 Ok, can we start please? Ok, all right. So, in the previous lecture we discussed recurrent and transient states, we discussed the notion of forbidden sub configurations, then we discussed the burning test, multiplication with identity test and then we showed oh I forgot what else did we discuss equivalence under topoline, equivalence classes under topoline and we discussed number of equivalent classes equal to number of recurrent states, recurrent configurations. Did we discuss anything else? Ok, only this much. Any questions about this? So, yes please. Yes, burning test is failing. Yes, burning test is not applicable in cases where the matrix is not symmetrical, why? You have to go through you know I can give a long answer, but I will give a short one. So, you go through the argument which we justified to explain burning and it depended on identifying forbidden sub configurations and you said that if this thing works, then you know there is a proof for saying this configuration is forbidden and if it is forbidden it will not burn and that depends on the burning rule and that just does not go through as if you can check for the particular example I gave 2 by 2 pi. So, then the argument is not working, it goes through steps and one of the steps fails, then the whole argument fails, but then you can modify it and the new test which is called the multiplication test is still working and so that is just a generalization of the burning test. So, there was a one sort of question which was asked which was that this counting of the eigenvectors under the topling rules or the counting of the vectors using phi was not fully clear. So, let me do that once more. So, what we say is that there are set of operators A i, A i operators commute with each other and they satisfy these equations A i to the power let me write it like this A i 1, A i 2, A i 3, A i 4 where the rule is this is i, this is i 1, this is i 2, this is i 3, this is i 4, these are the four neighbors. So, now I do not know too much about these objects A, A should be thought of as some abstract symbols and they satisfy these rules, they are some kind of operators. So, I can write A, A squared, A 1, A 1 squared, A 1, A 2, A 1, A 1 to the power 7, A 2 to the power 15 and so on. And I can multiply two of such operators A 1 to the power 8, A 2 to the power 3, A 9 to the power 7 whatever and the product is again one such thing. So, I construct all such products and form a group set which is equal to all products of A's ok. But these products I can use these reduction rules to reduce the power so that no of the particular power occurs more than 3. So, then the set of elements in here is finite and these closed under multiplication because if I multiply two of them the power will become more, but I reduce them till it becomes low and so this set is closed under multiplication ok. Then we say that ok. So, what is the smallest set of A's which I can use such that it is still closed ok. And so that number is lower than 4 to the power n and how yes sir how much lower it is we do not know yet ok. So, it is some size, but I do not know the size I want to determine the size of that set. So, can I use these product rules to determine the structure of A matrix which I do not know in great detail just now. So, in principle A was some big matrix I do not know it very well, but I want to determine the I want to diagonalize A. So, what can I do it? So, the answer is that we know that these A's are commuting with each other they can be simultaneously diagonalized. So, let me imagine guess there is an eigenvector phi 1 phi n which is a simultaneous eigenvector of all A i ok. So, then A i acting on this phi 1 phi 2 phi n gives me e to the power i phi i times the same vector I do not know if there is only some one such vector they may be 15 such vectors I do not know what are the values of phi yes now, but let us imagine that there is a vector like this and it has this eigenvalue why did I write the eigenvalue as e to the power i phi i because I know that the eigenvalues have to be modulus 1 ok. So, now, what can I say about this? So, I apply these reduction rules and all the phi i's have to follow this equation still right. So, I have to have. So, this implies e to the power delta i j phi j summation over j equal to 1 for all i otherwise these equations will not hold and this then implies that delta i j phi j summation over j is equal to twice pi m i where m i are some integers another way of just looking at this problem is that suppose there is a simultaneous eigenvector phi then A i acting on phi just gives me a number times phi. So, I will call it a number A i. So, the last time I wrote A i acting on phi vector is equal to A i. So, these A i's are matrices these A i's are just numbers, but applied on a particular vector this matrix becomes a number and these equation becomes an equation in numbers complex numbers. So, the same equations which were first discussed as operator equations I should think of equations involving complex numbers these are polynomial equations involving complex numbers. So, can I solve these equations together and that is the solution. So, these equations thought each think of each A i as a complex number and I want to solve this equation then I write A i equal to e to the power phi i and phi i have to satisfy this equation I take the log of the equation it becomes a little bit easier to handle and the solution of this is phi j is equal to delta inverse i j m j phi summation over j for some m i twice phi phi. So, it gives you some information about what are these vectors, but I do not know everything about them, but then we said that at least we can count how many distinct Eigen vectors you can form and there was some longish argument, but it said number of distinct to the power i phi i set of numbers is determinant of delta, because we just said that there is some all the solutions form some kind of vector in the space of phi's and we looked at the vectors within a block minus pi 2 plus pi to the power d to the power n and this is the number and this is the answer. Then we went back to the original problem and we said oh we have a space of configurations they we make an equivalence class of all configurations and then the operators A i take you from one configuration to another, but for each equivalence class there is at least one recurrent configuration. In the first row you know that is clear, but I did not assume anywhere that the sum of all the rows is 0. Sum over all the row elements for each row is 0 only if the boundary not on the boundary if the site is not on the boundary. So, there is no problem it is what you say is correct, but we have never used the fact that the things add up to 0. So, far so good now I want to determine determinant of delta that is what the first thing we want to do today. So, let us take the case where you have an L by M rectangle and I want to I write the corresponding delta matrix it is l m by l m matrix and I want to find the determinant. So, what we will do a little bit more we will determine all the Eigen vectors and Eigen values of this matrix. So, the solution is done by inspection. So, delta i j psi j m n is equal to lambda m n psi i. So, this is the Eigen value equation where psi i m n is the Eigen vector of the matrix and this is the Eigen value. Formally it looks like del squared psi equal to lambda psi it is the same it is just discrete version of this Laplacian equation and we are solving the discrete Laplacian. And the answer is that I just remember Jackson's book and chapter 1. He actually does not discuss discrete equation, but psi of x y m n. Let us change notation Eigen functions are psi i m n and m n x y is the notation clear m n is some index x y is the argument of the wave function and this is equal to sin m pi x over l plus 1 sin n pi y over l plus 1 there is a root 2 by l root 2 by m and corresponding lambda m n is equal to 4 minus 2 cosine m pi over l plus 1 minus 2 cosine n pi over m plus 1. Bad notation I think we will have to stick with it is this clear by inspection you write down this vector and check that it works on the solution. Because I remembered that the solution of this equation on a square is sin m pi x sin n pi y and I try the same solution with now x and y are discrete numbers integers and it is still working this wave function vanishes at x equal to 0 and y equal to 0 and it vanishes at x equal to l plus 1 and m plus 1 by construction. And so it satisfies the boundary conditions and inside it satisfies the differential equation everything is working. And the number of distinct solutions is so m takes values 1 to l n takes values 1 to m. So, there are l m distinct eigenvectors distinct solutions everything is done. I even wrote down the normalization constant from memory the point being that sin square average is half and 2 by l takes care of the 2 takes care of the half and root l takes care of the fact that you sum over everything you should get 1. So, now my lambda m n is known and the determinant of delta is equal to product over m equal to 1 to l product over n equal to 1 to m lambda m n which is equal to just write it in boring detail because it is 4 minus 2 cosine pi m by l plus 1 2 cosine pi n over m plus 1 same kinds of brackets. So, this is an explicit formula for determinant of delta. Yes sir. No this is for the discrete problem they are the same that is the that was the proof you have to check. I wrote them down by inspection based on some guess work, but you go and check that it actually works. Yes sir yes actually that is a good point when I write this stuff like this I involve cosine of pi by 13 and I do not know cosine of pi by 17 depending on l n m and it is not at all clear that when you multiply all these numbers you will get an integer. So, it is a very interesting problem to see you know one thing is you can say that I will work out because mathematics is consistent the I did not make any mistake the other is to be able to see directly that this answer is actually an integer. So, how do I see that I see that by writing e to the power i pi by l plus 1 equal to omega 1 e to the power i pi over m plus 1 equal to omega 2. Then of course, this cosine is omega 1 to the power m plus omega 1 to the power minus m and that one is omega 1 to the omega 2 to the power some powers of omega are involved omega 1 and omega 2 when you multiply everything then all kinds of powers. So, this will become some product over m n sorry when I multiply out the product it will become something with some powers of omega to the power m i omega 1 to n omega 2 to the power m, but the only property I know about omega is that 1 plus omega plus omega square is equal to minus 1 and if you use this property all the terms in this product will cancel out and you will get only the integer parts left that is a miracle and it is good to be surprised at the miracle and you can sort of look at it further which I will not do here, but so the fact that this result is an integer is a result which follows from this thing called Galois theory where you write down the solutions of polynomial equations. After all there was some polynomial equation which was the characteristic equation of the determinant I wrote down which was an equation with integer coefficients and then you sum up all the roots of that integer equation and the answers are sny symmetric functions of the coefficients of the equation and all of them were integers and this turn out to be integers. The sum of the roots is much simpler function than the roots themselves. So, cosine of m pi over l plus 1 is very bad function for general values of l, but summation over m equal to 1 to l of this is a very nice and simple function which is just minus 1. So, we use this fact and the whole thing simplifies and you get an integer and that is what we get. Then we have some product the first terms are 4, the second terms are on the first cosine and some on the third cosine and some multiplication of these two. And then another term is 2 cosine theta for example and 2 cosine beta for example to the 2. And this multiplication is in fact the partition function of the dimer model. So, let me paraphrase your argument simply it says that in the dimer model the correlation function actually has a very similar form like this. And there we know that the answer is an integer and the same mechanism is working here whatever it is. I do not think I fully followed your argument for the rest no, no, no I actually did, but you know I am saying that even if I did not follow your argument for the rest the fact is that this kind of product form is very familiar in statistical mechanics. When you solve the partition function for the dimer model you get this result. When you solve the partition function for the Ising model you also get some such function form it is a product over 1 minus cosine minus cosine that kind of form. The partition function is this form and the log of the partition function is the sum over such log of such terms. So, we are kind of comfortable we are doing something which looks familiar like other models and so it is nice to be able to see this much yeah very good. So, in fact there is a good deal of symmetric polynomials mathematics and Galois theory which one can go off into to understand this partition function and the symmetries of this partition function and stuff like that and I will not do that. You know that is a very interesting direction, but that is sort of much more pure mathematics than I like to do and I will not get in that direction, but it is an interesting field it is a very interesting activity. So, let me just mention. So, let us just say the result is an integer. Now, what happens if l equal to 2 m equal to 2 then determinant of delta is equal to there are four terms. So, it is 4 minus 2 cosine this is pi by 3. Minus 2 cosine pi by 3 4 minus 2 cosine 2 pi by 3 minus 2 cosine pi by 3 into 4 minus 2 cosine pi by 3 minus cosine 2 pi by 3 into 4 minus 2 cosine 2 pi by 3 minus 2 cosine pi by 3 2 cosine 2 pi by 3, which I can evaluate, because I remember cosine of pi by 3. So, this is 2 into 4 into 4 into 6, which is equal to 192 last time I calculated ok. So, that is the number of recurrent configurations for the 2 by 2 sin pi. The point I am trying to make here is that one should not be scared of these formulas. They look a bit scary, but you put in the numbers they come out nice and simple and you know for 2 by 2 I got 192, for 3 by 3 there is a bigger number I do not fully remember ok, but we actually used mathematical to figure out the answer for determinant of 10 by 10 and it comes out as a big integer. And then what was interesting was to find the factors of this integer, because this is an integer, but what we are concerned with sometimes are the subgroups of the this number and subgroups the order will be the factor of this. So, for example, this was a prime number then the answer would be very small, but it turns out sorry the answer will be that it has no subgroups it has only trivial subgroups, but it turns out that typically these integers this product can be broken into subparts where each part is an integer. So, the whole thing usually has it is a let us say it is a 20 digit integer and it will have 15 factors or yeah something like that large number of factors you do not expect some 20 digit integer to break up into 20 factors. So, how come there are so many factors? I am sorry there is a lot of symmetry in the problem which we have not explicitly discovered yet and it will be nice to figure that out right now we will not do it, but let us go on. So, what so we have determined determinant of delta what is it good for oh well I can write g is equal to delta inverse. So, that is written as summation over m n psi m n 1 by lambda m n psi m n yes sir. No no no that is certainly true, but I am saying if I do not know even that result I only see this product somebody has given me this formula that there is a product like this is this an integer this is the question I am asked then how do I answer I just have to use the properties of this product and. So, I would say that yeah yeah yeah it is a I use this Galois symmetry or something something and then I will be able to infer that this is an integer even though I will not calculate it fully no it does not imply, but it suggests and I am saying that it is interesting to look at the consequences of this observation does not imply ok very good. So, this I can write down I will not write down the full formula, but let me write it in some detail g x y x prime y prime is equal to summation m n 1 root 2 by l root 2 by m root 2 by l root 2 by m sin pi m x over l plus sorry sin m x over l plus 1 sin pi m x prime over l plus 1 sin pi n y over l m plus 1 sin pi n y prime over m plus 1. So, that is an explicit formula for g of course, it involves some signs and some products and I you know if I am very smart I can do this simplify this formula, but for the moment let me not bother ok. However, in the problem we asked last time what was involved was summation of g x y x prime y prime over x y x prime y prime. So, in this formula that is easy to do because you know I just have to sum this one and this one and this one and this one separately and they are easy to do and so, this can be written simply is just summation m n I guess it says odd even terms will vanish odd terms will not vanish and then it will be some simple looking formula 1 by 1 important part was missed into 1 upon 4 minus 2 cosine pi m by l plus 1 minus 2 cosine pi n by m plus 1. So, sometimes the formulas are a little bit trickier, but they are not so hard and you should not have to be scared of them that is the message I mean yeah they look worse, but they are so, let me make a comment professor Ruhani is not here, but he said that you know the same pile corresponds to c equal to minus 2 conformal field theory and the minus 2 field theory was very bad theory because it had negative norm and all kinds of bad things were happening. But I am saying that we are studying all this problem and we have not encountered anything awful everything was very concrete and clearly explicitly done. So, whatever formal difficulties there are with the CFT approach are problems of that approach and not problems of the problem per se the sand pile problem is very clear and well defined the CFT theory for the sand piles is very useful, but it is a complicated theory it involves some mental gymnastics which I am not willing to do just now. So, it is useful to realize you know where is the problem why is the c equal to minus 2 I do not know I do not even know what is c. So, far we have not defined c no no no I think please I am pinpointing the difficulties the difficulty is you know when you have a field theory description you work with continuum fields. The continuum fields are difficult to define the discrete fields are much easier they have all kinds of singularities infrared divergences ultraviolet divergences we have managed to stay away from all those problems and that is why we are doing well. So, far all right. So, now we have done I have to write one more equation. So, usually people like to study l m large and then log of delta is equal to summation over m n log of 4 minus 2 cosine m pi over l plus 1 minus 2 cosine n pi over m plus 1 because the answer was the product and I have just taken the log it becomes a sum that is all I did, but now I can take the large l m limit and this goes to l square l times m into integral d theta d phi 2 pi 2 pi log 4 minus 2 cosine theta minus 2 cosine theta. So, the number of recurrent states increases as exponential of the size of the lattice and this is the coefficient of the proportionality this integral is less than log 4 is actually like 3.2 if you work this out the number is called you know it is just it is a number and it is around 3.2. So, the number of recurrent states goes like exponential 3.2 to the power l m which is much less than 4 to the power l m which was the total number of possible states the number of recurrent states is exponentially smaller than the number of all possible stable states to begin with, but still grows exponentially. So, firstly this thing becomes summation over l, but now if l is large I am doing log sum. So, I call this theta. So, this summation is over some discrete values of theta we change very little form each term. So, I can replace that summation by an integral over d theta and the summation over m can be replaced by an integral over d phi. So, this is this number is log of 3.2. So, this becomes 3.2 to the power l m and it is nice to notice that this number is much smaller than the size of the stable configurations with positive heights. So, this is the number of recurrent states goes like this that is correct this is the log of the determinant ah log of the determinant yeah sure ok very good what else. So, it turns out that this function like this form is very familiar in equilibrium state mac like as I said the Ising model partition function looks like this Nahid told us that the dimer partition function looks like this and there is something called the spherical model where the partition function also looks like this and so on. So, it is good to explore these a little bit. So, one place where the determinant delta appears ok. So, this part we have done determinant of delta is equal to number of spanning trees. So, there is a very famous theorem called the matrix tree theorem and I guess it is kind of attributed to Kerkow. So, it is very classical theorem and that is useful. So, first I should define what is a spanning tree. So, suppose you have a graph let us take some general graph some difficulty confusion can you help us no no no if there is a real problem you ask then there might be other students who are having the same question there is no problem I am happy to answer they must have been you were having very agitated discussion go ahead there is no question it was a private discussion alright. So, suppose I have a graph like this. So, there is an object called the spanning tree which is a it is a set of edges on I draw a tree graph on this. So, now I will draw the spanning tree. So, the red edges if you can see form a spanning tree of the graph what is it doing it is a tree it goes through the you know it is a graph on the it is a tree on the it is a set of edges which are occupied other edges are left empty the set of edges have the property that they go through all the vertices the tree goes through all the vertices and there are no loops. So, that is called a spanning tree now given a spanning tree I guess there is more than one way I can draw a spanning tree because I could have connected this node here by this edge and not by this one and so on there are many spanning trees. So, how many spanning trees there are that is a good question and people have tried to worry about it and the answer is the following. Let us first give the answer the matrix tree theorem is that define the edges in c matrix edges in c matrix of the graph which is for our case it will be called still let us may use call it delta. So, delta will be the diagonal entries will be the coordination number and the off diagonal entries will be minus 1 when there is a link and there will be 0 otherwise I did not write down the whole thing you know it will take me three lines to write but I just spoke it out the edges in c matrix is whose diagonal entries for each site it is the coordination number it can be different at different places the off diagonal entries are minus 1 if the sites are connected by an edge and it is 0 otherwise. So, I write down the matrix this matrix has 0 determinant because as was pointed out sometimes if all the row sums are 0 because the diagonal entries are degree and minus 1 minus 1 how many equal to the degree then row sums of this matrix are all 0. So, it is determinant is 0 that is not going to work. So, I so what I do is number of trees is equal to determinant of delta prime delta prime is obtained from delta by removing one row and column just take any row and column remove it and calculate the determinant of the rest now it is no longer 0 and now it is the same whichever row and column you delete no it is not obvious, but it is a it is a statement you follow the statement the proof of the statement you will have to read up I am trying to point out that this is Kirchhoff's theorem. So, you know it is very old 1850 like that and one should be familiar with it at some level little I do not know mass I just have a graph with some edges no. So, I am let us go back and let us just look at this problem as stated there is a graph there is and that is all I know I do not know any sign I do not know any mass on the graph I want to count how many spanning trees there are that is the question forget the you know of course, it is related to the sand pile problem, but we were forget about the sand pile in the beginning we only look at the graph and counting the spanning trees on the graph. So, let me just give some cultural background. So, there is a notion by Airdish who said that there is you know there are all kinds of theorems in mathematics and there are all kinds of proofs and a given theorem can have a large number of proofs, but some proofs are very beautiful and very nice and other proofs are not so good. So, then he said that there is one book by God where the nicest proofs are written in the book all the other proofs are missing. So, you know he if you give him a theorem and you know give a proof he will say is the proof nice or not nice is it sort of good enough to be in the God's book or not in the God's book ok. So, that was some criterion for judging the beauty of proofs in mathematics and so he used to do this, but later on some set of people whose names I have forgotten actually decided just let me finish my story then I will get back to you. No I think it is a useful story for you to know. So, some people decided to collect as nice proofs in mathematics. So, they consulted lot of other mathematicians in the end they converge to some book of 20 proofs which are very nicest proofs in mathematics whole of mathematics and they collected them in a book which said that the first 20 proofs in the God's book you know God did not intervene it was selected by people, but they selected 20 greatest proofs that was you know it was very appreciated people liked the book then there was a sequel in which they added some 20 more proofs in the mathematics which are really beautiful and one of the extra in the second edition of the book or the second supplement volume 2 sequel of the book one of the proofs was metric string theorem it has lot of generalizations to graphs which are directed undirected multi directed and all kinds of stuff, but you know one of the proofs was by on the metric string theorem one of the 20 greatest proofs in mathematics. So, I am saying that you should read up the original proof somehow it is very nice and maybe you will appreciate the beauty of the proof and if I present it here maybe I will garble it up very badly. So, I will not do it now you I can take your question yes delta prime was obtained from delta by removing one row and column any you pick it gives you the same answer no any one row and one column the remaining matrix is of size n minus 1 cross n minus 1 is the same whichever matrix you pick whichever row and column you pick is that right. So, it is a remarkable theorem how did Kirchhoff came across it the reason I am stressing on all this is because Kirchhoff's laws are familiar to everybody originally Kirchhoff was studying electrical networks and. So, we are discussing let me write relation to the resistor problem to the resistor network. So, what is the resistor network problem there is a set of resistors I am given all the resistances R 1 R 2 R 3 R 4 R 5 R 6 R 7 then I apply some voltage across two nodes and I want to calculate how much current will flow in this network that is the problem this problem is familiar to everybody. So, Kirchhoff gave a solution to this problem he said that you should say that there is a current going in each node and the Kirchhoff's laws tell you that you know the divergence of the current has to be 0 and the current is related to the voltage difference and so on and so forth and you get a set of linear equations and you solve them and that gives you the current. And typically that is where the text books stop about Kirchhoff's laws, but Kirchhoff actually went further he actually told you how much will be the effective resistance between two nodes A and B. So, I want to write down that answer and that you should be able to appreciate. So, he said that it is better to work with conductances than resistances. So, I will work with sigma i equal to 1 upon R i. So, each node is specified by its conductance and I do not want so much trouble I am sort of lazy yes now. So, these will be called sigma 1, sigma 2, sigma 3, sigma 4 these are the conductances of the parts and I pick any two nodes A and B and I want to calculate sigma AB which is the effective conductance between A and B which is the ratio of current to voltage ok. So, if the formula he wrote was this one Spanning I should write a little bit better W t divided by t prime W now it has become very obscure. So, he said they take the network drop all possible spanning trees on the network. So, one spanning tree looks like this this one has a weight sigma 1, sigma 4, sigma 3. So, write down a term sigma 1, sigma 4, sigma 3 then there is another way I can connect them which is like this. So, this one is called sigma 1, sigma 2, sigma 3 and there are two more terms when corresponding to erasing this one when corresponding to erasing that one. Now, you do the same thing, but you put these two A and B you collapse them together just short them. You get a new graph it may have multiple edges, but if it has multiple edges and you want to go from here to here you pick only one of them and draw all possible sums over this graph. So, now these are collapsed. So, I will get sigma 1, sigma 3 plus sigma 2, sigma 3 I guess that is for this one and then you can do others ok. So, the final result is a ratio is a rational function of the sigmas ok. This result looks rather obscured to me just now. So, let us check I remember this graph ok. So, this is sigma 1, this is sigma 2, this is A, this is B, sigma AB is equal to sigma 1, sigma 2 over sigma 1 plus sigma 2 that should be familiar ok. It is called the series combination of resistances. Looks a little bit less familiar because of the sigmas, but you know that is the way it is written. I can also do it for this one sigma AB is equal to sigma 1 plus sigma 2 because when you collapse them then you cannot draw any edges because they are already collapsed. So, you have to write one in the denominator ok. Now, the general proof I am not going to give, I have given the statement of the theorem it is like this. The key point which is important for us is that what you got to do to solve the resistor problem is to study spanning trees on the network. To each tree you give a weight corresponding to the product of edge weights and sum over all of them and that gives you the partition function some stuff which is also sum over all possible configurations we call that partition function of the trees or some such thing and study those properties. So, they have been studied for a long time in the trivial problem of this one, but a more interesting problem is this one. Suppose you have a set of resistors like this on a square lattice each resistor is one ohm and then I just take two nodes which are adjacent put some plus 1 volt on one and minus 1 volt on the other how much current will flow through this network that is the question. That is a home assignment problem number 5 ok, you have to calculate the effective resistance with or effective conductance between two adjacent nodes on the square lattice infinite square lattice with all resistances equal to 1 ohm. It is a problem you should could have done in your BSE you did not do it then. So, you should do it now it is very instructive and useful and you know it will help you in your future life sometime or the other. So, one should do it it is not immediately related to my course actually it is related, but not fully related all you say ok alright. So, we have done Kirchhoff's theorem, but the connection to my problem is not yet clear except through this indirect way that is the same determinant of delta was showing up ok, I should make one point. So, how do we connect it firstly to my problem. So, I I had this graph and I formed the delta matrix for my senpai. What you have to do is you have to imagine that there is a sink side there is a boundary which is called sink and you connect the stuff to sink sides like that. Now, you consider the. So, the sink is only one they are all shorted. So, there is only one side on the outside which is the sink side. Now, you write the augmented big matrix and calculate the spanning trees on this bigger graph, but then when I have the choice of removing one row and column I will remove the sink. Then I will get back my old delta and the determinant of delta will be the number of spanning trees on this bigger graph with sink side added. So, a trivial case one by one lattice I had this I had four recurrent states, but if I make the sink side then there is a you know the sink side is this one this the whole thing is one side and there are four spanning trees you can connect like this like that like that and. So, the number is four everything is working check a small check you know it is good to make this elementary checks. Now, we want to do something else which is called. So, it turns out that the spanning tree problem is also known to be related to another problem which is well established standard problem in statistical mechanics which is called the Potts model. So, I want to relate this problem to the Q state Potts model in the limit Q goes to 0 whatever just one minute yeah the matrix tree theorem and register networks and now we want to do something called number 4 which is connection the spanning tree problem Q state Potts model in the limit. So, this is sort of mouth full of words, but anyway we are going to define what is the Q state Potts model and then we will take the Q goes to 0 limit and then show that it is the same as the spanning trees problem and by inference it is the same as the sand pile problem that we are studying. So, we define we start a graph we define a graph like that some general graph. So, the vertices are called i j and if there is a link between the vertices that will be called v i j it has a strength there is a real number v i j attached to each link i j. Now, what I imagine is that there is a spin at each site and it is coupled to other spins by these links question. So, I will write a Hamiltonian for this. So, the spin takes values 1 2 3 there are discrete states 1 to Q and the spin takes these Q distinct values. And the Hamiltonian of the system is for the moment we will write it like this minus j summation delta sigma i sigma j minus j i j. So, between any two links there is a this delta function is 1 if it is sigma i equal to sigma j it is 0 otherwise. So, the two spins have the same state then there is an energy minus j minus j i j that is the contribution to energy otherwise it is 0. You can take all j i j equal to j let me do that for the moment for the moment we will set all j i j equal to j and later on we will say o, but if the bonds are different then everything goes through and we will work like that. For simplicity of notation I am setting all j i j equal if there is a bond if there is no bond then it is 0. So, we will put this away put this outside and this becomes a sum over edges which are the non trivial edges of the graph. So, now I want to write you know this is very standard partition function is equal to e to the power minus beta h sum over configurations of spins and. So, if you have n sides there are q to the power n distinct states and we sum over all of them and. So, this partition function can be written as summation over configurations product over edges 1 plus v delta sigma i sigma j where v is equal to e to the power beta j minus 1. So, what happens first I sum over configurations, but for each configuration there is a weight what is the weight if the bonds are distinct if the sides at the two ends of the bonds are distinct the weight is 1. If the sides at the end of the bond are in the same state the weight is a little bit more it is 1 plus v instead of 1. So, I write it like this this delta function is 0 then it is 1 otherwise it is this v is equal to e to the power beta minus 1 for us v is a positive number just now. So, now what I can do is that I can expand this product and I have a graphical representation of the sum. So, I will write that a particular term may be present or not present depending on if delta is kept or not kept. So, a graph may look like this you know this was my original graph, but so the first term I have is that I do not keep anything. So, it will be 1 plus then there will be 1 edge kept like this and nothing else is kept plus a graph like this plus for each is this point clear this is very standard, but I am trying to ensure that you people know this one. So, in the graphical expansion you just expand this out and each term is represented by a graph where an edge may be present or not present and you sum over all possible such graphs and the graphs are how many such graphs are there total number of such distinct graphs 2 to the power number of edges right is that clear to everybody the number of graphs is 2 to the power number of edges I take this graph then there will be 8 16 possible graphs depending on each edge is kept or not kept. So, I look at any one of these terms any one of these 2 to the power n terms. So, that looks like this the set of edges together will form disjoint clusters yes. So, this e to the power beta h is equal to 1 product over edges e to the power beta j sigma i delta sigma i sigma j because the Hamiltonian was a sum over terms I can write it as a product over e to the power beta j i j this product over i j and then this thing can be written as is equal to product over i j 1 plus v delta sigma i sigma j because either sigma j is 0 or 1 and then I adjust this value of v so that it gives the correct answer for both choices. What about graph very good so what I am doing is I am expanding this product. So, it is a product over edges how many terms are there if the number of edges is e then there are e terms. So, I expand this out each is 2 terms there will be 2 to the power e different terms in the expansion that is what I am saying that you look at one of those terms that will involve either keeping this term or this term. So, it is 1 plus v 1 2 delta let me write it like this 1 plus v 1 3 delta 1 3 1 plus v 5 7 delta 5 4 7 so on. So, I keep this term here and this term here and this term here I make a choice either I keep this one or this one. So, that will be represented by an edge present or not present if an edge is present I am keeping this term if an edge is not present I am keeping that term that is the room then the weight of a graph with given edges is just the product of weight of each edge which we will call v i j times delta function. So, now I have this graph I have the sum over 2 to the power e terms, but now I can do the summation of yes there is a summation over configurations of spins that summation is now easy to do why because within a cluster it says that spin here should be equal to spin here equal to spin here. So, there is only one choice for each, but total q choices for the whatever value it has. So, the summation over spins is equal to summation over. So, this is equal to summation over 3 graphs no edge graphs q to the power number of clusters product i j over edges. So, that is the that is the sort of basic expansion when I expand this out and sum over spin states the number of states allowed with these delta functions will be q to the power number of clusters because each cluster can be in a distinct q possible states. And if there are two sides which are one side is not coupled to anything that is called a cluster of size 1. So, very good we have got this far and the weight attached to this is edges which is q it is very good now I have. So, this thing is a z of q v it is a polynomial in q and v. It is a polynomial in q and v the partition function of the pots model is a polynomial in q and v for any given graph, but now comes this great insight of Kastelian not pots Kastelian said that oh once I have this partition function it is a polynomial in q then it can be defined for arbitrary values of q where q need not be an integer. So, you can define a pots model for q equal to 0.5 and it is a well defined model because this partition function is a polynomial in q put q equal to 0.5 it gives a nice value there is no problem. So, pots model can be defined for arbitrary values of arbitrary real positive values of q I guess you could extend it to negative values of q or complex values of q let us not do it yes integer positive integer. Now, we have extended the definition to all. So, we will forget all the derivation we will start with this definition the partition function is defined by this formula. So, what you can say is that suppose you have a system you can construct different clusters on it by occupying bonds at random to each cluster you stretch a weight which is a Boltzmann weight of the cluster and then this is a partition function over those different possible clusters. So, it is a partition function of clusters of various types I do not want to think of it as q distinct states of spins which was the starting point, but anyway the key point about this is that of course, when you put q equal to 2 this partition function reduces to the ising model. When you put q equal to 1 yeah percolation q equal to 1 I think that is very strange if there is only one state what is there to sum over nothing to do no, but all the states every spin is only in one state then what they know sum to do I am very comfortable. If there is one state per site then what is the partition function there is no partition function it is just one state per site that whole system has only one state. So, the question is that in what sense is q equal to 1 the percolation problem because the q equal to 1 seems to be a trivial problem while the percolation problem is perhaps a non trivial problem. So, what is the connection very good. So, that is the point the point is q tending to 1 is not as trivial a problem as q equal to 1 and. So, we look at the q tending to 1 and that is the percolation problem q equal to 1 by definition is sort of trivial we do not want to discuss, but q tending to 1 may not be so trivial very good. So, we want to do q tending to 0. So, now, we have defined pot's model for general q. So, what happens in the pot's model is that you get graphs which may have. So, this z q v is a partition function it starts with the term which is 1 which is nothing, but then q to the power n sorry may be I start with 1 I expand the partition function in powers of q. So, there is a q to the power 0 plus q to the power 1 plus q to the power 2 no there is no q to the power 0 there is only a q to the power 1 the minimum number of clusters can be 1. So, you sum over all configurations with only one cluster then sum over all clusters with 2 clusters so on. So, that is these terms I look at the coefficient of q to the power 1 that is called z 1 and it is only a function of v now and q squared z 2 which will be a polynomial in v still a complicated polynomial in v and so on. So, that is part of my definition there is a there is at least one cluster in the whole system. So, it is q to the power number of clusters so the first term is q to the power 1 next term is q to the power 2 and the maximum term is q to the power n no because I that was the last term because it was n clusters when there is a only isolated vertices with no bonds the weight is q to the power n and power of v is 0. But, for me I am expanding in powers of q so it is the last term it is there, but it is the last term not the first term. Now, comes a interesting idea that you take the z q v and usually we can do the following we can take log z q v divided by volume of system limit volume goes to infinity and that gives me minus f q. So, you can work it out or you can show that if you take this kind of a problem work with systems of bigger and bigger sizes. Then the partition function as defined will grow exponentially with the volume of the system is a very general thermodynamic proof of this result which is given in rule, but we are not going there I am just sort of giving a plausibility argument that this z will increase this v is the edge ratio. So, let us call it small v and this partition function for something which has n nodes will grow exponentially with n for well behaved system. Like I take a lattice model like this and n is the number of nodes in the system then I make a bigger lattice bigger lattice calculate z for each of them then the log z will increase linearly with n. We check that you know it one of the terms was q to the power n so it perhaps increases like that. So, this is equal so we define this f q v f q v now is a function of q and v and now I want to study the case where q goes to 0 because q can be continuously varied I can make it very small I can make it 0.0001 and look at what is the partition function at this value. So, what happens to the partition function when I put q very small. So, there was this term and there was this term, but the second term contribute much less than the first term. So, when q goes to 0 the only terms which contribute to the partition functions are the terms where there is only one cluster. So, everything should be connected and then, but it may have you know it may have this kind of structure. So, it has different powers of v in it, but since everything is connected everything they should be at least v n minus 1 edges there. So, it will be v to the power n minus 1 plus v to the power n plus v to the power n plus 1 all that kind of mesh and. So, I do not want to look it yet even that and. So, we will take we will state now obvious result limit as q tends to 0 v tends to 0 z n q divided by v is equal to partition function spanning trees on the lattice. So, this has been nice, but let us sorry this is the ok. So, I divide by v because I am letting sorry I divide by q because I am letting q tend to 0 and this partition function itself will go to 0 as q goes to 0, but the lowest term is order q. So, when I divide by q only the term where there is everything forms one cluster revive and then I let v go to 0 then the lowest term will be v to the power n minus 1 T n v to the power n minus 1 gives you that they should be at least n minus 1 links, but if there are more links then it has small weight because v goes to 0 and T n is the number of spanning trees on the graph T n is an integer. So, so if I want to calculate T n that was my quantity of interest because that was equal to my number of recurrent configurations I should calculate partition function for parts model for q let q go to 0 and take some limit and that is the answer. So, let us go back z q v is equal to summation over clusters q to the power number of clusters v to the power it is actually number of edges. Sometimes it is written like this it is a nice notation number of clusters summation over configurations q to the power c v to the power e and summation is over what is that e is a set of edges which are occupied and there are different possible choices of the subset of a occupied edges there are 2 to the power number of total number of edges choices. So, the summation is over all possible sets e for the graph v to the power e is the product of the weights of all the edges. So, if the edge weights are different v 1 v 2 v 5 you just multiply those edge weights and c mod c is the number of clusters it is just a notation it is the size of some set and c is the set of clusters and the size of the set is number of clusters and q to the power number of clusters v to the power e where e is a set and v to the power e by our notation means the product of the weights of all the edges in the set and then you sum over all the edges. Now, you are saying that they should be a term like 1, but know that lowest power of q is 1 in this not 0. So, the term starts with q to the power 1 if you expand in powers of v you can have a term which is v to the power 0 empty set, but empty set here corresponds to the power n it is not 1 it is q to the power n it is n distinct clusters. So, the weight is q to the power that is my definition it is the number of clusters the number of clusters is n then if there are no edges present then there each side is a different cluster the number of clusters is n and the weight is q to the power n yes sir. Why do we take the limit because I was interested in the spanning trees problem I did not want to work with graphs like this where there are no loops also present. So, in order to get rid of the loops present I get v tending to 0 limit then the loops drop out no because q to the power n is actually very high order in q and I do not worry about it. But q to the power n depends on q of course q q to the power n when I let q tend to 0 the edges are this number of terms is 2 to the power n, but the term without v is q to the power n not 1 is that clear to everybody yes. So, very good let us look at this problem just this reduced problem here. So, I take a graph like this you know I do not like to study very big problem there is just a graph with 4 edges what is the partition function for this graph. So, I can check that there are 16 terms in this. So, this will be z of this. So, there is one term like this where nothing is present that is called q to the power 4 plus then there is a term where there is one edge present like that some edge like this. So, let us write it first in graphs then we will write it in weights there is some term like this plus some term like this plus some term like this plus some term like this plus some term like this plus some term like this plus that clear I have not missed anything. So, this one is called q to the power 4 this one is called q cubed v and there are 4 of them this term is called q squared v squared and there are how many of them are there how many terms of this type I can write down 4 very good and then this one 2 q squared v squared times 2 plus 4 q v cubed plus q 4 that is it sorry q 4 we already wrote it is q v 4 are the chair now ok bottle also ok now ok ok. So, my time is kind of up and it is not fully up I have 10 minutes. We take a limit that q goes to 0 and then we come up to a situation that there is only a cluster but it is not a cluster we cannot distinguish between that cluster and full of the models. So, if you the way I have described this problem everything is working for finite graphs and it is well defined and well behaved what you are describing will only work in the thermodynamics large size limit or some such thing and I do not want to invoke that now I am describing finite systems you know 4 by 4 sand pile or some such thing I do not want to take the thermodynamic limit now most of the formulas or all the formulas are working for finite sizes and I do not take the thermodynamic limit first thermodynamic limit should be taken last no. So, I am saying whatever question you are asking answer it for this graph simple graph then what is the statement you will make not much no no no if the question answer disappears because the question disappears for finite graphs their questions you can ask. So, actually that comment is not trivial let me state it now suppose you have the sand pile model you can define it on finite graphs we that is what we have done and then you take the limit of large sizes. But people can do it the other way they say let us define an infinite lattice and then there is a sand pile there is a height at each site and then can I define a steady state of the pile and it becomes a very tricky problem and it is actually non trivial non easy not infinite volume limit does it exist I can go into some head spin because because sometimes the avalanche is last forever or some such thing and all those problems are not there if you work with finite systems. So, I want to get rid of all those problems and not even ask the question even the question of what happens in the infinite limit should be asked only after you have worked out the finite size cases not first yes. So, if you put the z q v and put q equal to 2 it gives you the exact partition function for all finite graphs and in the thermodynamic limit just put q equal to 2 the v is called the high temperature expansion parameter 10 hyperbolic beta j and so this gives you the partition function of the ising model exactly if you put q equal to 2 it gives you the exact partition function of the ising model ising type means what that is what we did no we started with q states per site and we gave an ising like interpretation to begin with, but then I when I have q equal to 0.1 it is not possible to define the model in terms of states per site then I have to define it through this summation this one it is a well defined summation it gives you configurations and weights over configurations, but the weights are non-local weights and they are not defined in terms of spins, but in terms of clusters no. So, I actually this one is of course, no edges this one is with one edge, but I wrote it like this, but the bracket shows all graphs of this type which says that 4 you add up all the 4 terms. So, I wrote a weight 4 here they are distinguishable, but I have been lazy and I wrote only one the brackets are supposed to indicate whatever else I did not I was just lazy I should have written 4 terms I did not write sorry say it again yeah you can put. So, the instead of v you can write v 1 plus v 2 plus v 3 plus v 4 this 4 v will become v 1 plus v 2 plus v 3 plus v 4 if I have different edges yes. Oh, but those are terms in the Hamiltonian I did not have an edge between them are there any other questions no. So, this is very nice I believe I hope that you are actually following whatever is being done it is being done at a somewhat slow pace, but I felt that it is better to ensure that you understand what is being done and appreciate that is this moderately straight forward and not get the idea that there is some problem it can be solved, but you know I will read it up next time I do not have to understand it now I am sort of hoping that you will actually just read them once read the notes or whatever once more you will understand everything it is not. So, big deal and so, then even if we do not cover everything it is quite I am not too unhappy some details are left for the people who enjoy the subject more right. So, reading up original Kirchhoff's papers is a very nice literary exercise you know you may actually enjoy people read novels in addition to reading physics right. So, read Kirchhoff's paper as a novel not as a sort of physics exercise it is very educative ok. So, we stop here now.