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All:

Let $S_g$ be the genus-g orientable surface (connected sum of g tori), and consider

a symplectic basis B= {$x_1,y_1,x_2,y_2,..,x_{2g},y_{2g}$} for $H_1(Sg,\mathbb Z)$, i.e., a basis such that

$I(x_i,y_j)=1$ if i=j, and 0 otherwise, where I( , ) is the algebraic intersection of $(x_i,y_j)$,

e.g., we may take $x_i$ to be meridians and $y_j$ to be parallel curves within the same sub-torus. Does it follow

that every non-trivial (non-bounding) SCCurve in $S_g$ must intersect one of the

curves in B? I think the answer is yes, since, algebraically, every non-bounding curve

is a linear combination of elements in B. Is this correct? Can anyone think of a more

geometric proof?

Thanks.

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1 Answer

Does "SCCurve" mean "simple closed curve"? I'll assume it does.

If any (possibly self-intersecting) closed curve $C$ represents a nontrivial element of $H_1(S_g)$, then it will pair nontrivially with some element of $H^1(S_g)$. This pairing can be read off by intersecting with a dual basis of $H_1(S_g)$ (e.g. the one you've got), and so the answer to your question is yes. Your own argument can equally well be completed (though it'll also use Poincare duality in the background), but note that it's not on-the-nose true that "every non-bounding curve is a linear combination of elements in B" -- this only holds up to boundaries.

I don't know how you could have a more geometric argument when all you're asking for is a symplectic basis of $H_1(S_g)$. Perhaps it might interest you to draw a fundamental domain (a $4g$-gon) for $S_g$ and choose explicit representatives for your basis elements. Then, observe that for a curve to be nontrivial, its image in this fundamental domain must cross through the boundary. That should yield a pretty convincing "proof by picture".

By the way, in the future please take the time to TeX your question. This will make it much easier to read.

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