One of my favorite unsolved problems in mathematics. Maybe it's best to start with
Euler's formula. Euler's formula tells something about triangulated spheres.
So maybe I have to tell you what a triangulated sphere is.
Well, intuitively it is a sphere but it is made
combinatorial, so that every face is a triangle.
So you want to attach bunch of triangles in such a way that the entire shape is
essentially identical to a sphere. You can continuously move it to form it, make it round. So maybe the
easiest
example is a simplex.
So a simplex has four vertices 1, 2,
3, 4. And if you just look at the boundary of it, the tetrahedron, then it is essentially a sphere. You can
continuously deform it to a sphere. Every face of it is a triangle.
In fact, you see 4 of them; 1, 2, 3 and 4. So, as you can imagine, there are many many different ways to
triangulate the sphere. Another example, maybe it's the second easiest example is the boundary of an octahedron,
is essentially a sphere. And I can go on - maybe the third familiar example is the boundary of an
icosahedron, where I glue together 20 triangles to make a sphere.
So Euler's formula says something about triangulated spheres.
He noticed that if you stare at each one of these triangulated spheres,
say the 2-dimensional spheres, as, as we are considering now.
And if you first count the number of vertices, so if I do the tetrahedron
I see four vertices: 1, 2,
3, 4. And then I'm going to count the edges: 1, 2, 3, 4, 5, 6. And then I count the triangles,
there are 4 triangles in a tetrahedron, so I'm counting
0 dimensional faces, 1 dimensional faces, and 2 dimensional faces.
So let me call them f0 and
f1
and f2. And what Euler did was to compute the alternating sum,
f0 minus f1 plus f2. Which is, in our case, four minus six plus four,
which is equal to two.
So let me do the same computation for the octahedron. The alternating sum of
the number of faces of various dimension, then I get 6 minus 12 plus 8,
which is equal to 2. As you have guessed this is a general pattern.
No matter how you
triangulate your
two-dimensional sphere, the alternating sum of the number of faces is always equal to 2.
Okay, so that was Euler's observation. So let's just imagine how
this
wonderful, but seemingly accidental pattern will continue in higher dimensions. So you can certainly think about 3-dimensional spheres
instead of two-dimensional spheres. And certainly you can triangulate them and there are
infinitely many different ways of triangulating them. And you can do the same. Now you have
four numbers to think about instead of three; you have vertices, edges,
triangles,
and many many tetrahedrons.
So the triangulated spheres in the next dimension will be a way to make a sphere
by gluing together many of the tetrahedrons. The next dimension is always hard to imagine, so
before making the correct guess let's, let's go to even a lower dimension.
So that was the picture for two dimensional spheres. But let's this time imagine one dimensional spheres.
Triangulated one dimensional spheres. So triangles in one dimensions are supposed to be
just straight lines.
We know very well how typical triangulated one-dimensional spheres look like.
This is one triangulated 1-sphere where I used five vertices and five edges.
So my numbers will be f0 equals 5, and
f1 equals 5. And the Euler's relation,
in this case, will look like this. And again the number 0 that we see here is
independent of the triangulisation of the 1-dimensional sphere that I have chosen. For example,
if you have used 7 vertices
you are forced to use 7 edges too,
because you want the entire thing to be a 1-dimensional sphere, which is a circle.
So it - Euler's relation in one dimension is simply saying that f0 minus f1 is 0. So
let's go even lower.
(Brady: Even lower? Alright?)
Yes, and think about zero dimensional spheres. So this is a little bit tricky if you have never thought about zero dimensional spheres;
but what is really a sphere? A sphere is supposed to be the,
something that lives in our space
which consists of all points that are of the same distance from the
origin, or whatever point that you have chosen.
So the zero dimensional sphere
should live in a 1-dimensional space, which is the line,
and if I collect all points that are of, say distance 1 from the origin, then I would get exactly two points. Zero
dimensional sphere is two points And that's, well, in my sense
that's a triangulated sphere. And there is really one number, f0, the number of vertices. And that number of vertices should be always 2.
So in this case the Euler's formula will look like at zero equals 2. - (Brady: Oh it's gone up again-)
That's right. So I add 2, in dimension zero,
zero in dimension 1, and
two in dimension 2. So we can imagine what would be the next number in, say in dimension 3.
(Brady: What is it gonna- should I guess?)
Yeah, you should guess.
(Brady: I'm gonn- is this gonna oscillate? Is it gonna go back to zero?) - Yes, that is a correct guess. And in fact,
although it's hard to imagine
3-dimensional spheres,
for this
particular problem we can actually test it for our favourite triangulated 3-spheres.
Because there is a higher dimensional analog of the tetrahedron, so there is called a simplex and
there is a simplex in any dimension.
So if you take four points in general position in 3-space,
then you get the tetrahedron. And analogously if you take five general points in 4-dimensional space
then you will get a simplex. And we can
kind of imagine how it would look like,
especially if we force it to live in our familiar 3-dimensional space. So it will have
five vertices and
because the five points, which lives in 4-space are in general position,
every pair of the vertices will be connected by an edge.
So that's how the set of edges of a 4-dimensional simplex will look like if I
draw it in this piece of paper. And
similarly every triple of points that you see in this picture
will form a triangle, because the five points in 4-space are in general position. And every four
collection of the five points will
form a tetrahedral face of
that simplex. So we can compute these numbers F:
so f0 is
5,
and f1
will be 5 choose 2, as one of the binomial coefficients, which is 10.
And f2 will be 5 choose 3,
next entry in Pascal's triangle. And f3 will be 5 choose 4,
which is 5. So, let's test our prediction. If you
start from 5,
subtract 10, add 10, subtract 5 - then you get zero.
(Brady: It alternated like we said!)
That's right. And you can imagine how it would work. And the amazing
fact is that
something like this happens for all triangulated spheres in any dimension. You just
see whether you're living in an even dimensional space or odd dimensional space, and
then you know that no matter how you triangulate a sphere, if you
perform this alternating sum
computation you get either 2 or 0. The number either 0 or 2, depending on the parity of the dimension, is called the Euler
Characteristic. It's the Euler Characteristic of the sphere and is independent of the triangulisation. So that's an amazing fact
which
took many many years for mathematicians to
rigorously justify;
but we know now as a fact that no matter how you triangulate a sphere you always get either 0 or 2.
That was great, but you may not like this
alternating pattern 0, 2, 0, 2, 0,2. But there is another way of
expressing this pattern that does not care about the parity of the dimension. And
this is our starting point. So let me express the
Euler characteristic
in
another way. So for example
let me start from our octahedron example. 6 vertices,
twelve edges, and eight triangles in the octahedron. And what I'm going to do next is
to attach a string of 1s, so as to make it a triangular shape. And
next I'm going to do
something like Pascal's
rule for constructing his triangle, but using subtractions instead of addition.
So what do I mean? So I see 1 and 6
here, so I'm going to subtract 1 from 6 and write down 5
below. And
I do the same. I see 1 and 5, I subtract 1 from 5 and write down 4 here.
And then I look at 5 and 12 and subtract 5 from 12. I have a 7 here. I
do the same, I get 3, I
get 3,
and I get 1.
This sequence,
which you get in the end of this process, is called the H vector of the triangulated sphere.
And as you see there's a,
some pattern here. It is palindromic,
it's 1 3 3 1, but if you read it from the other direction is again 1 3 3 1.
And this was the case for the octahedron. So let's try to see whether this was an accident or not.
So let's maybe try the icosahedron. 12 vertices, 30 edges,
20 triangles. I'll attach a string of 1s on the other side of
the triangle and I subtract
and complete this triangle. I get 10 here,
19. 9 here, I get 9 here,
get 1 here.
Again, you see the same pattern. If you keep the bottom row
it is palindromic: 1 9 9 1.
(Brady: Okay! This seems such an arbitrary game to have played.) - That's right.
But at least you can see what's going on
on
one of these numbers. So let me just focus on the last one.
So the first one served up, this is built-in,
I just wrote down one here. But the fact that one appeared in the last spot was a miracle, sort of.
But what was happening? So I
started with f0 and f1 and f2, and
how did I get 1 in here?
I subtract one from f0,
and I subtracted f0 minus 1 from f1,
and I subtracted this from that.
And
the assertion is that this is equal to 1.
So if I write the same thing again, it is what we have seen before.
The Euler's formula. At least for 2-dimensional spheres,
Euler's formula is part of a broader pattern, which says that you have this palindromic sequence at the bottom of this
strange Pascal-like triangle. Well one thing that
people have noticed is that if you view
Euler's formula as a part of this broader symmetry, then this
dependency on the parity of the dimension really goes away. So let's try to do a similar game in
one dimension up. Maybe we do the
simplex with 5 vertices in 4-space. And we already have computed the number of faces.
It had 5 vertices,
10 edges, 10 triangles, and 5
tetrahedrons. And again I add a long string of
1s
on the other side. And I do the subtraction. I subtract 1 from 5,
subtract 1 from 4, subtract 4 from 10, and continue.
And this is what I get: 1 1 1 1 1. There's a palindromic sequence of a very special kind.
That's just do, oh, maybe one or two more examples, just to make sure that the pattern we are seeing here is not an accident.
So maybe let's do a higher dimensional version of the octahedron. Number of vertices,
that's 8. How many edges? You need to be a little bit clever to actually compute this number, and the answer is
actually
24. So f2, if you count the number of triangles in
this higher-dimensional octahedron you get 32. And if you count the number of tetrahedrons
you get 16. So let's do the same game, this time a little bit faster. 1 1 1 1 1 1 and
7, 6,
17, 5,
11, 15, 4, 6, 4, 1. And that's my end result. In this sequence,
that you see, is what is known as the H vector of th- this particular triangulated sphere.
But if you think about at least very small part of the symmetry, which
predicts that the last entry in your H vector should be 1, then
again, it is the Euler's formula
for this 3-dimensional sphere.
Because think about how you got this 1, starting from the original data.
You have your f0, you have your f0 minus 1; and this number is
f1 minus f0 plus 1, this number is f2 minus f1 plus f0 minus 1, and
this number here is
f3 minus f2 plus f1
minus f0 plus 1. And
the symmetry of the H vector dictates that this should be 1.
I can write this again, in an equivalent form, saying that the alternating sum of the number of faces of various dimensions
is zero for this
triangulated 3-dimensional sphere.
We're almost there. So if you think of Euler's formula in any dimension as
a small part of this
larger symmetry that you see in the H vector, in the bottom row of this subtraction
Pascal triangle, then the dependence on the parity of the dimension goes away.
Euler's formula really just says that in any dimension
the last entry of this triangle that you see should be 1. Which is just a very tiny part of the
entire symmetry. It's highly non obvious, but in a certain sense it's a natural symmetry.
So the other numbers are also interesting, and they are numbers of some higher dimensional space;
which is not a sphere, but is constructed from a sphere.
And the fact that the sequence is palindromic, for example
this 4 here is
equal to the other 4 that you see, gives you another formula on the number of faces of various dimensions,
which is different from Euler's formula.
It is something like Euler's formula
but it is really new. But the fact that the H vector of a triangulated sphere is always palindromic, it was
discovered then
somewhat justified about hundred years ago by mathematicians Max Dehn and Duncan Sommerville.
They didn't do this. They had a
somewhat more complicated way of expressing this
symmetry of what we now call H vector,
but later
mathematicians have realised that this is the best way to understand. And I think this particular way of
viewing the symmetry is due to Richard Stanley, who is also now in this building. The
sequence that you started with, the sequence that you put on
one side of the triangle, which counts the number of faces of various dimension;
they are called the F vector because we usually call f0, f1, f2, f3, f4 the face.
(Brady: And so why- and where do we jump to H?) - Well, there is something called the G vector too. And
it is really essentially related to
the unsolved problem that I promised you to
introduce. So this is a fact. We know several different ways of justifying the fact that this H vector is palindromic
in
whatever dimension for whatever triangulated sphere that you start with - if you produce
H vector in terms of the number of bases then you always see this symmetry. But from all these examples that we have
executed so far, you see another pattern,
other than symmetry.
So, for example, think of this particular H vector,
14641. Another thing that's apparent in the example is that this sequence
increases
up to middle, 1 and
4 is larger, and 6 is even larger. Of course you see the same sequence going the other way.
So this pattern is called a unimodality, so there is a single
mode in the sequence, which should happen
exactly in the middle. So say in 17 dimensions,
if you pick your favourite triangulated sphere living in that dimension,
write down the sequence of H numbers - say
17 many of them. Then you will see the stem somewhere in the relation, which says that your sequence is palindromic.
But another
pattern that
you must see is that it has a single mode in the middle.
This happens to be true for all the examples that we and all other mathematicians in the world
have tried, but nobody knows how to justify this. And this is the unsolved problem that I have
tried to explain you, and it goes under the name the g-conjecture. - (Brady: The g-conjecture?)
So G was the missing letter, and G is is the difference of the h's.
2 successive h's.
So the G here would be 3,
G here would be 2.
And the g-conjecture says that for every i,
gi is non-negative.
Which is another way of saying that the H vector
increases, at least up to the middle,
weakly increases. - (Brady: Weakly increases?) - So it it is possible, for example in this sequence here,
1 1 1 1 1,
it doesn't strictly increase but still the G vector is non-negative.
And the reason why I like this conjecture is that you can do the experiment fairly concretely, as we have done
right now, in explicit examples.
So whenever you come up with a new way to triangulate a sphere, you have your chance to disprove this conjecture.
[Preview] There- another reason why I love this conjecture is that not only is very elementary and concrete, it actually predicts a
very deep
analog with several different parts of mathematics.