Topological Groups Definition Let G Be a Group Which Is Also Furnished with a Topological Structure. We Say That G Is a Topologi

Topological Groups Definition Let G be a group which is also furnished with a topological structure. We say that G is a topological group if the function (x, y) 7→ xy−1 is a continuous map G × G → G. (We endow G × G with the product topology.) This is same as saying that group multiplication and inversion are continuous maps. Example: As determinant of a matrix is a polynomial function in its entries, GLn(R) GLn(C) are topological groups. Also restriction of continuous maps being continuous, subgroups of topo- logical groups, with their subspace topologies, are topological groups too. But interesting subgroups usually are closed subgroups. The subsets SLn(R) SLn(C) where the determinant is 1, is the inverse image of the singleton closed set {1} under the (continuous) determinant map, and hence is a closed topological subgroup. Exercise: Check that for a topological group inner automorphisms are home- omorphisms, and so are the maps G → G multiplication by any element. Prove that for any subgroup of a topological group, its closure is also a subgroup. Any subgroup H gives rise a partition of a group G in terms of its cosets and so one can furnishe G/H with the quotient topology. Theorem: For any topological group G, and an arbitrary subgroup H, the quotient map π: G → G/H is an open mapping: Proof : Taken an open set U ⊂ G. We’ve to show π(U) is open in G/H. As the latter is given quotient topology we need only to show that π−1π(U) is open. The inverse image of a point being the whole coset we see

π−1(π(U)) = [ Ug g∈G As multiplication maps are homeomorphism Ug is open for each g, and so being the union of all these open sets the set under question is open, proving that the map is an open map. Theorem: If H is closed subgroup of a topological group G then the quo- tient space G/H is Hausdorff. Proof: Consider the continuous map φ: G × G → G defined as φ(x, y) = x−1y. Then H being closed its inverse image is closed too. But x−1y ∈ H iff xH = yH. So we have {(x, y) ∈ G × G | xH = yH} is a closed set. Or that {(x, y) ∈ G × G|xH 6= yH} is open. But the natural map G × G → G/H × G/H is also an open mapping. So we have that {(xH, yH) ∈ G/H × G/H | xH 6= yH} is open. But this set is precisely the complement of the digonal, showing that the diagonal in G/H × G/H is a closed set which is a criterion for a space being Hausdorff. Actions: When G is a topological group and X a topological group we talk of continuous actions of G on X, which is the usual notion of acction of a group on a set with the the extra requirement that the acction G × X → X, (g, x) 7→ g.x is continuous. In this case the ’permutation X → X defined as x 7→ g.x is a homeomorphism. Clearly the isotropy subgroups would all be closed subgroups. But one cannot say that the orbit of every point is a closed set or an open set. The orbits of action are homeomorphic to the quotient spaces by the isotropy subgroups. We will stick to topological groups that are matrix groups. Theorem: SLn(R) is a connected topological group. We will prove this theorem by making use of the following general result. Theorem: If H is a connected subgroup of G such that G/H is also con- nected then G is also a connected group. Proof: Suppose G were not connected. Then there would exist two disjoint non-empty open set U, V such that U ∪ V = G. Denoting by π the map x 7→ xH, we see that G/H = π(U) ∪ π(V ). But we know that π is on open mapping and so we have G/H is the union of two nonempty open sets π(U) and π(V ). Now the assumption of connectedness of G/H implies that π(U) and π(V ) have some common point, say zH. But disjointness assumption on U and V means that, for this to happen there must exist x ∈ U and y ∈ V such that x−1y ∈ H. So the coset xH intersects both U and V in G. As U ∪ V = G provides a disconnection xH has to be disconnected (intersecting both of them). But xH is homemorphic to H. So this contradicts the hypothesis that H is connected. So G is connected. Now to prove the connectedness of SLn(R) we will use this theorem and employ induction on n. For n = 1, the group consists of the single point {1}, which is connected and path-connected and so induction starts. For a general n we claim that any nonzero vector can be sent to anyother non-zero vector by matrix of determinant 1. We will show that the special T n vector e1 = (1, 0, 0,..., 0) ∈ R can be sent to any nonzero vector by SL(n). (Check that this suffices). That given nonzero v ∈ Rn to get a matrix A with det A = 1 such that Ae1 = v But Ae1 is the first column of A. So this boils down to showing that one cna construct a matrix of determinant 1 starting with any given nonzero vector as the first column. This is easily checked. n Now we have shown that R − 0 is an orbit for SLn(R). The isotropy subgroup H is described as

H = {B ∈ SLn(R)| first column of B is e1 }

n We have shown that there is a homeomorphism SLn(R)/H → R − 0. It is seen easily that Rn − 0 is path-connected for n > 1. Now we will show H is also connected and by the above theorem this will prove the connectedness of SLn(R). Now let us go to the description of H: It consists of matrices of the form

 1 ∗ ∗ ... ∗     0  B =    0 C  0

Expanding the determinat along the first column, we see that The stars can be arbitrary real numbers, and C has to be a matrix of determinant 1 and size (n − 1) × (n − 1). So H is homeomorphic to the product space n−1 R × SLn−1(R). By induction hypothesis SL(n − 1) is connected and its product with the Euclidean space is connected completing the proof.