So, to come to the Karno's theorem, we have to first define something which is called as a reversible input, because the Karno theorem involves something called as a reversible input. So, if you want I mean we have not defined something called as a reversible process up till now, but if you want to define a reversible process, we can just take let us say two systems A and B and these two systems are totally isolated and they are only reacting with each other or interacting with each other. So, it is an interaction only between two systems, you can show that they are isolated maybe I will just put a very solid box around it. They are not going to interact with anything else. So, let us say that you know initially the states of this were A1 and B1, these were the initial states and after interaction some interaction they reached A2 and B2. So, basically A executed a process it went from A1 to A2 and B executed a process it went B1 to B2, but during that execution of the process A interacted only with B, B interacted only with A and with no one else. So, the interaction was limited to the two. Now, if it is possible without again bringing any external agencies and without again interacting with anyone else outside, if it is possible for the system A to get back to A1 the same state exactly and with an interaction with B and for B in the same interaction to go from B2 to B1 then we will say that the initial process was reversed and the initial process is reversible that is because without using anything extra out from outside you could exactly reverse the process and get back to your original state. So, no other external agencies, no other external interactions were used completely it was the first process was an interaction only between A and B, the second process also was an interaction only between B and A, the first process changed the state to something else for both of them, the second process changed the state back to exactly the original state and absolutely nothing else from outside was used there was no other interaction with any other system. So, this is what we call that is the first process is reversible because it would be reversed back to get the original state. You can as well say the second process is also reversible because if you went the other way you could have always come back the other way around. So, if you went from A2 to A1 that is a reversible process because you can get from A1 to A2. So, we are just saying that this entire thing is reversible. So, this is what we mean by reversible as far as the process is concerned. So, we will use the same thing with an engine now. Why not we say system and surroundings? System and surroundings. So, I need to figure out how much of the interaction I can measure. So, if you are going to consider a surrounding. So, if you can tell me exactly what is the effect in the surroundings and what is the. So, surroundings as we already said is nothing but another system B. So, as far as we are concerned in all our discussions we are only going to consider two systems system A and B. And for us whenever we say surrounding it is always system B where we can actually figure out what has happened. So, you if as long as you are able to. We can predict with respect to A. No, that is what I am saying as long as see you can call anything the surroundings as long as you can tell that yes for A you went back exactly to the original state and for B also B is whatever else you wanted to call the surroundings. So, it is a limited do not make it unlimited we do not want something which is unlimited have it limited and if that B could come back to the same state as what it was originally without interacting with something else then only we want to call this as reversible. So, you cannot make it the entire universe out there. Is it wrong assumption or equally good? No, no, I did not get it what you want to say. Suppose if I assume B as surroundings whether it is wrong assumption or equally good whether system A and B instead of. So, no, I will definitely say you have to have a limited system that is how you would like it to be because for something which is unlimited I do not know how you are going to measure what its initial properties where and what its final properties as long as see you can have this room and have it totally insulated and you say I do not want and this is entirely surrounding that is fine. But if you say you know it is unlimited the whole universe then you know then we have a problem because you really cannot measure the state of that system which is outside it is beyond you. So, we would like to have something which is limited between A and B where everything about B can be figured out after some time. So, if there is no way to figure out we do not want to you know study such systems. So, the whole goal is to whether see whether we can really you know figure out what has happened to A and what has happened to B and whether the B can exactly come back to its original state without interacting with a third party at all except A and A can come back to its original state by interacting only with B. Initial state without is it not reversible? Yes, we will not say it is reversible if B cannot get back to its original state we will say it is not reversible. So, see you know this for sure you know for example you can create ice from water you can melt it down but you can use another external help of the fridge to turn it back to ice. But I will say the entire thing you did was actually a reversible you know there are no reversibility associated with you took help of the refrigerator you took help of something else. So, just turning back system A to its original state is not going to call it reversible you have to ensure the same thing has happened with whoever else it interacted with. Another thing if A1 A1 A2 is reversible can we say B2 or B1 to B2 must be a reversible? No, this is the process say again now you are see this is the process between A and B. So, the process involved A and B. So, this process between A and B that process was reversible. So, do not think about the system being reversible this system is not being reversible we are talking about the process which is being reversible. So, the process was executed between A and B that process has to be reversible. As the result of process A1 becomes A2. Correct. If bring back A2 to A1 if we treat this as a reversible whether B2 to P1 must be a reversible or it may be a reversible. See again you are looking at the system we are looking at the process which sent from A1 to A2 and which at the same time made B1 go to B2. So, the reversible process is such that you can exactly reverse the process and B2 has to come to B1 and A2 has to come to A1. That is what we are defining by reversible. So, if B2 does not come to B1 when A2 came to A1 then the first process where A1 went to A2 and B1 went to B2 is irreversible. So, that is the process when we went from A1 to A2. So, see between A1 and A2 I can execute many processes the system A can go to A2 through many processes. So, I am not going to deny that I mean I can always go from A1 to A2 by any number of processes and get back from A2 to A1 by any number of processes. I am talking about a particular process only between A1, A and B where the process was such that A1 went to A2 and B1 went to B2 and I could exactly reverse it back so that B1 came back B came back to its original state and A also came back to its original state and the interaction was entirely only between the two people. So, the process was between A and B and that could be reversed. So, if you are going to think of a third agency that is not okay. If you are going to think only A1, A2 that is also not okay because as I told you that could be many processes between A1 and A2 and I am not talking of those processes. I am only talking of a process which led to this change between the two systems and whether you could get back the states original states of both the systems. So, this is how this is the only way I want to define it. I do not want to define whether A went back from A2 to A1. I want that both of them should go back to their original state without external third body coming into picture at all. That is the only way I want to define it as reversible. I mean otherwise you are going to run into problem. So, this is my definition of reversible. Is that okay? Okay. Yes. If you are always borrowing with respect to a particular system. Yes. If this is that means if you want to know whether the process is reversible or not. Correct. We have to observe the process plots with respect to both the systems. Yes. Both the systems. So, you have to have two systems and observe both of them. That is the only way you can go. And both plots we have to run and we have to check whether they are restored. So, you will realize that we have not a defined entropy but once we come to entropy you will realize that we are always going to have an isolated system. An isolated system will always involve some system and its surrounding but both systems should be isolated and we will only talk about such systems. So, this is our definition of reversible. So, if I have an engine. So, I go from Q1 to Q2. Sorry. I take in Q1 there is an output W and there is a Q2. So, if this is an engine then I define what is called as a reversible engine as something which can be run in the reverse way and exactly the opposite should happen. That is I should if it sent out Q2 to T2 I should be able to pick up Q2, execute the cycle by putting in a work input W and send the Q1 back to T1. Is that fine? So, at any stage if you want you know to draw an analogy between this and A and B you can say either say the system is A and T1, T2 and wherever this W went to their system B. So, when I did the reverse exactly you know T2 came back to its original state that is it had Q2 it you know gave out Q2. Someone got that W it left out that W someone got had left out Q1 it had dumped here it got it back exactly. And this cycle did exactly the reverse cycle and nothing happened to it. So, it is exactly a reversible process. So, this is what we call an exactly reversible engine. It ran in the reverse way picked up Q2 took W and you know dumped Q into the higher temperature reservoir. Now we have decided what is high and low. I would not mind using high temperature reservoir. Is that fine? So, this is what I call a reversible engine or a reversible heat engine because this is what I really want. Of course, you can see that you know this is really a refrigerator. We are picking up Q from a lower temperature reservoir adding W and dumping it into. So, the reverse the heat engine reverse is nothing but a refrigerator. Maybe you can you know reemphasize this point. So, if this is what is called as a reversible heat engine then the reverse heat engine is nothing but a refrigerator. And you can call this as a reversible refrigerator because it can run in reverse exactly the same way and after the heating. So, it works both ways the definition for reversible heat engine is a reversible refrigerator. There is absolutely no difference between the two. So, this is something you can say. So, now once I have defined what is reversible Carnot theorem or the Carnot version of the second law talks about reversible engines and other general engines. So, the statement is as such that if I have any heat engine then efficiency now I have defined efficiency. Efficiency of any heat engine will be less than or at maximum equal to efficiency of a reversible. So, now you know you cannot accept this as true unless we use the same logic and try to see whether you violate the Kelvin flag statement. So, that should be your next step. Every time you make an extra statement you go. So, when we made the closest statement we said yes this corresponds with the Kelvin flag statement. And since we were you know dealing only with the Kelvin flag statement every other statement of the second law you can show is actually going to correspond and you know ensure that yes this statement actually did mean that the Kelvin flag we are fine with the Kelvin flag and it is not violating the Kelvin flag. In fact, violating this statement about the Carnot theorem actually means violating Kelvin flag. So, that is the method we will employ. We will say if you do not believe this let us say that N of some general engine is more than N of the reversible. So, we will assert this and show that if you assert this you will actually violate the Kelvin flag statement. So, the trick is the same. So, this is your any engine or any general engine. So, this is what you are doing some q 1, some q 2 and some w. So, this is heat engine and I will call it x some heat engine x that we did not know about but it was some general heat engine and this is our heat engine which is reversible. So, this is T 1, this is T 2. So, I will put different labels here as a q 1 dash because it is not the same engine q 2 dash and w 2 dash is that fine. So, now my assertion means actually that 1 minus q 2 by q 1 is the definition is the efficiency of engine x. Is this ok? That is I mean that has to be the definition of efficiency because we have defined efficiency of an engine as work output by q in. So, definitely the efficiency of this engine is q 1 minus q 2 upon w sorry q 1 1 minus q 1 by q 2 q 2 by q 1 and I will write here this is heat engine reversible. This is I am not saying what it is it is some engine x, this is engine x, this is engine r, this is r means reversible that is what we will go by. So, in this engine you are outputting work w, here you are outputting I should not put this w 2 some w dash. So, n reversible is w dash by q 1 dash is 1 minus q 2 dash upon q 1 dash and n x which we did not know was just w upon q 1 is equal to 1 minus q 2 upon q 1. Is this fine? So, now n x or eta x is greater than eta r is what we have asserted because we want to see whether this statement can be correct. Now, that is correct if I compare these two let me go to the next page. Is this ok? This is I mean I am just using the definitions of efficiency. Is that fine? So, I am subtracting q 2 minus q 1 from 1, I am subtracting q 2 dash minus q 1 dash by 1. If this is greater than this that means q 2 by q 1 should be actually a lower number than q 2 dash by q 1 dash. Is that fine? I mean this is just mathematics. So, what I will say is you have the reversible engine just reverse it. This is your engine x it is outputting w sorry q 1, q 2. So, is this fine? All I have done is now reverse this. I know this fact that q 2 by q 1 is less than q 2 dash by q 1 dash. Is it ok? q 2 dash in this system is actually being absorbed now once I start running it in reverse I am taking it from here. Is that fine? So, if I run now I can always do the same. I will run this n cycles this as n cycle and I will ensure that let us say n times q 1 is equal to m times q 1 dash. This is my most usual trick that I am going to do. I have done one cycle in such a way the other cycle in such a way that the net interaction with t 1 is 0. Is this fine? So, how much ever I took from here in this I just sent it back. So, now if the bottom here I am you know this will become just n n will become n n. Is that fine? Now n q 2 dash is more than this. Is this fine? So, what is happening is there is no interaction here. There is this net device is taking more q from this reservoir t 2 and less q 2 is dumping. So, the net q that is it is absorbing is n q 2 dash sorry this is m m q 2 dash minus n q 1 n q 2. This is the amount of heat it is taking. Now you will realize that this w here it is coming out sorry here it is coming out and here it is coming in. Now if what is w? It will be just m times w. So, m times w will be nothing but n times q 1 minus n times q 2 correct and m times w dash I mean originally would have been just the difference between m q 1 dash what did I do? First equation of oh sorry this one is that fine. So, now you see that the net w because for the same q 2 q 1 q 2 was lesser you realize that this difference the n w would be more than the m w dash because this numbers are the same this number is less than this number. So, this number is definitely more than this number it is just again algebra. So, what I have achieved in total is that this work output is actually going to be less is going to be more than this work output. So, I have a net work output I am interacting only with T2 because my interaction with T1 has been brought down to 0. I am removing heat only from T2 and I am coming out with work. So, you see now that I am immediately violating the Kelvin plan statement because I have created a one T heating that is I am interacting only with one temperature reservoir removing heat from it and converting it exact and definitely you can see that this quantity this work output is going to be exactly the same as this difference in the cubes again from this straight forward algebra you have created completely whatever q you took from T2 you have completely converted into work and you interacted only with one heat reservoir. So, the moment I made this assertion that this is true you saw that you immediately violated the Kelvin plan statement. So, we say since we believe the Kelvin plan statement this assertion must be wrong and hence the Carnot theorem as it is stated that is the efficiency of any engine should be either less than or equal to a reversible heat. Is that fine? See there is something called a specific Carnot cycle and a Carnot engine we are not going there we are only talking about reversible heat engines that is all that is mattering right now. If the engine is reversible every other engine that is possible between the same two temperatures. So, what we are saying is take the same two temperatures run a reversible engine between the same two temperatures if I run another engine the efficiency of that engine has to be less than or equal to the other reversible engine is that fine. So, hence we have you know we have used the same logic of trying to compare with the Kelvin plan statement to prove our next statement. So, you know see just see how we are proceeding logically in this fashion. I can use the same argument for the refrigerator take a refrigerator this is W, this is Q2, this is Q1, this is T2, this is T1 let this be the reversible refrigerator reversible and let this be a general refrigerator some Q2 dash here I have put this as dash. So, between the same two temperature reservoirs if I have a reversible heat refrigerator then I will assert that the COP the coefficient of performance let me call this X this is R the COP of X will always be less than or equal to the COP of R if you want to prove use exactly the same steps the reversible refrigerator can be run as an engine you run it as an engine this time you can probably try to remove T2 out of it and you will see it is interacting only with T1 and there is a net work output with only interaction with T1. So, every time I make a statement to test whether it is valid or not the only way is try to create try to combine things and try to create something which will violate the Kelvin plan statement or which will say yes the Kelvin plan statement is being you know kept where it is that yes this is true if it is true then your current statement is true if your whatever assertion you made if you can prove that the Kelvin plan statement is violated then an assertion is wrong. So, as a corollary to this Carnot theorem you can say that the efficiency of a reversible heat engine between the same temperature T1 and T2 of any reversible engine is always the same why again the same thing take this there are two reversible engines this is reversible engine 1 this is reversible engine 2. So, both of them can be reversed. So, you know there is some Q2 dash here Q1 dash here W1 dash here Q1 WQ2. So, between the same two temperatures I have two reversible they are different reversible the trick is this choose the less efficient engine of the reversible engine and reverse it make it run as a refrigerator you will come back exactly to this previous situation that we have why am I saying this because I already done this and hence I know that if I run the less efficient engine in the other way this is here the reversible engine was a less efficient one. So, if I run the less efficient engine in the reverse way I get back exactly the same situation that I got two flights which means I am violating the Kelvin time stage is that fine. So, that means between any two temperatures if I have a reversible engine or you know I can have any number of reversible engines all of them should have the same efficiency by the same logic between any two temperatures if I have reversible refrigerator all of them should have the same CO2. So, we have come from one statement to another statement to another statement systematically trying to tell you know first we said Clausius statement yes we defined. So, first we said Kelvin Planck statement we said this is possible then this is not possible then we said if it is possible to run engine between T1 and T2 it is not possible the other way around. So, we decided that there is a hierarchy because between T2 and T3 if you can run an engine then between T1 and T3 you can definitely run engine. So, we decided what is high and low we decide we figured out that yes now all this satisfies the Clausius statement. So, there is no problem then we went to the Carnot theorem of the second law if we define what is the heat engine what is the reversible heat engine and showed that the efficiency of any engine has to be less than is equal to the reversible heat engine efficiency between the same two temperatures and efficiency of every reversible engine acting between the same two temperatures has to be the same is that fine I mean that was the stepwise way we went as between this. So, what we will do next is go and do what is called as the Clausius inequality and you will realize that again this argument is the same thing. I think before this we need to define in some way what are the temperatures or you know how we can mention what the temperatures of the system are. One way that this is done is you see here that if I run let me just go to the next page let me draw it. So, you see that you know the efficiency of this the first engine is 1 minus q2 by q1 efficiency of second engine will be 1 minus q3 dash upon q2 dash I can always run the second engine such that I match the q2 dashes is that fine. So, basically when I want to run an engine between t1 and t3 if I want to run an engine between t1 and t3 I already made this argument that you know you can actually run this engine so many cycles this engine so many cycles such that these q2 and this q2 dash are the same or m times this q2 is n times this k. So, basically then you would have created an engine which ran between t1 and t3 took some q1 which would probably be n cycles into q1 so some q1. So, I will just call it as q1 q2 and I will remove this dash here because I have already ensured that I have run it so many times that q2 and this q2 match. So, the net interaction is such that there is this is just 1 minus q3 by q2 for this engine 2 or they let me write it as between 1 and 2 in efficiency between 2 and 3 and you will realize the efficiency between 1 and 3 is just 1 minus q3 by q1 because that is the net effect. So, all you are saying always there is a q2 by q1 q3 by q2 q3 by q1 you can see that you know in this case if I just wanted q3 by q2 if I just multiplied it by q2 by q1 I will get q1 by sorry q3 by q1. So, you realize the significance of this so I can always you know basically because multiple runs that I can do with a cycle I can always create a system where you know between any two temperatures when I am running the engine all that really matters q2 by q1 that ratio seems to depend only on this temperature T1 T2 because of the way we are doing q2 q1 q2 q1 is just some function of T1 and T2 q3 q2 is the same function of T2 T3 and q3 q1. So, all I really need is the temperatures and I can figure out that yes you know I just because if I can do this kind of a multiplication and get this all I really needed as if I know between these two what is it at this two what is it I can tell you what should it be between T1 and T3. So, q2 by q1 is just a function of T1 and T2 same thing with q3 q2 same thing with q1 q3 and I can keep on extending this argument. So, I see that yes there is some relationship between q2 and q1 and T2 and T1 one way to express this relationship is by just by saying you know I where let me call this instead of T2 I just theta 2 where this is what I call as the temperature or this is the number that I want to give for the temperature that is as long as I can decide what is the reference temperature and I can decide some kind of a reference value. For example, all I have to do really is run a refrigerator between some arbitrary temperature that I do not know about and some reference temperature a good method for the reference temperature is always you know as we saw the triple point. So, keep at some temperature where you have matched the triple point of water assign it a value. Again you know we had decided we can assign any value you want but just for historical reasons you would want to assign let us say the triple point of water as 273.16 there is no harm if you had assigned it some other value but you can just assign 273.16 run your heat engine your reversible heat engine between this and this and you will get 1 minus q2 by q1 q2 by q1 is equal to T2 by q1 or theta 2 by theta 1.  So, I get these values I have decided that this is my reference temperature. So, theta 1 will be only q2 by q1 sorry it will be only theta rep multiplied by q1 by q2. So, why should I just choose this again I mean I could have chosen any numbers for that q2 by q1 as a function is just a function of those two temperatures. So, why choose this that is again maybe you can relate it to what you know as the Carnot cycle. So, the Carnot cycle is 4 step cycle is a 4 step cycle yes. Sir are we not running into the circular thing by first means the refrigerator and heat engine they we say that they are working in two temperatures. So, temperature we already know and then how we are deciding the reference temperature and then calculating the other temperature on the basis of working of a heat engine or refrigerator. No, I only know what is high and low I just want to assign them numbers. So, but q the reference of q is there. Yes, the reference of q is there. So, I am choosing the definition that high q and low q. No, I am just saying that between any two temperatures the efficiency of the reversible heat engine is the same. So, that means 1 minus q2 by q1 if I run any reversible engine between the same two temperatures it is always going to give me the same number. So, I am saying why not. So, that means that eta official. So, let me go back or let me just create a new one. So, between so some reversible engine here I can take any reversible engine you agree because it does not matter all the reversible engine between the same. So, between the same two temperatures my eta is a function it is irrespective of the heat engine. So, eta is a function only of this T1 and T2 because that was the only choice I made. Yeah that is what I want to see. So, we have it is we know that it is a function of temperature and then we are defining temperature. So, what I am saying is that q1 see finally eta efficiency is 1 minus q1 q2 by q1 correct. So, now I know that this is the definition of efficiency and I know this is true correct when I know I did not get it eta efficiency is a function of T1 and T2 is that okay. Yeah okay eta is defined like this is that also okay. Yeah okay. So that means if this this 1 minus q2 by q1 is a function of T1 and T2 is that fine because this is this is the same. So, we have not assigned any number to T1 and T2. We have not assigned any number. We have we can talk the P1, P2 also or K1, K2 also not P1, T2.  And then on that basis we are defining something which is called temperature. No see we have decided that this is a temperature reservoir with label A this is a temperature reservoir with label B. We know that we have already decided which is high and which is low. So that also we have figured out. So we have said A is higher than B okay. Now we say the reversible heat engine efficiency depends only on that A and B and nothing else and this definition of efficiency of a heat engine is irrespective of anything else.  I mean I do not know what it is okay. So some A and B this is all I know. Then I know that you know 1 minus q2 by q3 or you know I will call it if I called it A and B then I should better call it qB and qA because it was rejected to B and taken from me. Similarly here this is between B and C it is a function only of B and C correct. So I have realized now that if 1 minus qA by qB is a function only of A and B obviously this one is meaningless this qA by qB is a function of A and B that is what I have realized correct and I have realized qB by qC you know it is the function of B and C. I just notice that if I multiply this 2 I can get qA by qC and I know yes this is exactly what I wanted it is a function of A and C this is all I know. So I say if I have qA by qB as a function of A and B why not decide that qA by qB is temperature label A upon temperature label B why not. The question is you know if tomorrow you tell me this should be square I am perfectly fine. You can do all your calculations assuming square you do anything you want you are not going to face any problems. So this is a question of why not and I was just going to say why not by using the Carnot cycle where you know some kind of ideal gas scale was used and we got everything that is what I mean probably you have all done the Carnot cycle and you saw that there is some efficiency which comes out as 1 minus T2 by T1 and that was based on that that Carnot cycle was based on some ideal you know normally what you draw is the ideal gas Carnot cycle you know all those legs are ideal ideal. You assume I am added to adiabatic processes to isothermal processes and the temperature scale that you assume there was based on you know whatever you knew by thermometry which was maybe the ideal gas scale or you know mercury scale or whatever you want there was some scale we had decided and you got T2 by T1 and this is the only reason we just want to stick to our whole thing there otherwise tomorrow you want to square it I will tell you there is going to be no difference in the gas flow that is what is going to happen. First of all a heat engine which is operating between T1 and T2 yes and the next heat engine operating between T2 and T3 yes that is the case ultimately we will be able to operate a heat engine between T1 and T3 yes and we know efficiency of first heat engine which is equals to 1 minus Q2 by Q1 second 1 minus Q3 by Q2 and we know that first heat engine is able to operate between T1 and T2 in one direction but how we will be able to explain Q2 by Q1 solely a function of T1 and T2. I did not get it you want to know see we had said that for any reversible heat engine efficiency is between the same temperature the efficiency in material of that engine as long as it is reversible the efficiency is the same. So, the efficiency is somehow a function of T1 and T2 only the two temperature levels that is all we know what that function is we do not know is that what you wanted I do not know what you wanted is that. Efficiency is a function so the temperature are we taking due to the efficiency of expression of power at same. No I am not I am just saying the moment I decide two temperature levels the efficiency is the same because that is by Carnot theorem not by Carnot cycle by Carnot theorem according to the corollary of Carnot cycle efficiency of any reversible heat engine between the same two temperature is the same correct. So, that means the efficiency of a reversible heat engine between the same two temperatures it is depending only on T1 and T2. So, it is a function of T1 and T2 what that function is I do not know correct, but I know that the definition of efficiency is 1 minus Q2 by Q1 this definition is for a heat engine it is not for reversible here reverse nothing, but let me for the reversible engine if the all the efficiencies are same then Q2 by Q1 for all reversible engines between the same two temperatures will be the same correct Q2 may be more for some engine, but Q1 will be proportionately more and Q2 by Q1 will always be same because 1 minus Q2 by Q1 has to be the same is that okay. So, that means I know I mean that now I realize that the one Q2 by Q1 is a function of the two temperature level again what that temperature level is what that function is I do not know okay and I have no knowledge of fixing what that temperature level is it is only I am taking a help of that fact that I can just multiply it the temperatures also go in the same way okay if I had chosen square of that again it would not have matter okay. I just know that Q2 by Q1 multiplied by you know Q3 by Q2 would be Q3 by Q1 okay and as far as I am concerned between 1 and 3 for a reversible engine Q3 by Q1 is a function of T1 and T3 correct that is all I know okay so I will say why not say that Q1 by Q2 is T1 by T2 why not okay this is up to me if I want to start putting numbers okay so it is my decision now that I want to put some numbers okay if I know what is Q2 sorry if I know what is T2 I mean I fix the value I mean for everything for any kind of thermometer I need to fix a value okay if I fix a value this is unknown X I will put I will just have to run a reversible heating in between these two measure Q1 and Q2 automatically I will know T1 if I assume this is the definition you tell me tomorrow no I do not like this definition I can show you that I can have T1 square by T2 square multiplied by T2 square by T3 square sorry this should be 3 this should be 1 okay is equal to T3 square by T1 square okay and that I can say this is this is this is having the same multiplicative effect. So I will say okay let me call that you know Q1 by Q2 is sorry Q2 by Q1 is just T2 square by T1 square go ahead I mean I am telling you absolutely you will not lose anything in your future discussion only wherever there is a T you may have to put a square root T for the current kind of discussion I mean current kind of way otherwise you know you are going to see that there is going to be no difference. So this is going to be our definition okay we have decided it that like this because we know that yes this is consistent with we are not harming anything else if you want another decision which maintains the same kind of multiplicative thing you are free to do so okay but everyone has to agree to it okay and the only reason people agree to this is history again okay somewhere there was a Carnot cycle and that is what I was going to show next which is based on so called ideal gas scale sorry ideal gas and it used some temperatures T1 and T2 which are not defined like this but we have defined based on some thermometer that was known earlier where high and low were decided where you use probably a constant you know pressure gas thermometer you figured out some temperatures use 273 you got some ideal you got then where you got the efficiency of a Carnot cycle okay there in that system as 1 minus T2 by T1 where this T2 and T1 were based on the ideal gas scale not on our current scale okay so I can see that I can have a relation like this as T2 by T1 within my purview I can do this I could have also chosen square I could have chosen square root anything I wanted or I could have chosen some other function which is not which would still maintain the same kind of multiplicative thing okay so it was up to me to decide I say okay but if I choose you know this as that theta as the temperature you know it matches some old thing okay and it will keep my whole life easy because I do not have to change anything else as far as the definition of temperature so see every time you come up with something new thought to you know formalize things okay you do not want to disturb a lot of older order okay so when you realize that you are causing no harm you might as well do it okay so but you should remain well aware that you would have not changed anything by choosing something else which maintained the same thing okay it is the same as with Kelvin scale you know everyone wanted that the degree Kelvin should be the same as degree Celsius okay and in that foolishness we have come up with 273.16 I could have chosen triple point of what was 100 but then you know I have to remove all my old things I mean no harm would have caused as far as thermodynamics all calculations would have been correct but everyone would have got have to be used to a new scale where triple point is called 100 Kelvin or something from any other number you want okay it is up to us to decide so we just wanted that degree K should be equal to degree so it is the same philosophy we are continuing with let us use this when we know that yes it can be used okay and it maintain so that is why I mean do I do you want me to go through the Carnot cycle derivation because you have done it I mean it involves coming with one eta minus eta is equal to 1 minus e2 by t1 but remember there again those that e2 by t1 was not decided by anything which we did up till now because decided based on some temperatures case that you had decided based on thermometry which was probably the ideal gas scale or something and just because that t2 by t1 exist we have continued with this t2 and t32 we have decided that this is the okay way to go ahead okay but now once we have decided it you must throw that okay once we have decided this you can say if someone says what is the thermodynamic absolute temperature do not go and define it based on the ideal gas k do not use that constant pressure gas thermometer or the constant volume gas thermometer and define that okay you say that but this is very hypothetical you say if I want to define the temperature I must take a reversible heat engine run it between the temperature t which I do not know and my reference t ref which let us say is the triple point temperature okay and again you know I still have no other choice but you know to maintain things and say triple point of water I will just assign it to 73.16 k I have no other way again because I still want to maintain history okay and I will say I run measure q1 measure q2 of course you must first see whether you can run this like this it may be possible that that temperature is lower than t ref in which case you have to run it the other way around correct so you have to first test in which way the engine will run okay let us say it runs like this then you find out q1 and q2 you know that as long as you chose a reversible heat engine it did not matter q1 by q2 would be the same so I will just say q1 by q2 is t by t ref I know this because I have assigned it some number okay there is nothing to know about it you have fixed it given it some value you have measured this so this you can find out so you will say that this is the thermodynamic absolute temperature what you had earlier was the ideal gas absolute temperature okay but if you want to have a thermodynamic basis now you have a thermodynamic basis of assigning the temperature that is you chose the Kelvin Planck statement you went through a series of steps okay you figured out what is high and low and you figured out that the efficiency of a reversible heating engine does not change okay and you went ahead and said that q1 by q2 is a function of some temperature level so you say why not use t1 by t2 that is all you did okay and that why not the answer was again sticking to history nothing else okay yes give say grading to the temperature and Carnot cycle kind of gives a value to the temperature why are you saying that because so far we did not give any magnitude to the temperature correct but only now we have so see wait a minute do not bring the Carnot cycle in I did not use the Carnot cycle at all okay so Carnot cycle is a reversible cycle that is why it matches the conditions of a reversible heat engine so it just happens to be a reversible heat engine and hence it comes out the same so I do not really need the Carnot cycle to do this to assign the values that I want so but you are right I have decided the hierarchy not based on the Carnot theorem but I use the Kelvin Planck only I never use the Carnot I use the efficiency of the Carnot cycle and found out that q2 by q1 is a function only of temperature and that is where I decided that maybe this is what I can use to assign temperature okay so the Carnot theorem somewhere I use that the corollary of the Carnot theorem I use that the efficiency of all reversible heat engines are the same I use that to come up with you know why this is very useful that q2 by q1 is the same for the same two temperatures okay it is just a function of t1 and t2 and it looks like it can work that q2 by q1 is t1 by t2 why not use it so that is the way that we have gone based upon second law we actually have set a grading to the temperature you have set a grading to the temperature based on the second law and you decided to assign values by using by noticing the fact that reversible heat engine efficiency independent and this thing comes out like this that you could as well use t2 by t1 so that is all you can say so I think people are here so let me just


Thank you very much for coming to the second lecture of this series and today we'll continue what we have started yesterday and today we'll speak about relation between the Fourier interpolation formula which I have introduced the previous time and certain functional equation which we have obtained for the generating function. So today we will not speak about energy optimization and universal optimality but rather we will concentrate on the Fourier interpolation and so here is the I'll just briefly remind the theorem which I've formulated on a previous lecture so it's a joint result with Henry Kohn, Abinav Kumar, Stephen Miller and Danilo Ratchenko and it says let d and n0 be either 8, 1 or 24, 2 so we've worked out all the details for dimensions 8 and 24 however this formula also works in more general settings and so what we have proven at least for these two dimensions that there exists an interpolation basis of Schwarz functions we number them an, bn, an tilde, bn tilde and they are radial Schwarz functions on rd and here the index n it starts from this number n0 which is either 1 or 2 and so these functions they satisfy the following property so if we now any radial Schwarz function any point x in the dimensional Euclidean space then we will have an equation like this we can reconstruct the value of this function at point x from its values at square roots of even integers for the function itself it's first derivative it's Fourier transform and the first derivative of its Fourier transform so we'll have a formula like this and we call it the interpolation formula so I think for this particular n0 then the formula is unique but for example in already in dimension 24 we could make some choices for example we could start from 1 and then there would be some ambiguity here so and so here's another result which is somehow closely related to this one which it tells that we cannot only interpolate existing functions but we can actually construct Schwarz functions with the values at these points which we like which somehow don't so the only condition is that they don't contradict the growth growth conditions which we impose on all Schwarz functions so for this we you know the following consider the following set so it will be the set of all fast decaying sequences so vector space of all sequences which we start from 1 such that the sequence decays faster than any power of the index and so now what we show that we can take any four elements in this space and construct Schwarz function which will have these values at these special points so so again let d and n0 be as above and we consider the following maps so one of them would be map we call it xi it will be the map which goes from the space of radial functions into the space of fast decaying sequences to the power four and so the function is defined in the following way so it takes function f and maps it into the collection of its values at the square roots of integers so for the function itself for its first derivative for its Fourier transform oh yes for dimension 24 it will start with two yeah but somehow still I which was just with this ambiguity somehow if we shift the numbering somehow the space will not change essentially so so here n goes from zero to infinity and so the second function which we can consider is this one we denoted by capital phi so it goes in other direction so it starts with sequences and maps them into schwarz functions and this time if you have a sequence like this alpha n beta n alpha n tilde beta n tilde and here n again runs from n0 to infinity then such a sequence would be mapped into a function like this which is given by the sum of elements in the interpolating basis who is exactly this coefficient so and so then what happens that the map xi is an isomorphism of vector spaces and if we take the composition of xi and phi it will be identity map and the same way if we compose phi with xi this would be also identity map and so also what we have done on a previous lecture so we have considered a generating series which was formed from our elements of the interpolating basis so one of them was a function we actually split it into two functions so one of them was the function which formed by functions a n and b n and the second one was by the a n tilde and b n tilde and so here tau was a variable at the upper half plane and x was a variable in the euclidean space and what we have shown last time is that now these two functions they satisfy the following functional equation if we believe in our interpolation theorem then from the interpolation theorem we could deduce the following functional equations so and the equation was like this so first f of tau and f tilde of tau they were related by the following transformation and then the both functions they are let's say linearly periodic it means that if we take a second difference of each of these functions with respect to tau here I omit variable x just for the speed of writing so then we would have equations like this and what we have seen last last lecture so we denote this by this would be like a functional equation and so what we have seen that interpolation formula it applies functional equation but now our how are you going to prove the interpolation formula so what we are going to do we are going to somehow show that the this implication it also works in different direction so what we will see today so we are going to show that if we start with a functional equation so if we if we are able to construct a function which satisfies the functional equation and the number of analytic conditions which I will specify a bit later then we can prove the interpolation formula and so now before we can do this I'll need to introduce several notations so first of what I think first thing I would like to do I would like to introduce a semi-norms on the space of schwarz functions so here so let f be a radial schwarz function and so we will write d of f for the we'll denote like this the let's say the radial radial derivative of fx so if our function is radial so it means that it depends only on the absolute value of x and this d of f it just it means just taking a derivative of this function f0 the same as you apply earlier vector field yes yes yes and so now we can consider the following semi-norm on the space of schwarz functions so it will depend on two indices chi and l and it's also would work for only for radial schwarz functions so it will be a supremum over all x in rd so the cladial norm of x to the power k times the absolute value of the radial derivative of f k and l are non-negative integers and so this set of semi-norms actually defines the topology on the schwarz space so so the sequence of functions converges if and only if to some limit only if it converges with respect to each of these semi-norms and so here's one important fact which is I think it's very well known but I could not find who was the first observed something like this so the complex gaussians so the functions of this which look like this for some tau complex number in the upper half plane they yes oh yes so they they span a dense subspace in the space of radial schwarz functions and here is one a bit more technical statement but I think we'll use it later in our series of lectures so now if we fix any positive i or y so the same the same is true it means that that if we consider only gaussians with imagine with for tau with imaginary parts equal exactly to y and so now you can maybe we'll guess where I'm going with this so now what we would like to do we would like to prove our interpolation formula only for complex gaussians and then to say that this the same would work for all other schwarz functions because of the of the density of this subspace in the space space of radial schwarz functions something in terms of functions of upper half plane I'm sorry it doesn't give also some kind of identification of s right a rd with some kind of holomorphic functions on upper half plane all right may yes I'll give one problem here is that of course like it's we have this space it's spans radial schwarz functions but it's not in this functions they're not linearly independent so there is like in some ambiguities in here so probably probably if you need one to choose a function on the upper half plane we need to make some choices in here yeah so from here I see how it works like in one direction from something in the upper half plane we could get a schwarz function but maybe it's not at the moment not so clear for me how to go with the other direction and so okay so now I will formulate our other long theorem which tells us exactly what I promised you so that functional equation does imply interpolation formula but except so we need to find the solution with but what somehow is a bit difficult what makes the theorem long is that we also need to specify some analytic conditions to make sure that our formula works not only formally but also analytically and so the theorem is like this so now suppose that that there exist smooth functions f and f tilde from the upper half plane cross euclidean space into the complex numbers and suppose that these functions they satisfy the following properties so first if we look at them as a functions of tau variable then they are holomorphic and tau the second statement would be that if we look at them as a function of a second variable of variable x then they have to be radial in x and third condition so we want some balance on the semi-norms of this two functions considered as a functions again of second variable and so now for all non-negative integers k and l so we have that so now we consider this the k l semi-norm of this function viewed as a function of x and tau is considered as a parameter here so the same has to be true also for f tilde no no no it's just no it's just some it's a positive some positive number delta sorry yes not maybe not a great notation and so we have the same estimate for f tilde and this would be for some positive real numbers alpha k l beta k l delta okay gamma delta k l and so so now what comes next is a condition which somehow assures that our formula it starts either from one or from two so now we want somehow in the special case when k and l are both equal to zero what we want we want actually a stronger we want to have a stronger estimate so and this has to be true for tau in the strip from minus one to one so here for all points x and the euclidean space and we also want this beta zero zero to be a positive number strictly positive number and so the this was condition four and the condition five finally it is that f and f tilde they have to satisfy the functional equation so which is now written on the upper blackboard and so whenever these conditions hold then we claim that then f and f tilde they have expansions of this of the form which is now given at the upper blackboard refer to them maybe i'll put so they will have expansions like this so let me put like a star here and sorry a value of zero which is one of two so yes yes yes so it will come a bit later yeah so for the value of zero we need somehow we need to stress on this this condition so they will have expansions of the form like star and star tilde as it's given on the upper blackboard only with the here if this condition four holds then what we can guarantee that n zero will be equal to one and for some radial functions n b n a n tilde and b n tilde and moreover for every which is a radial schwarz function the following interpolation formula will hold it would be an interpolation formula like this only starting indeed with one and so also what we will if we are able to construct a function f like this then we will know that for fixed k and l then the radial seminars of the all this for radial schwarz functions a n and b n we know that they will grow at most polynomially in n and so now the last condition which we have to write is the one which assures that n zero will start not from one but from the for example from two so we have that so now find last part of the formulation to be like a one and a one tilde would be equal to be one equals to be one tilde equals zero so this will happen if and only if so the functions f and f tilde so that they not only go to zero as a imaginary part of tau goes to infinity but also go to zero fast enough so they have to be small o of the e to the minus two pi imaginary part of tau as imaginary part of tau goes to infinity in the strip where the real value of tau is between minus one and one and with x fixed and so now this is the the function which we need to construct so we see that somehow functional equation plays a role also we see that our functions they have to satisfy somehow this nice growth conditions and so these growth conditions they have to work so to say uniformly when we when we change our variable x and so whenever we are able to to construct a function like that then we will also prove the interpolation formula and by computing the Fourier expansion of this function we will obtain explicit formulas for the interpolation basis so maybe you have some questions at this point so then the next step what I would like to do I would like to prove this theorem so to prove the equivalence which is so now we have all the necessary ingredients for this so now let's suppose that we have a function with all these properties so our first step would be to show that if we have a function which satisfies these properties then it will have the free expansion of the form which we are looking for and so this is a quite easy step so step one it is so prove that we have the expansions and start tilde and so now this will follow of course the expansions they will follow from the functional equation and so from the last two equations in our system from this from the linear periodicity of f and f tilde so what we do so now we consider the function like this suppose that f satisfies the functional equation then we can consider a function like this and we look at it as a function of of variable tau and so now we see that this function of course it is holomorphic also we see that it is one periodic in tau so it is in variant under taking tau into tau plus one and the last condition that it goes to zero as imaginary part of tau goes to infinity so this follows from our condition four and so now the thing which we look at very we do a very simple thing so we go from strips like this by taking tau into that it's e to the 2 pi i tau we can map such a strip into a punctured disk and so the periodicity condition for this function that tells us that our function can be written actually as a function of this expansion so it's it's a usual free analysis so from here we see that we have an expansion like this and now because our function is holomorphic and it has only one potential singularity here at the zero we know that it is removable and so that our function already on a disk it tends to zero at this point so we will go to an expansion like this and so now what we can do we of course can do the this is how we have obtained our functions bn of x and we obtain functions an in a similar way only for this we have to consider a slightly different function obtained from f so to get the coefficients a and we have to consider function like this and now the again this function again it is holomorphic it is periodic in tau and it also also goes to zero as imaginary part of tau goes to infinity and therefore we see that f of tau x it has the expansion which we expected it to have and the same works for f tilde and so analogously the same works for f tilde so this was a rather easy easy part so i'm sorry for maybe getting you bored here but so now the next thing what we have to check we have to check that the coefficients which we have constructed they are actually schwarz functions as the functions of x and so so here our second step would be to check that an bn an tilde and bn tilde are radial schwarz functions and to do this what what we are going to do we are going to give a so given an integral representation of our Fourier coefficient eighth coefficients a n and bn in terms of the function f itself and then this way we could read properties of Fourier coefficients from the properties of of the function so so here we write an an and bn in the following the following integrals so we fix some positive number y and so we integrate this function so we just use the usual formula for the computation of Fourier coefficients of a periodic function and the same we do for bn only we remember that we like bn are not just Fourier coefficients but they are normalized and here again we are integrating over some horizontal interval from i y to i y plus one and so now we see that from part three of the hypothesis of our theorem which is just here above from here we will see that the semi norms of an and bn they are all bounded so they are finite and this already tells us that the functions an and bn they are schwarz functions so but now what we also want to somehow to prove is that they are not only bounded but that they grow at most polynomially as an index n grows and so for this we take we use a usual trick for analyzing Fourier coefficients of functions of moderate growth of the upper half plane so here what we do we take this number y which was allowed to be anything but now we take it to be one over n where n is the index of the Fourier coefficients and so from here we will see that the semi norms they grow at most polynomially with respect to the index n for k and l fixed and so now of course if we consider an tilde and bn tilde everything works analogously and so from here we see that the consequences of this is that the interpolation formula it converges absolutely so sorry not the interpolation formula but so the right hand side of the interpolation formula it will converges converges absolutely for every function f which is a radial schwarz function and so and now we also see that this the right hand side of the interpolation formula it will define a continuous functional in the space of schwarz functions it defines a continuous linear functional on the space of radial schwarz functions and so now finally we are ready to prove the interpolation formula but now the proof of the interpolation formula it will actually follow from the density of Gaussians in the space in the space of schwarz functions so now let's fix some x0 in the dimensional Euclidean space and we consider the following linear functional which goes from the space of radial functions radial schwarz functions for example into r and it is given by the following this may be we did not define the lambda of f to be the following functions here we take the right hand side of the interpolation formula at point x0 and we subtract the left hand side and so now the interpolation point at the interpolation formula at point x0 it is equivalent to the fact that this linear functional lambda vanishes and so now because we know that this linear functional it is continuous so it will be sufficient for us to check vanishing only on the set of complex Gaussians that lambda is continuous and therefore we know that now it suffices to prove that lambda of each complex Gaussian for example for all tau in the upper half plane but now we see that that this condition it is just equivalent to to the functional equation and so this finishes the proof and so here one more remarks and now we see that the solutions of our functional equation they will give rise to interpolation formulas on the other hand what we could also consider we could consider the this the homogeneous version of this functional equation where on the right hand side we have only zeros where we don't have this complex Gaussian but we have only only zeros and then such functions they will correspond to relations between values of Schwarz functions on the square roots of even integers as derivative and the same information from the Fourier side so yes so lambda it's a functional on the Schwarz space so it's continuous linear functional on the Schwarz space and the topology on the Schwarz space is given by the by this set of semi-norms so that gives you zero yes i want to prove the function zero check on the dense subset yes yes yes so suppose that that h is a solution h and h tilde are our solutions of the homogeneous equation and this homogeneous equations probably I will write them on the next blackboard and so the equations would be like this so h of tau so then it's we also know that from this two last equations that h will have a Fourier expansion like this so and okay so here again somehow but we want them to be homogeneous solutions but we also want h to satisfy this gross conditions which are sure that n starts from one okay so let's do it like homogeneous was this a special piece of the function equation no you so i just like the first condition i replaced it so there i had some function on the left hand side but here i replaced it by by zero and you forget about dependence on max yes yes homogeneous equation yeah no no no it's okay i know maybe it's not a great terminology but i think again in many areas we do have it like we have like differential equations and whatever equations if we have something on the left hand side we say that it is a homogeneous sorry yeah so yeah so this is kind of a functional equation but cn plus i tau square root of 2n maybe some numbers dn so e to the 2 pi i and tau so we know that we will have Fourier expansions like this so here i i did not spell out the conditions which ensure that we have Fourier expansions starting from one not from zero but i hope you will excuse me for this so we'll have Fourier expansions like this and so now we have that for just for the as a consequence of the proof of the corollary of the proof of the previous theorem it will be now then that for any radial Schwarz function in a d-dimensional space we have the following relation we will know that sum from we know that such a sum always will be zero so in a some sense we have if you remember in the beginning of our lecture we have introduced the maps xi and phi and so this tells us uh this formula describes for us a subspace of uh so we have this map xi which went from the radial functions into the space of fast decaying sequences and now we see that the so to say orthogonal complement of this space it will be somehow isomorphic to the space of so space of solutions of this homogeneous functional equation but here again of course we have to consider such as solutions and we expect these functional solutions also to have this nice uh moderate growth property as a function of tau and then we can assure that the Fourier coefficients grow at most polynomially and then at least yeah and such a pairing it will always make sense it will be always convergent and so what you said before it means that this equation homogeneous question does not have solutions no it has it has solutions again depending on d so for example for if we restrict it like this like in dimension eight it will have only trivial solution but for example if we start from n equals one in dimension 24 it will have probably one non-trivial solution up to linearity and somehow in principle everything what we are saying here it would work in uh any dimension uh and so in a somehow usually well as dimension grows so what we will see later that the space of such solutions is always finite dimensional and actually we can we have a very good description of it very explicit and the dimension of space it grows with dimension of the space of solutions grows with dimension of our Euclidean space we are considering so should we make a small break maybe for five minutes now after we've shown that interpolation formula is equivalent to existence of a certain nice solution of a functional equation so now the equation which we have how to solve this functional equation so both first how homogeneous and then of course also inhomogeneous so so the next our next step which probably will take us sometime probably more than just second half of this lecture so there's probably something I will continue in May so how to construct and f tilde and so for this what will be useful for us to look at f and f tilde as a function of of a of tau variable in the upper half plane and to study the modularity properties of these two functions so so now let me introduce some notations so let's consider the following group so we consider the group ps l to z so it's a group of all uh two by two matrices with integer coefficients uh module all the subgroup which consists of identity and minus identity matrix and so now this group it acts on on the upper half plane by the linear fractional transformations and because it acts on a upper half plane it also acts on functions from the upper half plane and so there is a one particular action which is given by the slash operator so what we do we fix the function f from the upper half plane into complex numbers let's fix an integer k then and let's fix one particular element of of the group so and probably it probably suggests to be even integer so now we define the slash operator it acts on functions in the following way so we have a action like this and so this action it it is actually an action of a group so if we have two elements of in sl to z then the product of two elements acts on function in the same way as we first act by the first element and then act by the second element and so now because we want to consider this a bit more general functional equations so another object which we would like to introduce would be the group algebra of the group psl to z so so let r be the group algebra and now of course the action of slash operator it also extends to this group algebra by linearity so slash operator of weight k now it is extended can be extended to the group algebra by linearity yes so it is a right action and so now also as we have written it so the slash operator it acts on any functions without any restrictions on the upper half plane but we would like to consider one particularly nice class of functions so namely the functions which have a moderate growth so now we say that a function f on the upper half plane it has moderate growth if there exist positive constant alpha beta and gamma such that the absolute value of this function f at point tau is bounded by the following quantity so what we insist that our function it grows at most polynomially as tau tends to the real line or as tau goes to infinity and it's somehow this definition that actually agrees with the definition of moderate growth which is given in the theory of symmetric spaces so so this should be true for all tau in the upper half plane okay so at some point we will actually need some functions which are not continuous and have this growth so I maybe I'll just here I will just save this notation for all functions maybe or maybe also it's not very good analytic properties but for holomorphic functions we will have this class curly space curly p so this would be the space of all functions from the upper half plane into complex numbers such that f is holomorphic and has moderate growth and so now an important property it is that if we apply a slash operator to a function then if our initial function had moderate growth then the new function also will have moderate growth so if f has moderate growth this will imply that f slashed by gamma also has moderate growth and so now with this after we have introduced the slash operator so now what we can do we can rewrite our functional equations so so here again so rather standard notation it is that we consider two particular elements of sl2z one of them it will be a matrix like this t it is a matrix like this so these two elements they act particularly nicely on the upper half plane so s is an involution and it sends tau into minus one divided by tau and t it acts on the upper half plane just by translations by one and so now the group psl2z it is generated by s and t these two elements and so for between these two elements there are two relations only two the first they already said that s is an involution so s squared is one in psl2z and the product of s times t it's also an elliptic element but of order three and these are the only two relations between the generators and so now what we can do now we can rewrite the functional equations for f and f tilde in terms of a slash operator so the f tilde can be written as here this last one is always from the group algebra okay so first it's true for any element of the group but also for the element of yeah because of this thing it's of course this set of functions of moderate growth or class of functions of moderate growth it is a linear space if we add two functions with this growth we get a function which cannot grow too fast either so and so now let's write this in terms of the action of a group algebra this would be an equation like this so this first equation that again expresses the periodicity properties of f the second one translates periodicity properties of f tilde and the last is the functional equation which we have obtained from the interpolation formula for the complex quotients and so here for the notation so here I use this upper index tau because f it's a function of two variables and so this is just to stress that I apply slash operator with respect to the variable tau the first variable and so now what we can do somehow it's a bit difficult to work with two functions simultaneously one thing which we can do we can remove for example function f tilde from this functional equation and transform it into functional equation which will contain only f and so we do it in a following way so we see that from the last equation we can see that f tilde it's actually equals to e to to a function like this and so now we can rewrite the first two equations in terms of f and so from here we get that f the first equation does not change and the second equation can be written in the following form so now this would be the functional equation we want to solve for so we and now we want to solve this functional equation for f as a function in the class p and so again if we just require f to b and p we could have some extras maybe the solution might not be unique but as we will see later somehow the solution of such a homogeneous equation it will always contain only five it will be always finite dimensional but we also remember that we have this extra gross conditions on f close to the cusps and this will give us the uniqueness of the solution and so now we what we would like to do we would like to consider the following ideal inside of the group algebra so we consider the right ideal and so this will be ideal which is generated by these two elements of the group algebra so it is generated by and so so we see that in some sense among the properties of this ideal they will determine for us the properties of the solutions of the functional equation and it turns out that this ideal is quite nice maybe better than we could expect from a random ideal in this ring so it has the following properties which will be later important for us so first is that if we consider the quotient of the group algebra by this ideal and look at it just as a vector space then this vector space will have dimension vector space over complex numbers then it will have dimension six and so here like one can we can observe immediately that six is smaller than infinity which is a good news for us so this gives us some hope about so for solving the functional equation a second rather nice property is that so the ideal i it is actually freely generated by these two elements so it is i is a free r module generated by these two elements and so for example because it's a free module what it gives us it gives us that now if we know the action of these two elements on f then we actually will know the action of any element of the ideal on f at least theoretically can you remind for the sake of physicists what a free r module means okay so it means that somehow if we take any combination here like this element times some element of r plus this element times another element of r and we ask whether it equals to zero and it will be equal to zero if and only if these two coefficients which we applied to the generators they both have to be zero so it means that we don't have any hidden relations between these two generators and so the last property which is probably a bit difficult for me to state but yeah so then so the if we because we have a we have this quotient which is a vector space and now our group psl2z and also the group algebra they will act on the on the right on this finite dimensional space so it means that we will have some for example we will have a six dimensional representation of psl2z associated with this ideal and now it turns out that this representation is quite nice so the so the representation of psl2z of on the module it has so I'm not sure if it's right terminology or not but it has what I call a polynomial growth and so let me explain this to you in a moment so it means that if we for example if we have a representation on six dimensional space it will give us this after some right choice of basis and dealing with the fact that this is actually a right representation so but we can for example transpose it and get a representation of psl2z in gl6 and for example so let's also assume that we have a standard basis chosen here so for each element in psl2z we can associate a six by six matrix and also have to transpose to make it left representation and not right representation and so then the lemma which we proved is that there exist some constants so which don't depend on anything so absolute constants c and n which are positive real numbers such that so each entry of this matrix sigma of element abcd is bounded by so it's polynomially it grows at most polynomially depending on this uh entries abcd and d of a matrix in psl2z here again so here abd is any matrix in sl2z and this property it will be important for us when we will try to to prove that the solution of the functional equation which we have found it does belong to the class currently p it should be this property something standard in geometric group theory so it's certainly known in geometric group theory but i for example i do i could not find like whatever for example very good definition for this very good name for this property so i don't know maybe if you know what is like a more correct name for this property yeah but it certainly is very natural and property and it's quite useful so and also maybe i should stress that somehow it's not a trivial property so if we take some random representation of psl2z we don't hope for property like this for example it could happen that we map our matrix at t into some element t into some matrix which has an for example eigen value which is not of absolute value one and then of course we can compute representation of t to the power n and the entries of such matrix would grow polynomially yeah but also i just can have a question from the people get local system on more the space of electric curves yeah does it get a variation of what structure is it kind of geometric yeah so now we'll see so so this representation we get that it is quite nice so i will yeah so so it is for for example the solutions of homogeneous equations they can be expressed in terms of classical modular forms so it is a quite nice representation so maybe i will spend last 10 minutes on giving you an idea how so how we proved this statement we actually proved it just by analyzing our representation and recognizing some pieces of it so our representation it consists of some very very nice pieces and for those pieces it's more or less obvious that this property holds so so now let's give an easy description of sigma and so for this we consider the following representations the first let row three be the symmetric power representation of dimension three and so this representation on the level of entries of the matrix it can be given in the following form so with one possible choice of bases so it's a symmetric power representation and so this actually is the unique representation of psl2z which can be extended to psl2r to say up to after an isomorphism this is the only one and so one way to understand it it's also the way psl2z acts on homogeneous polynomials of degree two and so another representation which we consider it would be so here we take a representation of sl2z into representation of dimension two and it's given by it's we will give yeah so it is this time it has to be sl2z no not okay so we define it by action on generators and so now this will be a two-dimensional representation and this representation it has actually a finite image so the kernel of this representation it will be just the group gamma two which is the principal congruent subgroup of sl2z so the subgroup of all matrices which are equal to the identity matrix modulo two and so the index of gamma two in sl2z is six and the image of p6 it is isomorphic to the dihedral group of a triangle so it will be just a group of symmetries of a triangle and so now we need another another object which would be a map from sl2z2z2 and so here with z2 so we have we also give it action on generators so the s is mapped into zero and t is mapped into this so we consider this as a row vectors and so now what we require we require that this map v it satisfies the cycle formula it is so it is related to this representation row two and so now from these pieces we can finally construct our six dimensional representation so our six dimensional representation that actually splits into a direct sum of two three-dimensional ones so one of them it will be the representation row three and the second one it will be a three-dimensional representation which is constructed from this finite representation row two and this co-cycle so now we define the representation so we define row from I think a priori it's from sl2z gl6 and it looks like this so it is matrix which looks like this so this is a three by three part and this is another three by three part so and so now we have the our representation sigma it is equivalent to this one so after some change of coordinates one minus one in the representation row two x not really okay so maybe I have to yeah maybe I have to double double check this what happens to okay so I remember that at some point it did work but maybe I have to double double check what what I does is this so of course this representation it should factor through psab is to be well defined for psl2z as well so and so for the but now the fact that we have such an easy presentation of our representation sigma it actually helps us solving for example the homogeneous equation and from this we can deduce that the solution of a homogeneous equation it has to be some combination of classical modular forms also quasi modular forms and logarithms of modular forms but so the somehow like this part the symmetric power representation it will be responsible for the quasi modular forms side and so this the cacycle which we will have here it will bring for us logarithms of modular forms and now for the next time so of course we want what we want to do the next time so next time we would like to so next in the next lecture we would like to solve the the homogeneous functional equation and probably before we do that we will also have a small introduction into the theory of classical modular forms and also our first our big goal would be to just to solve the functional equation with an trivial left hand side and so here we will solve our equation in the following way so we will construct our function f as in the following form it would be an integral from zero to infinity integral of this form and so here where k would be a function meromorphic function on two copies of the upper half plane and so this function it will be a solution of a homogeneous equation with respect to first variable and this will be true for all elements of the ideal i in the group algebra and also it will have the singularities of k will be only at the points where tau and z are psl to z invariant to each other and there will be a special condition on the residue at this point and so probably then it's all for today thank you of which uh you guys it is some six by six man i think it it is given in our paper we have to say what was the explicit form of g we also have to say what is explicit form for sigma because there have been some choices which we haven't made explicit on a blackboard we do it in our paper but they are rather arbitrary i would say so it's just so i'm gonna say that like for for a particular choice of basis in in this in the quotient space so this would be the the row would be just equal to to sigma