 So, now let me try and list out more as a summary of what all groups are there in the literature and what are the notations. Any questions? Leigh groups, general linear group of degree n is a set of invertible n cross n matrices under matrix multiplication. Matrices with real entries are G L N R and matrices with complex entries are G L N C. Subgroups of G L N sorry G L are S L N R or S L N C. Orthogonal groups are subgroups of G L N R or orthogonal groups have only real entries there. Simplactic groups I am sure you would have done symplectic groups in your classical mechanics group course where you do not only work with position you also work with momentum that is what we call it is a phase space right. So, you have. So, let me take a simple two dimensional phase space. So, you can have x and p x. So, that is a two dimensional phase space ok. If you are in three dimension it will become x, y, z and p x, p y, p z. So, it is a six dimensional phase space and then you are allowed to do here it will be a 2 cross 2 matrix, here it will be a 6 cross 6 matrix. So, what are the matrix transformations you can do such that your Poisson bracket is preserved right. What is Poisson bracket is 1, x and x is 0, p x, p x is 0 you know that ok. So, this is what you need to preserve under any transformation. So, I can call this. So, this one gives me x prime, p x prime. This Poisson bracket will go to x prime, p x prime under such a transformation such that this is going to be still one. This is what you have learnt. So, what are that set of 2 cross 2 matrices that set of 2 cross 2 matrices belong to a symplectic group ok. So, these m's are elements of symplectic group 2 and if they are real entries sp 2 r ok. What about here? m will be elements of sp 6 comma sp 6 r ok. So, this Poisson bracket being preserved what does it imply on these matrices? What does it imply on these matrices? m j m transpose have to be j for its other way round. So, m transpose anyway does not really matter what I call as m you can call it m transpose. What is j? If you remember if I generalize orthogonal transformation this j was replaced by some diagonal matrix with some of them positive some of them negative right remember. Generalize orthogonal matrices I called it as a matrix G by m comma n I mean that this one will be m of them will be of one signature n of them will be of other signature. Similarly, here these have to be even dimensional. So, let me write it it is always even dimensional phase space is even dimensional it does not work in odd dimension because every position you will have a momentum ok. So, this is to n what is this j such that your Poisson bracket is preserved I am sure you would have done it anybody knows is it diagonal or off diagonal off diagonal. So, j turns out to be 0 0 which is n cross n n cross n it is 0 ok. All the entries in this n cross n matrix do not take it to be orthogonal it is 0 and this one is minus identity and plus identity that is also n cross n. If you have not done this Goldstein has a chapter on canonical transformations please go and take a look at it ok. I am going to confine myself only to unitary groups in the rest of the lectures even though I will give you a formal aspects I am not going to look at other groups in detail ok, but anybody is interested this is the natural canonical transformations which you do in phase space in your classical mechanics course. The set of matrices belong to a group which is called symplectic group element which belongs to the symplectic group will always satisfy this condition ok. So, let me stop on this Lee group business yeah is that fine. So, some of these listing is just for completeness on the slide which I am showing that symplectic groups sp 2 n r are another subgroups of SL special linear groups they are nothing, but classical transformation canonical transformation in classical mechanics. M is called as a symplectic 2 n cross 2 n matrix, symplectic matrix if it satisfies this condition ok. I have summarized whatever I did on board on the slide ok. So, this also I have already given you a flavor that rotation of any spin j particle the corresponding matrix will be what j hat is a formal linear operator I have written number of parameters is 3 number of generators is 3 no change in that, but the dimension of this matrix will depend on what vector space it is going to act on. If the spin of the particle is j how many states it allows? It allows from plus j to minus j in steps of 1 decreasing steps of 1. So, the dimensionality of the vector space is 2 j plus 1 for spin half it is plus half and minus half j is half. So, 2 j plus 1 is 2, but in general it is a 2 j plus 1 vector space and the corresponding j operators have to be 2 j plus 1 cross 2 j plus 1 matrices which are irreducible representations acting on that vector space, but they still are SU 2 belongs to the SU 2 Lie algebra. If I give you a 3 cross 3 matrix and I give a Lie algebra, if I ask what is the group then you should know it is SU 2, but it is a 3 dimensional irreducible higher dimensional representation of SU 2 Lie algebra. So, what is the meaning of saying that it is an irreducible vector space? The group elements when it acts on this state any arbitrary state it should only mix amongst the 2 j plus 1 states should not take you out of that state. That is the meaning of saying that this vector space is irreducible vector space ok. I will give you an example, no, no, no, I will I will give you an example now, I will give you an example and then you will see what is happening ok. So, a simple example is that take 2 spin half particles ok. So, suppose I take up and down as one vector space, let me take another one, another up and down let me put a subscript 1 1 and 2 2 to remember that it is particle 1 and particle 2 just like we took momentum and position taking particle 1 and particle 2. For this you know these are the matrices which are irreducible representation, for this you know the same matrices are the irreducible. Now, I want to take a tensor product of these two. So, what will happen? I am sure you all know. What will happen to the matrices on it? You take this j 3 and take a product of j 3. What will this be? Minus 1, have I done it correctly? Yeah, I think I have done it correctly. Is this a reducible representation? It is a reducible representation and not an irreducible representation. What you can do is now we can do a projection, a projection to break this into a 3 cross 3 and a 1 cross 1. What is the vector space corresponding to this 3 cross 3? This will get reduced, the basis states also will get reduced to someone. So, it is a 3 cross 1 and a 1 cross 1. This is a binary basis which is obtained by taking tensor product and doing a projection to find the basis for the 3-dimensional irreducible representation of SU 2 algebra and 1-dimensional representation of the SU 2 algebra. Another way of saying is this is like singlet, it better be a 1-dimensional representation with the binary basis. We can start doing this for tertiary basis. So, I am connecting up what you did in the discrete groups and only thing here is that I am working with the Lie algebra, generators of the Lie algebra acting on the vector space. But if you want to look at the group elements, it is not going to do anything new because exponentiation will only again follow the same. So, whatever group operation I am going to do on this, if it is an irreducible vector space it is going to mix only this ok. So, that will not be the situation for your in general. So, what I am trying to say is that you cannot go from here to here, these two are non-talkative space. So, this is a reducible space which obtained by tensor product of two spin half particles and then what is the projector is what you will ask, what is the analog formula for the projector and that I will that is the club squadron coefficients which I will try and give you some flavor.