Projective representation

Definition
Let $$G$$ be a group. A projective representation of $$G$$ over a field $$k$$ is defined in the following equivalent ways:


 * It is a homomorphism from $$G$$ to the projective general linear group for a vector space over $$k$$
 * It is (up to projective equivalence) a map $$\alpha:G \to GL(V)$$ (viz,to the general linear group) where the images of elements of $$g$$ are ambiguous upto scalar multiples, and such that $$\alpha(gh) = \alpha(g)\alpha(h)$$ upto a scalar multiple.

if we let $$f: G \times G \to k^*$$ be the function such that:

$$\alpha(gh) = f(g,h)\alpha(g)\alpha(h)$$

then we say that $$\alpha$$ is a $$f$$-representation.

Two projective representations $$\alpha_1: G \to GL(V_1)$$ and $$\alpha_2:G \to GL(V_2)$$ over a field $$k$$ are termed projectively equivalent if there exists a vector space isomorphism $$F:V_1 \to V_2$$ and a function (not necessarily a homomorphism) $$\theta:G \to k^\times$$ such that for every $$g \in G$$ and $$v\in V$$:

$$F(\alpha_1(g) \cdot v) = \theta(g)(\alpha_2(g) \cdot F(v))$$

In other words, they differ by a scalar multiplication combined with a change-of-basis isomorphism.

Linear representations give projective representations
Every linear representation $$G \to GL(V)$$ gives rise to a projective representation, $$G \to PGL(V)$$, simply by composing the given representation with the quotient map $$GL(V) \to PGL(V)$$ (which involves quotienting out by the center). However, not every projective representation arises from a linear representation.

However, it is very much possible that different linear representations descend to the same projective representation. The following is in fact true:

Two linear representations are projectively equivalent if and only if one of them can be obtained from the other via multiplication by a one-dimensional representation.

In particular, all the one-dimensional representations are projectively equivalent to each other.

Projective representation gives a 2-cocycle
Let $$\alpha$$ be a projective representation. Then we can associate to it a 2-cocycle such that:

$$\alpha(gh) = f(g,h)\alpha(g)\alpha(h)$$

By the assumptions for a projective representation, this turns out to be a 2-cocycle from $$G$$ to $$k^*$$.

It turns out that projectively equivalent projective representations give 2-cocycles that differ multiplicatively by a 2-coboundary. Thus, any projective representation up to projective equivalence defines an element of the second cohomology group for trivial group action $$H^2(G,k^\ast)$$.

When is a projective representation equivalent to a linear representation?
A projective representation is projectively equivalent to a linear representation iff the 2-cocycle associated to it is a 2-coboundary. In particular, this means that if $$H^2(G,k^*)$$ (the second cohomology group) is trivial, any projective representation is projectively equivalent to a linear representation.

When $$k = \mathbb{C}$$, this is the same as the assertion that the group has trivial Schur multiplier (or is Schur-trivial).

In general, any projective representation of the group gives rise to a linear representation of its Schur covering group.