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Let's consider an interval $I\subseteq\mathbb R$, and let $\mathcal F(I)$ be the set of bijective functions $f:I\to I$ so that the graph of $f$ is a analytic curve in $I\times I$.

The set $\mathcal F(I)$, together with the function composition $\circ$, is a group. Let $\mathcal H(I)$ be its subgroup consisting in all analytic functions from $\mathcal F(I)$ having the inverse an analytic function (here and everywhere in this question by "inverse" and $f^{-1}$ we mean inverse w.r.t. function composition).

For example, if $I=\mathbb R$, the function $f(x)=x^3$ is invertible, and its graph is an analytic curve in $I\times I$, but $x^3\notin \mathcal H(I)$.

Q1. What do we know about the cosets of $\mathcal H(I)$ in $\mathcal F(I)$? Do you know some references?

Let $Q\in I$ and let $\mathcal H_Q(I)\subset\mathcal F(I)$ so that $f(Q)=Q$ and both $f$ and $f^{-1}$ are analytic on $I-Q$. For example, the function $x^3\in\mathcal H_0(\mathbb R)$, but $x^3\notin\mathcal H(\mathbb R)$. Another example is $x^2$, where $I=[0,+\infty)$ and $Q=0$.

The set $\mathcal H_Q(I)$ is a subgroup of $\mathcal F(I)$, and $\mathcal H(I)\cap \mathcal H_Q(I)$ is a subgroup of $\mathcal H_Q(I)$.

Q2. What do we know about the cosets of $\mathcal H(I)\cap \mathcal H_Q(I)$ in $\mathcal H_Q(I)$? Do you know some references?

What about the case when the graph is not necessarily analytic, but only smooth?

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Is "$f$ is analytic" the same as "the graph of $f$ is an analytic curve"? If so, then $H=F$, since a group contains the inverse of each of its elements. If no, then what is the difference? –  Gerry Myerson Feb 18 '12 at 8:14
    
@Gerry Myerson: For example, the function $f(x)=x^{1/3}$ is not analytic, although its graph is an analytic curve in $\mathbb R\times \mathbb R$. –  Cristi Stoica Feb 18 '12 at 9:47
    
$\mathcal{H}(I)$ is not necessarily a subgroup of $\mathcal{H}_Q(I)$, since for example, $x+1\in\mathcal{H}(\mathbb{R})$ but not in $\mathcal{H}_0(\mathbb{R})$. –  Steve D Feb 18 '12 at 15:28
    
@Steve D: Thanks for pointing this out, I updated the question accordingly. –  Cristi Stoica Feb 18 '12 at 19:01
    
This is hardly worth mentioning, but perhaps in the definition of $\mathcal{H}(I)$ the member functions themselves (in addition to their inverses) should be required to be analytic. –  Tiankai Mar 4 '12 at 5:35

1 Answer 1

up vote 1 down vote accepted

Let $f$ be a function in this group. $f$ fails to be analytic at the points where its "derivative" is infinite, and this is a finite set. Similarly $f^{-1}$ is not analytic where its derivative is $0$. So count the number of points where $f$ or $f^{-1}$ is analytic, and list them in order. This should be finite. Furthermore, list at each point the order of vanishing of the zero of the derivative of $f$, or the order of vanishing of the zero of the derivative of $f^{-1}$.

This seems to be sufficient information to determine the coset. Suppose we have two functions with the same list. We can multiply one on the right to move the non-analytic points to the same $x$ coordinate as the non-analytic points of the other. We then look at $fg^{-1}$. This is clearly analytic away from those points, and it is also analytic at those points - indeed, we can calculate its derivative as the nonzero finite ration of the $n$th derivatives of $f$ and $g$.

Similarly, two functions in the same coset must have the same list, up to reversal of the order.

This gives a complete description of the cosets.

Smoothness seems like it might be harder.

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Yes, that's it! Thank you very much for your time and for considering my question. –  Cristi Stoica Mar 15 '12 at 4:20

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