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If I have a linked pair of circles in $\mathbb{R}^3$, they can be unlinked in $\mathbb{R}^4$. Said differently, there is an isotopy in $\mathbb{R}^4$ between two strands which have been twisted, and two untwisted strands. Here is an ascii art picture!

a   b          a    b
|   |          |    |
 \ /           |    |
  \            |    |
 / \           |    |
|   |          |    |
|   |   ~~~>   |    |
 \ /           |    |
  \            |    |
 / \           |    |
|   |          |    |

I'll call this "the unlinking isotopy" (actually I don't know if it's unique, but I mean to use some chosen one throughout).

Now if I cross the two strands three times, I have two different isotopies to the once-crossed version. To clarify, let $t$ denote a single crossing, transposing $a$ and $b$. Then the picture above shows the first and last steps of an isotopy between $t^2$ and $t^0$. One can use this in two different ways to make an isotopy from $t^3$ to $t^1$.

My question is whether these two isotopies are "the same", and in what sense. Consider the surfaces swept out by these two isotopies. Can one be deformed (isotoped?) to the other, while keeping the boundaries fixed?

I think maybe this should be possible in $\mathbb{R}^5$ but not $\mathbb{R}^4$. Maybe this can be answered as some sort of higher-dimensional linking; if so, please explain!

Post script

My motivation is coming from Baez-Dolan's Tangle Hypothesis. In particular, I've wanted to give a geometric description for sylleptic monoidal structure in 2-categories (and in bicategories). This is an intermediate notion, between braided monoidal and symmetric monoidal, which doesn't appear for monoidal 1-categories.

I've been trying to think about this by imagining the unlinking isotopy as a surface in $\mathbb{R}^4$ -- does it intersect itself? The two isotopies from the triple twist to the single twist give two surfaces, and the geometric avatar of sylleptic structure is an isotopy between them. If I'm understanding the relationship correctly, this should be possible in $\mathbb{R}^5$ but not $\mathbb{R}^4$.

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    $\begingroup$ Hi Niles, could you be a little more specific? When you talk about isotopies of isotopies, what exactly do you mean. And could you say what you mean by "three times" in regard to twisting? Your braid on the left does not permute a and b -- I think most people would say that's two "half twists", or one "full twist". $\endgroup$ Jul 27 '19 at 5:04
  • $\begingroup$ Thanks for looking at this! I've attempted to clarify I'm asking about the surfaces generated by the unlinking isotopies. Being unfamiliar with the language, it may still be unclear. If I have a chance, I'll try to upload another picture from Baez-Dolan which might help. $\endgroup$
    – Niles
    Jul 27 '19 at 16:29
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Thanks, I understand your question now. I asked a similar question in an old paper of mine, but it was with knots rather than braids. In short, the answer is yes they are isotopic in $\mathbb R^4$ subject to certain carefulness assumptions. But if you are particularly fussy about the set-up there is a way in which the answer could be no.

The unknotting operation you see for braids is an aspect of what Fred Cohen might call the Freudenthal suspension map, for configuration spaces. Specifically, the pure braid group is the fundamental group of $C_n \mathbb R^2$, and you are looking at the natural inclusion $C_n \mathbb R^2 \to C_n \mathbb R^3$. The latter space is simply connected, so you get an unknotting operation.

But there's more you can say. This inclusion is null-homotopic in exactly two different ways. The idea is when you think of $n$ points in $\mathbb R^2$ included into $\mathbb R^3$, you can lift those $n$ points to prescribed distinct altitudes (the height function on $\mathbb R^3$ that has $0$-set the inclusion of $\mathbb R^2$. You then do the straight-line homotopy in the $\mathbb R^2$ factor, and push back into the plane. So that converts your inclusion $C_n \mathbb R^2 \to C_n \mathbb R^3$ to a map

$$C_n \mathbb R^2 \to \Omega C_n \mathbb R^3.$$

and this map is non-trivial on the fundamental group. I suppose Fred would write the map as $\Sigma C_n \mathbb R^2 \to C_n \mathbb R^3$.

By 'converts' I mean you can recover your original inclusion by evaluating the loop at its mid-point. Anyhow, one issue with this is it shows you there are precisely two distinct ways to untangle a given braid canonically in this co-dimension one setting. If you included $C_n \mathbb R^2 \to C_n \mathbb R^4$ there would be a circle of ways of untangling your braids, and they would all (individually) be isotopic, since the circle is connected. But in co-dimension one, the two ways of untangling are not isotopic.

Anyhow, back to your question. Basically you are concatenating a braid with an isotopy of a braid, in two different ways and you are asking if these operations commute. To make it a proper question you would have to be a little more careful settings things up -- this is the issue of loop spaces being homotopy-associative and not (usually) strictly associative. But I see where you are going.

In your case of three half-twists, if you use the same unknotting operation on the two braids, they are not isotopic. If you use the opposite unknotting operations, then they are isotopic. If on the other hand you concatenated four half-twists and were comparing the same unknotting operation on the top half vs. the bottom half, yes they would be isotopic.

The way to see the isotopy (or lack of it) is just to be careful about exactly what you are doing. When you have your 2-stranded braid that is the product of three half-twists, imagine a little red rectangle that engulfs two of the half twists. In that rectangle you can do the unknotting operation as the two points in the plane are the same at the top of the rectangle as at the bottom -- so the unknotting operation is defined. Now imagine sliding that rectangle down. You keep the dimensions of the rectangle the same. Provided your braid started off as a proper helix, regardless of where your rectangle is, the two points from the braid at the top of the rectangle will be in the same position (in $\mathbb R^2$) as the two points at the bottom of the rectangle. So the untangling operation is defined for this entire 1-parameter family. That said, if you have three half-twists there is the issue of which point in the braid gets pushed up into $\mathbb R^4$. For a half-twist the point that gets pushed up switches at the end of the braid, due to the representation $B_n \to \Sigma_n$. So when you concatenate three half-twists the unknotting operation on the top two strands is isotopic to the reverse unknotting operation on the bottom two strands. For the same reason, if this braid were a concatenation of four half-twists, unknotting the first two would be isotopic to unknotting the last two (provided you use the same unknotting isotopy).

If you are curious about my analogous question for knots, it's in my "Family of embedding spaces" paper, proposition 5.1 and the discussion immediately after it: https://arxiv.org/pdf/math/0605069.pdf

edit: and yes these are all isotopic in $\mathbb R^5$.

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  • $\begingroup$ Fantastic! I'll have to think more before I understand whether this matches what the tangle hypothesis predicts, but my first impression is that it does. And thanks for the reference, I will indeed want to check it out. $\endgroup$
    – Niles
    Jul 28 '19 at 12:33

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