1/2 Geometric Structures on 3 Manifolds
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Transkript: Englisch(automatisch erzeugt)
00:16
Thank you very much. Welcome back to the last series of lectures. There will be
00:24
very few proofs in these lectures. There will be some fairly honest summaries of proofs, but not very many accurate proofs. The plan is to try and get to a fairly accurate statement of what's called geometricization theorem during
00:43
the first hour, and during the second hour I'll talk about some constructions and properties of hyperbolic manifolds, which are a subclass. So I start with this rather mundane observation that was known to 19th century mathematicians
01:00
that in dimension 2, real dimension 2, every surface S, and by every I mean orientable and compact, closed, is topologically classified by its genus.
01:39
The genus is a number such that S is homeomorphic to a connected sum of that
01:45
many copies of the torus. T2 is the torus, and you do connected sums of tori with each other a number of times. You can also compute G from the
02:00
fundamental group. It's the rank of the pi 1 of S, if you abillionize it, and divide by 2. So that is all there is in the world of surfaces, and also it admits a constant curvature metric modeled on one of the following, exactly
02:42
one, the hyperbolic plane, the Euclidean plane R2, or the sphere S2. In fact there's just one example here, the sphere, one example here, the torus, and one example in, as many examples as genera G, are hyperbolic.
03:08
So the subject of this talk is what can we say in this direction about dimension 3, and of course it's much harder. First, there is no such topological
03:23
classification, not nearly, but Poincare still ventured a kind of a claim in this direction, so that's called the Poincare conjecture, that if the fundamental
03:51
group of your three manifolds, so I'm not going to call it S now, but M as in manifold M3 for the dimension, is trivial, then M3 has to be homeomorphic
04:08
to the 3-sphere. Again, M is an orientably closed 3-manifold, and as I was saying, a reasonable classification of such 3-manifolds is,
04:36
let's say, hard, but here are two statements. First, connected sum is still our friend,
04:49
namely, there is a unique prime decomposition, holds in dimension 3.
05:17
So prime decomposition means connected sum into things that are no longer connected sums of other
05:22
things, like for integer numbers. So it exists, it's unique, and that's
05:50
already far from easy. You can have it by combining results of Kneisser in, I think, the 30s, and later results by Milner. So in a sense, it's enough to
06:07
understand prime 3-manifolds, and about prime manifolds, here's what we can say, a prime 3-manifold has a unique, so that's isotopically unique, minimal
06:29
decomposition by incompressible tori, so those are tori T2, that find
06:55
themselves injected in a way that's also injected at the pi 1 level.
07:06
So if there are such tori, then you can cut along them, and what's left at the end is essentially unique, into pieces that are either a toroidal, meaning
07:31
no incompressible T2 any longer, or ciphered-fibered. I'm going to say now
07:45
what ciphered-fibered manifolds are. This was the first class of 3-manifolds that can be said to have been well understood. So ciphered-fibered, 3-manifolds
08:04
were the first understood class by ciphered in the 40s. Oh, and by the way,
08:23
this second statement about tori is due to JSJ, that's a very common way of calling it, J-Co. I'm going to use it. J-Co, Charlene, and independently Johansson. I think it's one NSS, but don't trust me.
08:52
Okay, so the ciphered-fibered manifolds are the ones that are ciphered-fibered.
09:06
If M comes with a partition into circles such that all but finitely many
09:25
of these circles lie at the center of a solid torus, so that means the
09:44
nearby circles just travel along the same circle and close up at the same time it does, are cores of S1 cross B2, the ball across the circle.
10:03
Just think of it as you have fibered your cylinder into segments and you close up. The remaining ones, or the exceptional fibers, being modeled on
10:35
S, and I'm going to draw the picture first, you do the same vibration of the
10:43
cylinder into segments, but you close up with some fraction of a turn. So B2 cross 01 modulo x1 is the same as e to the i, so I need to really keep
11:10
track of the of the rational number that I put here, k over n 2 pi x 0, so I
11:23
have to subdivide into n segments here and identify this point, the point number 0 with point number k, the color maybe. So as a result the fibers
11:41
away from the core have to have to travel n times before they close up, travel along the cylinder n times, while the central fiber travels only once.
12:06
The sign does matter, the sign of k over n, so k is in Z modulo n Z.
12:24
Okay, so there is a full classification of those, these are fully classified. The
12:41
space of fibers is a two-dimensional thing that's really an orbifold, right? If you travel inside the space of fibers there are two dimensions worth of ways of perturbing your fiber, and sometimes it becomes singular
13:00
and that means that this two-dimensional space does something weird. It's in fact an orbifold. It's a two-dimensional orbifold, namely it's
13:21
modeled on R2 and R2 modulo an order n rotation, an order n, let's call it n sub s,
13:43
this n sub s being the n that we see here, so there's one n sub s for each singular fiber. So the orbifold forgets about the k, about the numerator, but it knows how many times the fibers fold, and so this orbifold kind of forgets part of the data. We still have to put back the data in,
14:02
the numerator data, and there's a whole calculus of these things. Sometimes a space can have several orbifold realization, sorry, several cypher fiber realizations, and the calculus of these things is well understood. So when you cut with incompressible tori, you get something is boundary,
14:23
and is it part of the cypher, is the cypher same with the boundary or not? This torus here? No, it's just a local model, so I'm just saying that there's an atlas whose charts would be just tori, and with a, sorry, tori with a compatibility condition when they are good.
14:43
So I'm just asking about the pieces in this JSJ thing, when you cut with incompressible tori? Right, so there's a version of all of this with boundary, and when there's a torus in the boundary, in the JSJ business, this torus will be itself foliated by circles of the cypher vibration.
15:10
All right, this two-dimensional orbifold has a, so let's call it O, with an orbifold Euler characteristic,
15:32
and as a way of computing this, it's just the Euler characteristic of the underlying surface of O. If you forget that the singular points are singular, you just have
15:41
a topological surface in front of you. Call this O between bars, and you have to do something for each singular fiber, namely you subtract 1 minus 1 over NS over all the singular fibers.
16:01
So this is a number that makes sense to measure. It will show up a little bit later in the statement of geometrization. Now, here are some examples of cypher fiber spaces.
16:28
It's of course perfectly okay to have no singular fibers, like in the surface of genus G across a circle, or maybe the surface of genus G, then take the unit-tangent bundle,
16:44
T1, that's also fine. And also, you can look at the length, what's called the length space, so L sub, I put a fraction here, length space, and that's by definition S3 modulo,
17:04
so S3 is seen as the unit sphere in R4 modulo, the action of R 2 pi over N rotation matrix,
17:22
sorry, B, let me use A, 2 pi over A, and R 2 B pi over A, where R theta is the rotation matrix.
17:49
You act on S3 by such a composition of rotations, and the quotient is in fact a smooth manifold, that comes with a cypher fiber structure, in fact with many cypher fiber structures,
18:02
and there's a question about this in your exercise sheet.
18:24
Okay, now in 1980, Ferson proposed, so I'm going to give it as a definition,
18:48
a theorem, and a conjecture. I'm going to state all of these, I'm going to take a few blackboards. First, the definition,
19:05
a model geometry is a simply connected manifold,
19:22
since we work in three dimensions, let's say a three manifold X, with a smooth action transitive by a Lie group G,
19:51
such that stabilizers G sub X are compact,
20:12
no larger group, in the sense of inclusion, acts with compact stabilizers,
20:32
and also there should exist, so to rule out some cases which are not of interest for us, there should exist a compact quotient of the form gamma under X,
20:53
for gamma in G discrete. You mean no larger group acts spacefully,
21:01
otherwise you can just take the product of this. You mean no larger group acts? Well, by diffeomorphisms. You can take the group and add the compact factor, it acts trivially, so it's... Oh, right, right.
21:23
Thank you. Now, conspicuously, there is no metric in this definition, and that's on purpose. Namely, you could put a metric on X
21:42
that's invariant under the action. By compactness, you can take a metric and then average it. By compactness of the stabilizers, X carries a G-invariant metric,
22:08
possibly not unique. So, for example, I'm going to give examples, in fact the full list of model geometries in a moment,
22:22
but for example, if you take for X what's called SL2 tilde, which is also the unit tangent bundle of H2, well, universal cover of that,
22:41
then the G... Yes, and G equals the same. G equals...
23:16
So, this is the X.
23:20
So, what I'm talking about is, if I want to put a metric on this X, is how do I measure the distance between two unit tangent vectors to the hyperbolic plane. I want to travel from this unit tangent vector to that unit tangent vector, and let's say that traveling,
23:40
carrying it over along a segment by parallel transport costs me some price in euros proportional to the distance, and so I add this up along the path that I choose to travel, and rotating as I go costs me some other amount of, at this time in dollars, Canadian dollars, I guess.
24:04
And what matters is the combined cost to define the distance between the two points. How much do I have to pay in the base, and how much do I have to pay in terms of rotating my vector? And the actual geodesics will depend on the rate change between euros and dollars. So, the geodesics of these metrics do not look the same.
24:22
These are different metrics on the same manifold, but for our definition it's still the same model geometry. Now, the theorem is that there exist exactly eight model geometries,
24:57
namely, so I'm going to start with the constant curvature one,
25:04
H3, so this looks very similar to the two-dimensional classification that we had at the beginning, H3, R3, S3. In those cases, the dimension of G is always six.
25:30
The isometry group of hyperbolic three-space is six-dimensional. Then there are the product geometries, so H2 cross R, S2 cross R.
25:44
So, these are self-explanatory. You let the group of isometries of H2 act, and you also allow yourself to travel along the second factor. So, that's a four-dimensional group of isometries.
26:01
There's also the one I gave here, SL2 twiddle, and something called nil geometry, about which I will say more towards the end of this lecture. So, for those, dimension of G equals four,
26:22
and the last one called solved geometry, dimension of G equals three. So, the two that I have not really explained are the last two, and there will be pictures and explanations about them at the end of this lecture, if I'm lucky.
26:48
And also, for us, a geometric structure on M
27:02
is a diffeomorphism to a space of the form gamma under X, for gamma in one of those Gs.
27:22
So, now the conjecture. So, this was a conjecture of about 1980, and it was in fact proved, it's the geometrization theorem, my parallel one,
27:45
in about 2002. So, it says the following. If M is oriented, compact, prime,
28:08
then the JSJ pieces, Mi of M,
28:20
all have finite volume geometric structures. So, over a Gi xi, where it may depend on the piece.
28:45
More precisely, and I think I'm going to use a white blackboard for the more precisely part.
29:30
I mean, what's left when you remove those tori, those sub-surfaces. Those are sub-manifolds with boundaries.
29:41
I mean the interior, you remove the boundaries. More precisely, if, let's go,
30:04
if pi 1 of M is finite, and just to be fancy, I'm going to call this virtually trivial,
30:21
then xi, the model, equals S3. So, you see how this generalizes the Poincare conjecture that was here. It says if pi 1 is trivial, then Mi is the sphere S3.
30:41
Otherwise, if pi 1 of M is virtually cyclic, then the model has to be S2 cross R.
31:01
Pi 1 of Mi, thank you. And the corresponding model is called xi. If not, but it is still virtually abelian, then we are dealing with Euclidean geometry.
31:22
So, I think I'll call it R3. If not, but it's still virtually nilpotent, and we're dealing with the mysterious nil geometry. If not, but we're still virtually solvable,
31:46
then it has to be solved geometry, and through five of them. If not, then the discussion becomes twofold.
32:08
So, if there exists an exact sequence defining,
32:21
so the kernel is Z, the group the sequence defines is a finite index subgroup of pi 1 of M, of Mi.
32:43
So, if there virtually exists such an exact sequence, then two cases may arise. If Mi is compact, meaning Mi is in fact the only piece.
33:02
Whenever I had to cut along something, since I had to remove the boundary, that means it's not compact. If it's compact, then if the exact sequence splits,
33:21
then the model space has to be H2 cross R. If not, xi has to be the SL2 twiddle metric,
33:42
like on the unit tangent bundle of a surface. If Mi is not compact, then both geometries actually can live on the manifold in question.
34:00
So, that means both xi, I don't have to choose, both xi equals SL2 twiddle, and H2 cross R are valid.
34:23
I'm going to say a little bit more about how this can happen, but the remaining case, so if there is no such sequence, Mi was obtained from a compact manifold,
34:41
a closed manifold, by cutting it along some surfaces. So, if I had to cut at all, then the pieces are not compact. I've removed something. If I have not, then it is still compact. So, that's really a matter of whether I had to cut or not. And the splitting is up to finite index again, or just splitting?
35:02
The splitting is an actual splitting here. And when you say both are valid, does it mean both at the same time? Yes, I can put two different geometries, two different finite volume geometries on the same manifold,
35:22
and I'm going to say how this happens after I'm done with the statement. So, in the remaining case, so if there is no such exact sequence, in the remaining case, which is also the case where pi1 of Mi is irreducible,
35:47
a toroidal irreducible with infinite fundamental group,
36:06
then Xi is hyperbolic. I'm going to add some more little things to this discussion. One is that in those five cases,
36:30
automatically, Mi is compact, namely, M equals Mi, no cutting.
36:50
And the other thing is that observation,
37:01
quotients of models other than H3 and Sol are ciphered-fibered.
37:26
And ciphered-fibered means really well understood. In fact, if I do this in orange, then there is a discussion to be had.
37:42
In these cases, the Euler characteristic, the orbifold characteristic of the base of the ciphered-fibered space is positive. In the next two, it's zero.
38:00
In these two, it's negative. So all of these six cases are really well classified, they are ciphered. The remaining case other than hyperbolic, the Sol case, also has a nice property, which is,
38:32
what is the Sol manifold? The isometry group,
38:44
the trivial connected component of the isometry group of Sol is just you act affinely on R3 by matrices of the form e to the minus t in the first two coordinates and the translation should be by t in the third coordinate
39:08
and here you put whatever you like. So this is really R semi-direct R2 where R acts on R2 by compressing one direction and stretching the other.
39:25
So the picture to be had in mind is you have, let's say, a little square in the first two directions and here's a square that's isometric to it, looks like a rectangle up here
39:41
and if I push it down there, it looks like a rectangle with the other aspect ratio and everything in the way of horizontal translations is good. So that's a description of the isometry group of Sol and this should convince you, hopefully,
40:04
that mapping tori of anosof, what's called anosof,
40:21
endomorphisms, automorphisms, phi of the torus are Sol.
40:41
So what's a mapping torus is you take the torus, cross interval, and you glue the top to the bottom by an SL2Z identification and by assumption, this identification phi is anosof, meaning two distinct real eigenvalues.
41:08
So the way to realise such a mapping torus as a quotient of Sol space is to rotate the eigendirections of phi
41:21
until they coincide with the first two axes, the x-axis and the y-axis. Then your lattice, the Z2 by which you quotient R2 to get the torus looks like some sort of weird lattice in the horizontal plane and it looks exactly the same up here
41:50
after stretching and compressing by phi. So that's an example of a manifold with a Sol geometry and theorem.
42:03
In fact, all Sol manifolds are of this form
42:24
or quotients thereof by groups of order,
42:43
I think it's at most eight. So really we have a good understanding of all three manifolds that do not fall into the last H3 class. So I'm going to spend the remaining time discussing a little bit about the remaining geometries,
43:02
especially nil. But before that, another observation. If you look at the unit tangent bundle of the genus G surface,
43:26
that's also the quotient of the unit tangent bundle of the hyperbolic plane.
43:43
And this has SL2 twiddle geometry, as we mentioned. SL2 acts on the hyperbolic plane. Here's another circle bundle over the same surface,
44:00
while circle cross SG has H2 cross R geometry.
44:22
And these are different geometries. However, if I do this with a surface with a boundary, then, as mentioned in the theorem, the two collapse in a sense, and that's because it's in fact the same bundle. However, if I look at the unit tangent bundle
44:49
of the thrice punctured sphere, and I'm drawing the thrice punctured sphere as a hyperbolic surface, then this is in fact homeomorphic
45:02
to the trivial tangent bundle, sorry, the trivial circle bundle over the same surface, because, so these are homeomorphic,
45:20
and for the same reason, it has both geometries. And the reason they are homeomorphic is that the thrice punctured sphere has a field of directions. So that's a trivialization of the unit tangent bundle.
45:46
How do I do this? You take a, well, there are many ways of doing it, but basically you comb singularities of any field to the punctures.
46:09
Here's a, I guess we could do it that way.
46:28
If you can find a field of directions without singularities on the surface, then that's a trivialization of the tangent bundle. And so it doesn't matter if you look at this circle bundle as the trivial or the tangent bundle,
46:40
it has both. So is that supposed to take S1 cross the surface? S1 cross the surface, yes. So a tangent vector on that surface is a tangent vector on the surface, but it's also given by just go to a point and say how many radians away from the orange direction you are. So that's a trivial, just a number at every point.
47:02
OK, so that explains the phenomenon here. Now, I'm going to say another, not really a classification, but a statement about spherical geometries because there are some questions about this in the exercise sheet.
47:26
So S3, in a sense the first simplest case, is also famously the double cover of SO3,
47:41
the orientation cover of the group SO3. The isometry group, or the trivial component of the isometry group of S3 is given by multiplications
48:01
by unit quaternions on both sides, namely you take two unit quaternions,
48:22
you view the 3-sphere as the unit sphere in the algebra of quaternions, and the way these act on an element of S3 is just for some reason, I like to write it like this, B times U times A inverse, so that gives you a very concrete handle
48:41
on how to send the 3-sphere isometrically to itself. And in fact, for spherical manifolds,
49:03
S3, there's an equivalence. pi1 of M is abelian if and only if it's cyclic, if and only if it does not contain
49:23
minus the identity of S3, so the antipodal map, and equivalently, it's a length space. So that's the first class
49:41
of spherical manifolds that there are, and the remaining ones, other spherical 3-manifolds,
50:03
are quotients of, so I can write it S3 modulo plus or minus 1, maybe SO3, by subgroups,
50:22
so SO3, acting by right and left multiplication,
50:44
as above, by subgroups that have finite index, in fact that have index 1, 2, or 3,
51:03
in some product group, gamma prime cross gamma second. So you can put a gamma prime in SO3, a gamma second in SO3,
51:20
let their product act by right and left multiplication on SO3, and this will usually not be a free action, unless you're in one of those very special situations, because one of them,
51:47
so one of the two has to be cyclic, and the other one could be an icosahedron group, or tetrahedron group, or something. Now there's an exercise that does not really ask you to show this, but some weaker statements in the exercise sheet.
52:02
And okay, I still have some time to draw some pictures of nil geometry, which is the only one we don't have pictures for yet. So nil geometry, I'm going to draw it to use R3,
52:21
as the model space. The action on it will not be by affine transformations, it will be like some sort of polynomial transformations. Here's the picture. Isometry is, well, there's an exact sequence.
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The good news is, we can write 1R, the isometries of nil, a fiber over the isometries of the plane.
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You view this as acting on R3, with three coordinates, x, y, z, and that acts on R2, and these here are translations along the z-axis.
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Over here, I'm drawing tangent planes at every point.
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So they are horizontal on the y-axis, and as I move along the x-axis, they start to tilt, and they are sort of more vertical. They get more and more vertical as I move to the right,
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and over here they go down, they flip. Everything is invariant on the z translations. Above every isometry of R2, there is an isometry of nil space.
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The way it acts, let me show you how the stabilizer of 0 in R3 preserves the red quadric.
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How can I use red? z equals x, y. So that's a quadric that looks something like, goes down here, here it goes up,
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and it goes down on the other side. It contains both axes, and it's on the positive side when x and y are both the same sign, and it's negative the rest of the time. It's this saddle, and you have to imagine that as you rotate in the projection,
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you rotate the points in the vertical x, y coordinates, and the vertical lines of nil have to do a little dance by going up and down and up and down along the quadric. So this map gets sent to that map by rotation.
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So this vertical line to that vertical line to the next. Now what can we say? The pi inverse, so if we call this capital pi I guess,
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pi inverse of R2 translations is generated by yz translations,
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and x translations composed with a shear,
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yz shears. I guess what I'm trying to say here is you're allowed to translate the whole picture in the yz plane,
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but if you want to move into the x direction, then you have to push down, you have to basically send these squares to those squares. It tilts, so it kind of bends your space as you go, and you bend more and more as you move to the left. So this description should show,
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if you work it out a little bit carefully, this shows two things. One is that nil can be identified with one, one, so the Heisenberg group,
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I think you have to put it this way. So it's the Heisenberg group, but you have to add things that are not just multiplications in the Heisenberg group. Those are these funny rotations that I described.
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It shows this, and that the 1, 1 and 0 mapping torus has nil geometry.
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You take a vertical square here in the yz direction as a fundamental domain for your torus. You can identify opposite sides by translations, which are in the group. And then you push this in the x direction
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and do some shearing. You push it by n units, then you can do shearing with intensity n, and that gives you a mapping torus with this monodromy. Okay, I can stop here. Thank you.