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Identity theorem

  1. Apr 21, 2014 #1

    disregardthat

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    I'm slightly confused at the proof of this theorem, hopefully someone can help.

    Identity theorem: Suppose X and Y are Riemann surfaces, and [itex]f_1,f_2:X \to Y[/itex] are holomorphic mappings which coincide on a set [itex]A \subseteq X[/itex] having a limit point [itex]a \in X[/itex]. Then [itex]f_1[/itex] and [itex]f_2[/itex] are identically equal.

    The proof starts out with: Let G be the set of all points [itex]x \in X[/itex] having an open neighbourhood W such that [itex]f_1|_W = f_2|_W[/itex].

    Now, it seems like they assume without argument that G is non-empty. Why is G non-empty?

    Also, maybe I'm just confused about this, what does it mean that A has a limit point a in X? As I understand it, a is simply a point which cannot be separated from A by two open sets. Couldn't it be the case that A is a one-point set, doesn't every set have a limit point? In that case I don't see how G could be non-empty.

    EDIT: Ok, upon some thought I think I recall that a limit point [itex]a \in X[/itex] of A is a point which cannot be separated from [itex]A / \{a\}[/itex]. Is this right? In that case A must be infinite. Still, this doesn't resolve the question of the existence of such a W as described above.
     
    Last edited: Apr 21, 2014
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  3. Apr 21, 2014 #2

    micromass

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    So far I don't see them assuming that ##G## is non-empty. Can you give the entire proof here (or at least the next few steps) so that I can check.

    A limit point of ##A## is a point ##a## such that for each neighborhood ##V## of ##a## holds that ##V\cap (A\setminus\{a\})## is nonempty. So ##A## can't be a one-point set in particular. Since your space is ##T_1## (all singletons are closed), you can indeed deduce that ##A## is infinite.
     
  4. Apr 21, 2014 #3

    disregardthat

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    Continuation of proof:

    By definition G is open. We claim that G is also closed. For, suppose b is a boundary point of G. Then [itex]f_1(b)=f_2(b)[/itex] since [itex]f_1[/itex] and [itex]f_2[/itex] are continuous. Choose charts [itex]\phi : U \to V[/itex] and [itex]\psi : U' \to V'[/itex] on Y with [itex]b \in U[/itex] and [itex]f_i(U) \subseteq U'[/itex]. We may also assume that U is connected. The mappings

    [tex]g_i = \psi \circ f_i \circ \phi^{-1} : V \to V' \subseteq \mathbb{C}[/tex]

    are holomorphic. Since [itex]U \cap G \not = \emptyset[/itex], the Identity theorem for holomorphic functions on domains in [itex]\mathbb{C}[/itex] implies that [itex]g_1[/itex] and [itex]g_2[/itex] are identically equal. Thus [itex]f_1|_U = f_2|_U[/itex]. Hence [itex]b \in G[/itex] and thus G is closed. Now since X is connected either [itex]G = \emptyset[/itex] or G = X. But the first case is excluded since [itex]a \in G[/itex] (using the identity theorem in the plane again). Hence [itex]f_1[/itex] and [itex]f_2[/itex] coincide on all of X.
     
  5. Apr 21, 2014 #4

    disregardthat

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    So basically,
    why is [itex]U \cap G \not = \emptyset[/itex], and why is [itex]a \in G[/itex] by using the identity theorem in the plane?

    [itex]U \cap G \not = \emptyset[/itex] is actually fine since we have already assumed [itex]b \in G[/itex]. But I don't see why such a b must exist. It does exist if G is non-empty.

    So it boils down that the fact that [itex]a \in G[/itex].

    EDIT: Just a note: [itex]b \in G[/itex] is wrong at this stage of the argument, but b being a boundary point of G implies that [itex]U \cap G \not = \emptyset[/itex] anyway.
     
    Last edited: Apr 21, 2014
  6. Apr 21, 2014 #5

    micromass

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    In ##C##, you could apply the following argument:
    Let ##f## be zero on a set ##A## with a limit point ##a##.Then there is a sequence ##a_n\rightarrow a## such that ##a_n\in A##. By continuity, we have ##0=f(a_n)\rightarrow f(a)##. Thus ##a\in A##.
    Thus (in a neighborhood of ##a##):

    [tex]f(x) = \sum_{n=1}^{+\infty}\alpha_n (x-a)^n = (x-a) g(x)[/tex]

    where ##g## is a holomorphic function. Clearly, ##g(a_n) = 0##. Thus by the same argument, ##g(a) = 0##. By induction we can show that ##\alpha_n = 0## for each ##\alpha_n##.

    In a standard way, you can lift this result to the Riemann surface to prove that if ##f## and ##g## coincide on a set ##A## with limit point ##a##, then ##f## and ##g## coincide on a neighborhood of ##a##.
     
  7. Apr 21, 2014 #6

    disregardthat

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    Thanks, this does resolve the issue. (I don't think that it follows that [itex]a \in A[/itex] by the way, but it doesn't affect your argument)
     
  8. Apr 21, 2014 #7

    micromass

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    Yeah, so consider the maps ##g_1## and ##g_2##. Since ##\varphi## are homeomorphisms, we have that ##\varphi(a)## is a limit point of ##\varphi(A)##. Furthermore, it is easy to check that ##g_1## and ##g_2## coincide on ##\varphi(A)##. Thus by the identity theorem in the plane (or the proof I gave in my last post), we can deduce that there is a neighborhood ##W## of ##\varphi(a)## such that ##g_1\vert_W = g_2\vert_W##. Then you can easily check that ##f_1 = f_2## on ##\varphi^{-1}(W)## and that ##a\in \varphi^{-1}(W)##. Thus ##a\in G##.
     
  9. Apr 21, 2014 #8

    micromass

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    Oh right. I was mentally taking ##A## to be the set of all points where ##f_1## and ##f_2## coincide. But that wasn't given. Sorry.
     
  10. Apr 21, 2014 #9

    mathwonk

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    a key point, not given in the statement, is the connectivity of X. Other than that, the proof of non emptiness of G rests on the principle of "isolated zeroes" for non trivial holomorphic functions. I.e. a non trivial powers series, looks like z^k.g(z) where g(0) ≠ 0. Hence there is an isolated zero at z=0. By taking local coordinates near a limit point a, and subtracting f1 from f2, one sees that f1 = f2 on a nbhd of a. This is why the set G is non empty. They presumably took it for granted that this fact from one vbl complex analysis was known.
     
  11. Apr 21, 2014 #10

    disregardthat

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    That's right, a riemann surface is defined in my book to be connected.
     
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