MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a question about representation theory of finite group $G$ (you may assume that the field has char 0). If you have a morphism of representations $f\colon V\to W$ then you can restrict it on irreducible components of these representations and get (by the Schur's lemma) the set of scalar mappings $f\mid_{V_i}\colon V_i\to W_i$ (scalars determined unique up to the change of bases in $\Bbbk^n\otimes V_i$). At least we can know for each $i$ the sum of these scalars. Is there any way to compute or discover something about these scalars or this sum? I think the answer may be given in terms of some sort of scalar product of characters, but I have no direct link for this (I've read the Serre's book on representation theory and haven't find an answer).

This question comes from the topology. If you want to compute the functorial higher Massey coproducts on simplicial complexes you'll have to compute it first on simplex $\Delta^{m-1}$ (it means the computing the sequence of linear mappings $\delta_k\colon\bigoplus_{i=0}^m \Lambda^i(\Bbbk^m)\to(\bigoplus_{i=0}^m \Lambda^i(\Bbbk^m))^{\otimes k}$ with some relations between them). These Massey coproducts seem to be $S_m$-equivariant morphisms. If you want to get the Massey coproducts invariant under the baricentric subdivision $B\Delta^{m-1}$ then there are endomorphisms of the spaces of $\delta_k\colon\bigoplus_{i=0}^m \Lambda^i(\Bbbk^m)\to(\bigoplus_{i=0}^m \Lambda^i(\Bbbk^m))^{\otimes k}$ and you have to compute the eigenvalues and eigenvectors of such endomorphisms (I'm not giving any detailed motivation - but there is an article about it - I can give the link). So I'm just finding the the way to compute these eigenvalues and eigenvectors.

share|cite|improve this question

This should have been a comment, but got too long.

Even if we assume that everything is semi-simple and that the field is algebraically closed, as you seem to be doing, the scalars you are talking about are completely non-canonical.

Remark 1: On the left hand side, the $V$-side, restricting is the right operation, but on the right hand side it is not. There is no meaning of "restrict to a subspace of the codomain". Instead, you want to project. So write $W=W_1\oplus\ldots\oplus W_n$ as a direct sum of absolutely irreducibles. Then you have canonical isomorphisms $W_i \cong W/\bigoplus_{j\neq i}W_j$ and that's what gives you your functions $f_{i,j}$.

Remark 2: What Schur's lemma tells you is that each $f_{i,j}$ is either an isomorphism or 0. It also tells you that the space of isomorphisms is $GL_1(\mathbb{k})$. But this identification is non-canonical, because it relies on fixing a basis of $V_i$ and of $W_j$. So, you can arrange your scalars to be anything you like by simply rescaling your basis on either side. The sum of the scalars over $i$ is also a very bad "invariant".

Here is an example: let $V$ be one-dimensional and let $W$ be isomorphic to $V\oplus V$. Let me fix such an isomorphism and identify $W$ with $V\oplus V$ (in a non-canonical way!). Let the function $f$ be defined as $$ f:V\rightarrow V\oplus V,\;v\mapsto (v,v), $$ using this identification. If you project onto the first and the second summand, you get 1 for your scalar in both cases. The sum you were talking about is then 2. Now, firstly, I can change my mind about the identification between $W$ and $V\oplus V$ and, changing this identification, arrange $f$ to be $v\mapsto (2v,2v)$. Secondly, I can change basis on $V\oplus V$ by the matrix $\left(\begin{smallmatrix}1&-1\\1&1\end{smallmatrix}\right)$, so that $f$ actually lands in the first component w.r.t the new basis, so the sum of the scalars will be 1.

In your particular application, once you fix a basis on the LHS, there is a natural choice of basis on the RHS, so you can compute things with respect to that. But you should not expect a general formula in terms of characters, since characters are oblivious to change of basis.

share|cite|improve this answer
Ah, I understood. – zroslav Apr 25 '11 at 8:54
However, there are a lot of cases in reality (as far as some part of algebra can be called reality) where a representation is multiplicity-free, i. e. every (absolutely) irreducible representation occurs only once in it, and the scalars therefore become canonical. For example, the restriction of an irreducible representation of $S_n$ to $S_{n-1}$ is multiplicity-free. In this case we should be able to compute the scalars e. g. by pre-composing or post-composing our morphism with the projection on the irreducible component that we want (these usually can be nicely computed ... – darij grinberg Apr 25 '11 at 10:51
... as elements of the group algebra) and take the trace. – darij grinberg Apr 25 '11 at 10:51
@darij Even if the representation is multiplicity-free, the scalars are not canonical, unless there is a canonical way of fixing an isomorphisms between the isomorphic summands. If I tell you that $V$ and $W$ are abstractly isomorphic, then the question "what scalar does the map $f:V\rightarrow W$ correspond to" doesn't make sense until you fix bases on $V$ and $W$ in such a way that one basis is a scalar multiple of f(other basis). Once you have done that, you can talk about the scalar, but the answer will depend on the bases you have chosen. – Alex B. Apr 25 '11 at 11:14
In other words, the identification between the vector space Hom$_G(V,W)$ and the one-dimensional vector space $\mathbb{k}$ is non-canonical and depends on the choice of basis on the former. – Alex B. Apr 25 '11 at 11:17

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.