Topology: More on Algebra and Topology

We’ve arrived at the domain where topology meets algebra. Thus we have to proceed carefully to ensure that the topology of our algebraic constructions are well-behaved.

Let’s look at topological groups again. Our first task is to show that the topologies of subgroups and quotient groups commute.

Proposition 1. Suppose N is a normal subgroup of G. If H is a subgroup of G containing N then there’re two ways to obtain the topology on H/N:

apply quotient topology to G/N, then subspace topology to H/N;

apply subspace topology to H, then quotient to H/N.

The two topologies are identical.

This follows from the more general fact that if p : X → Y is an open quotient map, then for any subspace $Z\subseteq Y,$ the restriction to $q = p|_{p^{-1}(Z)} : p^{-1}(Z) \to Z$ is also a quotient map.

Once again, the fact that p is open is critical.

[ To see why, suppose V is a subset of Z such that $q^{-1}(V)$ is open. Hence $q^{-1}(V) = p^{-1}(Z) \cap U$ for some open subset U of X. Then p(U) is open. We claim p(U) ∩ Z = V. Indeed, if $x\in U$ and $p(x)\in Z,$ then $x \in U\cap p^{-1}(Z) = q^{-1}(V)$ so p(x) is in V. Conversely, if $y\in V \subseteq Z,$ then since q is surjective, pick $x\in q^{-1}(V) \subseteq U$ such that q(x)=y. Hence y lies in Z as well as p(U). ]

Isomorphism Theorems

For a group homomorphism f : G → H, we saw that this induces a continuous bijective group homomorphism G/ker(f) → im(f). What about the remaining two isomorphism theorems of group theory?

Proposition 2. Suppose H is a subgroup and N a normal subgroup of the topological group G. Then $HN = \{xy : x\in H, y\in N\}$ is a subgroup of G and H/(H ∩ N) → (HN)/N is a bijective continuous homomorphism of topological groups.

Proof.

The composition H → HN → HN/N is surjective and has kernel H ∩ N. Thus the resulting map H/(H ∩ N) → HN/N is a continuous bijective homomorphism. ♦

As the reader may suspect, the resulting map is not a homeomorphism in general. For example, consider G = R × R with subgroups N = Z × Z and H = set of all real multiples of (1, √2). Then H/(H ∩ N) = H is isomorphic to the real line. On the other hand, (H+N)/N is not homeomorphic to H. Indeed, by proposition 1, (H+N)/N inherits the subspace topology from $G/N \cong S^1 \times S^1$ and is a dense subset:

Next, we have:

Proposition 3. Suppose $N\subseteq H$ are both normal subgroups of G. Then we have an isomorphism of topological groups (G/N)/(H/N) → G/H.

Proof.

The canonical map G/N → G/H is continuous and surjective, with kernel = H/N, so we do get a bijective continuous homomorphism of groups (G/N)/(H/N) → G/H.

To prove that the reverse G/H → (G/N)/(H/N) is continuous, it suffices by universal property to show: composing with the quotient map G → G/H gives a continuous map. But this is obvious since it’s the result of composing G → G/N → (G/N)/(H/N). ♦

Summary.

When topology is taken into account, the first and second isomorphism theorems gives continuous bijective homomorphisms of the underlying objects, but the third isomorphism theorem gives an actual isomorphism.

Other Algebraic Objects

Let’s look at some other algebraic objects with topology added.

Example 1: Topological Ring.

Definition. A topological ring is a ring R equipped with a topology such that the addition and product maps are continuous.

What about subtraction? Fortunately, a ring has -1 so subtraction a–b = a + (-1)×b is continuous. Examples of topological rings include Z (discrete topology), Z[1/2] = {a/2ⁿ : a integer, n positive integer}, and the p-adic integers Z_p, which will be covered later.

Since we said nothing about the inverse map (x → 1/x) being continuous, the group of units R* may not be a topological group if we let it take the subspace topology from R. The usual trick is to embed R* → R × R via x → (x, 1/x) instead and give R* the subspace topology from R × R. Now inverse is merely swapping of two coordinates so it’s continuous.

Consistency of Inverse

If inverse on R* were already continuous as a subsapce of R, then we get precisely the same topology. Specifically, let T (resp. T’) be the subspace topology from R (resp. R × R) and let f : (R*, T) → (R*, T’) be the identity map.

From the universal property of subspaces (see exercise after prop. 5 here), f is continuous iff $i'\circ f: (R^*, T) \to R\times R$ is continuous. But this map takes x to (x, 1/x) which is continuous since inverse is continuous on (R*, T). Hence, f is continuous.
Conversely, the inverse of f is continuous iff $i\circ f^{-1}:(R^*, T')\to R$ is continuous; the latter map takes (x, 1/x) to x, which is clearly continuous.

Example 2: Topological Field.

A topological division ring / field, is one equipped with a topology such that the addition, product and reciprocal (x → 1/x) maps are continuous. [ The final map has to be restricted to the subspace of non-zero elements. ]

The division map (x, y) → x/y is also continuous since it’s a composition of continuous maps. The most common topological fields are R, C and the extensions of p-adic fields.

Example 3: Topological Vector Space.

A topological vector space over a topological field K is a vector space V equipped with a topology such that vector addition (V × V → V) and scalar multiplication (K × V → V) are continuous. Topological vector spaces are a huge topic in functional analysis, and they’re a generalisation of normed vector spaces.

One can show that if V and W are n-dimensional topological vector spaces over R (n finite), then V and W are isomorphic. However, things are far more complicated for infinite-dimensional vector spaces. In particular, two vector spaces with the same dimension can have different topological properties.

Example 4: Continuous Action of a Group

A topological group G is said to act continuously on a topological space X if the underlying group action G × X → X is continuous. Thus for each $g\in G,$ the group action $l_g:X\to X, x\mapsto gx$ is a homeomorphism. Indeed, the inverse is given by $l_{g^{-1}}$ which is continuous.

Thus, a group action gives rise to a group homomorphism G → Homeo(X), although the converse isn’t true.

The typical constructions on algebraic objects can be extended to that of algebraic objects with topology.

Products

For instance, if V and W are topological vector spaces over K, then so is V × W. Let’s show, as an example, that scalar multiplication m : K × (V × W) → V × W is continuous. The proof merely relies on the universal property of products, i.e. it suffices to show that composing with projections

$\pi_V\circ m : K\times V\times W \to V$ and $\pi_W\circ m:K\times V\times W \to W$

give continuous maps. But these maps take (c, v, w) to cv and cw respectively, and the result follows from continuity of scalar multiplication on V and W.

Likewise, one can show that if R and S are topological rings, then so is R × S.

Quotients

Quotients are fine thanks to the fact that for algebraic objects, the quotient map p : A → A/B is usually open, which allows us to use the algebraic quotient lemma. E.g. since the canonical map p : G → G/H is open for any subgroup H of G, so is R → R/I for any ideal I of R, as is V → V/W for any vector subspace W of V. We’ll look at two examples to show that the quotients induce continuous maps as well.

Example 1.

Let I be an ideal of R. We claim that the induced multiplication map:

$m' : (R/I)\times (R/I) \to R/I$

is continuous. To that end, we use the algebraic quotient lemma to show that the canonical map q : R × R → (R/I) × (R/I) is a quotient map. Hence m’ is continuous if and only if m’q : R × R → R/I is continuous. But m’q is identical to composing ring product R × R → R with canonical R → R/I, which is clearly continuous.

Example 2.

Consider a group action G × X → X such that the action of the normal subgroup $N\triangleleft G$ is trivial. This induces an action m’ : (G/N) × X → X which we claim is continuous. Since q : G × X → G/N × X is a quotient map, it suffices to show m’q is continuous. But m’q is precisely the original group action G × X → X, so we’re done.

Isomorphism Theorems

The three isomorphism theorems generalise in a similar manner. E.g. for a ring R, if $I\subseteq J$ are ideals of R, then (R/I)/(J/I) is isomorphic to R/I as topological rings. Indeed, from classical ring theory, we have an isomorphism for the underlying ring structure. For the topology, we look at the underlying additive groups; by proposition 3, (R/I)/(J/I) and R/I are isomorphic topological groups. Hence, the two structures are isomorphic topological rings.

Likewise, for any subspaces $W\subseteq W'$ of topological vector space V, (V/W)/(W’/W) and V/W’ are isomorphic vector spaces.

Optional Case Study: Connectedness of SO(n).

Here’s a well-known example from manifold theory: the proof that SO(n) is connected.

First we define O(n) to be the set of all n × n matrices M with real entries such that $M^t M = I.$ This is a group under matrix multiplication since:

$M, N \in SO(n)\implies (MN)^t (MN) = N^t M^t M N = N^t N = I$ and
if $M\in SO(n)$ then M^t is its inverse and $M^t \in SO(n).$

O(n) becomes a topological group if we provide it the subspace topology from $\mathbf{R}^{n^2}$ since the product and inverse maps are all given by polynomials in the matrix entries (inverse is particularly easy: it’s just the transpose).

Summary. O(n) is a topological group, called the orthogonal group.

Next, we look at the geometry behind O(n). Denote the column vectors of $M\in O(n)$ by:

$M = (\mathbf{v}_1 | \mathbf{v}_2|\ldots |\mathbf{v}_n)$ ;

then $M^t M = I$ just means

$\mathbf{v}_i \cdot \mathbf{v}_j = \mathbf{v}_i^t \mathbf{v}_j =\begin{cases} 1, &\quad \mbox{if } i =j,\\ 0, &\quad\mbox{if }i\ne j.\end{cases}$

We also write this as $\mathbf{v}_i\cdot \mathbf{v}_j = \delta_{ij},$ where δ_ij is the Kronecker delta function which returns 1 if i=j and 0 otherwise. In other words, M maps the standard basis to the orthonormal basis $\{\mathbf{v}_1, \ldots, \mathbf{v}_n\}.$ Hence a matrix in O(n) preserves the geometry of the Euclidean space by leaving distances and angles invariant.

Property 1. O(n) is compact.

Proof.

Indeed, $O(n)\subset \mathbf{R}^{n^2}$ is closed since it’s defined by explicit polynomial equations in the matrix entries $p_{11} = 1, p_{12} = 0, \ldots.$ On the other hand, O(n) is bounded since each column vector of $M\in O(n)$ has unit length and thus each entry of M is in [-1, +1]. ♦

Next, consider the determinant map det: O(n) → R* which is a continuous group homomorphism. Since:

$1 = \det(I) = \det(M^t M) = \det(M^t) \det(M) =\det(M)^2$

the image of det is {+1, -1}. The kernel is denoted by SO(n), called the special orthogonal group.

Note. On an intuitive level, the special orthogonal group SO(n) comprises of rotations in n-space, while the general orthogonal group O(n) includes the reflections, which have det -1.

Note that since det(O(n)) is not connected, neither is O(n). However, we wish to prove that SO(n) is connected, via the following steps.

Step 1: SO(1) is connected

Obvious since SO(1) = {1}.

Step 2: Let SO(n) act on the Euclidean space.

Thus, $SO(n)\times \mathbf{R}^n \to \mathbf{R}^n$ takes $(M, \mathbf{x})\mapsto M\mathbf{x}$ which is a continuous action. Let’s compute the isotropy group for x = e_n. This is the set of all $M\in SO(n)$ whose last column is e_n. From $M^t M=I$ we see that M must be of the form:

$M = \begin{pmatrix} N & \mathbf{0}\\ \mathbf{0} & 1\end{pmatrix}$ , where $N\in SO(n-1).$

Hence, this gives a bijective continuous map from SO(n)/SO(n-1) to the orbit of e_n. What might this orbit be? First, Me_n is the last column of M so it’s a unit vector. Conversely, any unit vector x can be extended to an orthonormal basis via the Gram-Schmidt process. Thus, the orbit is precisely:

$S^{n-1} = \{\mathbf{x} \in \mathbf{R}^n : ||x|| = 1\}.$

[ The superscript n-1 comes from the fact that the space is (n-1)-dimensional, even though the ambient space is Rⁿ. ]

In short, we have a continuous bijective map:

$\phi:SO(n) / SO(n-1) \to S^n.$

Step 3: The map is a homeomorphism.

Indeed, since O(n) is compact, so is SO(n) – being a closed subset. Hence, SO(n)/SO(n-1) is also a compact space. Now a continuous map from a compact space to a Hausdorff space must be a homeomorphism.

Step 4: The (n-1)-sphere S^n-1 is connected.

Consider the two hemispheres:

$S^+ = \{(x_1, \ldots, x_n)\in S^{n-1} : x_1 \ge 0\}$ and $S^- = \{(x_1, \ldots, x_n)\in S^{n-1} : x_1 \le 0\}$ .

The projection map $S^+ \to \mathbf{R}^{n-1}, (x_1, \ldots, x_n) \mapsto (x_2, \ldots, x_n)$ which drops the first coordinate is injective and continuous. The image is precisely the unit disc $D:=\{(x_2, \ldots, x_n) : x_2^2 + x_3^2 + \ldots + x_n^2 \le 1\}.$ Since S⁺ is compact, the projection map is thus a homeomorphism onto D and so S⁺ is connected.

By the same token, S^– is connected too. Since $S^+ \cap S^- \ne \emptyset$ , their union S^n-1is also connected.

Step 5: Assuming SO(n-1) and SO(n)/SO(n-1) are connected, so is SO(n).

More generally, if H is a subgroup of topological group G such that H and G/H are connected, so is G.

Indeed, suppose $G = U\cup V$ for disjoint open subsets. If U contains a point x, it must contain the entire coset xH; otherwise, $xH = (xH \cap U)\cup (xH \cap V)$ would be a partition of xH into disjoint open subsets, contradicting connectedness of xH.

Hence, U and V comprise of disjoint unions of cosets. So if p : G → G/H denotes the canonical map, we have $p(U)\cap p(V) = \emptyset.$ Since p is open, $G/H = p(U)\cup p(V)$ for disjoint open subsets p(U), p(V) of G/H. This must mean $p(U)=\emptyset$ or $p(V)=\emptyset,$ so U or V is empty.

This completes the induction, and we’ve shown that SO(n) is connected for each n.

Topology: More on Algebra and Topology

Isomorphism Theorems