Why are there always pairwise intersections in a Heegaard splitting?

Let $M=A\cup B$ be a Heegaard splitting, such that $\{\alpha_i\}_{i=1}^g$ is a set of boundaries for meridian disks of $A$, and $\{\beta_i\}_{i=1}^g$ is a set of boundaries for meridian disks of $B$ (Recall that meridian disks are those properly embedded in the manifold, and such that the union of their boundaries doesn't separate the boundary of said manifold).

Why is it true that we can always find a permutation $\tau$ of $\{1,...,g\}$ such that $\alpha_i$ and $\beta_{\tau(i)}$ intersect? Moreover, if I'm not mistaken, once we've found such pairwise intersections for a subset of the $\alpha$'s, we can complete it to the rest of the $\alpha$'s (so that in the end each $\alpha$ intersects some $\beta$, and vice versa). Why is this possible?

Edit: It appears that this is not true in general, however it is true, for example, in the case of the rational homology sphere. I would like to understand why in this particular case the statement is true.

Thanks!

• I don't think you always can find such a permutation. For example take the genus one splitting of $S^1 \times S^2$. You can take your $\alpha$,$\beta$ curves to be parallel (and hence non-intersecting) meridians of a torus.
– user45861
Nov 23 '13 at 19:49
• @Bambi; You are right. Is there some assumption (except for the obvious) we can add so that this statement would be correct? Nov 23 '13 at 21:16
• I think it should be true if your manifold is a rational homology sphere, i.e. the first homology is finite.
– user45861
Nov 24 '13 at 0:30
• @Bambi, that would certainly help since I mainly need the result for this case. Could you elaborate why it is true then? (I'll edit the question.) Nov 24 '13 at 0:53
• I don't have time now, but I'll write you up an answer later.
– user45861
Nov 24 '13 at 7:13

As I noted in the comments above, what you originally asked is not true - the standard genus one splitting of $S^1 \times S^2$ does not have an intersection between its $\alpha$ and $\beta$ curve. However, it does hold for rational homology spheres.

We take a Heegard splitting for $M$ with $\alpha$ and $\beta$ curves $(\alpha_i)_{i=1}^g$, $(\beta_j)_{j=1}^g$. We can consider this Heegard splitting as a handle decomposition of $M$, where the $\alpha$ curves are belt disks if 1-handles and the $\beta$ curves are are the attaching disks for the 2-handles. This means that $H_1(M)$ is presented by the matrix $A=(\alpha_i \cdot \beta_j)_{1\leq i,j\leq g}$, where $\alpha_i \cdot \beta_j$, is the algebraic intersection of the $\alpha_i$ and $\beta_j$.

In particular, if $M$ is a rational homology sphere, then $|H_1(M)|=|\det(A)|> 0$.

Consider the expression coming from the definition of the determinant (see wikipedia link below): $$|\det(A)|= |\sum_{\sigma \in S_g} \epsilon(\sigma) \prod_{1\leq i\leq g}\alpha_i \cdot \beta_{\sigma(i)}|,$$ where $\epsilon(\sigma)=\pm 1$ is the sign of the permutation $\sigma$. Now since $\det A \ne 0$, this shows there exists $\sigma \in S_g$ such that $\prod_{1\leq i\leq g}\alpha_i \cdot \beta_{\sigma(i)}\ne 0$. So for this $\sigma$, $\alpha_i \cdot \beta_{\sigma(i)}\ne 0$ for $1\leq i \leq g$. Since algebraic intersection non-zero implies geometric intersection non-zero, this is the permutation you want.

• In somewhat of a delay: I re-read your answer now, and although back then it seemed to me that I understood it, I am now confused -- could you add an explanation about why the first homology group is presented by this matrix? Mar 28 '15 at 11:31
• I can think of two ways to explain this. Both require non-trivial algebraic topology. The first is to consider cellular homology. This is what I'm using in the sentence beginning "If we consider the Heegaard splitting as a handle decomposition...". A handle decomposition is just a "fattened up" cell decomposition. In cellular homology the 2nd boundary maps counts how many times the attaching map of each 2-cell winds around each 1-cell (See Hatcher p.140, for example). When you fatten-up to handles, rather than cells, the attaching maps of the 2-cells become the $\beta$ curves and
– user45861
Mar 28 '15 at 23:46
• counting how many times they wind around the 1-cells is the same counting the intersection number with the $\alpha$ curves (which can be viewed as fattened up mid points of 1-cells. Thus the boundary map with respect this handle decomposition is given by that matrix.
– user45861
Mar 28 '15 at 23:50
• The other way to see this fact is to use this handle decomposition to give you a presentation for the fundamental group and then use the fact that $H_1$ is the abelianisation of the fundamental group (Hatcher, Theorem 2A.1). The presentation for $\pi_1$ is the one where you take one generator for each 1-handle, that is each $\alpha$ curve. You then add one relation for each $\beta$ curve, which just records each $\alpha$ -curve that the $\beta$ curve traverses. When you abelianise this relation you just end up with a relation corresponding to the rows of the intersection matrix.
– user45861
Mar 29 '15 at 0:00
• I hope some of this helped. Really both approaches need some pictures.
– user45861
Mar 29 '15 at 0:02