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

Is there an "elementary" (say ultrafilter-free) proof of the following fact: if $G$ is a compact (Hausdorff) topological group, if $g \in G$ is any element from this group, and if $P$ is a polynomial with integer coefficients without constant term, then the identity element of $G$ is a limit point of the sequence $n \mapsto g^{P(n)}$.

An other question: for which integer-valued sequences $u_n$ is the result above still true with $P(n)$ replaced by $u_n$, whatever $G$ and $g$ are?

share|improve this question
1  
Can you give a reference to any proof? –  GH from MO Oct 23 '12 at 12:32
3  
For example : this is a straightforward generalisation of theorem 7.2 in the following article of Vitaly Bergelson math.osu.edu/~bergelson.1/VBContempMathUltrafiltersEtc.pdf . –  user25235 Oct 23 '12 at 13:10
1  
@km: The mentioned Theorem 7.2 is a statement about a torus. I am interested in a reference for noncommutative compact groups as in your post. –  GH from MO Oct 23 '12 at 14:40
6  
By passing to the orbit closure of g, which is monothetic and hence abelian, one can assume G is compact abelian, hence the inverse limit of finite union of torii. From this it is not difficult to reduce to the torus case. –  Terry Tao Oct 23 '12 at 14:46
2  
@GH: when $G$ is a torus, one can conclude by a standard Weyl's sums argument. But by "a straightforward generalisation" I meant "replace the torus by $G$ throughout the proof" (since noncommutativity causes no trouble in our ultralimits). @Terry Tao: Thanks for the elegant reduction. Your argument using Peter-Weyl theorem works also in the noncommutative case (which could arise if one considers instead sequences $n \mapsto g^{P(n)}h^{Q(n)}$ - where the ultralimit argument still applies), in which case we only have to consider (a finite list of) compact Lie groups. –  user25235 Oct 23 '12 at 16:09

3 Answers 3

up vote 6 down vote accepted

If $\mathcal{H}$ is a van der Corput set of positive integers, then the closure of $\{ g^h\mid h\in\mathcal{H} \}$ contains the identity element. This generalizes the statement in the original post, because $\{ P(n)\mid n>0 \}$ is a van der Corput set for $P\in\mathbb{Z}[x]$ without a constant term.

By Terry Tao's remark above, it suffices to show that for any $0<\epsilon\leq 1/2$ and for any $\theta_1,\dots,\theta_n\in\mathbb{R}$ there is $h\in\mathcal{H}$ such that $\|h\theta_1\|,\dots,\|h\theta_n\|<\epsilon$. We follow closely the proof of Theorem 9 in Chapter 2 of Montgomery: Ten lectures on the interface between analytic number theory and harmonic analysis. By the earlier results in the chapter, there is a cosine polynomial $$ T(x) = a_0 + \sum_{h\in\mathcal{H}} a_h \cos 2\pi hx $$ with real coefficients, nonnegative values, $a_0<\epsilon^n$, and $T(0)=1$. Put $$ f(x):=\max(0,1-\|x\|/\epsilon), $$ and consider the expression $$ a_0 + \sum_{h\in\mathcal{H}} a_h f(h\theta_1)\dots f(h\theta_n). $$ It suffices to show that this expression exceeds $\epsilon^n$, because then $f(h\theta_1)\dots f(h\theta_n)\neq 0$ follows for some $h\in\mathcal{H}$. The Fourier expansion $$ f(x) = \sum_{k\in\mathbb{Z}}\hat f(k) e(kx) $$ converges absolutely, hence the above expression equals $$ \sum_{h\in\mathcal{H}\cup\{0\}} a_h \left(\sum_{k_1\in\mathbb{Z}}\hat f(k_1) e(hk_1\theta_1)\right)\dots \left(\sum_{k_n\in\mathbb{Z}}\hat f(k_n) e(hk_n\theta_n)\right)=$$ $$\sum_{k_1,\dots, k_n\in\mathbb{Z}}\hat f(k_1)\dots\hat f(k_n) \sum_{h\in\mathcal{H}\cup\{0\}} a_h e(hk_1\theta_1+\dots+hk_n\theta_n)=$$ $$\sum_{k_1,\dots, k_n\in\mathbb{Z}}\hat f(k_1)\dots\hat f(k_n) \sum_{h\in\mathcal{H}\cup\{0\}} a_h \cos 2\pi h(k_1\theta_1+\dots+k_n\theta_n)=$$ $$\sum_{k_1,\dots, k_n\in\mathbb{Z}}\hat f(k_1)\dots\hat f(k_n)T(k_1\theta_1+\dots+k_n\theta_n).$$ The term corresponding to $k_1=\dots=k_n=0$ contributes $\epsilon^n$, while all the other terms are nonnegative, hence we are done.

share|improve this answer
3  
I accept this answer since this is close to what I was looking for. It seems reasonable to expect that your proof extends to the general case (without applying the reduction given by Terence Tao) : given a neighbourhood $U$ of the identity, all what we need is a function $f$ with real, nonnegative fourier transform, with nonzero mean, and which vanishes outside $U$. I don't know if this is always possible. –  user25235 Oct 24 '12 at 6:41
    
@km: Thank you. In the noncommutative case it is not so clear how to make this approach work as one has higher dimensional representations in the spectrum. –  GH from MO Oct 24 '12 at 9:47

Take a look at Theorem C and Proposition 1.10 in the article Polynomial Extensions of Van Der Waerden´s and Szemeredi´s Theorems by Bergelson and Leibman. I think they prove a lot more than what you need and still the proofs are completely elementary (although quite long and involved).

share|improve this answer
3  
Thanks. There's an argument (also due to Bergelson I think !) which shows that Van der Waerden's theorem implies the theorem in my question : if $U$ is a small neighbourhood of the identity, $G$ has a finite covering by some $(g_iU)_i$, and any map $n\in \mathbb{N} \mapsto$ some $i$ with $g^{P(n)} \in g_iU$ defines a colouring of the integers to which Van der Waerden's theorem can be applied. On then concludes using equations like $(x+2y)^2-2(x+y)^2+x^2=2y^2$ and its generalizations. But since I've never considered Van der Waerden's theorem as "elementary" ... –  user25235 Oct 23 '12 at 16:31

For getting the every-point statement, at-least in the compact abelian case (see Tao's comment above), one can either prove it by harmonic analytic approach (Weyl's equi. criterion + van der corput trick, just like in GH's proof), or one can use a dynamical approach (either topological dynamics or ergodic theoretic approach) by using induction on the degree of the polynomial and a skew-product theorem à-la Furstenberg.

Notice that an almost-every point statement (wrt the Haar measure) is true in a much more general settings by Bourgain's ergodic theorem. [Bourgain's theorem says even more, as an ergodic theorem, for example it says something about repetitions to (nhbds of-) the identity].

share|improve this answer

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.