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As the answers to this question (I think) successfully argue, it is often more useful to consider manifolds as intrinsically, i.e. without reference to an ambient Euclidean space. I am wondering if I have found a simple example of this phenomenon.

Consider the torus trefoil knot:

It can be embedded into $\mathbb{R}^3$, as the above pictures hopefully illustrate. What is perhaps surprising, however, is that it is in fact homeomorphic to the regular torus in $\mathbb{R}^3$ (I think).

Question: Is it impossible to "unknot" the trefoil torus into a regular torus without leaving $\mathbb{R}^3$? I.e. does any homeomorphism between the two "factor through a higher dimensional space"?

Does this impossibility (if it is true) show the need for abstract manifolds? I.e. is this a concrete example where ambient spaces actually make things more confusing, rather than less?

The only homeomorphisms I can think of which "take place in $\mathbb{R}^3$" involve self-intersections, which, being "unphysical", I think implies factoring through a higher dimensional space. The same might be true for cutting the knot, "untying it", and then reattaching it at the same two places it was cut.

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Chill2Macht
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1 Answers1

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Your question doesn't quite make sense. The torus, $T_1$, as a manifold, and the trefoil, $T_2$, as a manifold, are homeomorphic because there's a bicontinuous map between them.

On the other hand, the pair $(R^3, T_1)$ and $(R^3, T_2)$ are not homeomorphic as pairs: there's no bicontinuous map from $R^3$ to $R^3$ whose restriction to $T_1$ gives a bicontinuous map from $T_1$ to $T_2$. (This requires some proof, of course.)

You can also think of $Q = S^1 \times S^1$ as a manifold; the torus $T_1$ that you've drawn above is the image of $Q$ under an embedding (a locally maximal-rank injective map $f_1$ from $Q$ to $R^3$), as is $T_2$ (with map $f_2$. Asking whether you can "untie" the torus amounts to asking if there's a homotopy, i.e., a continuous map $$ H: Q \times [0, 1] \to \mathbb R^3 $$ with the property that $H(p, 0) = f_1(p)$ and $H(p, 1) = f_2(p)$, and the property that $p \mapsto H(p, t)$ is an embedding for every $t$ and that $H$ is $C^1$ smooth (which prevents the so-called "bachelor's isotopy). There is no such homotopy (which again requires proof). [Alternative; you can ask for an ambient isotopy...and again no such thing exists.]

If you regard $\mathbb R^3$ as being in $\mathbb R^4$, though, you can ask whether there's a homotopy from $Q \times [0, 1]$ to $\mathbb R^4$ which looks like $f_1$ and $f_2$ at its two ends and is smooth and for each fixed $t$ is an embedding; the answer to that is "yes". So in that sense, you can untie a trefoil in 4-space, but not in 3-space.

John Hughes
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    Do you mean isotopy? (Possibly ambient isotopy, not sure if bachelor's unknotting applies to torus.) – anon Sep 13 '16 at 03:32
  • I guess I do. I'd forgotten the bachelor's unknotting. Alternatively, one could ask for a regular homotopy...but isotopy is probably better. I'll edit. – John Hughes Sep 13 '16 at 03:38
  • I hadn't ever thought of considering the pairs $(\mathbb{R}^3, T_1)$ and $(\mathbb{R}^3, T_2)$ -- this answer provides a lot of nice ways to think about this -- I appreciate it. – Chill2Macht Sep 13 '16 at 03:39
  • The two embeddings are homotopic, since homotopies allow the image to self-intersect in the interim between $H(-,0)$ and $H(-,1)$. Now that I think about it I don't think the bachelor's unknotting applies to the torus because it's not one-dimensional like a knot, so the embeddings shouldn't be isotopic. – anon Sep 13 '16 at 03:45
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    They're homotopic, but not homotopic through embeddings. I agree that the bachelor trick might not work...but it also might. :( So I was cautious and required a $C^1$ homotopy through embeddings. – John Hughes Sep 13 '16 at 04:43
  • @JohnHughes your condition dies not prevent the bachelor isotopy. You need to add that the isotopy us through objective immersions, which means here that the derivative in the $p$ direction is nonzero for each $p, t$. – Moishe Kohan Sep 13 '16 at 07:27
  • Since I've said it's an immersion, it has to have (for fixed t) maximal rank on each tangent space, which certainly means that the derivative in the $p$ coordinates is nonzero. I think that the missing condition is that the $t$ derivative has to exist, and must be continuous, and that the $p$ derivative must exist and be continuous (as a function of $t$, for fixed $p$, for instance, and vice versa). This prevents the bachelor thing, I believe. (It's certainly what Whitney used in the Whitney-Graustein paper.) So my $C^1$ should be enough...but maybe I'm missing something. – John Hughes Sep 13 '16 at 15:22