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Consider finite group G and its subgroup H, and representation of G in k[G/H] i.e. functions on G/H.

Question: What is known about the question: when k[G/H] is multiplicity free ? (Let us consider k - complex numbers for simplicity).

More generally consider $Ind^G_H V$ some induced representation of G from $H$, same question. (In case V= trivial representation we get previous question).

Are there some general conditions ? What is know about the cases G = S_n, A_n, GL_n(F_q) ?

Example: $S_{n-1} \in S_n$ is multiplicity free since G/H = C^n - standard representation.

Particular question $H=GL_{n}(F_q) \subset G= GL_{n+1}(F_q)$ is C[G/H] multiplicity free ? (This is true from GL(C) and leads to Gelfand-Zeitlin basis theory, well , may be it is not correct). More generally what about induced representations in this case ?

Morally multiplicity free means that $H$ is "big" subgroup of G, however I do not know any precise way to say what "big" means.

PS

I googled "Multiplicity-free permutation representations of the alternating groups", which promises survey on the topic, but but I have no access to this file:(

https://www.ideals.illinois.edu/handle/2142/22828

If someone can help ... would be great...

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6 Answers 6

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Here's a very simple elementary criterion:

A $\mathbb CG$-module is multiplicity-free if and only if its endomorphism ring is commutative

The reason is simple: if $W\oplus W$ (for some irreducible module $W$) is a direct summand of $V$, then the non-commutative matrix ring $\textrm{End}(W\oplus W) \cong M_2(\mathbb C)$ embeds into $\textrm{End(V)}$).

You are dealing with permutation modules, and for those there is a very explicit description of the endomorphism ring in terms of double cosets: $$ \textrm{End}_{\mathbb CG}(\mathbb C [G/H]) \cong \langle \sum_{h\in HgH} h \mid \textrm{ $g$ runs over double coset representatives}\rangle_{\mathbb C} $$

You just need to check whether this ring is commutative. One criterion is that $H\backslash G /H $ has cardinality $\leq 3$ (because any semisimple algebra of dimension $\leq 3$ must be commutative).

A reference is Curtis-Reiner: "Methods of representation theory vol. 1". In this book these endomorphism rings are called "Hecke algebras", defined on page 281 (the endomorphism ring above corresponds to the Hecke algebra $\mathcal H(G,H,1_H)$).

However, computations in GAP show that $\mathbb C [GL_3(3)/GL_2(3)]$ is not multiplicity-free, so I see little hope for you question about permutation modules for $GL_n(q)$.

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  • $\begingroup$ The endomorphism ring also has a simple description in terms of the action of G on G/HxG/H. The characteristic functions of the orbits can be viewed as matrices. They form a basis for the endomorphism algebra. $\endgroup$
    – Benjamin Steinberg
    Commented Aug 18, 2012 at 22:54
  • $\begingroup$ @Benjamin: Yes, indeed. The reason for that is that the orbits of $G$ on $G/H\times G/H$ are (canonically) in bijection with the double cosets $H\backslash G /H$. So the two descriptions will actually even give identical generating sets. $\endgroup$
    – Florian Eisele
    Commented Aug 18, 2012 at 23:16
  • $\begingroup$ Of course but the orbital approach gives an explicit description as a set of matrices. For example, the symmetric Gelfand pair condition says the orbital matrices are symmetric and an algebra if symmetric matrices is clearly commutative. I find this less obvious from the double coset view. $\endgroup$
    – Benjamin Steinberg
    Commented Aug 19, 2012 at 0:52
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    $\begingroup$ @Benjamin: I'm not sure I can convince you that it is "more obvious" in the double coset approach (I'm not even sure it is), but here's the argument in the double coset case: The symmetric condition is equivalent to the condition that $HgH = Hg^{-1} H$ for all $g\in G$ (i.e. double cosets are closed under inversion). It is then clear that the sums over double cosets are fixed under the standard involution on $\mathbb CG$, and therefore commute (this is elementary: elements fixed by an anti-automorphism commute). $\endgroup$
    – Florian Eisele
    Commented Aug 19, 2012 at 1:26
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    $\begingroup$ Ok I am convinced the translation is just as easy. $\endgroup$
    – Benjamin Steinberg
    Commented Aug 19, 2012 at 1:36
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Look up Gelfand pairs. You want (G,H) to be a Gelfand pair. A sufficient condition is that each orbit on G/HxG/H be symmetric (viewed as a relation). This is called a symmetric Gelfand pair. Other sufficient conditions are known.

Additional information. My favorite book on Gelfand pairs is Harmonic Analysis on Finite Groups: Representation Theory, Gelfand Pairs and Markov Chains by Tullio Ceccherini-Silberstein, Fabio Scarabotti, Filippo Tolli.

Here they study Markov chains on $G/H$ using that the endomorphism algebra is commutative. Specific examples coming from symmetric groups and wreath products can be found in the book. Also Weilandt's classical book on permutation groups has a chapter on centralizer algebras and the connection with orbitals is made clear.

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As Benjamin's short answer might suggest, the questions being asked are extremely broad and have already led to a vast amount of organized work on multiplicity-free permutation representations of various classes of finite groups. For instance, there has been substantial progress toward a classification for simple groups and some of their close relatives. In the case of symmetric groups, a fairly comprehensive answer has been reached which is based heavily on earlier work by Jan Saxl and others. For a recent paper (with many references), see:

Chris Godsil, Karen Meagher, Multiplicity-free permutation representations of the symmetric group, Ann. Comb. 13 (2010), no. 4, 463–490.0219-3094.

But for finite general linear groups (and others of Lie type), definitive results seem much harder to achieve though there is some relevant literature. In any case, it's impossible to pursue any of this systematically without good access to standard journals in areas like algebra and combinatorics (and it also helps to consult the MathSciNet database as I've just done). Certainly the problem has been around for a long time and has attracted a lot of attention, though I'm not aware of any single ideal survey.

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The fact (mentioned in Florian Eisele's answer) that the endomorphism algebra must be commutative gives a further necessary condition for $\mathbb{C}_H\uparrow^G$ to be multiplicity free: $N_G(H)/H$ must be abelian. For $\mathbb{C} N_G(H)/H$ appears as a subalgebra of the endomorphism algebra, coming from those double cosets labelled by an element of the normalizer.

In the case when $G$ is the symmetric group $S_n$, the problem of which permutation characters are multiplicity-free is completely solved for $n \geq 66$ by work of Mark Wildon: http://arxiv.org/abs/0903.2864

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  • $\begingroup$ Thank you very much ! Is normalizer somehow distiguished in whole k[H\G/H] ? $\endgroup$
    – Alexander Chervov
    Commented Aug 20, 2012 at 20:24
  • $\begingroup$ in the case when $G$ is a $p$-group in characteristic $p$, the endomorphism ring is a split extension of $kN_G(H)/H$ by a nilpotent ideal: I proved this in a paper called "Endomorphism algebras of transitive permutation modules for $p$-groups" in Archiv der Mathematik. $\endgroup$
    – M T
    Commented Aug 21, 2012 at 13:20
  • $\begingroup$ @FlorianEisele's answer. $\endgroup$
    – LSpice
    Commented Sep 28, 2020 at 21:42
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Just for contrast/complement to the other good (upvoted) answers: in various situations I've found it sufficient, and possible, to just ask about multiplicity-free-ness of an induced repn for a restricted class of representations of either the big group or the small group. In particular, if one restricts one's attention to principal series of the larger (presumably reductive, whether over a finite or infinite field), decomposition over a reasonable (large) subgroup turns into a Mackey (-Bruhat) double-coset (et seq) computation which often has a useful outcome (e.g., finitely-many double cosets, only one candidate supporting an intertwining?)

This certainly does not offer a systematic, general answer, but is meant to suggest that in the face of lack of general (affirmative) answer, things are sometimes sufficient for very specific applications.

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  • $\begingroup$ Thank you very much ! At the moment it is somewhat beyond my understanding but hopefully I will learn more... $\endgroup$
    – Alexander Chervov
    Commented Aug 20, 2012 at 20:30
  • $\begingroup$ By the way is GL(F_q)/Borel is mult.free ? Over p-adics it seems related to "multiplicy one" theorem which is often used... Is it correct ? $\endgroup$
    – Alexander Chervov
    Commented Aug 20, 2012 at 20:33
  • $\begingroup$ One would expect mult-free-ness of the induced repn of trivial from the Borel for $GL_n(\mathbb F_q)$, but one must be a little careful due to triviality of the "modular function(s)" which move things around in the p-adic case, for example. Still, now, unlike the p-adic case, we do have complete reducibility, and the (Gelfand's?) involutive-anti-automorphism condition mentioned in other answers succeeds in showing commutativity of the endomorphism algebra, via Bruhat decomposition, so we apparently have mult-free-ness! :) $\endgroup$
    – paul garrett
    Commented Aug 20, 2012 at 21:22
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Just for interest, I point out that it is well-known that for every finite group $G,$ we can make the group algebra $\mathbb{C}G$ into a $G \times G$-module, by setting $a(g,h) = g^{-1}ah$ for all $a \in \mathbb{C}G,$ for all $(g,h) \in G \times G.$ It is easy to check that this representation is multiplicity-free. In fact, calculating the trace with respect to the natural group basis for $\mathbb{C}G,$ we see that ${\rm trace}(g,h) = |C_{G}(g)|$ if $g$ and $h$ are conjugate in $G,$ and $0$ if they are not conjugate. Hence the character of $G \times G$ afforded by this representation is $\sum_{i=1}^{k} \overline{\chi}_{i} \otimes \chi_{i}$, where $\{ \chi_{i} : 1 \leq i \leq k \}$ are the irreducible charactes of $G.$ It is also easy to see directly that the algebra of $G \times G$ endomorphisms of this module is isomorphic to the commutative ring $Z(\mathbb{C}G).$ This way of looking at the group algebra as a $G \times G$-module ( or as a bimodule for $G$) was profitably exploited in modular representation theory by J.A. Green. But in the context of this question, it illustrates that the existence of a faithful multiplicity free representation puts very little restriction on the structure of the group. Notice that if we take the group basis for $G,$ the module may be viewed as a transitive permutation module, and that the stabilizer of the identity element $1_{G}$ is the "diagonal" subgroup $\Delta(G) = \{ (g,g) : g \in G \}.$ It is easy to check that the module is faithful if and only if $G$ has trivial center. Another theme related to the question is that of "models". For example, R. Richardson constructed monomial representations of the symmetric group which contain every irreducible complex representation just once. These are not quite permutation representations: one takes a certain sign character for the centralizer of each involution (up to conjugacy) of the symmetric group $S_{n}$, and induces that up to $S_{n}$. This can be done in such a way that the sum of these is multiplicity free, and contains every irreducible character. For the general linear group, there are also Gelfand-Graev representations.

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  • $\begingroup$ @Geoff Robinson. Thank you very much for yours answer. May ask. You write: "it illustrates that the existence of a faithful multiplicity free representation puts very little restriction on the structure of the group." I do not quite agree/understand and it suggests the following question: Does every G has H, such that G/H is mult. free ? $\endgroup$
    – Alexander Chervov
    Commented Aug 19, 2012 at 15:51
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    $\begingroup$ I think it would be better to ask for "multiplicity free and faithful". Then every Abelian group does, for example. What the example shows is that for ever finite group $G$ with $Z(G) = 1,$ the group $G \times G$ has a subgroup $H$ such that $(G \times G)/H$ is multiplicity free. $\endgroup$
    – Geoff Robinson
    Commented Aug 19, 2012 at 16:31
  • $\begingroup$ Okay, I agree. This make me think why GxG is somewhat special ? Can we modify the construction GxG e.g. twist or extent by some cocycle... whatever, such that that why can preserve mult.free with some modification of H ? Is there natural subgroup, such that GxGxG (or more) is mult.free and faithful ? Well, just curiosity, may be not worth to pay attention... $\endgroup$
    – Alexander Chervov
    Commented Aug 20, 2012 at 20:28

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