Existence of a square root of a functional equation

Question

We can define the iterates $f^{n+1}=f\circ f^n$ for a given smooth map $f:X\to X$, where $X$ could be a finite interval, the real line $\mathbb{R}$, or the circle $S^1$, or any general smooth manifold. What about the reverse direction? More precisely, a map $g:X\to X$ is said to be an $n$-th root of $f$ if $g^n=f$. There might be a standard notation for this.

For convenience, let's say $R_n(f)=\{g\in C^\infty(X):g^n=f\}$. The iteration map $g\in C^\infty(X)\to g^n$ is continuous. So the set $R_n(f)$ should be discrete in $C^\infty(X)$.

In the following we may take $n=2$ for certainty.

Question 1. Does there exist a square root for any function $f$? If not, what are the possible obstructions?

Question 2. Suppose there exists one solution. What/when could we say about the uniqueness, finiteness, etc about the solutions?

A trivial example: $X=\mathbb{R}^n$ and $f$ is the identity. Then there are at least two square roots: $g_1(x)=x$ and $g_2(x)=-x$. I am not sure if these are the only two solutions.

Edit: see link for the discussion in this special case.

For question 1 in the case when $X$ has nontrivial homology, we can consider the induced action on various homology groups of $X$. In particular, the action $[f]$ must admit a square root matrix.

Edit: See link for discussions about the exponential function and a useful link provided there.

Answer 1

An answer to a restriction of this problem:

Let $F$ be the space of strictly increasing twice-differentiable functions $f:I \mapsto I$ on an open interval $I \subset \Bbb{R}$ such that the first derivative of $f$ is strictly increasing. Then for each $f\in F$ there exists a unique function $g \in F$ such that $g^2 = f$.

This statement holds if you change either (or both) of the words "increasing" to "decreasing.

Answer 2

There are obstructions involving periodic and eventually periodic points. Suppose, for example, $f:X \to X$ has an odd number of $2$-cycles (pairs $a \ne b$ with $f(a) = b$, $f(b) = a$). Then $f$ can't have a square root.

Answer 3

This is a very broad and classical subject, depending on the class of functions that you consider. For example, you can consider germs of analytic functions at $0$, such that $f(0)=0$. If $|f'(0)|\neq 0,1$ then all fractional iterates exist, (as analytic germs), and are unique. Moreover $n$ can be any complex number, not necessarily an integer. If $f'(0)=0$, there is an evident obstacle. The case $|f'|=1$ is very complicated, especially when $f'(0)$ is not a root of unity.

The literature on the subject is really enormous. Here are just a few important works: Baker, I. N. Permutable power series and regular iteration. J. Austral. Math. Soc. 2 1961/1962 265–294.

Écalle, J. Théorie itérative: introduction à la théorie des invariants holomorphes. (French) J. Math. Pures Appl. (9) 54 (1975), 183–258.

Voronin, S. M. Analytic classification of germs of conformal mappings (C,0)→(C,0). (Russian) Funktsional. Anal. i Prilozhen. 15 (1981), no. 1, 1–17, 96.

If we are talking only about analytic germs, then the question has been completely solved only for the germs $f(z)=\lambda z+z^2$, the answer in terms of $\lambda$ is quite complicated. (J-C. Yoccoz was awarded a Fields medal for this.)