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
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idriskameni

Updated on December 12, 2020

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  • idriskameni
    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
      Derek Holt almost 5 years
      Since it is usually not irreducible, you might perhaps have tried a few examples yourself before asking.
  • idriskameni
    idriskameni almost 5 years
    And do you know any example with $\rho \neq \alpha$?
  • José Carlos Santos
    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
    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?