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Let $q$ be a prime power and $k$ a positive integer. What are the possible degrees of faithful projective representations of the projective special linear group $\mathrm{PSL}(k,q)$ (over the Galois field $\mathrm{GF}(q)$ and of dimension $k>1$) over the complex numbers? How about the symplectic group $\mathrm{Sp}(2k,q)$ (of dimension $2k\geq2$)?

If the above questions are too difficult, what are the lowest degrees of faithful projective representations of $\mathrm{PSL}(k,q)$ and $\mathrm{Sp}(2k,q)$ over complex numbers?

I am particularly interested in the two special cases: $\mathrm{PSL}(4,7)$ and $\mathrm{Sp}(6,2)$.

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  • $\begingroup$ If I remember my theory, the projective reps of $G$ are the linear reps of the covering group (Darstellungsgruppe)? If so, for $Sp_6(2)$ where the Schur multiplier is exceptionally 2, the list for faithful irreps is [ 8, 48, 64, 64, 112, 112, 120, 168, 280, 448, 512, 560, 720 ]. $\endgroup$
    – NAME_IN_CAPS
    Commented Aug 7, 2014 at 22:21
  • $\begingroup$ The minimal degrees of projective representations of finite simple classical groups have been determined by Pham Huu Tiep and Alexander E. Zalesskii (Minimal characters of the finite classical groups, Communications in Algebra 24, 2093-2167, 1996). According to talbe II there, the minmal degrees for $\mathrm{PSL}(4,7)$ and $\mathrm{Sp}(6,2)$ are 399 and 7, respectively. $\endgroup$
    – Huangjun Zhu
    Commented Aug 9, 2014 at 0:22

3 Answers 3

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If you are really just interested in particular examples, it might just be easier to read it off from the character table, which you can compute with $\texttt{magma}$. Since the groups are simple in this case, it is especially easy, since everything is faithful (except for the trivial representation) so you just want the first column. Here's some cheap $\texttt{magma}$ code which computes the generating function of the dimensions of all representations of a finite group $G$ for which $\texttt{magma}$ can compute the character table:

$\texttt{Q := RationalField();}$
$\texttt{P<x> := PolynomialRing(Q);}$
$\texttt{Z := Integers();}$

$\texttt{function dimensionsofrepresentations(G)}$
$\texttt{X:=CharacterTable(G);}$
$\texttt{sum:=0;}$
$\texttt{for i in [1..#X] do}$
$\texttt{sum:=sum + x^(Z!X[i,1]);}$
$\texttt{end for;}$
$\texttt{return sum;}$
$\texttt{end function;}$

$\texttt{> dimensionsofrepresentations(ProjectiveSpecialLinearGroup(4,7));}$

$\texttt{x^182400 + 4*x^159600 + 2*x^139650 + 2*x^137200 + 27*x^136800}$
$\texttt{ + 9*x^119700 + x^117649 + 56*x^115200 + 15*x^102600 + 3*x^100548}$
$\texttt{+ 48*x^98496 + 2*x^91200 + 2*x^69825 + 8*x^51300 + 2*x^50274 + 4*x^22800}$
$\texttt{ + 2*x^22400 + 5*x^19950 + x^19551 + 9*x^17100 + 2*x^2850 + x^2450}$
$\texttt{ + 3*x^2052 + 2*x^1425 + 2*x^1026 + 2*x^400 + x^399 + x}$

$\texttt{> dimensionsofrepresentations(SymplecticGroup(6,2));}$

$\texttt{x^512 + x^420 + x^405 + x^378 + x^336 + x^315 + 2*x^280 + x^216 + 2*x^210}$
$\texttt{+ 3*x^189 + x^168 + x^120 + 3*x^105 + x^84 + x^70 + x^56 + 2*x^35 + x^27}$
$\texttt{ + 2*x^21 + x^15 + x^7 + x}$

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  • $\begingroup$ I did that too, but then realized(?) the OP wanted projective reps, not linear reps... $\endgroup$
    – NAME_IN_CAPS
    Commented Aug 7, 2014 at 22:53
  • $\begingroup$ If so, I guess you want linear reps of $SL_4(7)$, and your polynomial is $q^{182400} + 6q^{159600} + 3q^{137200} + 24q^{136800} + 12q^{119700} + 56q^{115200} + 18q^{102600} + 48q^{98496} + 6q^{68400} + 8q^{49248} + 6q^{22800} + 3q^{22400} + 12q^{17100} + 3q^{400}$ The list (w/o multiplicity) is [ 400, 17100, 22400, 22800, 49248, 68400, 98496, 102600, 115200, 119700, 136800, 137200, 159600, 182400 ]. $\endgroup$
    – NAME_IN_CAPS
    Commented Aug 7, 2014 at 23:21
  • $\begingroup$ Are projective reps of $\mathrm{PSL}(k,q)$ (over the complex numbers) equivalent to the reps of $\mathrm{SL}(k,q)$? $\endgroup$
    – Huangjun Zhu
    Commented Aug 8, 2014 at 0:00
  • $\begingroup$ In the cases that $SL(k,q)$ is the (full) covering group of $PSL(k,q)$, I think so. en.wikipedia.org/wiki/Schur_multiplier $\endgroup$
    – NAME_IN_CAPS
    Commented Aug 8, 2014 at 0:25
  • $\begingroup$ The poster most probably doesn't have access to Magma. (well, the same computations can be done in GAP (gap-system.org), which is free...) $\endgroup$
    – Dima Pasechnik
    Commented Aug 8, 2014 at 8:41
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This paper, and references therein, might be appropriate.

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I suppose the full answer is known, starting from the famous paper by Green. See the book Finite Groups of Lie Type: Conjugacy Classes and Complex Characters by R.Carter.

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