 So, now we will see an application of the above result. So, let g be equal to g l n r then g is a group and has a topology. So, we claim that g is a topological group. So, let us check that. So, we need to show that g l n r cross g l n r to g l n r the multiplication map is continuous. So, now note that this has a product topology. So, in particular we know that the projection maps are continuous. So, we can look at the projection maps g l n r this is some p i right and let us say on g l n r we have these continuous maps which project on to the coordinates. So, let us call this q i j's right. So, given a matrix a this map q i j it sends it to a h. So, this implies that the composite of these two is continuous so therefore, we get that from g l n r cross g l n r r. So, a comma b maps to a i j or similarly b i j these are continuous and this multiplication map m the coordinates of m are polynomials in the functions a i j and b i j right. And we have seen that if we take a collection of continuous maps and we form any polynomial using those continuous maps. So, like yeah. So, then the resulting map function is continuous right. So, this implies that m is continuous. So, similarly let us take g l n r let us look at the inverse map this a goes to a inverse right. So, we know the inverse is given by this formula 1 upon determinant of a into adjoint of a transpose. So, now once again the determinant does not vanish on g l n r. So, 1 upon determinant is a continuous function and this adjoint of a transpose this is also each of the coordinates is a continuous function in the coordinates of this g l n r therefore, the product is a continuous function right. So, this implies that g l n r is a topological. Now let p be this parabolic subgroup. So, here we have a r cross r matrix here we have 0 and these can be anything right. So, we are looking at all p contained in g is equal to g l n r right be this subgroup. So, all elements in g cross in g which look like this right. So, clearly p is a close up right it is given by as the inverse image of the projection on to this this piece projection of 0 on the inverse image of 0 under the projection on to this piece ok. So, therefore, p is a close up group. So, this implies that. So, this thus the previous discussion g mod p is a post of topological space. So, now we want to identify g mod p we want to we want to give a nice description of the points of g mod p which is what we are going to do next. So, for that consider the map from g is 5 to r dimensional sub spaces right. So, let us call this set script g. So, here what is the map. So, let us take a matrix A and let us write A as v 1 the v i's are the columns of A up to v n right. So, since A is in g l n r. So, g is g l n r this implies that the column vectors of A are linearly independent. So, we just send it to the span of v 1 up to v r. Now, we claim that phi of A is equal to phi of B if and only if there exists p in p such that ok. So, let me just write t a matrix t in our parabolic subgroup such that B is equal to A times t. So, let us see this. So, if there is we take 2 meters is A and B. So, it is B is equal to A times t right. So, then then clearly phi of A is equal to phi of B right because as we if we take B equal to write as w 1 w 2 up to w n right. So, this is equal to b 1 b 2 up to b n times this parabolic subgroup this r cross r 0 ok we can have anything right. So, this will show that this is A times t right. So, this will imply that the w 1 up to w r linear combinations of v 1 up to v r right and both are r dimensional subspaces right. So, this implies that the span w 1 up to w r is equal to right. So, this shows that phi of A is equal to phi of B. Next let us assume that phi of A is equal to phi of B and then show that right. So, if phi of A is equal to phi of B. So, then we have that the span of w 1 up to w r is equal to the span of v 1 up to b r. Now, since the w 1 up to w n they span r n and v 1 up to b n also span r n right since w i's and v i's are basis for r n this implies that there exists a unique t we have a unique matrix t in g l and r which can be obtained as follows. So, what we do is we write each w i. So, w 1 w 2 w n we can write it as a unique linear combination of the vectors v 1 up to v n right. So, if I write w 1 as lambda 1 v 1 plus lambda 2 v 2 plus lambda n v n right. So, then the first column of t will be lambda 1 up to lambda n and this matrix t will be clearly has to be in g l and r because we can do the same we can replace the roles of w and v and the composition the resulting matrix t prime that we get that is going to compose with t to give identity right. So, is the unique way to express w i as a linear combination of the v j's. So, we just have to show that t is in t. So, note that since w i we are given this information right that implies that is in the span of v 1 through v r right. So, this implies that the ith column of t is going to look like this is r right and then everything will be 0. So, and this happens for i lying between 1 and r right. So, this clearly implies that t belongs to b. So, clearly so this is so this proves this claim we have proved this claim. So, let us call this 1 next the second point we want to show is clearly phi is surjective. So, as so phi is this map from g l a matrix a it goes to the span of the first r column vectors right. Now, given any r dimensional subspace of r n we can choose a basis for that r dimensional subspace and v 1 up to v r and we can extend it to a basis of r n and correspondingly we can get a matrix a right as any basis v 1 up to v r for an r dimensional subspace can be extended to a basis v 1 up to v n for r n right. So, which implies we can take the matrix a to be v 1 v 2 ok. So, this shows that phi is surjective so this implies that. So, when we put together 1 and 2 one can easily check that we have this map phi to script g right and this factor is through g mod p and this map is a bijection let us call this phi naught this easily follows from 1 and 2 right that phi naught is a bijection. So, thus the points of g mod p are bijection with r dimensional subspaces. So, which means so using this bijection we can give this space of r dimension subspace of r n this set g we can give g a to be g. So, we identify g we identify g with g mod p using this phi naught and a subset of g is open if and only if it is inverse image under phi naught is open. So, clearly so g mod p is a topology and this bijection using this bijection we can transfer the topology. So, this topological space the R dimensional planes and it is denoted R comma n. So, by the proposition we proved we know that g mod p is host of yeah we get that grasmanian R comma n is host of and further note that also. So, note that g l and R plus surgex onto this grasmanian R comma n right because why is that. So, given an r dimensional subspace. So, given an r dimensional subspace we chose a basis v 1 through v r for the subspace and we extended it to a basis. So, we get this matrix a. So, if the determinant of a is negative then we just take the last column and put a minus sign take instead of v n we just take minus v n right and we had seen right as g l and R plus is path connected this implies that this grasmanian is path connected. And similarly the Gram-Schmidt orthogonalization process implies that o n the orthogonal matrix surgex onto this grasmanian R comma n right. So, why is this. So, we have o n this sitting inside g l and R right and here we have this map 5. So, we claim that this composition is surjective. So, why is that if we take any r dimensional subspace. So, first so we can we can choose a basis v 1 up to v r then applying the Gram-Schmidt orthogonalization process. We can get an orthogonal basis orthonormal basis for this and then we can extend get orthonormal basis for the subspace and then we can extend this to an extend this extend to an orthonormal basis for R n. So, we get w 1 up to w n right. So, then the matrix a we need to take here is this w 1 w 2 right. Since the w i's are orthonormal we will get that a transpose a is equal to identity right which will imply that a is in o n right. And now so this g is actually equal to g mod p right. Now, as o n is compact right since it is a closed and bounded subspace of R n square. So, this implies that and o n surgex onto g mod p this implies that g mod p being the image of a compact space. So, thus we have proved that the Grassmannians is host of path connected. So, we can also construct we can also do a similar construction for complex Grassmannians. So, there we will take r dimensional complex subspaces of C n and in the same way we can show that this is host of in a similar way path connected. And before I end let me just mention that when r is equal to 1 the spaces Grassmannian 1 comma n are also known as projective spaces. So, if we take complex projective spaces yeah. So, depending on the underlying field if we take we can do this entire construction for the real numbers of a complex numbers. So, if we are yeah. So, that this is the real projective space in this complex space. So, we will end this lecture here.