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Alex Meiburg
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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?

Alex Meiburg
  • 1.2k
  • 6
  • 16