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.


3 Answers 3


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.

  • $\begingroup$ 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. $\endgroup$ Commented Jul 4, 2021 at 2:26
  • $\begingroup$ If you pick bases and stuff right, that's actually the same answer. :-) $\endgroup$
    – 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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .