So we are going to talk about floating bodies of convex compact N-dimensional sets.
N, the dimension, can be anything: 2, 3, 570426;
whatever. But for the purpose of today's talk
let's just stick with N equals two. That is, let's just look at such sets in the plane.
So let's look at some examples. For instance, the circle, the square or
the triangle. So those are convex, compact sets in the plane.
So what does that mean?
Compact means that the set is bounded. Bounded
means that there is no direction where you can wander off to infinity.
So for instance if we looked at something like so, right, and we'd be allowed to continue here
then we could march off to infinity here and that is not allowed. Now. What's the other ingredient that we have? It's convex,
so what does that mean?
It means that whenever we take two points in the set in question, like for instance here in the circle,
and we look at the line segment that joins the two, then we stay completely in the set. The same here,
we'll take two points, we look at the line segment that joins them,
we stay completely in the set. Same here; take two points that are in the set, we look at the line segment,
we'll stay completely in the set. So for instance what would not be allowed is something like so,
right?
Here if we take that point and that point
and we look at the line segment that joins the two then we are outside the set. So for such
convex compact sets, let's call them K, we'll look at the floating bodies.
So let's take K convex compact,
so everything that I am talking to you guys about can be done in any dimension, as I said.
But let's stay in the plane,
so we'll have here our K convex compact and then we'll also give ourselves a parameter, 𝛿.
So that's just a number, typically think of 𝛿 being small. And then we'll take a line. And if we were in
dimension three, instead of a line we would take a plane. And if we were in higher dimension we
would take what is called a hyperplane, but we'll stay in the plane
so we'll take a line L,
and then we'll take a line such that it cuts off a set of area 𝛿 from K.
So then we'll look at the intersection of this line with
the body K,
so we look at this intersection, and then we'll take the midpoint of this intersection. And then we'll take another line,
L, that cuts off 𝛿. We'll look at the
intersection and we'll take the midpoint. And we'll keep doing this.
So we'll look at all those lines, look at their intersection, take the midpoint and
then we'll look at - let's draw it in colour.
We'll look at the - we trace out the curve of these midpoints, and in that way we obtain
a set inside K. So this red set is the floating body,
let's give it a name. The floating body of K, let's call it K𝛿 bar.
So why's it called floating body? It's called floating body by the
Archimedean principle. We know that an object floating in water,
so if you have an object and here is the water surface, so an object floating in water
has always a set of equal volume above the water surface. - (Brady: These lines that you drew, each time)
(you were snipping off, like, a 𝛿. How many of these lines do you draw?) - Infinitely many. All
possible lines that we can draw that chop off 𝛿.
(Brady: My first instinct, without having seen this done, would be that all you do is make like a miniature version of the bigger set.)
(But that's not what happens?)
No, that's not what happens and we will see that when we look at, when we compute actually those for the various examples.
Then we will see that it's not what's happening. - (Brady: Let's do it!) - So for the circle, that's easy. So here we go. Let's
say the circle has radius 1. Then what do we have to do?
We'll have to look at all lines L that chop off a set of area 𝛿, and we'll have to take the midpoint of the
intersection. So because the circle has all the symmetries what will end up as our floating body is again a circle.
So we'll end up with this circle here. Which has now radius,
of course smaller than 1, and one can compute it.
It's actually 1 minus 1/2(3 over 2) to the 2 over 3(𝛿) to the 2 over 3.
So that's the radius of this floating circle.
Let's look at our other example,
which was the square. And
let's put it in such a way that it's centred at 0 and has side length 2,
so here is 1 and
here is 1. So if we want to get the equation of the floating body, or the boundary of the floating body,
let's put the x axis here, the y axis here, and then we can show that
actually it's enough to compute the shape of the boundary of the floating body just in the first quadrant here; by
symmetry we'll get the other ones. And we can show, ok its shape is something like so. And
then by symmetry we can complete the picture
so the floating body of the square will look like so. And we can write down,
let's write down the equation for this part in the first quadrant. And then this is simply 𝛿 over 2 is equal to (1 minus x)(1 minus y).
(Brady: That's not a circle in there?)
No, it's not a circle in there. And - but - what we also see is that it's not like, you know,
the original shape we started with. It actually looks different.
Let's look at the last one of our examples that we put up there, which is the triangle,
We draw the floating body, then we will get something like so.
So that is the floating body of the triangle.
So that looks very strange, now very different from the other ones. In particular the floating body of the circle and the square,
they were giving us again convex objects whereas this is certainly not convex at all. And that's actually
something we are not too happy about and we want to change that. So, how do we come upon
this strange shape for the triangle? Well
we remember what we have to do to get our floating body is this operation, that cutting of
sets of area 𝛿 with our lines, and taking the midpoints.
So for instance if we took this line we would get that midpoint,
and then if we moved with our line, like so, always cutting off 𝛿 and
taking the midpoints, then we would be moving along
this curve here and we would end up - well now my picture is not super good.
We would end up this line giving us, as a midpoint, that point here. And then if we kept moving
then the line would go down here and up there and that then means that we are actually moving backwards
along that curve here till we hit this point where the - it;
this point is the midpoint of the line
that chops off 𝛿 and is parallel to the site of the triangle. And then we go here and move up there.
So as I said, here we'll get something that is not convex. And that is not
what we like, because we want to use these floating bodies as a particular tool.
(Brady: So, Professor, it's not like it's a mistake?)
No, no, it's not a mistake. It's just so happens that,
yeah, that comes out using this definition where we take those midpoints.
(Brady: So the fact that the floating body is not convex is fine, it's just not good for you?) - It's just not good for our purposes,
for the purpose we want to use the flo- floating body for, which is to investigate the boundary structure of convex sets.
But it's perfectly fine, perfectly okay. - (Brady: Is it easy to predict what the floating body will look like?)
It's not always so easy to predict what the floating body will look like.
One can get an idea how it might look like,
what is in general difficult is actually to compute the equation of its boundary, to compute that exactly this is difficult;
it can be done, like for this example here in the plane - and it can be done for most examples in the plane,
but typically we are interested not just in, you know, dimension 2 or 3
but in really very high dimensions. And there it is difficult to exactly compute it.
But at least when the 𝛿 is small we can pretty much
determine how it looks like, very well. There was nothing wrong a priori with this,
you know, floating body for the triangle, using the Dupin
definition, but for our purposes it's not good. So we
want - because we do want to stay in the realm of convex bodies.
And what we'll do is to get something that we then call the convex floating body.
So this was brought up around 1990 and it goes as follows. Again, let's just look at the two dimensional situation,
so we have again our
convex compact set K.
And again
we'll take lines L that chop off 𝛿. And now we will take all such lines that chop off 𝛿,
that is we take away 𝛿-
(Brady: Same as last time.) - Same as last time but we don't bother about taking any midpoints,
we just chop off 𝛿 and then we'll look at what is left
after we have chopped off 𝛿. And what is left is a convex set and that is the convex floating body,
let's give it a name, let's call it K𝛿. - (Brady: How have you drawn that line if you haven't got midpoints to guide you?)
It's just you chop off and look at what's remaining. Like for instance, if you
let yourself guide again by
you know an
object floating in water, then we know the water
surface, what's above the water surface, is always something of equal volume,
then if you rotate this thing around
then the same
volume, it would be the same equal volume above the water surface. And
the floating body is just that part that remains always under water.
(Brady: It never got to breathe?)
So it never got to breathe or it's
something that always stays wet, if you want. So that what we do here will construct this wet part.
(Brady: So this part, the new floating body, is basically all of the points that never got a chance to be in 𝛿?)
Yes, so to speak, yeah. Never got a chance to be beyond there.
Now if we want to compare this floating body, this convex floating body, with the one that was given to us by Dupin
then we see that if
this Dupin
floating body,
K𝛿 bar we called it, is convex then
actually, the two coincide. Then K𝛿 is just equal to this K𝛿 bar. So for this example of the
circle, we would just get the same circle. For the example of the square we'll just get the same
object like so. And what would we get for the triangle? Well for the triangle, we don't have that this Dupin
floating body is convex, but this convex floating body is convex and what we'll get is simply these things here
without these strange parts. So that would be our
convex floating body for the triangle. - (Brady: And for every possible)
(convex compact set-) - Mm-hmm
- (my floating body will always turn out convex? I can't construct a shape that won't give me a-) - Correct, yeah. Mm-hmm, yeah.
(Brady: Good, I'm with you. Sounds fun!)
Good, yeah. It is!
So there's a convex floating body, but you might be thinking, well what's the point of that?
What does a mathematician use this for? The answer to that is in a second video, links are on the screen,
description - all the usual places.
Thank you for watching.
[Preview] ...this area
because we'll take - so this is K𝛿, area of the big one minus area of the small one,
so what remaining is this one. - (Brady: I call this the crust.) - We take the crust, good. I like that!