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Let $V:=\mathbb C \setminus \{1,2,...,n\}$ where $n\ge 3$ ; then is it true that $Aut(V)$ is a finite group of order at most $n!$ ? I can easily see from definition that $Aut(V)$ is a group but I am unable to get the finiteness and the estimate on the order of the group . Please help . Thanks in advance

NOTE : $Aut(V)$ is the set of all bi-holomorphic maps on $V$ . If possible , Please avoid using any version of Picard's theorem

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    What kind of automorphisms do you consider? $V$ isn't a group in any obvious way. – Wojowu Apr 11 '17 at 19:44
  • @Wojowu : I mean the set of all bi-holomorphic maps on $V$ –  Apr 11 '17 at 19:49
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    In that case, this group has two elements - note that by great Picard theorem it cannot have an essential singularity at infinity, so is a polynomial, and consider what degree it can have. – Wojowu Apr 11 '17 at 19:50
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    Hardly standard notation. Put the clarification in the question - readers should not have to dig through comments to help you. Help people help you. – Thomas Andrews Apr 11 '17 at 19:50

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Let $X = \{ 1, \dots, n, \infty \}$ denote the set of isolated singularities of $f$. We will argue that all the isolated singularities are removable ones or poles. To see why, let $z_0 \in X$ and choose a ball $B \subseteq \hat{\mathbb{C}}$ around $z_0$ which does not contain all the other singularities. Let $B'$ be an open ball contained in $V \setminus B$. Since $f$ is an automorphism of $V$, we have $f(B') \cap f(B \setminus \{ z_0 \}) = \emptyset$ and since $f$ is not constant, $f(B')$ is open. If $z_0$ is an essential singularity, Casorati-Weierstrass tells us that $f(B \setminus \{ z_0 \})$ must be dense in $\mathbb{C}$ which contradicts $f(B') \cap f(B \setminus \{ z_0 \}) = \emptyset$.

Hence, all the singularities of $f$ are (at most) poles so $f$ extends to a holomorphic function on $\hat{\mathbb{C}}$ which implies that $f$ is a rational function. Writing $f(z) = \frac{p(z)}{q(z)}$, we see that the only chance $f$ will map $V$ bijectively to itself is if $\deg p = \deg q = 1$ so $f$ is in fact an automorphism of $\hat{\mathbb{C}}$ (a Möbius transformation) which preserves the set $X$. Hence, we get a map $\operatorname{Aut}(V) \rightarrow S(X)$ where $S(X)$ is the group of permutations on $X$. If $|X| \geq 3$ then the fact that a Möbius transformation is determined uniquely by its action on three points shows that this map is injective so $\operatorname{Aut}(V)$ is finite of cardinality less than or equal to $(n+1)!$.

Here things get a little messy. Since the action of $f$ is determined by its action on three points in $X$, it is actually clear that $|Aut(V)| \leq (n+1)n(n-1) \leq n!$ if $n \geq 5$. So you are left with analyzing explicitly the cases $n = 3$ and $n = 4$ to see what happens then. An example of such analysis is provided in this answer.

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  • why should we have $\deg p=\deg q=1$ ? –  Apr 21 '17 at 06:35
  • @SaunDev: For a generic $w \in \mathbb{C}$, the preimage $f^{-1}(w)$ has size $\max(\deg p, \deg q)$ (this is the number of solutions of the polynomial equation $p(z) = wq(z)$). Since $f|_{V} \colon V \rightarrow V$ is one-to-one and onto, this forces $\deg p = \deg q = 1$. – levap Apr 21 '17 at 06:48