Exercise 1

Let  be a topological group with unit . Prove the following statements.

(a)

Let  be a subgroup of . Its closure  is still a subgroup of . If  is normal, so is .

Proof.

Suppose that . We want to prove that

Consider the continuous map

Then

is closed in .

Trivially,

Since  is dense in , and a closed set containing a dense subset contains its closure, we have

Therefore  for every , and hence

Furthermore, suppose that . We want to prove that

For every , define

which is a homeomorphism. Since , we obtain

Thus .

(b)

An open subgroup of  is closed. A closed subgroup of  is open if it has finite index.

Proof.

First, suppose that  is open. For every , the map

is a homeomorphism. Therefore every left coset  is open. Since

the set  is open. Thus  is closed.

Now suppose that  is closed and that

Choose distinct left coset representatives . Every set is closed. Hence

is a finite union of closed sets, so it is closed. Therefore  is open.

(c)

Assume  is connected. Let  be an open neighborhood of . Then  can be generated by , i.e.

Proof.

Recall that

Let

First, , , and . Moreover,  is open.

Let  and . Then

If , where , then

Therefore .

Also,

Each  is open, so  is open. By part (b),  is also closed. Since  is connected and , we conclude that . Finally, since ,

Therefore

(d)

Let  be a connected subgroup of . If  is connected, so is .

Proof.

Let

be the canonical projection. Suppose that  is both open and closed. We prove that  or .

Since  is connected, every left coset  is connected because it is the image of  under the homeomorphism

For every , the set  is both open and closed in . Hence

Therefore  is a union of left cosets of , and

Similarly,

Since  and  are open, the definition of the quotient topology implies that both  and are open. Thus  is both open and closed in .

Since  is connected,

It follows that

Therefore  is connected.

(e)

Assume  is connected. Then any discrete normal subgroup of  is in the center.

Proof.

Let  be a discrete normal subgroup of . For every , define

This map is well-defined because , and it is continuous. Since  is discrete, the singleton  is both open and closed in . Therefore

is both open and closed in .

Obviously,

so . Since  is connected, we obtain

Thus  for every , and hence . Since  was arbitrary,

Exercise 2

Prove that , , ,  are compact.

Proof.

(1)

Let , which is continuous. Then, so  is closed.

Using the Frobenius norm, for every ,

Thus , so  is bounded. Therefore  is compact.

(2)

Since  is continuous,  is closed. Also,, so  is bounded. Therefore  is compact. 3,4 is similar.

Exercise 3

Show that  acts transitively on . Identify the stabilizer of , and obtain.

Proof.

Recall that an action of  on  is transitive if, for every , there exists  such that .

First, define

We prove that for every , there exists  such that .

Extend  and  to orthonormal bases and, respectively. There exists  such that for every . In particular, . Thus the action is transitive.

Now,

Obviously, every  has the form

Since , we have . Therefore

Define

It induces a bijection

Hence

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Exercise 4

The group  has an induced topology from . A lattice in  is a free -submodule of  that generates  as a -vector space.

(a)

Prove that  is compact.

(b)

Prove that  acts transitively on the set of lattices in .

(c)

Prove that  is the stabilizer of the lattice .

(d)

Prove that any compact subgroup of  stabilizes a lattice.

(e)

Deduce that every compact subgroup is conjugate to a subgroup of .

Exercise 5

A topological group is called profinite if

where the limit (product) is taken over all normal subgroups  of  of finite index, and the topology on the limit is the subspace topology induced from the product. We endow each  with the discrete topology and the product with the product topology.

(a)

Let  be profinite. Prove that  is compact and totally disconnected.

(b)

Let  be a norm on . Prove that there exists  such that the only subgroup contained in

is .

(c)

Let  be profinite. Prove that any finite dimensional continuous -representation of  factors through a finite quotient of .