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This is a very naive question, but I'm trying to understand the difference between the various categories when it comes to embedding surfaces in 4-dimensional manifolds. The situation I'd really like to understand is when we have a surface neatly and properly embedded in the 4-ball $D^4$, so that its boundary lies in $\partial D^4 = S^3$ as a link. I'd like to understand some of the differences between assuming everything in the setup is just continuous, versus when we assume it is either locally-flat, PL locally-flat, or smooth.

For example, if we have a PL-locally flat surface, after arbitrarily small PL-isotopy we can arrange so it has a movie presentation with only max, min, and saddle points. So it seems like we can treat such an embedded surface much as we would a smooth surface with Morse function. Intuitively it seems that the theories of smoothly embedded surfaces and PL locally-flat surfaces should be equivalent in some sense, though I've never seen it dealt with in the literature. In particular it seems like we should be able to smooth PL locally-flat surfaces and vice versa, so that equivalence under the different types of isotopy is preserved.

Where I get really murky is when we drop the PL or smooth condition and consider topological locally-flat embeddings. Here things can be quite different, as there are knots which bound topologically locally-flat disks, but whose smooth slice genus is arbitrarily high. It seems like there's an awful lot more freedom once we drop smoothness, or even just the PL condition from PL locally-flat, even though we can use embedded PL locally-flat disks to $\varepsilon$-approximate any topological embedding.


With that rambling out of the way, my main questions are he following:

1) Are the smooth and PL locally-flat situation really that "close" to each other? Can surfaces in one category be approximated by surfaces in the other, so that equivalence under smooth and PL-isotopy is preserved? Or are their hidden subtleties I'm missing?

2) Is there anyway to understand the problems we encounter when we drop to the topologically locally-flat situation? For example, whenever the locally-flat condition is explained, the example of the cone over a nontrivial knot is invoked to illustrate a situation where it fails. Is there a similar example to understand (even just a single situation) where a topologically locally-flat embedding fails to be PL locally-flat or smooth?

Of course anything else which might help me understand the situation better would be greatly appreciated as well.

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  • $\begingroup$ I found this question to be in a similar spirit and having an insightful answer. $\endgroup$
    – M. Winter
    Apr 5 at 18:25

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Regarding (1), yes there's a unique (up to smooth isotopy) smoothing of a PL locally-flat embedding. You can similarly ensure that if your original surfaces are close in some PL compact-open sense, that their smoothings can be made to be close in the smooth C^1 compact-open sense.

Regarding (2), yes, there are knots that are topologically locally-flat slice, yet they are not smoothly slice. A knot with a trivial Alexander polynomial does the job. These examples aren't as "concrete" as your cone example as they depend on theorems of Freedman which involve infinite constructions.

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  • $\begingroup$ @Ryan: is there any simple way to show that these knots are non-slice? The methods I know of are: computing the maximal Thurston-Bennequin number, and using Rudolph's "slice Bennequin" inequality; computing Rasmussen's $s$; computing Ozsváth-Szabó's $\tau$. $\endgroup$ Mar 7, 2013 at 22:52
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    $\begingroup$ @Marco: I believe there are twisted Alexander module constructions for proving things to be not smoothly slice. Off the top of my head I don't know a reference but I thought there were some, if not known then conjectured? $\endgroup$ Mar 8, 2013 at 0:39
  • $\begingroup$ Everyone who has written above knows this very well, but in case someone stumbles across this and is confused. Twisted Alexander modules will not detect the difference between topologically locally flat slice and smoothly slice. One needs one of the methods referred to by Marco. Also, it's not true that any knot with trivial Alexander polynomial will do the job. There are nontrivial knots with Alexander polynomial one that are smoothly slice, e.g. $11n42$. $\endgroup$ Dec 29, 2021 at 12:43
  • $\begingroup$ Untwisted Whitehead doubles of trefoils will answer the question, but one has to be careful. One can use the left or right handed trefoil, and a $+$ or $-$ clasp. Two of the possible knots are topologically but not smoothly slice, while the other two also have Alexander polynomial 1 so are topologically slice, but their smooth status is still not known. $\endgroup$ Dec 29, 2021 at 12:49

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