Is decomposition of a representation of a group G over a vector space V into irreducible subrepresentations over invariant subspaces of V unique?

I am reading Group theory and Physics by shlomo steinberg. On page no: 59 it says that two representations with same charecter functions are equivalent. I am unable to follow the argument.

I think it follows from the fact that decomposition of a representation of a group G over a vector space V into irreducible subrepresentations over invariant subspaces W of V is unique. Can anyone help me in proving that once we have V = w1+w2+.......+wn, each of wj being an irreducible subrepresentation of G, there are no other invariant subspaces. Here + means direct sum.

Thank you

• It's not unique. Only the so-called isotypic components are unique. Think of representations of the one-element group. These are simply the vector spaces. The decomposition of a vector space into irreducible vector spaces (= 1-dimensional vector spaces) is definitely not unique. – darij grinberg Jul 8 '12 at 11:11
• What you want to say is that the isomorphism classes of the $w_j$ are unique. But it's not necessary to know this in order to follow Sternberg's argument. It is actually a corollary of Sternberg's argument. – darij grinberg Jul 8 '12 at 11:12
• The question is out of focus, since the groups in question have to be specified more narrowly along with the field over which representations (presumably finite dimensional) are being studied. In any case, the question is probably too elementary for this site and shows the need to start with a more basic introduction than Sternberg gives. There are lots of readable beginning texts on representation theory of finite groups, compact Lie groups, etc. Sternberg's book has its merits but is definitely not a basic text. – Jim Humphreys Jul 8 '12 at 13:31

You are misunderstanding Sternberg's argument. The argument goes like this (with my notations):

Consider two representations $V$ and $W$ of $G$ with the same character.

For any representation $S$ of $G$, let $\chi_S$ denote the character of $S$.

Take any decomposition $V=V_1\oplus V_2\oplus ...\oplus V_k$ of $V$ into irreducible subrepresentations.

Take any decomposition $W=W_1\oplus W_2\oplus ...\oplus W_l$ of $W$ into irreducible subrepresentations.

For any irreducible representation $P$ of $G$, we have $\left(\chi_P,\chi_V\right) = \left(\text{the number of }i\in\left\lbrace 1,2,...,k\right\rbrace \text{ such that }V_i\cong P\right)$ and $\left(\chi_P,\chi_W\right) = \left(\text{the number of }j\in\left\lbrace 1,2,...,l\right\rbrace \text{ such that }W_j\cong P\right)$. Since $\chi_V = \chi_W$, we thus have

$\left(\text{the number of }i\in\left\lbrace 1,2,...,k\right\rbrace \text{ such that }V_i\cong P\right)$

$= \left(\chi_P,\chi_V\right) = \left(\chi_P,\chi_W\right)$

$= \left(\text{the number of }j\in\left\lbrace 1,2,...,l\right\rbrace \text{ such that }W_j\cong P\right)$

for every irreducible representation $P$ of $G$. In other words, for every irreducible representation $P$ of $G$, the two lists $\left(V_1,V_2,...,V_k\right)$ and $\left(W_1,W_2,...,W_l\right)$ contain the same amount of entries isomorphic to $P$. In other words, the lists $\left(V_1,V_2,...,V_k\right)$ and $\left(W_1,W_2,...,W_l\right)$ contain the same entries up to isomorphism the same number of times (but of course, not necessarily in the same order). As a consequence of this, $V_1\oplus V_2\oplus ...\oplus V_k \cong W_1\oplus W_2\oplus ...\oplus W_l$. In other words, $V\cong W$, qed.

Nowhere did this use that the $V_i$ are somehow unique. (They are not unique as subspaces of $V$. They are unique up to isomorphism, and that follows either from the Krull-Remak-Schmidt theorem or from the fact that

$\left(\text{the number of }i\in\left\lbrace 1,2,...,k\right\rbrace \text{ such that }V_i\cong P\right) = \left(\chi_P,\chi_V\right)$.)

• Your reply is fairly comprehensive and satisfactory. Thanks – somitra Jul 11 '12 at 4:21

Take the trivial representation of your favourite $G$ on ${\mathbb C}^{2012}$. What kind of uniqueness are you talking about?

• By now it is clear that the decomposition is not unique. What is the relationship between two different decompositions. – somitra Jul 11 '12 at 17:57
• You will have a linear map, commuting with the $G$-action, that maps one decomposition into another one. – Bugs Bunny Jul 12 '12 at 8:45