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Every group acts on itself by conjugation $h \mapsto g h g^{-1}$. Respectively considering functions on a group we obtain a linear representation.

Question 1: what is known about this representation ? Its decomposition to irreducibles ? Its relations to specific group theoretic properties (e.g. number of invariants = number of conj. classes) ?

The question seems to be natural, so one may hope for some "nice answers". On the other hand, it might seem that everything should be known, however quoting a recent paper:

Surprisingly, almost nothing is known about the multiplicities of other irreducible modules in adjoint representation.

Question 2: Is there some "nice/any" description for specific groups like S_n, A_n, GL(n,F_q), UT(n,F_q) ... ?

The question makes sense for Hopf algebras, as well, so

Question 3: what is known about Hopf algebras ? Especially Drinfeld doubles of finite groups ? It might be the question is more simple (i.e. "nice" answer may exist) for these Hopf algebras (and more generally for those which correspond to "modular categories"), since a kind of "self-duality" ... (cannot be more precise ).


What is known: there are several papers which are concentrated around the so-called Roth's conjecture/question (disproved in general). Every irrep of G/Z(g) can be found in adjoint representation.

See the paper quoted above and references therein: Conjugacy action, induced representations and the Steinberg square for simple groups of Lie type , G. Heide, J. Saxl, P. Tiep, A. Zalesski:

Let G be a finite simple group of Lie type. We prove that every irreducible representation of G is a constituent of πG, unless G=PSUn(q) and n is coprime to 2(q+1), where precisely one irreducible representation fails.

Let me quote another paper: Lie theory of finite simple groups and the Roth property J. Peña, S. Majid, K. Rietsch where similar question treated with the framework of non-commutative geometry.


Informal remarks/thoughts/motivations:

1) It is clear that every conjugacy class in G gives rise to certain subrepresentation in adjoint representation.

2) One can check for S_n, where we have bijection between irreps between conjugacy classes and irreps, that irrep corresponding to given class can be found inside the adjoint subrepresentation mentioned in (1).

3) How far 1,2 are incidental ? Are there other examples ? Can this be related with the orbit method (for finite groups) - since the adjoint orbits are "infinitesimal conjugacy classes", and in some specific cases adjoint orbits = coadjoint orbits (semisimple Lie algebras). So we need certain non-degenerate "Killing form", is it similar to non-degenerate pairing for "modular categories" ? to "Killing form" in Majid's paper cited above ?

4) If we restrict to subrepresentation on a conjugacy class of element "x", then normalizer of "x" acts on this representation. It seems such data - conjugacy class and irrep of normalizer appears in the classification of irreps of Drinfeld double of a finite group - is it incidental ? Or may be Drinfeld double also acts here ?

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4 Answers 4

There is a nice character theoretic answer for finite groups, working over the complex numbers. The character of the "adjoint representation" of $G$ is the permutation character of the action $G$ on itself by conjugation. The multiplicity of an irreducible character $\chi$ in this permutation character is exactly the sum of the entries of the row corresponding to $\chi$ in the character table for $G$.

One further obvious remark is that the only irreducible characters that can occur with nonzero multiplicity are those whose kernels contain the center of the group.

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Thank you very much ! –  Alexander Chervov Jan 8 at 7:33

Let $G$ be a compact (Hausdorff) group. $G$ acts by both left and right translation on $L^2(G)$, making it a representation of $G \times G$, and as a representation of $G \times G$ a mild strengthening of the Peter-Weyl theorem asserts that $L^2(G)$ decomposes as a Hilbert space direct sum of the external tensor products $V \boxtimes V^{\ast}$ as $V$ runs over all irreducible representations of $G$. Restricting to the diagonal, we get the tensor products $V \otimes V^{\ast}$ as $V$ runs over all irreducible representations of $G$. There is an obvious trivial summand corresponding to the identity $\text{id}_V \in \text{End}(V)$, which is necessarily the only trivial summand since the trivial summand is precisely the center of $\text{End}(V)$, and other than that I don't think much can be said.

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Thank you, but I think more can be said. In particular each conjugacy class is stable under conjugation, so functions on each conj. class are subrepresentations. It seems interesting to understand what representation lives in such spaces. –  Alexander Chervov Jan 5 at 9:42
    
@Alexander: in the finite case, each such subrepresentation is of the form $\mathbb{C}[G/H] \cong \text{Ind}_H^G(1)$ for $H$ the centralizer of a conjugacy class. I don't know what can be said about these representations in full generality. –  Qiaochu Yuan Jan 5 at 21:19

Some notes on Qiaochu's answer:

We can sometimes compute the decomposition into irreducibles of $V \otimes V^*$ using character theory. The number of irreducibles in $V \otimes V^*$ (or rather the squared sum over isomorphism classes) is of course the second moment of the absolute value of the character of $V \otimes V^*$, which is the fourth moment of the absolute value of the character of $V$. If this is $<5$, we know exactly how to decompose it into irreducible representations.

The same idea works for semisimple Hopf algebras. By the Artin-Wedderburn theorem, a semisimple Hopf algebra, as a diagonal representation of itself, is equal to a sum of the tensor product (over the automorphism algebra) of each simple representation with its dual.

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Thank you. Can you give a comment/refrence for a "Hopf algebra, as a diagonal representation of itself, is equal to a sum of the tensor product of each simple representation with its dual". –  Alexander Chervov Jan 5 at 9:43
    
Well I meant to assume semi simplicity. So we can view it as a sum of matrix algebras over division algebras. Each $n \times n$ matrix algebra over a division algebra corresponds to a simple representation, which is just an $n$-dimensional vector space over that division algebra. If we tensor this with its dual (I guess we have to tensor it over the division algebra = the automorphism group of the module, not over the base field), then we get the matrix algebra. –  Will Sawin Jan 5 at 15:15
    
It seems "Hopf property" is not mentioned explicitly in your argument ? –  Alexander Chervov Jan 5 at 15:16
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Initially this tensor product is only equal to the matrix algebra as an $A \otimes A$-module, with one $A$ acting on the left and the other $A$ acting on the right. The Hopf algebra structure gives us a map $A \to A \otimes A$, allowing us to turn this into an $A$-module. This is exactly the same structure we use to take the left action and the right action and combine them to a single action, forming the adjoint representation. So the adjoint representation is equivalent to the sum of tensor products. –  Will Sawin Jan 5 at 15:17

A similar situation to the one mentioned by Qiaochu Yuan is when $G$ is a reductive algebraic group over an algebraically closed field $k$.

The action of $G\times G$ on $k[G]$ by $\rho_l\times \rho_r$ (the natural choice of an action that gives conjugation when we embed $G$ into $G\times G$ as the diagonal) makes $k[G]$ a $G$-module, which has a filtration where the factors are $H^0(\lambda)\otimes H^0(-w_0(\lambda))$ for each dominant weight $\lambda$, with multiplicity $1$ (if the characteristic of $k$ is $0$ then each $H^0(\lambda)$ is simple, and everything decomposes).

When we then view $k[G]$ as a $G$-module via the diagonal embedding, then each factor $H^0(\lambda)\otimes H^0(-w_0(\lambda))$ has a good filtration, where the factors can be computed since the characters of the $H^0(\lambda)$ are given by Weyl's character formula.

Going from this to finding the multiplicities of the composition factors of the module is probably not doable with current knowledge though (unless the characteristic is $0$), since we do not even know these multiplicities in general for the $H^0(\lambda)$.

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Thank you ! I am primary interested in char zero –  Alexander Chervov Jan 8 at 10:16

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