Both (Lie) group and Lie algebra cohomology are essentially part of a more general procedure. Namely, we take an abelian category $C$ with enough projectives or enough injectives, take a (say, left) exact functor $F$ from $C$ to abelian groups (or modules over a commutative ring) and compute the (right) derived functors of $F$ using projective or injective resolutions.

For example, if $\mathfrak{g}$ is a Lie algerba over a field $k$, we can take the category of $U(\mathfrak{g})$-modules as $C$ and $$M\mapsto\mathrm{Hom}_{\mathfrak{g}}(k,M)$$

as $F$ (here $k$ is a trivial $\mathfrak{g}$-module). Notice that this takes $M$ to the set of all elements annihilated by any element of $\mathfrak{g}$; this is not a $\mathfrak{g}$-module, only a $k$-module, so the target category is the category of $k$-vector spaces.

In the category of $U(\mathfrak{g})$-modules there are enough projectives and enough injectives, so in principle to compute the Exts from $k$ to $M$ we can use either an injective resolution of $M$ or a projective resolution of $k$ as $U(\mathfrak{g})$-modules. I've never seen anyone considering injective $U(\mathfrak{g})$-modules, probably because they are quite messy. So most of the time people go for the second option and construct a projective resolution of $k$. One of the ways to choose such a resolution is the "standard" resolution with

$$C_q=U(\mathfrak{g}\otimes\Lambda^q(\mathfrak{g})$$ but any other resolution would do, e.g. the bar-resolution (which is "larger" then Chevalley-Eilenberg, but more general, it exists for arbitrary augmented algebras). Applying $\mathrm{Hom}(\bullet,M)$ to the standard resolution we get the Chevalley-Eilenberg complex.

All the above holds for group cohomology as well. We have to replace $U(\mathfrak{g})$ by the group ring (or algebra) of a group $G$. The only difference is that there is no analogue of the Chevalley-Eilenberg complex, so one has to use the bar resolution. Probably, Van Est cohomology of a topological group can also be described in this way.

Both (Lie) group and Lie algebra cohomology are essentially part of a more general procedure. Namely, we take an abelian category $C$ with enough projectives or enough injectives, take a (say, left) exact functor $F$ from $C$ to abelian groups (or modules over a commutative ring) and compute the (right) derived functors of $F$ using projective or injective resolutions.

For example, if $\mathfrak{g}$ is a Lie algerba over a field $k$, we can take the category of $U(\mathfrak{g})$-modules as $C$ and $$M\mapsto\mathrm{Hom}_{\mathfrak{g}}(k,M)$$

as $F$ (here $k$ is a trivial $\mathfrak{g}$-module). Notice that this takes $M$ to the set of all elements annihilated by any element of $\mathfrak{g}$; this is not a $\mathfrak{g}$-module, only a $k$-module, so the target category is the category of $k$-vector spaces.

In the category of $U(\mathfrak{g})$-modules there are enough projectives and enough injectives, so in principle to compute the Exts from $k$ to $M$ we can use either an injective resolution of $M$ or a projective resolution of $k$ as $U(\mathfrak{g})$-modules. I've never seen anyone considering injective $U(\mathfrak{g})$-modules, probably because they are quite messy. So most of the time people go for the second option and construct a projective resolution of $k$. One of the ways to choose such a resolution is the "standard" resolution with

$$C_q=U(\mathfrak{g}\otimes\Lambda^q(\mathfrak{g})$$ but any other resolution would do, e.g. the bar-resolution (which is "larger" then Chevalley-Eilenberg, but more general, it exists for arbitrary augmented algebras). Applying $\mathrm{Hom}_{\mathfrak{g}}(\bullet,M)$ to the standard resolution we get the Chevalley-Eilenberg complex.

All the above holds for group cohomology as well. We have to replace $U(\mathfrak{g})$ by the group ring (or algebra) of a group $G$. The only difference is that there is no analogue of the Chevalley-Eilenberg complex, so one has to use the bar resolution. Probably, Van Est cohomology of a topological group can also be described in this way.

Both (Lie) group and Lie algebra cohomology are essentially part of a more general procedure. Namely, we take an abelian category $C$ with enough projectives or enough injectives, take a (say, left) exact functor $F$ from $C$ to abelian groups (or modules over a commutative ring) and compute the (right) derived functors of $F$ using projective or injective resolutions.

For example, if $\mathfrak{g}$ is a Lie algerba over a field $k$, we can take the category of $U(\mathfrak{g})$-modules as $C$ and $$M\mapsto\mathrm{Hom}_{\mathfrak{g}}(k,M)$$

as $F$ (here $k$ is a trivial $\mathfrak{g}$-module). Notice that this takes $M$ to the set of all elements annihilated by any element of $\mathfrak{g}$; this is not a $\mathfrak{g}$-module, only a $k$-module, so the target category is the category of $k$-vector spaces.

In the category of $U(\mathfrak{g})$-modules there are enough projectives and enough injectives, so in principle to compute the Exts from $k$ to $M$ we can use either an injective resolution of $M$ or a projective resolution of $k$ as $U(\mathfrak{g})$-modules. I've never seen anyone considering injective $U(\mathfrak{g})$-modules, probably because they are quite messy. So most of the time people go for the second option and construct a projective resolution of $k$. One of the ways to choose such a resolution is the Chevalley-Eilenberg complex, but any other resolution would do, e.g. the bar-resolution (which is "larger" then Chevalley-Eilenberg, but more general, it exists for arbitrary augmented algebras).

All the above holds for group cohomology as well. We have to replace $U(\mathfrak{g})$ by the group ring (or algebra) of a group $G$. The only difference is that there is no analogue of the Chevalley-Eilenberg complex, so one has to use the bar resolution, which is exactly the same as in the $U(\mathfrak{g})$ case. Probably, Van Est cohomology of a topological group can also be described in this way.

Both (Lie) group and Lie algebra cohomology are essentially part of a more general procedure. Namely, we take an abelian category $C$ with enough projectives or enough injectives, take a (say, left) exact functor $F$ from $C$ to abelian groups (or modules over a commutative ring) and compute the (right) derived functors of $F$ using projective or injective resolutions.

For example, if $\mathfrak{g}$ is a Lie algerba over a field $k$, we can take the category of $U(\mathfrak{g})$-modules as $C$ and $$M\mapsto\mathrm{Hom}_{\mathfrak{g}}(k,M)$$

as $F$ (here $k$ is a trivial $\mathfrak{g}$-module). Notice that this takes $M$ to the set of all elements annihilated by any element of $\mathfrak{g}$; this is not a $\mathfrak{g}$-module, only a $k$-module, so the target category is the category of $k$-vector spaces.

In the category of $U(\mathfrak{g})$-modules there are enough projectives and enough injectives, so in principle to compute the Exts from $k$ to $M$ we can use either an injective resolution of $M$ or a projective resolution of $k$ as $U(\mathfrak{g})$-modules. I've never seen anyone considering injective $U(\mathfrak{g})$-modules, probably because they are quite messy. So most of the time people go for the second option and construct a projective resolution of $k$. One of the ways to choose such a resolution is the "standard" resolution with

$$C_q=U(\mathfrak{g}\otimes\Lambda^q(\mathfrak{g})$$ but any other resolution would do, e.g. the bar-resolution (which is "larger" then Chevalley-Eilenberg, but more general, it exists for arbitrary augmented algebras). Applying $\mathrm{Hom}(\bullet,M)$ to the standard resolution we get the Chevalley-Eilenberg complex.

All the above holds for group cohomology as well. We have to replace $U(\mathfrak{g})$ by the group ring (or algebra) of a group $G$. The only difference is that there is no analogue of the Chevalley-Eilenberg complex, so one has to use the bar resolution. Probably, Van Est cohomology of a topological group can also be described in this way.