This is related to a homework question in a condensed matter course. For each noncontractible loop which can be drawn on a lattice wrapped onto a surface of genus $n$ we can define two operators. Each pair of operators allows us to construct an extra gauge-distinct degenerate ground state of the system from another pre-existing ground state. Therefore, if we know the order $m$ of the maximal set of pairwise non isotopic, noncontractible loops that can be drawn on the surface, we can construct $2^m$ states. My question then is how do we get $m$ from $n$.
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To clarify: you mean a $2$-manifold, a.k.a. a surface? – Sammy Black May 21 '15 at 19:58
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Yes, I just realized my omission and edited the title. – Jordan May 21 '15 at 19:59
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An $n$-torus is not a surface; it's a manifold of dimension $n$, namely $S^1\times\dots\times S^1$. Do you mean a surface of genus $n$ instead? – Jack Lee May 21 '15 at 20:07
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Have a look at the answers to this question: http://math.stackexchange.com/q/656346/6509 – Sammy Black May 21 '15 at 20:07
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1I think that the OP is referring to the $n$-fold connected sum $T^2 # \cdots # T^2$. – Sammy Black May 21 '15 at 20:08
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Thank you, I've tried to clarify my question. – Jordan May 21 '15 at 20:44
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You will need an extra hypothesis to obtain a finite bound. The trouble is that if you have one loop, you can "push it off" itself to get another loop, and then you can keep pushing off more and more copies of itself to get an arbitrarily large number of loops. Of course you might say "those are the same", but that needs to be defined. A mathematican would say that the two loops obtained from one by "pushing off" to get a second one are related to each other by "isotopy".
You might also wish to be specific that when "drawing" loops on the manifold, different loops must be disjoint, i.e. they must not have any points in common.
So perhaps your real question should be: what is the maximal number of noncontractible loops that can be drawn on the manifold, subject to the requirement that no two of the loops are isotopic and every two are disjoint?
In which case the answer is $3g-3$. The maximum value of $3g-3$ is obtained by a pants decomposition of $S$: each component of the complement of a maximal set of loops is a pair of pants. Proofs of these facts are pretty elementary exercises using the Euler characteristic.
Also, as suggested by the comments, the wording "$n$-torus" has an entirely different meaning to most mathematicians, so care is needed.
Lee Mosher
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And, for all you fans of The Hitchhiker's Guide to the Galaxy, if the question to the ultimate answer of life, the universe, and everything is of any concern, I notice that if the genus equals $15$ then the answer is $42$! – Lee Mosher May 21 '15 at 20:21
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Hmm I just tried this in my head and for a genus of four, I counted $10$ loops, not $9$. Am I overcounting somehow (read: overlooking isotopy)? – Cameron L. Williams May 21 '15 at 20:44
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Naively I would think that the answer is $\frac{n(n+1)}{2}$. Edit: Actually I think I figured out what the issue is. I think I'm looking at all classes of loops, ignoring the intersection condition. Omitting intersection I get your result. – Cameron L. Williams May 21 '15 at 20:51
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Thank you, I've edited the question to reflect your comments and indeed that is what I meant to ask. – Jordan May 21 '15 at 20:55