# Matrices whose pairwise products form a basis

Over the vector space of 2x2 matrices, the Pauli matrices $$I, X, Y, Z$$ form a complete basis. Each of these matrices square to $$I$$, and with the additional relation that $$Z = iXY$$, we see that every 2x2 matrix can be uniquely written as a quadratic polynomial in $$X$$ and $$Y$$, i.e. $$aI + bX + cY + dXY$$.

Much the same thing can be accomplished on 4x4 matrices, using gamma matrices: $$\gamma^0, \gamma^1 \dots \gamma^3$$ are 4 of them, and $$(\gamma^i)^2 = I$$. The 16 products we can form (including the empty product, $$I$$) form a basis for the vector space of 4x4 matrices. But this involves quartic terms like $$\gamma^0\gamma^1\gamma^2\gamma^3$$. By adding the 5th gamma matrix, $$\gamma^5 = i\gamma^0\gamma^1\gamma^2\gamma^3$$, every 4x4 matrix can be uniquely written as a quadratic polynomial in the $$\gamma^{0\dots 5}$$.

We can ask when this is possible. For $$n\times n$$ matrices, the space has $$n^2$$ dimensions; with $$k$$ 'generators', we have $$1 + k + k(k-1)/2$$ at-most-quadratic products. Thus this can only work for integer solutions of $$n^2 = 1 + k(k+1)/2$$. The above cases are solutions with $$(n,k)=(2,2)$$ and $$(4,5)$$.

The Diophantine equation can be solved by standard methods and has solutions at $$n = 2, 4, 11, 23, 64, 134, 373, 781, 2174, \dots$$, a sequence which follows the recurrence $$n_i = 6n_{i-2} - n_{i-4}$$. So, we might hope to find a nice set of 15 matrices that each square to $$I$$, and whose pairwise products form a complete basis of 11x11 matrices.

But, by the Hurwitz "1 2 4 8" theorem, we cannot have all these matrices anticommute, as occurred in the above two cases. This means such a set of matrices is probably of significantly less interest to physicists, and generally harder to work with.

Question: Is there such a set of 15 matrices (or, in general, is there always a solution for the $$n$$ above)? Is there a nice construction, or key property they all share, that is related to (but distinct from) the anticommutation relations we get in 2 and 4 dimensions?

• If we can't make the matrices anticommute anymore, then how do we deal with the fact that the two possible orderings of matrices in the quadratic terms can't be readily identified with each other anymore? Are we to pick one of the two orderings for each quadratic term at will? One could also think of alternative problem formulations where the quadratic terms are taken to be, say, commutators of the pairs of matrices ... Sep 18, 2019 at 0:35
• Did you look at higher dimensional $\gamma$-matrices? There are more than $15$ in the case $n=11$ but maybe some suitable subset works? Sep 18, 2019 at 7:21
• I don't understand the link with Hurwitz 's theorem. Hurwitz says that if you have an real algebra with identity having a quadratic form with real values commuting with the product, then its dimension is $1,2,4,8$. I don't see any quadratic form with real values here. Can you elaborate, please ? Sep 18, 2019 at 9:09
• @GreginGre, see page 3 of people.math.osu.edu/shapiro.6/BOOK/book1.pdf , or users.math.cas.cz/~hrubes/PDFs/AntiCom.pdf (where I got the reference). Hurwitz showed that if $k$ matrices of size $n\times n$ all anticommute, then $k \le 2\log_2(n) + 1$. Sep 18, 2019 at 18:38
• @quarague, I don't think there are enough gamma matrices. At least according to Wikipedia's article, they are a set of d matrices of size NxN, where N = 2^floor(d/2). If we wanted 15 gamma matrices, we would need N at least 2^floor(15/2) = 128-dimensional matrices. Gamma matrices also always anticommute, which as I mentioned, can't be possible in general. Sep 18, 2019 at 18:41