So, we can start. So, of course, we let us start with asking is the let me recapitulate the previous lecture. We started by you know defining these operators a i which are addition operators. So, we said you added site i and relax then we showed that these operators commute with each other. And then before that there was a matrix delta i j which specifies how the toppling occurs you know under toppling the under toppling at site i delta i i is the number of particles lost and minus delta i j is the number of particles gained at site j. So, this matrix specifies the toppling actually. And then we showed that a i product for all i a i to the power delta i j is identity because if you add delta i i particles at site i that will correspond to one particle added at all the neighbors in the simplest setting. And then using this group this operator structure we showed that the set of all the a i's form a group which looks like a torus in the dimensions. And then we said that the operators a i can be simultaneously diagonalized. And so you get there are eigenvectors which will be slightly change notation. Now there is eigenvectors which can be labeled by phi 1 phi 2 phi n such that a i phi 1 phi 2 phi n is equal to e to the power i phi i phi 1 phi 2 phi n. So, these are simultaneous eigenvectors of all the a i. And right now we do not know what is the value of phi i. So, phi i satisfy these equations twice pi delta i j phi j is equal to twice pi m i for all i. And so you can choose any set of integers m i and invert this set of equations. And you get once allowed set of phi i and then that set of phi i will give you a solution to the eigenvalue problem. So, what are these vectors? See the eigenvalues we have determined, but the vectors what are they? These are the eigenvalues of the translator operator on the torus because a i's are translation operators on the torus. These are eigenvectors of the translation operator on a torus. So, they are easily determined they will be exponential i k y i where y are the displacements along the torus. Is this point clear? Because I am sort of writing saying something without writing anything on the board. And I am hoping that that is good enough yes sir very good. So, now the point is this. So, we said that we start with all possible states which are 4 to the power n, but some subset of these states are these recurrent states. And then most of the discussion we will restrict ourselves to the recurrent states because the other states do not come into picture they do not matter. So, we do not worry about them. The group structure is only valid for the recurrent states. When you restrict yourself to the recurrent states then each a i has an inverse. If you do not restrict yourself to recurrent states then you can have it such that this you know there is some state here this is a i is some other state this is a i and this is some other state you know we drew it like this. So, at this point there is no a i inverse because two things are coming in and only one is going out. So, if I know this one I do not know which state I should apply so that I get a i there is no unique inverse, but we want a unique inverse. So, we get rid of this state and then in the restricted set of states on the ring there is a unique inverse any other question yes sir is the orbit of the operator a i. So, in our argument we are using only a 1 and orbits of a 1, but yeah you can choose some other operator because these are under a 1, but you use some other operator a 2. A 2 can take you a state from one state to some other state. So, I think it is nice to work this out in a rather trivial problem. See this structure appears to be rather abstract and complicated and I am trying to show you that there is nothing very abstract about it. So, let us take these very trivial sand pile there are only two sides 1, 2 and the rule we said yesterday was that four particles if they are there then they topple three of them leave and one of them goes to the right. So, this one I can do sort of like this start with a state which is 0 0 0 0 is the height at this side height 1 height 2 0 0 I apply a 1 what will I get 1 0 then I apply a 1 again 2 0 then I apply a 1 again 3 0 1 apply a 1 again 1 1 2 1 3 1 and then 0 2 1 2 and then 0 3 and then 1 3 and then 2 3 right and then 3 3 and then 1 0. So, you get a loop that is the operation of a 1 what is the operation a 2 doing on this. So, we kind of realize that the 0 0 state is transient it will not show up in the recurrent structure, but anyway under operator a 2 what happens to this one goes to 0 1 0 1 is here I should use a different method. So, what is the operation of a 1 what is the operation of a 2 doing on this. So, we kind of realize that the 0 0 state is transient it will not show up in the recurrent structure, but anyway under operator a 2 what happens to this one goes to 0 1 0 1 is here I should use a different color and 1 1 will go to 0 0 0 1 will go to 0 2 and 0 2 will go to 0 3 and 0 3 will go to 1 0. So, it will come here. So, it and this will go to 1 1 it traverses the swim orbit, but now 4 steps at a time, but since 4 is coprime to 15 it will actually travel the whole loop again and you will get a period orbit of period 15 again. So, in this case we get a 2 is equal to a 1 to the power 4 applying a 2 once is the same as applying a 1 4 times, because you know I jump 4 steps when I go to a 2 and vice versa a 1 is equal to a 2 to the power 4. So, it is straight forward and the other stuff is just a trivial extension of this result is not more complicated than this. So, now what we want to find is the number of this different solutions of phi I can construct number solutions phi I the set phi I which is phi 1 phi 2 phi together ok. So, we said that you can take these m I and give them any values and each integer will give rise to a solution, because phi I is equal to twice pi delta inverse times m and this is i j m j ok. So, if I look at the values of phi I in n dimensional space there will be some points it is very hard for me to draw n dimensional space, but schematically it looks like this I drew something. So, it forms a periodic lattice the set of allowed values of phi I will form a periodic lattice in the n dimensional phi space, because it is a sum of basis vectors you know m times I am is equal to summation m j e j in that m dimensional space. So, there are some basis vectors and it forms a big lattice of these if you in the phi dimension phi space you take a space and then add a unit vector you get another allowed point another allowed point and so on you can do it and all that ok. So, what is the what are the unit vectors they are just these ok. So, what is the volume of the unit cell in this phi space? So, so we say that values of phi form a lattice in n dimensions ok. So, I guess this result is I assume is known to everybody and I know that it is not, but once I say it maybe you will accept the fact that it is actually true. So, if I take three vectors v 1 v 2 v 3 in three space which are tilted in some way they are linearly independent, but they are some vectors what is the volume of the parallel pipette formed with these vectors? How many of you know the answer? It is some standard stuff taught in three dimensional geometry one only one no no no please the question is given a parallel pipette whose I sit at one vertex I look at the vectors in that they are these three vectors v 1 v 2 v 3 what is the volume of the parallel pipette in terms of these three vectors louder ok. So, it is called the box product all of you have seen it perhaps. So, the volume is equal to determinant of v 1 1 v 1 2 v 1 3 v 2 1 v 2 3 v 3 1 v 3 2 v 3 3 plus minus because the volume has to be for the moment volume is positive the determinant may be positive or negative depending on how you calculate it. So, the volume with the will be the modulus of the determinant is now this result seems ok it is immediately generalized to n dimensions. If you have an n dimensional parallel pipette all you got to do is to take v 1 the vectors v 1 1 v 1 2 v 1 3 sorry this is vector 1 this is vector 2 this is vector 3 you write vector v n form a determinant and that gives you the volume of the parallel pipette in n dimensions. As a trivial check if I if one of the vectors is a linear combination of the others then the parallel pipette volume becomes 0 also if I take up you know suppose I have these two vectors v 1 v 2 it forms this parallel pipette, but I take a different vector v 1 prime is equal to v 1 plus v 2 and form a parallel pipette with v 1 prime and v 2 parallel pipette form parallel pipette form with v 1 prime v 2 ok. The volume of the new parallel pipette is the same as the old one right there is a geometrical construction because v 1 prime is this the parallel pipette with v 1 prime will be this one, but it is equal to this one ok. So, very good so the volume of the unit cell in the phi space is the is equal to so the it is equal to this stuff. So, it is to pi to the power d determinant of delta inverse because these are the unit vectors no unit vectors I delta inverse I yes sir how do we know that i i's are linearly independent if they are not then the volume will be 0 I guess you follows no no no it follows that the volume will be 0 then my answer will fail and I suppose there will be something wrong with the original problem. What happens with the original problem is that in this case there are some sites or there is some some problem say that think keep on toppling forever and there is no steady state ok. So, in cases where you are guaranteed that from every side there is some probability that particle can leave then you can show that the I n there is no 0 I n value possible of delta I will not give that proof here ok. So, this is the volume of the unit cell. So, but we said that if you have a value of phi i then phi i plus 2 pi gives you the same physical state because we are actually looking at Eigen values of e to the power i phi i. So, the number of distinct Eigen values of phi is the number of points. So, number of distinct vectors Eigen vectors is equal to volume. So, it is the number of points in the cube minus pi to plus pi to the power d. So, I take I have with my phi dimension in this phi space and you know there are all these points inside, but there is a unit cell minus pi to plus pi in d dimensions I cannot draw it very well ok. But all the points outside these are equivalent to points inside. So, I only need to know what the volume of points inside the unit this pi cubed 2 pi cubed minus pi to plus pi to the power d ok. So, what is that? That is the volume of the cube divided by volume of the unit cell is equal to volume of cube divided by volume of unit cell ok. So, luckily the 2 pi cancel of course, they should cancel and then volume of the unit cell is determinant of delta inverse and so this is determinant of delta which is nice because delta was an integer matrix and the answer is some number which is a clearly an integer. All the 2 pi's which came in the middle have cancelled. Yes sir, number of distinct solutions of phi i. So, we counted how many points fall within minus pi plus pi to the power d. We said these are equal to 2 pi to the power d divided by volume of the unit cell volume of the unit cell was this and so this number is determinant of delta ok. So, there are that many distinct points. Now, what happens is that you can actually prove all these results from the definition that is what we are doing most of the time, but proving everything in gory detail is very tedious even for this simple problem. So, some of the results I will state and you will prove them yourself because they are easy to prove, but I will not prove everything here. So, if you force me I will sit down with you and I will give you the proof for every statement I make ok. So, here it turns out that this number of distinct eigenvectors is actually equal to then the dimension of the space I working which is the number of recurrent states we will prove this one actually. So, this number of distinct eigenvectors is we will show this number of eigenvectors distinct eigenvectors distinct right now we have shown distinct eigenvalues is equal to determinant of delta equal to the number of recurrent states ok. So, now the dimension of states space I am working in is the number of recurrent states because all the other states I throw away the number of recurrent states is determinant delta number of eigenvalues I have found is determinant delta. So, now, I have counted everything there is nothing missing and everything is in good shape I only have to prove this part that the number of recurrent states is equal to determinant of delta we have only shown the number of eigenvectors of something or the other is determinant of delta ok. So, very good. So, that is what I want to do next yes question ok recurrent and transient configurations. So, I start with my basic sand pile again and we construct some configuration with various heights now 1, 2, 0, 3 whatever. I give you a configuration and I ask you is this recurrent or is this transient can we get that configuration and tell whether it is recurrent or transient what is the test to distinguish between recurrent and transient configurations and. So, now, I give you a test and there is a. So, in a recurrent state cannot occur proof suppose you start with the configuration of the pile in which there are two adjacent sites and both of them are not 0 now then can I keep add anything anywhere and do it such that both of them become 0 you are allowed to put any other starting configuration anywhere and you come to this and produce a state in which both are 0 the answer is it is not possible why because whatever you do from outside you can only add particles into this and then you can topple particles when you topple particles this suppose you want to produce a 0 here then there should be a 4 here if there is an 8 you reduce it to 4 and then 4 has to be reduced to 0, but if 4 has to be reduced to 0 then it will topple 1 and we make 1 here. So, if you had made 0 here then this would have become 1 if now I want to make this 0, but if I try make to make this 0 I will have to add something here, but that will make this 1 1 try as you might you cannot produce two adjacent 0s if they are not there in the beginning ok. So, that is a provides a simple test you look at the configuration if somewhere you find two adjacent 0s that is a transient configuration throw it out. So, very good, but now I can yes, but now I can sort of add to this and cannot occur it is the same as before if I want to produce a 1 0 here if I want to produce a 1 0 at both ends both of them will throw in 1 to the middle. So, we will make it a 2 if you want to change 2 to 1 then you will have to make it into 5 and then throw out 4, but then they will produce something at the end. So, if you want to make this thing into a 0 then you will throw something here now if this has to become a 1 then it should have been 0 before, but 2 0s are not allowed in a recurrent configuration. So, 0 1 0 is also not allowed. So, this cannot occur and this cannot occur and you know you use the same method to prove that this cannot occur and 0 0 0 2 cannot occur because each will throw a 1. So, it will be at least 3 ok alright. So, but so these things are called forbidden sub configurations forbidden sub configurations they are not allowed to occur in the recurrent state. And there are sub configurations because these are just finite set of sides and they cannot occur anywhere on the big lattice whatever size you have. So, now can we have a list of these forbidden sub configurations. So, I produced you know I gave you this, but I suspect you can construct some more. So, you have a list of forbidden sub configurations how big is the list. So, we will try to keep the list minimal by which is meant that if you have this one this is forbidden we have shown then of course this one is also forbidden it will never appear. We do not write that one because you know it is taken care of by some other stuff smaller. If you can if there is a configuration which is forbidden, but you can erase one site and still keep it forbidden then we erase that site and produce a minimal list. Minimal forbidden sub configuration is one if you erase anything then it will become allowed is this point clear very good. So, what is the list of minimal forbidden sub configurations. So, sorry boundary should not be 0 yeah that is a you can construct once which are boundary is not 0 forbidden proofs are for each one of these proofs can be constructed the proofs can be constructed recursively start with a smaller one this is forbidden in order to produce this one. If I add something here then it and it produces a forbidden one then before that should have been a problem or some such thing, but anyway let us try to produce a systematic way to check for forbidden configurations. So, that method yeah. So, the burning test is the following you take the configuration you like you are asking about and then look at all the sites and burn any site whose greater than or equal to number of burn neighbors. Burn means erase. So, I will start with this 3 2 1 0 this site has 2 right now all all sites are unburned, but this site has height 3, but it has 2 unburned neighbors this site has height 3, but 2 unburned neighbors. So, it is burnt by which is meant that I just erase the site. Now, I look at this site it has height 2, but it has 2 unburned neighbors. So, this is also burnt this one has 2 unburned neighbors, but height is 1 sorry I cannot burn this one, but I will come here now this has height 3, but 2 unburned neighbors. So, it can be burnt and this can be burnt, but this cannot be burnt what about this one I can burn this one, because you know 3, but it has at most 3 and then this can be burnt and so on. So, you recursively apply this method to the full lattice see if at some stage everything gets burnt then it the configuration is recurrent if at some stage something is not burnt then it is forbidden or then the configuration is not recurrent. So, as a check suppose somewhere or the other there was 0 0 none of these 2 sites will get burnt, because you know here it will say the number of unburned neighbors is 1, but the height is 0 you cannot burn this one the other one you cannot burn they will strip put this one also we you check that if there is this configuration anywhere under this burning test it will not burn they will just stay this one cannot burn, because it has height 0, but it has 1 neighbor it has height 1, but it has 2 unburned neighbors. So, since all these sites in the set are unburned initially none of them will burn. So, the general definition of general definition forbidden sub configurations is that it is a subset of sites such that for each site is strictly less than number of neighbors in. So, these are these are subset f which is the forbidden subset you look for if all the neighbors in f are number of neighbors in f any the height is less than the number of neighbors in f then that particular set is a forbidden set again we can check you know that definition says this one is forbidden this one is forbidden this one is forbidden and so, it is the automatic generalization. So, the initial state is forbidden, because it is not recurrent it is forbidden to occur in the recurrent steady state it was there in the original model, but in the recurrent state it will not occur it is forbidden only in this limited sense yes sir very good. So, that is a good point yeah so, you can check or you can true verify that that is true you can do this operation in any order and you get the same result. So, you scan through the lattice you know I have this lattice I check once whichever I can burn I burn and then I do it again and I do it again if at some stage I find I am not making any progress I did not burn anything extra in the new scan then I stop and say that for in nothing is happening now there is something remaining if everything burns then it is a recurrent state it everything does not burn it is a transient state it does not matter which order you do it how you do it ok yes sir no no no it is a proof and I am giving you the I am giving you the statement and I am telling you that the proof is trivial and you should construct it yourself of course I give you the statement and you produce the proof how one will come with this idea is a different question ok I do not know I think this is how I came up with the idea I started with this observation that 0 0 cannot occur then I found that yeah 0 1 0 cannot occur then the question was what is the list of all possible forbidden configurations and so there is a list is infinite in principle for a big lattice and so then there was a question of characterizing the list and this is not a very profound observation but this strategy works and so that is done ok very good. So, this burning test the way I have defined is applicable for matrices where the toppling rules are symmetric in the sense the delta matrix is a symmetrical matrix if when the toppling occurs that this side it transfers one particle here when toppling occurs at this side it transfers one particle here if the matrix is unsymmetrical then you have to modify the algorithm in some suitable way. So, you have to work with directed graphs instead of undirected graphs and you have to distinguish between the in degree and out degree of a site you know how many sites are there with arrows coming into this site which are of some type or some such thing and then the whole proof goes through there is no particular difficulty but just one has to realize that what we are calling number of unburnt neighbors will be modified in a suitable way if you have unsymmetrical bonds in the sense some bonds have direction you can transfer one way but not the other way and so on. So, that part I will not do here what we will do instead is to redo this test in a different way. So, this is multiplication by identity test. So, I still have my lattice and we have these equations a i to the power 4 is equal to a i 1, a i 2, a i 3, a i 4 where the notation is i, i 1, i 2, i 3, i 4, i 1, i 2, i 3, i 4 are the four neighbors of i and there are n such equations. So, what I do here I multiply all these n equations and I get product over a i to the power 4 is equal to whatever is on the right. So, what now suppose I work with the recurrent set then a inverse is defined and if there are powers of a which occur on both sides I can cancel them. So, what will I have left multiply everything every side here will have power 4 on the left will have power 4 on the right. So, it will not show up it will cancel out side set the boundary will not cancel fully. So, you get so this equation becomes product over a corners squared over corners product over boundary a boundary equal to i. So, what that says is this clear this is just algebra yes I will repeat. So, let us take a 3 by 3 pi 1, 2, 3, 4, 5, 6, 7, 8, 9 and I want to multiply these equations for all these 9 equations and write down the answer. So, I will write down the answer a 1 to the power 4, a 2 to the power 4, a 9 to the power 4 equal to what is the right hand side for this one it will be a 2, a 4 what is the right hand side for the next one this is 1, 3, 5, a 1, a 3, a 5 write down all of those terms and then you will see that a 5 will occur 4 times in the right and 4 times in the left it can be cancelled. But a 1 will occur 4 times on the left, but it will occur twice on the right because once it will occur when you write this term and once it will occur when you write this term. So, it will 2 powers will be left. So, the answer is that the final tally of a's will be 2, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 2, 1, 1 and 0 everywhere else is equal to 0 is equal to identity. Yes, precisely this is only for the recurrent states this will work. So, for the recurrent states it says that this is an identity because all these equations are true this is an identity. So, it says that if you take any recurrent state and you add this stuff to it add 2 particles here 1 year, 1 year, 1 year, 2 year, 1 year, 1 year nothing in between and then topple everything and see what happens you get back the original state. What is the proof? Well if you add this stuff then something what the other must topple otherwise the whole thing is a forbidden configuration. If it topples you can check that under the relaxation process every site will topple exactly once when this site topples exactly once this topples, once this topples, once this topples, once it gets 2 particles and 4 particles leave. And so these extra 2 particles will leave and here this extra 1 particle will leave and whatever is the starting configuration you get back. So, this operator adding 2 here, 1 year, 1 year is an identity operator on the recurrent configurations and it is not an identity operator on the non-recurrent configuration. So, you take your test configuration apply this stuff and if it relax if you get back the original then it was recurrent if you do not get back the original it is not recurrent. All these results are generalizable to other cases we are just giving as a simple example that to the square lattice. So, that is called the multiplication by identity test, but it is exactly the same as the burning test why because I said topple is the same as burn. If this site I add 2 here then what will burn when will this site topple you add this to see the original configuration if the original configuration had at least 2 then it will topple. And then in my criterion I would have said burn it if the number of unburned neighbors is more than or equal to 2. So, it will burn. So, when a site can be burnt it will be toppled in the multiplication test and then it comes here it throws a particle here, but this thing goes away under burning the site is removed. So, then the number of unburned neighbors becomes less and the counting is different, but the test is the same. So, every time a site can be burnt it is the same as every time it can be toppled. And if all the sites are burnt then you get the same thing as every site topples once and so you have the same test yes sir. Because maybe in this case it is easy to define the inverse because we have a clear direction but when the problem is a bit harder how can you be sure that there is an inverse and how to define it. Which one? I do not know like the inverse because in this case you have A i there is an operation and you can define A i to the minus 1 by saying that there is a complete series of operation. But when the problem is a bit harder like I do not know there is no closed number of operation to go back to the beginning. You ok. So, that is the transient configuration when the configuration is not inside this loop. So, but in this case it is easy when the problem is not I mean there is no this configuration of. No, no, no. So, the mathematical proof is that on the set of recurrent configuration there is a unique inverse. And this identity holds you know we have proved that if you multiply with this you will get back the original stuff. If it is not working then you must not have been on the set of recurrent states. So, that is the multiplication by identity test. So, next so all these tests have a generalization to the case of unsymmetrical piles. You know you use the same algebra nothing changes multiply all the A's put them together you get something that operation is valid for unsymmetrical matrices also and everything goes through there is no problem. The burning test is a little bit less sophisticated. So, it does not work for unsymmetrical matrices automatically. So, you can easily produce counter examples where the burning test will not work, but this more general test will work. The reason being that in the equivalence which we showed between these two tests everything has to topple once, but suppose there is a site such that you know suppose I have a two site problem and there is a delta matrix which is the 7 minus 4 minus 1. I can produce a matrix like that it says that this site you know it can have i 1 to 7 and whatever sorry 0 to 6 and 7 will topple it send 4 things here, but if it sends 4 things to this site this site has only its topples will send only 2 things out then if it other site topples once this site topples twice. So, in the corresponding equation I will perhaps write this equation. So, the equation was a 1 to the power 7 is equal to a 2 to the power 4 a 2 to the power 2 is equal to a 1. Now, if I want to apply the old method I will square this equation and write a 2 4 is equal to a 1 squared and then it will work out and everything will go through, but if I just multiply the equations without squaring it then it may not work you know. So, there is a little bit of trick or little bit of kink involved in the application. So, the multiplication test is pretty good it works more generally then the burning test burning test works for symmetrical matrices, but not for unsymmetrical matrices. What has been said in the past is that if you have greedy sites which take in lot of sites in one toppling, but get rid very few in each toppling then the burning test by itself does not work and some variation of the burning test is required which we will not produce you can make it you know, but we will not do it is not it is not the burning problem of the day ok. So, now equivalent equivalence classes of configurations. So, the key result in the senpai model which makes all of these things work is that under toppling suppose you have a configuration z i it can go to some other configuration z i prime equal to z i minus delta i j if you topple at j no delta j i ok. So, now I imagine this configuration space of z and so this is also a lattice in n dimensions z 1 z 2 and all let us allow all possible heights including positive and negative heights and let us get rid of the toppling condition the toppling condition is that you can topple only if the height is bigger than 3 we will get rid of you can topple whenever you want ok. And we will also allow an extra thing which is that you can untoppel whenever you want untoppel means the inverse of that toppling process which says take one from each neighbor their height can become negative I do not care I just take it ok. So, given any point you can topple at any site or you can untoppel at any site you get another configuration and then you can topple at any site or untoppel at any site you get another configuration and so on. So, all the configurations which can be reached from a given configuration are said to be equivalent under toppling there is defined an equivalence class if you can go from i to j you can also go from j to i if you can go to i to j by toppling then you can go from j to i by untoppling ok. So, two configurations z and z prime are equivalent if they can be reached from each other by a sequence of topplings or untopplings I guess this should have been called in topplings or whatever, but it is called untoppling. So, now interestingly the key point is that still you cannot go from anything to anything ok. So, the number of there are configurations you can go from here to there, but you cannot go from here to here. So, the number of equivalent classes is finite not one sorry yes ok it is more than one the number of equivalent classes is more than one ok. So, how much is it? So, each stable configuration each sorry each recurrent configuration belongs to some equivalent class because you can start with some configuration and suppose I start with this spine let us take and the heights are minus 5, minus 3, minus 2, minus 7, 10, 11. So, very bad you know it has negative heights and very large positive heights, but I can make all heights positive by untoppling everywhere once when I untoppled here then I make this minus 4 and make this 9, but make this was minus 5 it has become minus 1 then I untoppled once more then will become plus 1 because 2 particles from outside will come in. So, when you untoppled the boundary you introduce particles from outside and. So, the number of particles increases and if you do enough untopplings at the boundary the number of particles will become positive everywhere and then you topple and get rid of them until the heights become less than 4 and. So, you will get into a recurrent state. So, from any configuration you can go into a recurrent configuration ok. Now, can you have two recurrent configurations which are equivalent to each other under toppling the answer is it is not possible you cannot have two recurrent configurations which are two recurrent configurations which are equivalent under toppling ok. So, we need to prove this one I can leave the proof as an exercise or I can do it here let me leave it as an exercise. It is not a big deal you think about it for 3 minutes you will come up with a proof if not you can read the proof in the notes it is given there ok. So, let me skip the proof for the moment because I want to focus on something else. So, it turns out that in each of these equivalence classes there is exactly one recurrent configuration. So, the number of equivalent classes is equal to the number of recurrent configurations because you take any configuration you topple and un topple by un toppling you can make all the heights positive when you make all the then you decrease the height by toppling and it will go into a recurrent configuration. Because if there is a configuration where every height is bigger than 4 then it can be reached from every recurrent from some recurrent configuration by adding and so then if you topple then you go back to another configuration which is still recurrent ok. So, that is the proof ok. So, I think lot of these results are in the end very elementary, but you have to still convince yourself that you did not miss out anything which is important, but it is sort of difficult for me to do everything on the board here. So, I am leaving something says an exercise yes, but you can reach of course, you can un topple everywhere you know. So, topple everywhere 50 times all the heights will become negative that is not a problem you can go to a configuration which is everywhere negative or everywhere positive and so on and so forth, but you can also go to a configuration which is recurrent and there is actually one unique recurrent configuration in each equivalence class to which you can go yeah by un toppling and so on. So, the statement is precisely this that in each equivalence class equivalence class there is exactly one recurrent configuration ok. So, the number of equivalence classes is also equal to the number of recurrent configurations ok. So, now, I want to count the number of equivalence classes ok. So, what is the number of equivalence classes ok. So, now, it is a repetition of some old argument. So, suppose there is a configuration what is the other this is I am working in the z i space did n dimensional space of integer points each of which labels a height configuration ok. And z i can now go from minus infinity to plus infinity all of them. So, now, this site I can topple by delta 1 or delta 2 or delta 3 or delta n that is a toppling process is equivalent to a translation in this by a vector delta 1 ok. And there are n such vectors and all the sites which can be reached from a given site form a super lattice on this space and so then set of equivalent points equivalent points. So, the equivalent points forms a super lattice is the word which physicists use. So, I am using it super lattice with basis vectors delta i. Delta i are just the row vectors of vector delta and the matrix delta. So, now, can I determine the volume of the parallel pipe it volume of the unit cell in this problem I do not know we are working in n dimensional space and integer points in n dimensional space. And I am making some equivalence between different points in the n dimensional space and they are connected by unit vectors then they form. So, then that is all there is. So, what is the volume of the cell in the super unit cell in the super lattice is the same determinant of matrix formed by delta. So, the volume of cell volume of cell in this volume of the unit cell this super lattice is equal to determinant of delta is the same argument if those are the basis vectors you take the determinant it gives you the volume. And now of course, it is also an integer no problem and it is also equal to the number of recurrent configurations. So, the number of recurrent configurations is equal to determinant of delta. So, that is I think our number of recurrent configurations is equal to determinant of delta appears with equal probability in steady state. So, now, we have a full characterization of the steady state we know which configurations are there which are not there and what is the probability of each one is equal. So, now you have you know you may like to ask something else like you know there is a pile in the steady state what is the mean height of the site at some particular site that becomes a problem you know I have given you a measure in the space of all configurations a uniquely well defined measure, but now you have to find some averages under this measure that is often a hard problem you know like in most equilibrium state make problem that is what we do we find the measure which is the Boltzmann measure, but calculating averages is not trivial. So, I guess the same thing happens here the measure is fairly straight forward, but given the measure if you want to find the mean height or you want to find the mean size of toppling or some such thing that may not be so easy. So, that is the work we have to do, but at least one step has been reached namely that we started with this problem at least the steady state is well characterized after that requires more work some of which can be done some other things cannot be done yet. So, that is a problem of you know I give you a model which is well defined like the Potts model or something or the other, but I cannot calculate the critical exponent of the Potts model in three dimension or some such thing. So, that is a technical problem it is important, but it is a technical problem which may be we can address may be other people in the past did not manage and you can do it now or some such thing. So, what is the time now? So, 15 minutes I guess what we would like to do is to get some other rather easy results which one can derive. So, let me define g i j is equal to average number of topplings at i given i 1 adds a particle at j in the steady state. It is some kind of a measure of the amount of disturbance I create by adding a particle how many topplings occur at side j when I add a particle how many topplings occur at i when I add a particle at j. So, I guess the measure is complicated you know you will have a configuration the lots of states and then I will add something it may topple it may not topple and further topplings may occur may not occur all that kind of complication I want to not worry about it. What I realize is that after all the toppling is done on the average every side has the same height as before under the steady state condition that is by definition true. So, what happens is that in steady state mass balance condition says that inflow must be equal to outflow. So, we should have g i j 4 times g i j is equal to g i 1 j plus g i 2 j plus g i 3 j plus g i 4 j where we again use this i i 1 i 2 i 3 i 4 these are the four neighbors this is the number of particles which this is for j not equal to i. If I go to any other side i which is not equal to j then number of things coming in must be equal to number of things going out. So, this is the number of things going out this is the number of topplings at i is toppling throws out four things how many things come in on the average i 1 side will topple g i 1 times and it will throw in g i 1 things and the i 2 guy will throw in g i 2 things and. So, this equation is exacted just mass balance actually this one does not use any of the toppling conditions. So, it is valid for all kinds of sand piles including the non abelian ones is just a mass balance condition. I guess you have to put in the fact that exactly four things leave each time a toppling occurs because that is part of this construction otherwise I may not be able to do this mass balance exactly. But once I put in this fact that exactly four things leave then this condition is true, but when i equal to j then g i i. So, I will write I guess more things leave then come in at the origin because at the origin or at side i sorry at side j I have added a particle. So, I have to take care of this fact and. So, I put in a delta i j and then I get rid of this condition. So, the g equation more generally satisfies the equation delta g equal to identity and g as a matrix is equal to delta inverse. So, you write the Laplacian as a matrix take its inverse that gives you the g matrix which is now interpreted as the propagator which is the average number of topplings at i when you add a particle at j. If I know the average number of topplings at i when I add a particle at j then I can certainly calculate the total number of topplings when I add a particle at j by just summing over i. And then I can average over j if that is the way I was planning to add particles and I will get the average number of topplings when I add a particle. So, let us write that down average number of topplings when I add a particle. At random equal to g i j summation over j summation over i divided by n. So, that is not once I have calculated g then this one is not much tougher to calculate. Yes. Steady state is by definition the set of recurrent states with correct probability nothing else. No steady state is a collection of recurrent configurations with suitable probabilities configurations are you know if you take a snapshot that is a configuration. Steady state is a movie of the stuff in the long time steady state is not one snapshot, but it is a long time average of snapshots. I think that point is important to just the notation, but the terminology there is a configuration and there is a steady state. Steady state is not a configuration it is a set of configurations with various probabilities. So, this we can do maybe I should do it I will do it anything else I should do before no. So, let us do this first no I have other things to do I will do this, but may be tomorrow exact calculation of this g i j can be left as an exercise, but may be I will take 5 minutes to do this on the board it is not such a big deal. So, I will do it, but let us do something else which was left over from previous lecture in the sense that in the notes it is done before the subsequent stuff which we will do later. So, in the notes what I have done which you know here when I was writing the notes I was organizing the material in some way and I thought that it was useful to put them in before then after and. So, right now I am doing it before then you know I could do it in the 8th lecture, but I am doing it in this one ok. So, other models of SOC. So, the point is this that senpai model is a model of SOC, but it is not the only possible model you can make and there are other possible models one can study with equal justification. We study senpai model saying it is a simple model to understand, but may be there are other models which are also simple which we can equally well study. So, that should be kept in mind and whenever we study this in further you can see ask if similar things can be said for other models ok. So, let us discuss them. So, there was actually a interesting issue at one time what is the simplest model of SOC and various proposals were given and I think that is a personal opinion that the simplest model is this one it is called loop erase random works. So, first let us define the model that then we will understand how it is a model of SOC. So, you can take you can work in d dimensions, but we are taking 2 d. So, there is a lattice we are taking the square lattice you can work with triangular lattice whatever is your weight and particle does a random work that is a very standard well studied model. So, it takes a step to a neighbor with equal probability ok. So, it takes a step we will remember how it took the step then it went here then it went here then it went here and here. So, the x the only rule added is that if a particle comes back to a site which it has listed before then it has formed a loop maybe it comes like this then as soon as it forms a loop that loop is erased and. So, this is erased and it looks like this then it goes and goes and goes and goes and goes and comes here then this loop is erased and then it goes and goes and then this loop is erased and so on. So, what we have is a loop erase random work that is the name given to the problem. So, what can one say about this one well suppose you work on a finite lattice L by L lattice then the worker will keep on working, but we will keep on erasing loops what is left at any time has no loops. So, there will be a finite number of configurations they occur with different probabilities. So, there is a well defined steady state we will say that the length of the loop length of the work is my measurable it is the quantity I look at and that sometimes I drive it by increasing it by one at each time step, but it decreases in random ways because sometimes some loops are formed sometimes no loop is formed sometimes a very big loop is formed sometimes small loop is formed and so the relaxation the addition is addition of one length. The relaxation is the loop is erased and it occurs over different sizes and there are some events with very big size some events with very small size and what is the probability that you will have a loop erasure of length L. The answer is that it is a power law in L and it is a non trivial power law. So, then that looks like a model of SOC you have produced a non trivial power law starting with this very simple model one particle thing around very good. So, one of the problems which makes this of course, professor Ruhani has already said something about loop erased box in his first talk maybe he will discuss it in some length later because you know that was one of the things, but anyway this problem is well studied in various contexts including even in conformal field theory you can learn something about the properties of loop erased box their fractal dimension the probability of loop of size L or that kind of stuff, but let us just say that suppose I ask a simple question what is the probability that you are at R at time t. So, this answer is the same as in the suppose you did not erase the loops you will steal the probability of being at R at time t is unchanged. So, it is the same as probability of R t with no loop erasure R t is the position at time t R t is the position at R at time t. So, some properties of this system are immediately obvious R squared average will go like t for example, for infinite lattice and it will go as L squared for finite lattice is the second result obvious I have a finite L by L lattice and let us say the work reflects when it reaches the boundary then it will be equally likely to be anywhere on the lattice at large times and so the R squared average will be L squared I can calculate the exact probability I can calculate the coefficient in front of this. So, the problem is a tractable problem, but it has some features which are less obvious in particular the distribution of loop sizes is not so immediately obvious. So, it makes a good example of SOC it has burst like events it has slow driving it has everything that we look for. So, this is called the Takayasu aggregation model. So, the problem is defined like this you take a again d dimensional space, but we will take one dimensional space d equal to 1 sorry or higher and so the rule is that if you have a d dimensional space we add one particle at each side at every time step there is a discrete time state evolution you start with everything empty and you add one particle at each side at every time step and what then what happens is that you add a particle and then each particle jumps to its nearest neighbor one of them at random. So, there is it has three steps one is addition of particles and there is a jump each particle jumps to one neighbor and then there is a third is a aggregation. Aggregation says that if two particles jump to the same place then they join up and then they never separate they become a single particle of a mass two then next time some mass will come and add to this you know three and then this three will jump together to a random neighbor and it keeps on going like this I keep on adding one particle at each side and then this repeats at each time step. So, what happens in this system the total mass in the system keeps on increasing forever linearly with time. So, there is no steady state in the system however if I just sit at one site and I ask what is the probability that where I am now the mass is m that mass distribution of masses actually has a probability distribution because there is some probability that you know this site of course, this site the particle jumped out, but none of its neighbors jumped into this one. So, there is a finite probability that just before the new particle is added after all the jumps have occurred after all the aggregation has occurred the mass at this site is actually zero then there is a finite probability the mass is actually one there is a probability the mass is exactly two. So, you can calculate the distribution of masses at one point and it will be a non-trivial distribution probability of m at a site at time t goes to p of m infinity which is a non-trivial distribution which looks like this m this is probability of m and it kind of looks like this and there is a tail which as bigger times the tail becomes bigger and bigger because bigger and bigger masses become possible, but the beginning part does not change and if you keep m fixed and let t go to infinity you get a finite limit which is denoted by this and this quantity is a power law. So, this can be called to be an example of SOC how can you call it an example of SOC there is not even a steady state well what you can say is that suppose I sit at one site I do not care what is happening everywhere and I will say that if something comes to me on my site with mass bigger than 100 then that is a big event that is like an earthquake, but if something occurs with some mass comes with size 200 then it is a bigger earthquake right. So, if you choose any threshold then there are some events which occur with bigger than that threshold and there is a power law distribution of sizes of these things and so this is a model of SOC. Yes sir. No all of them all sites at all sites jump to neighbors in parallel at one time step I add one grain to each site and next site every that grain adds to the existing grain and becomes a clump it becomes a mass m and then the whole mass jumps to a neighbor and it happens at every site and then the masses combine and then I add something in the jump again and then it goes on like this. So, this is an example of a system in with finite drive rate where the drive rate is finite because normally people study SOC with slow drive where the drive rate is really slow or some such thing this is occurring at a finite drive rate and you still see these power laws which are some kind of signal there are big events small events and there is a power law distribution of big events. So, that is an example of interesting model of SOC third model it is called the train model it is defined like this there is a one dimensional track and there are two trains compartments which are connected to each other by these couplings which are spring like and this one is called the engine and the engine moves with rate one. So, every time step it so there is a discrete lattice and so engine moves at rate one engine moves to write. So, the engine can move what happens if the engine moves is of course, this spring will become longer and we say that the spring can be stretched, but cannot be stretched too much once the length of the spring becomes three then it relaxes less than that it does not do anything there is the threshold relaxation. So, if length of spring is greater than equal to three it relaxes to a length which is one or two with equal probability it can relax to this way or this much, but once it relaxes to length two of course, it pulls the compartment behind to keep that length, but once this compartment is pulled of course, this length will be stretched, but it was initially one and this moved at one unit this become two then it does not do anything stops and then you know, but if this goes very far then eventually the trains just goes along very clear simple model. So, what happens if this model if you just study with single engine and no compartment that is a very trivial dynamics nothing happens, but suppose there is a compartment behind and that is all there is then x one as a function of time will do this it will go like this it will evolve discreetly like that x two will be initially here then it will remain here then it will jump and jump like that sometimes it stays put sometimes it jumps by one unit sometimes it jumps by two units. So, it is easy to analyze it is not very hard. So, what happens if you have a train of length three or two compartments firstly whatever happens to the second compartment does not affect the first compartment first compartment will keep on going the same way it was going before now you only have to study what happens to the second compartment and it also undergoes the jerky motion and it turns out that you can study this model in some detail now because of this reduction property that if you get rid of the third forget about the details of the third compartment then the problem is easier to analyze. So, I start with one two three four five like that and then it turns out that if you look at the compartment behind very often it does not move at all and then it moves in big burst by a large amount and then it gets stuck for a long time and then it moves and then stuck and moves and its movement is in burst and the size of these bursts can become very big if that length of the train is very big and. So, it is an example of SOC in the sense it shows burst like relaxation for compartments in the far back. So, it is a tractable model of SOC and seems to be an interesting model it has been kind of studied by simulations and it is not well understood theoretically it cannot be analyzed fully for arbitrary lengths of train and so, that is an interesting problem and there was a number four which is actually called number zero in the old one which is called variance of sand piles. So, you can take other kinds of sand pile models with different toppling rules you know if the height is so much then all height is equally distributed to all the neighbors the height can be real numbers and not integers the topplings can be stochastic with some probability you do this with some probability you do that all those kinds of models can be studied and they are examples of SOC. The particular point about this first two examples is that they are actually turns out in the end that they are variants of this billion model which we discussed one can show they are actually equivalent to it in some sense. So, it is not immediately obvious from the definition, but once you study it once you analyze it that you find that oh these models are the same they are different garbs garments you have the same person wearing different dresses you know. So, it is the same model actually I do not know about the third model, but it is perhaps also a variant of one of those models that has not been established that is for you to show. So, variance of sand piles I will stop here now the at any sheet can we have it back. This model is the number of the length of jump for the spot right and in the first two problems the length of the loop for the spot sometimes it is big.


Okay, so thank you very much for your time and just a short introduction of this series. Although, although Roma are the largest ethnic minority in Europe, they still experience high level of racism and have to deal with many societal issues. Additionally, Roma are being represented in a very stereotypical way in the media and artwork as well. But the Roma Theatre has more than a century long past with active professional theatre groups in many European countries, but unfortunately they are still hardly known. The Roma Heroes Theatre Festival, initiated by the Independent Theatre Hungary, is the only international Roma actor encountering the world, which has been organized since 2017. Some of the present artists and their work are introduced in our series and today my guest is Richard O'Neill, the writer of the play The Hardest World. Hi Richard, how are you? Hello, I'm very well, thank you. Okay, so my first question to you is just very quickly, could you please summarize briefly what the play is about? The play is based on a real life happening up in Scotland and there is a long history in England and Scotland, because there is no physical border between England and Scotland, so Romani people have travelled between Northern England and Scotland as my family did for many years. The Scottish Gypsy people, the travelling people of Scotland were treated incredibly badly as they were in England and one of my friends and dear colleagues wanted the Scottish Government to apologise for all of the things that had happened in the past, some of the terrible things and she had this campaign and tried and tried and tried to get the Scottish Government to apologise and that gave me the idea for the play about this woman who has this campaign and she is desperate just to get this apology and it's called The Hardest Word because we often say that that little word sorry is the hardest word to say and really that's what the play about, how far she's prepared to go, a mild-mannered but strong-willed woman, how far she's prepared to go to get that apology. Thank you, it sounds very interesting, I already saw it in Budapest two or three years ago as far as I remember it was, you were then here in Budapest. Okay, Richard what is your art creed, what are your principles in your work when you are creating art? The first thing in any of the work that I do is I have to tell a really good story, that's where my family heritage is in storytelling and it always has to be a really really good story, it has to come from the heart first and then the head afterwards, it has to be something that I feel passionately about so if I'm going to write something, create something I have to feel it, it has to be a really powerful feeling for me otherwise I don't think it really gets across to the audience so that's my starting point that it has to come from here, I have to feel it and then the head writes it. And who and what inspires you during this process, in general in your life? So many people, the old traveling people because I grew up with the very old English and Scottish Romani people who were sort of almost around at the turn of the century some of those so I was a child 50 years ago and some of them had been born in the previous century so they had this amazing tales, stories of struggle and triumph but it was always about the family and I think you know when we talk about the Roma people in Budapest, we talk about the Roma people in different parts of the world, we are in fact one big Roma family and all of those different Roma people from all the different groups, they inspire me because we haven't given up you know of all the things that have happened to us throughout history, we're still here and we're still making art and we might be doing it virtually at the moment but we're still here and we're still doing it and you know every time I see somebody do a flamenco dance, every time I see somebody sing, every time I see somebody tell a story from the Roma tradition just inspires me to keep going and to do even more. And what do you think about the representation of Roma in general in the art world also in media? Well first of all in media it's generally negative and that's you know that's the media for most people I think they want to show the negative, the sensational and because we don't have many Roma reporters, we don't have many people in the media for us then it tends to just be very stereotypical as you mentioned and it tends to be negative. We always make a good negative story you know that that's just the nature of it and I think because we don't have the protection of that a lot of other groups have then people feel as though they can just have a go. In the arts it's a really interesting one because our culture has inspired so much art from Van Gogh through to writers, other painters, musicians and yet we haven't been the makers of that art so we've been there and we've been used you know our culture has often been appropriated by others and it's just so wonderful now that I see so many young people coming through writers as dancers and just thinking you know as artists and and we are doing our own arts and that's just one of the best things. So we've got a really terrible history of it but that's changing and the independent theatre in Grifes are one of the most amazing groups who are you know in the past three years previously but in the past three years just doing amazing things getting our work as you are around the world to share it with other people. Thank you and I'm connecting to this question what does Roma theatre mean to you? The Roma theatre means so much to me because they asked people, Romani people from all over the world all over Europe specifically to go to Budapest to do this amazing professionally performed theatre with a director who is Roma with actors who are Roma but also non-Roma actors as well and to put our work on the world stage and for a small theatre company I think that's just absolutely amazing you know it would be like trying to go to the moon in an old car you know people would think it's not possible but you know they they built the most amazing rocket ship with this theatre and we're flying now so they mean I think they mean everything in terms of you know my plays have been performed in England but to have the play the hardest word that we talked about performed in Budapest in Hungarian and to be watched by the British ambassador you know it's just wow fantastic and their work is ongoing you know that this isn't this isn't the end for them this is just the beginning. What does Roma theatre mean to you do you think it's necessary or important to have Roma theatre? The Roma theatre is really really important because it's the one place in the whole of Europe where we as Roma artists can have our work professionally performed by a really really superb team so it actually means everything to me because when we went over there together and I took my play the hardest word and we rehearsed it and we workshopped it and there was such an amazing buzz with so many people came to watch it and to have one of your plays that you've written in England and to go over to Budapest with all of the other Roma people from around Europe and then you watch your play being performed in Hungarian and you're sat next to the British ambassador it's one of those wow moments and so it's really really important I think probably in the future Roma artists the younger ones coming through will look back and they'll look at the Roma theatre and they'll see the independent theatre they'll see what an amazing job they've done and how I think especially from 2017 how things have just changed for the better how positive they are how hard working they are and the results they've had and you know and just as you and I are talking now that work is going around the world and it continues to go around the world and they're commissioning new work you know I'm working with them on a new piece at the moment so to me artistically I think for a Roma artist the independent theatre are just brilliant they're just the best really fantastic thank you lastly I would ask you what are your future plans or goals I think I just want to keep on improving as an artist you know as a writer I want to write more I want to get more Roma material out into the world whether that's on stage whether that's on TV or film you know I mean one of the things I'd really like to do is to get some of my work onto TV get some of my work into film write more plays to work with more Roma people as a as a collective and see if we can actually you know raise the profile and open some doors for younger Roma artists coming through you know we've got to create this this this Roma space where people can come in and you know give us that give us the latest stuff you know I'm I write about the past I write about the present and I think we need to start writing about the future more as well so hopefully all of the work that we're doing will allow us to do that so I'm very positive at the moment about the future and what we can do and the part that I can play in it yeah definitely thank you that sounds great so thank you for the discussion now you can see the performance called the hardest word created by Richard O'Neill and I just stopped recording and


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It's the morning of truth. Will I be able to have cold AC while also being able to then turn the heat on? Which means we fixed it, or did we Did we fail? I have a feeling it's gonna work out. This morning I'm gonna head over to the parts store. I'm gonna replace, not replace, but I'm gonna return this door that we didn't end up using. If you saw yesterday's video, we replaced the essentially the motor to the blend door. It came in the butt if you didn't see that. And then this is the leftover refrigerant. There's three bottles, and I think they're like 20 bucks, 27 bucks each. They're insanely expensive. We used seven or eight to fill this thing up. So my fingers are crossed. The valves inside the engine. I think we fixed the problem because check it out. It's about 40 degrees. Disregard this one down here. This one's not working. It's about 40 degrees and let's for example go from cold to hot and see if anything happens. Yep. Yep. Hot air is now coming out right now. You won't be able to see it on the thermometer right away because the metal is still cold, but as you can see the temperature is increasing. It's now 80 degrees coming out of air and the heat is on. And this is getting hot. So I now have a 100% functioning air conditioning system still on vent. We'll just hit it to cool. This should blow out cold air. It jumped all the way up to about 95 degrees before I switched it over. Now I would say the air coming out of here is about 70. And then if you hit the AC on then it you know creeps down into the 40s, so it's fixed. It is fixed. All right, I might have spoke to you soon. I got it on cold AC and it is only blown 80 degrees. All right, so we're back. We're checking things out and by we I mean batch yet again. See what?  Yeah. That's telling you that it's charged. So I'm thinking that electric switch is the one that it's not cutting in. Remember I told you that if the air Conditioning got working it'll tell you the temperature it is outside. Yep. And then what is it? 92 degrees right? So it means it's charged. Yeah. This one's not hooked up. All right. Okay, let's give it a whirl. I got a card. I'm sure it's that electric switch on that because it's not cutting the AC down in the air. You can see the clutches is turning. And the clutch is right there and it's not pulling out. And when the clutch engages that's when it operates well that's when the belt powers the AC to turn on. What are you gonna do with the switch right there?  Yeah. All right. All right. He's wiggling it around.  It kind of engages and stops. All right. He got it to engage. Yeah, now it's working. I want you to know that I'm out here looking at your truck and the Daytona 500 on. I just want you to know that. I just want you to know that. I know that and I appreciate you. That's why I got you like five pounds of meat, beef, pork and chicken. How about it? You call it a deal? We're gonna be pig.  You got a deal? Yeah, we're a deal. Look at that. We're already down to 60. Look at that. Magic the Magician does it again. I don't even know how to repay you, man. Like seriously. Yeah, we already told you that. Well, through this magic of storytelling, it's the next day. I took a shower and we're gonna go to the part store. This is on order for tomorrow. Today's a holiday. It's President's Day. We need to figure out what part that is and get that ordered. And we have about, let's say, 40 minutes to get to the store and get that part ordered. So we've got a boogie. Good news. We made it about a quarter mile down the dirt. We might have fixed it. We might have fixed it. Loose connection. Loose connection, he says. Looks like this thing is a little loose or frayed or something. So what are you doing in there? Terminals, they loose up because they're pins, right? So these here little things, I don't say they're perfect. They're far from perfect. Hey, you just bend them pins a bit and then put it back in. We'll clean it up with some brake foam. All right. It could be it because you did fiddle with this and then it immediately started working.  Oh, yeah. Yeah, you can see there those, this end of it's kind of bent back a little bit. Sometimes you just cut that end off there. This is the new clutch, right? Yeah. Cut that end off there. But the thing is that if they put new ends on, they don't bend these a little bit.  So nothing really gets accomplished is what you're saying? Pretty much. Okay, now start it up and we'll see if this thing keeps running. All right, here we go. We're on Max AC. Cool, fans on high. You ready? He's fiddling with it. Survey says, I know that face. It works. All right. Yep, problem solved. We'll see. We said that the last three days. We did say the last three days. So I think we're going to go for a test drive now. Oh, God.  Yes. Or Canada. Well, seems to be working. We got ice cold air coming out here 40 degrees. If we turn this to warm, say for some strange reason, it goes from a 90 degree day to a 20 degree day. We got hot air now. And Badge the Magician did it again. Put it back on. Put on mix. All right. Mix. Mix will give you a dash, right? It should. It's not. Feel it now. That's pretty hot. Look at it. It's up to 80 degrees. There's no air out of this one or this one? Yeah, but it's... Oh, the dash might not. The dash doesn't work on that. Mix. There, feel that. You only get floor and you only get a very little bit out of here, right? Oh, yeah, because we're not going fast or what? No, well, that just doesn't do it. The dash ones are mostly air conditioned. So now, if you pull a jox and you put it over the AC, it's not cold yet. It's not cold yet, jox. Yeah. And it's still blowing hot air, isn't it? Yeah. You've got to wait a couple of minutes for it to cool. He turns his back over and it's cold. There we go. See? Look at that. That's frozen right now. It's almost instantaneously cold. It's like this is better than it was brand new. Yeah, it is, actually. Now, you're going along cold, right? Now, remember, I told you about AC? Yeah. Okay, now put it here. Right? Yeah. And it warms up, right? Oh, it turned off now. That's right. No, it's coming through the AC and it's using AC to warm it up. Because remember when I told you Ford straight-liner and them used air-conditioned to keep the moisture off the windows, right? Uh-huh. And that's what they do. And if you put that on, like if you put it and start it up cold in the morning, you'll feel the air coming out of here right away because the AC is on. Because it's faster to warm up the AC than it is to warm up the heaters. And then you turn it back to cool. Look at this. This is crazy. It's crazy because it works. Look, it's cold already. Yeah, it is. It's cold 40-degree air already. Yeah, it works.  No one else. Thank you so much. I can proudly say proudly and proudly say Badge Fix Badge Fix AC, everybody. Badge, finally. After a couple years of having this bus paying $2,500 getting ripped off in Utah. Badge, finally. True.  But Badge was the only one to fix it. Unbelievable. Unbelievable. And I was just commenting. Clare Bear probably said different. He probably said I fix it too good.  But now we're going to tackle this I was given a 190-degree backup camera by someone at the RTR and I think I have his business card somewhere but I don't know where it is. So I just want to say thank you for the kind gesture and the kind gift and this little guy is a lot smaller than that little guy and yeah, it's a better wide-angle camera but no one doesn't do so well. Alright, here's reality as I know it right now. I have to hang on to this little plug while Badge feeds me the other wires. That's the old one. The new one's right there. Hopefully this is long enough, huh? Yep. I don't like this fiberglass insulation right there. Not to be a complainer but it makes your hands itch like crazy. Got the red connected. The yellow is right here but I need the connector which is over there now. Alright, if I had four hands I could have filmed what I did but I basically just connected these new wires. One's yellow, one's red and then I covered them with electrical tape. And then I put the key to the on position and look at how much better hang on. Badge, we're upside down. There we go. Look at how much better the quality of the camera. Is it working? Yeah. Yeah, that's hilarious. Dude, the camera is so good. Is it working really good? Yeah. This is what the new backup camera looks like. It's way higher in quality. Actually, it's way higher in quality and it has this red like the warning. The closer you get it's red, yellow, blue, green. Let me just show you guys the brand. Like I said, I forget the guy's name who gave it to me. I think I have his cards in place. But this, here's the box for it. Here's the serial number. If anybody needs a really good backup camera, this is a premium backup camera. This is a Tika backup camera with waterproof HD rear view camera and 170 degree wide angle. There we go. There's another thing right there. And then it says all the little things on it. This is really good. This is what I was looking for in the beginning. There we go. This is who gave me the camera. Steve Nobler, thank you very much. This is a gift. Not to share too much information about Steve, but he has a history experience in this kind of stuff that was partly his profession. Badges scream out at our final adjustment. It looks good dude. Yeah, it looks good. Yeah, it looks great. And there we have it. A functioning backup camera in their badges. Working away as always.