Projective representations of loop groups If $G$ is a Lie group and we take its loop group $LG$ why do we deal with projective representations of $LG$ and central extensions thereof? Where does the extra complexity come in to require us to consider this extra, very complicated, step?
 A: Well, we are usually concerned about these exotic objects at the algebraic level. They pop up when we consider spinor representations of the orthogonal group. Lets write $\gamma_{\alpha}$ for the generators of the Clifford algebra which are represented by $\Gamma_{\alpha}$. They satisfy
$$\Gamma_{\alpha}\Gamma_{\beta}+\Gamma_{\beta}\Gamma_{\alpha}=2\eta_{\alpha\beta}$$
where $\eta_{\alpha\beta}$ is nondegenerate. (If it's degenerate, we have a Grassmann algebra and not a Clifford algebra, but it's not terrible.)
So we have commutators of the generators of the Clifford algebra. We will leave it as an exercise for the reader to determine
$$[\Gamma_{\alpha},\Gamma_{\beta}]=\text{something quadratic}$$
and
$$[\Gamma_{\alpha},\Gamma_{\mu}\Gamma_{\nu}]=2\eta_{\alpha\mu}\Gamma_{\nu}-2\eta_{\alpha\nu}\Gamma_{\mu}.$$
Now we expect commutators of quadratic terms to look like
$$[\Gamma_{\alpha}\Gamma_{\beta},\Gamma_{\mu}\Gamma_{\nu}]\sim a\Gamma_{\sigma}\Gamma_{\tau}+b\cdot\textbf{1}$$
and wouldn't it be lovely if we could make this an equality and ditch the constant multiple of the identity operator $b\textbf{1}$? This would make the representation of quadratic guys a closed algebra!
We find if we write an arbitrary element of these generated algebra as $a\cdot\textbf{1}+b^{\mu\nu}\Gamma_{\mu}\Gamma_{\nu}$ (where we use Einstein summation convention) we have an algebra, provided one condition on the matrix $b^{\mu\nu}$ namely we need
$$ b^{\mu\nu}=-b^{\nu\mu}.$$
This says we have a "representation" of the orthogonal Lie algebra. Is it a representation? Yes and no. No, it's technically not a "vanilla representation"; but with all this extra structure, we have a central extension of the orthogonal lie algebra. 
Wouldn't it be lovely, after all this work, if we could do the same thing to other Lie algebras? That's how we get started with losing ourselves in the jungle of loop groups, its central extensions, etc.
A: Projective representations are needed in quantum mechanics, and I think that is the main motivation behind studying projective representations. The point is that in QM, it is rays in a Hilbert space which have physical meaning. So the symmetry group is required to act on and transform these rays rather than vectors.
