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I've been reading about the Artin Spin operation. It's defined as taking the classical $n$-knot ($S^n\hookrightarrow S^{n+2}$) to an $(n+1)$-knot. For the $1$-knot case (in $\mathbb{R}^3$), I reproduce the procedure in knot spinning, p. 8,

  1. We manipulate a knot $K$ so that all but a trivial arc $a$ lie in the upper half space $H^3=\{(x,y,z)\>|\>z\geq 0\}$. We remove the interior of $a$.

  2. We rotate $H^3$ around $\mathbb{R}^2$ in $\mathbb{R}^4$, inducing a parameterization $(x,y,z)\mapsto (x, y, z\cos\theta, z\sin\theta)$

-- Question -- How is this operation similar to the suspension functor (on a topological space $X$) $\Sigma X\equiv X\wedge S=\frac{X\times S}{X\vee S}$?

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  • $\begingroup$ Why should this be related to suspension, other than that there is a circle involved in both? $\endgroup$ – Mike Miller Feb 25 at 21:07
  • $\begingroup$ @MikeMiller Its a valid counter-suspicion. I had a hunch that they might be related in some way because the spin operation takes the knot embedding from $S^n\hookrightarrow S^{n+2}$ to $\Sigma S^n\hookrightarrow \Sigma S^{n+2}$ $\endgroup$ – Matt Cuffaro Feb 25 at 21:44
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This question is answered in section 4 of my first paper (with Alex Suciu)

Klein, John R.; Suciu, Alexander I. Inequivalent fibred knots whose homotopy Seifert pairings are isometric. Math. Ann. 289 (1991), no. 4, 683–701.

The paper is available here:

http://www.math.wayne.edu/~klein/inequivalent_knots.pdf

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  • $\begingroup$ I will need to read the section further, but at first glance this looks like it helps. $\endgroup$ – Matt Cuffaro Feb 25 at 21:41
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    $\begingroup$ In the notation of the section: if you take $M$ to be $S^1$ with standard inclusion into the 3-sphere, your will get classical Artin spinning. $\endgroup$ – John Klein Feb 25 at 22:32
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Although this is not exactly an answer to your question, it answers a slightly different question in a strong affirmative.

In my "A family of embedding spaces" paper I show that Litherland's spinning operation (a strictly more general spinning operation than Artin or Fox spinning) can be identified with the "connecting map" for the Pseudo-isotopy fibration sequence of long knot embedding spaces. Another way of saying this is that the fibrations produce natural maps

$$\Omega K_{n,j} \to K_{n+1, j+1}$$

where $K_{n,1}$ is long embeddings of $\mathbb R$ to $\mathbb R^n$, and $K_{n+1,2}$ would be long embeddings of $\mathbb R^2 \to \mathbb R^{n+1}$. These are a type of spinning map, but in the flatten-out context of long knots.

The caveat is that, this is only strictly true for smooth embeddings in co-dimension $3$ or larger.

In the co-dimension two case, there are analogous maps and they produce Litherland spinning, but the domain is really the free loop space. You get a relation with pseudo-isotopy theory, but the pseudo-isotopy long exact sequence is only onto the null-concordant subspace of the knot space.

The paper also identifies a pair of these maps with Haefliger's construction of his smooth non-trivial $S^3$ in $S^6$.

This is more or less the content of section 3.

There is one other connection to suspension and embedding spaces identified in the paper (but it no longer relates to spinning constructions).

The inclusion $\mathbb R^n \to \mathbb R^{n+1}$ gives an inclusion

$$K_{n,1} \to K_{n+1, 1}$$

I know of two (perhaps homotopically-distinct) null-homotopies of this map. If you edit the map accordingly, it gives you an extension

$$\Sigma K_{n,1} \to K_{n+1,1}$$

and I've often wondered if this map is null-homotopic. If you look at the map through the eyes of the evaluation maps (Goodwillie-Weiss calculus) this map becomes Freudenthal suspension. So from the perspective of rational homotopy theory you would expect this map to possibly be null.

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  • $\begingroup$ I appreciate this and will read your paper. Does this mean there is some sort of spectrum object $\Sigma^\infty K_{n,1}$? $\endgroup$ – Matt Cuffaro Feb 26 at 13:48

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