Extension of a topological group

In mathematics, more specifically in topological groups, an extension of topological groups, or a topological extension, is a short exact sequence $$ 0\to H\stackrel{\imath}{\to} X \stackrel{\pi}{\to}G\to 0 $$  where $$H, X$$ and $$G$$ are topological groups and  $$i$$ and  $$\pi$$  are  continuous homomorphisms which are also open onto their images. Every extension of topological groups is therefore a group extension.

Classification of extensions of topological groups
We say that the topological extensions
 * $$0 \rightarrow H\stackrel{i}{\rightarrow} X\stackrel{\pi}{\rightarrow} G\rightarrow 0$$

and
 * $$0\to H\stackrel{i'}{\rightarrow} X'\stackrel{\pi'}{\rightarrow} G\rightarrow 0$$

are equivalent (or congruent) if there exists a topological isomorphism $$T: X\to X'$$ making commutative the diagram of Figure 1.



We say that the topological extension


 * $$0 \rightarrow H\stackrel{i}{\rightarrow} X\stackrel{\pi}{\rightarrow} G\rightarrow 0$$

is a split extension (or splits) if it is equivalent to the trivial extension


 * $$0 \rightarrow H\stackrel{i_H}{\rightarrow} H\times G\stackrel{\pi_G}{\rightarrow} G\rightarrow 0$$

where $$i_H: H\to H\times G$$ is the natural inclusion over the first factor and $$\pi_G: H\times G\to G$$ is the natural projection over the second factor.

It is easy to prove that the topological extension $$0 \rightarrow H\stackrel{i}{\rightarrow} X\stackrel{\pi}{\rightarrow} G\rightarrow 0$$ splits if and only if there is a continuous homomorphism $$R: X \rightarrow H$$ such that $$R\circ i$$ is the identity map on $$H$$

Note that the topological extension $$0 \rightarrow H\stackrel{i}{\rightarrow} X\stackrel{\pi}{\rightarrow} G\rightarrow 0$$ splits if and only if the subgroup $$i(H)$$ is a topological direct summand of $$X$$

Examples

 * Take $$\mathbb R $$ the real numbers and $$\mathbb Z $$ the integer numbers. Take  $$\imath $$ the natural inclusion and $$\pi $$ the natural projection. Then


 * $$ 0\to \mathbb Z\stackrel{\imath}{\to} \mathbb R \stackrel{\pi}{\to}\mathbb R/\mathbb Z\to 0 $$


 * is an extension of topological abelian groups. Indeed it is an example of a non-splitting extension.

Extensions of locally compact abelian groups (LCA)
An extension of topological abelian groups will be a short exact sequence $$ 0\to H\stackrel{\imath}{\to} X \stackrel{\pi}{\to}G\to 0 $$ where $$H, X$$ and $$G$$ are locally compact abelian groups and  $$i$$ and  $$\pi$$  are relatively open continuous homomorphisms.
 * Let be an extension of locally compact abelian groups
 * $$ 0\to H\stackrel{\imath}{\to} X \stackrel{\pi}{\to}G\to 0. $$
 * Take $$H^\wedge, X^\wedge$$ and $$G^\wedge$$ the Pontryagin duals of $$H, X$$ and $$G$$ and take $$i^\wedge$$ and  $$\pi^\wedge$$   the dual maps of $$i$$ and  $$\pi$$. Then the sequence
 * $$ 0\to G^\wedge\stackrel{\pi^\wedge}{\to} X^\wedge \stackrel{\imath^\wedge}{\to}H^\wedge\to 0 $$
 * is an extension of locally compact abelian groups.

Extensions of topological abelian groups by the unit circle
A very special kind of topological extensions are the ones of the form $$0 \rightarrow \mathbb T\stackrel{i}{\rightarrow} X\stackrel{\pi}{\rightarrow} G\rightarrow 0$$ where $$\mathbb T$$ is the unit circle and $$X$$ and $$G$$ are topological abelian groups.

The class S(T)
A topological abelian group $$G$$ belongs to the class $$\mathcal S (\mathbb T)$$ if and only if every topological extension of the form $$0 \rightarrow \mathbb T\stackrel{i}{\rightarrow} X\stackrel{\pi}{\rightarrow} G\rightarrow 0$$ splits
 * Every locally compact abelian group belongs to $$\mathcal S (\mathbb T)$$. In other words every topological extension $$0 \rightarrow \mathbb T\stackrel{i}{\rightarrow} X\stackrel{\pi}{\rightarrow} G\rightarrow 0$$ where $$G$$ is a locally compact abelian group, splits.
 * Every locally precompact abelian group belongs to $$\mathcal S (\mathbb T)$$.
 * The Banach space (and in particular topological abelian group) $$\ell^1$$ does not belong to $$\mathcal S (\mathbb T)$$.