 So, next as an application of quotient topology, we will define the Grassmannians. So, let us begin. So, let G be a group and such that the set G also. So, the underlying set G also has a topology. So, we say that G is a topological group if the two maps. So, we have since G is a group, we have the multiplication map which is x comma y maps to x y. This is the product in the group and we also have the inverse map x goes to x inverse. If these two maps are continuous. So, this question makes sense because G is also there is an underlying there is a topology on G and therefore, G cross G has the product topology. So, we can ask if both these maps are continuous and if both these maps are continuous then we say that then we call G a topological group. So, let us begin by making a few observations about topological groups. First if G is a topological group left translations. So, these are the maps L A from G to G and A of G is defined to be A times G right translations. So, these are the maps R A from G to G defined as R A of G is equal to G times A. These are continuous these are continuous right. So, let us see why this happens. So, we can look at G cross G to G we have the multiplication map right and here we can look at A times G. So, inside G cross G. So, this inclusion is given by A comma and when we multiply we get A G. So, this composite so this is homomorphic to G right. So, this composite is L A. So, since this inclusion is continuous and multiplication is continuous as G is a topological group. So, this implies that L A is continuous. Similarly, we can take G cross A sitting inside G cross G and this is homomorphic to G right and to see that the right translation R A is continuous. So, the inverse of L A is. So, the left translation is a bijection and the set theoretic inverse is exactly L A inverse right. So, this implies that L A and R A. So, similarly for R A R homomorphism similarly the inverse of I since the inverse map is a bijection I square is equal to identity. So, this implies that the inverse map is also. So, left translations right translations and inverse are all homomorphisms. The next observation we want to make is let U be the collection open subsets containing the identity element for A in G the collection A U. So, here A U is equal to the left translate of this set U is the collection of open sets. So, clearly since L A is a homomorphism L A of U is an open subset and it contains A because the identity is in U yeah and conversely if we take an open subset W which contains A. So, then this will imply that A inverse of W contains E and since this is equal to L A inverse of W right. Since L A inverse is a homomorphism L A inverse of W is going to be in this collection over here. So, therefore, A inverse W can write as A times inverse of W right. So, therefore, even this W is obtained is in this collection. So, similarly U times A is the collection by the same reason containing. So, let us make this third observation which is a lemma which is a very useful lemma. So, let V containing identity P an open subset then there is an open set which contains the identity such that the set U times U this is equal to all those A B with A and B in U. So, this is equal to U square right. So, we have U square is going to be containing. So, give it any open set V which contains the identity we can find a smaller open set U such that U square is contained in V. So, for instance if we take the real line under multiplication I am sorry under addition and we take the neighborhood 0. So, if we take any if we take an open subset which contains 0. So, let us say B 0 epsilon right this V then we can take U to be B 0 epsilon by 2 right. So, the multiplication over here is just the addition. So, if we take any x comma y in U then clearly x plus y is in V. So, let us prove this. So, consider the multiplication map from G cross U G this is the multiplication map this is which is continuous. So, this implies that the inverse image of V is an open subset and it contains this element right. In particular there is a basic open subset which contains this element and it is contained in M inverse V. So, this implies there exists U 1 containing E and U 2 containing E such that U 1 cross U 2 is contained in M inverse V. So, we just take U to be U 1 intersection U 2 right. So, then U cross U is containing M inverse V which implies that U times U is contained in M. So, let us take let us prove make the fourth observation let V be an open set element x in G. So, then there exists an open set U which contains the identity such that U is equal to U inverse and. So, U inverse is the set of U inverse where U belongs to U yeah or equivalently this I of U. So, U is equal to U inverse and U x U is containing. So, the proof is very similar to the previous lemma. So, let us prove this. So, proof. So, consider the map G cross G cross G to G right. So, we can first project to the first two coordinates ok sorry not this is this a comma b comma c maps to a b comma c which maps to a b c right. So, this map is continuous this is easily checked right. So, we restrict it to. So, we restrict this to the substrates G cross x cross G. So, this composite is then a comma x comma b maps to a x b right. So, this implies that. So, let us call this composite map f this is homomorphic to G cross G. So, since f is continuous. So, this implies that f inverse v is open. So, again this implies that there exists f inverse v is open and it contains the element identity x identity. So, this implies that there exists open sets U 1 containing identity and U 2 containing identity such that U 1 cross x cross U 2 this x is really playing no role over here right. So, we could have simply considered the map G cross G to G given by a comma b maps to a x b right. And once you prove that this maps continuous we can do this entire argument. So, is contained in f inverse v right. So, we take U naught to be equal to U 1 intersection U 2. So, we take U naught first. So, this will imply that U naught cross x cross U naught is contained in f inverse v which implies that U naught x U naught is contained right. But U naught may not have the property that U naught inverse may not be equal to U naught. So, to rectify this we take U to be equal to U naught intersected U naught inverse which is equal to U naught intersected I of U naught right. Since, I is a homomorphism I of U naught is also an open subset. So, both these contain the identity. So, the intersection is an open subset which contains identity right and it is easily checked that. So, then it is easily checked U is equal to I of U and obviously, U x U is going to be contained. So, now that we have these lemmas in place we will prove this interesting ok we need one more lemma let H be contained in G be a closed subgroup. So, what this means is that H is a closed subspace of G in the topology on G and also H is a subgroup. So, let X be an element which is not in H ok. So, then there exists U contained in containing identity open such that U is equal to U inverse U x U intersected H is empty. So, let us prove this. So, since H is closed and X does not belong to H right. So, this implies there exists an open set V and we can simply take the complement of H containing X such that V intersected H is empty ok. So, 4 this previous lemma that we proved this lemma this 0.4 implies there exists U open which contains identity such that U x U is contained inside right. So, this implies that U x U intersected H is empty. So, using these preliminary results we prove the following proposition. Now, let H contained in G be a closed subgroup. So, then G mod H with the quotient topology. So, let us recall that G mod H is equal to G mod equivalence where X is equivalent to Y every subgroup defines an equivalence relation X is equivalent to Y if and only if Y inverse X belongs to H. So, we have this equivalence relation on G and G is a topological space. So, we can look at G mod equivalence and give it the quotient topology and our claim is that the space is host of. So, let us prove this let us take 2 points in the in G mod H which are distinct. So, B 2 cos X. So, then this Y inverse X it does not belong to H right. So, choose a neighborhood E of identity open such that U is equal to U inverse and U Y inverse X times U intersected H is empty. So, we can use we can do this using the previous lemma right. So, note that this happens. So, note that U Y inverse X U intersected H is empty if and only if X U H intersected Y U H is empty ok. So, this is an easy set theoretic check. So, let us do this. So, suppose let us assume this let us assume this is empty and prove this that this intersection is empty if not then there is X U 1 H this is equal to Y U 2 H 1 H 2 right. This implies that U 2 inverse Y inverse X U 1 is equal to H 2 H 1 inverse right this implies that U since U 2 is in U U 2 inverse is also in U because U is equal to U inverse intersection H is not this is a contradiction. So, similarly we can prove the other way let us assume this is empty and let us prove this. So, if this is not empty then there is U 1 Y inverse X U 2 is equal to H. So, this implies that X U 2 is equal to Y U 1 inverse H, but this implies that X U H intersected Y U H is not which is a contradiction. So, thus we see that this holds ok. So, now if pi from g to g mod H do not know the natural map then note that pi inverse of pi of X U H is equal to X U H right. In fact, one easily checks that pi inverse of pi of A is equal to A times H right. So, this is equal to union of H in H A times H right for every subset right. So, this is an easy check which I will leave it to you and using this check. So, what we will get is this check implies pi inverse of pi of X U H is equal to X H multiplied with H right, but H times H is equal to H since H is a subgroup. So, this is just equal to X U H. So, similarly pi inverse of pi Y U H is equal to Y U times H. So, now as X U H is equal to union H in H X U H right and these sets are open and right translations by elements of H are homomorphisms. So, each of these sets is open and we are taking a union of open sets. So, this implies that this set is open right which implies by the definition of quotient topology pi of X U H is an open subset g mod H which contains X H this coset right. Similarly, pi of Y U H is an open subset which contains the coset Y H right. So, that is we have. So, we have our X H here we have our Y H here we have constructed two open subsets of these two in the quotient topology and we claim that these are disjoint we claim that these are disjoint if not. So, let us assume they are not disjoint. So, what we have we have pi of X U H intersected pi of Y U H is non-empty right. Since pi is surjective that will mean that when we take pi inverse of pi of X U H intersected pi of Y U H is non-empty. This implies that pi inverse of pi of X U H intersected pi inverse of pi of Y U H is non-empty. So, this is non-empty as pi is surjective, but this is equal to X U H and this is equal to Y U H as we saw and this intersection is empty which is a contradiction. So, thus these two open neighborhoods and so, this shows that G mod H is sourced off. So, this completes the proof of the proposition.