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Let us denote by the symbol $\mathcal{G}$, a group of functions $f: \mathbb{R} \rightarrow \mathbb{R}$ (with the composition operation) that is additionally closed under all affine change of variables of the form (homothety):

$$ h(x) = mx, m>0$$

In other words, I would like the following property to hold for any affine maps $h$ (of the above form): $$hGh^{-1} \in G$$

Intuitively such a group is a group of functions that is invariant under all rescalings.

A simple example of such a group is the group of fractional linear transformations (FLT) (with real coefficients), namely the group $\mathcal{S}$ consisting of that $f(x) = \frac{ax+b}{cx+d}$ where $a,b,c,d \in R$.

My questions are:

  1. Do such groups have a name?

  2. What is the classification of all such groups? (with the properties of $f'(x) >0$ and $f\in C^3$ if possible)

  3. Is there a general way of constructing such groups or putting this question in a general context?

Thank you in advance to all those who respond,

E(up)lio M.

.

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3 Answers

up vote 2 down vote accepted

If $G$ contains the linear group consisting of all functions of the form $h_m(x) = mx$ then obviously it satisfies your conditions.

Conversely, by the definition of the derivative,

(*) $lim_{m \to \infty} (h_m^{-1} f h_m(x)) = f'(0) x$

So the closure $\bar G$ of your group (in some appropriate topology) contains all linear functions of the form $h_m$ where $m = f'(0)$ for $f \in G$. So if you add to your conditions that the group $G$ be closed, and that the set of derivative values $f'(0)$, $f \in G$ consists of all positive real numbers, then $G$ does indeed contain the linear group.

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That's an interesting observation. –  Euplio M. Apr 23 '12 at 16:51
    
@Lee: Did you assume that $f(0)=0$ in obtaining (*) above? –  Euplio M. Apr 23 '12 at 21:15
    
Yeah, I think you are right. So this is more a class of examples than anything else. It would not say anything about the extreme opposite case where $\bar G$ has no elements whose graphs pass through the origin. –  Lee Mosher Apr 24 '12 at 11:55
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Let $G$ be any group of functions. Let $H$ be the group of all compositions of functions of the form $f(mx)/m$ where $m>0$ and $f\in G$. Then $H$ is a group of the sort you're looking for, and all such groups arise in this way.

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$H$ need not be closed under composition. Unless you mean for $H$ to be the group generated by all functions of that form? –  Lee Mosher Apr 23 '12 at 2:22
    
Lee Mosher: I did in fact mean this and dropped a few words along the way. I'm fixing this now; thanks for catching it. –  Steven Landsburg Apr 23 '12 at 2:52
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For what its worth, there seem to be $2^{2^{2^{\aleph_0}}}$ many such groups (or at least it is consistent with ZFC that there are this many). It is easy to see the somewhat lesser claim that there must be at least $2^{2^{\aleph_0}}$ many such groups, since we may simply close a single bijective function under composition and conjugation by affine functions. Since there are only continuum many affine functions, this leads to a group of size at most continuum, and so at most continuum many functions can lead to the same group this way. But there are $2^{\frak{c}}=2^{2^{\aleph_0}}$ many bijections. For the larger claim of $2^{2^{2^{\aleph_0}}}$, consider the forcing to add $2^{\frak{c}}$ many mutually generic bijections $f:\mathbb{R}\to\mathbb{R}$, without adding reals. The groups generated by any subset of these functions in the corresponding forcing extension, closing under composition and conjugation by affine functions, will be different by mutual genericity, and so there will be $2^{2^{\frak{c}}}=2^{2^{2^{\aleph_0}}}$ many different such groups in the forcing extension. I expect that one can get rid of this forcing argument by more a more careful counting, and just prove outright that there are $2^{2^{2^{\aleph_0}}}$ many such groups. For example, all one needs is a family of $2^{\frak{c}}$ many bijective functions $\mathbb{R}\to\mathbb{R}$, such that omitting any one of them $f$ and closing the remaining functions under composition and conjugation by affine functions, does not generate $f$; that is, a maximal independent family for the generating process in your question. In this case, every subset of that family will generate distinct groups with your property, giving $2^{2^{2^{\aleph_0}}}$ many such groups. The forcing argument shows that it is consistent with ZFC to have such a large independent family, but probably one can just prove it outright.

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