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Let $G$ be a group and $\mathcal T$ be a locally compact topology on $G$ such that for any $a\in G$ the functions $x \mapsto ax$ and $x\mapsto xa$ are continuous on $G$.

How to prove elementarily that:

  1. There's some neighborhood $U$ of $1$, the identity element of the group $G$, such that $\overline{UU}$ is compact.
  2. There's some neighborhood $V$ of $1$, such that $\overline{V^{-1}}$ is compact.

?

Assume $\mathcal T$ is Hausdorff.


By elementary I mean not to use Ellis or other strong theorems in (semi-)Topological Groups Theory without proof. Using only theorems in pure Topology or pure Group Theory is what I call elementary. For example I appreciate using any result in Uniform Spaces; because a locally compact Haudorff Space is uniformizable. Also according to this thread it seems there's a minimal uniformity on $G$.

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  • $\begingroup$ Are you allowed to use the fact that locally compact semi topological groups are topological groups? Link: groupprops.subwiki.org/wiki/… $\endgroup$ – Brian Rushton Apr 27 '13 at 2:49
  • $\begingroup$ that result does not have an elementary proof I think. $\endgroup$ – user59671 Apr 27 '13 at 4:35
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I had some attempts to obtain elementary proofs of (1) or (2), but I failed. Maybe these proofs should not be easy, for instance, in the case when (1) and (2) elementarily imply that the group $G$ is topological.

Since I am a specialist in paratopological groups, not semitopological, I propose a sketch of a proof that each locally compact Hausdorff paratopological group (that is, a group with jointly continuous multiplication) with (2) is topological. It is based on simple ideas and manipulations with the neighborhoods, but needs some background. So below I shall intensively cite our paper [BR], which you can look for details.

Following my teacher I say that a paratopological group $G$ is saturated if for any neighborhood $U\subset G$ of the unit the set $U^{-1}$ has nonempty interior in $G$.

Now let $U$ be an arbitrary neighborhood of the unit of the given group $(G,\tau)$. Since the set $\overline{V^{-1}}$ is compact, there is a finite subset $F$ of $\overline{V^{-1}}$ such that $\overline{V^{-1}}\subset FU$. Then $V\subset U^{-1}F^{-1}$ and therefore the set $U^{-1}$ has a nonempty interior. Thus the group $G$ is saturated.

Given a paratopological group $G$ let $\tau_\flat$ be the strongest group topology on $G$, weaker than the topology of $G$. The topological group $G^\flat=(G,\tau_\flat$), called the group reflexion of $G$, has the following characteristic property: the identity map $i:G\to G^\flat$ is continuous and for every continuous group homomorphism $h:G\to H$ from $G$ into a topological group $H$ the homomorphism $h\circ i^{-1}:G^\flat\to H$ is continuous.

The group reflexion $G^\flat$ of any abelian Hausdorff paratopological group $G$ is Hausdorff. Moreover, in this case the topology of $G^\flat$ has a very simple description: a base of neighborhoods at the unit in $G^\flat$ consists of the sets $UU^{-1}$ where $U$ runs over neighborhoods of the unit in the group $G$ (such the groups we call 2-oscillating). A bit later it was realized that the same is true for any paratopological SIN-group, that is a paratopological group $G$ possessing a neighborhood base $\mathcal B$ at the unit such that $gUg^{-1}=U$ for any $U\in\mathcal B$ and $g\in G$ (as expected, SIN is abbreviated from Small Invariant Neighborhoods). Unfortunately, Hausdorff paratopological SIN-groups do not exhaust all paratopological groups whose group reflexion is Hausdorff (for example any separated topological group has Hausdorff group reflexion but needs not be a paratopological SIN-group).

In [BR, Pr. 3] we showed that each saturated paratopological group $G$ is 2-oscillating. For this purpose fix any neighborhood $U$ of the unit $e$ in $G$. We have to find a neighborhood $W\subset G$ of $e$ such that $W^{-1}W\subset UU^{-1}$. Find an open neighborhood $V_1\subset G$ of $e$ such that $V_1^2\subset U$. Since $G$ is saturated, there are a point $x\in V_1$ and a neighborhood $W\subset G$ of $e$ such that $x^{-1}W\subset V_1^{-1}$. Then $W^{-1}x\subset V_1$ and $W^{-1}\subset V_1x^{-1}$. We can assume that $W$ is so small that $x^{-1}W\subset V_1x^{-1}$. In this case $W^{-1}W\subset V_1x^{-1}W\subset V_1V_1x^{-1}\subset V_1V_1V_1^{-1}\subset UU^{-1}$.

Now we are able to prove that the topology of $G$ coincides with the topology of its group reflexion $G^\flat$. For this purpose it suffices to show that each sufficiently small open neighborhood $U\in\tau$ is open in $G^\flat$ too. Let $W\in\tau$ be a neighborhood of the unit with compact closure in the group $(G,\tau)$. Choose a neighborhood $W_1\in\tau$ of the unit such that $W_1W_1\subset W$. Since the group $G$ is saturated, there exist a point $x\in W_1$ and a neighborhood $W_2\in\tau$ of the unit such that $W_2\subset W_1$ and $xW_2^{-1}\subset W_1$. Then the set $A=xW_2^{-1}W_2\subset W_1W_2\subset W_1W_1\subset W$ has a compact closure $\overline A$ in the group $(G,\tau)$. So the restriction $i|\overline A$ of the identity map $i:G\to G^\flat$ is a homeomorphism. Let $U\subset W_2$ be an arbitrary neighborhood of the unit it the group $(G,\tau)$. Then $xU$ is a open subset of $A\subset \overline A$. Hence $xU$ is an open subset of $\overline A$ as a subspace of $G^\flat$. Thus $xU$ is an open subset of $A$ as a subspace of $G^\flat$. Since the group $G$ is 2-oscillating, the set $A$ is open in $G^\flat$. Therefore the set $xU$ is open in $G^\flat$ too.

Update: You won’t believe me, but just now Katya Pavlyk decided to sent to me the original Ellis paper [E]. :-D The claim “locally compact Hausdorff paratopological group has (2)” is the next to the last step (Lemma 4) of the original proof of Ellis Theorem. Moreover, Katya has a question concerning the proof of this claim too. :-)

Remarks to the paper. It seems that:

– in the next to the last sequence in the proof of Lemma 4 should be “$x^{-1}\in\overline{E^{-1}_{m+1}}$” instead of “$x^{-1}\in E^{-1}_{m+1}$”;

– in the last sequence should be “$x_n^{-1}\in x^{-1}U$” instead of “$x_n^{-1}\in x^{-1}U^2$”;

– in the proof of Theorem, “$U’$” means “$X\backslash U$”;

– in the proof of Theorem, should be “$U’$” instead of “$\cal U’$”;

– in the proof of Theorem, should be “$\{e\}=$” instead of “$e=$”.

References

[BR] Taras O. Banakh, Alex V. Ravsky. Oscillator topologies on a paratopological group and related number invariants // Algebraical Structures and their Applications, Kyiv: Inst. Mat. NANU, 2002, 140--153.

[E] Robert Ellis, A note on the continuity of the inverse, Proc. Amer. Math. Soc., 8 (1957), 372-373.

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  • $\begingroup$ I have a simpler proof that paratopogical and (2) imply $G$ is topological using this: math.stackexchange.com/questions/370936/… thanks for sharing these interesting results. $\endgroup$ – user59671 May 5 '13 at 15:35
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    $\begingroup$ @CutieKrait: You are lucky. :-) See the update of my answer. :-) $\endgroup$ – Alex Ravsky May 7 '13 at 16:17
  • $\begingroup$ .. lucky that you are a member of MSE! :) thanks. I tried to find Continuity and homeomorphism groups when I saw it is refereced in the pdf you linked. I also tried to prove them using compactifications; if a Hausdorff compactification of $G$ is a paratopological group then (2) is trivial. Also it may be much easier to prove (1). $\endgroup$ – user59671 May 7 '13 at 16:47
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    $\begingroup$ As far as I know, there is no natural compactification of a paratopological group, in the opposition to precompact topological group, where this compactification should be the Raikov completion. There is a problem with the extension of the operations from the group onto its compactification. For instance, it seems, when we try to extend the operations even from a topological group $G$ onto $\beta G$, the $\beta G$ will be a topological group iff $G$ is pseudocompact. $\endgroup$ – Alex Ravsky May 8 '13 at 6:15
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    $\begingroup$ There should be no compact paratopological group $\hat{\mathbb R}$ which has $\mathbb R$ as a subgroup, because in this case $\mathbb R$ should be precompact with respect to the left uniformity. $\endgroup$ – Alex Ravsky May 8 '13 at 10:17

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