Let $(\pi, V)$ be a representation of $G$ with character $X$. Prove that if $\langle X, X\rangle=2$ then $V$ is the sum of two irreducible representations

I was under the impression that the inner product of a character by itself is $1$.

Could you give me some help on this one please? Thanks

  • 2
    $\begingroup$ The inner product of an irreducible character with itself is $1$. Also, the characters of non-isomorphic irreducible characters are orthogonal. The claim follows from those bits and the (hopefully known to you) fact that any representation is a direct sum of irreducible ones (assuming we are talking about complex characters of finite groups). $\endgroup$ Mar 25 '16 at 13:41
  • $\begingroup$ Note that if the base field is not algebraically closed, it can happen that $V$ is irreducible with $\langle X, X\rangle = 2$. In this case, when passing to the algebraic closure, $V$ will decompose as a sum of two one-dimensional representations. Most people do character theory over an algebraically closed field, though, as in the accepted answer . $\endgroup$ Aug 30 '19 at 4:03

I'm not sure whether this holds for infinite $G$ or infinite-dimensional $V$ but I'll answer the question for when both are finite.

The inner product of a character by itself is 1 if and only if the representation is irreducible.

First decompose $V$ as a sum of distinct irreducible representations $V_i$: $V=a_1V_1 \bigoplus a_2V_2 \bigoplus ... \bigoplus a_nV_n $ where $a_iV_i$ means the direct sum of $a_i$ (integer) copies of V_i. Then we have that

$$\langle X, X\rangle= \langle a_1X_1 + ... + a_nX_n, a_1X_1 + ... + a_nX_n\rangle = a_1^2\langle X_1, X_1\rangle + ... + a_n^2\langle X_n, X_n\rangle = a_1^2 + ... + a_n^2 = 2 $$

since irreducible characters are orthonormal, i.e. $\langle X_i, X_j\rangle = \delta_{i\,j}$. The only solution to this is $a_1 = 1$, $a_2 = 2$ and $a_i = 0$ for other values of $i$. Hence $V=V_1 \bigoplus V_2$ and so V is the sum of two irreducible representations.


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