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Let $G$ is some group and $U$ is an irriducible $G$-module over the complex numbers. Now if $V$ is a $G$-module of dimension 1, I would like to prove $U \otimes V$ is an irriducible $G$-module.

My knowledge of tensor products is lacking..

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How is the tensor product defined? Specifically, how is the action of the group on the tensor product defined? (note: This is meant to be hints towards a solution, not just idle questions). –  Tobias Kildetoft Apr 8 '13 at 17:13
    
Do you mean $U\otimes_{\Bbb C}V$? –  Berci Apr 8 '13 at 17:13
    
@Berci I'm not familiar with that notation –  user68654 Apr 8 '13 at 17:19
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Take a closer look at how the structure as a $G$-module is given, consider a submodule and see that this gives you a submodule of $U$ in a nice way. –  Tobias Kildetoft Apr 8 '13 at 17:21
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@Berci only as vector spaces, not as $G$-modules. –  Tobias Kildetoft Apr 8 '13 at 17:25

3 Answers 3

up vote 3 down vote accepted

Suppose that $\{u_1, u_2, \dots \}$ is a basis for the irreducible $G$-module $U$. Since $V$ is $1$-dimensional, $V = \mathbb{C}v$ for some nonzero $v \in V$.

Then, a basis for $U \otimes V$ is $\{u_1 \otimes v, u_2 \otimes v, \dots \}$. Any $G$-submodule of $U \otimes V$ will be of the form $W \otimes V$, where $W \le U$. Why? How does $G$ act in the tensor product?

As $U$ is irreducible, either $W = 0$ or $W = U$. As a consequence, either $W \otimes V = 0 \otimes V = 0$ or $W \otimes V = U \otimes V$. Therefore $U \otimes V$ is irreducible.

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If $V$ has dimension $1$ then it is invertible in the sense that there's another module $V'$ such that $V \otimes V'$ is isomorphic to the trivial representation. (A good exercise is to work out what this $V'$ is.) If you have a submodule $W \subset U \otimes V$ then $W \otimes V' \subset U$, so any submodule of $U \otimes V$ gives a submodule of $U$ with the same dimension.

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Another, definitively less elegant, method to solving this is to recall that a representation $\rho:G\to\text{GL}_n(\mathbb{C})$ is irreducible if and only if $\langle\chi_\rho,\chi_\rho\rangle=1$, and that if $\rho,\psi:G\to\text{GL}_n(\mathbb{C})$ are a pair of representations then

$$\chi_{\rho\otimes\psi}=\chi_\rho\chi_\psi$$

With this in mind, let $\rho:G\to\text{GL}_n(\mathbb{C})$ be an irreducible representation and $\psi:G\to S^1$ a character. Then,

$$\begin{aligned}\langle\chi_{\rho\otimes\psi},\chi_{\rho\otimes\psi}\rangle &= \frac{1}{|G|}\sum_{g\in g}|\chi_{\rho\times\psi}(g)|^2\\ &= \frac{1}{|G|}\sum_{g\in G} |\chi_\rho(g)|^2|\chi_\psi(g)|^2\\ &= \frac{1}{|G|}\sum_{g\in G}|\chi_\rho(g)|^2\\&=1\end{aligned}$$

where we used the fact that $\chi_\psi(g)=\psi(g)\in S^1=\{z:|z|=1\}$ for all $g$, and the last step was achieved using the fact that $\rho$ was irreducible.

This is much less elegant, and much more difficult than is needed, but it's always useful to remember this brute-force method of attacking irreducibility over $\mathbb{C}$.

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