# A basic question about group representation

Let $G$ be a finite group. If $\chi : G\to \mathbb{C}$ is a one dimensional representation, and let $\rho: G\to GL_n(\mathbb{C})$ be an irreducible representation of dimensional greater than 1. It's easy to verify $\chi \cdot \rho$ is also an irreducible representation. My question is, are $\rho$ and $\chi\cdot\rho$ the same irreducible representations?

If the answer is YES, then it seems that this contradicts to the fact that every representation is a direct sum of irreducible ones.

If the answer is NO, then it contradicts to the following fact: Define group $G_{2k-1}$ to be generated by $g_1, g_2, \ldots, g_{2k-1}$ satisfying $g_i^2 = -1$ and $g_i g_j = - g_j g_i$. The group has $2^{2k-1}$ one dimensional representations, and has two $2^{k-1}$ dimensional representations.

Added later: In group $G_{2k-1}$, the number of conjugacy class is $2^{2k-1}+2$. By the counting formula, there exists $2^{2k-1}$ one-dimensional representations, and two $2^{k-1}$ dimensional irreducible representations. But it's quite confusing to me, since $\chi\rho \not= \chi'\rho$ as long as $\chi \chi' \not= 1$. There should be much more $2^{k-1}$ dimensional irreducible representations.

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$\rho$ and $\chi\rho$ are not necessarily the same irreducible representation, and I don't see why you think this contradicts every representation being a sum of irreducibles. –  Gerry Myerson Feb 28 '13 at 0:41
Sorry, it's a typo. If $\rho$ and $\chi\rho$ are the same irreducible representations, then it contradicts with the fact that every representation is a sum of irreducible ones. –  user42212 Feb 28 '13 at 0:43
Why? ${}{}{}{}$ –  Qiaochu Yuan Feb 28 '13 at 0:52

The answer to this is no in general but it depends. Take for example the sign and the standard representations of $S_3$. Their characters are respectively $$\chi_{sgn} : e \mapsto 1, (12) \mapsto -1, (123) \mapsto 1$$ while $$\chi_{std} : e \mapsto 2, (12) \mapsto 0, (123) \mapsto -1.$$ So $\chi_{sgn} \cdot \chi_{std} = \chi_{std}$ and this is a case where the answer is yes.

Now for $S_4$ there is again the sign representation $$\chi_{sgn} : e \mapsto 1, (12) \mapsto -1, (12)(34) \mapsto 1, (123) \mapsto 1, (1234) \mapsto -1$$ but there is also another irreducible representation with character $$\rho : e \mapsto 3, (12) \mapsto 1, (12)(34) \mapsto -1, (123) \mapsto 0, (1234) \mapsto -1.$$ So in this case $\chi_{sgn} \cdot \rho$ is a new irreducible representation, so the answer is no.

EDIT: To answer Sean Ballentine's comment below, multiplying and comparing characters here is enough to answer the question. Indeed, if $\rho_1 : G \to GL_n(\mathbb{C})$ is an irreducible representation (with $n > 1$) and $\rho_0 : G \to GL_1(\mathbb{C})$ is a 1-dimensional representation, then $\rho_1 \cdot \rho_0 : G \to GL_n(\mathbb{C})$, defined using ordinary multiplication as $(\rho_1 \cdot \rho_0)(g) = \rho_1(g) \cdot \rho_0(g)$, is an irreducible representation with character $\chi_{\rho_1 \cdot \rho_0} = \chi_{\rho_1} \cdot \chi_{\rho_0}$. Since two irreducible representations are "the same" iff they have the same character, my examples are pertinent. This is a standard trick when you want to reconstruct an incomplete character table for some group : multiply each new character you find by the known 1-dimensional characters and you might find new characters (new irreducible representations), as in the case of my second example.

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In his question he multiplied a character and a representation, not two characters. –  Sean Ballentine Feb 28 '13 at 1:25


So by definition $\xi\cdot\rho: G\to GL(V)$ is $$\xi\cdot\rho(g)(v) = \xi(g)\rho(g)(v).$$ Here you are just multiplying the vector $\rho(g)v$ by the complex number $\xi(g)$. How can we answer the question about whether the two representations $\rho$ and $\chi\cdot\rho$ are the same (isomorhpic)? We can try to calculate the characters of both representations. If the characters are equal, then the representations are isomorphic.

Say that the character of $\rho$ is $\nu$.

Then $$\tr (\xi\cdot\rho(g)) = \tr(\xi(g)\rho(g)) = \xi(g)\tr(\rho(g) = \xi(g)\nu(g).$$ You see that the characters are the same if and only if $\xi(g) = 1$ for all $g\in G$ for which $\nu(g) \neq 0$.

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