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

When I first met the concept of "characters" of topological groups in Pontryagin's book "Topological groups", it was defined as follows:

Let $G$ be a topological group. A character of $G$ is a homomorphism of topological groups from $G$ to the torus $T=\mathbb{R}/\mathbb{Z}$. Here, the torus $T$ has the induced topology from the usual topology of the real line $\mathbb{R}$.

I found the same definition on Wikipedia. So, I think this is a standard definition of character. But, in a lot of modern articles, it seems to me that characters are defined as follows:

Let $G$ be a topological group. A character of $G$ is a homomorphism of topological groups from $G$ to the discrete additive group $\mathbb{Q}/\mathbb{Z}$.

Clearly, these two definitions cannot be the same for all topological groups. However, if $G$ is a (discrete) finite group, then the two definitions agree.

Questions:

  1. What kind of conditions on a topological group $G$ one needs in order to identify the two definitions of character?

  2. If $G$ is a "profinite group", then do the two definitions agree? If the answer is yes, then how can one prove it?

Please give me any advice.

Later

I found a way to answer the second question. One only needs to show the following: for any profinite group $G$ and any continuous homomorphism $f:G \to T$, the image of $f$ is finite. This statement can be shown as follows. The torus $T$ has an open neighborhood $U$ of $0$ which contains no non-trivial subgroup of $T$. Since $G$ is profinite, there exists an open normal subgroup $H$ of $G$ satisfying $H \subset f^{-1}(U)$. This implies $H \subset \ker(f)$. So, the map $f$ factors through the finite group $G/H$. This implies that the image of $f$ is finite.

Hence the two definitions of character agree for any profinite group.

share|improve this question
    
The image of a profinite group under any cts finite dimensional representation is finite, in particular under 1-dimensional reps. –  Benjamin Steinberg Jan 19 '12 at 18:36
    
I'm just curious-- could you give an example reference to where the 2nd definition is used? –  Matthew Daws Jan 19 '12 at 22:15
    
Dear @ Benjamin Steinberg: What is "cts finite dimensional representation"? I could not find the definition in Google. –  Hiro Jan 19 '12 at 22:24
    
Dear @ Matthew Daws: I saw it for example in Neukirch's "Algebraic Number Theory." But he was dealing with only "profinite groups". –  Hiro Jan 19 '12 at 22:28
    
"cts" means continuous. –  MTS Jan 19 '12 at 22:28

2 Answers 2

up vote 8 down vote accepted

I am assuming all groups we are talking about are locally compact and commutative.

The two definitions indeed do ageree on profinite groups. To prove it, you have to check that the functors $Hom(-,\mathbb Q/ \mathbb Z)$ and $Hom(-,\mathbb R/ \mathbb Z)$ both transform limits of compact groups into colimits of discrete groups. That's routine.

The two definitions also agree on discrete torsion groups, hence on all groups which contain a closed and open profinite subgroup such that the quotient is torsion. I guess that this is exactly the class of groups on which the two definitions agree.

share|improve this answer
    
Thank you for your advice. I could prove that the first functor Hom(-,Q/Z) has such property, but I have difficulty in proving the statement for the latter functor Hom(-,R/Z). I could show that the natural map colim(Hom(G_i,R/Z)) -> Hom(lim(G_i),R/Z) is injective, but cannot show that it is surjective. Will you please tell me how to do it? –  Hiro Jan 19 '12 at 21:36
    
I found a way to show surjectivity... Let f:G->T be a character. The torus T has an open neighborhood U of 0 s.t. U contains no subgroup of T except for {0}. Take an open subgroup H of G contained in the inverse image of U. Then, H is contained in the kernel of the character f. So, f factors through the finite group G/H... Thanks for your advice! –  Hiro Jan 19 '12 at 23:25

It is easy to miss the point that in the second definition $\mathbb Q/\mathbb Z$ is required to be discrete in Hiro's question.
Hence even if a continuous morphism $f:G\to T$ has image in $\mathbb Q/\mathbb Z$ it cannot automatically be considered as a continuous map $f_0:G\to \mathbb Q/\mathbb Z$ and so might not be a character in the second sense.

An example for this failure is to take $G=T_{tors} =\mathbb Q/\mathbb Z\subset T$ , the torsion subgroup of $T$ with its induced topology from the circle and for the character $f $ (in the first sense) the inclusion $f:G\hookrightarrow T$.
Even though $G$ is torsion, the corestricted morphism $f_0:G\to \mathbb Q/\mathbb Z$ is not a character in the second sense, since it is not continuous.

However if a character on a compact group $G$ happens to have values in $\mathbb Q/\mathbb Z$,so that both definitions can be compared, then its image in the circle is finite and the two concepts coincide.
Xandi explains in his answer that this is always the case for profinite groups.

share|improve this answer
1  
N.b. The group $G$ in the 2nd paragraph is not locally compact. @georges: If $G$ is compact, its image under a character isn't necessarily discrete in the circle... –  Xandi Tuni Jan 19 '12 at 18:49
    
Dear @ Xandi: 1) Neither Hiro nor I ever mention local compactness. So it is true that my $G$ in the second paragraph is not locally compact, but this does not affect my answer . 2) It is of course true that a character on a compact group does not have discrete image. I meant a character with values in $\mathbb Q/\mathbb Z$, so that the image has two induced topologies which can be compared. I modified my badly written last paragraph thanks to your comment: thanks a lot. –  Georges Elencwajg Jan 19 '12 at 21:12
    
Dear @ Georges Elencwajg: Thank you for your answer. Since any proper closed subgroup of the torus T is a finite discrete cyclic subgroup, what I have to show is the following statement: "for any profinite group G and any continuous homomorphism f:G -> T, the image of f is a proper subgroup of T". This statement cannot be true without the hypotheses that G is profinite, since we have a counter example G=T. However, I do not know how to use the profinite assumption. Will you please tell me how to show the statement above? –  Hiro Jan 19 '12 at 21:47
    
Dear Hiro, everything you write is correct. The sad truth is that, much as I would like to help you, I don't know how to prove the statement you ask about: that's why I wrote "Xandi explains..." I hope he will answer the question you ask in your comment to him. –  Georges Elencwajg Jan 19 '12 at 22:05
    
Dear Georges, thank you for your immediate response. I greatly appreciate your kindness. Please lend your hand next time, too! –  Hiro Jan 19 '12 at 22:16

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.