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Stream: theory: algebraic topology

Topic: Homotopy question

JR (Jun 20 2025 at 15:05):

Glancing at Bartosz's blog post referenced here

If we want to allow for extreme shrinkage, as with a torus shrinking down to a circle, we have to relax the identity conditions.

Drawing a torus in 3-space, I can easily visualize how to shrink a torus to a circle by shrinking along one factor, but not the other. Am I alone in my limited spatial intuition, or is there a topological obstruction at play here?

John Baez (Jun 20 2025 at 16:38):

The solid torus is not homeomorphic to a circle, but it sounds like you're correctly imagining a continuous map from the solid torus to a circle which is a homotopy equivalence.

Kevin Carlson (Jun 20 2025 at 17:27):

Or, alternatively, a homotopy from the standard embedding $S^1\times S^1\to \mathbb{R}^3$ to a map which is constant on one factor. Since $\mathbb R^3$ is homotopy-theoretically trivial ("contractible"), any two maps into it from any domain are homotopic, so you have to think about homotopies that don't move a space around in $\mathbb R^3$ to start to get at the intrinsic topology.

Kevin Carlson (Jun 20 2025 at 17:27):

More simply, you can certainly imagine continuously contracting a circle to a point in the plane, but you can't imagine doing so without moving any of its points off the original circle.

JR (Jun 20 2025 at 19:35):

Right, but I'm saying that a homotopy to a circle is easier to visualize along one factor than the other. Is there some invariant (or more generally, reason) that captures this?

Mike Shulman (Jun 20 2025 at 20:06):

Are you talking about a solid torus or the surface of a torus? A solid torus is not a product space, and is homotopy-equivalent to a circle, but only in one "direction". A surface torus is a product of two circles, and as such admits two projections to a circle, but neither is a homotopy equivalence. In the latter case, one of the projections is indeed easier to visualize once you fix an embedding of the torus in 3-space; the other is abstractly equivalent but looks more natural if you embed it 'inside-out' in a different 3-space.

John Baez (Jun 20 2025 at 20:33):

Mike Shulman said:

Are you talking about a solid torus or the surface of a torus?

Bartosz's blog article is trying to explain homotopy equivalence and how it's more general than homeomorphism, so when he says "torus" he must mean "solid torus" - since only a solid torus, not a torus, is homotopy equivalent to a circle. He's not a topologist, and this is a hand-wavy introduction to a more carefully written post:

In topology, we say that two shapes are the same if there is a homeomorphism– an invertible continuous map– between them. Continuity means that nothing is broken and nothing is glued together. This is how we can turn a coffee cup into a torus. A homeomorphism, however, won’t let us shrink a torus to a circle. So if we are only interested in how many holes the shapes have, we have to relax our notion of equivalence.

He's taken the old "coffee cup and doughnut" joke and made it less accurate by replacing "doughnut" by the more technical sounding but less technically correct term "torus".

Mike Shulman (Jun 20 2025 at 21:24):

That makes sense. But I wonder if Bartosz's use of "torus" for "solid torus" is what has confused JR, since JR is talking about the "two factors" of a torus which doesn't make sense in the solid case.

John Baez (Jun 20 2025 at 21:32):

Yes, possibly.

Oscar Cunningham (Jun 20 2025 at 21:35):

Mike Shulman said:

A solid torus is not a product space

Isn't the solid torus the product of a circle and a disk? Sorry I don't mean to nitpick, I just want to check that I'm not losing my mind.

Mike Shulman (Jun 20 2025 at 21:37):

Haha, yes, you're right. I should have said, it's not a product of two circles.

Mike Shulman (Jun 20 2025 at 21:37):

(Of course, if we really want to get into it, every space "is a product space", namely of itself and a point...)

Oscar Cunningham (Jun 20 2025 at 21:38):

Thanks, that makes sense.

Joe Moeller (Jun 21 2025 at 13:25):

I thought the question was more about the fact that it’s easy to imagine contracting a torus (the proper one, hollow) to a circle which preserves the donut-hole circle, but much harder to imagine contracting to the inner-tube circle. This seems to imply a bias in the $\mathbb R^3$ embedding that shouldn’t be intrinsic to the torus. This reminds me of the fact that I think in a projective space you can deform an embedding to itself which swaps the two circle factors. You can visualize this as increasing the radius of the innertube radius to infinity and then it like wraps around to becoming the donut hole. I would be interested in hearing more math added to my descriptions. Also I believe there’s an animation of this torus flipping, which it would be cool if someone found.

JR (Jun 21 2025 at 15:22):

Mike Shulman said:

That makes sense. But I wonder if Bartosz's use of "torus" for "solid torus" is what has confused JR, since JR is talking about the "two factors" of a torus which doesn't make sense in the solid case.

I'm not confused about the solidity. I am wondering precisely why the embedding of a vanilla 2-torus relates to the visualization being inequivalent with respect to factors in the way that it is, and I can't put my finger on the reason.

John Baez (Jun 21 2025 at 15:38):

Maybe this is what you're thinking about:

The standard embedding of $S^1 \times S^1$ in $\mathbb{R}^3$ has an asymmetry between the two factors that goes away when we take the one-point compactification of $\mathbb{R}^3$ and get $S^3$. Then we get an embedding of $S^1 \times S^1$ in $S^3$ such that the complement $S^3 - (S^1 \times S^1)$ is the disjoint union of two solid tori. One of them should be thought of as $S^1 \times D^2$, and the other $D^2 \times S^1$.

John Baez (Jun 21 2025 at 15:41):

We can move the torus $S^1 \times S^1$ around in $S^3$ in such a way that the arbitrariness of the "inside" and the "outside" becomes clear:

.

Oscar Cunningham (Jun 23 2025 at 08:28):

You can also see what the torus would look like if you were to pause that video just at the halfway point, so that it symmetrically divides the space in two:
image.png
(The torus passes through the point at infinity.)

John Baez (Jun 23 2025 at 13:12):

Nice!

Another way to think about it: if we describe a 3-sphere in $\mathbb{R}^4$ by

$x^2 + y^2 + z^2 + t^2 = 2 \qquad (x,y,z,t) \in \mathbb{R}^4$

then we get a nice torus in the 3-sphere from the equations

$x^2 + y^2 = 1 \text{ and } z^2 + t^2 = 1$

The point $(x,y)$ can be any point on the unit circle in the $x,y$ plane, and $(z,t)$ can be any point on the unit circle in the $z,t$ plane!

JR (Jun 25 2025 at 12:59):

John Baez said:

Maybe this is what you're thinking about:

The standard embedding of $S^1 \times S^1$ in $\mathbb{R}^3$ has an asymmetry between the two factors that goes away when we take the one-point compactification of $\mathbb{R}^3$ and get $S^3$. Then we get an embedding of $S^1 \times S^1$ in $S^3$ such that the complement $S^3 - (S^1 \times S^1)$ is the disjoint union of two solid tori. One of them should be thought of as $S^1 \times D^2$, and the other $D^2 \times S^1$.

Exactly what I was looking for, thanks! I am curious if you know of references that discuss this in more detail, or if there is a surgery-theoretical way to look at all this--just because I'm seeing products of spheres and disks in two different ways.

John Baez (Jun 25 2025 at 13:55):

Later I gave a specific formula for embedding a torus in $S^3$ whose complement consists of two solid tori as discussed. This torus is actually flat in the sense of Riemannian geometry! It's called a Clifford torus, and you can read more here:

• Wikipedia, Clifford torus.

This business about $S^1 \times S^1$ being a boundary in two different ways is indeed used in surgery: we can chop out an $S^1 \times D^2$ in a 3-manifold and sew in a $D^2 \times S^1$, and this is just a special case of something that works in any dimension. You also see it a lot in Morse theory. There's a lot to read about both those subjects.

John Baez (Jun 25 2025 at 14:07):

The thing that works in any dimension is that $S^n \times S^m$ is the boundary of both $D^{n+1} \times S^m$ and $S^n \times D^{m+1}$, and if you glue $D^{n+1} \times S^m$ and $S^n \times D^{m+1}$ together along this common boundary you get a copy of $S^{n+m+1}$. There are formulas for this that are completely analogous to the formulas I gave in the case $n = m = 1$.