# Conjugacy of projective representations

Given characters of the Schur covering group of $G$ of the same degree, how does one tell if the projective representations (as homomorphisms from $G$ into $\operatorname{PGL}$) are conjugate in $\operatorname{PGL}$?

For some reason I had assumed one simply checks for equality of characters, but this does not appear to be the case for $A_4$ into $\operatorname{PGL}(2,7)$. The (Frobenius) characters are not equal. The first rep has trace 6 on an element of order 3, and the second has trace 5 on that element, yet the two projective representations are conjugate in $\operatorname{PGL}(2,7)$.

I suppose that the elements themselves are only defined up to a scalar multiple, and so the trace can be changed by any multiple of 2, but surely not all projective representations of the same dimension are conjugate in general?

Edit: Karpilovsky seems to suggest one just checks equality of characters, as in theorem 3.3 on page 16 of vol 3 of his Group Representations book. Maybe theorem 3.5 is enough?

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math.stackexchange.com/questions/66353/… has the matrix reps –  Jack Schmidt Oct 19 '11 at 1:44

Karpilovsky's result is not useful here. It deals with GL-conjugacy, not PGL-conjugacy.

We setup two projective representations of $G=A_4$ that have the same defining cocycle $\alpha$, whose $\alpha$-characters are different, but such that the representations are conjugate in the projective general linear group.

Notice that the $\alpha$-characters are not class functions, but that one could choose $\alpha$ so that $\chi_1$ becomes a class function.

## The example


Define $\rho_1: A_4 \to \PGL(2,R)$ by $\rho_1((12)(34)) = \m{ 0 & 1 \\ -1 & 0 }$ and $\rho_1((123) ) = \m{ \ze & 0 \\ -1 & \zi }$.

Define $\rho_2: A_4 \to \PGL(2,R)$ by $\rho_2((12)(34)) = \m{ 0 & 1 \\ -1 & 0 }$ and $\rho_2((123) ) = \m{ 0 & -\zi \\ 1 & -\ze }$.

Note that for $x=\left[\begin{smallmatrix}\ze & 1 \\ -1 & \ze \end{smallmatrix}\right]$ we get $$\rho_1((12)(34))^x = \rho_2((12)(34)) \quad\text{and}\quad \rho_1((123))^x = \zi \cdot \rho_2((123))$$ so that the projective representations are conjugate in $\PGL(2,R)$. Note that all matrices mentioned so far are invertible over $R$.

Explicitly we get the following values (the choice of signs defines the cocycle; we've chosen signs for $\rho_1$ and $\rho_2$ to give the same cocycle, $\alpha$):

$$\tiny\begin{array}{r|c|ccc|ccccccccc} & () & (12)(34) & (13)(24) & (14)(23) & (123) & (132) & (124) & (142) & (134) & ( 143) & (234) & (243) \\ \hline \rho_1 & \m{ 1 & 0 \\ 0 & 1} & \m{ 0 & -1 \\ 1 & 0} & \m{ -\ze & -\zi \\ -\zi & \ze} & \m{ -\zi & \ze \\ \ze & \zi} & \m{ -\ze & 0 \\ 1 & -\zi} & \m{ -\zi & 0 \\ -1 & -\ze} & \m{ \ze & -1 \\ 0 & \zi} & \m{ \zi & 1 \\ 0 & \ze} & \m{ -1 & \zi \\ -\ze & 0} & \m{ 0 & -\zi \\ \ze & -1} & \m{ -1 & -\ze \\ \zi & 0} & \m{ 0 & \ze \\ -\zi & -1} \\ \rho_2 & \m{ 1 & 0 \\ 0 & 1} & \m{ 0 & -1 \\ 1 & 0} & \m{ \ze & -\zi \\ -\zi & -\ze} & \m{ -\zi & -\ze \\ -\ze & \zi} & \m{ 0 & \zi \\ -1 & \ze} & \m{ \zi & -1 \\ \ze & 0} & \m{ 0 & \ze \\ -1 & -\zi} & \m{ -\ze & -1 \\ \zi & 0} & \m{ 1 & -\ze \\ 0 & \zi} & \m{ 1 & \zi \\ 0 & \ze} & \m{ \ze & 0 \\ -\zi & 1} & \m{ \zi & 0 \\ \ze & 1} \end{array}$$

$\chi_i(g) := \operatorname{tr}(\rho_i(g))$ has values:

$$\scriptsize\begin{array}{r|c|ccc|ccccccccc} & () & (12)(34) & (13)(24) & (14)(23) & (123) & (132) & (124) & (142) & (134) & ( 143) & (234) & (243) \\ \hline \chi_1 & 2 & 0 & 0 & 0 & 1 & 1 & -1 & -1 & -1 & -1 & -1 & -1 \\ \chi_2 & 2 & 0 & 0 & 0 & \ze & \zi & -\zi & -\ze & -\ze & -\zi & -\zi & -\ze \end{array}$$

Note that these characters are not equal, but the representations $\rho_i$ are conjugate in $\operatorname{PGL}(2,R)$. The lifts of these representations to $\operatorname{GL}(2,R)$ are non-isomorphic representations of $\operatorname{SL}(2,3)$.

For reference, $\alpha$ is:

$$\tiny \begin{array}{r|r|rrr|rrrrrrrr} \alpha & () & (12)(34) & (13)(24) & (14)(23) & (123) & (132) & (124) & (142) & (134) & ( 143) & (234) & (243) \\ \hline () & 1&1&1&1&1&1&1&1&1&1&1&1\\ \hline (12)(34) & 1&-1&-1&1&1&-1&1&-1&-1&-1&1&1\\ (13)(24) & 1&1&-1&-1&1&-1&1&1&-1&1&-1&-1\\ (14)(23) & 1&-1&1&-1&-1&1&1&1&-1&-1&-1&1\\ \hline (123) & 1&1&1&-1&-1&1&1&1&1&-1&-1&1\\ (132) & 1&-1&-1&1&1&-1&1&1&-1&1&1&1\\ (124) & 1&1&-1&-1&-1&1&1&1&-1&1&-1&-1\\ (142) & 1&-1&-1&-1&1&-1&1&1&-1&-1&1&1\\ (134) & 1&1&-1&1&1&1&1&1&1&1&1&-1\\ (143) & 1&1&1&1&1&1&-1&1&1&1&1&-1\\ (234) & 1&-1&1&-1&-1&1&1&1&1&-1&1&1\\ (243) & 1&-1&1&1&1&1&-1&-1&1&-1&1&1 \end{array}$$

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At any rate, $\alpha$-characters do not look terribly useful or well-behaved. They are only defined up to $B^2(G,R^\times)$, but I don't see a way of writing them down succinctly in anything resembling a square character table. Furthermore, they don't even capture the notion of projective equivalence, so I'm not sure what good they are. Any use of them seems simpler to achieve using ordinary characters of the covering group. –  Jack Schmidt Jun 18 '13 at 1:44

Just an update: One needs to choose special cocycles $\alpha$ so that the projective characters are class functions, as in Haggarty–Humphreys (1978). Then Theorem 1.3 gives the desired result: two representations $\rho_i$ of $G$ with the same (special) cocycle are projectively equivalent if and only if the projective characters $\chi_i$ satisfy $\chi_1 = \lambda \chi_2$ for some linear character $\lambda$ of $G$.


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