# what is the tensor product $\mathbb{H\otimes_{R}H}$

I'm looking for a simpler way of thinking about the tensor product: $\mathbb{H\otimes_{R}H}$, i.e a more known algbera which is isomorphic to it.

I have built the algebra and played with it for a bit, but still can't seem to see any resemblence to anything i already know.

Thanks!

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Hint :

(1) Show that the map $$H \otimes_R H \rightarrow End_R(H), x \otimes y \mapsto (a \mapsto xay).$$ is an isomorphism of $R$-vector spaces (I don't know the simplest way to do this, but try for example to look at a basis (dimension is 16...)).

(2) Denote by $H^{op}$ the $R$-algebra $H$ where the multiplication is reversed (i.e. $x \times_{H^{op}} y = y \times_{H} x$). Denote by (1,i,j,k) the usual basis. Show that the map $$H \rightarrow H^{op}, 1 \mapsto 1,i \mapsto i, j \mapsto k, k \mapsto j$$ is an isomorphism of $R$-algebras

(3) Show that the map in (1) $$H \otimes_R H^{op} \rightarrow End_R(H), x \otimes y \mapsto (a \mapsto xay).$$ is an isomophism of $R$-algebras

(4) Find an isomorphism $H \times_R H \rightarrow M_4(R)$ of $R$-algebras.

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+1: This is very likely more helpful than an obscure reference to Brauer groups! – Jyrki Lahtonen Oct 30 '11 at 17:17

The order of the class of Hamiltonian quaternions in the Brauer group $Br(\mathbf{R})$ is two -- the class of $\mathbf{H}$ is the only non-trivial element. Therefore it is its own inverse. The product in the Brauer group is (essentially) the tensor product. So this tensor product represents the trivial class in the Brauer group, and thus we end up with a matrix algebra of the correct size: $$\mathbf{H}\otimes_{\mathbf{R}}\mathbf{H}\simeq {\mathcal M}_{4\times4}(\mathbf{R}).$$ Finding an explicit isomorphism may take a while :-)

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This week's finds no 211 may be helpful... – t.b. Oct 30 '11 at 14:16
Dear @t.b. That is a nice, unexpected reference! – Georges Elencwajg Oct 30 '11 at 17:27