4
$\begingroup$

How can we calculate the 4-genus of a link L? The 4-genus is defined to be the minimal genus of orientable surface bounded by L in B^4. Is there any routine method to calculate that?

Especially, any good idea how to calculate the 4-genus of a 2-bridge link? Thanks.

Improve this question
$\endgroup$
1
  • $\begingroup$ the 4-ball genus is fairly easy to compute for positive knots. $\endgroup$
    – Mohamed Ait Nouh
    Commented Feb 19, 2017 at 5:41

4 Answers 4

Reset to default
5
$\begingroup$

You might get interesting lower bounds using Rasmussen's s-invariant, and his calculations of the KR homology of 2-bridge links.

Or you might not. As Ryan says, there are various ways of bounding genus above and below, but unless you get lucky and two of them match, it can be hard to tell. The 4-genus of torus knots was an open problem for a scandalously long time (if my history is right, it was first proven in the 90's!).

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thanks. I think KR homology is more interesting because it can give a finer lower bound for 4-genus. As for Rasmussen's s-invariant, it equals the signature when the link is alternating, which seems kind of trivial. $\endgroup$
    – Megan
    Commented Feb 9, 2010 at 5:05
  • 1
    $\begingroup$ I completely disagree. Remember, there aren't that many alternating knots. Separating knots into alternating and not-alternating is like separating the world into bananas and not-bananas, to steal a famous quote. $\endgroup$
    – Ben Webster ♦
    Commented Feb 9, 2010 at 5:23
6
$\begingroup$

Take a look at Paolo Lisca's papers "Rational balls and the ribbon conjecture" and "Sums of lens spaces bounding rational balls". He determines which 2-bridge knots are slice and the concordance order for all 2-bridge knots.

But no there's no known effective procedure to compute 4-genus.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thanks a lot. They are really helpful but don't apply to the case I'm interested in. It seems that not the concordance order of ALL 2-bridges knots K(p,q) is determined but only for those with p odd. And the knot I'm studying has p=208, which is neither odd nor a square. $\endgroup$
    – Megan
    Commented Feb 7, 2010 at 21:06
3
$\begingroup$

Rasmussen and Lee's results say that the $s$ invariant of a 2-bridge knot will be just equal to the signature of the knot. So you can compute the signature of your knot to get a lower bound (there are extremely rapid ways of doing this from an alternating diagram). Unfortunately the only decent way to get an upper bound that I know of is by spotting a smooth surface! Good luck.

Remember that $s$ might not be the best you can do. In particular, among alternating knots, the figure 8 knot ($4_1$ in Rolfsen) has vanishing $s$ invariant (for example, because it is torsion in the concordance group) and yet it is not even slice if you allow your surfaces to be locally flat, let alone smooth.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Aha! Thanks. I'm reading your paper "a slice genus lower bound from $sl(n)$ KR homology" these days. $\endgroup$
    – Megan
    Commented Feb 10, 2010 at 18:08
  • $\begingroup$ That's nice! However, I think that you won't get any better bounds on 2-bridge knots with the KR bounds than with the signature or with standard $s$ or $\tau$. This is because if the bounds are a homomorphism to Z from the concordance group and are tight for the positive knots, it forces them all to agree for alternating knots. So the possibility that the KR bounds are better since the $E_2$ homology doesn't have to be thin is not really a possibility after all... $\endgroup$
    – Andrew Lobb
    Commented Feb 10, 2010 at 18:46
  • $\begingroup$ Thanks. I'm not quite sure what you denote by $\tau$. Is it the Thurston-Bennequin invariant? I thought KR might be a better one since it's more sophisticated than many other invariants.But it may be a good news for me that KR doesn't give a better bound. I don't need to be involved in the tough computation of KR any more. $\endgroup$
    – Megan
    Commented Feb 11, 2010 at 6:08
  • $\begingroup$ Sorry - I meant the bound on the slice genus coming from Heegaard-Floer knot homology. $\endgroup$
    – Andrew Lobb
    Commented Feb 13, 2010 at 0:59
3
$\begingroup$

(This is really a comment on the answer relating to concordance order.)

Since your p is even, then your 2-bridge knot is actually a link. So, while it makes sense to ask if it's a slice or ribbon link, asking about its concordance order doesn't make sense.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .