QUESTIONS
- does this arithmetic check out?
- 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 $(P \otimes Q)(R \otimes S) = PR \otimes QS$ $$ i^2 = j^2 = k^2 = -\mathbf{1} \\ ij = k = -ji, \\ jk = i = -kj,\\ ki = j = - ik, \\ $$