 So, let us start. So, I am going to confine now to SU2 group. SU2 group is now by now you should all know the usual notation of group notation. S denotes the group element should have determinant plus 1, U is for the unitary groups, 2 is to denote the first lowest non-trivial dimension of the representation of the vector spaces two-dimensional, ok. The corresponding matrices will be 2 cross 2 matrices and the number of constraints if you subtract 2 cross 2 matrices, unitary matrices means it should satisfy unitarity condition and determinant equal to 1. So, you will get the number of parameters which is number of real elements in the matrix, ok. So, there are only three that is why you have theta vector is formally written to remember that there are three components theta 1, theta 2, theta 2, ok. So, U of theta this is for SU2 group once it is the group I try and put it to be capital S capital U and 2. So, this will have an exponential form it will be an exponential form I by h cross theta dot the generators I said no number of parameters will always be equal to the number of generators. In the case of rotation I wrote it as orbital angular momentum here I am going to write it as a total angular momentum which includes spin angular momentum and orbital angular momentum of a particle, ok. So, this will be your general element where J's are the generators this cap is generally written for a linear operator or you can even write the matrix representations for it depending on which vector space it is going to operate, ok. So, this is a familiar algebra of the generators algebra is denoted by small s small u and 2, ok. This is SU2 Lie algebra. So, how do you define this explicitly the way to see it is that you could take two multiply two such matrices, ok. I am not going to do this for you take two such matrices, ok. What is the group property required if you take a product of two such matrices it should again be an SU2 matrix, but the parameters are not just going to be additive when will it be additive if you do a rotation about an axis which is the same then you can add it up, ok. But if you do rotation about theta 1 vector which is in a different direction and theta 2 vector in a different direction it will give you again a unitary matrix which depends on phi, ok. This is also a vector, but it will depend on theta 1 and theta, ok. It will be what exact functional form I will not know, but depending on what is the value I can find it, ok. So, use this use the infinitesimal piece and keep up to order theta 1 theta 2 and see what you get and you will end up getting this algebra for the generator just like I would for the orbital angular momentum you can do this for this case, ok. You see that the dimensionality of the vector space depends on 2 j plus 1 dimensional vector space. So, I said that j hat is 2 j plus 1 cross 2 j plus 1 matrix acting on vector space which is given formally as you all know this you write it as j and m we will come back to why we add two numbers here, but this m value. So, this m takes values from minus j minus j plus 1 up to j. So, there are 2 j plus 1 states on which this 2 j plus 1 cross 2 j plus 1 matrix act. So, those are if j is half which is your lowest two dimensional vector space which is up spin and down spin. So, someone interestingly pointed it out that in your discrete groups the character tables tells you that you have only up to certain dimension irreducible representations, right. So, like if you take a tetrahedral group you cannot get a 5 dimensional irreducible representation. In the case of continuous groups for ij you still get a irreducible representation, but that irreducible representation acts on the character table here, but we will see how to get irreducible representation. Just because I write a 3 cross 3 matrix it need not be an irreducible, ok. Always if you are able to find a matrix with block diagnosis then it is not an irreducible. But analog of the character table you can allow all possible dimension irreducible representations for the continuous group, ok. So, why are we using this state is something which you have learnt in quantum mechanics that is what is called as a Casimir operator. What is a Casimir operator? Casimir is bilinear in generators, ok and it should commute with all the generators of the liquid. So, for example, this j dot j is bilinear in generators, right. I can write it as j 1 squared plus j 2 squared plus j 3 squared and j dot j you can see with any of the j i's it is 0 for all i's. So, this tells you that you can write a state which is a simultaneous eigenstate of j dot j as well as any one of the j's. The reason why any one of the j's is because amongst the j 1, j 2, j 3 the commutator is non-zero. So, you cannot simultaneously diagonalize j 1 as well as j 2 or j 3. One of them can be simultaneously diagonalized with this one. In all the quantum mechanics textbooks we use a choice which we call it as j z which is this is just a choice. You can write a new book or a new you know set of notations where you can call j x to be your axis where this is happening, but usually you are following even in spherical coordinate z to be the axis which and we choose the j z eigenvalues to be m, right. What I mean by this is that j z on j m will be m h cross j m, is that right and j dot j on j m you can find the number, but if you find it explicitly it turns out to be this. Both are eigenvalue equations, but if you do j x on j m it is not proportional to j m. Similarly, j y is not proportional to j, is that clear? Because of this property and my choice that we are going to take j as to j z to be simultaneously commuting with j dot j. So, this convention which we write j m is a simultaneous eigenstate of a Casimir operator. This is what we call it as a Casimir operator. Casimir operator is one which commutes with all the generators of the group. I am confining to SU 2 group, SU 2 group have j 1, j 2, j 3 as the generators. Casimir operator will commute with all the and Casimir is constructed using bilinear of the generators. And then we make a choice and write those states which are simultaneous eigenvalue equations. So, simultaneous eigenstates of this. In fact, using this you can also see the matrix representation of it for every j. This I constructed it for j equal to half by showing that it is 1 and minus 1. If you take j equal to 1, you can construct what is the matrix for it. So, if you want to find j z which is a 3 cross 3 matrix, then you can rewrite it as because it is an eigenvalue equation. If m is j, it will give you plus 1. If of course, I have suppressed the h cross, if you want I can put the h cross. And if m is 0, so j m is running from minus j to plus j. So, if m is j it is 1, m is 0 it is 0, m is minus j it gives you minus 1 for j equal to this corresponds to j equal to 1 ok. So, some of the notations j equal to 1 is 2 j plus 1 dimensional matrices ok. So, the degree of the matrix is 2 j plus 1. So, j equal to 1 this will be 3 dimensional for j x j y ok. Is that clear? Similarly, using these you can see it is off diagonal and you can work out what are those off diagonal matrices for every representation. You do not need to every representation I mean different j's corresponds to different representations just like any reps, 2 dimensional irreps, 3 dimensional irreps. So, each one is 2 j plus 1 dimensionally rep and for different representations you can use these equations to construct the matrix representation. Is that clear? So, bilinear in generators bilinear means product of 2, 2 linear operators. Take a product of 2 linear operators, quadratic. This is the quadratic cousin. Is that clear? For example, j dot j. In every group you will have a set of matrices matrix representations for the generators which will be diagonal ok. So, if you go to SU 3 for example, you want to construct the generators with certain properties for unitary group special unitary group. What is the properties? It has to be traceless and Hermitian. This is traceless and diagonal also. If I go to SU 3, so for SU 2 if I am going to do I am going to do only with the lowest non-trivial dimension. For SU 3 the lowest non-trivial dimension is 3 cross 3 matrix. So, there you can allow one more diagonal matrix which is traceless. Is that right? So, if I write the diagonal matrices. So, SU 3 how many generators are there? SU 3 number of parameters, real parameters is 8, number of generators is 8 and I am going to denote it as lambda 1, lambda 2 up to lambda i just like j 1, j 2, j 3 you will have 8 such matrices. Any element g belonging to SU 3 will be rewriteable as exponential of I by this is the group element which is an exponential of the generators and there are 8 parameters and 8 generators and the lowest non-trivial dimension has to be dictated by this. So, you can write lambda hat the lowest one just like we wrote for SU 2 as poly matrices. Here you have a set of matrices which are 3 cross 3 matrices okay, lambda i's are 3 cross 3. Team is to find traceless Hermitian matrices 3 cross 3 which is traceless whatever you write here you understand what I am saying right. The generators have to be traceless and Hermitian and from there you found that the number of parameters will be 9 and then you traceless Hermitian will have number of parameters 8 right. So, it is 1, 2 which are real then these are complex conjugates. So, this is 2, 4, 6, 8 complex numbers have 2, but this element is it is not independent. So, two reals a, b, z1, z2, z3 are the independent parameters required to specify a traceless Hermitian matrix. So, they are 8 number of real elements these are complex. So, 6, 7, 8 okay is that clear? So, 8 real elements that is why 8 parameters. Want to construct the analog of j z which I did for SU 2 the poly matrix sigma z was diagonal. How many diagonal matrices I can construct for the lowest non-trivial representation of SU 3? So, tell me how many are possible? First of all inside SU 3 SU 2 will be a subgroup okay that is something which you should know that any element of SU 3 you could write the elements which is so SU 2 is definitely a subgroup of SU. So, the conventional generator which you have for sigma z which was plus 1, 0, 0, minus 1 this should sit inside the SU 3 generator as one diagonal jet. So, let me call the generator for conventional reason just like we call that as J 3. So, lambda 3 should be a 3 cross 3 matrix for SU 3. So, this is an element of SU 2 algebra. So, this 3 cross 3 will be 1. So, this one is an element of SU 3 this is traceless transformation can be concerned one more is the next question. So, it turns out that for 3 cross 3 you have one more diagonal generator which you could construct because of this 3 cross 3 freedom otherwise you would not have been able to do it is 1, 0, 0, 0, 1, 0, 0, 0, minus and then you go to SU 4 now tell me what is the procedure by induction these two should sit inside SU 4 you will definitely have rank 2 test sorry you will definitely have two diagonal matrices and you could also construct one more ok. So, that is the chain process by induction where I have told you that there is one diagonal generator for SU 2 and that is sitting inside this and then you can construct one more. In the literature we call those diagonal generators with the subscript 3 and subscript 8 for these lambda matrices incidentally these like poly matrices the name which is given to this this was initiated by Gelman when he was looking at the particle data ok. So, just like your periodic table there is also in particle physics you can call it as a particle zoo protons neutrons are simple ones, but there are other particles he just looked at the charges looked at their angular momentum and he saw their masses and he grouped them. And he said these groups are looking like irreducible representations I should find a theory to explain it. So, he started working on this extension of this unitary SU 2 group to SU 3 and he exactly fitted every particle with whatever nature has told us in a beautiful fashion. We will we will see this to appreciate how much group theory actually speaks physics ok. So, in that sense since he started doing this I think this these are the matrices which now is well known as Gelman matrices just like poly matrices these are called Gelman matrices. So, recently he passed away he was the one who got the noble prize for this quark model saying that protons neutrons are all made of three quarks. So, three quarks in our group theory language should be seen as tensor product of three primary basis and look at the irreducible representations of a tensor product of three fundamental three initial representations ok. Three reps will give you the composition and then you break it up and you will see we will do this just to see how the clarity on SU 3 particles is exist.