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 the restriction to 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 is open. Hence for some open subset *U* of *X*. Then *p*(*U*) is open. We claim *p*(*U*) ∩ *Z* = *V*. Indeed, if and then so *p*(*x*) is in *V*. Conversely, if then since *q* is surjective, pick 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 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 and is a dense subset:

Next, we have:

Proposition 3. Suppose 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. Atopological ringis a ringRequipped 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*/2^{n} : *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 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 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 the group action is a homeomorphism. Indeed, the inverse is given by 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

and

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:

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 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 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 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 This is a group under matrix multiplication since:

- and
- if then
*M*is its inverse and^{t}

*O*(*n*) becomes a topological group if we provide it the subspace topology from 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 theorthogonal group.

Next, we look at the geometry behind *O*(*n*). Denote the column vectors of by:

;

then just means

We also write this as 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 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, is closed since it’s defined by explicit polynomial equations in the matrix entries On the other hand, *O*(*n*) is bounded since each column vector of 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:

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, takes which is a continuous action. Let’s compute the isotropy group for **x** = **e**_{n}. This is the set of all whose last column is **e**_{n}. From we see that *M* must be of the form:

, where

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, *M***e**_{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:

[ The superscript *n*-1 comes from the fact that the space is (*n*-1)-dimensional, even though the ambient space is **R**^{n}. ]

In short, we have a continuous bijective map:

### 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:

and .

The projection map which drops the first coordinate is injective and continuous. The image is precisely the unit disc 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 , their union *S*^{n-1 }is 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 for disjoint open subsets. If *U* contains a point *x*, it must contain the entire coset *xH*; otherwise, 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 Since *p* is open, for disjoint open subsets *p*(*U*), *p*(*V*) of *G*/*H*. This must mean or so *U* or *V* is empty.

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