Abstract Nonsense

Crushing one theorem at a time

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


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 kth 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.


April 29, 2010 - Posted by | Algebra, Group Theory, Topological Groups, Topology, Uncategorized | , , ,

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