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.

  • For your trivial example, even restricting to linear maps, you have the set of diagonalizable matrices with eigenvalues $\pm 1$. This is not a discrete set in the usual topology. – S. Carnahan Feb 23 '16 at 16:16
  • @S.Carnahan You are right. Then it came to my mind that rotations on the plane by $\pm\pi/2$, reflections around some axis, etc. Maybe the linear ones are too special. – Pengfei Feb 23 '16 at 16:20
  • @LoïcTeyssier It would be very interesting if the equation $g^n=h^n$ boils down to finding all the $n$-th roots of Id. I can see this when $g$ and $h$ are commutative. Is this also true in the general case? – Pengfei Feb 23 '16 at 16:48
  • Ah, you're right, sorry for this. Let me delete my previous comment... It remains that Question 1 has a lot of variations around this site. – Loïc Teyssier Feb 23 '16 at 17:05
  • @LoïcTeyssier I tried unsuccessfully before posting the question. Several related posts are now linked to the right side of this page after posting. As you said, this topic has been extensively discussed on MO. – Pengfei Feb 23 '16 at 17:11
up vote 1 down vote accepted

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.

  • Two questions: 1. Does some initial condition need to be specified e.g. the value of the function or its derivative at some point inside the interval? 2. Do you have a reference for the proof of the restricted problem? – sobasu Mar 20 at 17:23

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.)

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.

Your Answer

 
discard

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.