# Identity theorem

1. Apr 21, 2014

### disregardthat

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

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

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

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 $a \in X$ of A is a point which cannot be separated from $A / \{a\}$. 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
2. Apr 21, 2014

### micromass

Staff Emeritus
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.

3. Apr 21, 2014

### disregardthat

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 $f_1(b)=f_2(b)$ since $f_1$ and $f_2$ are continuous. Choose charts $\phi : U \to V$ and $\psi : U' \to V'$ on Y with $b \in U$ and $f_i(U) \subseteq U'$. We may also assume that U is connected. The mappings

$$g_i = \psi \circ f_i \circ \phi^{-1} : V \to V' \subseteq \mathbb{C}$$

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

4. Apr 21, 2014

### disregardthat

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

$U \cap G \not = \emptyset$ is actually fine since we have already assumed $b \in G$. 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 $a \in G$.

EDIT: Just a note: $b \in G$ is wrong at this stage of the argument, but b being a boundary point of G implies that $U \cap G \not = \emptyset$ anyway.

Last edited: Apr 21, 2014
5. Apr 21, 2014

### micromass

Staff Emeritus
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 $a$):

$$f(x) = \sum_{n=1}^{+\infty}\alpha_n (x-a)^n = (x-a) g(x)$$

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 $a$.

6. Apr 21, 2014

### disregardthat

Thanks, this does resolve the issue. (I don't think that it follows that $a \in A$ by the way, but it doesn't affect your argument)

7. Apr 21, 2014

### micromass

Staff Emeritus
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 $a\in G$.

8. Apr 21, 2014

### micromass

Staff Emeritus
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.

9. Apr 21, 2014

### mathwonk

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.

10. Apr 21, 2014

### disregardthat

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