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Prove there does not exist a holomorphic function $f: \mathbb{C}\setminus [-1,1]\longrightarrow \mathbb{C}$ such that $f^2(z)=z^2-1$ for all $z\in \mathbb{C}\setminus [-1,1]$.

I really do not know how to solve... Thank you a lot.

mrf
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Shiquan
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2 Answers2

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There does exist a branch of $\sqrt{z^2-1}$ on $\mathbb{C}\setminus [-1,1]$ (since that is connected, there are exactly two branches).

Firstly, $G_1 = \mathbb{C}\setminus (-\infty,1]$ is simply connected, and $z^2-1$ has no zeros in $G_1$, hence there are two branches of $\sqrt{z^2-1}$ on $G_1$. Let $f_1$ be the branch taking positive real values on $(1,\infty)$.

Secondly, on $D_1(0)$, let $h$ be the branch of $\sqrt{1-w^2}$ with $h(0) = 1$, i.e.

$$h(w) = \sum_{k=0}^\infty (-1)^k\binom{\frac{1}{2}}{k}w^{2k}.$$

On $G_2 = \mathbb{C}\setminus \overline{D_1(0)}$, let

$$f_2(z) = z\cdot h\left(z^{-1}\right).$$

Then $f_2(z)^2 = z^2\cdot h(z^{-1})^2 = z^2(1-z^{-2}) = z^2-1$, so $f_2$ is a branch of $\sqrt{z^2-1}$, and $f_2$ takes positive real values on $(1,\infty)$. Also, $G_1 \cap G_2$ is connected, hence $f_1 = f_2$ on $G_1 \cap G_2$ by the identity theorem. Therefore

$$f(z) = \begin{cases}f_1(z) &, z \in G_1 \\ f_2(z) &, z \in G_2 \end{cases}$$

is a holomorphic branch of $\sqrt{z^2-1}$ on $G_1 \cup G_2 = \mathbb{C}\setminus [-1,1]$.

Daniel Fischer
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EDIT: I made a mistake in my computations, and the map does factor through, and, as pointed out in Daniel Fischer's answer, a square root does exist.

You can use the fact that the map $p(z)=z^2$ is a covering map from $\mathbb C\setminus \{0\} \rightarrow \mathbb{C}\setminus{0}$ . Given a function $f(z)$, a square root of a function $f(z)$ is a lift of $f$ by the covering map. This layout gives you topological obstructions for the existence of the square root as a lift; the induced map on the respective fundamental groups must satisfy that $\pi_*(\mathbb C-[-1,1])\subset \pi_*(\mathbb C)=0$, which is not satisfied here .

This When does a complex function have a square root? gives an excellent extended answer from which I learnt a lot.

Community
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user99680
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    This is wrong. There is a branch of $\sqrt{z^2-1}$ on $\mathbb{C}\setminus [-1,1]$. The branch cut joins the two simple zeros of $z^2-1$, so when analytically continuing a local branch, we get back where we started when winding around the cut. – Daniel Fischer Mar 27 '14 at 11:42
  • @DanielFischer: What is wrong with the argument then? I guess I had my spaces mixed up and I didn't realize that the lift actually exists. – user99680 Mar 27 '14 at 17:27
  • @Ren Shiquan: my comment was right on the idea, but wrong on the execution; the map does factor through, so the square root does exist, as pointed out in the post below. – user99680 Mar 27 '14 at 17:28
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    Let's write $g(z) = z^2-1$, then we have $g \colon \mathbb{C}\setminus [-1,1] \to \mathbb{C}\setminus {0}$ and the induced map $g_\ast \colon \pi_1(\mathbb{C}\setminus[-1,1]) \to \pi_1(\mathbb{C}\setminus{0})$ (ignoring the base points). $g_\ast$ is basically multiplication by $2$. The condition for the existence of a lift is that $p_\ast(\pi_1(\mathbb{C}\setminus{0})) \supset g_\ast(\pi_1(\mathbb{C}\setminus[-1,1]))$. Since $p_\ast$ is multiplication by $2$, that is the case here (the images are equal). – Daniel Fischer Mar 27 '14 at 18:30
  • @DanielFischer: yes, thanks, that's what I thought; I had the order of my spaces mixed up. So I used a correct premise to reach an incorrect conclusion. – user99680 Mar 27 '14 at 18:37
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    Dear @DanielFischer I think you get the inclusion the other way round: the condition foe the existence of a lift isn't that $g_(\pi_1(\Bbb C \setminus [-1,1]))\subseteq p_(\pi_1(\Bbb C \setminus{0}))$ ? – Luigi M Jan 06 '18 at 16:10
  • @LuigiM Right, thanks. Fixing that. – Daniel Fischer Jan 06 '18 at 16:25
  • @DanielFischer no, thanks to you for your insightful comments! – Luigi M Jan 06 '18 at 16:43