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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?

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    $\begingroup$ Can you give a reference to any proof? $\endgroup$
    – GH from MO
    Commented Oct 23, 2012 at 12:32
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    $\begingroup$ 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 . $\endgroup$
    – user25235
    Commented Oct 23, 2012 at 13:10
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    $\begingroup$ @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. $\endgroup$
    – GH from MO
    Commented Oct 23, 2012 at 14:40
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    $\begingroup$ 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. $\endgroup$
    – Terry Tao
    Commented Oct 23, 2012 at 14:46
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    $\begingroup$ @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. $\endgroup$
    – user25235
    Commented Oct 23, 2012 at 16:09

3 Answers 3

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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.

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    $\begingroup$ 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. $\endgroup$
    – user25235
    Commented Oct 24, 2012 at 6:41
  • $\begingroup$ @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. $\endgroup$
    – GH from MO
    Commented Oct 24, 2012 at 9:47
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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).

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    $\begingroup$ 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" ... $\endgroup$
    – user25235
    Commented Oct 23, 2012 at 16:31
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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].

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