# Tensor product of two irreducible representations is not irreducible

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Take the standard representation $$\rho$$ of $$S_3$$ on $$V=\{(z_1,z_2,z_3)\in\mathbb{C}^3\,|\,z_1+z_2+z_3=0\}$$:$$\rho(\sigma)(z_1,z_2,z_3)=(z_{\sigma^{-1}(1)},z_{\sigma^{-1}(2)},z_{\sigma^{-1}(3)}).$$Then $$\rho$$ is irreducible, but $$\rho\otimes\rho$$ is not ($$S_3$$ has no irreducible representation whose dimension is greater than $$2$$).

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### idriskameni

Updated on December 12, 2020

• idriskameni almost 3 years

I have seen that if $$\rho: G \longrightarrow \text{GL}_{\mathbb{C}}(V)$$ and $$\alpha: G \longrightarrow \text{GL}_{\mathbb{C}}(W)$$ are two irreducible representations of a finite grup $$G$$, then its tensor product representation $$\rho \otimes \alpha: G \longrightarrow \text{GL}_{\mathbb{C}}(V \otimes W)$$ is usually not irreducible.

Can anyone tell me an example?

• Derek Holt almost 5 years
Since it is usually not irreducible, you might perhaps have tried a few examples yourself before asking.
• idriskameni almost 5 years
And do you know any example with $\rho \neq \alpha$?
• José Carlos Santos almost 5 years
Sure. The group $S_4$ has two and (up to isomorphism) only two irreducible representations of dimension $3$, $\rho$ and $\rho^\star$, and no irreducible representation of dimension $9$. Therefore, $\rho\otimes\rho^\star$ is reducible.
• ancient mathematician almost 5 years
Come @idriskameni, if you take an irreducible of highest degree and tensor it with some other irreducible of dimension greater than $1$ that's not going to be irreducible, is it?