# Abstract Nonsense

## Topological groups (Direct Product and Product Spaces)

Up until now we’ve discussed how the subspace topology reacts with subgroups and how the quotient topology interacts with the quotient group, it seems like a natural progression to then discuss how the product topology reacts with the direct product. So, as in the past we firstly need to verify that the two do agree, namely:

Theorem: Let $G,G'$ be topological groups. Then, $G\times G'$ is a topological group with the direct product group structure and product topology.

Proof: Clearly $G_1\times\cdots\times G_n$ is a set with both a group theoretic and topological structure and so it remains to show that the map

$\alpha:(G\times \cdots\times G_n)\times (G\times \cdots\times G_n)\to G\times \cdots\times G_n$

given by

$\alpha:\left((g_1,\cdots,g_n),(g'_1,\cdots,g'_n)\right)\mapsto (g_1{g'_1}^{-1},\cdots,g_n{g'_n}^{-1})$

is continuous. But, note that

$(G\times \cdots\times G_n)\times (G\times \cdots\times G_n)\approx (G\times G)\times\cdots\times (G_n\times G_n)$

And that

$\alpha_G\times\cdots\times \alpha_{G_n}:(G\times G)\times\cdots\times (G_n\times G_n)$

given by

$\alpha:\left((g_1,g'_1),\cdots,(g_n,g'_n)\right)\mapsto\left(g_1{g'_1}^{-1},\cdots,g_n{g'_n}^{-1}\right)$

being the product of continuous maps is continuous. The conclusion immediately follows. $\blacksquare$

There are canonical topological epimorphisms from $G_1\times\cdots\times G_n$ into $G_k$. Namely:

Theorem: Let

$\pi_k:G_1\times G_n\to G_k:(g_1,\cdots,g_k,\cdots,g_n)\mapsto g_k$

Then, $\pi_k$ is an open topological epimorphism.

Proof: The fact that it’s open, surjective, and continuous follows since topologically $\pi_k$ is merely the canonical projection of a product space onto it’s $k$th coordinate. Thus, it remains to show that it’s a homomorphism. But, this is a routine calculation. $\blacksquare$

Theorem: Let $\pi_k$ be as above. Then,

$\ker\pi_k=G_1\times\cdots\times\underbrace{\{e\}}_{k}\times\cdots\times G_n$

from where it follows from a previous theorem that

$\left(G_1\times\cdots\times G_n\right)/\left(G_1\times\cdots\times\{e\}\times\cdots\times\right)\overset{\text{T.G.}}{\cong}G_k$. $\blacksquare$

Now, so we can use it we should mention in passing that:

Theorem: $G_1\times\cdots\times G_n\overset{\text{T.G.}}{\cong}G_{\sigma(1)}\times\cdots\times G_{\sigma(n)}$ where $\sigma:\{1,\cdots,n\}\to\{1,\cdots,n\}$ is any bijeciton.

Proof: Obvious. $\blacksquare$

In fact, that is all I wanted to mention about them. Unfortunately the interesting stuff about the direct product of topological groups that is interesting isn’t “accessible” and would take too much time to really discuss. Thus, for once I have a post which is less than one-thousand words. Enjoy.

Advertisements

April 29, 2010 -

No comments yet.