# Third cohomology group of a simple Lie algebra

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 ?

• This is true for an absolutely simple real Lie algebra, but false for a simple, not absolutely simple one, in which case the third cohomology has dimension 2.
– YCor
Jul 5, 2019 at 5:35
• For a semisimple Lie algebra in char. $0$, let $E$ be the space of invariant sym. bilinear forms, and $\eta$ the given map from $E$ to $H^3(\mathfrak{g})$. The observation that $B$ is well-defined (i.e., that for any $B\in E$, $(x,y,z)\mapsto B(x,[y,z])$ is a 3-cocycle) seems to be due to Chevalley-Eilenberg, 1948, and they also proved the injectivity of $\eta$ (and hence the nonvanishing of $H^3$ if $\mathfrak{g}\neq 0$). Koszul (1950) then proved surjectivity of $\eta$. The space $E$ was known before, e.g. it's 1-dimensional if $\mathfrak{g}$ is absolutely simple (Killing or E. Cartan?).
– YCor
Jul 5, 2019 at 5:41

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 $$\mathfrak{g}\neq\{0\}$$.

• I think you're proving the injectivity of the map into $H^3$. But the question rather addresses surjectivity, which I think is in Koszul's 1950 paper rather than the Chevalley-Eilenberg one.
– YCor
Jul 5, 2019 at 5:50
• Dietrich, surjectivity of the map from invariant symmetric bilinear forms to $H^3$ is true for every semisimple Lie algebra over a field of characteristic zero. For instance, for a simple, not absolutely simple Lie algebra over $\mathbf{R}$, both spaces have dimension 2.
– YCor
Jul 5, 2019 at 16:23

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.

• For the dimension of the cohomology group $H^3(\mathfrak{g},K)$ see the comments at this question. It depends on the field. Jul 2, 2019 at 9:37
• @DietrichBurde "it depends on the field": it depends in which sense. One good thing about taking (co)homology groups is that it commutes with taking field extensions; in particular Betti numbers are absolute invariant. But indeed the statement "a simple Lie algebra has 1-dimensional $H^3$" is not correct, because "simple" is not an absolute invariant.
– YCor
Jul 4, 2019 at 21:27
• @YCor I mean " its dimension is the number of simple factors of the complexification", so for the field of complex numbers, a simple Lie algebra has $1$-dimensional $H^3(L,\Bbb C)$. Jul 5, 2019 at 8:18