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Stream: learning: questions

Topic: Étale space of analytic functions is Hausdorff

Peva Blanchard (May 10 2024 at 08:52):

This is not a question.

Here I am answering a side quest from @David Egolf 's thread.

Puzzle: Show that the étale space $\Lambda(F)$ associated with the sheaf $F$ of analytic functions on $X = \mathbb{R}$ is also Hausdorff.

The topology on $\Lambda(F)$ is generated by the following base of open subsets. Given $(U, s)$ with $U$ an open subset of $X$ , and $s \in F(U)$ a section of the analytic sheaf, define $[U, s]$ to be the set of germs $[s_x]$ for all $x \in U$ . An open set is $\Lambda(F)$ is an arbitrary union of these basic open sets.

Let $a, b$ be two distinct points of $\Lambda(F)$ . I.e., $a$ (resp. $b$ ) is the germ at $x$ (resp. $y$ ) of some analytic function $s$ (resp. $t$ ) with an open domain $U$ (resp. $V$ )

$\begin{align*} a &= [s]_x \\ b &= [t]_y \\ \end{align*}$

Peva Blanchard (May 10 2024 at 08:54):

Case $x \ne y$

First, consider the case $x \ne y$ . Because $X = \mathbb{R}$ is Hausdorff, there exists two open subsets $U', V'$ such that

$\begin{align*} x \in U' &\subseteq U \\ y \in V' &\subseteq V \\ U' \cap V' &= \emptyset \end{align*}$

Let $A = [U', s_{|U'}]$ (resp. $B = [V', t_{|V'}]$ ) be the open subset of $\Lambda(F)$ associated to the restriction of $s$ (resp. $t$ ) to $U'$ (resp. $V'$ ). Then

$\begin{align*} a &\in A\\ b &\in B\\ A \cap &B = \emptyset\\ \end{align*}$

Peva Blanchard (May 10 2024 at 08:58):

Case $x = y$

Let $W = U \cap V$ , and

$\begin{align*} A &= [W, s_{|W}] \\ B &= [W, t_{|W}] \\ \end{align*}$

Those sets are open neighbourhoods of $a$ and $b$ respectively.

We claim that $A \cap B = \emptyset$ . Assume, by contradiction, that there exists a germ $c \in A \cap B$ . This means that $s$ and $t$ have a common germ $c$ at some point $z \in W$ .

$[s]_z = [t]_z = c$

In particular, $s$ and $t$ have the same power series at $z$

$\sum_{n\ge0} s^{(n)}(z)\frac{(T - z)^n}{n!} = \sum_{n\ge0} t^{(n)}(z)\frac{(T - z)^n}{n!}$

Because $s$ and $t$ are analytic functions, this implies that they agree on some open subset of $W$ . Moreover, the unicity of analytic continuation implies that $s$ and $t$ actually agree over $W$ . Hence

$a = [s]_x = [t]_x = b$

But we assumed $a$ and $b$ to be distinct, whence a contradiction.

Peva Blanchard (May 10 2024 at 08:59):

Final remarks

The crux of the argument seems to rely on the fact that, for every point $x \in X$ , the map

$\begin{align*} \phi_x : \Lambda(F)_x &\rightarrow \mathbb{R}^{\mathbb{N}} \\ [s]_x &\mapsto (s(x), s^{(1)}(x), s^{(2)}(x), \dots) \end{align*}$

is injective.

Peva Blanchard (May 10 2024 at 09:07):

Now, I am wondering if this injectiveness characterizes the Hausdorff property.

Let's consider an arbitrary sheaf $F$ over $\mathbb{R}$ , for which there is a way to make sense of the $\phi_x$ 's (e.g., maybe a sub-sheaf of some sheaf of distributions).

Can we prove that the proposition " $\Lambda(F)$ is Hausdorff" is equivalent to "the $\phi_x$ 's are all injective"?

John Baez (May 10 2024 at 13:18):

The only problem with proving a general proposition like that is that for an arbitrary sheaf $F$ on $\mathbb{R}$ we don't have any natural way to define the $\phi_x$ functions. We could try to study which sheaves do have such functions, and then try to prove the proposition you want. But I think there's a necessary and sufficient condition for $\Lambda(F)$ to be Hausdorff which is very general, simple to state, and easy to prove! It's this:

Conjecture. Suppose $F$ is sheaf on a topological space $X$ . Then $\Lambda(F)$ is Hausdorff if and only if for every open $U \subseteq X$ and every $s_1, s_2 \in \Lambda(U)$ , the set of points $x \in U$ for which the germ of $s_1$ equals the germ of $s_2$ is closed in $U$ .

John Baez (May 10 2024 at 13:19):

I'm calling it a conjecture because I haven't checked it, but a special case is proved here.

John Baez (May 10 2024 at 13:20):

In the case where $F$ is the sheaf of analytic functions on $\mathbb{R}$ , we use the conjecture by showing that for any two analytic functions defined on some open set $U \subseteq \mathbb{R}$ , the set on which their germs are equal is closed in $U$ . The conjecture then implies $\Lambda(F)$ is Hausdorff.

John Baez (May 10 2024 at 13:22):

In the case where $F$ is the sheaf of smooth functions on $\mathbb{R}$ , we use the conjecture by finding two smooth functions on $\mathbb{R}$ for which the set on which their germs are equal is not closed. We can use the function $0$ and the function

$f(x) = \begin{cases} 0 &\text{if } x \le 0 \\ e^{-\frac{1}{x^2}} &\text{otherwise} \end{cases}$

We can check that these two functions have equal germs at $x$ for $x \lt 0$ , but different germs for $x \ge 0$ .

The conjecture then implies that $\Lambda(F)$ is not Hausdorff.

Peva Blanchard (May 11 2024 at 21:52):

This proposition is way more neat, indeed!

Here is a proof (I think).

Forward direction

Assume $\Lambda(F)$ is Hausdorff. Let

$D = \{ x\in U | [s]_x \ne [t]_x \}$

We show that $D$ is open. Fix $x \in D$ . Since $\Lambda(F)$ is Hausdorff, there exists two disjoint open subsets $S, T \subseteq \Lambda(F)$ such that

$\begin{align*} [s]_x &\in S \\ [t]_x &\in T \\ S \cap &T = \emptyset \end{align*}$

By definition of the topology on $\Lambda(F)$ , this implies that there exists open neighborhoods $V, W \subseteq U$ of $x$ such that

$\begin{align*} [s]_x &\in [V, s_{|V}] \\ [t]_x &\in [W, t_{|W}] \\ [V, s_{|V}] \cap [W, t_{|W}] &= \emptyset \end{align*}$

The last condition further implies that

$[V \cap W, s_{|V \cap W}] \cap [V \cap W, t_{|V \cap W}] = \emptyset$

In other words, $x \in V \cap W \subseteq D$ . I.e., $D$ is open.

Reverse direction

Assume that for any open subset $U$ , for any sections $s, t \in F(U)$ , the set $\{ x \in U ~|~ [s]_x \ne [t]_x \}$ is open. We prove that $\Lambda(F)$ is Hausdorff.

Fix two germs $a = [s]_x$ and $b = [t]_y$ in $\Lambda(F)$ , and assume that they are distinct, $a \ne b$ .

If $x \ne y$ , then, since $X$ is Hausdorff, there exists two disjoint open subsets $V, W$ containing $x$ and $y$ respectively. Then

$\begin{align*} [V \cap dom(s), s_{| V \cap dom(s)}] &\text{ is an open neighborhood of } a\\ [W \cap dom(t), t_{| W \cap dom(t)}] &\text{ is an open neighborhood of } b\\ \end{align*}$

Because $V$ and $W$ are disjoint, these open sets are also disjoint.

Assume now that $x = y$ . Let $D = \{ x \in U ~|~ [s]_x \ne [t]_x\}$ . Then $x \in D$ . Since $D$ is open, there exists an open neighborhood $V \subseteq D$ of $x$ . Then

$\begin{align*} [V \cap dom(s), s_{|V\cap dom(s)}] &\text{ is an open neighborhood of } a\\ [V \cap dom(t), s_{|V\cap dom(t)}] &\text{ is an open neighborhood of } b\\ \end{align*}$

and, by definition of $D$ , these two open subsets of $\Lambda(F)$ are disjoint.

Therefore, $\Lambda(F)$ is Hausdorff.

Peva Blanchard (May 11 2024 at 22:04):

I think my wanderings about analytic functions and stuff is due to the fact that I was trying to capture the argument "if a property $P$ is true at some point $x$ , then it is true on a neighborhood of $x$ ".

But this is actually pure topology: this argument is true exactly when the subset characterized by $P$ is open w.r.t to the topology of $X$ .

John Baez (May 12 2024 at 09:08):

Peva Blanchard said: