Projective representation

This term is related to: linear representation theory
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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.

Facts

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.