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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.

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  • $\begingroup$ Yes, for finite-dimensional Lie algebras the (co)homology can mechanically computed using linear algebra (using the Chevalley-Eilenberg complex). In addition in the presence of a grading in an abelian group it can be computed degree by degree. Actually, for an $n$-dimensional Lie algebra, it immediately follows that the $n$-th Betti number is $\le 1$, and equals 1 iff the Lie algebra is unimodular; this applies in particular to $b_3(\mathfrak{sl}_2)=1$. $\endgroup$ – YCor Apr 15 '18 at 23:48
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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 $H^3(\mathfrak{sl}_2,k)=k$.

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  • $\begingroup$ Thanks. Yes, I overlooked the crucial part that $k$ is seen as a trivial $\mathfrak{sl}_2$-module. $\endgroup$ – TheGeekGreek Apr 16 '18 at 11:23
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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 $L$.

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  • $\begingroup$ Thank you for your answer. However, I "need" this for a short presentation about the cohomology of Lie Algebras, and there is a very limited terminology I can use. So maybe I should edit my question (nonetheless anice answer). $\endgroup$ – TheGeekGreek Apr 15 '18 at 20:13
  • $\begingroup$ Dietrich: $H^3(L,k)\simeq k$ is false if $L$ is simple but not absolutely simple. For $L$ semisimple in general (char 0), its dimension is the number of simple factors of the complexification. $\endgroup$ – YCor Apr 15 '18 at 23:41
  • $\begingroup$ @YCor Yes, you are right, thank you. I was thinking of $K=\mathbb{C}$. $\endgroup$ – Dietrich Burde Apr 16 '18 at 8:08

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