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

We would like very much to know the answer to the following question:


Let $\|\cdot\|$ be any norm on $\mathbb{Z}^d$ and let $W(\mathbb{Z}^d)$ be the group of all bijections of $\mathbb{Z}^d$ such that $$\|g(j)-j\|\leq C_g,$$ for some constant $C_g$ which depends only on the element $g\in W(\mathbb{Z}^d)$. Consider the Hilbert space $L^2(\{0,1\}^{\mathbb{Z}^d},\mu)$, where $\{0,1\}^{\mathbb{Z}^d}$ comes with the standard Bernoulli measure $\mu$.

We are looking for a sequence of functions $f_n\in > L^2(\{0,1\}^{\mathbb{Z}^d},\mu)$ with $\|f_n\|_2=1$ and $$\|g.f_n-f_n\|_2\rightarrow 0, \text{ > for every } g\in W(\mathbb{Z}^d),$$ $$\|f_n\cdot\chi_{\{\omega_j\in > \{0,1\}^{\mathbb{Z}^d}:\omega_0=0\}}\|_2\rightarrow > 1.$$ where $\chi_{\{(\omega_j)_{j\in \mathbb{Z}^d}\in \{0,1\}^{\mathbb{Z}^d}:\omega_0=0\}}$ is the characteristic function of the cylinder set $\{(\omega_j)_{j\in \mathbb{Z}^d}\in \{0,1\}^{\mathbb{Z}^d}:\omega_0=0\}$


Motivation:


The existence of such a sequence for all $d$ would disprove the conjecture of Katok, that the interval exchange transformation group contains a free subgroup.


What we know about the question above:


In the joint work with Nicolas Monod, here, we showed that for $n=1$ the following function satisfy the properties above: $$f_n(\omega)=e^{-n \sum\limits_{j\in \mathbb{Z}} \omega_j e^{-\frac{|j|}{n}}}=\prod_{j\in Z} a_j^{\omega_j},$$ where $a_j=e^{-n e^{-\frac{|j|}{n}}}$. We are interested in extending this result to higher dimensions. Note that the function above is the product of functions of independent i.d. random variables.

Let $G< W(\mathbb{Z}^d)$ be a finitely generated subgroup of $W(\mathbb{Z}^d)$.

In addition to above (in collaboration with Nicolas Monod and Mikael de la Salle) we know that the existence of the functions with the property above in the class of functions which can be written as the product of functions of independent i.d. random variables is equivalent to a certain property of the Schreier graph of the action of $G$ on $\mathbb{Z}^d$. Let me give more details on this.

The Schreier graph of the action of $G$ on $X$ with respect to $S$ is the graph with vertices $X$ and with an edge between $x$ and $y$ for each $g \in S$ with $g x=y$.

We say that an infinite graph $G=(V,E)$ satisfies a Sobolev inequality rooted at $x_0 \in X$ if the value at $x_0$ of any $c_0$-function on $V$ is bounded by the $\ell^2$-norm of its gradient, i.e., there is a constant $C>0$ such that $$\|f\|_{c_0(V)} \leq C \sum_{x \sim x' \in V} |f(x') - f(x)|^2.$$

We can show that

The functions $f_n$ with the property above can be found in the class of products if and only if the Schreier graph of the action of $G$ on $X$ with respect to $S$ does not satisfy a Sobolev inequality.

Moreover, for $d=1,2$ and $G$ be a finitely generated subgroup of $W(n)$ with symmetric generating set $S$. Then the Schreier graph of the action of $G$ on $\mathbb{Z}^d$ does not satisfy a rooted Sobolev inequality. However there are subgroups in $W(\mathbb{Z}^3)$ such that their Schreier graph satisfies Sobolev inequality.

To summarize above, we can find a sequence of functions in the class of products with the above property only in cases $n=1,2$.

Any suggestions on potential examples of functions that can do the higher dimentional cases?

share|improve this question
    
@Kate: How does it relate to subgroups of the group of interval exchange transformations? –  Misha Jun 7 '12 at 17:22
    
Misha: to answer shortly your question, we did similar (to the paper in arxiv) embeddings of IET into a certain semidirect product which acts on some set with torsion stabilizers and the above produces the mean on this set which is a.i. under the action. –  Kate Juschenko Jun 7 '12 at 22:12
add comment

1 Answer

up vote 4 down vote accepted

This is my first visit to MO, so I have to apologize for making this an "answer": it is just a comment on Nik Weaver's suggestion.

I don't think that NW's conjecture is much different from the original question. Indeed, you comment that it is different because f should work against all permutations g. But in fact you restrict to those g that have C_g bounded by a universal constant (say, 2). This is a very small set, it is even compact in the natural topology relevant to the action on Z^d (which is the action underlying the whole question). Therefore, it is not so different from considering finitely many g's and thus from the original question.

Anyway, this is my intuition, it is not a mathematical statement. Nicolas Monod

[Edit after reading NW's argument]: at first sight, it seems that your reduction to your conjecture is indeed just exploiting that the set C_g less than a constant is compact (approximation argument).

share|improve this answer
    
It could be a valuable exercise to try to prove my reduction using compactness. Good luck! –  Nik Weaver Jun 10 '12 at 13:22
    
Nik, I would very much appreciate if you keep your answers to the question available online. Also I would like to thank you for spending your time on the question and new ideas about it. –  Kate Juschenko Jun 10 '12 at 14:55
    
The "bounty" automatically selected your answer and accepted it, Nicolas... –  Kate Juschenko Jun 15 '12 at 16:35
1  
Oops, sorry about that, it's an unintended consequence. –  Nicolas Monod Jun 17 '12 at 19:54
add comment

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.