For an orientable surface of genus $n$ what is the maximal number of noncontractible loops that can be drawn on that surface? 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 $n$.
 A: 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.
