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Given the two matrices: $\sigma_i$ and $\sigma_j$ we can construct a Clifford algebra based on the anti commutator rule: $$\{\sigma_i,\sigma_j\}=\delta_{ij}1$$ where $\delta_{ij}$ is the Kronecker symbol. The question is: if the matrices are $(N\times N)$ and their elements are Natural numbers, how many matrices vs. $N$ can I find satisfying the anti commutator equation? I would appreciate any suggestion.

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I think there something missing, since $\{a,b\}$ might be a matrix and $\delta_{ab}$ is $0$ or $1$. And since there are plenty of natural numbers, I expect plenty of matrices, so might be interested in some basic ones, aren't you? – draks ... Mar 5 '12 at 12:50
@draks: sorry. I forgot to say the elements should be not greater than $M$ with M positive integer – Riccardo.Alestra Mar 5 '12 at 12:53
Do you actually want $\{\sigma_i,\sigma_j\} = 2 \delta_{ij} I$? Else the answer is simply "none". The anticommutator $\{\sigma_i,\sigma_i\} = 2 (\sigma_i)^2 = I$ by assumption. So that $\sigma_i^2 = \frac12 I$. But if $\sigma_i$ has integer entries, so must $\sigma_i^2$. So it cannot have $\frac12$ as entries along the diagonal. – Willie Wong Mar 5 '12 at 13:28
Assuming you mean what Willie said, there will be a lot-- the group $SL(n,\mathbb Z)$ acts on the set of all such via conjugation. – Eric O. Korman Mar 5 '12 at 13:28
I've never heard of natural number matrices being used with Clifford algebras, but then again I know that Clifford algebras and such matrices are important in combinatorics, so maybe there is a bridge somewhere. How did you come across this particular question? – rschwieb Aug 7 '13 at 16:49
up vote 3 down vote accepted

To briefly synthesize the comments by Willie Wong and Eric O. Korman above, there are no (natural number) solutions to the original equation you posted unless you meant $\{\sigma_i,\sigma_j\}=2I\delta_{ij}$. Eric's comment applies if we are talking about integer matrices, and it's true that once you find an integer matrix solution $(a,b)$, then $(xax^{-1},xbx^{-1})$ is a solution for every invertible integer matrix $x$, of which there are many. Some of the $x$'s are even filled with natural numbers, and as I found out at this MO post, they are all permutations.

My contribution is that even if you adopt the change to $2I\delta_{ij}$, you will still not have any natural number matrix solutions.

If $\sigma_i$ and $\sigma_j$ both have natural number entries, then $\sigma_i\sigma_j$ and $\sigma_j\sigma_i$ also have natural number entries. But {$\sigma_i,\sigma_j\}=0$ implies $\sigma_i\sigma_j=-\sigma_j\sigma_i$. The left hand side has nonnegative entries and the right hand side has nonpositive entries, so both sides are the zero matrices.

But then $\sigma_i\sigma_j=[0]$ implies $[0]=\sigma_i\sigma_i\sigma_j\sigma_j=2I\cdot 2I=4I$, a contradiction. So, there are zero natural number matrix solutions.

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