9
$\begingroup$

Let $G$ be a finite group and $(\chi_i)$ its irreducible characters. Then $\forall i,j,k, \exists!n_{i,j}^k \in \mathbb{N}_{\ge 0}$ such that $$\chi_i \chi_j = \sum_k n_{i,j}^k \chi_k.$$

Let the tensor multiplicity of $G$ be $m(G):= \max_{i,j,k} (n_{i,j}^k)$; for ex. $m(A_5) = 2$ and $m(A_6) = 3$.

Theorem: Let $q > 5$ be a prime-power, then $$m(\mathrm{PSL}(2,q)) = \left\{ \begin{array}{ll} 2 & \text{ if } q \text{ even,} \\ 3 & \text{ if } q \text{ odd.} \end{array} \right.$$ proof: see (for example) the generic computation of $(n_{i,j}^k)$ for $\mathrm{PSL}(2,q)$ in my talk at 30:00. $\square$

This post is about the converse of above theorem.

Let $G$ be a non-abelian finite simple group with $m(G) \le 3$.
Question: Is it true that $G \simeq \mathrm{PSL}(2,q)$ for some prime-power $q$?

It was checked by GAP for $|G|<10^7$ (see Appendix).

Remark: The classification below suggests that the family $(\mathrm{PSL}(2,q))$ is the only one, among the usual infinite families of (non-abelian) finite simple groups, where the tensor multiplicity is bounded above. This would provide an other interesting characterization of this family.

Appendix

We improved our way to check as suggested by Mikko Korhonen in comment. Let $G$ be a finite group, $d_+(G)$ be the maximum among the degrees of complex irreducible characters of $G$, and $\Sigma(G)$ their sum. They are exactly $27$ non-abelian finite simple groups $|G|<10^7$ not isomorphic to some $\mathrm{PSL(2,q)}$. The classification below lists them ordered according to $d_+^2(G)/\Sigma(G)$, which is less than or equal to $m(G)$.

Computation

Each element of the list reads as $d_+^2(G)/\Sigma(G)$, its numerical approximation, $G$ and $|G|$.

gap> classification(10000000);
[ [ 256/63, 4.06349, PSU(3,3), 6048 ], 
[ 1125/208, 5.40865, PSU(3,4), 62400 ], 
[ 13/2, 6.5, PSL(3,3), 5616 ], 
[ 18432/2107, 8.74798, PSU(3,7), 5663616 ], 
[ 175/18, 9.72222, A7, 2520 ],
[ 2048/203, 10.0887, PSL(3,4), 20160 ], 
[ 175/16, 10.9375, A8, 20160 ], 
[ 279/25, 11.16, PSL(3,5), 372000 ], 
[ 729/64, 11.3906, PSp(4,3), 25920 ], 
[ 3025/218, 13.8761, M11, 7920 ],
[ 8281/484, 17.1095, Sz(8), 29120 ], 
[ 10368/553, 18.7486, PSU(3,5), 126000 ], 
[ 7225/271, 26.6605, PSp(4,4), 979200 ], 
[ 321489/11210, 28.6788, PSU(3,8), 5515776 ],
[ 43681/1464, 29.8367, J_1, 175560 ], 
[ 704/21, 33.5238, M12, 95040 ], 
[ 23328/683, 34.1552, A9, 181440 ], 
[ 5472/155, 35.3032, PSL(3,7), 1876896 ], 
[ 28224/709, 39.8082, J_2, 604800 ],
[ 16384/319, 51.3605, PSp(6,2), 1451520 ], 
[ 321489/5356, 60.0241, A10, 1814400 ], 
[ 152100/2519, 60.3811, PSp(4,5), 4680000 ], 
[ 21175/258, 82.0736, M22, 443520 ],
[ 173056/1875, 92.2965, G(2, 3), 4245696 ], 
[ 20800/207, 100.483, PSL(4,3), 6065280 ], 
[ 775/7, 110.714, PSL(5,2), 9999360 ], 
[ 28672/227, 126.308, PSU(4,3), 3265920 ] ]

Below is the alternative classification suggested by Goeff Robinson in comment, considering $d_+(G)/c(G)$, with $c(G)$ the class number of $G$:

gap> classification2(10000000);
[ [ 16/7, 2.28571, PSU(3,3), 6048 ], 
[ 13/4, 3.25, PSL(3,3), 5616 ], 
[ 75/22, 3.40909, PSU(3,4), 62400 ], 
[ 35/9, 3.88889, A7, 2520 ], 
[ 81/20, 4.05, PSp(4,3), 25920 ],
[ 5, 5., A8, 20160 ], 
[ 11/2, 5.5, M11, 7920 ], 
[ 31/5, 6.2, PSL(3,5), 372000 ], 
[ 32/5, 6.4, PSL(3,4), 20160 ], 
[ 192/29, 6.62069, PSU(3,7), 5663616 ], 
[ 91/11, 8.27273, Sz(8), 29120 ],
[ 72/7, 10.2857, PSU(3,5), 126000 ], 
[ 176/15, 11.7333, M12, 95040 ], 
[ 12, 12., A9, 181440 ], 
[ 340/27, 12.5926, PSp(4,4), 979200 ], 
[ 209/15, 13.9333, J_1, 175560 ],
[ 16, 16., J_2, 604800 ], 
[ 256/15, 17.0667, PSp(6,2), 1451520 ], 
[ 81/4, 20.25, PSU(3,8), 5515776 ], 
[ 228/11, 20.7273, PSL(3,7), 1876896 ], 
[ 390/17, 22.9412, PSp(4,5), 4680000 ],
[ 189/8, 23.625, A10, 1814400 ], 
[ 385/12, 32.0833, M22, 443520 ], 
[ 1040/29, 35.8621, PSL(4,3), 6065280 ], 
[ 832/23, 36.1739, G(2, 3), 4245696 ], 
[ 224/5, 44.8, PSU(4,3), 3265920 ],
[ 1240/27, 45.9259, PSL(5,2), 9999360 ] ]

Code

classification:=function(n)  #for Geoff way: classification2:=function(n)
    local it,LL,g,A,L,l,dmax,S,c,cc;
    it:=SimpleGroupsIterator(2520,n);; #2520=|A_7|
    LL:=[];;
    for g in it do
        A:=StructureDescription(g);;
        if Length(A)<6 or List([1..6],i->A[i])<>"PSL(2," then #PSL(2,q) excluded
            L:=CharacterDegrees(g);;
            l:=Length(L);;
            dmax:=L[l][1];;
            S:=Sum(L,i->i[1]*i[2]); #Goeff way: Sum(L,i->i[2]); 
                    cc:=dmax^2/S;;
            Add(LL,[cc,Float(cc),g,Order(g)]);;
        fi; 
    od;
    Sort(LL);
    return LL;
end;;

Recall that $A_7$ is the smallest non-abelian finite simple group not isomorphic to some $\mathrm{PSL}(2,q)$.

Improve this question
$\endgroup$
5
  • 4
    $\begingroup$ Let $\Sigma(G)$ be the sum of the degrees of complex irreducible characters of $G$. Here is another question. For $G$ finite simple and not isomorphic to $\operatorname{PSL}_2(q)$, does there exist an irreducible character $\chi$ such that $3 \Sigma(G) < \chi(1)^2$? Then in the tensor product $\chi \otimes \chi$ some irreducible occurs with multiplicity $> 3$. There are upper bounds for $\Sigma(G)$ in the literature, maybe for $G$ of Lie type taking $\chi$ to be the Steinberg character would suffice. $\endgroup$
    – Mikko Korhonen
    Commented Jan 28, 2021 at 6:23
  • 1
    $\begingroup$ @MikkoKorhonen: The answer to your question is yes for $|G| \le 10^6$, where the smallest possible $\chi(1)^2/\Sigma(G)$ is $256/63 \simeq 4.06$ (given by $G = \mathrm{PSU}(3,3)$ only), so that $4 \Sigma(G) < \chi(1)^2$ is also true. $\endgroup$
    – Sebastien Palcoux
    Commented Jan 29, 2021 at 7:53
  • 2
    $\begingroup$ As an even cruder bound, it is (more than) enough to find an irreducible character with $\chi(1) > 3k$, where the simple group $G$ has $k$ conjugacy classes. It is a conjecture (originally formulated in a more general form by D. Gluck, mainly motivated by solvable groups) that a simple group $G$ should have an irreducible character $\chi$ with $\chi(1) > |G|^{\frac{1}{3}}.$ There are probably only finitely many families of simple groups of Lie type with $k = k(G) > \frac{|G|^{1/3}}{3}$ ( which include ${\rm PSL}(2,q)$), and finitely many alternating groups satisfy that inequality. $\endgroup$
    – Geoff Robinson
    Commented Jan 29, 2021 at 22:45
  • 1
    $\begingroup$ @GeoffRobinson: yes, in fact, let $G$ be a non-abelian finite simple group with $|G| \le 10^6$, let $\chi$ be an irreducible character of maximal degree and let $k$ be the class number. Then $\chi(1) \le 3k$ if and only if $G$ is isomorphic to $\mathrm{PSL(2,q)}$ for some $q$ or $\mathrm{PSU(3,3)}$, see the added list in the post. I guess it is also true without retriction on $|G|$. $\endgroup$
    – Sebastien Palcoux
    Commented Jan 30, 2021 at 11:24
  • $\begingroup$ @GeoffRobinson: the point is that PSL2 is the only (usual) infinite family of non-abelian finite simple groups on which $m(G)$ is bounded above. And it turns out that this above bound is $3$. $\endgroup$
    – Sebastien Palcoux
    Commented Feb 1, 2021 at 15:34

2 Answers 2

Reset to default
5
$\begingroup$

The check of $\textrm{max}\{\textrm{deg}(\chi)\mid \chi \in \textrm{Irr}(G)\} > 3 k$ could be extended to more finite simple groups (including all sporadics) by using the character tables which are available in GAP:

simpnames:= AllCharacterTableNames( IsSimple, true, IsAbelian, false);
LL := [];;                      
for nam in simpnames do  
  Print(nam," ");        
  t := CharacterTable(nam);   
  d := Maximum(List(Irr(t),Degree));  
  k := NrConjugacyClasses(t);    
  Print([d,k,Float(d/k)],"\n");  
  Add(LL, [d/k,Float(d/k),nam, Size(t),d,k]);
od;                         
Sort(LL, {a,b} -> a[4] <= b[4]);   
Filtered(LL, a-> a[1] <= 3);

From the result I would guess the following statement:

G is non-abelian simple with max. character degree $\leq 3k$ iff G = PSU(3,3) or PSL(2,q).

Sketch of proof:

  • G sporadic: checked above (GAP, resp. ATLAS)

  • G alternating: (note that $A_5\cong PSL(2,4)$ and $A_6 \cong PSL(2,9)$ are among the exceptions) Symmetric group $S_n$ has representation for partition (2,2,2, ...., 2,2 or 1) and the hook formula and Stirling approximation of $n!$ yield that its degree grows with $n$ like $e^n$, while the number of conjugacy classes of $S_n$ (= number of partitions of $n$) grows like $e^{const \sqrt{n}}$.

    (The numbers for the alternating group differ at most by a factor $2$).

  • G of Lie type of rank l:

    • they always have the Steinberg representation of degree $q^{N}$ where $N$ is the number of postive roots of $G$, which is roughly $N \sim l^2$
    • their number of conjugacy classes is bounded by $< c q^l$ for some constant c

    (so only rank $l=1$, $PSL_2(q)$, or very small $l$ and $q$ will give exceptions)

So, asymptotically the statement looks ok, one needs to be a bit more precise to get the list of small exceptions.

Improve this answer
$\endgroup$
0
4
$\begingroup$

Liebeck and Pyber [Journal of Algebra 198, 538-562, Th. 1] showed that a finite simple group of Lie type of (untwisted) rank $r$ over $\mathbf{F}_q$ has at most $(6q)^r$ conjugacy classes. On the other hand, the Steinberg character has degree $q^{(d-r)/2}$ (where $d$ is the dimension of the associated algebraic group), so as soon as $q>6$ and $d>3r$ (which holds except for $PSL_2$), we have a character with $\chi(1)>3k$, and Geoff Robinson's argument applies.

Improve this answer
$\endgroup$
4
  • 1
    $\begingroup$ Indeed, I've added it (it's the dimension of the algebraic group). $\endgroup$
    – Denis Chaperon de Lauzières
    Commented Jan 30, 2021 at 13:57
  • $\begingroup$ Very good! So it remains to consider the twisted case, the alternating groups and the sporadic groups. $\endgroup$
    – Sebastien Palcoux
    Commented Jan 30, 2021 at 17:23
  • 1
    $\begingroup$ The result of Liebeck and Pyber does apply to the twisted groups of Lie type, and I think that the Steinberg character has the same degree also for twisted groups; in that case, the argument also applies to twisted groups. $\endgroup$
    – Denis Chaperon de Lauzières
    Commented Jan 31, 2021 at 7:44
  • 1
    $\begingroup$ Small detail: for $q < 6$, the same argument works if $d > 7r$. Then check separately the finite number of groups that occur for $q < 6$ and $d \leq 7r$. $\endgroup$
    – spin
    Commented Feb 2, 2021 at 4:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .