# Tensor Product of Quaternion Algebras is a Real Matrix Algebra

I found following isomorphisms in Cor. 13.24 in the book "Topological Geometry" (by Porteous): $$H \otimes H(2) \cong R(8)$$ $$H(2) \otimes H(2) \cong R(16)$$

($$K(n)$$ is the algebra of all real n x n matrices with entries from the algebra $$K$$. By 'algebra' I mean a unital associative ring.)

I cant find proofs for them. I am guessing that in general, $$H(k) \otimes H(l) \cong R(2^{k+l})$$ and that this can be proved inductively by first proving that $$H \otimes H = R(4)$$. Are my claims true? Can you provide some proofs?

I am aware of the representaion of $$H$$ as a subalgebra of $$R(4)$$ (https://en.wikipedia.org/wiki/Quaternion#Matrix_representations). But I am not sure if that can help here.

Thanks to @runway44 for their answer. I found a different way to set up the isomorphism $$H\otimes H \cong R(4)$$, while working with Clifford algebras. Thought I should share it as well.

(In the following, $$R_{p,q}$$ denotes the (Universal) Clifford Algebra of $$R^{p,q}$$)

Following a procudure similar to Prop. 13.24 of the book ("Topological Geometry" by Porteous), we can show recursive relation $$R_{p,q}\otimes R_{0,2}\cong R_{q,p+2}$$. Specializing this, we have $$R_{0,2}\otimes R_{0,2}\cong R_{2,2}$$. Then we recognize $$R_{0,2}$$ as $$H$$ and $$R_{2,2}$$ as $$R(4)$$ (the first isomorphism is easy to see, and the second one is due to the general rule $$R_{n,n} \cong R(2^n)$$, proved earier in the chapter).

There are a number of rules to consider here. (All $$\otimes$$s will be over $$\mathbb{R}$$, of course.)

• $$\mathbb{K}(n) \cong \mathbb{K}\otimes\mathbb{R}(n)$$. Both sides can be seen to be comprised of $$\mathbb{K}$$-linear combinations of elementary matrices. I'll let you turn that into an explicit isomorphism if you want.

• $$\mathbb{R}(m)\otimes\mathbb{R}(n) \cong \mathbb{R}(mn)$$. The isomorphism is given by taking Kronecker products of matrices. To see how this is an algebra homomorphism, note the Kronecker product $$A\otimes B$$ produces the matrix representation of the linear transformation (also denoted) $$A\otimes B$$ acting on $$\mathbb{R}^m\otimes\mathbb{R}^n$$ with respect to the obvious choice of basis $$\{e_i\otimes e_j\mid \substack{1\le i\le m \\ 1\le j\le n}\}$$, and $$\mathbb{R}^m\otimes\mathbb{R}^n\cong\mathbb{R}^{mn}$$.

• The tensor products $$\mathbb{K}_1\otimes\mathbb{K}_2$$ for normed division algebras $$\mathbb{K}$$ are given by

$$\begin{array}{c|ccc} \otimes & \mathbb{R} & \mathbb{C} & \mathbb{H} \\ \hline \mathbb{R} & \mathbb{R} & \mathbb{C} & \mathbb{H} \\ \mathbb{C} & \mathbb{C} & \mathbb{C}^2 & \mathbb{C}(2) \\ \mathbb{H} & \mathbb{H} & \mathbb{C}(2) & \mathbb{R}(4) \end{array}$$

To determine the above table, we can reason as follows:

• $$\mathbb{R}\otimes\mathbb{K}\cong\mathbb{K}$$ for any $$\mathbb{K}$$, which leaves the lower $$2\times2$$ corner above
• By the Chinese Remainder Theorem, we can determine $$\mathbb{C}\otimes\mathbb{C}$$ to be $$\begin{array}{cl} \mathbb{C}\otimes\mathbb{C} & \cong \mathbb{C}\otimes\mathbb{R}[T]/(T^2+1) \\ & \cong \mathbb{C}[T]/(T^2+1) \\ & \cong \mathbb{C}[T]/(T+i)\oplus\mathbb{C}[T]/(T-i) \\ & \cong \mathbb{C}\oplus\mathbb{C} \end{array}$$
• We may view $$\mathbb{H}$$ as a module over $$\mathbb{C}\otimes\mathbb{H}$$ by having $$\mathbb{C}$$ act from the left and $$\mathbb{H}$$ from the right, i.e. defining $$(a\otimes b)x:=ax\overline{b}$$. Every such transformation we get is $$\mathbb{C}$$-linear, so the module structure yields an algebra homomorphism $$\mathbb{C}\otimes\mathbb{H}\to\mathbb{C}(2)$$, which we can say is onto by checking it turns a basis into a basis, which means it is an isomorphism because of dimensions.
• Similarly, $$\mathbb{H}\otimes\mathbb{H}$$ acts on $$\mathbb{H}$$ in an $$\mathbb{R}$$-linear way.

I'll let you work out the explicit details of these points as exercises if you want (or I'll elaborate upon request I guess). In particular, the last two isomorphisms restrict to so-called isomorphisms

$$S^3\to\mathrm{SU}(2), \qquad S^3\times_{S^0}S^3\to \mathrm{SO}(4).$$

These rules are sufficient for what you're asking. In particular,

$$\begin{array}{cl} \mathbb{H}(m)\otimes\mathbb{H}(n) & \cong (\mathbb{H}\otimes\mathbb{H})\otimes\big(\mathbb{R}(m)\otimes\mathbb{R}(n)\big) \\ & \cong \mathbb{R}(4)\otimes\mathbb{R}(mn) \\ & \cong \mathbb{R}(4mn). \end{array}$$

You may be exploring these things to classify Clifford algebras. To finish having a full set of rules for doing that, you'll want the recursive rules for how adding $$(2,0)$$ or $$(1,1)$$ or $$(0,2)$$ to the signature of $$C\ell(p,q)$$ affects its isomorphism type. Also see the "Clifford clock" described by John Baez, IIRC.

• Thanks for the answer! Another nice way to see $C \otimes H \cong C(2)$ would be to see $H$ as Pauli Matrices, which are complex 2 x 2 matrices. Commented Jul 4, 2021 at 2:26
• If you pick bases and stuff right, that's actually the same answer. :-)
– anon
Commented Jul 4, 2021 at 3:13

$$\mathbb{H}$$ is a finite central simple $$\mathbb{R}$$-algebra. The tensor product of two finite central simple algebras is still simple.

Now consider the homomorphism

\begin{align*} \phi: \mathbb{H}\otimes\mathbb{H}&\rightarrow \text{End}_{\mathbb{R}}(\mathbb{H})\cong \mathbb{R}(4)\\ a\otimes a'&\mapsto (x\mapsto ax(a')^{-1}) \end{align*}

Since $$\mathbb{H}\otimes\mathbb{H}$$ is simple, the kernel must be $$0$$. By comparing the dimensions, we see it is also surjective.