# can quaternions be expressed in terms of tensor products?

QUESTIONS

1. does this arithmetic check out?
2. if so, is there a geometric interpretation?

note: my aim was to try to find a very simple but non-trivial example which might help me begin to understand tensor products. i apologize if this little exercise is nonsense - or if the question is a duplicate of one already dealt with. my arithmetic is unfortunately always rather error-prone. at first it seemed to work, then, after trying to tidy up the notation, it didn't and i thought i was barking up a non-existent tree. but finding the scraps of paper a couple of days later, after checking again it looked OK. so rather than flounder around any further i would appreciate input from someone with a more mature understanding.

for $v \in \mathbb{R}^2$ if $v \to vB$ is a reflection in the $x$-axis, and $v \to vC$ is a reflection in the line $x=y$ then the product $v \to vBC = vA$ is a counter-clockwise rotation by $\frac{\pi}2$. these transformations may be represented as matrices:

$$A = \begin{pmatrix} 0 & 1 \\ -1 & 0 \\ \end{pmatrix} \\$$ $$B = \begin{pmatrix} 1 & 0 \\ 0 & -1 \\ \end{pmatrix} \\$$

$$C = \begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix} \\$$

with these definitions, and using $I$ to signify the identity matrix, we have the multiplication table:

$$A^2 = -I, B^2 = I, C^2 = I \\ AB = -C \\ BC = A \\ CA = -B \\$$ using the tensor (kronecker?) product over $\mathbb{R}$ if we now set: $$\mathbf{1}=I \otimes I \\ i = I \otimes A \\ j = A \otimes B \\ k = -A \otimes C \\ \\$$ then assuming that with all matrices $2 \times 2$ we have $(P \otimes Q)(R \otimes S) = PR \otimes QS$ $$i^2 = j^2 = k^2 = -\mathbf{1} \\ ij = k = -ji, \\ jk = i = -kj,\\ ki = j = - ik, \\$$

• Without checking your calculations, I just want to point out that your tensor products are in ${\rm M}_2(\mathbf R) \otimes_{\mathbf R} {\rm M}_2(\mathbf R)$, which is isomorphic to ${\rm M}_4({\mathbf R})$. It is well-known that the quaternions can be realized inside ${\rm M}_4(\mathbf R)$; see the section about matrix representations on the Wikipedia page for quaternions. Therefore it would not be surprising that you can find a realization of the quaternions $i$, $j$, and $k$ inside ${\rm M}_2(\mathbf R) \otimes_{\mathbf R} {\rm M}_2(\mathbf R)$. – KCd Jul 15 '14 at 2:39
• thank you KCd, i will check out the reference – David Holden Jul 15 '14 at 2:40
• Comparing the representation of quaternions as $4 \times 4$ real matrices in the Wikipedia page on quaternions with the formula for Kronecker products (that is the concrete matrix expression of tensor products of matrices -- see the Wikipedia page on Kronecker products), the connection with your notation $A$, $B$, and $C$ is that $1 = I \otimes I$, $i = B \otimes A$, $j = A \otimes I$, and $k = C \otimes A$. – KCd Jul 15 '14 at 2:47
• @KCd this looks like a very useful subject for me to investigate, leading in several directions i have heard of but never really come to grips with. i've just seen, for example (in en.wikipedia.org/wiki/…) that what i have termed I, A. B, C are closely related to what are called the "Pauli spin matrices" - which would be I, A, iB and iC (with i=$\sqrt{-1}$ ) which gives me a handle on the still (to me) rather mysterious notion of Lie Groups, here the Lie group SU(2). it will take some time to absorb this info! – David Holden Jul 15 '14 at 6:01
• Of which quaternions do we speak, of W.R. Hamilton’s (where = = = −1), or ? – Incnis Mrsi Nov 9 '14 at 14:25