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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1How 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 08 '13 at 17:13
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Do you mean $U\otimes_{\Bbb C}V$? – Berci Apr 08 '13 at 17:13
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@Berci I'm not familiar with that notation – user68654 Apr 08 '13 at 17:19
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@TobiasKildetoft I can see $U \otimes V$ is a $G$-module.. but I can't seen to find a way to make it irreducible – user68654 Apr 08 '13 at 17:20
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3Take 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 08 '13 at 17:21
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@TobiasKildetoft, I may have said too much... – Sammy Black Apr 08 '13 at 17:23
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@SammyBlack In my opinion not by a great deal at least. You have left the most important part to the OP, which seems like a fine way to do it. – Tobias Kildetoft Apr 08 '13 at 17:24
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If the tensor is over $\Bbb C$ (what else could it be?), then isn't it simply the case that $U\cong U\otimes V$ for $\dim V=1$, and the $G$-action is obtained by this isomorphism? – Berci Apr 08 '13 at 17:25
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1@Berci only as vector spaces, not as $G$-modules. – Tobias Kildetoft Apr 08 '13 at 17:25
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@user68654 : related : http://math.stackexchange.com/questions/745262 – Watson Jun 08 '16 at 13:31
3 Answers
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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 $\mathbb{C}$.
Alex Youcis
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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.
Sammy Black
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@SimonSMN $W \leq U$ denotes $W$ being a submodule of $U$. This notation is often used with subgroups, vector subspaces, etc., basically, anywhere where we want to indicate that it's not only subset containment $W \subseteq U$. – Sammy Black Oct 03 '23 at 21:41
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All right. I'm studying representations and subrepresentations. I did not manage to solve not the question in the way you indicated. I instead followed Tobias Tildetofts comment. In general, is there any relation between $U\bigotimes V$ and $W\bigotimes V$ for $U$ a subrepresentation of $W$? Because I don't see it. The latter is a quotient of the free module of functions $W\times V\to R$ whereas the former is a quotient of the free module of functions $U\times V\to R$. – Simon SMN Oct 04 '23 at 06:19
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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.
Noah Snyder
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