Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am asking for clarifications of the basic definitions in the representation theory of a subrepresentation and a character of a subrepresentation.

Given a represenation $\rho:G \to Gl(V)$ and a subrepresentation $W \subset V$, is $\rho_V(g) = \rho_W(g)?$

I think it should be true because a subrepresentation of $(V,\rho)$ is a pair $(W, \rho)$ such that $W \subseteq V \wedge \forall g \in G, \forall x \in W, \rho(g)(x) \in W $. So $G$ is mapped to the same group of matrices. If $W$ a proper subset of $V$, then only the basis changes and $\rho_V = \rho_W$ always but $V \neq W$.

One corollary says: if $V$, $W$ are representations of $G$, then $V \cong W \iff \chi_V = \chi_W$.

By definition the character is $\chi(g) = Tr(\rho(g))$. So the subrepresentations of some representation should have the same characters. But it could be that $V \neq W$ and therefore $V \ncong W$ although $\rho_V = \rho_W$.

So where is the problem in my understanding and the definitions given?

share|improve this question
add comment

1 Answer

up vote 3 down vote accepted

Your problem is that even though the same matrix can represent an element in both $GL(W)$ and $GL(V)$, it's trace depends on which way you want to think of it. For example, consider the representation of $\mathbb R$ on $V=\mathbb R^2$ given by $$ t \mapsto \operatorname{diag}(e^t, e^{2t}). $$ Then a subrepresentation is given by $W = span(1,0)$. Then $\chi_V(t) = e^t + e^{2t}$ but $\chi_W(t) = e^t$.

EDITED for more details


There is a slight abuse of notation in that really you want to think of the subrepresentation as the composition $G \to GL(V) \to GL(W)$. Now write what the matrix representation looks like when $\rho(g)$ is thought of as an elemento f $GL(W)$.

Also, maybe it will help to think more abstractly about the trace. If $V$ is an $n$-dimensional vector space and $T: V \to V$ is a linear map, then you can define it's trace as follows. Let $e_1, \ldots, e_n$ be a basis for $T$. If $T e_j = \sum_k T_{jk} e_k$ then, by definition, $$ \operatorname{tr} T = \sum_j e_{jj}. $$ Now you can think of $T$ as the $n\times n$ matrix whose entries are $T_{ij}$ and then the trace is the sum of the diagonal elements.

In the example I gave, when you consider the representation on $V = \mathbb R^2$ you can take the basis (1,0) and (0,1) and the matrix for $\rho(t)$ is just $\operatorname{diag}(e^t, e^{2t})$. Now consider this matrix as a map from $\operatorname{span}(1,0) = W \to W$. We can choose $(1,0)$ as a basis for $W$. Then $\rho(t) (1,0) = (e^t, 0) = e^t (1,0)$ so the trace is $e^t$. In other words, the $1 \times 1$ matrix that represents $\rho(t) \in GL(W) = GL(1) \simeq \mathbb R$ is just the number $e^t$.

share|improve this answer
    
I know that $W$ is spanned by one vector only. But how did you get $\chi_W(t) = e^t$? You are not changing the matrix $\rho(g)$ and still $\rho_V(g) = \rho_W(g)$ So the trace should be the same. –  David Toth Mar 7 '13 at 14:03
    
Let me know if my edit helps. –  Eric O. Korman Mar 7 '13 at 14:27
    
Your great answer reveals my weakness in ability to compute the trace of the matrices under different bases. How would we compute the trace of a matrix representing $(1 2 3)$ of $S_3$ in a permutation representation in a subrepresentation spanned by a) $W_a=span((1,1,1))$, $W_b=span((1,0,-1),(-1,1,0))$? Thanks a lot. –  David Toth Mar 7 '13 at 16:06
1  
@DavidToth : For $W_a$, $(123)(1,1,1) = (1,1,1)$ the trace is 1 (the coefficient on $(1,1,1)$). For $W_b$ we have $(123)(1,0,-1) = (-1,1,0)$ and $(123)(-1,1,0) = (0,-1,1) = -(1,0,-1) - (-1,1,0)$ so the trace is 0 + -1 = -1, with the 0 coming from the coefficient of (123)(1,0,-1) on (1,0,-1) and the -1 coming from the cofficient of (123)(-1,1,0) on (-1,1,0). –  Eric O. Korman Mar 7 '13 at 16:32
add comment

Your Answer

 
discard

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.