Does there exist an open connected set in the complex plane on which the identity function has an analytic square root but not an analytic logarithm?

  • $\begingroup$ Sorry, I am unable to use your hint. Kindly elaborate. $\endgroup$ – Kabo Murphy Jan 22 '16 at 4:47
  • $\begingroup$ If I have an analytic logarithm I can define an analytic square root from it. I am asking for the converse. An analytic square root does not give me an analytic logarithm in any simple way. $\endgroup$ – Kabo Murphy Jan 22 '16 at 8:13

Suppose $U \subseteq \mathbb{C}$ is a connected open set on which an analytic square root can be defined. Then it follows easily that $U$ cannot contain the origin, and every closed curve in $U$ must have even winding number around the origin. But any closed curve in the plane with nonzero winding number around the origin contains in its image a simple closed curve with winding number one around the origin, so $U$ cannot have any closed curves with nonzero winding number around the origin, and hence an analytic logarithm exists on $U$.

See also this question.

  • $\begingroup$ "... any closed curve in the plane with nonzero winding number around the origin contains in its image a simple closed curve with winding number one around the origin ..." – I think that is intuitively clear to me, but could you give a hint how to prove that rigorously? $\endgroup$ – Martin R Jan 23 '16 at 17:34
  • $\begingroup$ I just noticed that the solution to my question is easy: it is a consequence of the proof of Riemann Mapping Theorem. For example we can take w_0=0 in the proof of that theorem in Rudin's Real and Complex Analysis. Thus the region is necessarily simply connected. $\endgroup$ – Kabo Murphy Jan 25 '16 at 7:50
  • $\begingroup$ I withdraw my last comment. Riemann Mapping Theorem requires existence of square roots for all holomorphic functions that don't have zeros. I Don't understand Jim Belk's argument. Apart from the question raised by Martin, I also don't understand the last step in Jim's Argument. $\endgroup$ – Kabo Murphy Jan 25 '16 at 8:14
  • $\begingroup$ I agree that my argument skips over some important details. Unfortunately, I don't have time to work on filling these in, so I've placed a bounty on the question in the hopes of attracting someone else to give a more complete answer. $\endgroup$ – Jim Belk Jan 26 '16 at 17:15

Here's a topological proof, along the lines of what Jim Belk proposed. I have tried to make it as precise as possible. Let's denote the punctured plane as $\Bbb C^* := \Bbb C \setminus \{0\}$, and throughout argument of a complex number will be $[0, 2\pi)$-valued.

I claim that the analytic square root is defined over $U$ if and only if $U \subset \Bbb C^*$ is a connected open subset such that $U$ does not intersect image of some proper ray $\gamma : [0, \infty) \to \Bbb C^*$ with $\gamma(0) = 0$.

First note that the map $f : \Bbb C^* \to \Bbb C^*$ given by $f(z) = z^2$ is a holomorphic double cover. For the forward direction, as $\Omega = \Bbb C^* \setminus \gamma([0, \infty))$ is a simply connected domain, $f|_{\Omega} : f^{-1}(\Omega) \to \Omega$ must be the trivial cover as every covering space of a simply connected domain must be trivial. This implies existence of a biholomorphism $g : \Omega \to f^{-1}(\Omega)$ which defines a section of the cover, i.e., $f(g(z)) = z$ for all $z \in \Omega$. $g$ defines the required analytic square root over $\Omega$.

For the other direction, suppose that $U$ intersects every proper ray emanating from the origin. In particular, it intersects the radial lines $\theta = \theta_0$. For any point $p = r_0\exp(\theta_0) \in U$ consider the open sector of the plane $S_\delta(p) = \{r_0\exp(\theta_0) \in \Bbb C : r_0 - \delta < r < r + \delta, \theta_0 - \delta < \theta < \theta_0 + \delta\}$, where $\delta$ is very small, so that $S_\delta(p_0) \cap U$ is a connected open neighborhood of $p_0$. Starting from a point $z_0 \in U \cap [0, \infty)$, iteratively mark down a sequence of points $\{z_k\} \in U$ such that $z_{n+1} \in S_\delta({z_n})\cap U$ and $\text{arg}(z_{n+1}) > \text{arg}(z_n)$ by moving a constant angular distance counterclockwise from $z_n$ to $z_{n+1}$ while staying in the $\delta$-sector. Eventually stop the algorithm for some $N$ when $\text{arg}(z_0) - \text{arg}(z_N)$ is less than the constant we're moving by, so $z_0 \in S_\delta(z_N)$. Since for each pair $z_i, z_{i+1}$ (where indices are modulo $N+1$) lies in $S_\delta(z_i) \cap U$, a connected open subset of $\Bbb C$ and therefore path-connected, there is a path $\gamma_i$ joining $z_i$ and $z_{i+1}$ contained in $S_\delta(z_i) \cap U$, hence in $U$.

Take the path $\gamma = \gamma_1 * \gamma_2 * \cdots * \gamma_N$ obtained by concatenating these individual paths. This is a loop $\gamma : [0, 1] \to U$ with basepoint $z_0$ which has winding number $1$ about the origin, therefore represents the generator of $\pi_1(\Bbb C^*$). As $f$ is a double cover, the map $f_* : \pi_1(\Bbb C^*) \to \pi_1(\Bbb C^*)$ is multiplication by $2$, once the identification $\pi_1(\Bbb C^*) \cong \Bbb Z$ has been made. $[\gamma] \in \pi_1(\Bbb C^*)$ does not belong to image of $f_*$, therefore $\gamma$ must lift to a path which is not a loop by the covering map $f$, but by path lifting lemma it must join the two distinct points in the fiber $f^{-1}(z_0)$. This implies the covering space $f : f^{-1}(U) \to U$ is connected, and therefore cannot even have a topological section, let alone a holomorphic one. Phrased otherwise, we have proved there is no analytic square root defined over $U$.

This implies if $U \subset \Bbb C^*$ has an analytic square root defined over it, it must miss a proper ray emanating from the origin. Since analytic logarithm is simply a (locally defined) section of the holomorphic covering map $\exp : \Bbb C^* \to \Bbb C^*$ given by the exponential, we can play out the same argument in the first half of the proof of the previous claim: if $U \subset \Omega := \Bbb C^* \setminus \gamma([0, \infty))$ for some proper ray $\gamma$, then $\exp : \exp^{-1}(\Omega) \to \Omega$ must be the trivial cover by simple-connectedness of $\Omega$. Therefore there is a holomorphic section $g : \Omega \to \Bbb C^*$ such that $\exp(g(z)) = z$ - this $g$ is precisely (some branch of a) holomorphic logarithm defined over $\Omega$. Therefore the analytic logarithm is defined over all of $\Omega$, hence in particular on the subset $U \subset \Omega$, as desired.


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