# square root of diffeomorphism of R: does it always exist?

Let $f:\mathbb R\to\mathbb R$ be a smooth, orientation preserving diffeomorphism of the real line.
Is it the case that there always exist another diffeomorphism $g:\mathbb R\to\mathbb R$ such that $g\circ g = f$?

Note: it is relatively easy to show that a continuous $g$ exists, but I am not managing to find a smooth (i.e. $\mathcal C^\infty$) solution of the above equation.

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This is reminding me of the statement that the exponential map from $Vec(S^1)$ to $Diff(S^1)$ is neither locally injective nor surjective (from "Loop Groups"). –  Allen Knutson Mar 26 '11 at 23:57
It's not even locally injective?!? How do you see that? –  André Henriques Mar 28 '11 at 18:52
I looked it up (Proposition 3.3.1 of Pressley and Segal's book): The rotation by $1/n$ is in the image of the exponentional map $exp:Vect(S^1)\to Diff(S^1)$ and its preimage under that map is huge. It contains all $1/n$-periodic vector fields whose integral over $S^1$ is $n$. [Note: here I'm using the model $S^1:=\mathbb R/\mathbb Z$] –  André Henriques Apr 10 '11 at 19:59

The answer is no, assuming that you seek an orientation preserving square root.

(I see unknown's answer appeared while I'm typing. I don't quite understand it at the moment but the construction looks different.)

Consider a diffeomorphism $f$ such that $f(0)=0$, $f(1)=1$, $f(t)>t$ for all $t\in(0,1)$, $f'(0)=a>1$, $f'(1)=b<1$ and furthermore $f$ is linear in a neighborhood of 0 and in a neighborhood of 1. Let $g$ be a $C^1$ square root of $f$. Then obviously $g(0)=0$, $g'(0)=\sqrt a$ and moreover $g(t)=\sqrt a\cdot t$ in the neighborhood of 0 where $f^{-1}$ is linear. Indeed, suppose that $g(t_0)=ct_0$ where $c\ne\sqrt a$, then $$g(a^{-k}t_0) = g(f^{-k}(t_0)) = f^{-k}(g(t_0)) = ca^{-k}t_0 .$$ Since $g(0)=0$ and $g(a^{-k}t_0)=ca^{-k}t_0$, there exist a point $t$ between $0$ and $a^{-k}(t_0)$ (and hence arbitrarily close to 0) such that $g'(t)=c$, contrary to continuity of $g'$ at 0.

Similarly, $g(1)=1$, $g'(1)=\sqrt b$ and $g$ is linear near 1: $g(t)=1+\sqrt b(t-1)$. So we know $g$ near the endpoints. Furthermore there is a compatibility condition: take $t_0$ close to 0 and a very large $n$ such that $f^n(t_0)$ is close to 1. Then $$f^n(\sqrt a t_0) = f^n(g(t_0)) = g(f^n(t_0)) = 1+ \sqrt b(f^n(t_0)-1)$$ Obviously a generic $f$ does not satisfy this (and hence doest not have a square root), since the $f$-orbits of $t_0$ and $\sqrt a t_0$ are essentially independent.

More formally, modify $f$ near some point $t'=g^{2k+1}(t_0)$ which is far from 0 and 1, so that the resulting function $\tilde f$ equals $f$ outside the segment $[f^{k}(t_0),f^{k+1}(t_0)] = [g^{2k}(t_0),g^{2k+2}(t_0)]$ and $\tilde f(t')\ne f(t')$. The new function satisfies $\tilde f^n(t_0)=f^n(t_0)$ but $$\tilde f^n(\sqrt a t_0) \ne f^n(\sqrt a t_0) = 1+ \sqrt b(\tilde f^n(t_0)-1) ,$$ hence $\tilde f$ cannot have a $C^1$ square root.

A similar argument shows that diffeomorphisms without square root are dense among diffeomorphisms with at least two fixed points.

EDIT: If $f$ has exactly two fixed points, then it does not have an orientation reversing square root as well. Because such a square root $g$ would have exactly one fixed point, and all other fixed points of $g^2$ come in pairs of the form $\{t,g(t)\}$. So $g^2$ must have an odd number of (or infinitely many) fixed points.

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There is an example in "Kuczma, Marek; Choczewski, Bogdan; Ger, Roman: Iterative functional equations" that (I believe, unfortunately I only have incomplete access to it at the moment, so I am not certain) shows that this does not have to exist in general (Ex 11.4.1).

However the example is preceeded by: 'we wish to exhibit two phenomena a) diffeomorphism with no (smooth) interative roots' b) [...something on convex functions...] so I am quite sure this answers what you are asking for.

The example is this:

Fix $0 < s < 1$ and $0 < x_0 < x_1 < 1$ such that $x_1 < \sqrt{s} x_0 / s$. Take any convex $f \in C_1 (R)$ such that $f (x) = sx$ for $x \le x_0$ and $f(x) = \alpha x + \beta$ for $x\ge x_1$ where $\alpha = (1-\sqrt{s}x_0)/(1-x_1)$ and $\beta= (\sqrt{s}x_0 - x_1)/(1-x_1)$. Suppose that a function $\varphi: R \to R$ satisfies $\varphi \circ \varphi =f$ and $\varphi$ is $C_1$ or convex. [...] This leads to a contradiction.

Of course the book contains an argument why there is a contradiction.

Since the intersection of the assumptions of this example and your question seem non-empty, this shows that the answer to your question is 'no', and that there even does not exist a $C_1$ solution.

Possibly there are easier accessible sources or simpler arguments in your situation, but I don't known this.

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Here this problem (and some more general) are studied for the case $f$ has only one fixed point. In general, one can get $C^1$ square roots, but for higher smoothness there are obstructions when the derivative at the origin is the identity which are studied there.

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