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This is a question asked by a student in my lecture. After drawing pictures for awhile, I thought it was a good one. I seek a nontrivial example of a pointed Heegaard diagram $(\Sigma,\mathbf{\alpha},\mathbf{\beta},z)$ of a closed, oriented 3-manifold that fails to be weakly admissible in the sense of Heegaard Floer homology.

What I call the trivial example is the diagram of $S^1\times S^2$ consisting of two parallel circles in the torus, with $z$ appropriately chosen so that there is a positive domain given by an annulus, bounded by these circles, whose index is 0. However, there are no generators of $\widehat{CF}$ in this diagram. For all I know, any Heegaard diagram that has generators is weakly admissible!

Lipshitz's index formula (Lemma 4.11 of http://arxiv.org/abs/math/0502404) seems to imply that a domain in $\widehat\pi_2(\mathbf{x},\mathbf{x})$ that violates weak admissibility either needs lots of acute corners (assuming acute corners are even available to periodic domains, it seems their negative contribution to the index should be more than canceled by $2n_{\mathbf{x}}$), or high genus (with similar cancellation coming from any entries of $\mathbf{x}$ in its interior).

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  • $\begingroup$ Just to clarify, you're looking for a Heegaard diagram which is not weakly admissible but has generators? $\endgroup$ May 28 '14 at 14:18
  • $\begingroup$ Yes; thank you for your fine response. $\endgroup$ May 29 '14 at 4:02
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If you have any 3-manifold $Y$ with positive $b_1$, and any (generic) Heegaard diagram $\mathcal{H} = (\Sigma, \alpha, \beta)$ for it, you can place the basepoint $z$ so that $(\Sigma, \alpha, \beta,z)$ is not admissible.

In fact, consider any domain $P_0$ in $\mathcal{H}$ whose boundary is a nontrivial sum of $\alpha$- and $\beta$-curves (this would be called periodic, if the usual definition of periodicity didn't require that the multiplicity of the basepoint be 0), and let $m$ be the minimal multiplicity of $P_0$, attained at a region $D$. Consider $P = P_0 - m\cdot \Sigma$, and place the basepoint $z$ in $D$. Clearly, $P$ is a nonzero periodic domain in $(\Sigma, \alpha, \beta,z)$ which has only non-negative multiplicities.


Here's what happens for the "standard" admissible-looking diagram for $S^1\times S^2$ you refer to. (I'm linking to an SVG picture, that I can't convert at the moment -- if anyone wants to edit, feel free to do it). The "standard" periodic domain is the difference of the two bigonal regions. Adding the whole torus and placing the basepoint in the 0-multiplicity bigon yields the linked picture. Colour indicates multiplicity: white means 0, lighter gray 1, darker gray 2.


EDIT: as pointed out by the OP in a comment below, the domain above has index 2, while we'd like to have an index-0 periodic domain. I think I have an example for this phenomenon (again, a diagram for $S^2\times S^1$):

The Heegaard diagram

Consider the domain $P_0$ shown in the picture below (yellow means multiplicity -1, white multiplicity 0, and gray multiplicity 1), and generator $x$ corresponding to the two thick black points, and the green basepoint $z$. The index of $P_0$ as a domain in $\pi_2(x,x)$ is -2 (the Euler number is the Euler characteristic of the doubly-pointed torus, which is -2), and the multiplicity at $x$ is zero. If you add the entire surface, you get a periodic domain of index 0 with only non-negative multiplicities.

Notice that there is a lot of redundancy, both in the number of generators and the genus of the diagram, but I wouldn't even try to formulate a question where this is addressed.

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    $\begingroup$ I believe the index of this domain is 2, so it is not the requested example if you are thinking of the original definition of weak admissibility (all positive index-0 periodic domains are trivial), though Sarkar-Wang do define their admissibility as the stronger ``all positive periodic domains are trivial" and this is an example of an inadmissible diagram in that sense. $\endgroup$ May 29 '14 at 4:26
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    $\begingroup$ Ah, yes, I should have seen that: you increased the genus of your domain by 1 by stabilizing at a point inside it, then isotoped so the generator would have an entry outside the domain. $\endgroup$ Jun 3 '14 at 16:26

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