 Okay, so this is lecture 27 and before we start a discussion on compact metric spaces we will see a nice application of what we have seen so far. So today we will prove a very interesting theorem, so SON is connected. In fact SON is also path connected but that takes a little bit more work and so in this course we will prove that SON is connected. Before we start the proof it is perhaps worth appreciating that the definition of SON is somewhat complicated so recall that SON is all those matrices A in M and R such that A transpose A is equal identity and determinant of A is 1. So to show that if you try and show directly that SON is path connected then it would probably be quite hard. So let us begin the proof. So first consider this map from SON to the sphere Sn-1 okay before that so let E1 in Rn denote the column vector C0 0 okay and consider the map from SON to the N-1 sphere given by we take a matrix A so let us say N is greater than equal to 2 so we take a matrix A and it gets mapped to AE1 right. So we are taking this matrix A in SON and this is getting sent to the first column right. So a priori it is not clear that the image lands in Sn-1 but let us just check that so AE1 so we need to like a priori this is a map from SON to Rn and we need to check the image lands in Sn-1 right. So SON has the subspace topology from M and R and this is just a projection projection on some coordinates right so therefore this map is continuous so this map is continuous right so let us compute the norm of this AE1 the inner product of AE1, AE1 this is equal to so this is the standard inner product inner product on Rn given by V, W is defined to be W transpose times V right. So when we use this so we get this inner product is equal to A times E1 transpose times AE1 which is equal to E1 transpose A transpose AE1 but now A is in SON which means A transpose A is identity so this is equal to E1 transpose E1 which is just E1 right. So this implies that the image lands Sn-1 the norm of this vector AE1 that is 1 right. So we have this SON we have this continuous map to Rn and the image actually lands inside Sn-1 and since Sn-1 has a subspace topology from Rn this implies that this map we have defined SON to Sn-1 is continuous ok. Next let us consider the subgroup H equal to SON-1 sitting inside SON as follows 1 0 0 here 0 here and here we have this matrix SON-1. So we claim that AE1 is equal to BE1 so if you call this map phi so what we are saying is phi of A is equal to phi of B if and only if B inverse of A is in H. So let us check this so if B inverse of A is in H right. So this implies that B inverse of A is equal to H for some H in H this implies that A is equal to B times H simply by multiplying on the left with B. So now we apply both these on E1 so this will imply that AE1 is equal to B times H E1 but now note that H E1 is simply equal to E1 right because when we apply this a matrix of this type on E1 we get back E1. So this implies AE1 is equal to BE1 ok. So now let us prove the converse conversely suppose AE1 is equal to BE1 then multiplying with B inverse on both sides then this implies that B inverse A applied on E1 is equal to E1 ok. So we let C be the matrix B inverse A right so then CE1 is equal to E1 right. So this implies that C is a matrix which looks like this so CE1 is equal to E1 which means the first column looks like this and the others can be anything. Now as A and B are in SON this implies B inverse A is equal to C is also in SON right and from the condition C transpose C is equal to identity. So if we write C as columns C1, C2 up to CN right. So this C transpose C is equal to identity so this becomes C1 transpose C2 transpose CN transpose into this column C1 C2 to CN is equal to identity. So this implies CI transpose Cj is equal to 0 for i not equal to j. So if we now so but this CI transpose Cj that is just the inner product CI. So this implies that the inner product Cj comma C1 is equal to 0 for j not equal to 1 right. But C1 is this column vector 1 0 0 right and Cj so C1 is this column and let us say let us take C2 C2 will be this column. So when we take the inner product C1 and C2 we get this entry this first entry over here right. So since the inner product is 0 this implies that this entry will be 0 and similarly this entry will be 0 and similarly all these entries will be 0. So as C1 is this from this we conclude that C is this matrix looks like this and here we have whatever right. So this implies that C belongs to H. So thus B inverse A belongs to H which is what we want to prove okay. Next we claim that the map SON to SN minus 1 is subjected okay. So recall that given a vector V with norm V is equal to 1. So V is in Rn right we may extend it or the normal basis V is equal to V1 V2 up to Vn of Rn right and so now let A be equal to matrix V1. We write these vectors as column vectors and we let this matrix A. So A is an n cross n matrix right. So then clearly so this matrix A transpose A is going to be equal to this matrix obtained by VI taking the inner products of VI and VJ. This is an easy check right and this is clearly equal to identity because the VI is they form an orthonormal basis right. So if determinant of A is equal to minus 1 then we simply replace the last column by negative of that then A prime be the matrix V1 V2 up to minus Vn right. So then we easily check that A prime transpose A is equal to identity and determinant of A prime is equal to 1 right. So moreover the first column of A prime is V1 is equal to V is equal to V right. So this proves that so this is my phi this proves that phi is subjective and we need one more ingredient. So let us look at that. So if A is in SON then consider the map from SON to SON. This is given by left translation by elements of A. So precisely this map is it sends a matrix B in SON to A times B. So it is easily checked that if A and B are in SON then obviously A times B is in SON. If SON is a subgroup of GLNR. So let us check that this map is continuous. So to show this map is continuous. So note that SON has a subspace topology from MNR. So it is enough to check that this LA after we compose with this inclusion right. So this composite is continuous right. But this composite map this dotted arrow it factors like this to LA right. So both these triangles commute. We can take this matrix A. This matrix A is fixed and we can define left multiplication by A on MNR itself yeah. So here also it is A goes to A right. But clearly this map is continuous because all the coordinates they are just some linear combinations of the coordinates of AB are simply linear combinations of the coordinates of B. So therefore this horizontal map is continuous. As a result when we look at this map that is continuous because we have just restricted this continuous map to SON. Which means the dotted arrow is continuous which means the map LA which we started with that is continuous right. So similarly LA inverse so let me just write yeah. So let us call this map AB LA let us call this LA tilde yeah. So to show LA is continuous enough to show B compose LA is continuous right. But B compose it is enough to show B compose LA is continuous because SON has a subspace topology from MNR right. But B compose LA is equal to this dotted arrow A which is equal to LA tilde compose C right. And LA tilde is continuous because LA tilde it is enough to check what the coordinate functions of LA tilde are continuous. And so and C is just a restriction of LA tilde to SON this implies LA tilde compose C is continuous. So this implies LA is continuous okay. So similarly LA inverse is continuous. But clearly we have LA compose LA inverse is equal to identity which is also equal to LA inverse compose LA right. So thus so we have LA is continuous and LA is a bijective continuous map and it is inverse it is clear is L of A inverse and that is also continuous yeah. So thus LA is a homomorphism okay. So now with these ingredients we are ready to prove that SON is connected. So we will prove the theorem by induction on M. So the base case is N equal to 1. So in this case SON is simply this is one element which is obviously connected. So assume we have proved that N is greater than or equal to 2 and we have proved that SOK is connected for okay. So let us assume that N is greater than or equal to 1 for K lying between 1 and N okay. So now we will show SO N plus 1 is connected okay. So consider the map SO N plus 1 to SN right. So note that N as N is greater than or equal to 1 we have N plus 1 is greater than or equal to 2. So therefore we can consider this map which we analyzed before. So if SO N if SO N plus 1 is disconnected then we can write it as the disjoint union of 2 non-empty disjoint non-empty open sets and all and so also closed. So let us pick any element of SO N plus 1 right and let H be the subgroup SO N contained inside SO N plus 1. So A is in SO N plus 1. So A is either in U or it is in B. So assume that A belongs to U right. So then we can take this subset AH and we can write it as AH intersected U disjoint union AH intersected B. So as LA so LA is from SO N plus 1 to SO N plus 1 is a homomorphism and we have H sitting over here LA sends it to AH sitting over here right. So LA is a homomorphism. So therefore it is a homomorphism this implies that LA restricted to H this map from AH to AH is also a homomorphism. So and as H is connected so H is homomorphic to SO N and we are assuming by induction that SO N is connected. So as H is connected and AH is homomorphic to H this implies that AH is also connected right. So thus one of these two open sets has to be empty but A belongs to this set. So this implies that this set over here has to be empty or in other words AH is completely contained inside U right. So we have proved that we have proved that if A belongs to U then AH is completely contained inside U right. Similarly if B belongs to V then BH is completely contained inside okay. So now let's consider this map from SO N plus 1 SM right. So U is closed SO N plus 1 is compact implies U is compact right which implies 5U is compact in SN yeah which implies that 5U is closed right. So similarly 5V is also closed. So as P is subjective this implies SN is equal to P of U union P of V. So we claim that this union is disjoint is disjoint. If not there exists some vector V in P of U intersection P of V right. So this implies that there exists A in U and B in V such that phi of A is equal to AE1 is equal to V is equal to BE1 is equal to phi of P right. But now we know that this happens this will imply that B inverse A belongs to H which implies that A belongs to B of H and since B belongs to V, B of H is contained in U contained in V right. So which is which is a contradiction because we assume that A is in U and U and V are disjoint. So thus we have written if SO N plus 1 is disconnected then SN is going to be P of U disjoint union P of V right. We can write it as a disjoint union of non-empty closed subsets which means it will be also be the disjoint union of non-empty open subsets which means SN will be disconnected yeah. But this contradicts but we have seen before that SN is connected the fact that SN is connected. So thus SN plus 1 is connected. So this proves this completes the proof here. So we will end here.