I'm confused about the relationship between strong admissibility and weak admissibility for pointed diagrams in Heegaard Floer theory. For reference, here are Ozsváth-Szàbo's original definitions:

A pointed Heegaard diagram is called strongly admissible for the ${Spin}^c$ structure $\mathfrak{s}$ if for every nontrivial periodic domain $\mathcal{D}$ with $$\langle c_1(\mathfrak{s}),H(\mathcal{D})\rangle=2n \geq 0,$$ $\mathcal{D}$ has some coefficient $>n$. A pointed Heegaard diagram is called weakly admissible for $\mathfrak{s}$ if for each nontrivial periodic domain $\mathcal{D}$ with $$\langle c_1(\mathfrak{s}),H(\mathcal{D})\rangle =0,$$ $\mathcal{D}$ has both positive and negative coefficients.

Here's a concrete case that puzzles me: If $c_1(\mathfrak{s})$ is torsion, then it evaluates to zero on every homology class. In that case, it seems that a diagram is

  • strongly admissible for $\mathfrak{s}$ if every nontrivial periodic domain $\mathcal{D}$ has some positive coefficient, and
  • weakly admissible for $\mathfrak{s}$ if every nontrivial periodic domain $\mathcal{D}$ has both positive and negative coefficients.

If these conditions are to coincide for $c_1(\mathfrak{s})$ torsion (or at least for "strong" to be stronger), then it seems like any nontrivial periodic domain with positive coefficients must also have negative coefficients. Is this true?


1 Answer 1


Yes, this is correct. It is spelled out explicitly on page 20 of András Juhász' A survey of Heegaard Floer homology.

Edit: it was not clear to me that you were also asking why, for torsion spin$^c$ structures, strong admissibility implies weak admissibility. Here's a possible proof: if $\mathcal D$ is periodic, then so is $-\mathcal D$. Hence, both $\mathcal D$ and $-\mathcal D$ have some positive coefficient, i.e. $\mathcal D$ has both positive and negative coefficients.

  • $\begingroup$ Thanks, the edit really clears things up -- and my question was admittedly unclear. I've accepted the answer since it fully handles the torsion case I asked about, but I'm also curious about the non-torsion case (where it's pointed out that strong admissibility still implies weak admissibility). My follow-up question won't fit here, so I've edited it into the original post. Would you mind taking a look? $\endgroup$
    – hfquestion
    Commented Feb 9, 2016 at 14:03
  • $\begingroup$ Ah, I think I see it now. Weak admissibility only requires a nontrivial periodic domain to have positive and negative coefficients if $c_1(\mathfrak{s})$ vanishes on it for the reasons you've explained, and makes no demands of periodic domains seen by $c_1(\mathfrak{s})$. Strong admissibility makes the same requirement of periodic domains on which $c_1(\mathfrak{s})$ vanishes, but now also requires periodic classes on which $c_1(\mathfrak{s})$ is positive to have some sufficiently positive coefficient (and doesn't care when $c_1(\mathfrak{s})$ is negative on the domain). Correct? $\endgroup$
    – hfquestion
    Commented Feb 9, 2016 at 14:17
  • $\begingroup$ This does sound correct, indeed. $\endgroup$ Commented Feb 9, 2016 at 22:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.