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Differentiable Automorphisms of ℂ

  1. Jan 23, 2012 #1
    Are there any nontrivial differentiable automorphism of the complex numbers? I know there are many automorphisms, but I could only find one article that discussed them. I didn't read the entire thing, but it mentioned that AC is often necessary to construct them, but I didn't see whether it said that was always the case.

    Obviously, there's the identity; and then there's the conjugate which is trivial and continuous but of course not differentiable. My intuition is that any others would have to be pretty crazy, possibly not even integrable, but I haven't been able to prove that. Any insights?
  2. jcsd
  3. Jan 24, 2012 #2
    Did you see this article?

    http://mathdl.maa.org/images/upload_library/22/Ford/PaulBYale.pdf [Broken]

    It's a very readable analysis of complex automorphisms. His theorem 4 says that any complex automorphism other than the identity and complex conjugation must be discontinuous.

    It occurs to me that there's an easy proof. Any automorphism f must fix the rationals Q. That's because f(1) = 1 and so for any natural n, f(n) = f(1 + ... + 1) = n*f(1) = n.

    And n*f(1/n) = f(1/n) + ... + f(1/n) = f(1/n + ... + 1/n) [n-times] = f(n/n) = f(1) = 1. So f(1/n) = 1/n.

    Then for m/n rational we have f(m/n) = f(1/n + ... + 1/n) [m times] = m*f(1/n) = m/n.

    So f fixes the rationals. Any continuous function on the reals is determined by what it does to the rationals, so if f is continuous it fixes the reals. But the only complex automorphisms that fix the reals are the identity and conjugation. So those are the only possible continuous automorphisms, and the rest (given by AC) are discontinuous.

    ps better throw in f(-1) = -1 so you can get to the negative rationals. f(-1)*f(-1) = f(1) = 1 so f(-1) must be either 1 or -1. But f(1) = 1 and f is injective so f(-1) must be -1.
    Last edited by a moderator: May 5, 2017
  4. Jan 24, 2012 #3


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    There are none besides the identity. To see this, note that an automorphism of C will fix Q. So a continuous automorphism of C will fix R (because Q is dense in R), and consequently will be either the identity or complex conjugation. And only the former is (complex) differentiable.

    An alternative way to see this is to note that any differentiable automorphism of the plane is in particular a biholomorphism C->C so must be of the form z->az+b for some a,b in C with a nonzero (a cute exercise). The only such map that's a ring homomorphism is the one with a=1 and b=0 (another cute exercise).
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