# Which holomorphic maps $f:\mathbb{H}\to\mathbb{H}$ satisfy $f(z+1)=f(z)-1$?

Let $$\mathbb{H}$$ denote the upper half-plane. Which holomorphic maps $$f:\mathbb{H}\to\mathbb{H}$$ satisfy $$f(z+1)=f(z)-1$$ for all $$z\in\mathbb{H}$$?

My guess is that none exist. I think there might be orientation issues or something.

One observation that I made is that $$e^{2\pi i f}$$ is $$1$$-periodic. I was not able to get any use out of this though. I am quite stuck.

Do such maps exist? And if not, then why not?

• $f(z)=-z$ is fine. Commented Jun 1 at 18:17
• $-z$ does not send $\mathbb{H}$ to $\mathbb{H}$. Commented Jun 1 at 18:18
• What is $\mathbb{H}$ ? Commented Jun 1 at 18:18
• Upper half plane Commented Jun 1 at 18:18
• Also @RyszardSzwarc your example is essentially why I thought it might be an orientation issue. The only map that I could think of that satisfies this property and preserves $\mathbb{H}$ was $-\overline{z}$, but this is antiholomorphic rather than holomorphic. Commented Jun 1 at 18:25

There are no such holomorphic functions. The proof is largely topological. Consider the quotient map $$z\mapsto \exp(2\pi i z), \mathbb H^2\to D^*=\{w: 0<|w|<1\}$$, where $$D^*$$ is the quotient of $$\mathbb H^2$$ by the action of the group $$\mathbb Z$$ generated by the translation $$g: z\mapsto z+1$$. Since $$f$$ satisfies $$f\circ g= g^{-1}\circ f$$, the map $$f$$ descends to a holomorphic self-map $$h: D^*\to D^*.$$ The latter induces a homomorphism $$\pi_1(D^*)\to \pi_1(D^*)$$ sending the generator to its inverse. Hence, taking the counterclockwise oriented circle $$C=\{w: |w|=1/2\}\subset D^*$$, we see that the winding number $$w(h(C))$$ of $$h(C)$$ around $$0$$ equals $$-1$$. By the argument principle, since $$w(h(C))=-1$$, we get $$Z(h) - P(h)=-1$$, where $$Z(h)$$ is the number of zeroes of $$h$$ in the open unit disk $$\Delta=\{w: |w|<1/2\}$$ and $$P(h)$$ is the number of poles of $$h$$ in $$\Delta$$. However, since $$h$$ is bounded, it has no poles. This is a contradiction since $$Z(h)\ge 0$$.

Holomorphic functions mapping $$H$$ into itself are called Pick functions and have the following integral representation

$$f(z)=\alpha z+\beta +\int_{-\infty}^{\infty}\frac{\gamma z+1}{\gamma -z}\mu(d\gamma)$$ where $$\mu$$ is a bounded positive measure, $$\alpha\geq 0$$ and $$\beta$$ is a real number. The theorem is due to Nevanlinna. For your $$f$$ we have $$f(z) +z=g(z)$$ where $$g$$ as period one and is also a Pick function with the same $$\mu$$ and $$\beta$$ but where the $$\alpha$$ component is replaced by $$\alpha+1>0.$$ If we watch $$t\mapsto h(t)=g(i\epsilon +t)$$ clearly $$h$$ is the sum of non zero affine function and of a bounded continuous function and cannot be periodic since it is not bounded.

• Perhaps $h(t)$ should be $g(t+i\varepsilon).$ Commented Jun 2 at 15:28
• Oh thanks. I am making the correction. Commented Jun 2 at 15:32
• For reference: en.wikipedia.org/wiki/Nevanlinna_function Commented Jun 2 at 22:50
• I also like this answer, and I have upvoted it. If I could accept multiple answers I would. Thank you for posting it! Commented Jun 3 at 17:16

Suppose such a function $$f(z)$$ exists; we will arrive at a contradiction. First note that $$g(z) = f(z) + z$$ has period $$1$$ and thus can be expanded in the form $$g(z) = \sum_{n=-\infty}^{\infty} a_n e^{2\pi i n z}$$ You can see this by looking at $${\displaystyle h(z) = g\bigg({\log z \over 2\pi i}\bigg)}$$, an analytic function on $$\{z: 0 < |z| < 1\}$$ and using the Laurent series expansion of $$h(z)$$. Note that $$g(z)$$ also maps $$\mathbb{H}$$ to itself. This precludes $$h(z)$$ from having either a pole or an essential singularity at $$z = 0$$ because in either case the range of $$g(z)$$ will then not be contained in $$\mathbb{H}$$ due to the behavior of $$h(z)$$ near $$0$$, which translates into corresponding behavior for $$g(z)$$ for large Im $$z$$.

Thus the only remaining possibility is that $$h(z)$$ has a removable singularity at $$z = 0$$. This means that $$a_n = 0$$ for all $$n < 0$$ and we may write $$f(z) = -z + \sum_{n=0}^{\infty} a_n e^{2\pi i n z}$$ If we move upwards along the imaginary axis $$f(z)$$ will eventually lie outside of $$\mathbb{H}$$ since the series tends to $$a_0$$. Thus we have a contradiction and no such $$f(z)$$ exists.

• I thought there should be a rather elementary proof along these lines, but failed to see the idea of excluding a singularity at $0$. Well done. Commented Jun 3 at 4:05
• I have a question about this answer. I agree that $h$ cannot have an essential singularity at the origin. I don't immediately see why $h$ cannot have a pole at the origin though. You say this is "because $g$ maps $\mathbb{H}$ to itself," but there are plenty of self-maps of $\mathbb{H}$ that exhibit this behavior (e.g., $z\mapsto -1/z$). I guess I can see it if you quotient the target space by $z\mapsto z+1$ and descend to the punctured disk like in Moishe's answer. Is that would you had in mind? Commented Jun 3 at 17:26
• @4plus4man Suppose $h(z)$ had a pole of order $k$ at the origin. On a small circle centered at the origin, $h(z)$ wraps around the origin clockwise $k$ times. In particular the range of $h(z)$ will leave ${\mathbb H}$. Translated in terms of $g(z)$, on a horizontal line Im$(z)= N$ for large enough $N$, $g(z)$ will leave ${\mathbb H}$. Commented Jun 3 at 17:38