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



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


$\left(\text{the number of }i\in\left\lbrace 1,2,...,k\right\rbrace \text{ such that }V\_i\cong P\right) = \left(\chi\_P,\chi\_V\right)$.)


