Suppose I have a closed genus $g$ surface. I can pick a canonical homology basis for the surface by picking $g$ "A-cycles" $a_1,\ldots,a_g$, and then $g$ "B-cycles" $b_1,\ldots,b_g$, represented by simple closed curves, all disjoint except for each pair $a_i$ and $b_i$ intersecting at 1 point.

However, there are many such choices of basis. I'd like to know whether there is an algebraic way of classifying all the choices. The concrete question I'm most interested in is whether we can classify the choice of A-cycles, or:

Given a $\mathbb{Z}^g$ subgroup of $H_1$ (expressed in a particular canonical basis), how can one tell if it can be generated by $g$ homologically independent disjoint simple closed curves? Can I classify all such subgroups?

  • $\begingroup$ Do you know how to compute the intersection form on $H_1$? $\endgroup$ May 20, 2014 at 17:19
  • $\begingroup$ Yes, in Riemann surface language at least. But does it tell me enough? 1) How can I tell if a homology class has a simple closed representative? 2) If I have several such homology classes, with all intersection numbers 0, does it follow that I can choose representatives which are all pairwise disjoint? $\endgroup$ May 21, 2014 at 15:33
  • 2
    $\begingroup$ For 1), ams.org/journals/proc/1976-061-01/S0002-9939-1976-0425967-3/… shows that this is true iff the homology class isn't divisible by a positive integer greater than $1$, and for 2), yes, this is part of the justification for the term "intersection form." $\endgroup$ May 21, 2014 at 16:59
  • $\begingroup$ Thanks for the reference. On the second point, after quite a bit of thought it is far from clear that one can pick disjoint simple representatives in general. Indeed, take $a_1+2a_2$ and $a_1$ on a genus 2 surface. If you can pick disjoint simple representatives, cut the surface along them and use them as canonical homology generators. But then no choice of B-cycles can allow you to span $H_1$, which seems to be a contradiction. $\endgroup$ May 24, 2014 at 11:10

1 Answer 1


EDIT : Just to make sure that this is clear, this answers the more difficult question asked in the comments. It also gives a complete answer to the original question.

In general, this is a subtle question and I doubt there is a simple-to-state answer. See the paper

Edmonds, Allan L. Systems of curves on a closed orientable surface. Enseign. Math. (2) 42 (1996), no. 3-4, 311–339.

for a discussion of some of the difficulties.

But one answer to a restricted version of your question is given by the following fact. Fix a closed orientable surface $S$. Say that a set $\{\vec{v}_1,\ldots,\vec{v}_k\}$ of elements of $H_1(S;\mathbb{Z})$ is isotropic if $i(\vec{v}_i,\vec{v}_j)=0$ for all $1 \leq i,j \leq k$ and unimodular if the $v_i$ form a basis for a $k$-dimensional direct summand of $H_1(S;\mathbb{Z})$.

Prop : Let $\{\vec{v}_1,\ldots,\vec{v}_k\}$ be a set of elements of $H_1(S;\mathbb{Z})$. Then there exist disjoint simple closed orientable curves $\{\gamma_1,\ldots,\gamma_k\}$ with $[\gamma_i] = \vec{v}_i$ for $1 \leq i \leq k$ such that $S \setminus (\gamma_1 \cup \cdots \cup \gamma_k)$ is connected if and only if the set $\{\vec{v}_1,\ldots,\vec{v}_k\}$ is isotropic and unimodular.

For a proof of this, see Lemma 6.2 of my book-in-progress "Lectures on the Torelli group", available here. You also might find the discussion of realizing symplectic bases in Chapter 2 enlightening.


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