 So, first thing is that as I said there are L fundamental weights how do I find the L fundamental weights? Basically it is out of those simple roots which you have you take the dot product of the simple roots with some arbitrary vector and that should satisfy this kind of an ortho orthogonality condition ok. So, using this and knowing the alphas you can actually fix what are the fundamental weights. There should be if there are L simple roots there should be L fundamental weights and that is given by this expression which involves the dot product between the simple roots and the fundamental weights. So, you will it will turn out that the first fundamental weight is this highest weight and this I think we worked it out last time this turns out to be half and root 3 by 6 in the normalizations of these Gelman matrices. So, this was the highest weight that is the one option for the fundamental weight as shown in the slide and then you can find the other option which is half 1 minus root 3 by 6 ok. So, incidentally if you see this expression if you see this expression you do see that it is the second Eigen value which has become negative right. There is a mirror reflection of this which you can draw, draw the mirror reflection of it anything downstairs will become upstairs anything upstairs will become downstairs. So, how does it look? So, I will have a point here and two points here. So, this will be your half and minus and this will turn out to be your highest weight state and you can call this sometimes people call this as D bar. What is bar denotes? It is the antiparticle of the it is an anti quark diagram. So, this will turn out to be a D bar and then this will be a U bar and that will be an S bar. So, the highest weight here which I will call it as mu 2 will be the fundamental weight. So, these two are the only both are three dimensional. Interestingly for SU 3, this is three dimensional, this one is also three dimensional and this I am going to denote it by this diagram, dimension of this diagram is 3, this I am going to denote it by this diagram, dimension of this is 3 and these two are the fundamental diagram to compose. You can compose now this fundamental diagram with itself ok. So, you can start doing composition or tensor product of two fundamental quarks, you can do that. You can do composition of anti quarks, you can also do combinations. What is the combination? I can do a quark and an anti quark. This is quark, this is anti quark. These are the three possibilities in the SU 3. SU 2 you do not have this diagram you have only just spin half particle is a fundamental object up and down and you take tensor product of only spin half particles clear. You just take multiplication of spin half particle in the case of SU 3, you have these combinations. That is why when I combine these two, you have to make sure that it is an it is just adding things in such a way that the diagram formula which is here is not violated. So, what will be the possibility for this? It has to be allowed to be this and this one will be anything else is possible as a diagram respecting that kind of a situation. What about here? You can allow this is one possibility. You can have all the three here and a single one and what else? Anything else? Diagram should be such that if I put a 1, a 2, b 1, b 2, a 1 and a 2 has to be anti symmetric. b 1 and b 2 has to be anti symmetric. There is no relation between a 1, a 2 to b 1, b 2. So, when you combine diagrams, you can put the entries in such a way that here you see a 1, a 2, b 1, b 2, b 1 and b 2 are anti symmetric to a 1, a 2. Here it is b 1 is symmetric to a 1, b 2 is symmetric to a 2 does not matter because there is no symmetry between a 1, a 2 to b 1, b 2, but there is a symmetry amongst a 1, a 2 amongst b 1, b 2. I cannot violate that when I am putting those diagrams. So, if you start drawing the diagrams, these are the only three possibilities which I can get. What about this one? Incidentally, a quark and an anti quark together is a composite object. It is called mesons. In nature, you see a composite particle which is a bound pair of quarks and anti quarks which is called mesons. To see that bound state, you have to take a tensor product of fundamental objects. One is a quark, one is an anti quark and they decompose into irreps. It will be the dimension here. It is 3 into 3 which is 9. This is 3-dimensional, this is also 3-dimensional, this is 9-dimensional. This 9-dimension is a reducible representation. Same here, 3-dimensional quark, this is 3-dimensional quark. It will be a 9-dimensional die quark. You have a 2 quark object, but it is reducible representation and that 2 quark reducible representation breaks up into irreducible representations. What are these dimensions? Now, you can work it out. For SU 3, can you work out this anyways 3? This has to be 6. Is that right? So, 4, 3, 4 and then 2, it will be 6. So, this will give you 6. So, this will give you in dimension language, it is 6 plus 3 because it is vertical. Sometimes in books, they put a 3 bar that it is associated with this diagram. So, if you have a quark into anti quark, you will have one vertical and then one which is this. These are the two possibilities. So, there is no meaning to, this is just a diagrammatic way of representing it that any anti quark will be represented by a vertical column of n minus 1 boxes of SU n. So, for SU 2, there is no distinction. SU 3, that is an anti quark representation because both share the same dimension. If you go to, if you take this vertical 2 box and work it out the dimension for SU 4, you will not get 3 or you will not get 4 also. So, it is not an anti quark representation. Anti quark representation will be always a vertical column with n minus 1 box. That is the, you should not think of this diagram as if it is a composite of 2 quarks already, that is what is your confusion, right. Yes, mathematically both are equivalent, but when you want to compare it with nature, the way to see it is that this. See, you know when you take an electron, if you make it into a positron which is an anti particle, your electric charge gets an opposite sign. This you all know. So, similarly if I take this u quark, this should become a u bar with an opposite charge. That is what will happen here, ok. So, that is why this is like a diagram wise by doing this. It could be interpreted as a bound state of 2 quarks, but this diagram conveys that it is an anti quark representation. Is that clear? So, just fall back only on the SU 3 when you are getting confused with the abstract notation, ok. So, let me get to the slide back again. So, for arbitrary representation of SU 3, any rep, the highest weight of an irrep will have number of single boxes A of them, number of double vertical boxes B of them, ok. So, this is summarizing my young diagram way of looking at it. Single box denotes defining representation of SU n, where n denotes the number of basis states that 1 0 0, you know that way of writing it, you can have n of them. 2 for SU 2, 3 for SU 3 and so on. There are n minus 1 fundamental weights. I illustrated for SU 3 that there are 2 fundamental weights. In principle, you can find for SU n, n minus 1 fundamental weights, n minus 1 is also the rank of the SU n group. So, you can have n minus 1 fundamental weight. Pictorially as an young diagram, we denote this n minus 1 fundamental weights as single box, double vertical box, triple vertical box and so on. Any irrep will be given by an young diagram involving the corresponding diagrams corresponding to the fundamental weight. So, this is what I said and now I am going to get on to SU 2 for clarity on tensor products which you all know and how to see it from the group theory and also the young diagram language. So, SU 2 irreps we denoted by J which is your highest magnetic quantum number which we call it as a highest weight and then the magnetic quantum number and these belong to a vector space V j that subscript j to denote that the dimension of that vector space is 2 j plus 1. There is something which you all know and if you take tensor product of 2 irreps, let us take spin J 1 and spin J 2. You will write it as a decomposed one involving mod J 1 minus J 2 in steps of 1 up to J 1 plus J 2. I am sure you all done this, but for the sake of completeness as a group theory I will take a tensor product of this is tensor product of 2 fundamental representation 2 spin half and that gives you 2 irreps and those 2 irreps have to be what are the if you compose 2 spin half what do you have to get? You have to get spin 1 and a spin 0. One of the diagram should be spin 1, another diagram should be spin 0. So, that is the way to see the diagrams and the corresponding dimension should also multiply and add up that is what happened here 3 into 3 was 9, 6 plus 3 is 9. This is what you have done even in the earlier case. I remember when you did C 3 V or C 4 V if you took a 2 dimensional irrep E cross E, E cross E is 4 dimensional reducible representation and then you broke it up into A 1 plus A 2 plus E which is also 4 dimension total, but it is 1 plus 1 plus 2. So, such a decomposition in the context of E groups can be written at least for the unitary groups in this young diagram language and you can see that the number of irreps will be 2 for composing 2 defining representation, ok. So, composing 2 defining representation is a reducible representation, but then diagrammatically you can break it up into irreducible representation to find the dimension you use this point. So, the dimensions also will match this something which you have done in SU 2, but this can be done for the SU 3 and higher groups. So, this is what I am explaining it here. Take a tensor product of single box with single box. This combination will be a reducible left hand side is a reducible representation and you can break it up into 2 irreducible representation. What do you have to check? The dimension of this times the dimension of this, what is dimension of this for SU n? It is n dimension for this is n. So, n squared you have to make sure that this plus this is also n squared. Maybe you can check it by using the Hooke formula for an arbitrary n, ok. The Hooke formula also I have written here. Only modification is there is a non-trivial numerator which I explained already. So, the non-trivial numerator is that you place n n plus 1 in the boxes along the first row and decrease the number by steps of 1 along the vertical columns and take the following diagram. So, for SU 3 you will have 3, 4 along the row and column which I explained already and it gave you 8 dimensions and that 8 dimensions are the 8 states which you can write it out using the 3 fundamental states of SU 3. Keeping track that along a row you can have a symmetry, but along a column it should be antisymmetric. So, you cannot put the same state both you cannot put u and u here. If you put a u here the rest either it can be a d or an s that is all I am saying, ok. So, now tell me what is the young diagram denoting spin j representation of SU 2? What will be the diagram? So, spin j dimension is what? 2 j plus 1, right because you write it as j m, m will take values minus j up to j. So, this number of states is 2 j plus 1. Are you all with me? Which diagram will give me 2 j plus 1 for the SU 2 group? SU 2 group you can only put highest weight as, highest weight as some number of boxes times mu 1, right. Vector is also no meaning because it is just one component. So, this can be written as a 1 boxes. This is the diagram which I have written for SU 2, right. This is an irrep of SU 2 whose highest weight is a 1 number of fundamental weight, there is only one fundamental weight. And for that one fundamental weight I would have put a single box, a 1 of the fundamental weights you will have a 1 box. What is the dimension of this? Can you work it out for SU 2? Somebody what will the last box be a 1 or a 1 plus 1? First box is 2, second box is 3, a 1th box has to be a 1 plus 1. What is the hook number for this? It is a 1 plus 1, a 1 goes on up till 2 in the numerator, denominator is a 1 factorial 1, 2. Are you all with me? What is the dimension? A 1 plus 1, right. So, for a spin j representation what is the diagram you should draw? A diagram with how many boxes? This dimension is a 1 plus 1. I want this to be 2 j plus 1. What is the diagram? Spin j diagram will be a single row with how many boxes. So, each of these diagram is a higher spin representation of SU 2, where the dimension tells you how many states are there and this one will be just a 1 plus 1, which is 2 j plus 1. The number of boxes has to be equal to 2 j plus 1.