Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I am not specialist on Topological Group Theory, I apologize if this is a trivial question.

Question. If $G_1=G_2$ are amenable topological groups what additional hypothesis we have to consider on the group, in order to prove that $G_1\times G_2$ is amenable ?

Following Leinster, in this question Why are abelian groups amenable?,
"The direct product of two amenable groups is amenable. This isn't exactly trivial, but the measure on the product is at least constructed canonically from the two given measures."
So discreteness of the $G_1$ and $G_2$ are enough to prove that $G_1\times G_2$ is amenable and also we do not need to suppose that $G_1=G_2$.

Looking for a proof, in more general cases, I found the following statement:
"... direct product $G_1\times G_2$ of two countable amenable groups can not be amenable."
in the paper, On Subadditive Processes on Direct Product of Countable Amenable Groups by Seyit Temir - Publications De l'Institute Mathématique (2002), 119-122.

Since the author did not mention if the example needs two different groups and I have no access to the paper containing this information, I decided to post this question.

I would be grateful if you could point me out some references discussing about this problem.

share|improve this question
3  
I'm confused-- extensions of amenable groups are amenable-- that is, if N is a normal subgroup of G and both N and $G/N$ are amenable, then so is G. And trivially $G$ is a normal subgroup of $G\times G$, with the quotient being $G$... –  Matthew Daws Mar 18 '11 at 10:43
    
Hi Matthew, of course your observation also works for different factors $G_1$ and $G_2$, now I would like to understand what is Temir's point. –  Leandro Mar 18 '11 at 15:29
2  
I have just had a look at the paper emis.de/journals/PIMB/086/13.html and I am confused, verging on appalled. I will perhaps have a closer look later, but my current feeling is that the author has misunderstood something. The list of references is a bit suspicious too. –  Yemon Choi Mar 18 '11 at 17:28
    
Thank you Yemon. I agree with you about the suspicious. Anyway now I am curios about the general case. Because the proof I cited in the comments does not seems to be straight foward generalized for general topological groups. If you remember a reference with this proof, could you please post it here ? –  Leandro Mar 18 '11 at 17:57
1  
MathReview for that Temir paper: ams.org/mathscinet-getitem?mr=1997617 There is no comment made on the staggering claim that it is unknown if the product of two countable amenable groups is amenable, and the MR in some ways appalls me even more than the original paper –  Yemon Choi Mar 18 '11 at 23:36

2 Answers 2

up vote 3 down vote accepted

The wikipedia article is quite clear on the subject:

http://en.wikipedia.org/wiki/Amenable_group

share|improve this answer
    
Hi Igor, thanks for your reply. I looked at wikipedia article before post this question, but I thought that the statement are made for discrete amenable groups, am I right ? I also checked the references given in this article to be sure about the right hypothesis, except the book F.P. Greenleaf, Invariant Means on Topological Groups and Their Applications because we don't have it here, but I did not find this statement. –  Leandro Mar 18 '11 at 14:55
3  
Actually, this was not my impression. The wikipedia article is quite careful, and this statement seems to be made for all amenable topological groups. –  Igor Rivin Mar 18 '11 at 16:12
    
The product of two locally compact amenable groups is definitely (locally compact and) amenable. Whether some subtlety arises in the non-locally-compact case, I am not sure off the top of my head –  Yemon Choi Mar 18 '11 at 17:28

Say that a topological group $G$ is amenable if every continuous affine action of $G$ on a non-empty compact convex subset of a locally convex topological space, has a fixed point. With this definition, it is completely standard to prove that, if $N$ is a closed normal subgroup of $G$ such that $N$ and $G/N$ are amenable, then $G$ is amenable.

share|improve this answer
    
Thanks for the answer Alain. –  Leandro Apr 19 '11 at 21:33

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.