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.

  • 1
    $\begingroup$ $\rho$ 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. $\endgroup$ – Gerry Myerson Feb 28 '13 at 0:41
  • $\begingroup$ 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. $\endgroup$ – user42212 Feb 28 '13 at 0:43
  • $\begingroup$ Why? ${}{}{}{}$ $\endgroup$ – 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.

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

I just wanted to write out a bit of what @jef808 has already said in a different way. $\newcommand{\tr}{\text{tr}} \renewcommand{\xi}{\chi}$So you are saying that you have two irreducible representations $\chi: G \to \mathbb{C}^\times$ and $\rho: G\to GL(V)$ of a finite group $G$. You are saying that $\dim\xi = 1$ and that $\dim\rho > 1$. You are also saying that $\chi\cdot\rho$ is irreducible.

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 $\nu(g) \neq 0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.