My high school physics teacher joked that a topologist couldn’t tell his coffee cup from his donut. To this extent I understand topology. But I remember hearing that one can mathematically prove the following:
Given two sheets of paper of identical size, one on top of the other, take the top sheet, crumple it to whatever size/shape and place it anywhere on the bottom sheet. It is guaranteed that at least one piece of the top paper is over the same spot on the bottom that it was before crumpling.
Observe a still cup of coffee. Stir without splashing (which would change the topology of the surface - I get that) at whatever rate for however long using whatever motion. Wait for the motion to still. When it does, it is guaranteed that some part of the coffee will return to where it was before the stirring started.
I just can’t penetrate on what would these proofs be based. Is there a way for a mere mortal to grok this? Are there other cool findings of topology?
Okay, topology is the study of continuous functions. What do we mean by continuous? There’s been a lot of work put in to boil it down to this: start with sets, and define what it means for points to be “close”. Continuous functions are maps between sets that “preserve closeness”.
Yeah, that’s a bit fuzzy. You don’t want to see the real version right now. Luckily, this is good enough for us.
Both of your examples are actually versions of the same theorem in two and three dimensions. First, background.
Imagine drawing out a closed loop in the set. For the moment, think of a region of the plane and our usual inuitive idea of “closeness” of points. Now, there are many different loops you could draw, but let’s say that if you can deform one loop smoothly into another without leaving the set that they’re essentially the same. If the region is the inside of a circle, say, then all loops can be shrunk down to a point. If there is a single hole in the region, then there’s one different loop for each integer: the number of times the loop wraps clockwise around the hole.
Now, if I think of a continuous map from one region into another, then loops in the first set get carried along into loops in the other. Continuous functions on regions give rise to functions on the sets of loops. Further, if you apply one continuous map followed by another, the resulting map on loops is the composition of the two resulting maps. This property is called “functoriality”, and it’s the backbone of a lot of modern mathematics.
Anyhow, think of a circle in the plane and its interior. There’s obviously a continuous map from the circle to its interior (each point just goes to itself. subsets are continuous functions), but can there be one from the interior to the border that leaves the border points where they are? Well, if there were then the map sending the border to the circle and back to the border would just leave the border points as they were, so the function it induces on loops is just the identity. But the function from loops on the border to loops in the interior sends all integers to a single element, and the function our proposed map would induce sends that single element back to only one integer, so the composition of the loop maps can’t be the identity. Thus, no such function from the filled circle to its boundary exists. Incidentally, the fact that it was a circle doesn’t matter – any region without holes would behave the same way.
Now, imagine crumpling up a sheet of paper and setting it on another. This defines a function sending a points on one sheet of paper to the point below it on the other. Further, if you move just a little on the crumpled sheet (following the crumples) you’ll move just a little on its image. So we have a continuous function from the sheet to itself. The question is: can we do this in a way that the function never sends a point to itself? If we can, then for each point x in the sheet we can draw a line from its image f(x) through x and on to the edge of the sheet. This defines a continuous function g(x) sending each point in the sheet to a point on the edge, and g sends points on the edge to themselves. But we just established that no such function exists, so every map from the sheet into itself must have a fixed point.
The coffee example is pretty much the same, but in three dimensions and using little spheres rather than loops in the space.
In addition to the examples you give, there’s also the “hairy billiard ball theorem”: imagine a ball with hair sticking out of it at every point. It’s impossible to comb the ball smooth; there will always be at least one hair sticking straight out.
Ding… the light bulb starts to glow… just enough for me to see I was about to take a sip from my donut. I look forward to re-reading this post, as well as the links everyone has provided.
To Thudlow Boink: That explains my perpetual bad hair day. I’m sure there are great applications of this for salons and troll-doll manufactuers!
I prefer the definition that “A topologist is someone who can’t tell his ass from a hole in the ground, but can tell his ass from two holes in the ground”.
As a side-note on the Hairy Ball Problem, it depends on the number of dimensions. If your surface is even-dimensional (as, for instance, the 2-d surface of a sphere), then you can’t comb it, but if it’s odd-dimensional (as, for instance, a circle or a hypersphere) then you can. It’s easy to see this for S[sub]1[/sub], the circle: You just comb every hair clockwise, or every hair counterclockwise. The combings for higher dimensions are a bit more complicated, but (for odd dimensions) always possible.
On the fixed-point theorems, the one-dimensional cases are easier to understand. Suppose instead of crumpling up pieces of paper, we take a piece of string and arrange it into a swirl on the surface of a table. At least one point on the string will be directly above where it started off. In this case, you can show this using the Intermediate Value Theorem: Roughly, some parts of the string are left of where they started out, and some parts are right of where they started out. So there’s some point on the string in between those points that’s neither right nor left of where it started out.
Consider the portion of the flat sheet that lies directly below the crumpled sheet (the ‘shadow’ of the crumpled sheet, so to speak). The equivalent portion of the crumpled sheet must lie entirely above this flat portion. Then consider the ‘shadow’ of the smaller crumpled portion… and so on. It’s an infinite regress, proving there must be a point which coincides on both sheets.
This example really bothers me. I guess there is an underlying assumption here that if you stir “without splashing” that coffee molecules that started out next to each other will stay next to each other… otherwise this would not represent a continuous function? In the real world, it seems like this would not be the case, or else your milk and sugar would never dissolve, right?
Oh wait, the coffee isn’t made of molecules, is it? It’s “mathematically ideal coffee,” a continuous substance without discreet elements. Ok, now I get it. I guess the example isn’t meant to apply to the real world, right?
The crumpled sheet of paper is a bit like the mountain climbing puzzle;
You set of at 9AM on Saturday and begin climbing a mountain; you reach the summit at 4PM and camp there overnight.
The following day, you set off to descend, but you don’t leave until 10AM. Because it’s nearly all downhill though, you get back to base camp by 3PM.
There is one point on the mountain that you were at the same time on both days; there has to be. To visualise it, change it into two people; one ascending and one descending, but on the same day and that point is where they bump into each other.
So imagine the two sheets of paper as being identically coloured with a smooth gradient, fading from blue to green along the long axis and from (the point gradient colour) to white along the short axis; crumple the paper and place it in any orientation on the uncrumpled sheet; no matter how the blu>green gradient lies on the other blue>green gradient, there will be a point where the two colours are the same; no matter where the colour>white gradient lies on the other, there will be a point where the two intensities are the same.
If it doesn’t converge to a point, then there’s an extra step to the problem: You then have to show that any isometric function from a 2-d region to itself contains at least one fixed point. Fortunately, once you get past the verbiage, this is fairly straightforward, since all such functions are rotations, and therefore have a pivot point.
A thread with a title like this has to drag me out of my usual lurking.
Fixed-point theorems are fun. Here’s another one: Assuming that temperature and air pressure vary continuously over the globe, there must be two antipodal points on the earth with exactly the same temperature and exactly the same pressure.
Then there’s the Ham Sandwich Theorem: Given any three finite volumes in space (say, two pieces of bread and a slice of ham), there is a plane cutting all three exactly in half. (I.e., no matter how sloppy you are, there is still a way to cut your sandwich exactly in half. Who says topology isn’t practical?)
I’ve long been fascinated with Topology–ever since I was introduced to it in Junior High math class. You can imagine my delight in Syntax class when its application to that subject became evident for me.
No, the centroids are the points of vanishing first moment; a plane through the centroid does not partition a generic volume into two equal subvolumes.
Perhaps I used the wrong term. Doesn’t every volume have some point, such that any plane through that point divides the volume into two equal volumes? Of course, I suppose that the existance of such a point is itself a topological result, but it seems like it’d be a much more interesting one than the special case presented by the Ham Sandwich Theorem.
> I’ve long been fascinated with Topology–ever since I was introduced to it in
> Junior High math class. You can imagine my delight in Syntax class when its
> application to that subject became evident for me.
No. Think of an equilateral-triangular flat plate, for example (work in two dimensions because it’s easier to describe). The equipartitioning line perpendicular to each side is the one through the other vertex, so if there is such a point it must be at the intersection of these lines: that is, at the centroid=incenter=circumcenter of the triangle. But the equipartitioning line parallel to each side must cut the triangle 1/sqrt(2) of the way from vertex to base (so that the area of the cut triangle is half the total), and the centroid is 2/3 of the way down.