Projective representation: Difference between revisions

Latest revision as of 01:59, 16 February 2013

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 $α : G \to G L (V)$ (viz,to the general linear group) where the images of elements of $g$ are ambiguous upto scalar multiples, and such that $α (g h) = α (g) α (h)$ upto a scalar multiple.

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

$α (g h) = f (g, h) α (g) α (h)$

then we say that $α$ is a $f$ -representation.

Two projective representations $α_{1} : G \to G L (V_{1})$ and $α_{2} : G \to G L (V_{2})$ 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) $θ : G \to k^{\times}$ such that for every $g \in G$ and $v \in V$ :

$F (α_{1} (g) \cdot v) = θ (g) (α_{2} (g) \cdot F (v))$

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 $G \to G L (V)$ gives rise to a projective representation, $G \to P G L (V)$ , simply by composing the given representation with the quotient map $G L (V) \to P G L (V)$ (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 $α$ be a projective representation. Then we can associate to it a 2-cocycle such that:

$α (g h) = f (g, h) α (g) α (h)$

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 $H^{2} (G, k^{*})$ .

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 $k = C$ , 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.

@@ Line 5: / Line 5: @@
 * It is a homomorphism from <math>G</math> to the [[projective general linear group]] for a [[vector space]] over <math>k</math>
-* It is a map <math>\alpha:G \to GL(V)</math> (viz,to the [[general linear group]]) where the images of elements of <math>g</math> are ambiguous upto scalar multiples, and such that <math>\alpha(gh) = \alpha(g)\alpha(h)</math> upto a scalar multiple.
+* It is (up to projective equivalence) a map <math>\alpha:G \to GL(V)</math> (viz,to the [[general linear group]]) where the images of elements of <math>g</math> are ambiguous upto scalar multiples, and such that <math>\alpha(gh) = \alpha(g)\alpha(h)</math> upto a scalar multiple.
 if we let <math>f: G \times G \to k^*</math> be the function such that:
@@ Line 13: / Line 13: @@
 then we say that <math>\alpha</math> is a <math>f</math>-representation.
-Two projective representations are termed ''projectively equivalent'' if at any <math>g</math>, they differ multiplicatively by a scalar matrix.
+Two projective representations <math>\alpha_1: G \to GL(V_1)</math> and <math>\alpha_2:G \to GL(V_2)</math> over a field <math>k</math> are termed '''projectively equivalent''' if there exists a vector space isomorphism <math>F:V_1 \to V_2</math> and a function (not necessarily a homomorphism) <math>\theta:G \to k^\times</math> such that for every <math>g \in G</math> and <math>v\in V</math>:
+<math>F(\alpha_1(g) \cdot v) = \theta(g)(\alpha_2(g) \cdot F(v))</math>
+In other words, they differ by a scalar multiplication combined with a change-of-basis isomorphism.
 ==Facts==
-===Linear representations are projective representations===
+===Linear representations give projective representations===
 Every [[linear representation]] <math>G \to GL(V)</math> gives rise to a projective representation, <math>G \to PGL(V)</math>, simply by composing the given representation with the quotient map <math>GL(V) \to PGL(V)</math> (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===
@@ Line 29: / Line 39: @@
 By the assumptions for a projective representation, this turns out to be a 2-cocycle from <math>G</math> to <math>k^*</math>.
-It turns out that ''projectively equivalent'' porjective representations give 2-cocycles that differ by a 2-coboundary.
+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]] <math>H^2(G,k^\ast)</math>.
 ===When is a projective representation equivalent to a linear representation?===
@@ Line 37: / Line 47: @@
 When <math>k = \mathbb{C}</math>, this is the same as the assertion that the group has trivial [[Schur multiplier]] (or is [[Schur-trivial group|Schur-trivial]]).
-In general, any projective representation of the group gives rise to a linear representation of its [[universal covering group]], if such a thing does exist.
+In general, any projective representation of the group gives rise to a linear representation of its [[Schur covering group]].