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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
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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

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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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