Exercise 7.7.3 in Weibel (computation of $H^3(\mathfrak{sl}_2,k)$ via Chevalley-Eilenberg complex)

Question

This exercise can be found in an introduction to homological algebra by C.A. Weibel.

Use the Chevalley-Eilenberg complex to show that $$H_3(\mathfrak{sl}_2,k) \cong H^3(\mathfrak{sl}_2,k) \cong k$$ where for simplicity $k$ is a field of characteristic $0$ and as usual, $\mathfrak{sl}_2$ denotes the space of traceless matrices in $k$.

I am only interested in the cohomological case, i.e. showing that $$H^3(\mathfrak{sl}_2,k) \cong k.$$ However, I am quite lost. I mean, the best way is to use the definitions, so we have the complex $$ \mathrm{Hom}_k(\Lambda^2\mathfrak{sl}_2,k) \overset{d}{\to} \mathrm{Hom}_k(\Lambda^3\mathfrak{sl}_2,k) \overset{d}{\to} \mathrm{Hom}_k(\Lambda^4\mathfrak{sl}_2,k)$$ where $$\begin{align*}df(x_1,\dots,x_{n + 1}) =& \sum_i (-1)^{i +1}x_if(x_1,\dots,\hat{x_i},\dots,x_{n + 1})\\ &+\sum_{i < j}(-1)^{i + j}f([x_i,x_j],x_+,\dots,\hat{x_i},\dots,\hat{x_j},\dots,x_{n + 1})\end{align*}$$

Also, I do know a basis of $\mathfrak{sl}_2$, so it should be enough to calculate the above on basis elements. Am I right? How would one proceed to calculate the cohomology?

Edit. This is for a short presentation of cohomology of Lie algebras at my university and thus I cannot introduce other terminology than the main onces (it was a course about Lie algebras and Lie groups). Since the topic before covers Hochschild-Serre spectral sequences and things like that I am not even sure if it is possible to calculate the homology only using the definition and some linear algebra.

Answer 1

This can be done by hand very quickly using just linear algebra. First note that $\wedge^4 \mathfrak{sl}_2=0$ as $\dim \mathfrak{sl}_2=3$, so $H^3(\mathfrak{sl}_2,k)$ is the cokernel of $$d: \hom(\wedge^2 \mathfrak{sl}_2,k) \to \hom(\wedge^3 \mathfrak{sl}_2,k) $$

Since $\wedge^3 \mathfrak{sl}_2$ is spanned by $f\wedge h \wedge e$ (where $f,h,e$ is the usual basis of $\mathfrak{sl}_2$), to understand $d(\alpha)$ for $\alpha \in \hom(\wedge^2 \mathfrak{sl}_2,k)$ we need only compute $d(\alpha)(f\wedge h \wedge e)= \alpha( d(f\wedge h \wedge e))$. But you can check directly, using the definition of $d$, that $d(f\wedge h \wedge e)=0$. It follows that the map in the displayed equation above is zero and so $H^3(\mathfrak{sl}_2,k)=k$.

Answer 2

For another nice construction see the article On Lie algebra crossed modules by Friedrich Wagemann. For all complex simple Lie algebras $L$ we have $H^3(L,\mathbb{C})\cong \mathbb{C}$, by using the interpretation as crossed modules. Section $4.1$ is for $L=\mathfrak{sl}_2(\mathbb{C})$. For a field $K$ of characteristic zero, and $L$ semisimple, $\dim H^3(L,K)$ is the number of simple factors of the complexification of $L$.