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If, with a bit abuse of notation, $U\notin R\bigotimes{R}\bigotimes{R}\bigotimes{...}$ (i.e., regardless how long you clebsch up $R$, $U$ won't appear in the expansion), I call $U$ unsociable with respect to $R$ (in the group $G$). Clearly everything else is unsociable with the $1$ irrep, since (duh) $1\bigotimes{1}\bigotimes...=1$. For finite groups there can be nontrivial (ahem) cases: For $G=C_{3v}$ (the symmetry group), $A_2\bigotimes{A_2}=A_1$ and thus $E$ is unsociable w.r.t. $A_2$.
But what is the deal with Lie groups? E.g. is the defining irrep of $E_7$ unsociable w.r.t. the adjoint one?

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I've heard (but not understand) that cuspidal irreps (of say G(F_q)) are those which you cannot get from tensoring principal series... So might be cuspidal are unsociable with principals... I would be happy if some expert can comment more on this... –  Alexander Chervov Aug 14 '12 at 12:06
    
@Alexander: As Burnside's theorem shows, your recollection is not quite correct. Essentially, the cuspidal characters of a finite group of Lie type are elusive because they don't show up in the ordinary process of induction from easy characters of parabolic subgroups. This is what prompted the more sophsiticated approach of Deligne-Lusztig. –  Jim Humphreys Aug 14 '12 at 18:11
    
@Hauke: Be careful about the language "defining irrep" here, which isn't helpful. Whether the notion of "unsociable" will be helpful I won't try to predict. –  Jim Humphreys Aug 14 '12 at 18:32
    
@Jim Humpreys You are right, thank you ! I misunderstand the claim. It is claimed that cuspidal does enter decomposition of induced reps coming from tensor product of subgroups... imsc.res.in/~amri/html_notes/notesch4.html#x7-260004.1 "4.2. Cuspidal representations. The cuspidal representations of GLn(Fq) are those which are disjoint from all representations of the form π′∘ π′′, where π′ and π′′ are irreducible representations of GLn′(Fq) and GLn′′(Fq), where n = n′ + n′′ and n′ and n′′ are both positive." –  Alexander Chervov Aug 16 '12 at 5:55
    
I mean cuspidal does NOT enter... –  Alexander Chervov Aug 16 '12 at 5:55
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4 Answers 4

up vote 12 down vote accepted

It follows from a theorem of Burnside/Steinberg that if G is a finite group then the tensor powers of a module V contain all irreps if and ony if V is faithful for G.

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Or, to put it another way: an irrep $U$ occurs in some tensor power of $V$ if and only if $\ker(V) \leq \ker(U)$. In Isaacs' character theory book, this theorem is attributed to "Burnside-Brauer", by the way. –  Frieder Ladisch Aug 14 '12 at 12:34
    
Burnside did it over C and R. Steinberg (no relation to me) did it for semigroups over any field. Rieffel extended it to bialgebras. –  Benjamin Steinberg Aug 14 '12 at 13:29
    
THX! So it's even in the Wiki? :-) In any case, since by accident I found LiE (www-math.univ-poitiers.fr/~maavl/LiE/form.html) I never have to annoy MO with nagging questions for practical details of Clebsch-Gordan series again. Or so I hope :-) –  Hauke Reddmann Aug 15 '12 at 9:36
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This is meant as an extended comment (sometimes correction) to things said in various answers and comments. As a public service I'll try to fill in some of the history and references. I recall some of this coming up previously on MO (exercise: find it).

In the framework of representation (or character) theory of a finite group $G$ over $\mathbb{C}$, Burnside led the way in his 1911 treatise (Chap. XV, Thm. IV): given a faithful representation of $G$, all irreducible representations (or characters) occur in some tensor power. A proof of this is given in the 1962 Curtis-Reiner book (32.9); a later addendum to that book refers also to the slightly later work of Brauer and his student Steinberg. Note that faithfulness is an obvious necessary condition here, not emphasized in most sources.

Steinberg, Complete sets of representations of algebras, Proc. Amer. Math. Soc. 13 (1962), 746-747, offered a simpler proof of Burnside's theorem. Then Brauer, A note on theorems of Burnside and Blichfeldt, Proc. Amer. Math. Soc. 15 (1964), 31-34, provided his own version of Burnside's theorem with a number of nice refinements: limiting how many tensor powers are needed, limiting the size of the characteristic 0 field, commenting on what remains true in prime characteristic. Both of these papers are available online at the AMS website.

Meanwhile, in his 1946 book Theory of Lie Groups (Chapter VI), Chevalley treated compact Lie groups and essentially provided a variant of Burnside's theorem. But here it's essential to use both the faithful representation and its contragredient (dual). The proof is in the context of the ring of representative functions, Tannaka duality, Peter-Weyl theorem.

In a later short note, G.I. Lehrer provided a much more algebraic proof of Chevalley's theorem (again for compact Lie groups): Produits tensoriels de grepresentations de groupes de Lie, C.R. Acad. Sci. Paris 275 (1972). He also refers to Steinberg's paper.

From compact Lie groups it's not so far to the finite dimensional representation theory of a semisimple Lie algebra or algebraic group in characteristic 0. But Burnside's theorem and its variants take their sharpest form for groups, whereas a single Lie algebra may belong to simple Lie or algebraic groups or the same Lie type but different isogeny type. This conceals the requirement that one start with a faithful representation.

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@Jim, M. A. Rieffel, Burnside’s theorem for representations of Hopf algebras, J. Algebra 6 (1967), 123–130 proves the analog for finite dimensional bialgebras. Both Steinberg and Rieffel work over any characteristic because they don't rely on characters. Steinberg really proved his result for semigroups (not necessarily finite). He showed if V is a faithful module for the semigroup S, then the tensor algebra of V is faithful for the semigroup algebra of S. If S is finite one immeiately deduces each irreducible is a constituent of a tensor power. –  Benjamin Steinberg Aug 14 '12 at 18:34
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Solution 1 to exercise: mathoverflow.net/questions/10126/… Solution 2: mathoverflow.net/questions/18194/… –  Benjamin Steinberg Aug 14 '12 at 18:58
    
@Benjamin: Excellent solutions (but the question won't be on the exam). –  Jim Humphreys Aug 15 '12 at 20:36
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Let $G$ be the simply connected group of type $E_7$. Then its center is $\mu_2$, the cyclic group of order $2$. Of course, it acts trivially on the adjoint representation, hence the same holds for all components of all tensor powers of the adjoint. But the action of the center on the defining representation is nontrivial, hence it is unsociable with the adjoint.

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The proof of Burnside-Brauer adapts easily to give the following result for Lie groups:

Let $G$ be a compact semi-simple Lie group and let $R$ be a faithful irrep of $G$. Let $U$ be any irrep. Then $U$ occurs in $R^{\otimes M}$ for some $M$.

Some preliminary comments:

Since $G$ is compact, the Lie algebra of $G$ is the direct sum of a semi-simple Lie algebra and an abelian Lie algebra; so the semi-simple hypothesis just says that there is no abelian summand.

Since $G$ is compact, we may assume that $R$ is a unitary representation, by the standard averaging argument. Since the representation is faithful, we may thus view $G$ as a subgroup of a unitary group. Let $n= \dim R$ and consider $G$ as a subgroup of $U(n)$ from now on.

Let $Z$ be the center of $G$. By Schur's lemma, $Z$ acts on $U$ and $R$ by scalars; let $\psi_R$ and $\psi_U$ be the characters by which $Z$ acts on $R$ and $U$. Since $G$ is semi-simple, $Z$ is discrete. Now, $Z$ lies in the center of $U(n)$, which is $S^1$, so $Z$ must be a cyclic group; say $Z \cong \mathbb{Z}/m$. Let $\psi_U = \psi_R^h$. What we will actually be showing is that $\mathrm{Hom}(U, R^{\otimes (h+mN)})$ is nonzero for all sufficiently large $N$.

Proof sketch: Let $\mu_G$ be Haar measure, normalized so that $\int_G \mu_G=1$. Then $$\dim \mathrm{Hom}(U, R^{\otimes (h+mN)}) = \int_G \chi_R^{h+mN} \overline{\chi_U} \mu_G$$ where $\chi_R$ and $\chi_U$ are the characters of $U$ and $R$.

Since $\chi_R$ is the character of a subgroup of $U(n)$, we have $|\chi_R(g)| \leq n$, with equality only when $g$ is of the form $e^{i \theta} \mathrm{Id}$. Such $g$'s are precisely the elements of $Z$. So, for large $N$, the integral is dominated by the contributions from small neighborhoods of the points of $Z$. For $z \in Z$, we have $\chi_R(z)^m = n^m$ and $\chi_r(z)^h \overline{\chi_U}(z) = n^{h+1}$. If you work out the asymptotics of the integrals, using the method of steepest descent, you get the contribution from each $z \in Z$ is of the form $$c n^{mN} N^{-\dim G/2} (1+O(N^{-1/2}))$$ for some $c>0$.

In particular, for $N$ large, all these contributions are positive and $\dim \mathrm{Hom}(U, R^{\otimes(h+mN)})$ is nonzero. $\square$

Commment 1 I'd love to know whom to credit for this trick. I came up with it in this blog thread and used it again here but I'm sure it's not original to me.

Comment 2 If $G$ does have an abelian factor, life is harder. The obvious counterexample is that, if $G$ is $S^1$, the irrep $R$ is $\theta \mapsto e^{i \theta}$ and $U$ is $\theta \mapsto e^{-i\theta}$, then $U$ is not a summand of $R^{\otimes n}$ for any $n \geq 0$. More subtly, let $G = U(3)$, let $V$ be the standard three dimensional irrep and let $R=\mathrm{Sym}^2 V$ and $U = \bigwedge^2 V$. Then $U$ is not a summand of $R$, and the matrix $e^{i \theta} \mathrm{Id}$ acts on $R^{\otimes n}$ by $2n \theta$ and acts on $U$ by $2 \theta$, so $U$ is not a summand of $R^{\otimes n}$ for $n>1$.

Comment 3 If $G$ is a complex semi-simple group, the same result holds for $G$ by the standard nonsense relating complex and compact representations.

Comment 4 If $G$ is a real semi-simple group, life is more confusing. For example, let $G = SL_3(\mathbb{R})$, let $V$ be the standard three dimensional rep of $G$, and let $R = \mathrm{Sym}^3 V$. Then $G \to GL(R)$ is injective, but the corresponding complex representation $SL_3(\mathbb{C}) \to GL(R \otimes \mathbb{C})$ is not. One should be able to formulate a statement here with a bit more care, but I'm not going to do it.

Comment 5 I have no idea what the answer is if $G$ is not semi-simple.

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@David: Note my reference to Chevalley, where the dual representation is also used. Your approach will need this extra ingredient at some point. Concerning your Comments 4 and 5, I'm unaware of further work in those directions (especially the last). –  Jim Humphreys Aug 14 '12 at 18:23
    
Putting in the dual rep will allow you to discard the semi-simple hypothesis (and replace it by either reductiveness or compactness), see my answer to mathoverflow.net/questions/58633 –  David Speyer Aug 14 '12 at 18:44
    
I agree that looking at $R^{otimes m} \otimes (R^{\vee})^{\otimes n}$ is more useful than looking just at $R^{\otimes m}$, but its not what the OP asked for. –  David Speyer Aug 14 '12 at 18:45
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Instead of assuming $G$ semisimple you can assume (after assuming WLOG that $R$ is unitary) that the image of $G$ in $\text{Aut}(R)$ is contained in $\text{SU}(R)$. Then $R^{\ast} \cong \Lambda^{d-1}(R)$ where $d = \dim R$ and your argument from the other thread works. –  Qiaochu Yuan Aug 14 '12 at 19:16
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