 Thank you. So, the original title had the word Taichmuller in it. It said hyperbolic surfaces and Taichmuller theory. That was mainly because it sounded fancy. A Taichmuller space, in case you don't know what it is, it's the space of deformations of hyperbolic surfaces. So, since I will talk about deforming hyperbolic surfaces, it doesn't make sense to say I will talk about Taichmuller space. However, if you don't really talk about the uniformization theorem like Bertrand did, you are not really allowed to say Taichmuller. I think primarily this first lecture will be an lecture on basic hyperbolic geometry, real hyperbolic geometry. So, you can see it as a background for several of the other lectures. I think the other is mini-course, third mini-course, possibly the other mini-course is too. I will try to start really slowly by building a model of the hyperbolic plane where you don't really need to know anything except basic plane Euclidean geometry. So, hyperbolic geometry has just a few words about the history of the subject. It's often written about including in popular science books because it starts really early if you can put Euclidean's name at the beginning. In Euclidean's times, people would think about space as something given to us by nature. Euclidean says, look how beautiful, there are all sorts of systematic things that this space verifies and you can study them, see how they imply each other and so on. And then around, there were many centuries and then I can call the names of Lambert and Borlii, Lobachevsky, Gauss, that's all around 1800 where people toyed around with these logical constructs that you find in Euclidean. Euclidean's actions are not so easy to work with. If you really look at them in the text, they are kind of clunky and don't look like the nice building blocks that we like to work with but still the idea was there and they started to realize there might be some weird counter examples. So, around this time, there's a strong connection with the development of logic and finding out whether Euclidean's actions are redundant or things like this. And then this weird counter example started to become the basis for many, many, many other things. So, around the time of Riemann, I should give the names of Klein, Poincaré, also Hilbert. It's partly arbitrary the names I give here, but let's say in the late 1800s, it had become a basic object. A basic object that you generalize in many, many different directions. And the question of how it relates to the space that we live in has kind of lost its object. So, here's a weird counter example and then later on it becomes a basic example. So, we mathematicians tend to shy away from thinking about the kind of confusion that must have been around in those days, but it's interesting to think about this. Okay. So, as promised, I will try to build a model, a hyperbolic plane, assuming as little as I can. And the, so it will be the Poincaré model and I will build other models at the end of the lecture too. The basic, there's one property of Euclidean plane geometry which is called, I think, the power of a point with respect to a circle, relative to a circle. So, let's call the point P and the circle C. And it's the fact that whenever you have a point and a circle, and then you draw two lines through the point intersecting the circle, that gives you four intersection points. Let's call them A, A prime, B, B prime. Then there's a number which depends only on the point and the circle, not on the two lines, which is P A, P A prime, the distance, distance from P to A and distance from P to A prime equal to the other product, P D, P D prime. And that number is sometimes called the power of the point with respect to the circle. How do we see that that's true? Proof, if you divide things out, you see that it amounts to proving that the triangles A, P, B and B prime, P, A prime have the same angles. Let me draw these two triangles, one in blue, one in red. So, these two triangles have one angle in common, it's the angle of P. There's a certain angle here. And you see that this property of product says that P B is longer than P A by the same factor that P A prime is longer than P B prime. That says the two triangles look the same of this K. And why do they? Let's take the center of the circle here. They have one common angle, they're done already. So we need to check something about the other angles. We only need to check one of the other two angles because the sum of the angles is Pi. So let's check that the angle here, A, B, B prime, is the same as the angle out there. B prime, A prime, and then away from P. And what we do is draw the radii to A, A prime, B prime, and subdivide all those sectors in half, another color for the subdivision. Then we give names to all these numbers, X, X, Y, Y, Z, Z, W, W. Those are eight angles. And we can compute the angle here. The angle A, B, B prime is going to be Pi over 2 minus W plus Pi over 2 minus Z. And the angle out at the last angle is the complement Pi over 2 minus X plus Pi over 2 minus Y. So when we put everything to the same side, we get plus Y plus Z plus W equals Pi, which is true because double that is 2 Pi, so we are done. So that used to be taught in high school a long time ago and then things went wrong. So depending on the country you come from, you may have seen it long ago or more recently. So a consequence of this construction is that I can talk about inversions and what they preserve. Let's define inversions. Well, yes, I should really redo this picture when P is inside the circle and then some of the Pi would shift around but that's not a big set of cases to create. You have to be a tiny bit more careful than this. Well, sometimes the center may not be inside this quadrilateral. Maybe B, B prime, A prime are all in the same half circle and then either I have to count some of the angles with sign or I have to... So, okay. Inversion in a circle. By definition it's going to be in a circle C. It's still in the plane. It takes, it goes from R2 minus, let's say it's centered at O. R2 minus O, actually the center is P. And it takes any point A to the point A prime of radius R. The point of the ray P A at distance R squared over distance P A from P. So it's really, you place yourself in polar coordinates and you invert the radius. You keep the angle the same and far points become close and close points become far. As a consequence of the property, inversions, if I call this map inversion in C, I sum C takes circles, let's say circles not going through the point P to circles. And I make an exception for the ones that contain P. Why is that? Well, if I have... Let's check it first for a circle C and a point. If I invert a circle with the radius, sorry, whose tangents to the point are exactly the power of P with respect to the circle. Now this is going to take any A, so if there's R, and A will go to A prime just as in the picture. So it works if the two circles are perpendicular, basically. And then it also works for other circles because if I have a bigger circle out there, I can scale it down, invert and scale down again. That will be the same as dividing by some number, taking the inverse and dividing by the same number again. That's the same thing as just taking the inverse. So it's enough to treat the case of a circle that's perpendicular to C, indeed. Circles perpendicular to C are taken to themselves. For other ones, use a scaling argument. One over distance is the same as alpha times one over alpha distance. And once again, I should also treat the case where the circle goes around. The circle I'm inverting goes around the center of inversion. That's treated by a scaling by a negative number. So I can now talk about homographies or Mobius transformations or Mobius maps. You don't see the world homographies a lot in the English-speaking world, but I kind of like it. It's a standard word in French. The reason I like it is, first, it's a single word as opposed to two words. Second, I know how to pronounce it. I'm never sure about Mobius. And third, it doesn't have a person's name in it, which means it has a better chance of describing something. It preserves some things in the picture. So let's see. The map where, oh, one more thing I wanted to say was instead of removing P, I could have added a point at infinity. Then P would have gone to infinity. Infinity would have gone to P. And I could have delta, well, end lines. Instead of taking all circles, not through P, not through P, I take all circles, but I also add lines. And here, again, circles, not through P, become circles and lines. So a circle that does go through P inverts to a line. And a line inverts to a circle through P. Unless it's a line through P, then it inverts to itself. So you can convince yourself that 1 over Z bar is the inversion. As a map from C to C is the inversion in the unit circle. Therefore, for all A, B, C, B complex numbers, I map taking Z to AZ plus B over CZ plus B, which up to solving some simple equations, I can also write Z goes to a constant P plus another constant Q divided by Z minus yet another constant R. I can always be written in this form. P is the limit at infinity and Q and R, whatever they need to be, is a composition similarity by inversion by similarity. Z goes to Z minus R, that's the translation. Then I invert and conjugate, that would be an inversion. Then I unconjugate, because I didn't want to conjugate in the first place, but unconjugating is again an isometry, that's a similarity, multiplying by a complex number Q is another similarity. It rotates and scales and adding a constant P is another similarity. So it has this form. Observation, let's see. By definition, a complex homography, maybe I should say orientation preserving here, is a map of the form C with a point at infinity to itself and I claim that they form a group. This is a very famous thing, but I'll check it anyway. I'll claim the maps, oh sorry, one more definition. Right, otherwise it maps everything to a point. It's non-zero and in fact I can scale ABCD by the same number so it only makes sense to require that it's not zero. It's the same map if I scale the four numbers. So the theorem is up there. If I take PGLL2 of C, so 2 by 2 invertible matrices with complex entries up to scalar multiplication and take this to the set of homographies, if I take this map or another map that's defined similarly, PGL2R to homographies preserving the circle R with a point at infinity, or if I take PSL2 of R to homographies preserving this circle oriented, whenever a homography sends a circle to itself it may or may not preserve the orientation of the circle. So all these maps given by ABCD goes to AZ plus D over CZ plus D are group isomorphisms. And the reason, multiplicativity, let's say, I want to compute ABCD operating on ABCD operating on Z. That's just A times AZ plus B over CZ plus D plus B. Write it like this. And reducing to the same denominator and collecting, I can write this as AA plus CB, something like this. CA plus DC plus LCB plus ED. And this should be the product of matrices ABCD, ABCD, something like this. So that's just a verification. The first line of orientation preserving. I'm sorry? Yes, I won't say orientation preserving every time. I'll assume it. I'll just pretend it's part of the definition for the purpose of this talk. So the only thing we have to check is that if it preserves R, then we may as well assume that all entries are real. If it preserves R and is oriented, then this AD minus BC is positive. Turns out AD minus BC is the numerator that you get when you differentiate a real homography. So it indicates the direction of an ontonicity and checking that it takes R to R if and only if ABCD are proportionalism exercise. I won't go into this. So by this time I can define the hyperbolic plane in a sense. The hyperbolic plane H2 is R plus I times R positive. So the upper half plane of the complex plane. Endowed with the automorphism group PSL2R. Any proposition? That was already in Bertrand's talk. It's called this group G. I mean, I sense not done defining the hyperbolic plane because I haven't told you what the distance is. The distance should be preserved by the group and in fact the group does preserve a certain distance. dx squared plus dy squared over y squared where complex numbers given by its coordinates x and y. So how can we see that this is the case? Well, we could do it by a formula applying a homography to the differential form here but we can also check it essentially without computation by saying look, this homography is a composition of the standard inversion in the unit circle and of things of the form z goes to az plus b. If I write p plus q over z minus r and I only have to check that the z bar here for p plus q over z minus r with q negative is the correct thing to say. q negative is what it means to be an orientation preserving. I just have to check that adding r preserves the form. That's clear. It doesn't do anything. I have to check that adding p doesn't preserve the form. That's also okay. I have to check that multiplying by q preserves the form. That's also okay because if I multiply both x and y get scaled but y gets scaled so everything gets scaled by the same number. I just have to check that the standard inversion in the unit circle in the unit circle. If I look at inversion in the unit circle up close then it's just a symmetry. It takes a point at distance 1 plus epsilon from the origin to a point at distance 1 minus epsilon from the origin so it preserves the metric near the circle and then by the scaling argument, it also works on all the circles concentric with the inversion center. You can see it in that way and save computations. Another proposition would be g acts transitively and isotropically on h2. Transitively, again, you want to take the point i to the point x plus yi. What you do is multiply by y and then add x. All of these are homographies so it's transitive. And isotropically, I need to show that we can rotate that the stabilizer of a point rotates the plane around that point. Now I'll do this by saying, well, I can apply a complex homography to the whole thing. So let's prove isotropy. Let's apply a complex homography to the whole thing. We have the upper half plane. Here's i. You can send this to the unit disk. i being sent to, let's say, zero. So this will be z minus i over z plus i, I guess. That sends the point i to zero, clearly, and that sends all the reals to complex numbers of modulus one. So it does do this. It's a homography. I can rotate the circle around. So this is isotropic. We have a clear group of rotations. We try in particular homographies acting here. And I can send everything back. So the proof of isotropy is essentially contained here in this picture. And a consequence of this proof is that circles of H2, what is a circle of H2? It's a circle. Well, sorry, I haven't said what the distance is yet. Well, sorry. The distance is what you get when you integrate this infinitesimal metric. So circles for the induced distance function, that's the set of points at a given distance from a given point, look like circles in the complex plane, right? Circles look like circles. Circles go to circles. Centers do not go to centers. So here, if I take all the points at a given distance from O, it's going to give me some scaling of the unit circle. That's an example of a circle centered at O. But if I send it back, it will look like something like this. It looks a lot bigger above than it does below. But things that are up there look big by nature. If I'm inside this hyperbolic plane, all these little drawings are really isometric to each other. They look smaller and smaller as we go down. If it's twice closer to the horizontal line, it has to be twice smaller. But it's still the same for the hyperbolic distance. So what I'd like to do is now to give you a sense of how you can compute distances or angles or whatever you're interested in inside this plane. We don't really have coordinates to work with, like we have in R2. Sorry, there is over there. We can nevertheless compute in a systematic way. I mean, we do have coordinates, but they are not as convenient as Cartesian coordinates for the plane. So formulas, hopefully useful formulas, I claim that cache of the distance, well, so before getting there, just one word about some interesting special subgroups, and there's a trap here. Here's a special subgroup of PSL2R. So the group G contains the group of rotation matrices, and I purposely write them cosine of theta over 2, cosine of theta over 2, sine of theta over 2, and minus sine theta over 2. So that's slightly unusual to, well, you may ask yourself why I'm putting the 2 here, and I claim this is a rotation of angle theta, the point I, so the square root of minus 1 in H2. You just have to get used to seeing these two here. Even though it looks like a rotation by theta over 2 radians, as it acts on the hyperbolic plane, it acts by the double. And of course, when you take theta equal to 2 pi, this becomes minus 1, minus 1, 0, 0. But again, we are in PSL2, so it's projectively just the identity. Doesn't that one go clockwise? I think it is correct this way because if we look at how it acts on 0, where does this take 0? For small theta, it goes to sine theta over 2, it goes to the right. So that's the second trap in this formula. The minus has to be there, which may be the unfamiliar place. So that's one group. Here is another. E to the lambda over 2, E to the minus lambda over 2 is a translation of length lambda along the axis 0 infinity. I should have said what geodesics are. If you try to minimize the cost of going from a point on the imaginary axis to another point on the imaginary axis in terms of this dx squared plus dy squared over y squared, then clearly don't waste any time varying in y squared. You should better go straight up. It would be the same cost, the same vertical cost, and you save the horizontal cost. So clearly vertical lines, vertical rays are geodesics for the metric that we wrote. And since we can act by homographies, all the half circles terminating on the real line are also geodesics. And you get geodesics between 82 points. So here's another interesting group of isometries. Take cos lambda over 2, cos lambda over 2, sinh lambda over 2, sinh lambda over 2. And this will be a translation of length lambda along the axis minus 1, 1. So it looks a little bit like the first group I wrote, except there's no minus here. It's cos sinh sinh, and you don't have to remember where to place the minus. So I claim under these formulas that the cos of the distance from i to the point a, b, c, d times i that may be a basic thing to be interested in. Let's just think of it as the Euclidean norm of the matrix over 2. Why is that? Well, indeed. So let's call this the matrix A, sorry, the matrix M. And let's give names to those groups down here. So this will be compact group K. This will be, let's call it capital A. Indeed, the formula is true M in the group A. Namely, if I have a diagonal matrix here, I have a diagonal matrix. It sends i to a position, lambda units of length above i. So I should find cos of lambda. And you can check that if I apply the, if I take the half Euclidean norm of this matrix, I do get cos of lambda. It's e to the lambda plus e to the minus lambda over 2. It's true for M in A. And preserved by multiplying either on the right or on the left by matrices in K. If I multiply the matrix M by a rotation matrix on the left, then the Euclidean norm of the first vector doesn't change. The Euclidean norm of the second vector doesn't change. So the whole formula doesn't change. If I multiply on the other side, it's the same thing with lines, which is respected. And K happens to be the stabilizer of i. So if I ever have a matrix M acting on the upper half plane of this, here's i. I can find here's where it sends i. Here is a point at the same distance from i as Mi. So that's a circle. And what I can do is I can send i to the same position as Mi by applying an element of A followed by an element of K that we rotate. And I might end up here with the wrong orientation, but that's with the wrong direction, pointing in the wrong direction away from i. But I can solve this by precomposing, multiplying on the other side by another element of K. So check that K, A, K is in fact all of G. So that was one formula. Another one that I like to give is, let's see, the cosh of the distance between the line 0 infinity and A, B, C, D acting on the line 0 infinity. That's like, it looks like the determinant, except it's not. It's A, D plus B, C. It's sometimes called the resultant of the matrix. Again, you can see it in the same way. So here's the line 0 infinity. Here's its image under M. Indeed, it's true for M in the second group I gave, A prime. Namely, I translate by a distance lambda away from along this line minus 1, 1 perpendicular to 0 infinity. And it's preserved under multiplication by A on the left or right. So if I multiply by a diagonal matrix on the left or right, A and D get multiplied by two numbers, inverses of each other. So our B and C, and if I multiply on the other side, it's still true. So I can multiply on the left and right by this group A, which is the stabilizer of 0 infinity. OK. I'd like to give this kind of tedious to produce all these formulas, but I'd like to convince you that you can play with them and find out any quantity you may be interested in. So I'll give one more example of this. Let's say you have a triangle of sides P, Q with an angle theta here, and you're wondering what the distance R is, the third side. What you can do is, let's say you start with the hyperbolic plane. You apply E to the P over 2, E to the minus P over 2 diagonal matrix that moves the point I, P units up. Then you can rotate. The effect will be... So let me draw in red where the initial point goes. Sorry. Do I have red? Let me draw in green where the initial point goes. Here it is. Here it is. Then I'm rotating away just in case I need red. Maybe it's more visible. Let me draw in red. That's the point I, right? So I is still there, but the red point has gone up. Then I is still there, but the red point has rotated. Maybe it's here now. It's still there, but we multiply by the E to the minus Q over 2, E to the plus Q over 2. So we scale everything down, and we get a tripod like here. So P, theta, Q. So we get this product of matrices taking the point I to another point, and it bears exactly the same relationship to the original point I. It slices at the same distance as do those two vertices of our mystery triangle. So then apply this formula to the product of these three matrices. And we could do it. It's an exercise that gives you a nice formula. Clearly you could extend this by taking any path that you're interested in and find the distances. Okay, so I will, in the remaining time, I have something like 15 minutes left, I will discuss some of the other models of H2 and how we can relate them. So let me call this chapter a quadric models of H2. So let's say we are in R2. We have a unit circle. I mentioned at one point that the upper half plane up to a homography is the same as the unit circle. So we can ask, given a circle C, any circle has an equation of the form x squared plus y squared plus something affine. So let's say 2ax plus 2dy plus C equals 0. I can always put it in this form because it's x minus one coordinate of the center squared plus y minus the other coordinate of the center squared equals some radius squared. So we have this form. So on what condition on ABC is this circle perpendicular to the unit circle? Suppose we are at an intersection point. So x squared plus y squared equals one and x squared plus y squared plus 2ax plus 2dy plus C equals 0. Then the direction of the circle through this point is given by the differential of this function. So perpendicularity means the differential of x squared plus y squared is perpendicular to the differential of this formula of x and y. If and only if, so that's called 2x dx plus 2y dy perpendicular to 2x plus 2a dx plus 2x plus 2b dy. So in coordinates dx dy this means 2x x plus a plus 2y y plus a equals 0. I think of dx dy as vectors in the bay, orthogonal vectors in the basis of differentials. And I just pair them for the canonical scalar product. And this you can recognize as just the sum of the, let's see, that's 2x a, that's x squared plus 2x a plus y squared plus 2y a plus x squared plus y squared. So it's minus C plus one. So long story short, this equation for a circle defines a circle perpendicular to the unit circle if and only if the constant term here is one. So I'm done doing computations, now I will draw pictures. In other words, let B the surface given by Z equals x squared plus y squared. That's not a type, it's really Z on Z squared, maybe. So in R3, that's a paraboloid. What this says is that intersected with Z less than one projects vertically down to the unit disk. Tight Z equals one, I intercept exactly the unit circle. So that's fine. And H intersected with any plane of the form Z equals a x plus b y. Maybe two here. Minus one projects a circle C perpendicular to the unit circle. So let me, in other words, if I take a point, if I take the point zero zero minus one here, then we use color for the vertical projection. Whenever I take a plane through this point zero zero minus one, it's going to intersect the paraboloid along a certain conic. I really want to draw the full conic, so it will intersect the parabolic along a circle like this and if I project the circle down, it will look like a circle perpendicular to the unit circle. So what do I see if I place my i at this point zero zero minus one? Well, these planes, these circles in the paraboloid will look like lines to me. I see them in the plane they live in. So if I place my i down here, I'm going to think that H2 is a disk with a bunch of straight lines as geodesics. So this part, okay. And if I place my i down at infinity, then I'm going to see, in perspective, I'm going to see the true vertical projection, the true Poincare model. So this whole paraboloid model is, of course, closely related in the sense of projectively identical with the hyperboloid model, which is maybe more common. In fact, I can build such a model for any quadric in P3 that I like and that is a sphere. So more generally, take any quadric in projective P3 space, while it has to have the right signature. So it has to be essentially one circle, a sphere of the torus. That's just a sine problem. Take your favorite point outside of it. It's going to see a disk. And this equator will be a copy of the Klein model, so this projective model, or the Klein model of H2. And the near hemisphere will be a copy of the Poincare model. And the far hemisphere is another copy of the Poincare model. And the lines in all these models are just the intersections of the given surfaces, upper hemisphere, lower hemisphere equatorial plane with the planes through P, P being that point. Okay, so just a brief preview of what I will be talking about tomorrow. I will be describing quotients of the hyperbolic space by discrete subgroups and be saying a little bit about their deformation space and a little bit about the mapping class group, which is another discrete group related to those quotients. Thank you.