At around the one minute mark of this video by Professor Tadashi Tokieda, (See Link: https://www.youtube.com/watch?v=CJBfpvWBmSs&list=PLTBqohhFNBE_09L0i-lf3fYXF5woAbrzJ&index=13)

Professor Tokieda makes the claim that it is only in $\mathbb{R}^n$ where $n>5$ that one can unknot a knotted torus. He claims that this is because a torus is of dimension 2, and the surface spanned by a moving torus is of dimension 3. Hence, it is only in $\mathbb{R}^n$ where $n>5$ that this torus can be unknotted.

I have quite a few difficulties when trying to understand this claim. Intuitively, if every coordinate of the torus is given by $(x_1,x_2,x_3)$ then simply by adding one more degree of freedom, $(x_1,x_2,x_3,x_4)$ for example, one can prevent any two arms of the torus from intersecting and thus the torus can be unknotted in $\mathbb{R}^4$.

Is there something that I am missing? Could someone help clarify my misgivings here.

  • $\begingroup$ Apparently the formula that lead to $2+2.m<-1$ was tracted in a previous video? $\endgroup$ Nov 4, 2017 at 8:00
  • $\begingroup$ Yes, it is given in a previous video- but he gave a very loose heuristic based proof borrowed from linear algebra. The proof (silently) assumes that each manifold is a vector subspace but I do not think that a torus($T^2) is a vector subspace. $\endgroup$
    – Kurosu
    Nov 4, 2017 at 8:04
  • 3
    $\begingroup$ Hm. It seems the spcific knot given in the lecture can be unknotted in $\Bbb R^4$. Imagine the fourth dimension as colour, and parts of different coulour can pass through each other. You can make a small part "red" and the rest "blue" and pass through $\endgroup$ Nov 4, 2017 at 8:11
  • $\begingroup$ I think, what he means that the following are equivalent: (1) $k\ge 5$ and (2) every two surface knots in $R^k$ are isotopic. $\endgroup$ Nov 4, 2017 at 18:11

2 Answers 2


In order to un-knot the toroidal knot presented in the video, we essentially need to make the transformation from the left to the right part of the following immage:enter image description here

Once we have managed this, it is merely a matter of unwinding the remaining twist, and that is obviously possible.

Note that the above are only two-dimensional images of a three-dimensional situation. However, illustration "tricks" such as lighting, shadows, and occlusion were used in its production such that you can imagine and mentally visualize the full 3d scene. To add a fourth dimension, we need yet another trick: We assign a real number to each point of the surface and identify it with a wavelength in the visible spectrum. In other words, we use colours to represent the fourth dimension. So in a first step, we embed our knot into 4d space by putting it into the "red hyperplane":

enter image description here

Next, we deform it so that a part of our choice moves to the "blue hyperplane". Continuity dictates that the colour changes smoothly from red via orange, yellow, and green to blue: enter image description here

Now the blue part can be moved past the red part, even though it looks as if it passes through it. Any points that occupy the same 3d coordinates during that move would only occupy the same 4d coordinates if they had the same colour, but that doesn't happen: enter image description here


  • $\begingroup$ Although you used the specific know from the video, the argument seems to apply to any knot. Any knot can be untied if you're allowed to pass one piece through another (changing an overcrossing to an undercrossing), and you've shown how to do that when a fourth dimension is available. $\endgroup$ Nov 4, 2017 at 15:16
  • $\begingroup$ I definitely agree with you, (wonderful pictures by the way) but it doesnt quite answer my question. Professor Tokieda consistently uses the same argument (the one that argues for the necessity of 6 dimensions to unknot the torus) in his lectures- both the one i linked and later ones. i'm concerned he may be correct and our intuition and explanation for four dimensions may be wrong or perhaps not what he had in mind $\endgroup$
    – Kurosu
    Nov 4, 2017 at 16:34

If Tokieda said that tori do not unknot in $R^5$, he was wrong.

Theorem 1. If $M^k$ is a smooth $k$-dimensional manifold (possibly disconnected) then any two smooth embeddings $M^k\to R^{2k++2}$ are isotopic. (This is due to Whitney, if I remember it correctly, with a proof similar to the proof of Whitney embedding theorem.) In particular, there are no nontrivial surface links in $R^6$.

Theorem 2. If $M^k$ is a connected $k$-manifold, $k\ge 2$, then any two smooth embeddings $M^k\to R^{2k+1}$ are isotopic. (Wu, 1958) In particular, every 2-torus unknots in $R^5$, contrary to the cited claim.

Theorem 3. There are nontrivial 2-component smooth links in $R^5$ whose components are 2-tori. (This is a variation on the Hopf link construction.)

Theorem 4. There are smooth knotted 2-dimensional tori in $R^4$. (Hudson, 1963)

You can read more in this survey by Skopenkov:

Embeddings and knottings of manifolds in Euclidean spaces.


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