Symplectic Basis on Sg and Non-Trivial Curves

[Bacle]

Hi, All:

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

a symplectic basis B= {x1,y1,x2,y2,..,x2g,y2g} for H_1(Sg,Z), i.e., a basis such that

I(xi,yj)=1 if i=j, and 0 otherwise, where I( , ) is the algebraic intersection of (xi,yj),

e.g., we may take xi to be meridians and yj to be parallel curves. Does it follow

that every non-trivial (non-bounding) SCCurve in Sg 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.

 

[pat_connell]

Yes, this is correct. To prove it geometrically, just consider any non-bounding simple closed curve C in the surface Sg, and note that C must intersect itself at least once (as it is not contractible). That intersection point must be a point where C intersects one of the curves in B.