 Terima kasih. Ini adalah kali pertama saya di ICTP. Dan ia telah menjadi pengalaman yang hebat. Saya ingin berterima kasih kepada ibu bapa yang meminta saya. Semua orang yang berada di sini. Saya sangat gembira untuk memperkenalkan oleh Prof. Kulkarni, Ravi. Ternyata saya ada beberapa hubungan dengan ICTP dari awal kerjaya saya. Apabila saya kembali ke Singapura pada tahun 1988, ia masih diperkenalkan sebagai negara yang berkembang. Dengan UN, saya rasa. Saya sangat gembira dengan apa yang ICTP membuat. Kita cuba mempromosikan negara yang berkembang. Saya melihat beberapa tahun di Singapura. Saya memperkenalkan dan tidak membuat negara yang berkembang. Sesetengah kawan saya cuba membantu saya dan kata, kita patut melakukan sesuatu. Mereka adalah sebuah teori. Dan sebab sesuatu, kerana berat, kami telah menunjukkan sebuah buku dari keperluan berada di ICTP dalam sebuah teori keadaan dari veli gemetrik. Ia diperkenalkan pada tahun 1990. Dan ada sebuah artikel yang hebat oleh Prof. Kulkarni. Dan kami hanya memandangannya. Mereka berasal beberapa penutup berdasarkan pemerintahnya dan saya sangat menghargai karir saya. Saya rasa ini sangat bagus untuk datang ke sini. It's really nice to come back here and sort of know that, you know. Right, so it's really nice to come back here and see people who plays such a key role at the beginning of my career. Well, 25 years old. So anyway, today I'm going to give a... I'm giving the last talk so I try not to make it too tiring for you all. I'm going to be talking about identities on hyper surfaces. And I will try and give you the basic ideas behind the proofs of these identities without going too much into the technical details. Okay, here's my plan. Okay, so minus one was to thank the organizers. Then I give an introduction. I wrote that down because I'm no longer a young postdoc. I'm 55 years old and if I don't write it down, I'll forget. So as the last speaker, I think I would like to, on behalf of all the participants, the other lecturers and all, thank the organizers for doing such a wonderful job. So in particular... So I should mention their names. I probably get it all wrong. So Jose Ciade of Pepe, John Parker, Todd Drum, and then Bertrand Deron, who left us for the time being. Not permanently. Okay, so that takes care of point minus one. So I'm going to give a quick introduction. I'm going to introduce four sets of identities and then give you a quick idea about how really if you look at it the right way, the proofs are quite simple. Okay, and then, well, the last set is something I did with Feng Luo which just appeared last year, I guess. Okay, so since this is a conference on a geometry of discrete action, I really wanted to tie it in. Am I talking with the title of the conference? So the discrete groups come from the topology of the surfaces. You look at the fundamental group of a surface, it's a discrete group. You can also look at some other discrete groups like the automorphism group of the fundamental group, or since it's a surface, you can look at a mapping class group. Now, my talk is about identities on surfaces and somehow they're related to these two sets of discrete group action. The first set of discrete group action is when you think of the fundamental group of surface as a subgroup of PSL2R and then you get a hypoic surface. Actually, the identities turn out to be related to these other group action which is the mapping class group action on the type molar space, which is the space of all hypoic structures. And there's a vapid symmetric on this type molar space and that's preserved under this action and then you get a modular space. And it turns out that some of the identities have very, very important applications in that case as well. But I won't go too much into the second interpretation. So, my main aim is really to give you... I mean to talk about the last identity which is an identity I did with Feng Luo which appeared last year. But in the process, I really want to encourage all of you. I didn't participate in a problem session but I want to encourage all of you to sort of go out and find your own identities basically. So, I think if you have the right point of view, then sometimes a nice idea comes to you in the middle of the night or something and you can sort of push it ahead and then sort of really get some interesting mathematics. So, there are four sets of identities like I say Bas Mahjong, McShane, Bridgman and then the ones I did with Feng Luo. I'll discuss the actual identities themselves first and then I'll give you some generalizations and I'll end up with the sketch of the proofs. Okay, we start with Bas Mahjong. So, the idea here is I have a hyperic surface with a boundary, something like this. Now, on this surface, I can look at a set of geometric objects which are called the ortho-judasics and let's orient the ortho-judasics. So, what's an ortho-judasic? It's a judasic arc which starts from the boundary and ends at the boundary and is perpendicular to the boundary at both the starting and ending points. So, it could look something like this. It could go around a little bit go around a little bit over here and then it comes over here. So, this is an example of an ortho-judasic. Now, the set is a countable set, an infinite countable set and for each ortho-judasic you can associate since I started with hyperic structure I can associate a length to this ortho-judasic. We call this set of lengths the ortho-spectrum. So, Bas Marjan in 93 prove this quite a beautiful identity says that if I sum over the set of all ortho-judasics so alpha ranges over the set of all ortho-judasics and then there's this function B of the length of alpha and the function is very simple it's just a 2 log cotangent of x over 2. So, let me write it down. Bas Marjan is white. Over the set of all ortho-judasics of this function of the length of the ortho-judasic is equal to L where L is the length of the boundary. So, in this, in the example I drew there were four boundary components each one has length L1, L2, L3, L4 so the sum is equal to the sum of L1 L1 up to L4 and I guess I won't write out the function. It's a very simple function and in the course of the prove you see how the function appears. The second identity I want to talk about is the bridgement. Let me write the bridgement. So, the setting is exactly the same. You start with a hyperic surface with judasic boundary and then you look at the set of all ortho-judasics but what bridgement did was he found another function which arise BR of alpha and I sum up overall the ortho-judasics and this is equal to the volume of the unit tangent bundle which I can write as 2pi times the area of S. So, let me move to the next page. So, this bridgement's identity you're summing over the same same set of objects on the hyperic surface but the function now is different on the right-hand side and the length of the boundary you get the volume of the unit tangent bundle so in particular this thing here doesn't depend on the length of the boundary at all. You take the area of the surface and then for each point you take the whole set of 2pi direction so you just multiply by 2pi and that gives you the volume of the unit tangent bundle. The third identity which is of course my favourite and I'll just do it I'll write it first in the most simple case which is a case for hyperic surface with one cast. Now if I have a cast over here I can find pairs of ortho-judasic sorry pairs of simple closed-judasics alpha and beta such that together with the cast this forms an embedded pair of pens. And the action identity now says that the sum overall such pairs alpha and beta so let me draw it this way of 1 over 1 plus e to the l of alpha plus l of beta over 2 is equal to half so I wrote it as half let me write the 2 on the other side So this is the action identity and I want to stress again that this is an infinite sum simple closed-judasics alpha and beta which sort of cuts out embedded pair of pens together with this cast Now in all of these identities that you see over here you see that you require some kind of boundary and this you will see in the proof you either need a cast or you need a boundary component or you need a kind of cone point So while I guess one of the questions I was interested in was whether you could get an identity for a closed surface I gave a talk about this some 5 years ago without actually proving anything and someone in the audience got very excited and said that he could on interesting and started working with me and so we sort of managed to prove the result So this is a 4 and this will be I'm not saying this because I guess the point is that you should just if you find a talk interesting talk to a speaker and maybe something comes out of it So let me have a surface closed surface So the identity looks like this sum of f of p the sum of g of t is equal to the volume of the unit tangent bundle this is my surface s and what do I mean by p and t so p are all the embedded pairs of pens and t is all the embedded taurai So I'll just draw 2 examples of embedded pairs of pens and taurai So you could have a for example over here you could have a this could be a p you have embedded pair of pens you could have an embedded taurai there are infinitely many of them that's a good question so okay I'm going to skip some of the generalizations but f and g are this I'm sorry so so first of all it looks kind of terrible if you work with it enough you realize you start to love it this is not so terrible but they've got some loveable points so let me just say something about why it looks a bit more complicated because if you have a pair of pens there are several geometric invariance involved in a pair of pens there are 3 boundary components so you have to take all this into account and then actually if you look at the formula over there something turns up which is Roger's dialog which is very similar to what happens to a bridgeman identity and I forgot to write down sorry maybe let me write this now this is not it's falling apart but I think this should be correct so this was a bridgeman function and I'm thinking I'm thinking properly embedded in the sense that all the 3 boundary components are distinct torah you could think of it as lots of possible different pairs of pens not quite the same thing you have to sort of work slightly a little bit differently for that because the g of t is kind of a sum you know you see that there's a sum involved in it an infinite sum involved in the g of t whereas the f of t is just a finite sum sorry because you can decompose the torah pair of pens in infinitely many ways okay so let me just explain something of this again Roger's dialog which is interesting because that's one of these functions that turns up in all kinds of weird places number theory in calculating the volume of hyper tree manifolds and so on the second thing that sort of if you look at the terms over there it looks kind of complicated it involves the cross ratio of certain points okay so they're really cross ratios they're not as bad as it looks okay you know you have a 3 whole sphere pair of pens and then if you look at this picture inside the universal cover inside h2 there's a cross ratio of certain endpoints and these are what is appearing inside Roger's dialog okay so these are the sets of identities that I'm going to try to talk about 1, 2, 3 and 4 and how are they sort of related in terms of the proof the basic idea of the proof is very simple so I'll tell you what the idea is since this is really the okay I have to show you this page because this thing is recorded people who have worked on this may go and read it so I should give credit to these people so Bas Mahjong's identity by the way, works for higher dimensions as well McShin's identity was generalized into surfaces with boundary judasic boundary by myself and Wong and Chang to cone surfaces and then in various other context punctured surface bundles by Boldich Akiyoshi, Miyachi and Sakuma for quasi focusing representations also by Boldich for Hypolycrim Manifers from Day Surgery by myself, Wong and Chang for close juniors to surface by McShin himself for hitching components so I think there were some talks on hitching representations so this was done by Laburi and McShin and complex Hypolyc representations by Kim Kim and Tan but this two kims, one of them is the same as the two kims in one of the previous the other one is not okay Do, Norman Do and Nobri for Dubric as well and then Sakuma and Lee Lee and Sakuma so anyway, there's a fair amount of work I mean to say done on the McShin identity but since this is a talk with a lot of young people but in particular Pepe I want to say so when McShin prove this identity it was quite beautiful and kind of remarkable but it's kind of an oddity it's like what's it for and so Miss Kani when she prove a generalization use it as a very important tool in calculating the very bits and volumes of the modular space and also in calculating the asymptotics of the growth the length of super close jodasic on the surface so this was maybe I would say this part was one half of a feels metal spinning work and actually did some other things so anyway you can do something and it's so interesting and you never know how useful it can become later so for Bridgeman he also generalize it to higher dimensions with Jeremy Khan and then I just want to say for all identity for closed surfaces you actually can also do it for surfaces with boundary you can also do it for non-orientable surfaces and there are some possible applications actually what time do you want me to end? 20% I'll try and end a quarter pass anyhow okay I'll try and end a quarter pass okay there are these four sets of identities and there are some generalizations and it's all interesting and useful how do you go about trying to prove your own identity so the basic this is the basic idea and I'm going to write it out because it's the basic I'm going to clean this up you choose some set X with a certain measure I mean if you like think of it as a circle just an interval or something like that and then the measure of X should be finite here so whatever it's the length of the circle then you try and decompose this set X in an interesting way so you just write X as so as a kind of a countable union of disjoint subsets just you know there are many pieces countable many pieces but usually what happens is this part over here doesn't cover all of X and you have another piece over here okay and this piece over here which I just call Z typically it's a very complicated piece okay I mean if you're in dynamical systems Z is the part where all the dynamics is happening but as far as identity is concerned Z plays no role because the Z will have measures 0 okay so when you try and compute the measure of X then you just take the sum of the measure of Z plus the sum of the measures of the WI but then this part will have measures 0 even though it's a very complicated set and then this part over here for part 2 okay for the B which is actually a big part of this kind of problem WI I'm going to I apologize to the experts in the view or rather but I'm going to start with example minus 1 so example minus 1 is my set X is just the interval from 0 to 1 with the measure 1 and then what's my set Z my set Z is just the set 0,5,3,4,7,8 I just realized you could have taken for all you like 0,5,1,3 1,4 and so on okay so what do you know about this set Z this set Z is a countable set so it has measures 0 on the other hand it divides this interval into countably many other open intervals so you just need to sum up all the measure of all the open intervals plus 0 gives you 1 in this case if you choose Z in such a way it's 1,5,1,4,1,8 and so on if you choose the points 1,5,1,3 1,5,1,3 you get a slightly more interesting identity I just realized I mean you don't really have to prove anything right you don't have to try and use some geometric series or anything to prove I mean this is the identity you just need the fact that Z is a countable set and so has measures 0 so we go to example 0 in example 0 we do a cantor set construction so my blackboard technique is terrible and I remove the middle third and then I remove the middle third we all know this and so on what's left over is a cantor set so the set Z is the cantor set but the rest consists of this open intervals you see there's a big one which has length 1,3 there are 2 smaller ones which has length 1,9 and there are 4 smaller ones which has length 1,27 and so on so what we have really is the measure of X which is equal to 1 is the measure of the cantor set plus the sum of the measures of all the intervals okay and then you get again a geometric series like this but this example is a little bit more complicated because the set Z already is complicated it's a cantor set I mean it's not obvious that this thing has measures 0 I mean in some sense you need to prove that this thing has measures 0 to get the identity but this is an exercise we often give to our students prove that this cantor set has measures 0 and then has the big measures 0 and you do it and although this is a very simple example essentially the basmah jen and the matching identities are just this okay the difference is that the way that you do the decomposition of the set is a little bit more geometrically motivated and because of the geometry the functions come up from the geometry okay so we do example 1 now which is the basmah jen so what is the set X that I want and the set X I want is basically the set of all unit tangent vectors emanate base at the boundary which are perpendicular to the boundary right so X is a set of all unit tangent vectors perpendicular to the boundary what is the measure of this set well the measure of this set is just the length of the boundary because basically at every point then what do you do well if I give you a tangent vector v I can just exponentiate the tangent vector so you just exponentiate the tangent vector you use it to create a geodesic and you just do it so you think of it as shining some laser beam which on your surface and it goes along what can happen when you do that but this can happen either in a finite time it hits the boundary again or the other thing that can happen is that it goes on and on without ever hitting the boundary okay what do you think is the likelihood that you will never hit the boundary again come on I'm dying over here Elisha you can give me zero X so why is it so essentially you have to be extremely unlucky not to hit the boundary again you know and you can see this quite clearly if you go to the universal cover what happened to my erasers I'm slowly erasing the things I need but never mind when you go to the universal cover so let me say this is a boundary component and this is a fundamental sort of interval component then you have all these leaves of all these boundaries and basically what happens is that in order to sorry in order to continue on indefinitely you really have to point towards a limit point and the set of limit points looks like a cantor set and it has measure zero okay by so if you have a group what type do you call this I don't know the set of limit points is a very small set it has measure zero so that will be your Z so the set of directions where you continue on indefinitely will have measure zero and that's this set Z but what happens if you hit the boundary you may hit the boundary tangentially or I mean not quite in an orthogonal way but you can shift it a little bit and you see that this thing will be homotopic so if I started over here I went around and somehow I sort of hit this a little bit off here I can shift this a little bit and if you shift it a little bit you get the same curve up to homotopy but eventually you can get the shortest representative so every every V for which gamma V is finite is homotopic to some unique alpha I you sort of shorten you sort of shorten the curve while moving the two end points on the boundary so everyone is homotopic to some unique auto-judicic alpha I and we can now define W alpha I will be the set of all V inside which acts such that gamma V homotopic to alpha I so what I just say is that this set X now decomposes into this set of measures 0 disoan union with the W of alpha I and you can then calculate the measure of W of alpha I if you look at the universal cover picture again you will see that basically you're going to be trying to measure something like this you're going to try to measure this interval over here it only depends on this is a bad drawing but it only depends on this length over here which is the length of the alpha I this is a a very very basic hyperjompere computation which maybe you did last week you do a calculation and then you just get the vast margin identity so that's the vast margin identity so it's sort of like you're standing over here I'm going to do what you do you're standing over here and you're looking down and you sort of see the boundary so you may see one of this sorry, I start from here I see this boundary component now that boundary component projects onto some piece over here onto this project on a geodesic so that's why I want to use oriented geodesics to keep everything clean yeah so it's just basically the projections of this geodesics onto the other geodesic and then you calculate this thing and then you sum it all up and you get the vast margin identity you can sort of see how to do this in higher dimensions and so on I mean this is the basic idea okay I hope you all got this because this is really the easiest case but this was a paper the young people you can get a very nice result without necessarily having a lot of technical expertise so you can do this and how do you get the margin identity then well what you do is that instead of projecting a laser light okay you're going to do exactly the same thing you're going to do exactly the same thing and you start from for simplicity do it from this picture assume there's only one cusp so I'll take something over here which is a horocycle and I look at a set of all unit tangent vectors perpendicular to this coming out so what can happen so basically these things are coming from the cusp do not think of yourself as what happened in a bus margin you go on and on but now what you do is you think of yourself as building a wall instead of shooting a light beam if you're building a wall then if you hit yourself again you cannot continue I like to give this top in China where there are great walls so you just go along and then maybe you come back like this and then the moment you hit yourself you stop so again I'm going to ask this question out of all the possible unit tangent vectors what is the probability that you end up with something that has infinite length so in order to get infinite length you sort of have to keep avoiding yourself and if you know hyper geometry if you have this surface it's very hard to avoid yourself you sort of end up with smaller and smaller intervals so this is really sort of so the answer to the question is the probability that you go on and on and create a simple a simple curve of infinite length the probability of that happening is zero and this comes from the Berman series result so I should write it down so there's going to be a set z which is set of all v let me call this g of v it's infinite and this set here has a measure zero by the Berman series okay so we throw it away we don't worry about how complicated the set looks it's a horrible set looking thing you just throw it away it's not going to be relevant to the identity what's left over with geodesic g of v which hit themselves so when a geodesic hits itself it's going to create some kind of topology over here you can look at the regular neighborhood of this geodesic when you let a regular neighborhood of this geodesic you're going to have two other curves which together of this thing over here forms a pair of pens I've drawn it in this case the two curves will be one curve over here the other one will be homotopic to beta so you have to look at a small regular neighborhood and then you get this thing but then you can always because it's a hyperbolic surface you can take geodesic representative so when you do that which is this picture over here you see that this thing here this curve g of v is embedded inside a pair of pens so you can now instead of doing some global analysis over the surface you can now restrict yourself to a local analysis on pairs of pens so we just look at pairs of pens so if I draw a pair of pens what happens this is mine so I'm not going to clean it off you have a pair of pens over here and then and there's one sort of direction where it starts from the cast and ends in the cast then somewhere around here you can go this way and then you there's a kind of a limiting case where you spiral around this so let me call this alpha and call this beta there's one thing over here which spirals what happens when you're in between here and here if you're in between here and here you'll be going around here and then you self come back and hit yourself so similarly there's something over here which sort of goes spirals there's a gap over here there's a gap over here where any unit tangent vector which lies inside this gap will produce a g of v which is a spine for this pair of pens it's going to so it will be associated to this pair of pens if there's the same gap on the backside of this thing there's one in front, there's one at the back but what happens if you go beyond here a little bit if you went beyond here a little bit it's going to cut the boundary component beta before it intersects itself you're going to go around this cylinder and you're going to go across the boundary component and you're going to intersect yourself when this happens by the time it intersects itself it will be the spine for a different pair of pens so it will correspond to a different alpha and beta different pair of so there's this thing over here and I'm going to call it w alpha beta for example to be the set of all v such that this is a spine for p alpha beta so let me call this p alpha this is the pair of pens determined by the alpha and beta then again what we have done here is that this thing over here which is x is the union of this with this set over here so there's one horrible set very complicated but it measures 0 and then there's a countable set for which you can compute the measure so how do you compute the measure of this I've more or less told you how to compute this thing what you do is you just you need to be able to compute these gaps so let me remind you how you get a gap and there's a limiting geodesic over here which spirals into beta there's a limiting geodesic over here which spirals into alpha spiral means asymptotic towards alpha and in between everything is going to create a spine so everything inside this interval over here corresponds to leaves inside w alpha beta so in a in a action identity you can do this and if you start out with a cast you get a action identity and then if you and you can also start out with geodesic boundary component there's no reason why you should start with a cast over pairs of pens pairs of pens which contain a cast so I'm looking at all pairs of pens which contain a cast as a boundary component and then there's going to be a measure associated with each one of this pair of pens which really depends on the local geometry over here in other words that only depends on the length of alpha and beta it doesn't depend on the rest of the surface sorry? ya ya ya ya ya exactly the same thing that's Mr. Kani's version of the McChina Denti okay I didn't write down the formula but let me write it out for you so if I have something over here of certain length delta and here okay so maybe I have alpha and beta over here the function becomes g of delta over 2 alpha over 2 beta over 2 where g of xyz is equal to log of e to the x plus e to the y plus z over e to the minus x plus e to the y plus z okay so there's a completely explicit function that you can write for this ya this is a minus x oh no, I'm taking over the halves over here because I didn't want to write all the halves inside here okay and although there might be a 2 in front of course just because there's a front side and a back side but basically this is the function okay but this is the basic idea so okay Bridgeman's identity, how does Bridgeman prove his identity so you need a new idea so I guess he was not aware of Bas Marjan's identity so what he did, I mean he came from a slightly different point of view he looked at the unit tangent bundle you look at the set of all possible points with all possible directions which is really looking at a unit tangent bundle of the surface so you don't start from the boundary you start from inside and when you start you start from everywhere you don't just choose a specific starting point you look at all possible starting points so if I give you a v over here a unit vector, a tangent vector then what you can do is you can exponenti it in a forward direction and a backward direction and you will get some geodesic okay so you can exponenti it in a forward and backward direction and you get some geodesic gamma v okay again the question is what is the probability that you do not hit the boundary okay I mean if you start from many points at the surface and the direction exponenti geodesic what is the probability that you somehow go on forever in a forward direction so I'll just tell the answer is again zero the set of such directions is very very small and this follows from a goddess of the geodesic flow so you see that a lot of things sort of going to saying you don't have to prove it anymore but the fact that z has measures zero actually it's not really obvious but it's been done so the z there's some set which has measures zero and we ignore that set but then what's left over will be things for which gamma of v is finite in other words it's going to hit the boundary in the forward direction and it's going to hit the boundary in the backward direction but if that happens you're almost back to the same thing as the bas margin identity because that thing is going to be homotopic to a ortho geodesic it's going to be homotopic to a ortho geodesic you can sort of shift the two end points a little bit and so every finite geodesic arc starting and ending in the boundary is homotopic to a unique ortho geodesic and how do you do a calculation over there well you can do the calculation as follows you choose a lift you choose a lift and then here's the ortho geodesics of alpha i you choose a lift to the universal curve and you're really looking for all unit tangent vectors so let me put this put a direction into here you're looking for all unit tangent vectors which sort of start from this part and end over this part so you have to look for things that look like this so you're looking for all tangent vectors for which if you're exponential in a forward and backward direction the starting point is over here somewhere on this geodesic and ending point somewhere on this geodesic and you have to do a calculation and Richmond spent about several pages doing this calculation 6 or 7 pages, i mean he's quite good at this things he eventually ended up with this the Richmond function which is sort of VR so x is equal to 4 times the Rogers dialog algorithm of second square i think x over 2 so Caligari who's even better at this calculation sort of did the calculations in half a page i guess using a slightly different point of view anyway but there's a certain more work involved after you get the idea so what do you do for closed surfaces in the last 7 minutes i'll try and indicate how you can get an identity for closed surfaces so if you sort of look back at what happens in a vast margin identity you had to have a boundary to get started because you were sort of decomposing the boundary of the surface if you look at Richmond's identity you also needed boundaries because otherwise if for example this was a cast or this was a closed surface or whatever there's no way to do anything and if you look at the action identity again you started from boundary component so the question then is if you had a closed surface with no boundary where do you start so when i first found out of Richmond's paper i guess i was quite excited and i was looking at it and then one night i guess it just hit me the idea was exactly what he did which is anywhere and everywhere you look at the unit tangent bundle but then if you understood the proof of the action identity you realised okay what you do is you start from anywhere and everywhere but you try and create some topology instead of exponentiating until it goes on forever and covers up the whole surface or whatever you try and stop whenever you hit yourself you sort of build a wall so that you can create some kind of topology and that's the basic idea so you start from anywhere you take a unit tangent vector v and you're going to you're going to exponentiate this in both the forward and backward direction and for simplicity we do it in equal speed okay so you have two themes of people building a wall one in the forward direction and one in the backward direction then what happens well there's a kind of a i don't have coloured chalk but there are teams of people so one red wall and one white wall whatever but anyway you just continue until you hit you hit the wall either built by yourself or the opposing team then the other team will continue on until it hits the wall so what do you create when you do this typically you create some kind of graph embedded in the surface which is of euler characteristic minus 1 basically you create something homotopic to two loops okay but of course there's a possibility that you went on and on and you build this really long wall and you never really hit yourself and you just manage to go on and on you chose this point in this direction where you just went on and on and it's not likely to happen in fact it's the probability that you started with a point in the direction where you have this infinite length wall is zero and this follows from either a godicity or the geodesic flow or it follows from the Berman series argument basically there's just very few possibilities there's almost no chance of not hitting yourself from measure theoretic point of view there's no chance of not hitting yourself okay so you do and so that set again like I say it's a really complicated looking subset of the unit tangent bundle we throw it all away because the set has measure zero what's left over will be sets of unit tangent vectors V such that let me call this thing over here G of V okay so there's V and then this object over here is this graph such that G of V is a finite graph but when you have something like this then you can do the same trick as you did for the Maxine Identity because you can look at the regular neighborhood of this thing over here if you look at the regular neighborhood of this it's either a three-hole sphere or one-hole torus you can do that and then for example you might get some of the trios here this boundary of the regular neighborhood tight and then you have to prove that somehow when you pull this tight this graph over here still lies inside the embedded pair of pens or punctured torus that's not very difficult but is it obvious to you Mahat which this one it will not shrink away from this thing here it shrinks away from this thing but so what's the point over here you then reduce yourself to a local analysis I mean you have a surface of genus, high genus G everything's really complicated but really when you try and do this thing over here you only need to look at this embedded pairs of pens or embedded punctured one-hole torus so if I just look at one of these pairs of pens which lies embedded inside here with judasic boundary over here like this the question becomes for which tangent vectors unit tangent vectors V will I generate a spine so maybe it might go like this a spine for this pair of pens so the way I drew this over here this particular V maybe generated a spine but sometimes you hit the boundary of the pair of pens before you were able to hit yourself because you've gone beyond that boundary and then what you've done that corresponds to a different pair of pens embedded pair of pens so this gives you a decomposition of the unit tangent the unit tangent bundle of this surface into one horrible piece Z which has measure 0 and a countable union of pieces which are sort of indexed by the embedded pairs of pens and the embedded computing this thing here it's a pen this was a thing I couldn't do so when I got this idea I thought what the hell I'll just give this talk anyway and I think oh it's almost impossible to really get a formula for this thing it looks terribly complicated but Feng Luo was really good he just sort of zoomed in and together we managed to get a formula so that's where this identity comes from this is the F, this is the G and you have to really you need to do some work and calculate these functions F and G and then if you do that you can get a formula so my point in some sense is that you need a good idea and you need a little bit of technical expertise and you want to put it all together and you need a lot of luck so that this thing may be useful to someone else so I'll stop here