 And my plan is a little bit sloppy, but if you want to have a few keywords of what's going to come up, I wrote some of them on the blackboard here as a substitute for a plan. So first I would like to do a little thing that Bertrand asked me to do in his talk, or maybe wondered for what reason I hadn't done. It's a famous property, in fact, that in a hyperbolic triangle, when you have angles alpha, beta, gamma, then the area of the triangle is pi minus the sum of angles. That's kind of the star property that you study when you begin in hyperbolic geometry. So what I claim, the way I'm going to prove this, is let's extend the edges in this fashion. So I'm going to cover the edges, so we see which extends which. And I claim that the areas in those three triangles are in fact alpha and beta and gamma. And that the big triangle, with all three points out at infinity, has area pi, so by a subtraction that will give you the relationship. So I only have to prove the identity here for one of these shapes. So this is a shape bounded by two lines meeting at infinity, so I'll promptly send this point out to infinity, and there's an angle here, pi minus alpha. So in fact, if I move up to the upper half plane, picture is going to look something like this. There's the real line. Here's a bit of the unit circle. Here's the region I want the area of. Here is our angle alpha, here's pi minus alpha. And let's see, I'm going to choose a moving point on the circle arc here. So it's going to be a value theta. Here's, so it's called the point theta. So this is a height sine theta. And I'll take a little element of arc length here. So that's Euclidean arc length, d theta. And when I look at the vertical strip above this element of length, it has a Euclidean width sine theta, d theta. So let's say that we're going to compute the area, the infinitesimal area of this vertical strip and then integrate over theta from zero to the value alpha. So this area is equal to the integral between zero and alpha. So here I have this vertical strip. It extends from height sine theta up to infinity. Let's just write this as an integral from sine theta to infinity of dT over T squared. So this T squared is the conformal factor between the Euclidean metric and the hyperbolic metric. It's the T squared that you see when you give the metric in the hyperbolic half plane, dx squared over dy squared divided by y squared times sine theta d theta times, that's the width, right? Okay, now this sub integral is just one over sine theta. If you can see that. So we have in fact the integral between zero and alpha of the function one. And that's the same as alpha. So we're done. Now, this little property for hyperbolic triangles lets you prove by gluing triangles together and canceling out multiples of two pi and so on. Let's you prove what Bertrand referred to as the Gauss-Bonneille formula. So I'll just give the general version of this that you can build. Let's say you have a surface. It's a hyperbolic surface. It has some boundary components. It has some boundary components with angles. Let's say you call them alpha i. And then it also has some genus out here maybe. And the area of this is going to be two pi times two times the genus minus two plus the number of components of the boundary. I call s the surface. Plus a contribution coming from the angles like here. Like we have in the triangles. So I have to add a sum over the boundary of the numbers pi minus alpha i. And again, it's an interesting exercise. It won't teach you a lot if I do it at the blackboard, but if you have never seen it, it's interesting to fit small triangles like these together and check using the Euler relationship. In other words, the definition of the genus. And canceling out two pi. This is always true. Okay, so in order to introduce pairs of pants, hyperbolic pairs of pants, the compositions, I have to talk a little bit about the deformation space of the right-angled hexagons. So geometry-angled hexagons. So I claim that for any positive numbers of ABC, there exists a right-angled hexagon whose odd-numbered sides have lengths A, B, and C. So maybe I can use a color convention. So there exists a right-angled hexagon of side lengths A, B, and C. And moreover, it's unique up to isometry. And when I say side lengths, implicitly it would mean alternating sides, not consecutive sides, but the right sides. So we can see this in a very concrete way. This is what the proof is going to be a picture. Here's a line. Here's a segment perpendicular to it. Take it up length A. Here, far enough, is another segment also perpendicular. Take it up length B. And at the other end point of those two segments, I can draw perpendicular lines. And all of this fits into the hyperbolic plane. And it's just a matter of continuity. You can adjust the length here, adjust L to achieve any positive C. And C being the distance between those two lines out there. If we move the points A and B together, towards each other along the original line, after a while the two lines above will crash into each other. So I need to move for them apart and get C equals 0, and then I can increase to any value I like. There's nothing more to it. So that's very nice. There's no triangle inequality between the side lengths of these hexagons. In fact, it's popular to call this a hyper elliptic, a hyper-ideal triangle. Hyper-ideal triangle is a synonym for a right-angled hexagon. And you may... Well, I can maybe justify the name in the following way. When you have a triangle or a polygon, here is H2, and here's the boundary of H2. Well, I can have a polygon with a vertex. That's the edges of the polygon meeting out at infinity. And that's an interesting case, called sometimes an ideal vertex. But when they don't meet, when they don't meet, then there's always a common perpendicular line like so. And I followed the color code. This vertex and ideal vertex were green. And this current perpendicular should also be green. And it always weighs the role of a vertex. So, yeah, this is called a hyper-ideal vertex. Yes, if you want the formula, I can give it to you. No, I can literally give it to you. Let me do it. So there is a formula, and it's in fact the one I left as an exercise last time by saying you could multiply these three matrices to get the mystery distance between two ends of a triangle. You do this, except it's a hyper-ideal triangle, and the formulas are the same. So what you find is in this is A, this is B, this is C. Then the length L here satisfies Cosh L. This is Cosh A, Cosh B, plus Cosh C over Cynch A, Cynch B. And if you did the exercise or whatever of last time, you should have found a formula very similar to this one, except some of the signs were inverted or something. There was probably not a... Anyway, some of the Coshes must have been a Cosh, and some of the signs must have been inverted, but that justifies the terminology. In the end, it plays the same role in the formulas up to putting an H here and there once in a while. So what are these hyper-ideal triangles good for? Well, I can double them and get the so-called famous pairs of pants. So double along, let's say, the red boundary, and you get pair of pants, and then you're on your way to building actual hyperbolic surfaces by gluing those together. So I'd like to draw a little zoo of hyper-ideal triangles and right-angled hexagons to give you a sense of what can happen because this picture tends to be deceptively simple. So here's the zoo. Here is hyper-ideal triangles. Here's one. It's an H2, but seen from a large distance. So whenever you see such a strip, I draw it with a pretty thin, and in fact it's even thinner than it looks. It collapses together exponentially fast before increasing again on the other side. So that's extremely thin here. And then there are those fat regions, and this is a perfectly good hyper-ideal triangle. Which are red and which are blue is kind of irrelevant. There's still a higher right-angled hexagon, so let me put three of them in red and three of them in blue. But yeah, I'm really drawing two of them, the red version and the blue version, will double up to different pairs of pants. So that's one. Here's another one. And here is another one. Let's see. Here's another one. Once in a while you find a perfectly decent hexagon, but it could also degenerate. I try to be systematic in a sense, in the ways they can degenerate. They can do this. Or they can have a big fat region here. Or they could do, let's see, one, two, three, four, five, six, seven, eight, nine, three, four, five, six. So there's one missing. This one. Let me circle the big fat regions. And what's left uncircled is extremely thin. So here are two. This one has one. This one has two. This one has one. And this is reduced to one big fat region. So there's those nine examples of possible degenerations, and in fact it's twice that many because you put double along the red boundary or the blue boundary. So there's a great variety of possible pairs of pants, despite the weight that we always draw them. And to drive this point home, I'll submit a little riddle to you. Here are a number of pairs of pants. It's extremely thin out there. I can't really draw it because it's exponentially thin. Here's one. Here's maybe another one. This boundary component is to be left alone, but the long ones are to be glued up together. Maybe I can draw one that has just one extremely long. It doesn't matter. I have many of these, and all the long boundaries are glued together. Their lengths are supposed to match up, and the short colored boundary components are left alone. And the riddle for you is to figure out what the surface looks like. Where are the short geodesics in the resulting surface? That's what it means. We need to understand the surface. We need to understand where the short geodesics are. A billion of long ones, but they're just kind of... So if I glue this all up together, I may get something that has a very short geodesic somewhere, and I may get something that does not. And how do I find out? And the non-answer to this question is that it's extremely hard. It's extremely hard to figure it out just looking at the long pairs of pants, and in fact it's extremely sensitive to how you glue things up. If you glue with a slight amount of twist, if you... When you glue up this boundary length to that boundary length... Sorry, this boundary component to that boundary component, provided they have the same length, you still have a one-parameter set of ways of doing it, and the intrinsic geometry of the resulting surface will be extremely sensitive to this twist that you do. So essentially you can't know what the resulting surface will look like without doing it, or in other words, without doing some tough computations. Let me let you in on one picture, one more picture. If you do this in the simplest possible way, where you have one... Just one surface like this, it has two boundary components of the same length, and you glue them together. You still have this amount of twist to play with. Maybe you glue each point to the closest one on the other curve. In that case, what you're going to get is some once punctured torus. It's very thin out there. It has an extremely short geodesic here. That's what you get if you really glue each point to the closest point. If you do a small amount of shearing before gluing, then I'm not promising anything, but if you shear by exactly one half of the length of this long curve, then there will be a short curve again. And why is that? Well, I can find... It's extremely thin down here, but it's already pretty thin out here. I can find a point here that's very close to its friend on the other curve. And walking around by half the length, there's another pair like this up here. And this, if I glue up with one half turn, then these two green segments will glue up together and that will give me a short curve in the resulting surface. So it's just going to look something like this again. Is there an upper bound to... Right, so that's something I do plan to talk about later. There's a shortest curve in every hyperbolic surface, and it cannot be very long if you have fixed genus and maybe fixed outer boundary length. That's why I fixed the length of the outer boundary. There's always a fairly short curve somewhere. In fact, you can see this in terms of area. We said that the total area of the surface is essentially its genus. So if you let a little ball blow up and up somewhere in the surface, it will bump into itself before it has engulfed more than ye many units of area and that gives you a non-trivial loop there. So there's a bound to how long the shortest curve can be. And exceptional surfaces have a very short curve somewhere. So this is a special surface in a sense because it has an extremely short curve. And again, if I twist by half the boundary length, I will already have a very short curve. This is e to the minus t. If t is the total length, this is e to the minus t width here. This is e to the minus t over 2 width. So that's already pretty short. If I turn by one-third of a turn, the same story applies again. I have somewhere out here curves of length e to the minus t over 3. e to the minus t over 3. And these are one-third of a turn apart. Then one more third of a turn puts me here. That's extremely short. And the next one goes up here. So again, if I twist by one-third of a turn, there's again a very short curve. So the length, if I plot, if I make a plot where the horizontal axis rotates between 0 and 2 pi, the amount of twist before gluing, the vertical axis is a c-stall, or the c-stall is another name for this shortest geodesic, then pretty much of the time, if I take a very long curve, then much of the time it will be a reasonable surface with a fairly large c-stall. But for every rational, so at pi, there will be a dip here to a very small value of pi. And at 2 pi over 3, there will be a slightly less dip. So at every rational value, I should expect a dip like this, and the dips will grow deeper and more isolated from each other the longer this boundary component L is. And that's the sense in which the behavior of the surface is very complicated in terms of how we glue things up. And if there are many pairs of pants, it's, of course, much worse. Okay. Yes, there's a nice formula. Yes. Well, it depends on ABC, but not in a very interesting way. No, here's why. Take the surface, the reasonable surface that you want to get, and take an extremely long curve, an extremely long, simple, closed curve in this, slide along it, and you get a pair of pants like that. No, but there will be a number of curves, I say. So when I say that when you, if L is very large, it will make your animal here. Then for any how you twist, there will be a very short curve. No, because maybe you'll hit this surface. This surface has a very long curve somewhere. If you slide along it, then you get, so it's complicated, life is complicated. Now, you may have noticed something, which is that we use this doubling procedure where the system is bounded above just in terms of the genus. You can probably say that then your amount of twists should be close to some rational and then quantify this. It can be a large rational. Large rationales can appear, but only for extremely long L, right? You can probably work out a statement like that. I cannot give one to you, but yeah, that makes sense. So this pair of pants that we have been playing with, they have a symmetry by construction. They are obtained by doubling. And you could wonder, maybe there exist other pairs of pants that do not have this symmetry, right? That are not constructed by doubling. Well, there aren't, because if I have a pair of pants, I can find the common perpendicular to any pair of boundary components. It has a length A, B, and C. So by uniqueness, both sides are isometric. I do have to say this, right? Otherwise, there might be some magical pairs of pants. Now, the hyperbolic, the hyper elliptic involution, let me call them rank two surfaces. So this is based, okay, there's a formula that may have shown up in the other talks, but I'll give it anyway. Let's take a matrix, I mean pgl2 of r, and by plus or minus one, I mean it's a representative that has determined either one or minus one. Then there's this property that the trace of m is one-half times the cos of translation length of m as a hyperbolic isometry, over two. If the determinant of m is one, if m preserves the orientation of h2, it's the same thing with a cinch. I should put absolute value here because m is known only up to sine, right? So I can make sense of the absolute value of the trace but not of the trace. This becomes a cinch if m reverses the orientation. So the first case was just translation along an axis of h2. The second case is you translate, but you flip at the same time. You flip the two sides. I think you wanted two at the front, not the half. Thanks. Yeah, I wanted the half on the other side. Two. Thank you. And it's also going to be cosine, sorry, two cosine of the rotation angle if m is a rotation. So never mind the formulas. If you do memorize the formulas, memorize that there's a trap here. That's once in a while you have to take a cinch. I fell into that trap a number of times. But if you don't memorize them, just memorize that the trace tells you the nature of the transformation. So in particular, trace less means it's an involution. Pi rotation or axial symmetry. Zero can arise either here or there. And if it's zero for an orientation preserving thing, then it's a half turn rotation. If it's zero for an orientation reversing thing, then it's a reflection in a line. OK. Now take AB in PGL2R and let's say, suppose AB minus BA is invertible. That's kind of the generic case, right, for a pair of elements. Then I claim m conjugates B to their inverses. So how do we see this? Proof. So I have to prove that m a m inverse equals a inverse. In other words, let's multiply by a on one side. That's a m a equal to multiplying by m on the other side m. Now let's see, a m a. That's a times AB times a minus a times BA times a. However, a has a certain relationship with its square. We know that a squared equals lambda a plus mu times the identity for some pair of numbers, lambda and mu. That's a characteristic polynomial. So I can turn this into lambda times AB a minus AB a plus mu times BA minus AB. Here I'm recovering m or minus m, in fact. So that proves what I wanted to prove, right? Interestingly, there will always be an element. We don't know what its determinant is. Maybe it reverses orientation. Maybe it doesn't. But it sends each of the two generators, A and B, to their inverses. So let's look at this in terms of pictures. Consequence, rank two hyperbolic surface. So hyperbolic surface H2 modulo, a free group on A and B, has some kind of involution. And if you remember, those are the surfaces that a non-trivial involution. Those are the surfaces that Todd drew at one point in his last lecture. There will be, in fact, a pair of pants, which is the whole point of this buildup. There's a pair of pants. There will be some non-orientable surfaces. Here's the one-hold Mobius strip. I still don't know if I should say Mobius, Mobius, or it's not just in the transformations. Here's another surface that's a one-hold Klein bottle. You can nowadays purchase Klein bottles made of glass. There's a guy in California who makes them. It's all due to a translation mistake. Those used to be called in German Kleinsche Fleche. It's Klein surface. What would you call them? And somebody mistook this for Fleche, which is what a bottle is called in German. Never mind. Then there is the third surface. Todd drew it this way, and I'm drawing it in the same way for good reason. So these are the three surfaces. Here is their common four surfaces. They're common involution axes. So I just rotate by an angle of pi around this skewer. And that will exchange the top and bottom of the two pictures. In each of the cases, I see a right-angled hexagon above and a right-angled hexagon below. That's the one-hold Torus pair of pants. And so I see very concretely what the involution does, and it could have proved the same thing by observing that they're always decomposing two hyperideal triangles with the same side lengths, ABC, and so therefore the two halves can be inverted because they have the same shape. OK. How am I doing with time? We started at half past, is that correct? So it has been used implicitly, I think, in a number of talks. So the composition of a surface S pairs of pants can be realized geodesically if S is hyperbolic. In fact, that's a negative curvature property. You don't need constant curvature. In negative curvature, you can always find geodesics that will stay disjoint when they have to. So the way to see this is, again, I will try to make this just a picture. Here's our surface S. Here's a pair of pants, the composition that you like, where you can lift this up to the universal cover, getting some curves. And the point is that in the universal cover, if you straighten up all the curves and they did not intersect initially, then after straightening them out, they still don't intersect. So it's a cyclic order relationship on the boundary. So we're good. And then we can project back down to a nice... That point is the same. Yes, that's what we mean by straightening up. You keep the same endpoints, but you take the geodesic that has the same endpoints on that. And of course you should ask, why are the two endpoints distinct whenever I have a curve and I lift it? Why is the lift... It's supposed to look the same. This is supposed to be a hyperbolic surface with a squiggly-looking pants decomposition. I'm just making the claim that we can make the pairs of pants look nice. Without deforming? No. I just straightened out the curves in a given surface. So I'm sorry that my surfaces look so different. I should say it's isometric to this new surface, except the pants now look nicer. Both surfaces have a given hyperbolic metric and it's the same, right? So that's something you need to know, but once you have this, you can study the complex, so the space of all pants decomposition. And so here's where I wanted to prove that the bear's lemma or the bear's constant. So this happens exactly in one case. It's when A and B... Sorry, I'm going to assume A is a hyperbolic translation, so lambda inverse. Translate along an axis. Never mind about the other cases. In that case, the commutator AB minus BA won't be invertible exactly when B also has an axis and it has an endpoint in common. And then in that case, the quotient is never a surface. It's always ugly. It could be ugly for different reasons, but in that case, it's definitely ugly. That's not hard to see, by the way. A... So if B is x, y, z, w, then AB minus BA is going to... B minus BA will have zeros here and there. And here is z, and here is, I think, minus y. That's what happens. Project up to scalar multiplication. That's what happens when A is diagonal. And you can see it because the... multiplying on the left will multiply the first line by lambda and multiplying on the right will multiply the first column by lambda. So the thing at the intersection of the first line and first column becomes zero. Same thing for that entry. So this is non-invertible exactly when one of y or z is zero, which means that B is triangular, which means it fixes zero or infinity. Sorry for the digression. Yes, that was the answer. You said the quotient is ugly. Well, there's a set theoretic quotient. But it's not... I think the correct words are it's not Hausdorff. So the burst constant, you fix the genus. For each genus G, there is a constant positive such that any hyperbolic surface, or let's say hyperbolic metric SG, the genus G surface, has a balanced decomposition of total length at most C. So let's see. We've already given the first part of the proof, which is how do we find the first curve to cut along? The first curve for the balanced decomposition. I want a curve that's not too long. So I'm going to do like we did before. We know that the genus G surface has area not too much, so we embed a ball somewhere, that the ball blow up, it will bump into itself after a while, just for area reasons, and that gives us a short curve. So point one. There exists a systole. Some... So the systole has bounded length. And then we'd like to do the same and over and over again and exhaust and get a nice balanced decomposition. The problem is that the new balls that I have growing the surface will grow in a surface with boundary. Now that I have cut the systole, I have some genus and maybe the two halves that I had cut previously. And now if I try to increase the ball in radius somewhere, it might exit the surface before it bumps into itself. And so the area growth function is not what you might think it is. There's a lot of area getting lost outside the boundary. So we need to prove something slightly stronger. So what we'll do is revise the very statement we are proving and say in any genus and any number of boundary components, if I fix the number of... If I fix the length of the boundary components, then there's an upper bound on the systole of that surface. So improved statement. Surfaces of genus G with K boundary components of length at most the number C, so that's for any G, K and C have systole, have bounded systole. And clearly it's hopeless. This is the statement I want to prove. Clearly it proves the initial theorem because if I have this, then I can cut along a systole that gives me a new surface with maybe less genus, maybe more boundary components, but never mind. It falls into another instance of the improved statement. I can cut along the new systole. At each iteration the statement gives me a bound on the new systole that I get, and I just need a fixed number of steps anyway until I get to a bunch of pairs of paths. So this implies the main statement by induction. And the other point I want to make before proving it, the other point I want to make is that I clearly cannot get rid of the dependence on C. Namely, there exist surfaces with extremely long systoles even in given genus. In fact, I have even drawn them for you in the zoo. In the zoo there were things like this, and if those long arcs are the arcs, if I double along the short arcs, then the long arcs become the boundary components, and that's a surface in which all the curves are extremely long. If I want to do a non-trivial loop in that surface, I have to travel along some number of edges of this underlying graph, and they'll go very long. So I cannot get rid of the existing surfaces with long systoles. It's just that they have long boundaries in the first place. Okay, so proof of the claim. If genus is zero and you have a lot of boundary components, you still need to cut a large number of times before you get pairs of pants. In fact, the number of pairs of pants that you need is known in advance from the genus of the surface view. It's just a matter of area. So you want to cut, for instance, if you have a pair of pants, you can... Yeah, you should stop once you have a pair of pants. You put it aside, then you have less area and you can use induction as part of the whole strategy. Maybe I was a little bit sloppy in saying by induction, but that's really how you... There will be curves, yeah. Unless you have a pair of pants, there will be curves along which to cut. In fact, they will have boundary length. So proof of the claim, when... Let's see, I could expand a... So instead of expanding a ball outwards from a point, you are going to expand a uniform neighborhood outwards from a boundary component. It will bump into itself after a while for reasons of area or from... I should say from the whole boundary. I should have said something before, which is... Let me back up a bit. The first point is there exists a number independent of anything such that the components of boundary S have this joint embedded uniform neighborhood of length, let me call this the outer length, at least kappa. So what I mean by outer length is... Here's your boundary component. Here's the surface. It does something, has some genus. And as I take a uniform neighborhood of it, it's bounded by two loops, the original boundary, and this other loop that's not geodesic, slightly longer, and that's the outer boundary. So if I have a long boundary component, there's nothing to do. I just take the component itself and maybe push it inwards a little bit. But if I have a very short one, then I may have to dig into the surface quite a bit. So the content of the theorem is that if I have a very short boundary component in S, then I can expand a uniform neighborhood of it and it will reach at least kappa units of outer length before it bumps into anything bad, like itself or another boundary component or whatever. That's the content of this first point. And the reason is, well, we lift to the universal cover. In the universal cover, we see H2. We have this very short boundary length. And anything bad that I could bump into has to stay between two iterations. So here's a point P. Here's gamma P, where gamma is the loop around the boundary. And anything bad has to happen out there. If ever there was a line boundary loop of the surface that I was worried about, it would have to come in and come out again of H2 in between the two perpendicular raised from P and from gamma P. Otherwise, pushing it forward, it would intersect itself, which never happens. So everything happens pretty far out and for that reason, I have the boundary. So that's sometimes called the color lemma. It's an important tool that comes up a lot of the time. So once we have this kappa number, we can, so, second point, expand these, all these neighborhoods, the UI. I expand, I take those uniform neighborhoods for all the boundary components, and then I do it independently for all of them, and then I expand all of them simultaneously. The new neighborhoods UI of T, neighborhoods, will bump after a bounded amount of time. And that's again for reasons of area. Since I have at least kappa units of length and expanding further, we'll eat up more area at rate, at least kappa. It's kappa times the width, the extra width by which I blow things up. So this eats up area at the definite rate. So therefore, after a bounded amount of time, the things bump together again. Bump in bounded time. And what I can do then is whenever, let's see, if I have a boundary component here and a boundary component there, and they bump together, so their neighborhoods bump together, that gives me a curve to work with, right? Follow from one boundary component to the next, follow around, and back out. And now there's a bound on the length of this thing, because, well, sorry, I should have made this a UI. So here's UI, UI. Here's UJ. After a little bit of further expansion, they bumped. And now this curve that goes around UI to UJ, around UJ, and back has bounded length. The back and forth trips cost a bounded amount of time, the extra time it took to bump. And the trips around UI and UJ are bounded. So that's when it bumps into another boundary component. If it bumps into the same, that's also fine. I can write here. This is something bumping into itself. And I can just take this loop, or maybe that loop. Okay. End of the theorem. So we get this burst constant. And what I'd like to talk about next is, I'll give you a working definition of Taichman-Respace and of the mapping class group. So let's see. The definition is let S be a topological smooth surface of genus G. And let's consider the following set. It's the set of all hyperbolic metrics. Let's say they are smooth hyperbolic metrics on S. Divided out by the isotopes from S into itself. That's called Taichman-Respace. The statement I made before that a given pants composition can always be realized geodesically gives you the following statement, observation. There exists a natural map from Taichman-Respace to plus, so positive reals to the N, where N is the number of curves in my chosen pants composition. So for every C1, CN curves of a pants composition, there exists a natural map taking the type of space to the N tuple of the length. Taking G, the metric on S, so taking the pair S and G to the N tuple of G lengths of C1. So that's a completely natural map. It's subjective because I can choose the boundaries of any pair of pants, whichever way I want. If I take lengths, I can realize pairs of pants that have these boundary lengths and glue them together. So that's fine. So pi is subjective. In fact, it's a vibration of fiber. So let's see what do I get? What are all the hyperbolic metrics that have a given set of boundary lengths as their cuff lengths? So the curves in a pants composition are sometimes called the cuffs. I think the cuff is this or maybe this boundary of a pair of pants. So what are all the hyperbolic metrics with a given set of cuff lengths? Well, I can twist by any number. Like in the example that was on this blackboard at the beginning, I can twist by any number the gluing between two adjacent pairs of pants or between a pair of pants in itself. So the fiber X again homomorphic to R to the N parameters so continuous twists. So at this point, if you've never seen this before, you may be wondering what happens if I do a full twist. Isn't that the same surface in some way? Well, in some way it is isometric to the old surface, but it's not isometric as what we call a marked surface. The isometry is not in the isotopic class of the identity of the surface. So that's something you have to wrap your mind around if you've never seen it before. These are two isometric surfaces, but they secretly want to be mapped to one another in a way that's not the isometry. That's the sense in which they are different. So the space, the isometric space is homomorphic to R to the 2N consequence. The isometric space is a ball of dimension 2N, which if you work out the Euler characteristic is 6g96, but there are many surfaces in it that are isometric to each other and in fact there's a group acting on it that sends surfaces to isometric surfaces. So that's my next definition. Maybe you could just mention the names of fenchial and Nielsen. Right, right. I'm sorry I should have said that. Those lengths and those twist parameters are numbers called the fenchial-Nielsen coordinates. I'm sorry. So this particular realization of the isometric space which is the easiest proof that it's a ball is called fenchial-Nielsen after fenchial-Nielsen. Let's see. So the new definition that's kind of a dual or whatever to this one is of a mapping class group, Cg of S. My definition is the group of self-homomorphisms of S divided, it's a quotient group, rather self-isotopes. So not necessarily at this point. You may want it to, certainly not in this talk but in case you want to apply any of this safely to any of the other talks, let's say orientation. Another thing I should say is if S has boundary I should take homomorphisms that fix the boundary point-wise and isotopes that also fix the boundary point-wise. If I have a whole, just a cusp in the surface, just an isolated point and there's nothing to do there. Right? So this quotient thing is a group. Maybe that's not even obvious. But let's see, if we have, or does that mean it means the subgroup I'm dividing out by is, I should end pretty soon, right? So I should divide out by a normal subgroup. So why is this normal? Suppose S to S is an isotope. So we can find a continuous path from the identity to phi. So let's call this phi 0, phi t, phi 1. It's the conjugate of an isotope. If alpha from S to S is a homomorphism then I claim that alpha phi, alpha inverse is an isotope. Why? Well, alpha phi t, alpha inverse interpolates the identity. The phi t interpolates between the identity and phi and conjugates of phi t to operate between the identity and the conjugate of phi. So there's nothing deep there. I wanted to prove a few things about the mapping class group but let me just state them and then we'll be done. So facts. The mapping class group, which is not a group of mappings, you cannot find actual realizations of your homomorphisms that multiply consistently to each other. So you really have to divide out by this thing. There's no section, in other words. Acts properly discontinuously on time and space. I may actually prove this in the first five minutes of next time. Here's another fact. The group is generated in twists and I'm stating this mainly for the purpose of having used the word day in twists. Those day in twists are those full twists that I was talking about that are the easiest elements to realize. In fact, good news. They generate everything. And I was planning to prove that G, the commutator, sorry, that's hard. That's day in the 30s but redone in a different way by licorice in the 50s or 60s. There's no really easy proof of that. But the commutator, once you know this, there's a nice proof that the commutator of the mapping class group is a mapping class group itself, which is a way of one of the ways in which it's a nasty group. It's a complicated group. Not very friendly but there's a lot of research going into the what it does to type in the space. Thank you.