Is it possible to find matrix solutions to the following : $$\left(\sum_1^m M_k x_k\right)^n=\left(\sum_1^m x_k^n\right)I_d$$ where $M_k$ are the desired $d \times d$ matrices (no restriction on $d$) and $x_i$ are indeterminate variables; For n=2 the gamma matrices satisfing $M_i M_j + M_j M_i = 2\delta_{ij}$ work; so in a way this is a generalization of these to larger $n$.

$\begingroup$ Any motivation ? $\endgroup$– Denis SerreOct 1, 2015 at 6:38

2$\begingroup$ Look at orthogonal space time codes by Tarokh, Jafarkhani, Calderbank and Seshdri $\endgroup$– user76479Oct 1, 2015 at 7:55

$\begingroup$ @Arul, It looks like they use orthogonal designs to define codes. Their examples would have provided solutions for the $n=2$ case if the orthogonal designs are also symmetric $A^T=A$ in their notation which doesn't look like it's the case. Interesting connection though. $\endgroup$– unknownOct 1, 2015 at 17:39

$\begingroup$ @secretlyfamous Can I ask your application? $\endgroup$– user76479Oct 1, 2015 at 18:23

$\begingroup$ @DenisSerre,Arul; At this point I'm interested in the question for its own sake; it did come up in some calculations I was making (in optics); the closest thing to these is this (en.wikipedia.org/wiki/Hong%E2%80%93Ou%E2%80%93Mandel_effect), but the connection (if any) is highly speculative $\endgroup$– unknownOct 1, 2015 at 19:34
1 Answer
Yes, these are called Generalized Clifford Algebras. The earliest reference I could find was an article by Yamazaki from 1964.
An explicit construction is given by Morris
For example, with $m=2$ and $n=3$, we find as a solution $M_1 = \begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 1 & 0 & 0\end{pmatrix}$ and $M_2 = \begin{pmatrix} 1 & 0 & 0\\ 0 & e^{2 \pi i/3} & 0\\ 0 & 0 & e^{4\pi i /3}\end{pmatrix}$

$\begingroup$ thanks for the references; glad to see that someone has looked at this before. I'll go through the references in more detail, but I think I can see how defining $M_i M_j=\omega M_j M_i$ with $\omega^n=1$ would lead to solutions; although it might be a subset of all possible solutions. Still, there's plenty to work with here... $\endgroup$– unknownOct 1, 2015 at 19:28

$\begingroup$ as a final comment, the paper by Morris is as complete as can be; I'm now able to get explicit matrix solutions for any $n,m$. $\endgroup$– unknownOct 2, 2015 at 0:24