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How can we reason to show that an entire function that's invariant under two rotations of the plane, must be constant ? Assume the rotations are around different axes, and by rational multiples of $\pi$.

Concretely, let $f$ be an entire function. Let $\varphi(z)=e^{i\theta\pi}z$, $\theta\in \mathbb{Q}$, be a rotation around $0$, and let $\psi(z)=1+e^{i\eta\pi}(z-1)$, $\eta\in \mathbb{Q}$, be a rotation around $1$. How to show that if $$f(\varphi(z))=f(z),\;\;\; f(\psi(z))=f(z)$$ for all $z\in \mathbb{C}$, then $f$ must be a constant function.

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$\cos(2\pi z)$ is invariant under rotations by $\pi$ around both $0$ and $1$. Did you mean to exclude that case? – Robert Israel Feb 28 '13 at 8:30
@RobertIsrael Thanks. I did not originally mean to exclude it. But what if ? suppose we exclude these $\pi$ rotations ? – Teddy Feb 28 '13 at 8:36
Then you get $\cos(2 \pi z^2)$ invariant under rotations of $\pi/2$. There isn't an obvious notion of degeneracy at play here; I don't think there is any real merit to excluding counter-examples ad-hoc as they're found; either seriously rethinking the question, or accepting there are counter-examples. – Hurkyl Feb 28 '13 at 10:22
@Hurkyl Your example only works for rotations around zero, not around $1$. – Teddy Feb 28 '13 at 11:02
up vote 5 down vote accepted

Let $\mathscr{C} \subset \mathbb{C}$ be the unit circle. For any $\omega \in \mathscr{C}$, define two maps: $$ \phi_{\omega} : \mathbb{C} \ni z \mapsto \omega z \in \mathbb{C}\,\,\,\text{ and }\,\,\, \psi_{\omega} : \mathbb{C} \ni z \mapsto 1 + \omega (z - 1) \in \mathbb{C} $$ If $f$ is an entire function on $\mathbb{C}$ invariant under the coordinate transform $\phi_{\zeta}$ and $\psi_{\eta}$ for some $\zeta, \eta \in \mathscr{C}\setminus \{1\}$, i.e. $$ f\circ\phi_{\zeta} = f\circ\psi_{\eta} = f$$ $f$ will be invariant under their inverses $\phi_{\zeta}^{-1} = \phi_{\zeta^{-1}}$ and $\psi_{\eta}^{-1} = \psi_{\eta^{-1}}$ and hence under $\phi_{\zeta}\circ\psi_{\eta}\circ\phi_{\zeta^{-1}}\circ\psi_{\eta^{-1}}$, i.e. $$f\circ\phi_{\zeta}\circ\psi_{\eta}\circ\phi_{\zeta^{-1}}\circ\psi_{\eta^{-1}} = f$$

But $\phi_{\zeta}\circ\psi_{\eta}\circ\phi_{\zeta^{-1}}\circ\psi_{\eta^{-1}}$ is the map

$$\mathbb{C} \ni z \mapsto z - (1-\zeta)(1-\eta) \in \mathbb{C}$$

which is a translation.

If at least one of $\zeta, \eta \ne -1$, say $\eta \ne -1$, then by a similar argument on $\zeta, \eta^{-1}$, we find $f$ is invariant under another translation:

$$\mathbb{C} \ni z \mapsto z - (1-\zeta)(1-\eta^{-1}) \in \mathbb{C}$$

It is easy to check these two translations are linear independent from each other. As a result, $f$ will be a doubly periodic entire function.

The fundamental domain of the lattice formed by the two translations is a compact subset of $\mathbb{C}$. Since $f$ is continuous, $f$ is bounded on it. By double periodicity of $f$, $f$ is bounded over all $\mathbb{C}$. By Liouville's theorem, $f$ is a constant.

Conclusion: If an entire function $f$ is invariant under rotations around two points, some of which are rotations with angle differ from $180^{\circ}$, then $f$ is a constant.

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If the orbit of some point is dense in the plane, then any continuous function invariant under the two rotations must be constant.

When this does not happen, I think the question is no longer really about complex analysis -- this is a question about group theory and plane geometry. Specifically, you want to know the subgroup of Euclidean motions of the plane generated by the two chosen rotations, and you want to understand how they can act on the plane.

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your reasoning seems very nice. can you please explain your answer little bit more. – GA316 Aug 20 '14 at 8:40

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