I'm confused by the relationship between the cocycle condition in Lie algebras vs Lie groups.
For Lie groups, a 2-cocycle is defined (e.g. here) as a map $\Phi : G \times G \rightarrow \mathbb{F}$ such that
$$ \Phi(x,y) + \Phi(xy,z) = \Phi(y,z) + \Phi(x,yz) $$
It seems that it must also be symmetric.
On a Lie algebra a 2-cocycle $\phi$ (e.g here) is instead defined as an antisymmetric bilinear map satisfying
$$ \phi(x,[y,z]) + \text{cyclic permutations} = 0 $$
Could someone point me in the direction of details of the relationship between these two maps? Is there one or are they just two totally different types of cohomology? I've tried playing with them (e.g. making an antisymmetric form from $\Phi$), and seeking references, but no luck.
I'm reading about this in the context of projective representations (group 2-cocycle) and central extensions (algebra 2-cocyle). Weinberg's (II.2.7) notation for the group 2-cocyle in the context of a projective rep $R$ is
$$ R(x) R(y) = e^{i \Phi(x,y)} R(xy) $$ which makes $\Phi$ kind of look like an algebra-y thing.
It seems you can have a nontrivial projective representation in the presence of a nontrivial group 2-cocyle. This article by van Est states that the cohomologies are isomorphic if the group is simply connected. Weinberg shows you can still have a projective representation when the group 2-cocycle is trivial, if $G$ is not simply connected. How does that relate to the formation of central extensions using the algebra 2-cocycle? Will either of the cohomologies tell me absolutely whether I can have a projective rep? I'm a bit at sea.
Thanks a lot!
