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.

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 ${\mathcal {S}}(\mathbb {T} )$.