I'll state the theorem I am posing up front, and then explain why I think this theorem appears to be true. I am asking if anyone can prove it, or knows references to where it is proved. Please, forgive me if it is trivial, or if a counter example is trivial too.

If $\phi(s,z) : \mathbb{C}_{\Re(s)>0} \times \mathbb{C} \to \mathbb{C}$ is holomorphic in both variables, and $$\phi(s_0,\phi(s_1,z)) = \phi(s_0 + s_1,z)$$ then necessarily $\frac{\partial^2}{\partial z^2} \phi(s,z) = 0$, so that $$\phi(s,z) = e^{qs}(z-z_0) + z_0$$ for $q,z_0 \in \mathbb{C}$ or $$\phi(s,z) = z+cs$$ for $c \in \mathbb{C}$

I ask this question because

A) I've never encountered a counter example in an extensive study of semigroups.

B) I can prove on a whole bunch of occasions if $\phi(a,z) = f(z)$; for $\Re(a) > 0$, for specific $f$; implies $\phi(s,z)$ cannot exist.

Examples include $f = \sin,\cos,\exp,\exp(p(z)), p(\exp(z))$ where $p$ is an arbitrary polynomial. If $f$ has a super attracting fixed point, no such $\phi$ exists. If $f^{\circ n}$ has fixed points $f$ doesn't have, then no such $\phi$ exists. In all these cases, no such solution exists for the orbits of $f$ either. I'm wondering if this is a universal trait. That, in some sense, the theorem above can be a kind of Liouville theorem.

Let's say that $\phi$ satisfies $f$, if $\phi(a,z) = f(z)$ for some $\Re(a) > 0$. To highlight the similarity to Liouville's theorem, the original theorem can be restated as

If $f:\mathbb{C} \to \mathbb{C}$ and some semigroup $\phi$ satisfies $f$, then $f= mz+b$ for some $m,b \in \mathbb{C}$.

Supposing the pair $f(\cdot)$ and $\phi(s,\cdot)$ don't have to map $\mathbb{C}$ to itself, that $\mathbb{C}$ is weakened to the unit disk $\mathbb{D}$, and $f(0) = 0$ with $|f'(0)| \neq 0,1$, then there always exists a $\phi$ satisfying $f$. It seems though, as soon as we lift from $\mathbb{D}$ (or any simply connected domain biholomorphic to $\mathbb{D}$) to $\mathbb{C}$ it fails.

I can show the result in a restricted form, which I also think is interesting

If $p:\mathbb{C} \to \mathbb{C}$ is a polynomial, and there exists a semigroup $\phi$ satisfying $p$ then $p(z) = mz+b$ for some $m,b \in \mathbb{C}$

My main avenue of approach for deriving a contradiction has been to consider the Weierstrass product. It follows that if $\phi(s_0,z_n) = z_n$ then $\phi(s,z_n) = z_n$, so $\phi$ has a countable list of fixed points invariant on our choice of $s$, allowing us to say

$$\phi(s,z) = z + e^{g(s,z)}\prod_{n=0}^{\infty}(1-\frac{z}{z_n})e^{-p_n(z/z_n)}$$

where $p_n(z) = \sum_{j=1}^n \frac{z^j}{j}$. I've been fiddling around with this as it seems like a smart idea. Since $\lim_{s\to 0}\phi(s,z) = z$, and we have a semigroup property, it seems reasonable to think we can show that for all $\epsilon >0$ there exists a $\delta > 0$ such that $|\phi(\delta,z)-z| < C_\delta e^{|z|^{\epsilon}}$ then by Hadamard

$$\sum_{n=0}^\infty \frac{1}{|z_n|} < \infty$$

which greatly reduces the candidates for $f$ that can be satisfied by some $\phi$—$f$'s fixed points have to be sufficiently spaced out, so to speak.

Overall, I'm lost on the general case, and seem to be only able to prove it in specific circumstances. I think it says something rather profound about multiplication and addition, that, for another reason, they are incredibly special. Help, comments, suggestions, edits, anything is welcome, thanks.

  • $\begingroup$ Does e. g. $\phi(s,z)=\frac z{1-c s z}$ fall into your cases? $\endgroup$ Mar 26, 2017 at 20:16
  • 2
    $\begingroup$ @მამუკაჯიბლაძე I was going to mention these types of semigroups, but I thought it would thicken the question too much. Essentially, $\phi : \mathbb{C}_{\Re(s) > 0} \times \hat{\mathbb{C}} \to \hat{\mathbb{C}}$, it is a semigroup on the Riemann-sphere, but not on $\mathbb{C}$. As you can see, this $\phi$ has poles in $z$, so it is not holomorphic on $\mathbb{C}$ and fails the requirements of the theorem. $\endgroup$
    – user78249
    Mar 26, 2017 at 20:35
  • $\begingroup$ I see, thanks, interesting, sorry for not noticing this. $\endgroup$ Mar 26, 2017 at 20:41
  • $\begingroup$ No problem, happens to the best of us :) $\endgroup$
    – user78249
    Mar 26, 2017 at 20:45
  • $\begingroup$ I guess "holomorphic flows", "entire flows" will lead to more feedback than "semigroups". For instance see: Brian A. Coomes, Polynomial Flows on $\mathbb{C}^n$, Trans. A.M.S. Vol. 320, No. 2 (Aug., 1990), pp. 493-506 $\endgroup$
    – YCor
    Feb 3, 2020 at 17:02

1 Answer 1


Set $f_t(z)=\phi(t,z)$. Notice that entire functions $z\mapsto f_t(z)$ all commute with each other. I. N. Baker proved that if $f$ is a non-affine entire function with a repelling fixed point then the set of entire functions with commute with it is countable. MR0147650
Baker, Irvine Noel Permutable entire functions. Math. Z. 79 1962 243–249.

On the other hand, in a later paper he proved that for every non-affine entire function some iterate has a repelling fixed point. MR0226009
Baker, I. N. Repulsive fixpoints of entire functions. Math. Z. 104 1968 252–256.

Combination of these two theorems settles your question. Indeed, start with $f_1$. Then $f_n$ with some positive integer $n$ has a repelling fixed point. Then the set of entire functions which commute with $f_n$ is at most countable, contradiction.

Once we know that all $f_t$ are affine with respect to $z$, the rest is easy: commutative sub-semigroups of the affine group are just what you listed.

EDIT. Second theorem of Baker was based on the Ahlfors Islands Theorem, which was considered "heavy machinery" at that time. Since then both Baker's proof and the Ahlfors Islands theorem were very much simplified. For an elementary proof of the Baker theorem see, for example,

MR1782673 Berteloot, François; Duval, Julien Une démonstration directe de la densité des cycles répulsifs dans l'ensemble de Julia. Complex analysis and geometry (Paris, 1997), 221–222, Progr. Math., 188, Birkhäuser, Basel, 2000.

  • $\begingroup$ This is the kind of heavy machinery I was looking for, thanks so much! $\endgroup$
    – user78249
    Mar 27, 2017 at 2:03
  • $\begingroup$ Heavy machinery is involved only in the second theorem of Baker. However it has been dramatically simplified since. $\endgroup$ Mar 27, 2017 at 2:06
  • $\begingroup$ I guess I meant complex dynamics machinery, which my naive mind sees as 'heavy.' I'm much more familiar with complex analysis, and don't have that great of a purview of complex dynamics. I was looking at this problem purely analytically, which I knew was a set back. After reading some related papers, I realize this question needed to be attacked using complex dynamics, no wonder I couldn't construct a proof. Thanks again for the great literature! $\endgroup$
    – user78249
    Mar 27, 2017 at 20:15

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