,x_3) \\ & = -g([x,x_1],x_2,x_3)+g([x,x_2],x_1,x_3) -g([x,x_3],x_1,x_2)\\ & -g([x_1,x_2],x,x_3)+g([x_1,x_3],x,x_2)-g([x_2,x_3],x,x_1) \end{align*} By the invariance the first three terms sum up to zero. Hence also the last three terms sum up to zero, i.e., we obtain \begin{align} 0 & = -B([[x_1,x_2],x],x_3)-B([[x_1,x_3],x],x_2)+B([[x_2,x_3],x],x_1). \end{align}

We claim that $g$ is not a $3$-coboundary. Assume that $g=d_2f$, and recall that \begin{align*} (d_2f)(x,y,z) & = -f([x,y],z)+f([x,z],y)-f([y,z],x). \end{align*} Now we use the fact, that there is a linear map $T\colon \mathfrak{g}\rightarrow \mathfrak{g}$ such that $$ f(x,y)=B(T(x),y)=-B(x,T(y)). $$ Then the identity $g(x,y,z)=(d_2f)(x,y,z)$ gives \begin{align*} B(x,[y,z]) & = -B(T([x,y]),z)+B(T([x,z]),y)-B(T([y,z]),x) \\ & = B([x,y],T(z))-B([x,z],T(y))-B(x,T([y,z])) \\ & = B(x,[y,T(z)])-B(x,[z,T(y)])-B(x,T([y,z])). \end{align*} Since $B$ is non-degenerate, this implies that $$ [y,z]=[y,T(z)]-[z,T(y)]-T([y,z]) $$ for all $y,z\in \mathfrak{g}$. This is equivalent to $$ [ad(z),T]=ad(z)-ad (T(z)) \in ad(\mathfrak{g}) $$ for all $z\in \mathfrak{g}$. Hence there exists a $t\in \mathfrak{g}$ such that $$ [ad(z),T]=[ad(z),ad(t)]=ad([z,t]) $$ for all $z\in \mathfrak{g}$. Since $ad$ is faithful it follows that $$ T(z)=[z,t]-z $$ for all $z\in \mathfrak{g}$. Hence we have $$ f(x,y)=B(T(x),y)=B([x,t],y])-B(x,y). $$ Evaluating again $g=d_2f$ and using $(1)$ yields \begin{align*} B(x,[y,z]) & = B([x,y],z)-B([[x,y],t],z)+B([[x,z],t],y) \\ & - B([x,z],y)-B([y,z],t],x)+B([y,z],x) \\ & = B([x,y],z) - B([x,z],y) +B([y,z],x) \\ & = B(x,[y,z])+B(x,[y,z])+B(x,[y,z]). \end{align*} This gives $2B(x,[y,z])=0$ for all $x,y,z\in \mathfrak{g}$. Since $2\neq 0$ and $[\mathfrak{g},\mathfrak{g}]=\mathfrak{g}$ we obtain that $B$ is identically zero. This is a contradiction, since $B$ is non-degenerate and $\mathfrak{g}\neq\{0\}$.

Answer 2

Thanks to the reference in @Dietrich Burde's answer, I can now finish the proof :

Take $\mathfrak g$ a simple Lie algebra so that $[\mathfrak g , \mathfrak g ] = \mathfrak g$.

Take $\omega$ an invariant 3-cocyle, we define a bilinear form on$\mathfrak g$ by $\langle x, y \rangle = \omega(x, a,b)$ when $[a,b] = y$.

Claim :This formula for $\langle x, y \rangle $ does not depend on the pair $(a,b)$ :

Proof : write $x = [c, d ]$ then : \begin{align} \omega([c,d], a, b) =& \omega(c, [d, a ] , b) + \omega([c,a],d, b)\\\\ = & - \omega([c, a], d , b) - \omega(c, d, [b,a]) + \omega([c,a],d, b)\\\\ = & - \omega(c, d, [b,a]) \end{align}

We also see from the formula that the bilinear form is symmetric.

Now, because of the invariance of $\omega$ the bilinear form is invariant.

The form $\omega$ is then a multiple of the Killing form. We are then left to show that this cocycle is not a coboundary. This follows from the theorem of Chevalley Eilenberg, that there is only one invariant cocycle representing a cohomology class.