OK. Now we're ready to finish proving the main theorem, classification of closed surfaces.
And here are the four lemmas which I will be using, which are put here for easy reference.
So we start with a polygon with a complete set of edge identifications,
then we want to show that the corresponding surface is one of these.
So we start with a word, and Case 1
is that there exists a letter
which appears
twice in the word, both as x, not as x and x inverse. Or,
once as x inverse, which of course would be a bit stupid, because we can
redefine it to be x, and then this case visibly reduces to that case.
So this case is the case of non-orientable surfaces.
OK, so our word is of the form— so up to cyclic permutation
we assume that it starts with an x, so x and then some word and then x and then some other word.
So it does exactly have that form x A x B. And by this lemma,
so let's name these Lemmas 1, 2, 3, 4. By Lemma 1,
we have that the surface, which is the one we care about,
is, OK, so we rewrite it as x x and then this thing. And then
that's the same thing as a connected sum of RP2 with some other surface.
And now we may invoke induction on the length of the word
so by induction on the length of the word, this is a shorter word.
We may assume known the result of the theorem or this.
So by induction
on length of the word
this is, well, either S2 or a bunch of T2s or a bunch of
RP2s.
So we have three sub-cases.
And we get that our surface is homeomorphic to
either S2, so RP2 connect sum S2,
but connect sum S2 does nothing because if, S2 I remove a disc, then
that just, I just have a disc and you then put, you're doing nothing. So that's just RP2.
Or RP2 connect sum
n copies of T2, which by Lemma 4
applied n times is the same thing as a connected sum of
two n plus one copies of RP2.
Or this is
RP2 and n copies of RP2
in which case that's just n plus one copies of RP2.
So in all cases, we're good
and we have shown using the induction hypothesis that our surface is one of these.
So let's do the Case 2. So Case 2. Now we're assuming that no letter appears twice the same,
like every letter appears x and then x inverse.
So this is the case of orientable surfaces.
So now I'm going to assume that there exists
two letters
x and y
which are interlocking.
And what do I mean by interlocking?
I mean that they appear, if you draw your word cyclically, they appear not like that,
but like that. So something x and then y and then x
inverse and then y inverse. So that's what it means that they're interlocking.
Or if you draw them on the surface of a circle,
then you have the x and the x inverse pair somewhere,
then the y and the y inverse pair somewhere. And if you draw a line that connects them,
these two lines cross. That's what it means to be interlocking. OK.
So we have our word like that. And we may as well,
by cyclic permutation, assume that it starts with x. So here it is.
So it is x and then A and then y and then some other part
and then x inverse and then something and then y
inverse and then something. And now we're going to apply Lemma 2 a number of times.
So first, we're going to apply Lemma 2.
So Lemma 2 says that between an x and an x inverse I can perform a cyclic permutation.
So here between x and x inverse I can perform a cyclic permutation to bring y close to x inverse.
So I have B A y x inverse. And here, that's also between an x and an x inverse.
It's between x inverse and this other one, which is x, you can imagine it here.
And so now again, I'm going to put this y inverse close to this. y inverse D C.
And then finally, I'm going to apply once more the same Lemma 2
but now to this chunk here, which is between this y and this y inverse. And
by doing that, I'm going to do it cyclic permutations so that x is close.
So I have x y x inverse y inverse.
And then the rest is x, well, it's B A D C, which I can for example put here. Yep.
So the manifold associated to this word is homeomorphic to the manifold associated to
this word is homeomorphic to the manifold associated to this word.
And these are all instances of Lemma 2.
And what is that?
Well, I have here the word for a torus, and here some other word.
So this is the same thing as the manifold associated to the word,
this is the same thing as the torus, T2, connect sum with the manifold associated to B A D C. And
once again we may assume by induction
on the length of the word and also we know that this is,
this is not creating any x xs or x inverse x inverses.
So we're still in the orientable case. This is either S2 or a bunch of T2s.
So, well, if I connect sum T2 with S2, that's just T2,
and if I connect sum T2 with a bunch of T2s, that's
bunch of T2s. So this is either T2
or a bunch of T2s. And so great, we're again,
well, so here in this Case 1, we found that
we are one of these. In Case 2 we found that we're one of these.
And now Case 3 is the case when there are no interlocking letters, so...
And if there are no interlocking letters, if I draw
the letters on the boundary of a polygon
just like I did here, and I draw, whenever I see an x and an x inverse,
I draw a line connecting them, these lines never cross.
And so we have a pattern that looks a bit like that. And in particular
we have the closest arcs, the arcs that are sort of closest— the smallest arcs
that correspond to edges that are in cancelling position,
we have y y inverse. So there exists letters to which
one can apply Lemma 3.
And this brings us
to a smaller word, which again has no interlocking letters, so you cancel,
cancel, cancel, cancel, cancel until you're left with nothing,
well, with the smallest possible
word, which is just x x inverse if you admit bigons, or if you don't admit bigons
because you want edges to remain straight, it's going to be like x
x inverse y y inverse, which is S2.
So, so this is Case 3.
And that's it. That is the proof of this theorem using these lemmas.
And that is also the end of this sequence of videos. Thank you for watching.