- #1
Bacle
- 662
- 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.
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.