Good morning. In today's lecture, we will be studying Sturm-Liouville theory, which will take us from the domain of ordinary differential equations to the domain of approximation theory. So, in between the two theories, this Sturm-Liouville theory forms the bridge. Now, recall that till now, a lot of our discussion has been in the context of initial value problems with boundary value problems getting very small share of our focus. In the context of Sturm-Liouville theory, we concentrate on boundary value problems in a particular manner. Consider this simple boundary value problem y double prime plus twice y equal to 0 with both the boundary values of y at x equal to 0 and x equal to pi given to be 0. Everything homogeneous, the differential equation as well as the boundary conditions. We try to solve this boundary value problem first the ODE. As we try to solve this ODE, we know that we will construct the auxiliary equation, which will be m square plus 2 equal to 0. So, m will be found to be plus minus root over 2 and based on that, we will form this solution, general solution of this differential equation. Up to this point, it is fine. Then, we will try to supply these boundary conditions to determine the values of a and b. Now, as we force this condition at x equal to 0, y is 0. Then, at x equal to 0, this term will vanish. So, we will find that b cos 0 is 0. Now, cos 0 is 1. So, it will be b equal to 0. So, from the first condition, we find b equal to 0. So, putting that, then the solution reduces to only this much, this y x is a sin x root 2. Then, we apply the second boundary condition. At x equal to pi also, y is 0. So, as we put x equal to pi, we find that a into sin pi root 2 is 0. Now, sin of pi root 2 is not 0. So, a has to be 0. Now, if we put a equal to 0 here, then we find that y is equal to 0. This is the only solution possible, which we knew before we started the entire process. So, that is, in this particular situation, with these boundary conditions, this differential equation has only the trivial solution and no non-trivial solution. In a similar manner, we try to solve another boundary value problem, in which everything remains the same, except that this 2 changes to 4, and we consider this boundary value problem. Exactly the same steps will give us, in this case, this solution, y equal to a sin 2 x in which a can be arbitrary. Now, you find that in this case, we could find only the trivial solution. There was no non-trivial solution at all. In this case, we have got infinite number of non-trivial solutions. For every value of a, we have got a non-trivial solution. So, what makes the difference? Between this case and this case. In a more general setup, we can consider this boundary value problem with k here, where earlier we had 2 in one case and 4 in the other case. Now, we say that for which values of k, we will be able to find non-trivial solutions of this boundary value problem. We repeat the same steps and find that with k less than or equal to 0, there is no hope of finding a non-trivial solution. Why so? Because with k negative, we find that this will give rise to an initial equation, an auxiliary equation, which is m square minus 1 equal to 0, m square minus something equal to 0, not necessarily 1. So, m square minus something equal to 0 and that will mean that m will have 2 real solutions, m will have 2 real values and there will be exponential solutions. Now, if you find an exponential solution, which will rise from the 0 value at some point, then it will never return to 0 value again. Therefore, with negative k, we cannot hope for a solution. With 0 k also, there will be no non-trivial solution, because in that case y double prime will be 0, which means that y prime will be constant, which will mean that y will be a linear function. And a linear function also, if it rises at some point from 0, then it will never come back. So, with k equal to 0 also, we do not find any solution for this, except that trivial solution, which is y equal to 0, which will always remain. Now, we consider positive values of k. Now, if k is positive, then we can call it nu square. Now, we find that with nu equal to plus minus 1, plus minus 2, plus minus 3 etcetera, we can find solutions of this b v p and those solutions will be a sin nu x. So, in the earlier case, with k equal to 2, we could not find a solution. With k equal to 4, we could find a solution, because this was the case, which fits there. So, we find that for quantized values of k, that is 1, 4, 9, 16 and so on, we get non-trivial solutions of this b v p and corresponding values of nu are 1, 2, 3, 4 etcetera. Now, for the purposes of recording these solutions, we include 0 to include the trivial solution also in this family. So, nu equal to 0 will give the trivial solution, nu equal to 1 will give us a sin x, nu equal to 2 will give us a sin twice x and then similarly, a sin twice x and so on will be the non-trivial solutions. All of these solutions are bound at the two values at 0 and pi. This is like a string, which can stay tight or which can have this kind of a shape or this kind of a shape and so on. So, the different shapes of the string will be, one is this and there is this and there is this and there is this and so on for values of nu as 0, 1, 2, 3 and so on. Now, the size, the amplitude of the solution is here a, as that can be very continuously you can have the same solution copied several times with different scale factors. So, that is simply the scale factor and that scale factor tells us that for the single that amplitude that shows us that for the same value of nu among these positive integer values you can have infinite solutions every solution for every value of a. Now, we ask these questions that for what values of k or Eigen values, does the given BVP possess non-trivial solutions and the answer to that in this question is k equal to 0, 1, 4, 9, etcetera. And the second question is what are the corresponding solutions, we call them Eigen functions. These particular values of k are called Eigen values and these particular corresponding solutions are called Eigen functions up to arbitrary scalar multiples which is here in the amplitude. Now, these two questions look very similar to the algebraic Eigen value problem and which is a v equal to lambda v for which we ask the question that for which values of lambda this system possesses solutions, non-trivial solutions and what are the corresponding solutions v. So, this is the reason that we get a problem of this kind which we will be calling as the Eigen value problem that in the case of this problem we called it the algebraic Eigen value problem. This qualifier, this adjective algebraic was used to basically qualify this particular problem from the Eigen value problem of the differential equations which we are studying now. And as this particular boundary value problem framed in the form of an Eigen value problem in this manner has a direct resemblance with the algebraic Eigen value problem which is this. Similarly, we also have a resemblance of the Hermitian matrix or in the case of real numbers the symmetric matrix with what we call in the literature of differential equations as the self adjoint differential operator. And that is the next preliminary idea that we try to develop before going into the actual stream liable theory. So, for considering for figuring out what is this self adjoint c of a differential operator, we again start from some simple notions. Consider this second order homogeneous ordinary differential equation linear homogeneous equation and in the corresponding first order equation the way we try to solve the equation the Leibniz equation in that we had y prime plus p x y equal to something. The homogeneous version would be y prime plus p x y equal to 0 and there the way we try to find the solution was to find out an integrating factor f with which we can multiply the left hand side to make the entire left hand side the derivative of something. A similar exercise if we try here then let us say let us see where do we reach. We say that is it possible to multiply this entire differential equation with something some integrating factor f such that this entire stuff f into this entire thing turns out to be the exact derivative of an expression which is which involves up to first order derivatives in y first order derivative in y. That is can we find out a function f and an associated function g such that f into y double prime plus f p y prime plus f q y turns out to be the exact derivative of a first order differential expression of y in which y prime and y will be involved and y double prime will not be there exact differential of this. So, we know that if we can do that then the coefficient of y prime in that will be f because its derivative being this will mean that there must be a term in which we will have f y double prime. So, that is why this term must be f, but what about the rest of the terms will they oblige to match these terms correctly.  So, we try to differentiate it exactly and find f y double prime matching this first term plus f prime y prime which is here plus from here g y prime which is here plus g prime y which is here. Now, as we try to match this first term with this as we try to match this entire expression with this entire expression the first term matches what about the others coefficient of y prime here it is f p here it is this. So, we say equality will require that this is equal to f p and then here coefficient of y here it is g prime here it is f q. So, equality will require this. Now, what we have got in this line any line we have got two differential equations in the two functions that we were asking for that makes sense two unknown functions which you would like to determine and for that two differential equations coupled yes this and this are coupled. So, we say that for solving two unknown functions from two differential equations we can try elimination that is we try to our focus is f the function which is supposed to turn out to be the integrating factor. So, we try to eliminate the other function g x. So, how to eliminate that we can differentiate this and there g prime will appear there will insert this value of g prime. So, as we differentiate this we get f double prime plus g prime. So, f double prime plus g prime. So, for g prime we put f q.  So, those two things we bring to the this side this is minus p f prime this is minus p prime f. So, the two parts two components of the derivative of f p with negative sign has come to this side. Now, see what does this mean this means that it will be possible to find f x and g x such that f x into this will be turn out to be the exact derivative of an expression like this if the function f x satisfies this. Now, what is this this is again a second order differential equation in f which is of the same type as the original differential equation. This differential equation is called the adjoint of this differential equation. So, that means the adjoint of this differential equation is that other differential equation which an integrating factor has to satisfy such that that integrating factor multiplied with this will reduce it to the exact derivative of an expression like this. Now, in that in terms of helping in the solution process of this differential equation did this analysis make any contribution really no because this was a second order differential equation in which coefficient of y prime was unity here also this is also a differential equation in which the coefficient of f prime f double prime is unity. Here the coefficient of y prime was p here the coefficient of f prime is minus p something similar here the coefficient of y was q here the coefficient of f is q minus p prime there is no reason to believe that this will be simpler compared to this in general. So, in terms of procedure to solve the differential equation this discussion did not help, but the concept that we develop further in this line of argument will help in some other place. So, the next question that we ask is that what are the properties of this adjoint for example, till now we have seen that the adjoint of this differential equation turns out to be this where the coefficients of y double prime and f double prime are one in each case in both the cases. If the original differential equation had a coefficient p x here then the adjoint will have p 1 of x which is just the negative of p.  So, this much we have got. So, this is the adjoint of this differential equation where p 1 is minus p and q 1 is q minus p prime. Can we find the adjoint of this differential equation adjoint of this differential equation will be another such differential equation phi double prime plus p 2 plus p 2 phi prime plus q 2 phi equal to 0 in which p 2 will be minus p 1 and q 2 will be q 1 minus p 1 prime exactly like this.    Now, you find that these 2 cancel these 2 cancel and you get back q. So, what we have got? We have got that the adjoint of this turns out to be phi double prime plus p phi prime plus q phi which is exactly the same as this differential equation. In this the name of the unknown function was y here the name of the unknown function is phi, but they are the same differential equation. So, what do we establish out of this? We say that the adjoint of the adjoint of a second order linear homogenous equation is the original equation itself. The way you use to say that the transpose of a matrix you can find and then you can say the transpose of the transpose of a matrix is the original matrix. So, that transposition or the conjugate transposition have has the same relationship as the adjoint would is here. So, adjoint of the adjoint of a second order linear homogenous equation is the original equation itself. Now, transpose of the transpose of the matrix is the original matrix. Then you ask that which matrix is its own transpose and as answer you found that it is a symmetric matrix or which matrix is the conjugate transpose of itself as answer you found that it is the Hermitian matrix. A similar question you can ask here when is an ODE its own adjoint. So, for asking this question you find that you have this differential equation and you say that if this differential equation is to be its own adjoint then its adjoint which is this equation with p 1 and q 1 like this. Then you will find that p and p 1 has to be have to be equal, but p and p 1 are negative of each other. So, for being equal as well as negative it has to be 0. So, that means that this kind of a differential equation can be its own adjoint only in the trivial case when p is 0. So, if p is 0 then you will find p prime is also 0 and q 1 is equal to q in that case it will be a self adjoint that is the ODE will be its own adjoint. So, with p x equal to 0 in this case every differential equation every second order differential equation like this with p x equal to 0 will be self adjoint this is trivial, but what about something into this that is if you have this as the differential equation f y double prime plus f p y prime plus f q y equal to 0 that is if this differential equation is multiplied with a factor f f of x then you say that under what condition this will turn out to be self adjoint. That means after multiplying this standard form of the ODE with the integrating factor when we can say that now it is self adjoint even if p is not 0. And for that we ask the question in a slightly different manner first we try to ask what is the adjoint of this equation and then after finding the adjoint we compare that adjoint equation with this and then ask under what condition these two are actually the same differential equation. So, we ask this refresh question of this what is the say when we want to find out what is the adjoint of this. So, we if we can look for the differential equation with the adjoint has to satisfy then we have reached some level of answering this question. So, we refresh the question and say what is the ODE that phi has to satisfy if phi multiplied with this makes it an exact derivative like this phi multiplied with this whole thing phi f y double prime plus phi f p y prime plus phi f q y equal to exact derivative of a first order derivative expression like this first order expression. So, the first term of that must be phi f y prime. So, that its derivative can account for this term plus something into y. So, we compare terms the derivative of this is exactly derivative of this derivative of this will involve two terms one is psi y prime and the other is psi prime y. So, psi y prime must be equal to this. So, phi f p must be equal to the this will produce another term for that matter the derivative of this is phi f y double prime plus derivative of phi f into y prime. So, derivative of y prime phi f into y prime. So, that will have with itself this part. So, phi f derivative into y prime plus psi into y prime. So, phi f derivative plus psi should equal the coefficient of y prime here. So, that is this and the coefficient of y from here will be psi prime and that must be equal to phi f q that is this. So, these two conditions must be satisfied. Now, again our question was what is the ODE that phi has to satisfy if it is going to work as the integrating factor for this. So, psi is an extraneous function which we had to accommodate to satisfy the form. So, here in this system of two differential equations in two unknown functions phi and psi we want to get rid of psi. So, we differentiate this and in that where psi prime appears we insert this. So, eliminating psi we differentiate this we get second derivative of phi f plus psi prime in which case we insert this is equal to derivative of this which is this. So, this is the differential equation which phi must satisfy if it is going to make this an exact derivative through multiplication that is the adjoint of this differential equation will be this differential equation which phi must satisfy. So, let us expand it and write it in proper form.  Now, we collect the terms together second derivative terms here only one first derivative term one from here and the other from here that is twice f prime minus f p phi terms without any derivative.  Now, you say that this is the differential equation which phi must satisfy to play this role that means this is the adjoint of this equation. Now, we ask this question that is this self adjoint it will be self adjoint when it is adjoint which is here turns out to be the same differential equation could be because here the coefficient of y double prime was f here the coefficient of y double prime is f same. Now, if the other coefficients are also equal then this differential equation and this differential equation will be exactly same when will that happen when f p is equal to this which will mean twice f prime equal to twice f p that means f prime equal to f p that is f prime equal to f p and if f prime is equal to f p then here you see this is f p. So, if it is f prime then f prime prime f double prime then this f double prime and this f double prime will cancel each other and what will remain is f q which is the same as here that means this satisfaction of this simple condition single condition will ensure that this differential equation turns out to be exactly the same as this differential equation. But the way we derive this we know that this differential equation is actually the adjoint of this. So, that means that this differential equation will be its own adjoint if this condition is satisfied that is if f prime is equal to f p that means if the coefficient of the y double prime is differentiated to get the coefficient of y prime then we have a self adjoint second order differential equation. Same thing happened when we tried to solve the Leibniz equation in that case we had y prime plus p y equal to something and we multiplied it with f and then we said f y prime plus f p y equal to something and then this left side we had the condition that f prime must be equal to f p the same condition we get here. So, we can cast a second order differential equation into the self adjoint form by multiplying with f x which is this the same thing which we found as the integrating factor in the case of Leibniz equation. So, when we have a second order differential equation like this and we want to convert it into the self adjoint form all that we need to do is to get this function p integrate it and put that integral over the exponential in the index and get e to the power integral p d x and that turns out to be the f f of x which we should multiply this to convert it into the self adjoint form. So, once we multiply this equation with this particular f x then we find that the here we have got f and here we have got f p and then it is certain that the derivative of this f will turn out to give us first this and then into the derivative of this, derivative of this is p and that p is simply sitting here.  So, to cast the o d e into the self adjoint form all that we need to do is to multiply the equation in this form with this f x. So, in that case we find that these two terms together will become the exact derivative of an expression like this f y double prime will come here and then f prime y prime will come here because here we will get f p which is same as f prime. So, this entire term these two terms together will be sitting here the exact derivative of this kind of a combination plus whatever comes here into y. So, this will be the general form of the self adjoint differential equations that we are going to study. Now, there are two small working rules which we can immediately appreciate and remember one is to determine whether a given o d e is in the self adjoint form we simply check whether the coefficient of y is the exact derivative of the coefficient of y double prime. So, if the coefficient of y prime is the derivative of the coefficient of y double prime if it is not so and we want to convert it into self adjoint form then first we convert it into the standard form by dividing throughout with the coefficient of y double prime if it has any non trivial coefficient which is not one. So, to convert an o d e into the self adjoint form first we obtain the normal form the standard form by dividing the entire equation with the coefficient of y double prime after that whatever appears as the coefficient of y prime call that p x and then multiply this new equation throughout with this integrating factor this f and the resulting equation is in the self adjoint form. Now, with this self adjoint form what can we do with this self adjoint form in our hand we have certain an interesting theory which is the basic issue in stream level theory. This way when we define the equation and in the next step we are going to expand this and have a lot of other terms involved in it including the so called Eigen value. In the place of f x now we will have this small r x and in the place of this capital r x we have this large thing in which this lambda that is getting involved here will be termed as the Eigen value and this equation which is a little bigger version of the previous equation that we studied is called the stream level equation in which the functions p q r etcetera have certain properties. What are those properties first of all p q r and r prime we need all of these to be continuous over the closed interval a b and p x must be a positive definite function on a b and r x is also a positive definite function on this interval a b. In some of the variations of the stream level problem this also is needed to be on the closed interval, but there are situations where it is required only in the open interval because at a and at b r x is allowed to be 0 in one version of the stream level problem. So, only this much is required for the general case in which this is the stream level equation. Now with this stream level equation we can define three different classes of stream level problems with different kinds of boundary conditions. If we define the boundary conditions in this manner that is at x equal to a at one boundary we have got a condition like this that is this linear combination of y and y prime is 0 and at the other boundary this linear combination of y and y prime is equal to 0 in which both a 1 and a 2 should not be 0 together that is a 1 and a 2 together is a non 0 vector one of them could be 0, but not both because if a 2 is 0 that will simply mean that at that boundary the boundary condition is y a that is y a equal to 0. Similarly if a 1 is 0 that will simply mean that at that boundary y prime a is 0, but if both a 1 and a 2 are taken to be 0 then this will mean nothing. So, therefore a non trivial boundary condition is given when a 1 and a 2 are not both 0 at the same time. So, a 1 a 2 is a non 0 vector similarly b 1 b 2 is a non 0 vector. So, if the boundary conditions at two ends are like this then what we have got with this stream level equation is a regular stream level problem it is a boundary level problem. Now in a very specific case where this function r has equal values at both the boundaries at a and b in that case another special kind of boundary conditions also will define a stream level problem and that is if y a and y b are equal similarly y prime a and y prime b are equal. So, this will define what is called a periodic stream level problem this makes sense direct sense in a situation where the variable under question say x is a cyclic variable like the angular position theta on a cylinder. So, in that kind of a situation r 0 and r 2 pi will be same. So, in that case this is a natural condition y 0 and y 2 pi is same y prime 0 and y prime 2 pi will be same. So, this is the periodicity involved. So, over that interval a to b if we try to define the problem then this will be the boundary condition which is periodic in nature. So, we define we can define a periodic boundary problem with this kind of a condition which is fundamentally different from this kind of a condition and this will define a stream level problem the periodic stream level problem when r a is equal to r b that is this function has same value at the two end points of the domain interval. In a singular stream level problem which is a very special critical kind of case in which if this function r a is 0 at one boundary say at x equal to a then to define the stream level problem at that boundary we do not need a boundary condition. Similarly, if r b is 0 then at that boundary at x equal to b we will not need a boundary condition and in that case while solving this boundary problem we will just look for bounded solutions over a b that is solutions y of x we will look for in this interval which are bounded functions of x that is which are continuation bounded do not get undefined that kind of a solution we will just look for over the interval. Now, with this differential equation with this kind of boundary conditions or with lack with no boundary conditions in the case of singular problem as we define a boundary level problem like that what are the solutions of such a boundary level problem defined as the stream level problem and what are the applications of them the great property of the solutions of such a boundary level problem is the orthogonality and the way we have defined the problem itself the proof of the theorem turns out to be quite straight forward. If y m and y n are two Eigen functions or solutions of the stream level problem defined as such defined as earlier corresponding to distinct Eigen values lambda m and lambda n that is corresponding to different lambdas lambda m and lambda n if these are the two Eigen functions two solutions then they turn out to be orthogonal with respect to each other with the weight function p x here. So, this is the definition of orthogonality of two functions y m and y n over this interval a to b with weight function p x. So, mutually the two functions will be orthogonal if they are the solutions of the stream level problem corresponding to different Eigen values distinct Eigen values. Now, this can be proved quite easily what we do for that is that if y m and y n both are solutions of that boundary level problem that we have defined. That means they are in particular the solution of the differential equation also and that will mean that y m is a solution of the differential equation corresponding to Eigen value lambda m will mean this and when we multiply this with y n we get this. So, this multiplied with y n is here and this term with multiplied y n has been taken on the other side of the equality. Similarly, y n is a solution of the differential equation with Eigen value lambda n that will be this and this equation we multiply throughout with y m and then we get this. Now, simply we subtract these two as we subtract these two equations q terms will cancel out and we will have lambda m minus lambda n into p y m y n that is the integrand here. So, we will get this and on this side we will have this minus this. So, that means this will be positive that is here and this will be negative which is here and in between these two terms are actually same we have added and subtracted.  Similarly, these two terms together will give us the derivative of this r y n y m prime. So, together the four terms will be the derivative of this whole thing and that is equal to this. Now, we integrate both sides from a to b and this are this is constant. So, this stays outside the integral and we have got this which is equal to this is exact derivative of this bracketed expression. So, it is integral will be this bracketed expression with the limits at x equal to a and x equal to b subtracted that is that entire expression at x equal to b minus that entire expression at x equal to a. Now, note the three kinds of boundary condition in the regular stream variable problem we had this boundary condition a 1 y m plus a 2 y m prime at x equal to a is 0 and a 1 y n plus a 2 y n prime at x equal to a is 0. Since y m and y n both are solutions of the stream variable problem. So, that condition at x equal to a is satisfied for satisfied by both y m and y n. So, they constitute these two equations and now since a 1 and a 2 are both not 0 that means this system of two equations has a non trivial solution which means that this coefficient matrix is non singular this coefficient matrix is singular which means y m y n prime minus y n y n prime at x equal to a is 0. Similarly, the other boundary condition satisfied by y m and y n will mean that this is also 0 that means this is 0 and this is 0 and that means this entire side is 0 and since lambda m and lambda n are distinct eigenvalues they are not equal that means this is non 0 which will mean that this integral is 0. Now, in a singular stream variable problem r a and r b both are 0 both are same. So, now r a and r b are both 0 in a singular problem in periodic problem they will be same. So, r a and r b both are 0 in a in a singular problem or if the singular problem has a singularity only at one end and not at the other then at whichever end it has a singularity that is say only r a is 0 then at x equal to a we do not need a boundary condition. If r b is 0 then at that end we do not need the boundary condition that means if this is 0 then we do not need this to be 0. Similarly, if this is 0 then we do not need this to be 0 if at both ends of the domain we have r x equal to 0 that is r a and r b both 0 then we do not need boundary conditions at either of the ends. Anyway, this will this entire expression will become 0 in a periodic stream variable problem r a and r b are not equal, but are not 0, but they are equal to each other r a is equal to r b and in that case y a and y b same y prime a and y prime b same will constitute the periodic boundary condition which will mean that r a and r b are same and this bracketed expression and this bracketed expression will turn out to be the same and in that case also though individually the terms are not 0, but their difference will vanish. So, in all the three cases of the stream variable problem defined regular singular as well as periodic we find that this right hand side vanishes and lambda m not being equal to lambda m lambda n that will mean that this integral will vanish in each of the three cases. So, we find this and this shows us that y m and y n are orthogonal functions with respect to the wet function p x over this interval and that shows that if the two Eigen functions two solutions of the stream variable problem that we take corresponding to distinct Eigen values then they must be orthogonal to each other. Similar is the situation of the Eigen value algebraic Eigen value problem of the symmetric matrix when we took the two Eigen vectors corresponding to distinct Eigen values and found the two Eigen vectors must be orthogonal similar is the case here. You can extend this and also say that Eigen values of a stream variable problem are necessarily real and the way you prove that is also similar to the algebraic Eigen value problem you assume a complex Eigen value and then force the Eigen function corresponding to that to satisfy the stream variable problem and then you can show that the imaginary part turns out to be 0 the way we found the corresponding result for symmetric matrices. Now, as an example of this orthogonality let us take the Legendre equation which we studied in the previous lecture previous lesson you will recall that this is the Legendre equation in which case this is already in the self adjoint form why because the derivative of this coefficient of y double prime is minus 2 x which is exactly the same as the coefficient of y prime. So, it is already in the self adjoint form and these two terms together can be written as the derivative of 1 minus x square into y prime and that is exactly what we will do when we consider that in our context here. Say we take the Legendre equation which is this the two terms having y double prime and y prime we have put together here and k k plus 1 is sitting in the place of lambda the Eigen value here in place of r x we have got this function in place of q x we have got 0 in place of p x which is here we have got 1 and in place of lambda we have got k k plus 1. Now, if we find two solutions of this corresponding to different values of lambda that means different values of k k plus 1 lambda m is m m plus 1 lambda n is n n plus 1 then we can write down the corresponding solutions as y m and y n and continue with a with analysis which is similar to the general case that we considered here. But we can also observe that this turns out to be a singular stream variable problem over the interval minus 1 to 1 because this function r x 1 minus x square is singular is 0. This is 0 at x equal to minus 1 and x equal to 1 that is at a and b a equal to minus 1 b equal to 1 at these two values this function is 0. So, over this interval from a to b this differential equation without any boundary conditions defines a singular stream variable problem and as such the Legendre polynomials p 0 p 1 p 2 p 3 p 4 etcetera that we will get will turn out to be mutually orthogonal with the weight function p x in this case it is 1. That means that other such families of solutions or Eigen functions of stream variable problems will turn out to be mutually orthogonal with respect to the weight function p x which appear in the place here in the general stream variable equation format. Here that function is 1 that means the Legendre polynomials will turn out to be orthogonal with respect to 1 that means we will have this integral equal to 0 for every case of m not equal to n. And if so then we already know that Legendre polynomial of order k turns out to be a k th polynomial equation polynomial expression and then if we have p 0 as 1 p 1 as x p 2 which is this. So, then we will find actually p 2 is 1 by 2 of 3 x square minus 1 as we found in the previous lecture. So, then we can see that 1 can be expressed in terms of p 0 then x can be expressed in terms of p 1 x square can be expressed as a linear combination of p 2 and p 0 x cube can be expressed as a linear combination of p 3 and p 1 and so on. That means that all ordinary polynomial terms 1 x x square x cube x to the power 4 etcetera can be expressed as linear combinations of p 0 p 1 p 2 p 3 p 4 etcetera.  If these turn out to be expressed in terms of the Legendre polynomials up to that order that will mean that suppose we have gone up to till this point. Then we say that p 5 that will define will be orthogonal to each of these expressions because this is a linear combination of p 0 this is a linear combination of p 1 that is up to this 1 x x square x cube x to the power 4 all these are linear combinations of p 0 p 1 p 2 p 3 p 4. Then p 5 will be orthogonal to p 0 p 1 p 2 p 3 p 4 as we have found from here and that means p 5 will be orthogonal to any linear combination of these 5 Legendre polynomials p 0 to p 4 which will mean that p 5 will be orthogonal to all these 5 functions and any linear combination of themselves. That means that p 5 will be orthogonal to all polynomials up to the fourth degree similarly p 7 will be orthogonal to all polynomial expressions of the sixth degree and so on. So, p k will be orthogonal to all polynomials of degree less than k. So, this gives us a handle to represent the polynomials or other functions in terms of linear combinations of Legendre polynomials. Other such families of Eigen functions also can be used for similar purposes and that is the great use of the solutions of stream level problems or Eigen functions of a stream level problem. Now, this case of real Eigen values we can omit the proof of this, but you can try to establish this small result in the same manner as we did in the case of symmetric matrices. Now, we come to the most important application of the Eigen functions of stream level problems and that is for function representation. So, Eigen functions of an S L problem gives us a convenient and powerful tool to represent and manipulate fairly general classes of functions. So, for function representation if you select a family of functions y 0 y 1 y 2 y 3 etcetera which are continuous functions over an interval which are mutually orthogonal. Then a representation of a function over this interval could be made as a linear combination of this family. Of course, this will be an infinite series in general and this kind of a series is called generalized Fourier series which is analogous to the representation of a vector as a linear combination of several basis vectors orthogonal basis vectors. Now, how to determine the coefficients? To determine the coefficient a n we can do a similar thing as we could do in the case of ordinary vector spaces and that is we take the inner product of this f with the n th function in this for which we are seeking the coefficient. And if we construct this inner product that will in the case of functions the inner product is given like this which we have studied earlier in the context of function space. Now, if we substitute the expression the summation expression for f x here, sum of a m y m then we will find that this integral will give us a series of inner product like this. Now, since y m and y n are orthogonal corresponding to m not equal to n therefore, in this sum only that particular term in which m and n are equal will survive all others will become 0. So, we will get this and y n norm turns out to be the square root of the inner product of y n with itself which is this. Now, these a n which we get out of the division of this inner product with this norm square turns out to be the Fourier coefficient. Now, other than the orthogonal family of functions y 0 y 1 y 2 y 3 we could also say that we look for orthonormal family of functions in which all functions are mutually orthogonal and each of the basis functions has a unit norm. In that case from the same y 0 y 1 y 2 etcetera we could have defined phi 0 phi 1 phi 2 etcetera like this which will be an orthonormal basis and in that case the Eigen function expansion in orthonormal basis will be given like this. Now, the question arises what C 0 C 1 C 2 etcetera will give us the best representation of the function and that is very interesting result that is if we say that we are looking for an expansion of this kind and then up to capital N terms if we take a finite sum a finite number of terms if we keep in hand then we will get this phi n that is sum up to capital N. Then in that case the error will be this actual f minus phi n norm square and when we try to put that effects and this phi n here and try to find out with which values of alpha 0 alpha 1 alpha 2 etcetera this norm will be this error norm square will be minimized. Then we find that the first order condition del e by del alpha n equal to 0 will give us alpha n equal to C n that means the previous method of finding the coefficients based on the inner products will give us that particular composition for a finite number of terms in the series which gives us the least square error and that is why the least square approximation or as it is called the best approximation in the mean involves this orthonormal coefficient orthonormal components and the coefficients based on the inner product as we have seen earlier. Now using the Fourier coefficients you can find out the error the least error that you get as this. This is something like the vector error that is for this vector if you try to represent with the help of several vectors in the space several basis vectors then if you take a partial sum not taking all the basis vectors then the error that remains will turn out to be this square plus whatever component minus whatever components you have accounted for. So, the size of the function square minus whatever components you have accounted for will turn out to be the error square that has remained. Now since this is greater than equal to 0 therefore you find that the sum of the squares of the coefficients is always less than equal to the square of the function norm and this is called the Bessel's inequality and if you if the number of terms capital N tends to infinity then the sum converges to the function norm and that is called the Perseval's identity. Now we can omit these steps and the most important result out of this entire Eigen function expansion is that this is the Perseval's identity. So, we say that an orthonormal set of functions on an interval is said to be complete in a class of functions or to form a basis for it if the corresponding generalized Fourier series converges to the mean in the mean to the function that is as you include more and more components as you include more and more terms the sum of the terms converges to the actual function then you say that for all functions in that class then you say that that set of basis functions is complete in that class of functions. And here is the great use for the solutions of a stream removal problem that is the generalized Fourier series in terms of Eigen functions of a stream removal problem turn out to converge for all continuous functions with piecewise continuous derivatives that is functions of this kind with which are continuous and the derivatives also are continuous piecewise for all such functions the solutions of a stream removal problem turn out to form a complete class of basis functions. And therefore, they serve to expand the arbitrary functions of this class in order to represent and manipulate them in the coming two lectures we will consider a few cases of such representations and that will include Fourier series and its generalizations in terms of Fourier integrals and Chebyshev polynomials. Thank you.


Hi, thanks. It's great to be here. Several of you here, I realize, have heard a version of this talk before, despite the result being like seven months old only. I guess we all go to the same conferences. But Maxime and I, so I'm going to talk about new joint work with Maxime Wolf. And we decided we would take the opportunity here the fact that we're giving two talks to an audience who we hope will be very interested to use both talks to give more details of the proof, explain something that maybe you can take home and use to do other stuff with. So I have the pleasure today of giving more surrounding motivation and ideas and tools than I usually would in such a talk. And Maxime tomorrow will try and get into a piece of more detail, I think. So that said, also having more time and flexibility, your questions are extra welcome today. So ask away. OK, so what are we interested in? I want to understand just rigidity of group actions. So let's start from a very broad setting and narrow down. I want to understand representations of a discrete group, so gamma is some discrete group. I should write bigger, excuse me. So here's our setup. I have gamma, some discrete group, think finitely generated. And I have G, some topological group. And I want to understand representations of gamma into G and how flexible and rigid they are. So the context that you're maybe used to is the case where G is a lead group or a linear group, and this is classical representation theory. Or for dynamicists, G is maybe a group of homeomorphisms of some manifold or diffeomorphisms of manifold if you're doing smooth dynamics. And this is giving you possible group actions on the manifold. And there's lots of ways that you can say how flexible or rigid these are. I'm going to take a strong definition of rigidity. So provisionary definition. So we might have to come back and adjust. A representation of this sort is rigid if, well, so I want to say that all deformations of this, even on a global scale, all deformations are just trivial. So what do I mean by trivial? How about attained by conjugation in G? That's always something you can do. So to say that this is trivial means that if I take the space of all representations, gamma G, I want to quotient by the conjugation action of G. So to say that the only deformations in here just came from this, which I have now killed, should be saying if rho inside of this space, when I pass to the quotient, is an isolated point. It's a very strong notion, but I will give you some examples in a minute of where this actually occurs. Bit of a problem. I just sort of very brazenly took a quotient to be like, I don't care about conjugacy. And what I've come up with is probably a very terrible space. So issue, even in the nicest cases, this quotient is typically very highly disgusting, like non-Hausdorff say. Let's do an easy example. How about gamma? It's just a cyclic group. And G is classical example SL2C. It's a familiar space to many people in this room. So now this space really is just homomorphisms of Z into SL2C. So that's just SL2C up to conjugation in SL2C. And so I'm just talking about conjugacy classes here. That's not very sophisticated, but already there's an issue. Here's a conjugacy class. Things that are upper triangular, where T is not 0. And here's a different conjugacy class. These are not conjugate. But I cannot separate them by open sets. Up in SL2C, before I quotient, I can take T as close to 0 as I want. And you can imagine that if I put in some more complicated group, this only gets worse. So I don't like non-Hausdorff spaces, because I don't know how to put coordinates on them and things like that, because we put coordinates in R or something. So let's fix this and take a bigger quotient. So does that line mean something to some people? But I won't put it there. That's meaningless. So here's the fix. Is define a character space x gamma g to be just abstractly the largest Hausdorff quotient, gamma g mod g by conjugation. This is a thing you can do using undergraduate topology. Any topological space has largest Hausdorff quotient with the property that a map to another Hausdorff space factors through this one. Great. And so now this is the thing that you could possibly hope to work with. And we'll go back to our definition and fix it up a little bit and just take this out and put this in. And now I'm saying that there is no, all the deformations are just by conjugacy as far as you could ever hope to measure in any kind of reasonable way.  So for an example in SL2C, in the case where your group really is this, this is just, or I should say four, this is the character variety. So it's better than being Hausdorff. It actually has the structure of an affine variety. So there's all kinds of algebraic structure. And this actually is true for SLNC representations. And for a larger class of league groups, it's the quotient you get from geometric and variant theory. And the example that we're interested in, so I'll edit this to SLNC so you think it's very good. So it's just captured as parameterized by characters and representations. It's as nice as you could hope for. What I like is dynamics of group actions. And there's one other context in which this is beautiful and it makes sense. And this was perhaps sort of known or understood, but Maxime and I had to write down the details because it hadn't been done. So proposition. If your group is the group of homeomorphisms, I can't do any manifold. That would be an interesting question. But let's take a nice one manifold. Let's take the circle. Orientation preserving homeomorphisms of the circle. Gamma, anything. So any discrete group. This thing that I presented as some weird abstract quotient of things up to conjugation in so far as you can understand them. This has a nice interpretation. This analog of the character variety is representations to semi-conjugacy. So if you haven't seen thought of semi-conjugacy in this context before, probably this goes without saying, but just in case we're doing quickly. If I have, so conjugacy means there's some homeomorphism of the circle so that H composed with row of gamma. Maybe I have row one and row two and they're conjugate. Is the same thing as row two of gamma composed with H for all gamma. Gamma, that's conjugate. H is a homeomorphism. Semicondugate means that I don't require H to be a homomorphism. Just take H from the circle to the circle to be, should not use my fingers. Continuous, so I'm gonna let you collapse intervals down to points. So continuous and degree one and I want you to weekly preserve cyclic order. Okay, so you can collapse intervals to points but you can't go back and forth.  Okay, so this is an example of two representations being semi-conjugate. So it's like conjugate but slightly weaker and I'll say that semi-conjugacy is the equivalence relation generated by this. So I'll say that in this case, so if this happens for some H, say row one is equivalent to row two and semi-conjugacy is the equivalence relation this generates. So as I've stated it, it wasn't symmetric but really the symmetric version of this is you have two actions and if you can kind of collapse a wandering interval in one and maybe blow up some points also to get the other, blow up an orbit or collapse a wandering interval down that's all you need to do to pass within a semi-conjugacy class. So it's pretty easy to check that semi-conjugacy classes are connected and they're the closure of conjugacy classes and so that's how you start to recover this. Okay, but to say that this is exactly the largest house-door quotient on the nose is a little bit more work. But this is a natural notion and it's been studied for a long time. It turns out nicely though that it just fits into this context. Okay, great. Okay, so this is what I'm interested in rigidity and flexibility of actions. Here's a familiar context. Here's a kind of a nice way to talk about this in the group action setting. How about some examples of things that are actually rigid? So probably the most famous example of rigidity. I wanna keep up that definition there. So let me start again. Yes. Where does this come from? Yeah, yeah. This is a quotient you have example of this matrix today. There's the P of the E1. Yeah. The E1 non-separated is not a power space. How does it begin when you have this extension? What happened? How does? No, whatever, you have this quotient. You have the trouble that they ended up in your group separation. So it's not quotient, so I should identify these two points. You have the, well, okay, so when you have this extension, what happened to those two guys? When I have this further quotient or? Yeah, so you have the largest count or? Yeah, yeah, yeah. So if I'm really taking just the SL2C, so ham, Z, SL2C, mod, this and I take the largest house torque quotient. That is all I identify. Now I'm just, yeah, traces, it's now the space of traces of matrices. So X, yeah, Z, SL2C is in natural correspondence by a trace with, ah, so this is exactly what, and so more generally, if I add some other group, I should now recover traces of all the elements. If I have SLNC, I should look at characters, and I mean this is, you can prove that is exactly what happens, this is the largest house torque quotient. Yeah, cool. So one of the points of this talk is I'm kind of cooking this to try and draw parallels between the linear representations case and the groups acting on the circle case, okay, which is a theme that I've been motivated by for a long time is that dynamics of group actions with no regularity should, you know, a priori not be anything like linear representation theory, but if you look very hard, and you do a lot of work there, these surprising parallels that come up, and that's the whole point of this talk today. Okay, so let's draw another parallel. Let me do an example of rigidity in this lead groups case, how about Z, my favorite theorem since a long time, many of you will say this is master rigidity, but the form I'm gonna state it, I think is older and just do the kalabi, okay. It says the following, if my gamma is the fundamental group hyperbolic manifold of dimension at least three. So in other words, I can think of gamma inside of SON1, the group of isometries of hyperbolic space as something that where the quotient is compact and that will recover, I guess the unit tangent model of my manifold. So this is co-compact lattice, so IE. Then I just told you that gamma sits inside this lead group, this natural inclusion of gamma into SON1, this geometric one that comes here is rigid in the sense that it's an isolated point in this character space. So that's kind of the geometric version or the linear groups version and kind of the paradigm for all of this. IE groups acting on manifolds version V. So if I wanna recover this, I wanna substitute candidate for a theorem, I wanna substitute for being a co-compact lattice in a lead group. Well, here's the best substitute definition. Let's call representation of gamma into a group of homeomorphisms of M, geometric. This is not a lead group, it's not finite dimensional in any kind of way, it's not even locally compact. This is a disaster. How am I gonna try and recover something like this? I'll just artificially insist that there's a lead group in there. So it's geometric if it factors through an inclusion of gamma into some lead group as a co-compact lattice. Okay, so like this example here, and then I want G to act on M in a natural way. So G, I'm gonna require to be a connected lead group to act on the manifold transitive lead. So that I don't, I see all of M. Okay, in other words, M should be a homogeneous space for G. Okay, that's the definition. Gamma sits in the axon M in a way that sees a lot of geometry. Okay, it's a co-compact lattice in a group that makes this a homogeneous space. Maybe the example zero of this is you can put the fundamental group of surface of genus G, at least two. Okay, it's one dimension down from the clubby's theorem over there into PSL2R or SO21, if you like that notation better, which acts on the circle by homeomorphisms where this action is by mobius transformations, if you like, or if you think of this as the isometries of hyperbolic space and you take the Poincaré disc model, this is the action on the boundary. So this composition, if I call this row, that's an action of a surface group on the circle, which is geometric because I cooked it up that way. I put this inside of here, put a hyperbolic structure on the surface, if you like. It's now, its fundamental group is acting on the universal cover. The hyperbolic plane by isometries in a way that this is a lattice in this group and this acts nicely on the circle. I like this example particularly much because you might have run across it without this space in the middle. This is the sense, this is a hyperbolic group in the sense of Gromov and this is really the action on the Gromov boundary of the group, which is by homeomorphisms. There's no extra structure there. You're doing geometric group theory. This is all you've got and oh, nice. What a nice example of something that satisfies the definition of geometric. See, secretly there's a legal side there. Okay, great. So we have a definition, what's the theorem that goes along with it? An older theorem, due to Matsumoto from, I guess it's about 30 years old, that was from 1990. It's a remarkable statement that example zero is rigid. I.e. If you take this representation, it is an isolated point in the character space for actions of the fundamental group on a surface on the circle. In other words, any deformation even globally is semi-conjugate, dynamically essentially the same to the original action you started with. At the time, I don't think anyone thought that one would ever have sort of a master rigidity-ish theorem for things with no regularity and no group structure or anything in the target. This is the part in the talk where I have to tell everyone who studies surface groups in PSL2R and likes Teichmuller space, which is the space of all representations here up to conjugacy and it's large and it's interesting and it's rich, that I had just killed it entirely because all of these examples where you go into PSL2R first are conjugate in the group of homeomorphisms of the circle. Okay. So, although when you quotient by PSL2R you get something big and interesting. If I take the quotient by dynamism, topological dynamics if I care, but things up to topological conjugacy, that's now done. Okay, so this is not a contradiction in mathematics, this theorem makes sense. This example is one example and it is an isolated single point. Great. Okay, so here's my whole setup. What's the rest of the talk? It's a rigidity theorem for all the other geometric examples on the circle. So, rigidity of all other geometric actions on the circle. Let me tell you what all the other geometric examples are. So here's a fact. Exercise. Okay. If G, so I need, what do I need in this definition? I need to tell you I'm gamma, I want to put it in the homeomorphisms of the circle, but it's supposed to factor through some lead group. What are the options for the lead group? Let's be systematic. Okay. If G acts on the circle transitively, there's not very many options. G is either, I already gave you PSL2R as an example. There's the obvious example of just rotations.  So I could lift the action of PSL2R to that cover and I'll get all the other lifts will just be some central extension of PSL2R by the finite deck group. Okay. So, or there's extension of PSL2R by some finite cyclic group, central extension, okay, in which G sits, okay. And if this is acting by Mobius transformations on this little circle, this is acting on the K-fold cover. Okay, so it's just the group of all the lifts. Okay. This is a fun thing to try and prove. What's the key here is that this, the circle is not very big. A lead group inside of its group of homeomorphisms can be at most three-dimensional, okay. So these are all the geometric examples and lattices in these. Well, this one's already compact. Here you have surface groups and here you can have either surface groups or things that are virtually surface groups. So the fundamental groups of surfaces, of G at least two, are the only candidates here. And the theorem, okay. One direction was worked from my thesis in 2014. And the other direction is with Maxime Wolf from last year, okay. It's the statement that actions of surface groups on the circle, an action of a surface group on the circle is rigid, namely an isolated point in this character space. If and only if it's one of the examples I just gave you, it's geometric. So secretly there's a lead group there. Could look like Matsumoto's example. It could be one of these central extensions. And those are exactly the ones that have no non-trivial deformations. Oh, and of course these arrows now make no sense. Okay, so the easy direction was the earlier one. Geometric implies rigid. Here you have specific constructions. I give you a surface group sitting in here. I know what its action is. I can write down matrices if you like, whatever. You wanna prove that there's no interesting deformations of this. So this is something you start attacking by hand and saying, well, if I systematically tried to do these things, nothing can happen. There's a bunch of machinery to do this. And this is the truly remarkable example that that's an example. At that point I conjectured out of optimism that I had found all the rigid examples because I tried to find more and I couldn't, you know, brute force could conjecture by lack of creativity. This direction is, how does this go? Someone who hands you a mystery representation about which you know nothing. You just have some group action. Okay, oh, but you do know one thing. If you tried to change it, you would fail. From that, you're supposed to recover a lead group in which your surface group happens to sit as a lattice. So this is the direction that we're gonna be talking about that required an invention of a bunch of techniques for doing these things. Questions so far? Well, remark, it's really, really easy to produce lots of non-rigid, non-geometric examples. Okay? So from Gamma, homey OS, I'll even say actions of, I'll say surface groups for concreteness. This space is big. In fact, even this character space, it's big and it's kind of wild. It's easy to make lots of non-geometric examples. For instance, maybe you like working with representations of surface groups into PSL2R. That's a thing people like a lot. If you take one of those that doesn't come from a hyperbolic structure on your surface, then it's not rigid even in the PSL2R character variety or deformation space. So it follows from work of Bill Goldman that the non-tychmuller examples have a large deformation space. And in particular, you'll find elements that act by rotations. And as you perturb this, you can change the amount they rotate. And that's a conjugacy invariant, even a semi-conjugacy invariant in the group of homeomorphisms of the circle. So even if you just work within representations to PSL2R, you don't find any other candidates for things that might be rigid by topological conjugacy. So even with image in PSL2R, lots of this, lots of these. And if you wanna go outside of, I know how to write down matrices and do these things, here's another good way to make lots of crazy examples. Okay, so I guess one, you could take representations in the PSL2R. Or two, here's my favorite strategy. Take your surface, think of it geometrically, cut it in half somewhere. And this lets me think of my surface by what, Van Kampen? I can think of the fundamental group as an amalgamated pre-product of the fundamental group of this part and that part. Okay, but these are just both free groups. So this lets me write my fundamental group as a free group. That's a free group on, here I've done it as, let's write it properly, that's a free group on two generators. This one on four, I guess, and it's amalgamated over the cyclic subgroup generated by this curve. So what is this saying? Free groups, that's easy, just take any two homomorphisms of the circle, there's no relations to satisfy to any two homomorphisms of the circle can be the generators of this group. Similarly, pick whatever you want here and you're subject to a single constraint that the action of this group and the action of that group agree on this common element. But that gives you lots of room for variation and make this one do whatever you want and now just cook up the other side so that product of commutators that gives this curve agrees. So free groups are easy to make act and deform. I think that's interesting here that so I wanna start sort of warming up towards some ideas of the proof. And one thing that I sort of, I didn't realize until afterwards and when I went back and read Matsumoto's proof again is that there's ideas in here sort of well under the surface that parallel some global strategy of what we did in here. So let me tell you one of these very important tools. So a key tool, which is also another kind of beautiful analogy between the linear case and the surface group case. And in fact is part of how you see that representations up to semi-conjugacy is a house dwarf space and a reasonable corrosion. It's a following theorem, which is do tijis and rephrased in the way I'm gonna state it by Matsumoto. And it says that coordinates, there are natural coordinates on the space of actions of any group on the circle up to semi-conjugacy. Analogous to characters or trace coordinates of representations into SLNC. So taking SL2C to be concrete, if you have a representation of a group into there and you know the trace of everyone, theorem you know the, well I mean you know the representation up to conjugacy unless it was upper triangular basically. Okay, this space has same kind of thing. So what's the analog of character? I should have some conjugacy invariant function. This is given by the only dynamical invariant really of circle homeomorphisms, the Poincaré rotation number. Okay, so if you haven't seen this definition before, this is to any homeomorphism, you assign a number in R mod Z or in S1 if you like that captures the average amount it rotates points. And the theorem here is stated very imprecisely, slightly more precisely, it is that a representation from any group you like into the group of homeomorphisms of the circle is determined essentially. So there's some finite ambiguity there but nothing that we need to worry about for the proof by the rotation numbers of each curve gamma in gamma. And maybe it's worthwhile to say what I, when I meet by essentially, there's slightly more information that you need. Okay, so here's really what's going on. Let's take my favorite group, my favorite representation, the one that gives you a surface group in PSL2R, the Fuxian action, the nice boundary one. There we know the dynamics of things. Every element of my group acts as hyperbolic transformation has two fixed points and has sourcing dynamics. Having a fixed point is equivalent to having rotation number zero. So this representation gets assigned zero for everybody. Okay. It is highly non-trivial but unfortunately the trivial representation also gets assigned zero for everybody. So that's what essentially is saying, it said I was not being precise enough. Okay. What's the difference? It's determined actually by, so it's determined up to semi-conjugacy by the rotation numbers of not single elements. Okay, but comparing pairs. So what you need to know is the function that takes two elements and counts how their product wraps around the circle. So in PSL2R, if I have two things, maybe someone with an axis like this and then someone else with an axis like this in very strong dynamics and I take their composition. The composition will quite likely, if I start with some point here, this guy, the blue guy will move it all the way over here and then the black one will move it back to where it started. It will have a fixed point but one where if you track how it winds around the circle, it will have made a full rotation. Okay, that's somehow different than the trivial representation where everyone's the identity and the point doesn't move ever. Okay, so this is given by lifting these, so I wanna track how much something winds around the circle. This function is given by taking an element, lifting it to a homeomorphism of the line that commutes with, I'm thinking of the line here as a universal cover of the circle, so I'll lift it up to, if my circle is r mod z, this will be something that commutes with the deck group acting by translations. So it is z-equivariant homeomorphisms of r. I can take a lift of the other one. I can look at the average amount these translate points. So if I took their lifts with fixed points, I would get zero and zero and I wanna compare this to how the product does. So in the cartoon I just pointed at, this is exactly capturing the fact that this product winds around the circle once. I've lifted through the universal cover so I could keep track of my movement back and forth. So this is what you need to keep track of really that distinguishes the trivial representation from this folksy and guide, but it's framed completely in terms of essentially a rotation or translation numbers of elements. Great. Another question I wanna point out now before I go on is that I would very much like to know an analogy of this theorem if you increase the regularity here. So instead of homeomorphisms up to semi-conjugacy, let's study actions of a group by diffeomorphism, C1 even on the circle. And I can think of those up to C1 conjugacy. And that space I'm sure is not house dwarf and is nasty and terrible, although I haven't quantified how bad it could be. But we can say, okay, pass to the largest house dwarf quotient. What is that? And if you like smooth, replace C1 with smooth. I know no nice analog of this in the higher regularity, but it's mostly because I haven't tried. Or who knows, maybe it's reasonable to say or maybe it's extremely difficult. I think it's a fascinating question. Okay, so what does this all boil down to? It means that both us and Matsumoto are trying to understand the deformations of representations by understanding how much curves rotate points around the circle. We're looking, we know things are non-conjugate. Basically, as soon as this number changes for some pair of gamma one, gamma two. So I wanted to give the technical correct one, but you should think of this as being sufficient to just know rotation numbers of individual elements. If you're working locally, that's good enough to recover this. That point is the bad example I gave of the trivial thing and the spruxian thing are extremely far apart from each other. So working locally, you just need to understand rotation numbers of elements. Let me give you the two-step outline of what we do and maybe a cartoon of one of the steps and then leave it to Maxime tomorrow to go into more detail. So here's an outline, a very broad outline of the proof rigid implies geometric for surface group actions. Get some more space here. The first thing we noticed is that this theorem would be a lot easier to prove. If instead of just knowing something was rigid, we knew we can make arguments like this. We knew that all the elements looked like they had sourcing dynamics or we're secretly living in PSL2R or something like this. I was surprised very recently that going back to reading a Matsumoto's original paper of Fuxing's example is rigid. He actually spends two pages being like, let's pretend the image was just in PSL2R because that'll make my argument easier. And then he's using sources and things of hyperbolic elements and gets a nice conclusion and then goes back and says like, well, you can't expect this in general at all, but we'll try and do some deformations and say that this argument kind of sort of can be pushed through and five pages later of like technical misery nails it. In the penalty, we had the same idea. Well, if this was kind of looking like everything was just sort of nice hyperbolic dynamics in this topological sense, we could almost do that. So then came the hard work of reducing to this case. So part one is to say that suppose that row is rigid, we can conclude that after semi-conjugacy, I can, well, after semi-conjugacy, I can always pass to a minimal action, one where all orbits of dense. So I'll add rigid and minimal to kind of pick out my preferred semi-conjugacy class. I won't blow up anything I don't need to. Then, and this is the hard part, this is sort of the bulk of the work. Individual elements look like they sit inside of one of these lead groups. So then very locally, we look kind of geometric. There's a local geometric-ish picture. Precisely, if you take two curves on your surface, two elements of the fundamental group represented by simple closed curves, that have intersection number one. So if I have A and B are curves, here's maybe A, here's B. So it's up to some action of a mapping class group I can draw my curves like this, okay. Here's the dynamics of row A and row B. So then there exists some K, which is actually independent of row A, sorry, independent of A and B. So the same K works for all the curves on your surface. So that A and B each have two K fixed points, not necessarily fixed periodic points, okay. Half of which are attracting, half of which are repelling. Maybe this is the attracting one and the other repelling. So alternating, attracting and repelling periodic points, and they also alternate in the surface. So this is exactly the dynamical picture of how periodic points look in one of these lifts from an action of PSO2R, right. If you look at the geometric picture of this, that's K equals one. This is, these are two hyperbolic elements with crossed axes in the way we usually draw this picture. Okay, NFI. So there's an attractor, a repeller of this one, an attractor, a repeller, an attractor. If I lift this to a K-fold cover, I take any of the lifts, these will become periodic points alternating attracting and repelling. So here's the picture where K equals two, okay. So there's a local geometric picture. I say geometric-ish, because all we know is the kind of combinatorics of attracting and repelling periodic points. So this really is the the headache to go from nothing to, ah, a nice combinatorial picture. Step two is to just do a combinatorial argument and recover your surface. So step two is local to global argument, okay. So this is a statement that if row is rigid and satisfies the local picture on pairs of simple closed curves given by one, okay. Row is in fact geometric. And the idea of this is just to use very specific deformations, is we use very specific deformations, analogous to bending or doing an earthquake from Teichmiller theory, but really we're just, instead of elimination, we use a one single simple closed curve, okay. So it's almost as if you're putting a dain twist around the curve in a one parameter family. So bending from, say, the theory of clining in groups, you like. To say that if you're not in exactly the right configuration all your curves, so here's something only on pairs, but I really want to recover my entire surface, okay. To reconstruct the combinatorial configuration of all curves, all simple closed curves on the surface. So a hypothesis says that some pairs look good and now I want to say, well like now, but now I have a curve over here that's disjoint, maybe this one. And the geometric picture, it's supposed to have an axis that's disjoint, so in the lift I'm supposed to see, what if I see green attractor, then before I see the red repeller, I should see two points here, plus and minus. And then, oh, I have to go around a whole time and then I should see the other two lifts of this, plus and minus there. That's something that this hypothesis doesn't give me, and we have to recover this kind of information, okay. And we do this by producing, should this not be true, if this plus was actually somewhere over here, goodness forbid, we write down a specific deformation that moves it somewhere else, giving us the wrong combinatorics or the wrong rotation numbers for pairs of elements and say, oh wait, but it was rigid, that could not have happened. So the whole idea of here is to use specific deformations to say you had to be in the right combinatorial configuration. And once you know that the attractors and repellers of all your simple closed curves are in the right place, it's not so bad after that to say, your surface group had to be acting as it was in the geometric picture. But I have run out of time, and so hopefully that has given us kind of a starting point or an advertisement for Maxime's talk tomorrow. So I'll conclude with that. Like non-surface groups. So I, there's two ways I know to produce examples. Okay, so one is if you finite order elements, that will help you because, well, if you find an order, you're conjugate to a rotation. So you can play with this and do some things. So things like even SL2Z, free product of Z2 and ZM3Z is known to be rigid for an argument like that. I forget, I should cite who, I think Michele Tuestino and friends have written this down recently. In some work that they've been doing. But I'm gonna get the wrong people attached to the wrong theorems. And I think you can do is you could take a hyperbolic three manifold group from a three manifold that fibers over the circle and take the example, this is sort of, there's a nice geometric action of this group on the circle where the surface fiber subgroup is in PSL2R. And because the rest of your fundamental group normalizes this, then that forces that action to be rigid. So you can do sort of tricks like this. I don't know, so a question I would like to answer is that's an example of Thurston's universal circle construction. To what extent are the, but that's a very general construction that applies to a wider class of examples. Are these all rigid? What do they look like, if not? So it's trickier, because you have some choices in the construction more generally. So I think that's good, that's a good, it's a good long term project. In which part of the book do you use these coordinates? Yeah, so this is going, so first, well it's here everywhere. So first, the conclusion of this is that everything has a periodic orbit of period 2K. So first thing we have to show is that the rotation number of a simple closed curve is rational. And then that they're all the same denominator. And that's done by again producing an explicit deformation. We're saying, oh, if it was irrational, then I can do a particular thing and change it's rotation number. Oops, not rigid. So it must have been rational. So that's actually like a lemma zero is that all simple closed curves have rational rotation number. And then we want to put their, their put orbits in a good configuration. And that's Maxime's talk, definitely. So. Any more questions? So otherwise, let's get it again.


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