 Welcome to module 58 of point set topology part 1. We continue our study of topological groups. In the last module, you have already noticed that the topology of a topological group is regular. This we actually prove while proving even a stronger separability property for topological groups. Even authors are not satisfied with this much. In fact, there are two schools of, you know, mathematicians or we may say only topologists. One is stick to hostlessness, other one stick into regularity. So the host of people, especially the Burbaki oriented people, they would like to have every topological space host of. So right in the beginning they put this hypothesis, a host of space with a multibitiation blah blah blah. So we have not put that one. So let us see how you can guarantee hostlessness with minimal assumption, okay. So we have already noticed that the moment it is T0, it will be a moment it is T1, it will be host of T2, right. So now we would like to show that the T0-ness is just enough, okay. Since a diagonal map g cross continuous map g, h equal to g h inverse image of the singleton set is a diagonal in g cross g. So if singleton set is closed, diagonal will be closed so it will be host of that is that is clear. So T1-ness is enough, but even T1-ness will be guaranteed by T0-ness that is the first thing that we want to ensure today, okay. So let g be a topological group with its topology satisfying T0, I will say. Automatically it will be T3, all that we want to show is that it is T1. Then we have just now we have observed T1 implies actually T2, but even that is not necessary because we have already proved it is regular, regular plus T1 is T3, T3 implies T2 implies T everything else, right. So let us just prove that it is T1, okay. So what is the meaning of T0? T0 means given two distinct points, you may be able to find an open set around one of them not containing the other, but which one which will which it will contain you do not know that is the point. Whereas if you can be do sure that you can do it for both the point that is T1, but in the case of a topological group you do not have to worry about all the points you should just show that singleton e is closed then because of the translation homeomorphism are there all other points will be also closed, okay. This we have observed earlier. So all that we want to show that now is g minus e is open, okay. What does that mean? For each point g not equal to e we must find a neighborhood u, g which does not contain e. So that is all we have to do. The only problem is for some g we may find u, g which contains g but not e but so for some others it may be it may contain e but not u, g but not g, okay. If for all g we can find a u, g all g means for all g inside g minus e not equal to e. If we can find a open set u, g which does not contain e then we are done. So consider those g such that u, g contains e and not g, all right. For such g what you do put v, g equal to u, g intersection u, g inverse. See now just now I assumed that u, g contains e, identity element. So g inverse all u, g inverse, u inverse of that will also contain g, right. Symmetry, inverse is just under, image under symmetry, right, inverting all the elements. So e will be there on both of them. So e will be here. So v, g will be a neighborhood of e, right, so v, g is a neighborhood of e. And it will not contain g inverse, right. If it is g inverse is here then g will be here and where ours are. So g is not here, this is smaller than u, g are this one. So it will not contain g inverse either. So first it did not contain g but now it does not contain g inverse because it is the intersection of both of them. Therefore g of v is a neighborhood of g which does not contain e because if this g v contains e would mean that there is a g inverse inside v. g into g inverse is the only way you can get e, right. So that is not possible. So v, g of v does not contain e. So we have got a neighborhood of the first kind namely this now, this v will be inside g, okay. So that is the trick here, just t naught implies t1 and therefore t3. Now you may suspect that every topological group actually satisfies t naught as well. After you wanted to do it minimally, that may be the reason why so many good authors are assuming, let us assume for the half-dust match and so on. But that is not the case and there are big area of mathematics varying, you know topological groups are used. Those topological groups are not t naught. In particular I am giving you one example here, I cannot deal with all those examples. So especially useful in algebra, okay. My point is that there are such more general group, you know topological groups, interesting ones. So that is why we should keep this general definition. So here is the example. If you do not know any ring theory and so on, this is very elementary definition of rings and ideals I have used here, you can just believe it and believe it, okay. I have no time to explain what is a ring, what is a commutative ring with identity, what is a proper ideal and so on. If you know these things then what I am going to tell is very elementary. So you can just remember that some such thing was there and then go deeper into it when you come across it, okay. So now just concentrate on what I am saying. If you do not understand some terms here, I have no time to introduce them. Let R be any commutative ring with identity, just like integers, okay, just like rational number, rational number should be filled, just like integers, you can see. And I be a proper ideal in R. Ideals are right and time start, okay, let us start. Then the family B which is obtained by shifting i power n, x plus i power n, where x rings over all of R and n is natural number. So I am taking i, i square, i cube and so on, okay, there is a multiplication in the commutative ring. So I am writing i into i, i into i square and so on is a standard notation for when its ideal is there, these things will be also ideals, okay, look at all these x plus i power ns. They will form actually a cover for the whole of R because I am the x plus i power n. But they will form a base for a topology, let us call this as kappa, it is kappa in honor of Krull, it is called Krull topology, okay. So that topology, you have to check that it is topology, it is a base in the institute topology, okay, under that topology, addition in the ring, it is a commutative group, right, so addition in the ring becomes a topological group. So that will be continuous, both addition and subtraction are continuous, that is the meaning of that it is a topological group and we know that if you take this i n and x equal to 0 here, they will form a neighborhood of, neighborhood system for 0, intersection of all these neighborhoods, namely just i n, where n goes over this one, we know that it is equal to 0 bar, okay, this is a general fact about any topological groups. It follows that the Krull topology is given because singleton 0 is closed if and only this RHS is equal to singleton 0, the T1 is what singleton 0 must be closed, the singleton 0 is closed means 0 is equal to 0 bar, 0 bar is equal to this one, so this is the condition on i, the ideal i should have the property that intersections of all its powers, powers you know one included the other and so on, so that should become singleton 0 and that is a non-trail condition it does not happen always, the integers you can see that it happens, every non-zero element, okay, we will have to you know some power divides, if each take a prime p, if every power of p divides a number that number must be 0, okay, so in particular if you take the integers you can see that this happens, okay, alright, that is all I want to tell you about the Krull idea, Krull topology, there is another aspect in algebraic geometry, okay, what are called as algebraic groups, they are not actually topological groups, okay, you have to be careful because in the definition of algebraic group, you start with an algebraic variety G with a Zariski topology, namely close subsets are those which are given by vanishing of polynomial functions and then on G cross G you are not taking the product topology, it is not the product of the Zariski topology with itself but it is Zariski topology directly on G cross G, it just means that there will be a double number of variables and all polynomials in that has to be taken and so here we be careful that the group, you know topological group theory that we are developing in this sequence of lecture, you cannot apply them directly in the case of algebraic groups, there may be several parallel statements, okay, parallel definitions etc, you have to check each of them correctly properly and then only you can use them, some of them are even wrong, okay, so any none of them you would have proved because all our proof depends upon the topology on G cross G the product topology, okay, so having said that one let us do something in our own definition not Zariski topology now, a subgroup of a topological group G is called discrete, okay, discrete subgroup, if as a subspace it is discrete, it is already subgroup, there is no condition on group theory in algebraic condition, in the topology the subspace must be a discrete, discrete means what isolated and closed, okay, it must be a closed subgroup, it must be a closed subset and it must be, each point must be isolated, note that a closed subgroup H of G is discrete if and only if there exists a neighborhood U of H such that H intersection U is singleton E, as soon as the singleton E is isolated again by using translation you can show that all elements of H are isolated, take any H inside H, okay, you take the same neighborhood here H of U intersection little H of U intersection with H will be just a singleton H, so this is easy to see that once singleton E is isolated in H, H will be a discrete subgroup, only you do not know whether H is closed, so you have to put that closeness also, so why one is interested in discrete subgroups, it is very old notion, classically interest in discrete subgroups arose in the study of doubly periodic functions, okay, you can see that the exponential function is already periodic but that did not really give rise to study of discrete groups and so on. But the same thing when you, the doubly periodic functions inside complex plane, okay, there you have to start worrying about more general things and you know right now in C itself R2 itself it happens, so why many properties of this periodic function, okay, they can be deduced easily if you understand the discreteness of the periods, the set of periods becomes a discrete subgroup, that is how this is interesting. Since you may not know what is periodic function and so on, I will not elaborate on that one, this much motivation is enough, okay, so here is a easy lemma first of all and then we will improve upon this lemma, all non-trivial discrete subgroups of R are infinite cyclic, typical example is integer sitting inside R, okay, here I am looking at R as an additive group, so it is a topological group, additive topological group, right, so there Z is a subgroup which is discrete, so what this theorem says is everything is infinite cyclic, as soon as you take any non-trivial discrete subgroup it must be infinite, the proof is very easy, okay, it suffices to show that every non-trivial discrete subgroup is generated by one element, okay, obviously the additive group of real number, every element other than 0 is of infinite order, therefore this is all enough, that will be, that will be the group generated by that one will be infinite cyclic automatically, so what I start, I am looking for that generator, so p equal to infimum of mod R where R belongs to H minus 0, non-zero elements of H, look at the one which has least modulars, okay, what I want to say is there are exactly two of them, okay, you can check whether plus 1 and minus, minus of that will be also there because H is a additive subgroup, right, so look at the infimum of mod R, obviously this is bounded bureau, so infimum is redefined, but this is a discrete group, okay, so it is a discrete set of points inside R, right, so this infimum, this close also H minus 0 is closed, so this infimum will belong to H minus 0, therefore this T is positive and belongs to H, that means what, see if I take 0 less than T, T itself is inside H, you may say mod T, okay, mod T is inside, T R plus minus T is inside, right, and you take, change the sign, it will be also inside, so have you found an element like this, claim is that T generates H, okay, so for this all that you have to use is division algorithm, okay, T, 2T, 3T, 4T, they are there and you want to show that well, minus T, minus 2T, they are also there, but nothing else, that is what you want to know, right, so all of H is like that, so you divide by T, okay, any R can be divided N T plus S, where S, where N is some integer plus or minus, but S will be strictly less than T, you can always choose S to be positive, non-negative, okay, if you start with R inside H, T is already inside H, so N T is inside H, R minus N T which is S, that will be also inside H, but mod S is less than T, how can that be, so the only way it can happen is this S must be 0, not H minus 0, not in H minus 0, so S is 0 means R is intensity, where N is an integer, therefore H is generated by T, okay, now without much effort, the same idea can be generalized to any R N, only thing is you have to use now more linear algebra, not just real numbers, but linear algebra will have to be used, okay, not very deeper, very elementary linear algebra only, so let us see how, so statement is this one now, that V be a real vector space of dimension N, every non-trivial discrete subgroup of V is isomorphic to Z power M for some M less than or equal to N, okay, non-trivial I have assumed, so 1 is less than or equal to M less than, if N is 1 we have already proved it in the previous lemma, so the idea is to use induction, okay, induction and some linear algebra, let us assume this N is bigger than 1 and the statement is true for all vector spaces of dimension less than N, take the linear span V prime of H inside V, H is a subset of V, so you can take the linear span, that is sub vector space, vector subspace of V, if V prime is the whole of V, then by induction we are through, it is sorry, if V prime is not the whole of V, the subspace, okay, then the subspace has dimension less than N, so H is sitting there, right, it is a discrete subset inside V itself, so you discrete here also, okay, so you can apply the induction, so also without loss of generality we can assume that we are inside V prime equal to V, that is the case, what that means that the set of vectors H, they form a generating set for V, any generating set will admit a subset which is linearly independent maximal, that means a basis, at least this is true very easily for a finite dimension vector space, right, so that is one linear algebra I am using it now, okay, it follows that there is a R basis A contained inside H, okay, for the space V, so what is V? V is V prime, it is with the span of H, all right, so I could have made an elaborate statement in the theorem itself, but even this step will be useful to you, so you should remember the proof here rather than just the final statement, now let us go ahead, now the idea is same as dimension one case, what you do, T equal to infimum of norm H now, not just mod H, okay, where H runs over H minus 0, once again H is discrete, so H minus 0 is also discrete closed and all that, therefore this T will be positive, okay, and there is at least one element H1 belonging to H, so the name the infimum is attained, that is how I think, H belongs, that is that this T is equal to norm of H1, okay, you can now replace one of the elements of A by H1, okay, and assume that A is H1 union, some other Saphionite set, which is a basis for V, you can trade, how do you do that, remember write this H1 as something alpha 1 V1 plus alpha 2 V2 plus alpha n Vn, one of the alpha s must be non-zero, so let us say alpha 1, then instead of V1 you can put this H1 and keep other V2 with Vn as it is, that will be also basis, so this is trading, this thing is also part of linear algebra, which is used to prove that any two basis have same number of elements, right, I am just recalling you some linear algebra, that is all, so it follows that, see one element as, they are all elements of, remember they are all elements of H itself, but now I have put, started with a basis which is consisting of elements of H, now I have put this H1, which has what, what is the property, norm of H is minimal, there may be many elements, I have taken one of them and assuming that this is one of the generators, okay, now I just split up the whole thing, put V1 equal to R times A1, the linear span of the rest of them, H1 equal to H intersection V1, so I am taking a subspace which is one dimension lower, okay and then I am taking H1 H intersection V1, H is discrete inside V, H1 will be discrete inside V1, okay, so what you have is V is written as some copy of R spanned by H1, direct sum with V1 and H will be the infinite cyclic subgroup generated by this H1 inside R, so that is the lemma 1, direct sum with H1, this is a discrete group and this is the copy of Z, that is also discrete of course, so you get the direct sum decomposition like this because there is a direct sum decomposition of the whole vector space, also one checks that H1 is a discrete subgroup of V1 because it is intersection of H with V1, so by induction because the dimension has dropped down here, H1 must be isomorphic to some Zk for some k less than equal to n minus 1, add this one more component H1, what does it give you, H is isomorphic to this k plus 1, but that is intersection of k anyway, okay, so this is just a small beginning of the study of discrete subgroups, which is a large subject, you know there are books written on with title discrete subgroups of league groups, okay, let us go ahead with the study of these subgroups, take H to be a closed subgroup of G, okay, then the right cosets, the right cosets, similarly left cosets also you can take, given the quotient topology because they are the orbits of the action of H on G, right, cosets are what they are, they are the orbits of the action, left cosets, the right cosets depending upon which action you take, is given by that is given the quotient topology because it is a quotient set, okay, and this is called a homogeneous space, okay, what is the meaning of homogeneous space, you start with the topological group, take a closed subgroup, then look at G by H, there is a closeness is something as a mighty assumption here because non-closed subgroups are extremely badly behaved, you can always take G by H where H is also not closed, but you cannot do much topology on that one, okay, so I am using this notation G by H for right cosets, this also I will rate as G by H only, but I do not know how to rate it, this is very popular in league group theories and so on, so this is a left cosets, both of them they have to use, that is why they have cooked up this notation, observe that G acts on the left of G by H, on the right, on the right cosets it acts on the left, okay, this is the reason for the name homogeneous because action is transitive, what is the meaning of transitive, given any two cosets here, that is an element of G, so G of that is the other coset, so one point is taken to the other point, any two points are related by the action or the orbit space of this by this action will be just one single element, so that is the meaning of transitive, okay, such actions when you have such actions on a space, that space is called homogeneous, okay, that is the reason for homogeneity, so what is the good point of homogeneous space is all local properties, topological properties if they hold at one single point they will hold at all other points, just like topological groups have that property, the homogeneous space will also have that property, so here is an example here which we have just discussed already, right, consider the, consider the case when H is non-trivial discrete subgroup of n-dimensional vector space over the reals, as in just now we have considered all, okay, in the proof of this that theorem as well as the previous lemma, we have seen that the discrete subgroup H is generated by linearly independent elements v1, v2, vm, where m is less than or equal to n, right, extend this R basis, extend this to an R basis, v1, v2, vm, add some more to get an entire basis for the vector space v, okay, take e1, e2, en as standard basis for Rn, now what I am doing, I am going to map e1 to v1, et cetera, e1 to vm to get a isomorphism from Rn to v, okay, because they have the same number of elements after all, map ei to vi, i less than or equal to n to get an isomorphism from Rn to v, but this will have the property that if you take the subgroup generated by ei, that is z power n, see what are these, 1 comma 00, 0 comma 1 and so on, right, the lattice points will come, so they are, they will generate z power n, the direct sum of z with n times, the thread itself n times, let us call it as h, okay, no, sorry, phi of z n will be h automatically because they will be, they will be mapped to vi, so h should be generated by vi, that is the assumption, okay, so here we are denoting the subgroup of Rn generated by e1 e2 en by z power n, so that is h, consequently what you will get is gamma t n, this is a notation, s1 cross s1 cross s1, isomorphic to v by h, okay, I am sorry, I forgot to tell you that, suppose m is equal to n, then only this should be, okay, suppose m is equal to n, then phi of z n will be overlapped because that is the subgroup generated by all the v1 v2, v1 v2 vn because m is equal to n, then t n is nothing but s1 cross s1 cross s1 which is v by h, okay, v by h is this one, but how do you get this one, this is because this Rn by z n, Rn by z n is isomorphic to R by z, R by z n times, R by z you know is isomorphic to s1, e going to, t going to e power 2 pi i t giving you the isomorphism, R by z to s1, okay, so you know all these, these are called torus, you know each of them is called torus, s1 cross s1 is a standard torus in dimension 2, the same name is used in higher dimension also, it is a torus, they are all what, take any finite dimensional vector space and take some largest kind of discrete subgroup sitting inside there and that is a quotient. If you change subgroup, the topology does not change, but lot of geometry will change here, so that is why they are very very interesting these, okay, so these groups are very interesting, even the case of s1 cross s1 just n equal to 2, they are called elliptic curves, why because each group here how it is sitting inside R cross r will tell you a different story, different complex analysis is there, the complex analytically they will be different, so complex structure will be different, so each of them is a called a you know elliptic curve, in the general case when m is less than n what happens v by h will be t power m cross the rest of r n minus m, so you can do the direct, you know all this with, you could have seen this one in the proof of our theorem itself, up to m you have basis, you have extended it, extended part is r n minus m, so here it is already a quotient v by h is t power m, okay v1 by h is t power m where v1 is corresponds to the linear span of the first m elements v1, v2, vm, okay, so this is a general picture, so I have told you it is a complete general picture of any discrete subgroup of what of a finite dimensional vector space, okay, now here is some exercise which you can do easily, arbitrary product of tableaus or appreciate tableaus, all right, then some more exercise are there which are not all that easy, but if you keep solving them one by one then it is okay, in fact this one, the last third one here we have seen it in a different context, so this won't be difficult for you at all, okay and then comes the connectivity assumption, here you may have to spend little more time, but you know again try to solve them in that order, so slowly then you will get or you can solve all of them that is the whole idea, okay and then you can apply them to various different cases also, namely finally you can apply them to study these classical groups SON, SON plus 1 and so on, so there are some nice things here which is happening, so I have tried to motivate these examples you know by giving those elementary exercises so that you can solve these things easily, so finally these examples are the motivating examples for those exercises, okay, so rest of the work you can try to do this one, next time we will continue the study of topological vector spaces now, okay, that will be the last topic for this course, thank you.