 This second lecture will be only about the Last model h3 because if there's only one thing you remember from the first lecture. It has to be this it should be this But if you care about three-manifolds all you should really care about is hyperbolic three space Also, one of you was clearly paying attention because they noticed that I wrote eight here instead of infinity as I should have Okay, just I didn't really talk about I do not plan to say Anything about the proof of the geometrization theorem that I stated in the previous lecture the way parallel man prove it following an approach of Hamilton an idea of Hamilton was Okay, we are obviously not smart enough to find right down the Homogeneous metric on a manifold, but how about letting nature find it We put some smooth metric on the on the manifold and then let let a PDE Spread out the curvature and make it as nice and homogeneous as possible It turns out there's a natural quantity the rich a curvature that you can add to the curvature With a negative coefficient and and then it it will spread out the Irregularities of the metric This process might blow up But it will in fact exactly blow up it will find the interesting subsurface is the the tori and singular spheres where the where the Manifold needs to be cut so nature nature knows our PDE is know about the decomposition into elementary pieces okay, so Let's talk about h3 and I'm going to start by drawing a few a little cartoon here of h3 So h3 in the upper half space Model looks something like this the plane down here is P1 of C If I add the pointed infinity to it Acting on this is The isometry group of h3 which is also PSL to see That's all supposed to be a Reminders mainly but here are For example totally geodesic planes they looked like hemispheres over the over the boundary of the model here is a geodesic looks like a half circle Here's another geodesic it looks like a vertical line here is a a Plane at constant height It's called a horoscope horos here, and here is another horos here It's a sphere tangent to the boundary at at the point and if that point is taken out at infinity that the horse here starts to look like the like the This horizontal plane that I do so that's the first cartoon That I want you to have in mind and maybe there's another one That I will be using it's here's an ideal tetrahedron in h3 So it's it's the span the complex span of four points There's a deformation space of such things which is in fact the complex cross ratio of these four Points in P1 of C if you know what a cross ratio is and if I truncate this By a horosphere Notice that the whole sphere is Euclidean. There's a Euclidean metric preserved by the stabilizer of the horse here well, this horosphere intersects the Neighborhood of the When intersects the tetrahedron along a triangle that's in fact a Euclidean triangle and the picture if I send that point to infinity is Just like here P1 of C the four points one of them is at infinity and The red slice here now looks like a Horizontal slice by plane parallel to the to the floor I have a choice of seeing this triangle It's a Euclidean triangle either up here in hyperbolic space or maybe at the boundary because I see the same The same angles here by projecting down Also a carton to have in mind So I'm going to talk about some phenomena in hyperbolic three manifolds and How we can build many and what they look like? So this is the first Part will be about what's called thick thin the compositions and here's a theorem by Well, it's called the Margulis constant, so we can probably attribute it to Margulis there exists a positive number epsilon such that points in a I'm going to say any finite volume Hyperbolic three manifold This will always mean complete hyperbolic so a quotient of h3 by a group of isometries In any hyperbolic three manifold the the points at which The injecting injectivity radius is less than epsilon form a disjoint union of uniform neighborhoods of Two kinds of things either short geodesics of the quotient or second possibility Ranked two cosps that is to say ideal points of P1c fixed by some parabolic Parabolic is a transformation that acts on some holes here by translation if I put the holes here in the With its center at infinity. Those are very special transformation of the hyperbolic space and they can in fact arise in in a Even if I have a non-compact finite volume quotient So moreover This is I don't really want to make this part of the theorem, but it is very important morally to see To as an extreme case or a limit case of one Picture following Okay, what's the injectivity radius it's whenever you're in a manifold in a quotient You can ask yourself how big a ball can I embed around that point? Yes, in case one be a neighborhood of a geodesic Uniform neighborhoods off And when I know yeah, these are points at infinity But there's a reasonable sense in which we can talk about their uniform neighborhoods I have to talk about Boosman functions and the like And is it the finite essentially finite unit of such thing so if the volume is finite here Then yes, it's a finite union If not, obviously not in fact There's a version of this theorem where you don't assume finite volume But you have to say some things about rank one cosps as well. So I'm going for the simplest statement Right so I asked myself how big a ball can I embed around x and when the bomb ball starts to bump into itself or into an Imaging of itself in the universal cover and that's as big as the injectivity radius is so the the regions where the injectivity radius are small is small are kind of Simple and stay away from each other. That's the content of the theorem The sense in which two is a limit case of one is the following If you take a really short Okay, first of all a typical G in PSL2C is conjugate to Z maps to lambda Z where lambda is in Well, lambda is a complex number of modulus larger than one If I guess if the trace of my element is not in the interval minus 1 1 or something Minus 2 2 then the then that's what what it does it it acts on H3 by preserving a line an axis that I can choose to be between 0 at infinity It's vertical line up here and it translates along the axis and rotates by some amount And what I see in the boundary at infinity in the plane is is a similarity of the plane The the log of the ratio of the similarity is the length the length of the short geodesic so Let's assume that lambda is really really close to 1 in modulus that has some reasonable amount of rotation Then what I'm going to see is so 0 is here. Here's maybe a point of C here is G times C sorry P G times P. He's G 2 times P And it goes on and on it goes around the origin and After a while it will come back near the positive real axis But if the modulus is really close to 1 it will come back maybe here Right and go on and on and on and goes around the second time And comes back here goes on and on So that's what a typical really short geodesic does and What I want you to pay attention to is the fact that big powers can bring back the point very close to where it was So that's the sense in which 2 is a limit case of 1 namely if I if I look only at this region here I see something that looks very similar to a lattice of translations It's not quite a lattice of translations because these are really Rotations are on a point very very far away But by a very very small angle so in the limit it becomes translations That makes sense All right So the key to the proof I'm not really going to prove The The big thing the composition theorem, but you can make the following sketch work key idea In a league group so in let's say in PSL to see the commutator of Two really small elements Small elements small means close to the identity is even smaller That's a general idea. Yes The cost That's true in high for a hyperbolic surface But not for in fact that there will be an explicit example later on where the cost is a rank two lattice In fact, if the if the volume is either the group is Yes, that's what ranked. Yeah, it's ranked in this sense not not in the sense of Least theory Yeah, sorry. Yeah, if you chop up the cusp, it will be So it's rank in the sense of you look at the stabilizer of the cusp and that's a virtually a billion group you talk about the rank of that Rank can have some other meanings as well Okay The The commutator of two small elements is even smaller. What does it okay? What is a commutator a commutator is you apply your element a and yeah Then you unapply a but not quite a in fact a conjugate of a right. It's a times B a B inverse a Conjugate means the same thing, but done elsewhere. So you apply a Let's say you translate along its axis and then you Unapply something that looks like a except it's nearby So if you if you translate along your axis and then change the axis a little bit and translate back That kind of gives you a sense of why the what you get is really close to the entity identity even closer than a already was so in particular If it's even smaller it has to be trivial if for instance We start with the two smallest elements a B and we assume a B is discrete this this low injectivity radius business is something about elements that act by Think close to the identity and in fact they cannot have big They have to commute basically So I'm not being going to be more rigorous about this But maybe jack it up in into a more general statement in general in general Discrete subgroups of a league group generated by small elements must be Nill potent as abstract groups Here a billion Because we are in a very special the group So you can sort of what I described for small elements you can sort of use it to start an induction and show that the If small elements generate a discrete group that group has to be important Even better so you know it already right now This description Will be used in how it's called hyperbolic dain feeling a question for you How can we build many Compact hyperbolic three manifolds with such short curves unfolds, so here's Here's a strategy It would really be an example, but in fact an infinite class of examples, so I guess we An example that you can already call a theorem Do the following start With a cusp hyperbolic three manifold for example, I'm going to use this very Standard design example s3 the space minus Here's The figure eight not and for some reason I like to draw it thickened so so you can It's a torus we remove a solid torus from s3 not a solid torus. We call this m and I claim That this is homeomorphic to the gluing of the two following things So this is an ideal tetrahedron like the one I drew at the top right here except I have truncated its vertices so Clearing and truncation faces and I glue this you can this hyperbolic truncated ideal tetrahedron to another one and the map should respect the following I Can use orange it should always send a An edge not a truncation face mind you not the small ones, but the big ones Should be identified one of the of this tetrahedron to one of the other in such a way that the arrows match and if you Look into it. There will be only one way of doing that this if you have Seen the exercise sheet that's outside. It is not the example in the exercise sheet that's outside. It's a very related one But the arrows look a little bit different and I'm not okay some people claim that they can see how the tetrahedron the two tetrahedron here sit in the complement of the knot and Tile it I cannot I can tell you where the edges are supposed to go they go here and There and because it's such a nice manifold. I don't have to care which it's which or which direction it goes And so these edges get identified The faces get identified For example, here's a face a The face with the dots here is a face that has two simple arrows going out from a from a truncation vertex And the only one here that matches is over here And if you look at which way the third arrow goes then we are very fortunate because the the identification will in fact Work right I have to Turn it over as I do if I do it for the six for for the four pairs of faces then then the doing is that so There's a way that there's a What we call the cusp tiling the resulting cusp tiling is if I look at This face a here it gets glued to the face a prime there, right The truncation face gets glued not the face itself, but its boundary because the adjacent dotted face gets glued so there's really a A way in which the truncation faces will tile some sort of plane here if I want to be very Accurate how to put the a prime upside down I think right you have to Rotate the right that region to map it over So this defines a cusp tiling and oh I haven't really really convinced you yet that the this Complement here is a hyperbolic manifold, but in fact a Complete h3 structure results when both tetrahedra are Regular ideal regular ideal means its partner points are at infinity and regular means they have the full all the 24 symmetry group That should be clear because the the regular ideal tetrahedron tiles h3 just by reflection in its faces and What I'm defining here is Commence rubble with with this with this full reflection group of the regular ideal tetrahedron I keep It's a subgroup of the finite co volume group that stabilizes the Tiling by regular ideal tetrahedra in h3 Okay, so we start with such a manifold. It's not compact and I've said we wanted to build compact hyperbolic three manifold So we are going to perturb this in a certain sense that I try to make concrete and Usable so I Mentioned at one point that not a an ideal tetrahedron the shape of an ideal tetrahedron is the same datum as a Euclidean Triangle right namely what what is the Euclidean triangle? It's the shape of this truncation face of this Euclidean Truncation face something I have not said is that all four look the same If I truncate the face and the vertices of an ideal tetrahedron, then I get four Euclidean triangles for the price of one, but they in fact look all the same That's one of the symmetries of the cross ratio property So let's do the following whenever I have a Euclidean triangle I'm going to write a small label in this corner in each of its corners, namely if I can take Its vertices to zero one and the complex number Z by a similarity Then I write Z here and if you work it out, you're going to find that here I have to work So what is Z? It's the second segment Z minus zero over the first segment one minus zero So here I have to write Zero minus one over Z minus one or one over one minus Z and Here I have to write Again, if you work it out one Minus one over Z and write these little labels Then what we do is write all Labels in the cusp tiling into The cusp tiling So the cusp tiling I you could check this we're not going to do it now, but there's a there's a similar Exercise in the sheet you could check that the cusp tiling really does look like this like a tiling by regular ideal triangles meeting six at each vertex and I'm going to to Just write down what the resulting Labeled tiling looks like so here are the labels I've been in red Hey So A and B are the other numbers that go here want for each of the two tetrahedra I'm going to adjust their shapes. So the shapes are as follows a B a B I guess Then it's the same B B So this is in fact a fundamental domain of the tiling that we get namely the The four corners here are identified. Sorry These four get identified You can get that in this fundamental domain. I see exactly four A triangles and for B triangles That's normal because there are four truncation faces of the A tetrahedron and four truncation faces of B tetrahedron so that's the shape and Again to put this into context above the above each of these Euclidean triangles leaves an ideal Tetrahedron right and drawing only two of these but then they leave everywhere so Once we've done this We could do it as soon as we find tetrahedra. That's sort of fit together to Decompose another manifold if we had another good example I can see a constraint that weighs on the numbers A and B namely Since this is the same vertex as that by translation by periodicity I Can see that Multiplying all the all the numbers In the all the labels in those corners should give me one It's even more precise in that the complex arguments that they have should add up to exactly 2 pi Right, that's called the gluing condition And there's one gluing condition for each vertex So here is Z2 periodicity the gluing condition or constraint Says let's see I start with the bottom with the bottom for 1 minus 1 over B Times a Times 1 minus 1 over B again times 1 minus 1 over a and then I see B and 1 minus 1 over a again squared should be equal to 1 in complex variables So let's rewrite this to I guess a minus 2 plus a inverse B minus 2 plus B inverse equals 1 So I have two complex numbers a B a equals B equals e to the Sixth root of unity I pi over 3 is clearly a solution Right now because this then becomes Minus 1 times minus 1 This is still one only one condition for one vertex in fact if I look at all the other vertices I'm going to always write down the same The same condition in fact that's it's already expected that I should find the same condition at least one second time because this edge I Mean this point here is the endpoint of some edge Coming some blue edge coming from infinity and there's the opposite end point of the blue edge that has a representative somewhere in the tiling So I should see each equation twice But there's in fact one more redundancy in this case so other vertices Yield the same constraint There's in fact a very nice linear algebra that you can do to show to predict The number of independent constraints in terms of the number of tetrahedra and the number of cusps in this case There's two tetrahedral one cusp and the formula says there should be exactly one constraint on the two variables What about nearby solutions to the gluing equation nearby in the sense of close to the one given by the the ideal Given the the hyperbolic metric so if they are nearby that means the Locally triangles will still glue up right I will still be able to travel around an edge and find a nice developing map. That's That's defined at every vertex But this z2 periodicity has no longer any reason of being by translations. This was another miracle of e to the ipa over 3 that I get Horizontal periodicity by translation and vertical by another translation in general A prime B prime will define a solution a nearby solution a prime D prime will define a Developing map in variant under some Well equi variant under some abelian subgroup of The similarities so are I guess C star Semi-direct See the similarities of the complex plane the stabilizer of the point at infinity Usually usually valued in C minus some point omega Well, this looks like is Here is maybe my starting triangle my starting A and We know there's a vertical it's glued to another copy of itself above Glue to a copy of itself above. That's not necessarily quite a translation of a it's it's slightly rotated slightly scale and there's another one another one and who knows and And as I keep developing I mean you're going to use a for the I mean white for the copies of the a tile, but This extends and eventually I see another White triangle and then another one they get bigger and bigger as we copies of a that we see get bigger To the right in this case and smaller to the left and as a center omega here and the The lattice of translations has become a slightly different commutative group of similarities of the plane That preserves this tiling. It's not really a tiling because it wraps around so they If I travel far enough Upwards here. I will eventually find triangles that superimpose with other triangles in picture and so on so That's the That's what it looks like and I guess if I name I could name this one zero zero This is a One zero. This is a two zero and this would be choose one doesn't matter which a Zero one a one one a two one and here comes a Zero two and so I think I can I can coordinate eyes this these my just You know you using z two coordinates from over here and Wrapping them around now the key insight and that's a That's something again person realized Is that since? A B this value we started with is a generic point of the solution space to the bluing equation as We perturb a little bit. We can basically there will be locally a unique Way of ensuring that the pq copy gets superimposed exactly on the zero zero copy I'm getting there so this is a This is a transversality statement by transversality since A zero B zero is a smooth solution Is a smooth point defining equation for large pq pq there will be a unique a pq B pq Mear and zero B zero such that a Zero zero the tile the original tile coincides With So you have to imagine that this this nice sunflower really closes up in a in a In an exact way after turning around once and the and Tile that coincides with zero zero is Not just any tile but for large I'm going to say co-prime pq prime pair and irreducible vector of the lattice We'll go inside So then we can truncate off. What did I phrase this we can truncate off a all tetrahedra by a uniform neighborhood of the line omega infinity Right and because everything is symmetric around the point omega this what does a uniform neighborhood of these lines? Omega infinity look like it looks like a cone cone opening up at a constant angle from the from the line omega infinity and so Since a cone looks everywhere the same basically this will Truncate all tetrahedra in the same way. It doesn't matter which tetrahedra in the orbit you are looking at above Above each of these triangles leaves a Needs an idea of tetrahedral basically in the in the room and you're all truncating them in the same way because a cowl is a cowl So This truncation is equivalent and glue back in Omega is the point Omega It leaves in In C but more more more Generally in P1 of C C in infinity So there's this line connecting Omega to the point at infinity and the this the group of similarities Of C that preserves the tiling also preserves in H2 in H3 the line omega infinity Yes, some some group of rotations translations are on that line Is that kind of like developing the cost Yes, so it's a developing map of this combinatorial picture you forget you forget it had angles and everything you just keep the torus and You affect new shapes to them And that's called a developing map it maps to the to to see to the plaincy. It's complex structure So we do this truncation and glue back in Notice that the that these tetrahedra Some they're redundant because a zero zero is the same as a TQ and they they decompose everything In the cone inside the cone except the axis of the cone Above this point there's nothing Not sufficiently high above any any other point of the plane. There is some term tetrahedral that Covers that region so we glue back in a Solid torus in the corrosion to score curve is given by so is Is generated by whatever is left basically we've killed PQ we've killed the PQ translation in the in the stabilizer of the cusp So what's left is a similarity taking a zero zero to a P prime Q prime where a P prime Q prime is a component P prime Q prime Z Z times P prime Q prime plus Z times PQ is Z2 Right, so that In order to have all of Z2 that's this Z2 over here In order to get all of Z2 as a as a direct sum like this I had to assume PQ was primitive and then it has a complement and we can really see the this solid This solid torus is a quotient of the cone that I described uniform neighborhood of this axis by the rotation translation of Parameter P prime Q prime so let's say It's basically very short but has an angle that I have little control over it depends I mean when you have when you have The point PQ in your lattice Z squared then its complement P prime Q prime is Basically almost the same direction But it might look shorter Much shorter a little bit shorter. You don't you have a little control over its size So this this becomes a statement about the Close directions mean the the the length of the similarity is very short, but Uncontrolled Size means the angular translate angular component is very well is not not a priori small so Topologically what have we done topologically we have Taken the Compliment of this torus this this knotted solid torus and glued back in a solid torus that killed a slow PQ on The on the boundary of the knotted solid torus, and that's what the infilling is called. Oh, that's what sorry It's called infilling right in in the In this solid torus that that had the only drawing part of it, but that was tiled by Euclidean triangles We Have attached to the boundary of the compactification of M. So sorry torus. That's why the torus Killing the slope PQ If you want a cartoon here is M It has some topology. It has discussed That was really a torus times are and they've attached a solid torus to it With a map that that kills some slope, but that makes it compact The result is a compact hyperbolic three manifold called the dain filling so Theorem by Thurston, but I have basically proved it in this example Thurston say says if M is a cusped Let's say a one cusp for simplicity one cusp hyperbolic three manifold and then all All but finally many Dain feelings are still hyperbolic also hyperbolic with nearly isometric Nearly isometric thick part if If PQ is large PQ are the parameters of the dain filling And nearly isometric comes from the fact that that for large PQ the perturbation AB had to be only very small also Also the volume decreases under such feelings so ball of M PQ it may be a little bit counterintuitive because it might feel like we're adding something we're adding a sorry torus But in fact the volume goes down quickly So this can be proved Using something called Schleifle's formula That there's an exercise about the formula in the sheet Now this might look like a One little trick to generate some Compact hyperbolic three manifolds But it's in fact a very ubiquitous phenomenon That's the content of the last year and I want to mention so a theorem that's Almost an observation really my Thurston and Jorgensen under given volume There exists only finitely many topological types of thick parts of hyperbolic three manifolds so that's that It might look either deep or trivial depending how you think about it, but You're looking at the thick part so you can cover it by epsilon balls for some uniform epsilon and These balls will overlap in a bounded way So you can bound the number of balls that you will need and the possible combinatorial pattern of their intersections So there's a huge huge huge but finite number of things that can happen basically So you can make this rigorous, but it's it's really not nothing more to it and in fact everything Beyond these thick parts is the infilling you you take the thick part and then you do some something like this In So corollary The set of volumes of hyperbolic three manifolds Has is well ordered because basically everything is a dint filling on one of finitely many Thick parts under a given volume and the infilling always takes the volume down It's well ordered and in fact we know the type No, I do mean to include the cost ones So the cost ones will be the accumulation points of this of the in the volumes of the accurate the volumes of the And the type is omega to the omega so that means there's basically one hyperbolic manifold of smallest volume It's in fact known nowadays which one it is as one then there's a next one Smallest possible volume and the next one and the next one and the first accumulation point That's the smallest volume of the smallest one cost manifold And if you look a little bit further there will be a second accumulation point and a third accumulation point and so on and then The first accumulation point of accumulation points. That's omega squared. That's the first. That's the smallest two cost Manifold it goes on and on and on and you add accumulation possible accumulations point of accumulation points You add one word to that sentence each time you add one cusp and goes Up to this this ordinal here. Thank you. I'm over time you