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Symplectic Basis on Sg and Non-Trivial Curves

  1. Jul 24, 2011 #1
    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.
     
  2. jcsd
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