To address whether there exist two non-equivalent exotic structures on (\mathbb{R}^4) constructed from distinct topologically slice, non-smoothly slice knots (K) and (K'), we proceed as follows:

Key Background:

A knot (K \subset S^3) is topologically slice if it bounds a topologically embedded 2-disk in (B^4), and non-smoothly slice if no such disk is smoothly embedded. Free...

Short answer: This is open for the specific “topologically slice but not smoothly slice knot” construction. We do not currently know whether two different knots K and K′ of that type can produce nondiffeomorphic exotic R^4’s via the standard construction. It is conceivable (and consistent with present knowledge) that all such R^4_K could even be diffeomorphic.

Context and references:

• The constr...

Excellent question. This gets to the heart of how modern invariants in low-dimensional topology can be used to distinguish subtle structures.

The short answer is yes, you can find two such knots, and , that give rise to non-diffeomorphic exotic s. In fact, this method can produce an infinite family of pairwise non-diffeomorphic exotic s.

Here is a cons...

<think> Okay, I need to tackle this question about exotic structures on ℝ⁴ coming from topologically slice but non-slice knots. The user is asking whether there are two different knots that give non-equivalent exotic ℝ⁴s or if all such constructions yield the same exotic structure.

First, I recall the setup: For a knot K in S³ that's topologically slice but not smoothly slice, we can construct a ...

Let us denote by

  C (K)=B⁴∪(0–framed 2–handle attached along K)

and

  R⁴_K = int C (K).

If K is topologically slice, Freedman showed that ∂C (K)=S³ bounds a topological 2–disk, so C (K) is topologically B⁴ and therefore R⁴_K is homeomorphic to the standard ℝ⁴.
If, in addition, K is not smoothly slice, the slice disc has to be replaced by a Casson handle; the resulting smooth structure...

This is an excellent question about exotic smooth structures on arising from topologically slice knots.

The answer is yes - there are indeed infinitely many distinct exotic 's that can be constructed from topologically slice but not smoothly slice knots.

Here's the key insight: The exotic constructed from a topologically slice knot can be dis...

Exotic Structures from Topologically Slice Knots

This is a fascinating question in 4-manifold theory that deals with distinguishing exotic smooth structures on .

The Construction

Let me clarify the construction first: For a knot that is topologically slice but not smoothly slice, we can construct an exotic as follows:

1. Take a...

Yes – in fact the recipe you quote produces infinitely many mutually non-diffeomorphic exotic (\mathbb {R}^{4})’s.

How the construction works again
Let (K\subset S^{3}) be a knot.

1. Attach a (0)-framed 2–handle to (B^{4}) along (K); call the result
   (X_{K}=B^{4}\cup_{K}h^{2}).
   (\partial X_{K}=S^{3}_{0}(K)).

2. If (K) is topologically slice, its meridian bounds a loca...

No. In fact nobody knows how to tell two of these “knot‐exotics” apart, and it is quite possible that they are all diffeomorphic. Here is roughly what is known:

1. Construction via a Casson–handle.
   Given a topologically slice but smoothly non-slice knot K one builds a Casson–handle CH(K) in B^4 whose attaching circle is K. Freedman’s theorem says CH(K) is homeomorphic to the standard o...