# Group representations $\pi$ and $\rho$ are irreducible, but their outer tensor product $\pi\times\rho$ is not

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?

-
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). – Marc van Leeuwen Oct 6 '12 at 10:41

The kernel of the multiplication map $\mathbb{H}\otimes_\mathbb{R}\mathbb{C}\to\mathbb{H}$ is an invariant subspace.
 I am totally confused... Do you mean the multiplication of quaternions, thinking of complex numbers as a subset of quaternions? Isn't the kernel of that multiplication {0} (quaternions are an integral domain)? I think I got it all wrong :-( – sonjcy Oct 6 '12 at 8:58 Well, calculating dimensions, the kernel must have dimension $8-4$. Let $i$ live in $\mathbb C$ and $j,k,l$ in $\mathbb H$ (imaginary units), and assume $i==j$ is the correspondence. Then $j\otimes 1 - 1\otimes i$ is in the kernel. – Berci Oct 6 '12 at 9:09 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! :-) – sonjcy Oct 6 '12 at 9:28