4
$\begingroup$

Let $\mathbb{H}$ be the four-dimensional real vector space of quaternions, G multiplicative group $\mathbb{H}\backslash\{0\}$ and H multiplicative group $\mathbb{C}\backslash \{0\}$. Let $\pi$ be a representation of group G on the 4-dimensional real vector space $\mathbb{H}$ defined by $\pi(\alpha)\beta=\alpha\beta,\quad\alpha\in G,\,\beta\in\mathbb{H}$ and $\rho$ analogously defined representation of group H on the 2-dimensional real vector space $\mathbb{C}$.

This should be an example of a case when representations $\pi$ and $\rho$ are irreducible, but their outer tensor product $\pi\times\rho$ is not. I am trying to find a $(\pi\times\rho)$-invariant subspace of $\mathbb{H}\otimes \mathbb{C}$ that shows that $\pi\times\rho$ is reducible, but... I can't. Any hints?

  • 2
    It may be useful to note that failure for the outer tensor product of irreducibles to be irreducible is possible because the ground field $\mathbb R$ is not algebraically closed. If you try to do the example with $\mathbb C$ instead of $\mathbb R$ (you can equip both representations with a complex-linear structure) then the outer tensor product will be irreducible. But that is because the outer tensor product now has to be taken _over $\mathbb C$_, making it complex-$2$-dimensional, (and real-$4$-dimensional).2012-10-06

1 Answers 1

3

The kernel of the multiplication map $\mathbb{H}\otimes_\mathbb{R}\mathbb{C}\to\mathbb{H}$ is an invariant subspace.

  • 0
    I get it! I met the tensor product very recently, so I'm still doing some beginner's mistakes thinking about it. Thank you both veeeeeeeeery much! I never would have thought of this by myself! :-)2012-10-06