 So, I will continue with Lie algebra and then Lie groups as a warm-up of recapitulating whatever we were doing in the last lecture. So, Lie algebra is also vector space and the operation is going to be a commutative bracket operation which we call it as a Lie bracket and that vector space will have some basis states ok. So, if x and y belong to that vector space and the commutator of x and y is also in that vector space that is one of the properties. And then linearity any scalars lambda and mu could be complex or real depending on that you will call this as Lie algebra or complex Lie algebra. They will satisfy the commutator bracket or the Lie bracket will also satisfy this linearity property. And of course, the Lie bracket by definition if you interchange the two elements it is going to be negative of itself and the Jacobi identity is a cyclic property which is naturally obeyed if you write it out expand it out explicitly you will see that every term will cancel with another term. So, that the right hand side will become 0 because you are all familiar in Poisson brackets in classical mechanics and also the Lie algebra has this Jacobi identity property elements of the Lie algebra. If the Lie bracket is 0 for all the elements then you say that Lie algebra is abelian example was your translation group the Lie algebra was involving linear momentum and that was an abelian algebra. And then I went on to say that if you take any two elements this is the property 1 that it should give a linear combinations of the elements in that set because this is a commutator bracket which is antisymmetric you can show that when you interchange a centi this coefficient which is what we call it as a structure constant is going to pick up a negatives. It is antisymmetric in S and T indices. And Lie subalgebra is a subset of those elements in that Lie algebra just like we have studied Lie groups subgroups of discrete groups you also have subgroups of Lie groups you also have subalgebras of Lie algebras and the definition is that that subset if you write the commutator bracket within that subset it will always belong to that set that subset. So, a Lie subalgebra h is a subset of elements of the Lie algebra such that elements form that those subset elements itself form a closed algebra by closed algebra I mean that subset elements will satisfy this property. So, then you call it as a Lie subalgebra is that clear. And then I went on to say that there is the subalgebra could be an invariant subalgebra if you take any element outside that subset and take the commutator with the subset elements which is a subalgebra this commutator should get back to the subalgebra should be any element in that subalgebra. If that is happening then we say that the invariant sorry the subalgebra is an invariant subalgebra is that clear. I did not add two more points today I tried to add that also and if the commutator if you expand it is right if you take a b. So, this by the Lie algebra as I said I can write it as c the meaning as inbuilt that it is summation over t. But what you can see is this one you would have written it as x t these two are equal. So, what does that show that c t a b is minus of c t b. So, this is antisymmetric in the two indices. So, structure constants this is structure constants are antisymmetric with respect to a comma b indices is that ok. So, simple Lie algebra has trivial invariant subgroup is just an identity element or the whole group you should have a non-trivial invariant subalgebra. If you do not have a non-trivial invariant subalgebra then the Lie algebra is called as a simple Lie algebra. Whatever you are studying the angular momentum algebra, rotation algebra associated with the rotations all of them are simple Lie algebras. You cannot find a subalgebra you cannot find a invariant subalgebra non-trivial invariant subalgebra. Trivial is identity element and the whole set that is useless you have to find something which is non-trivial ok. So, in that sense all the Lie algebras which we are going to confine ourselves mostly here are simple Lie algebras. Then there is more variant if you find an invariant subalgebra you have to check whether that invariant subalgebra is an abelian subalgebra. So, if it does not have a non-trivial abelian invariant subalgebra then the group the algebra is a semi-simple algebra. So, these are some of the jargons which is which you will see in any books, but we are going to just confine to simple Lie algebras where there are no non-trivial invariant subalgebras. And this I went through and some students said I went through very fast. So, I thought let me just take you again through this because this is a concept which if you are the first time learners then it is nicer to get it understood ok. So, typically when you write 2 cross 2 you know invertible matrices, what do I mean by invertible matrices whose inverse exists right. Otherwise if you do not have an inverse existing then you are getting into problems clear. So, take a set of two-dimensional complex vector space and let the linear operators acting on those spaces are 2 cross 2 matrices. This is not new I have taught you in discrete groups that depends on the dimension of the vector space then you have operators whose matrix representations will be 2 cross 2 if it is a two-dimensional representation ok. So, the Lie algebra is formally it will have four elements and the four elements can be complex. So, G L 2 c denotes 2 cross 2 matrix with complex entries. In order to generate an arbitrary element you can take these four independent basis and the coefficients which you can write on an arbitrary element is a complex coefficient which means it has a real and imaginary ok. So, that is why I am saying that the corresponding vector space of these linear operators for the G L 2 c is four-dimensional complex vector space or eight-dimensional real vector space ok. It is equivalent to four-dimensional complex numbers will have real component in an imaginary component. There are four bases which is required to span the elements of G L 2 c. G for general L for linear 2 to denote 2 cross 2 matrices and C to denote the entries of the matrices can be complex ok. So, if I want to say what is it in the notation of real notation I will write E 1, E 2, E 3, E 4. On top of it I will write another four where one is replaced by I ok. Using those eight matrices I can generate any arbitrary general linear 2 cross 2 matrices with complex entries clear. So, in that sense it is a four-dimensional complex vector space. Then I went on to saying that I wanted to look at sub-algebras of these two cross 2 matrices with complex entries. One sub-algebra which I could look I am not given why I am looking at traceless Hermitian. Suppose I look at traceless Hermitian matrices for example, suppose I want to write down a matrix which is a 2 cross 2 matrix. Tracelessness will force you that this has to be opposite to this. Hermitian will force the diagonal elements to be real right and off diagonal elements should be complex conjugates of each other ok. So, essentially you have only three independent possibilities here. Earlier you had four independent possibilities because there was no restriction, but once I put a restriction you do see that I have to write only three independent bases clear. I cannot do anything more than that. So, this essentially tells me that S L 2 C this is what we call it as a S L 2 C algebra or the elements of the S L 2 C algebra involves set of two cross two traceless Hermitian matrices ok. And these matrices the number of independent bases you can have is a three dimensional complex vector space because you can still put complex coefficient. Top of it I want to see the sub-algebra of these S L 2 C's ok. So, S L 2 C's so you can write generators are your poly matrices which are the two with an I factor. S L 2 C's. Yeah, Z is complex. So, it has two more non-trivial entries in it that is the meaning of it ok. A is real, Z is complex. After you write any arbitrary element will be a linear combination of the basis stage which have complex coefficients you understand. So, that is why it will become either six real dimensional vector space or it will be a three complex dimensional vector space. Is that clear? So, sigma 1 sigma 2 sigma 3 did I make a mistake on this I? Is that right? So, 0 minus I I 0 minus I right ok. So, this is one set of generators all of them are traceless and Hermitian right. And then as I said that it will be you can also write S 1 which is I times sigma 1, S 2 which is I times sigma 2 and S 3 which is I times sigma 3 ok. So, it is six real dimensional vector space ok. What I mean by that is any 2 cross 2 matrix which belongs to S L 2 C we should be able to rewrite it as 1, 2, 3 some complex coefficients of these objects which I can call it as sigma 1 plus S 1 kind of sigma I plus S I. So, this will be a you can show that these will again be these are Hermitian and traceless any linear combination with coefficients will also make it Hermitian and traceless. So, it is not a problem. Now, you look at a sub algebra of this coefficient will be once I write it as this and this ok. So, I should be little careful here this plus sign is confusing. So, let me write it like this. So, it is coefficient C I. So, let me write 1 to 6 C I times these generators let me call them to be what is the best notation I can use. Let me call them as generators as G I let us see all the 6 are. So, it is C I times G I then you can keep C I to be real. This is probably a better notation adding those 2 may not be here I think with the common coefficient is not right ok. It is a 6 real dimensionals. So, these coefficients are real now 6 real dimensional vector space ok. So, these 6 are the ones which is going to generate for you the algebra of SL 2 C clear. So, the algebra of this SL 2 C involves sigma 1 sigma 2 sigma 3 you have S 1 S 2 S 3 and which one is a sub algebra if both of them are sub algebras of it you agree sigma 1 sigma 2 sigma 3 it closes amongst itself right sigma 1 sigma 2 S I sigma 3. So, this is a definitely a sub algebra it does not makes these 2. So, it is definitely a sub algebra and you can look at elements generated using the sub algebra. So, those are what I call it as elements which let me call it as some H which is 2 cross 2 will be summation over. So, these are A i sigma i. So, these are arbitrary elements involving only the subset of those le algebra elements of SL 2 C and they will generate for me your familiar angular momentum which you have studied which is going to give you an element of a le algebra which I am going to denote it as small s small u and the 2 element of this le algebra which is a linear combination of sigma 1 sigma 2 sigma 3 will be traceless will be hermitian and why we had introduced a u will be visible when I do the group elements. So, we will see why the u has been introduced here. In the earlier case I passingly said that tracelessness means determinant of the corresponding group element will be 1. So, I said that this is le algebra if you go to a group which is SL 2 C this will be an exponential of i any of these or linear combinations of them if there are 6 generators then you will have 6 real parameters. So, you can write it as summation over j A j times all these G i G j's G j's is what I am calling it here. So, this will be the element of the group which is obtained by exponentiating the generators of the le group which are going to determine your le algebra and this element must have determinant to be this has to be plus 1 that is the meaning of saying it is special linear group that is the important requirement. And once I put that condition on the determinant of an exponential I have to think of it as matrix representations. Determinant when you do a diagonalization if you do a diagonalization of this matrix. So, let us take a 2 cross 2 matrix let us take one element in the le algebra as 2 cross 2 matrix only thing what we remember is that it should be traceless and z star and z this is an element of le algebra S l 2 C you agree. Now, I want to exponentiate this let us take one element I am just taking an exponential of that and I want to say that this is an element of the group S l 2 C ok. Once I do a diagonalization here I find a matrix A e to the i x A gives me a diagonal matrix. So, let me call it as e to the i lambda 1 e to the i lambda 2 A inverse I can always do this on a given matrix. If I take the determinant of this determinant of this what is this mean? This is e to the power of i trace of lambda 1 0 0 lambda 2 you all agree this expression is the same as that expression ok. So, what I have tried to show is that the determinant of the matrix if it belongs to S l 2 C the determinant of this has to be 1 ok. So, this belonging to S l 2 C means that the determinant has to be plus 1. What does that reflect on the e to the i x it has to be traceless. So, that is why automatically they put in a small letter S and said I am looking at traceless matrices ok. So, this is the implication between the path from the Lie algebra to the Lie group the exponential map which I am doing and this clearly shows that this implies trace of x because you can map this e to the i x right. Trace of x is equal to 0 is this clear ok. So, now the additional thing is suppose I want to look at unitary groups S u 2 what do I mean by S u 2. So, let us now take this one the corresponding element of this e to the i h will belong to S u 2. Now, whatever I said your tracelessness condition will be satisfied again h should be traceless because the S denotes determinant has to be plus 1 ok. On top of it u denotes what just like in rotation orthogonality relation now u denotes unitarity. So, I have to make sure that e to the i h e to the i h dagger should be equal to identity that is unitary unitary matrices must satisfy. So, now you can because these two are same elements in the exponential you can add them. So, e to the i h dagger i will become minus i ok. So, minus i h dagger has to be equal to 1 what is that tell you h has to be this implies h should be h dagger that is Hermitian ok. So, I have just driven you through some complex path starting with arbitrary linear 2 cross 2 matrices with complex entries I tried to say what are the basis states which forms a vector space on which you can write a Lie algebra. Then I said subalgebras with determinant plus 1 will sit inside this general linear algebras and you can exponentiate these elements and if you impose that determinant is plus 1 then you can say that the Lie algebra elements have to be traceless ok. And then we also saw that from the generators of SL 2 C you saw that you also have a subalgebra and you can generate the corresponding Lie algebra for the subalgebra which is denoted as SU 2 why it is U is by exponentiating it and seeing it as a group special unitary group involving 2 cross 2 matrices which is determinant plus 1 and it should be unitary just given a unitary matrix to you with determinant plus 1 how many independent parameters you have given a 2 cross 2 matrices you have 4 complex entries and then you have to try and see putting a determinant condition unitary condition just like we did it for orthogonal groups you should do it for even for unitary groups if you do that you will find that there are only 3 independent parameters for SU 2 is that right clear traceless even in the generator language you can see you do not need to see in the group elements group elements also you can do 2 cross 2 matrices there are 4 complex elements determinant plus 1 is 1 constraint unitary is another constraint. So, unitary will have off diagonal diagonal everything and then the determinant will be one more constraint ok. So, you can check that out and see that the number of parameters is 3 for SU 2 yeah special unitary group SL we do not need Hermitian condition we do not need it, but if I want to look at SU 2 as a subalgebra there I am just trying to look at if you take a subalgebra then some of the properties have to be continued. So, in that sense here this turns out to be a Hermitian I do not have a reason why it should be Hermitian for SL in general only determinant has to be plus 1 or in the algebra it should be traceless yeah what is s 3. So, these are in the real direction base of writing it if I do not put this in I will take these 3 only and take the coefficients to be complex you understand what I am saying if I take it to be real then you see it to be anti Hermitian I agree with you, but this equivalent statement between complex vector space and real vector space is if it is an n dimensional complex vector space it is 2 n dimensional. So, if you put the i factor automatically it becomes anti Hermitian I agree with you yeah, but you start with only saying I am looking at a complex vector space and then these are the 3 generators with complex coefficients in that sense it will be even their coefficients can be complex right. So, then Hermitian things you need to worry about it I agree the generators are traceless and Hermitian that is all I can say. So, I do not think I need to put in the condition of Hermitian for the SL groups, but when I come to SU groups Hermitian comes up natural is that clear yeah any other question yeah that is a good point SL groups you only want determinant to be plus 1 there is no Hermitian requirement, but SU because of the unitarity you do see that you get that h and h, h should be equal to h dagger where h's are the generators of the Lie algebra. So, are you all syncing the fact now what is Lie algebra what is Lie group elements of Lie group are exponentiating elements of the Lie algebra and the Lie algebra elements if you take any two elements you do a commutator you have to get a linear combination which satisfies those same properties ok that is why it forms that Lie algebra is that clear I have given you a couple of examples just to clarify things for you. S once I put a letter S for the Lie algebra it is traceless which is equivalent to the group capital S to be determinant to be plus 1 that step is what I have shown you here that you can always diagonalize the matrix when you diagonalize this one will have eigenvalues because these elements are e to the i x I have written e to the i lambda 1 and e to the i lambda 2 are the two eigenvalues. So, determinant of this can be also written as e to the power of i. So, determinant is e to the i lambda 1 into e to the i lambda 2 if you take a trace it is lambda 1 plus lambda 2 it is e to the i lambda 1 into e to the i lambda ok. So, will also be some sub algebra of this S L yeah. So, only thing is there it is a orthogonality conditions and the coefficients have to be real and there will be some sub algebra of these S L groups. So, there you have to look at 3 cross 3 matrices even though I have considered 2 cross 2 you can do this for G L and C I will just mention those notation ok.