Related Question: Exotic Chains for Group Cohomology of a Complex Lie Group
Let's take the group homology of a affine algebraic group over $\mathbb C$ (with its discrete topology). The natural free resolution of $\mathbb Z$ over $\mathbb Z[G]$ is given by: $$\cdots\to\mathbb Z[G^3]\to\mathbb Z[G^2]\to\mathbb Z[G]\xrightarrow\epsilon\mathbb Z\to 0$$ If we want to compute $H_\ast(G^\delta,\mathbb C)=\operatorname{Tor}_{\mathbb Z[G]}(\mathbb Z,\mathbb C)$, then we just apply $-\otimes_{\mathbb Z[G]}\mathbb C$ to the above complex, and thus get: $$\cdots\to\mathbb C[G^3]^G\to\mathbb C[G^2]^G\to\mathbb C[G]^G$$ where $G$ acts on all factors by left-multiplication.
Now let's take a leap of faith and ask what happens if we replace $\mathbb C[G^k]$ with a natural counterpart, namely $U(\mathfrak g)^{\otimes k}$. The action of $G$ on $\mathbb C[G^k]$ by left-multiplication corresponds to the action of $U(\mathfrak g)$ on $U(\mathfrak g)^{\otimes k}$ by left-multiplication by $\Delta^{(k-1)}(\cdot)$ (this is the natural $U(\mathfrak g)$-module structure on a $k$-fold tensor product of $U(\mathfrak g)$-modules). Taking $G$-invariants of $\mathbb C[G^k]$ corresponds to tensoring $U(\mathfrak g)^{\otimes k}$ with $\mathbb C$ over $U(\mathfrak g)$ (which is just the same as taking $\mathfrak g$-invariants). Thus we are led to consider the complex: $$\cdots\to(U(\mathfrak g)^{\otimes 3})^{\mathfrak g}\to(U(\mathfrak g)^{\otimes 2})^{\mathfrak g}\to(U(\mathfrak g))^{\mathfrak g}\qquad(**)$$ Perhaps the first question to ask is: if we forget about taking $\mathfrak g$-invariants, is this complex exact? (if yes, then we are computing some derived functor). One should of course note the similarity to the Chevalley--Eilenberg complex which computes the Lie algebra homology $H_\ast(\mathfrak g,\mathbb C)$ as the homology of: $$\cdots\to(U(\mathfrak g)\otimes\mathfrak g^{\wedge 2})^{\mathfrak g}\to(U(\mathfrak g)\otimes\mathfrak g)^{\mathfrak g}\to(U(\mathfrak g))^{\mathfrak g}$$
Does anyone know what the homology groups of $(**)$ are, or if they have been studied before?