 So, let me begin with the notations which we started yesterday. What were the notations we are going to follow? We are going to take H is the notation for order of group. Typically we write order of group G by notation this is I am using H just for clarity mod G is what we would have written, but we are writing this as H. P is the number of classes, P is the notation which we are going to use for the number of classes. And classes we will denoted by C 1, C 2. So, do not confuse this C with your group structure. So, you can if you want you can put a some kind of curly C 1, C P, P classes are there that is 1. And then I also said that we will use gamma of G as a representation. Typically it could be an n cross n matrix representation and this will be of two types. It could be reducible and it could be irreducible. So, reducible in general irreducible we are going to denote it by gamma alpha of G and these are L alpha cross L alpha matrix representations. I am just using the short end notation that. So, the matrices will have dimensions L alpha by L alpha, number of rows and columns will be L alpha and L alpha that is the meaning of this. And how many alphas are there for this specific group? Alpha will take values 1, 2, someone number of classes, number of irreducible representations for the group will be equal to the number of classes is one of the postulates, ok. E reps is the short end notation I use for irreducible representations. Is this clear? Notations are getting clear, ok. And then what I am saying reducible means I basically can find an S and bring it to a block diagonal form, ok. So, this n cross n can break it up into pieces, ok and so on. The requirement is that suppose this is n 1 cross n 1, this is n 2, this is n 3, this is n 4 and so on. What is the requirement n 1 plus n 2 plus n 3, others are all 0. Reducible means you can find S to bring it to this form by similarity transformation. You can do an S, S inverse, you can find an S to bring it to this form. Here, no S, ok. Here you do not have any S, you cannot break it that is the meaning of this. So, now, this number should add up to it has to be n, right. The total matrix is n cross n, I am breaking up into blocks. The total block dimension should add up to be the dimensions you start with, ok. So, that is one. Nothing I said is that in principle I can rewrite this as summation over, ok, all the irreps. I am denoting the irreps by index alpha which goes from 1 to p, ok, a alpha, gamma alpha, ok, for every element. And someone was asking what is the meaning of this? The meaning of this is that if gamma 1 appears twice, you will suppose I say that let us look at n equal to 3, ok. For example, if you take n equal to 3 and let us say that a 1 is the number of times gamma 1 of g, this is an irrep notation occurs. And how do I denote it in the matrix form? This will be a gamma 1 of g, this will be a gamma 1 of g and this will be a gamma 1 of g. Are you all fine? The dimensions 1 plus, if suppose this is 1 cross 1 matrices, ok, let us take this to be 1 cross 1 matrix, then the dimensions adds up to 3. This is one way in which you can write this as a 1 times gamma 1 alone, 3 times gamma 1 is one possibility for this example. You could also have 2 times gamma 1 plus some other gamma 2, the only thing is that this should also be a 1 cross 1 matrix, this should also be a 1 cross 1 matrix. These two entries could be different that is why I put gamma 1 and gamma 2. This entry occurs twice, this entry occurs once. How will I show this? It will be again gamma 1 and gamma 2. This will also be 1 cross 1, 1 cross 1, 1 cross 1 for an example. So, this is the meaning of writing it as a sum of a alpha times gamma alpha, where gamma alphas are the fundamental piece by which you can construct any higher dimensional matrix representation. Which can be broken up into block diagonal form. Some of the reps can, irreps can repeat. Suppose the C-rep repeats, we say that a 1 is 3. Here the first T-rep repeats. So, a 1 is 2 and a 2 will be 1. Is that clear? It could also be that I get only a gamma 1 and a gamma 3, which is a 2 cross 2 matrix. That is also possible. That depends on what is this S operation and how it block diagonalizes. So, if this is 1 cross 1, the irrep 1 occurs once, the irrep 3 occurs once. So, the block will become a 1 cross 1 under 2 cross, this is 2 cross 2 and a 1 cross 2. You understand what I am saying? This notation, yeah, direct sum. I mean technically the direct sum, yeah. I agree. It is just that you people do not understand that notation as of now. So, I am using this. The meaning of this is the block diagonal fashion in which we write. It is not that some of the irreps can be repetitive. That is the meaning of this. Is this clear? Is the reducible, irreducible also clear? So, remember these two expressions, which I have written now. One is this. Let me write here. Gamma of G reducible will be rewritable in terms of this, where these are the irreps. And then what did we do? And we had set of properties, which follows for the. So, I showed these things for abelian groups. Number of classes is also ordered. So, you will get number of irreps to also be equal to order of the group. And then you need to find out the reducible representation also now I put in here, but you understand the notation. A alpha is the number of times an irrep alpha appears in the reducible representation, ok. Is this clear? Postulates were the first postulate I have already written here that the number of classes equal to number of irreps. And then I have one more property, which is summation over alpha l alpha squared equal to order of the group. So, l alphas are integers and squares of those if a positive integers squares of them is constrained by the number of elements in that group. And it is a very easy problem to find what are those integers there would not be more than one solution. You only strictly one solution, ok. And then I got to the postulate of the great orthogonality theorem. I am not here to. In fact, you can use this great orthogonality theorem to prove all these things. I am not proving it. I am saying as a consequence you get these as a product. You can do the proofs if you want, but as of now from the great orthogonality theorem you can prove this l alpha squared equal to h. And you can also show that if you take the trace of those matrices. So, these are matrices with M n element, the specific element. This M n means a specific element of that matrix multiplied with some other specific element of another irrep matrix. And if you take the sum over all the group elements of the M, it will satisfy this relation that it will be orthogonal alpha and beta they are not same you get it 0. If alpha and beta are same you also require the same elements M should be M prime N should be N prime. And then you have a factor which is h divided by the dimensions of those irreps. L alpha is the dimension for the irrep alpha, L beta is the matrix dimension for irrep beta. Is this clear? So, this is something which I do not want you to memorize. You can put it in your formula sheet. Everything you can put it in the formula sheet you do not need to memorize. And from this you can start playing around taking the trace. So, we defined character, character alpha of G as trace of gamma alpha of G, right. This is the notation we have. So, you can play around on this great orthogonality theorem by taking a trace. And then you can derive summation over G chi alpha of G, chi beta of G to be equal to h times delta alpha beta, ok. Is this clear? Are you all following? So, the characters are orthogonal. They are different irreps and you have to sum up over all these elements and then do that. Equivalently, since the character within the class is same, you could replace this by summation over classes 1 to p, number of elements in the class i. And then you have a character for the irrep alpha for a candidate in that class, which you can take as formally as C i and chi beta for the candidate in the same i at class. And this will also be exactly h times delta i. So, these two are equivalent statement. Instead of doing it over summation over group elements, you can do it by taking one element in the conjugacy class, but put the number of elements in the conjugacy class as a multiplicative. So, that is what I have written in the next equation, ok. So, now comes the writing the character table using these properties, ok. So, C 2 v was the first example we took. And here you had E, then you had a C 2, then sigma x z, which I will call it as sigma v and then I will call sigma v C 2, ok. So, this is nothing, but sigma x z I think this one sigma y is both are equivalent, but. So, first of all we need to determine how many alphas are there, ok. P for this case is it is an abelian group, number of classes is also equal to the order of the group P is h, ok. So, which means L 1, L 2 up to L h, correct. And we need to satisfy this condition. So, summation over alpha 1 to h L alpha squared has to be equal to 4, right. In this case it is 4. This is an abelian group, maybe I should just keep it to 4, ok. I am doing it for this abelian. So, what is the solution for this? What is the possibilities for this? Can you have a two dimensional matrix? Everything should be non-zero. There should be 4 non-trivial at least 1 cross 1. So, the solution is that you can have a gamma 1 which is 1 cross 1, gamma 2 which is 1 cross 1, gamma 3 which is 1 cross 1 and gamma 4 which is 1 cross 1. The 4 irreps have to be each of the irreps have to be only one dimensional, ok. So, let us do the C 3 v also here. Identity 1 conjugacy class as C 3, C 3 squared. The second conjugacy class as sigma v, sigma v C 3, sigma v C 3 squared, ok. So, in this case what is the classes, number of classes 1, 2 and 3. So, you have p equal to 3. So, you will have l 1, l 2, l 3. There will be 3 irreps. What are the dimensions? Summation over l alpha squared from alpha equal to 1 to 3 should add up to give you. Order of the group is 6. So, now, tell me what are the representations possible here? 1, 1 and anything else possible? Nothing, right. These are the only possible. So, you will have a one dimensional representation, another one dimensional representation, two dimensional representation. And then we can write the characters of these representation elements. I said that we are going to take this to be the unit representation, which is trivial representation. I am going to blindly put it as 1, 1, 1, 1, ok. Here also we will write the characters. So, now I am writing the characters here, ok. The entries which I am writing are characters, characters of which irrep to remember these irreps I have denoted it here. So, this one will also be 1, 1 and 1. And now I need to fill the complete table. Last time I tried to write the irreps by looking at a representation. Now I am going to use these identities to fix the table, ok. So, using these identities part of the properties we have seen that there are 4 irreps for C 2 v and there are 3 irreps for C 3 v. And one of the irreps in C 3 v is 2 cross 2 matrix. 1, 1, 1, 1 is like a trivial. See if you can put any group to have identity element. Every element to be identity element any multiplication table will be satisfied, right. So, it is like a trivial representation. As a matrix I can put 1, 1, 1, 1 and make the whole multiplication table to be satisfied. I agree. I agree. So, that is another way of saying this is at least I wrote last time. If you do a rotation about z axis the matrix representation is just 1, 1, 1, 1 that is another way of saying, ok. So, that is a trivial representation or unit representation as we call it. Next thing is I need to fix these things the characters. So, inside whatever I am writing I have to write the character for representation 2. I do not know what notation I used here. So, I have used character for alpha 2 for identity element, character for the irrep for the C 2 element and so on, ok. I need to fill this. I have not done it, yeah. How do we fill it? We just use this property. The product and sum it up over all elements should be orthogonal because one irrep is a unit representation, other irrep is a non-trivial representative. So, let us fill it. Only thing what you have to remember is for identity element is nothing you can do. Whatever is the character it has to be if it is 1 cross 1 matrix it is 1, if it is 2 cross 2 matrix it is 2. So, the character for the identity element you can write blindly. It is just dimension of that matrix, right, trace which will give you the dimension. So, this is actually L 1 is 1, L 2 is 1, L 3 is 2. So, similarly here you will have. So, this part is 1, 1, 1, no change. Now, I can play around with this. Thing is that property can be done for alpha refers to a row, beta refers to another row. If alpha is not equal to beta then those two rows have to be orthogonal. So, you have various conditions. What are the conditions? You will have a plus b plus c plus 1 has to be 0 and you can try and play around what all you can do because the products also we have to take. It is a 1 cross 1 matrix, entries should be either 1 or minus 1 and you can play around and try and fix this. So, so you can try and see that suppose I put this as 1 these two has to be minus 1. It is a simple way to check it. Here also I could put minus 1 and make one of them to be minus 1. Any other possibility I have missed out? These are the only non-trivial possibilities I can have right. So, basically what you are seeing is that this multiplied with this should also be 0. We did three of them last in the last class by looking at it as a x axis, y axis and z axis right. We also have a fourth one. I do not know which one is the fourth one, but this fourth one which I have is kind of very different, but still it is orthogonal to this as well as this and this. Is that right? Are you all with me? I have just used that property and I am just playing around with the numbers. What about here? What about here? Someone help me this should be a and b, it is not just a, you have to multiply by a and then 3 times b has to be 0. I am just multiplying this with this. Alpha is this one and beta is this one. I am using this one, this relation. So, n 1 is 1, n 2 is 2, n 3 is 3. There is also one more requirement. What is the other requirement? You could just take this particular row alpha and beta to be same, then you will have it has to be 6 ok. You have two conditions and then figure it out what is the best a and b have to be some integers, see which is the best one which will fit to this. So, which one will fit a squared equal to b squared has to be 1. Is that right? So, we will have a 1 very good. Now, let us come to this. So, just taking the square of this alpha and beta to be same will give you the square. So, that will be 4 plus 2 times a prime squared plus 3 times b prime squared. It has to be equal to 6 and what is the next condition? I could just play around with this 2 plus 2 a prime minus 3 b prime has to be 0. One more you can write 2 plus 2 a prime plus 3 b prime has to be 0 by multiplying this with this. Clearly, these two equation will tell you what is the solution b prime has to be 0. So, b prime has to be 0 and is it satisfying that other condition? Yes, a prime could be plus or minus 1, but then this condition will tell you minus 1. There would not be ambiguity even though I do not know how to prove it, but case by case when I take the discrete groups I will find that the character table is unique. So, whatever I have written today is a character table for C 2 v, one is an abelian group, C 3 v is another group which is not abelian and you do see that as you go to as you go from abelian, abelian will always have only 1 cross 1. What is the reason? It has to satisfy this condition and there will be the same number of elements in the abelian group. The only solution is all the L alpha has to be 1. You cannot put any L alpha more than 1 cross 1 matrices. So, abelian group the irreducible representation is always going to be 1 dimensional, but in the non-abelian you can have 1 dimensional, you can have 2 dimensional, it could also have more dimensional matrices. In this particular case with these orthogonality criteria it turns out that you will have irreps where there are two one dimensional, two different one dimensional irreps. One is a trivial unit representation, another one is a non-trivial representation which is also 1 cross 1, but you also allow for a 2 cross 2 non-trivial irreducible representation. Is this clear? So, that is what I have done. The same thing you could have done it for the symmetric group of degree 3 there is no change. How will you do it for the symmetric group of degree 3? This is isomorphic to symmetric group of degree 3, right. You can represent these classes by diagrams. This class has 3 1 cycles, this class has 1 3 cycle and this one is 1 1 cycle and 1 2 cycle. You are all with me? So, in the context of symmetric group these irreps can also be represented by the same diagrams with a different meaning. Here if I ask you how many elements are there for this diagram you know what to do, right. You have done this. You look at how many cycles are there and we have a formula. Similarly, unit representations are always denoted by this diagram for the symmetric group, ok. Do not confuse it with any other group. Not every group will have the symmetric group of degree n to do this. Whatever I am doing it for symmetric group of degree 3, you could do it for 4 5 and all, but C 3 v happens to be isomorphic. So, this diagram seems to match, but in general I cannot write any diagrams here, ok. For any discrete group, point group I cannot write a diagram, but at least for symmetric group I can write a diagram, ok. So, there is another diagram which is this and there is a diagram which is this. Now you can ask the question why should I use this diagram structure which is how do I know that looking at this diagram that it is two dimensional matrices. That is not obvious from here. I could have put this here, right. So, how do I know that this is two dimension, this is one dimension, ok. So, let me just review on that. So, this diagram is called symmetrisers. This diagram is called anti-symmetrisers, but you can also have mixed diagrams. Only requirement is that the number of boxes as it goes into the second row should be utmost equal to that that is the convention we are going to follow. You can have this. So, this is what I will call it as a mixed. Why mixed? It is not totally symmetric, it is not totally anti-symmetric, but it has some of them are symmetric, some of them are anti-symmetric and you mix and match, ok. So, these are called mixed irreps, ok. So, this is anti-symmetric irrep. This procedure is called symmetrisers. So, I can say that this is totally symmetric irrep, this is totally anti-symmetric irrep and this is some mixed irreps. Now, my claim is that dimension of three box is one which means that irrep is a one cross one representation. Then I said that dimension of this one is one, dimension of this one is two. This is what I am claiming. One of the ways you can see is that symmetric group of objects three, you are allowed to put in every box one of the objects. You can put object one here, object two here and object three here. That is it. There is no other possibility left. That is why it is one dimension. Here also you can put object one here, object two here, object three here. You will ask, can I not put object three here, object one and object two? But it is a totally symmetric or you can take it to be identical objects. You are doing total symmetry, whether you call the first box, second box, third box or third box, first box, second box does not really matter because it is totally symmetric. Similarly, the same thing. This is the only option. When it comes here, what happens? You can allow one and two to be symmetric and one and three to be anti-symmetric. This is one possibility. But you can also have one more possibility, which is one and three and one and two. Is there any more possibilities you can ask? But I am trying to tell you that if you try to do that, you can show it to be a linear combination of these two. It is not independent. So, these two are the two independent possibilities of putting the three objects in a mixed diagram. So, that is why it is two. But now, if I start doing this, this will be for some, let us say n objects going to become tedious. Again, like conjugacy class, I said there is a formula which will tell you if you had a breaking up of cycles, what is the number of elements in the conjugacy class? You had a nice expression, combinatoric expression. Similarly, there is also an expression for how to determine number of or the dimensions of the irrep for the permutation or symmetric groups of degree 3 or more degree n also, there is a formula which I am going to show. And then I will explain it. But this is the motivation. And the irreps can be shown exactly like the diagrams which you have here. But the meaning there is three one cycles, one three cycle, one one cycle and one two cycle. Here it is totally symmetric, totally anti-symmetric and a mixed representation. Mixed representation will be two dimensional, there will be a two cross two matrix and this these two will be one dimensional, that is why it is a one cross one factor. Is this clear?