# Why are all knots trivial in 4D?

A classical knot is defined to be an embedding $S^1 \to \mathbb{R}^3$ where $S^1$ is a 1-sphere or circle.

Embeddings $S^1 \to \mathbb{R}^4$ are usually not considered knots because they are trivial knots, i.e., they can be continuously deformed to $S^1$.

How can I show that such embeddings are indeed trivial knots?


1. Every continuous embedding $\kappa:S^{1} \to \Reals^{4}$ can be deformed (by an isotopy) to a smooth embedding. (That is, we may assume without loss of generality that $\kappa$ is a smooth embedding.) Let $K = \kappa(S^{1})$ denote the image.

2. The set of all lines through pairs of points of $K$, i.e., the set $\mathcal{C}$ (for "chords") of points of the form $$(1 - t)P + tQ,\qquad \text{P, Q in K, t real}$$ is the image in $\Reals^{4}$ of the three-dimensional manifold $K \times K \times \Reals$ under a smooth mapping. By Sard's theorem, the image is not all of $\Reals^{4}$. Pick an orthonormal basis $(\Basis_{j})_{j=1}^{4}$ of $\Reals^{4}$ such that $\Basis_{4} \not\in \mathcal{C}$. By construction, $K$ can be deformed (through smooth embeddings) to its projection to the three-dimensional subspace spanned by $(\Basis_{j})_{j=1}^{3}$. Explicitly, if $$\kappa(t) = \kappa_{1}(t)\Basis_{1} + \kappa_{2}(t)\Basis_{2} + \kappa_{3}(t)\Basis_{3} + \kappa_{4}(t)\Basis_{4},$$ then $$H(s, t) = \kappa_{1}(t)\Basis_{1} + \kappa_{2}(t)\Basis_{2} + \kappa_{3}(t)\Basis_{3} + (1 - s)\kappa_{4}(t)\Basis_{4},\quad 0 \leq s \leq 1,$$ deforms $K$ (through smooth embeddings) to its projection. That is, without loss of generality we may assume $K$ sits inside $\Reals^{3} \subset \Reals^{4}$.

3. Let $\phi:S^{1} \to [0, 1]$ be a smooth bump function supported in a "small interval" of the circle. Deform $\kappa$ to $\kappa + \phi \Basis_{4}$; that is, "(smoothly) lift" a small piece of $K$ along $\Basis_{4}$. The "portion of $K$ remaining in $\Reals^{3}$ is an open curve, which may be smoothly deformed in $\Reals^{3}$ to a standard interval; this deformation represents a deformation (through smooth embeddings) of $K$ to an unknot.

Alternatively, project the knot to $\Reals^{3} \subset \Reals^{4}$ as above, then remove crossings one by one: A knot crossing may be modeled by the $x$-axis and the $y$-axis "slightly lifted in the $z$-direction". Specifically, if $\phi$ is a smooth bump function supported near $0$, then the images of parametric curves $$(t_{1}, 0, 0, 0),\quad \bigl(0, t_{2}, \phi(t_{2}), 0\bigr) \tag{1}$$ represent crossing strands. The mapping $$\bigl(0, t_{2}, \phi(t_{2}) \cos(\pi s), \phi(t_{2}) \sin(\pi s)\bigr),\quad 0 \leq s \leq 1,$$ is a homotopy (through smooth embeddings) that converts the over-crossing (1) to the under-crossing. $$(t_{1}, 0, 0, 0),\quad \bigl(0, t_{2}, -\phi(t_{2}), 0\bigr).$$

• For "Alternatively", could there be an uncountable number of crossings, and would that block that solution?
– Yakk
Sep 8, 2015 at 13:50
• @Yakk: Good point; I was implicitly assuming the knot is "tame", and a tame knot can be projected to have only finitely many crossings. The very first step (deform to a smoothly-embedded knot) breaks down if the original knot is wild. Offhand I don't know whether every wild knot in $\Reals^{3} \subset \Reals^{4}$ can be isotoped to an unknot. Sep 8, 2015 at 15:04
• I think you should add in point 3 that the deformation in $\mathbf R^3$ is done while fixing the end points of the open curve. Sep 8, 2015 at 15:12
• @Marc: The idea was to retract the open arc in $\Reals^{3}$ not keeping the endpoints fixed, i.e., to "untie the knot" while allowing the "lifted" interval to "ride along", held out of the way along the $4$th axis. :) Sep 8, 2015 at 15:35
• @AndrewD.Hwang: I think I see what you mean. But it seems somewhat more complicated than what I proposed; you need to specify exactly how to "ride along", and address worries that this might produce a nontrivial knot outside $\mathbf R^3$. You might insist that at each point in time the lifted interval project onto $\mathbf R^3$ as a straight path between the end points. But then you can also modify your "untying" transformation by transforming each intermediate stage by a suitable transformation to get the end points to their fixed locations, and the remaining arc can just stay in place. Sep 8, 2015 at 16:04

I like to use color any time I go into $$4$$ dimensions. What I mean by this is that we can think of any point in $$\mathbb{R}^4$$ as a point in $$\mathbb{R}^3$$ with a color. Let's say we color a point black if the fourth coordinate is zero, with it fading to blue for negative numbers and red for positive numbers.

Any knot has crossings, and changing enough of them from over to under or vice versa will give us the unknot. (The minimum number of changes is known as the unknotting number.) So to show that we can always get the unknot, all we have to do is show that we can change any crossing we want.

At a crossing, if we want to pull the underarc of the crossing past the overarc, we cannot do it in $$3$$ dimensions. But if we first push the underarc into the fourth dimension, by making it fade to red and back to black, we will be able to slide it over. This is possible since they are not the same color (not occupying the same point in $$\mathbb{R}^4$$).

Thus any crossing can be switched, and hence, any knot can be made into the unknot.

• Nice visualization! Thank you for this answer. Aug 22, 2020 at 12:15
• This is a beautiful visualization of this question. However, correct me if I'm wrong, but are you not assuming that the knot was contained in $\mathbb{R}^3\times \{t\}$ for some $t$? (Your answer is still insightful even if you do, obviously, but perhaps I am wrong and you are not even assuming it). Sep 21, 2021 at 15:00
• @Cronus Yes, that is basically what I am assuming. But if the knot is not, the crossings locally can be perturbed slightly to lie in a "slice" of $\mathbb{R}^3$ where the argument holds. Usually this is thought of as making your knot in 3-space as usual, first, then considering an extra dimension, which is why I framed my argument this way. Sep 21, 2021 at 16:22

The other answers resolve the question for tame knots. Here I'll talk about wild knots. These are knots that don't locally around every point admit a tubular neighborhood.

First, we need to figure out what we mean by two knots being equivalent. In the tame case, the three standard equivalent definitions of knot equivalence are: 1) Locally flat isotopy. This is an isotopy of the emvedding of your knot that extends to an isotopy of a tubular neighborhood. This doesn't even make sense anymore because your knot no longer has a tubular neighborhood. 2) Ambient isotopy. This is an isotopy of homeomorphisms $f_t: S^3 \to S^3$ (or, in our case, $S^4$) such that $f_0$ is the identity and, if $i_0, i_1$ are our knots $i_j: S^1 \to S^3$, $i_1 = f_1 i_0$. This still makes sense for wild knots. 3) Homeomorphism equivalence. This is just an orientation-preserving homeomorphism $f: S^3 \to S^3$ such that $i_1 = f i_0$.

(2) and (3) are the same equivalence relation for wild knots (and tame knots alike). But it is not true that wild knots can be unknotted using these definitions of equivalence in $S^4$, because a wild knot can only be equivalent to a wild knot - any homeomorphism taking one to the other will automatically take a tubular neighborhood of one to the other!

So for all the standard definitions, no, a wild knot cannot be unknotted. We have one possibility left up our sleeve: 4) Continuous isotopy. That is, $i_0$ and $i_1$ are equivalent if they're homotopic through continuous embeddings $i_t$. This is normally a very silly equivalence relation. Every tame knot is equivalent in $S^3$ under this: pick an arc, and then slowly make the rest of the knot smaller and smaller until it's just a point. Counterintuitively, this really is a continuous isotopy, I promise! In fact, this works for any knot with a tame arc, so knots that are wild indeed. (It is conjectured that they are actually all equivalent.)

So the question: are wild knots in $S^3$ automatically continuously isotopic in $S^4$? And I would guess the answer is yes. After spending some thought on it and some googling I have been unable to produce an answer, but I am temporarily unable to try more. Here's what the proof should look like. 1) $K$ bounds a disc in $S^4$ - just take the cone with $K$ as the base. 2) Continuous embeddings of $D^2$ are continuously isotopic in $S^4$, because, well, obviously they are.

It may be clear that (2) is the tenuous part in the above argument. The corresponding statement for PL/smooth isotopy of PL/smooth embedded balls of any dimension in $S^n$ is true, so hey, this one probably is too. If anybody knows a reference please comment on this post.

The following theorem is implicit in the paper by Blankinship, where it is proven that for each $$n\ge 3$$ and $$q\in [1, n]$$, there exists a subset $$A\subset {\mathbb R}^n$$ homeomorphic to the (closed) $$q$$-dimensional cube and whose complement is not simply-connected. Note that this result also shows that not all 2-dimensional topological disks in $${\mathbb R}^4$$ are isotopic to each other.

William A. Blankinship, Generalization of a construction of Antoine, Ann. Math. (2) 53, 276-297 (1951). ZBL0042.17601.

Theorem. For every $$n\ge 4$$ there exists a nontrivial (wild) 1-dimensional knot $$K\subset {\mathbb R}^n$$.

Proof. Blankinship proves that for each $$n\ge 4$$, there exists a subset $$C\subset R^n$$ homeomorphic to the classical cantor set such that $$R^n\setminus C$$ is not simply-connected (the complement is necessarily connected by the Alexander duality). Now, take such $$C$$. I proved in my answer here that for every totally disconnected compact $$C$$ in $${\mathbb R}^n$$ (e.g. one as above), there exists a subset $$K\subset {\mathbb R}^n$$ homeomorphic to $$S^1$$ and containing $$C$$. Moreover, in the construction $$K$$ can be chosen such that each component of $$K\setminus C$$ is a smooth (open) arc in $${\mathbb R}^n$$. (There are at most countably many of these components.)

Lemma. $$\pi_1({\mathbb R}^n \setminus K)\ne 1$$.

Proof. Since $$\pi_1({\mathbb R}^n \setminus C)\ne 1$$, there is a smooth loop $$c: S^1\to {\mathbb R}^n \setminus C$$ which is not null-homotopic in $${\mathbb R}^n \setminus C$$. A priori, the image of $$c$$ intersects $$A=K \setminus C$$. However, by perturbing $$c$$ to be transversal to $$A$$, since $$n\ge 2$$, we can find a loop $$c_1$$ homotopic to $$c$$ in $$R^n \setminus K$$ and whose image is disjoint from $$A$$. Hence, $$c_1$$ is a homotopically nontrivial loop in $${\mathbb R}^n \setminus K$$ (since it is already homotopically nontrivial in the complement to $$C$$). qed

On the other hand, if $$T\subset {\mathbb R}^n$$, $$n\ge 4$$, is a trivial knot, $$\pi_1({\mathbb R}^n\setminus T)=1$$. Thus, the knot $$K$$ constructed above nontrivial. qed

What Andrew's answer proves that every tame 1-dimensional knot in $${\mathbb R}^4$$ (and, more generally, $${\mathbb R}^n, n\ge 4$$) is trivial, i.e. is ambient isotopic to a round circle. The knot constructed in my proof is wild.

In particular, step 1 in Andrew's answer breaks down for general knots.

On the positive side:

Theorem. If $$K\subset {\mathbb R}^3$$ is a topological knot, then $$K$$ is unknotted in $${\mathbb R}^4$$.

Proof. The key is to prove that $$K$$ is tame (or locally flat) as a knot in $${\mathbb R}^4$$, i.e. for every $$x\in K$$ there is a neighborhood $$U$$ of $$x$$ in $${\mathbb R}^4$$ and a homeomorphism $$U\to V\subset {\mathbb R}^4$$ sending $$U\cap K$$ to a subset of a straight line $$L\subset {\mathbb R}^4$$.

Once it is done, it follows from the main theorem of

H. Gluck, Unknotting $$S^ 1$$ in $$S^ 4$$, Bull. Am. Math. Soc. 69, 91-94 (1963). ZBL0108.36503.

that $$K$$ is unknotted in $${\mathbb R}^4$$.

Let $$f: K\to [0,1]$$ with exactly two point of minimum and maximum, $$f(a)=0, f(b)=1$$. Let $$\alpha, \beta$$ denote the closures of the components of $$K\setminus \{a, b\}$$. The function $$f$$ defines an embedding $$K\to {\mathbb R}^4$$, $$\phi: x\mapsto (x, f(x)), x\in K,$$ whose image is a knot $$K'$$ in $${\mathbb R}^4$$. Let $$a':= \phi(a)=a, b':= \phi(b)$$.

Step 1. $$K'$$ is ambient-isotopic to $$K$$ in $${\mathbb R}^4$$.

First, extend $$f$$ to a continuous function $$f: {\mathbb R}^3\to [0,1]$$. For each $$t\in [0,1]$$ define the "shear" $$h_t(x, s)= (x, s+tf(x)), x\in {\mathbb R}^3, s\in {\mathbb R}$$ which is clearly a homeomorphism. This defines an ambient isotopy of the knot $$K$$ to the knot $$K'=f_1(K)$$ in $${\mathbb R}^4$$.

Step 2. I will now prove a slightly easier result than unknottedness of $$K'$$ in $${\mathbb R}^4$$, I will show that the arc $$\alpha'=\phi(\alpha)$$ (and, similarly, $$\beta'=\phi(\beta))$$ is tame, equivalently, unknotted in $$R^4$$. Note that $$\alpha'$$ intersects each horizontal slice $${\mathbb R}^3 \times \{s\}\subset {\mathbb R}^4$$ in exactly one point, namely, $$(f^{-1}(s), s)$$. The straight-line segment $$I':=a'b'$$ intersects each slice $${\mathbb R}^3 \times \{s\}\subset {\mathbb R}^4$$ also at exactly one point, $$p_s$$. Thus, the shear $$h^\alpha: (x, s) \mapsto (x + (p_s- f^{-1}(s)), s), s\in [0,1]$$ is a homeomorphism $${\mathbb R}^3\times [0,1]\to {\mathbb R}^3\times [0,1]$$ sending $$\alpha'$$ to $$I'$$; this homeomorphism is the identity on the boundary of the slab $${\mathbb R}^3\times [0,1]$$. Hence, $$h$$ extends to a homeomorphism $${\mathbb R}^4\to{\mathbb R}^4$$ (by the identity outside of the slab). Linear interpolation again yields an isotopy $$h^\alpha_t, t\in [0,1]$$, of $$h$$ to the identity map. Hence, the arc $$\alpha'$$ is unknotted in $${\mathbb R}^4$$. Note that $$\{h^\alpha_t(\alpha'): t\in [0,1]\}$$ lies inside the convex hull of the arc $$\alpha'$$ in $${\mathbb R}^4$$.

The same construction applies to the arc $$\beta'$$ resulting in an isotopy $$h^\beta$$. The trouble is that we cannot combine these "straightening" maps of $$\alpha', \beta'$$. We obtain, however, that the knot $$K'$$ is tame in $${\mathbb R}^4$$, except possibly at the points $$a', b'$$. More precisely, $$K'$$ is the concatenation of two tame arcs, $$\alpha', \beta'$$. Since $$K$$ is ambient-isotopic to $$K'$$ in $${\mathbb R}^4$$, it follows that $$K$$ was tame in $${\mathbb R}^4$$ as well, except possibly at the points $$a, b$$. But the points $$a, b$$ in $$K$$ were chosen arbitrarily. Hence, $$K\subset {\mathbb R}^4$$ locally flat at every point, i.e. is tame.

• It is not easy for me to check all your claims at the moment, but, assuming they are correct, thanks for the general answer! FYI, Andrew admits in the comments of his answer that he assumed a tame knot. Feb 3, 2021 at 16:22