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Theorem: Let $U$ be an open set in $\mathbb{C}$, $f$ an analytic function on $U$, and $z_0 \in U$. Then $f$ has a power series expansion centered at $z_0$. If $r$ is a positive real number, and $U$ contains the disc of radius $r$ centered at $z_0$, then the radius of convergence of that power series is at least $r$.

Any analytic continuation of $f$ to a larger open set $W$ containing $U$ is unique, so the same result holds with $U$ replaced by $W$. In other words, the boundary of the disc of convergence of a local power series expansion of an analytic function is the point at which the given function ceases to be analytic.

The definition of an analytic function has a natural generalization to any topological field $k$ which is complete with respect to some absolute value (the main examples are $k = \mathbb{R}$ or a finite extension of $\mathbb{Q}_p$). For $U$ an open set of $k^n$, an analytic function $f: U \rightarrow k$ is one which has a local power series expansion about every point of $U$. This is defined in Serre, Lie Groups and Lie Algebras.

When $k = \mathbb{R}$, the theorem is false. Let $U = k$, and define $f: k \rightarrow k$ by $f(x) = \frac{1}{1+x^2}$. Interpreting $f$ as the restriction to $\mathbb{R}$ of a complex analytic function, we see that $f$ is analytic, and about $z_0 = 0$ has the power series expansion

$$1 - x^2 + x^4 - \cdots$$

which has radius of convergence $1$ (which follows from the Theorem and the fact that $\frac{1}{1+z^2}$ is meromorphic on $\mathbb{C}$ with singularities at $i, -i$.

What about when $k$ is a finite extension of $\mathbb{Q}_p$? Is the theorem still true?

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    $\begingroup$ Actually, if $p \equiv 3 \pmod{4}$, then $-1$ not a square in $\mathbb{Q}_p$, and the power series expansion of $f(x) = \frac{1}{1+x^2}$ about $0$ should again have radius $1$. $\endgroup$
    – D_S
    Jul 16, 2017 at 21:50
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    $\begingroup$ It’s true to the extent that the question makes sense, I think. Maybe a better example is the expansion of the exponential function, convergent for $v(z)>\frac1{p-1}$, where here $v$ is the additive valuation normalized so that $v(p)=1$. There the basic function is $\log(1+x)$, convergent for all $z$ with $v(z)>0$, and it becomes not one-to-one as soon as $0<v(z)\le\frac1{p-1}$. That’s what prevents the exponential function from being convergent when $v(z)=\frac1{p-1}$. And those roots of the log are not in $\Bbb Q_p$ except for $-1\in\Bbb Q_2$. $\endgroup$
    – Lubin
    Jul 18, 2017 at 23:37
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    $\begingroup$ "is the point at which the given function ceases to be analytic" and what does that mean exactly? $\endgroup$
    – zhw.
    Jul 19, 2017 at 23:13
  • $\begingroup$ Analytic continuation of complex functions being unique, you can conclude that there is a maximal connected $\Omega$ open set in $\mathbb{C}$ on which a given analytic function is defined. Taking a point near the boundary $\Omega$, the radius of convergence of the power series there stops at the boundary. $\endgroup$
    – D_S
    Jul 20, 2017 at 2:56
  • $\begingroup$ The problem is see in $\mathbb{C}_p$ is how to cover $|z| < 2$ by finitely many balls $|z-a| < 1$ ? This is how we prove in $\mathbb{C}$ that a function $f$ analytic on $|z| < 2$ and given by a power series on each of those finitely many balls is given by a power series on $|z| < 2$. $\endgroup$
    – reuns
    Jul 28, 2017 at 10:50

1 Answer 1

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The problem with carrying over your theorem from $\mathbb{C}$ to say finite extensions of $\mathbb{Q}_p$ is that the notion of "ceases to be analytic" cannot be carried over. In your definition of when a function ceases to be analytic you are considering a maximal connected open set on which a given analytic function is defined.

The natural analogue of $\mathbb{C}$ is $\mathbb{C}_p$, which is obtained by first taking the algebraic closure of $\mathbb{Q}_p$ (which is not complete) and then the completion of this algebraic closure, which can be proved to be algebraically closed (see the book of Koblitz). The $p$-adic valuation $v_p$ can be extended uniquely to $\mathbb{C}_p$ and we can endow $\mathbb{C}_p$ with the $p$-adic topology, by taking as a basis the balls $$B(a,r)=\{ x\in \mathbb{C}_p: v_p(x-a)>r\},$$for all $a\in \mathbb{C}_p$, $r\in \mathbb{Q}$. With respect to this topology, every nonempty open set of $\mathbb{C}_p$ is disconnected. The complement of $B(a,r)$, that is $$B(a,r)^c=\{ x\in \mathbb{C}_p : v_p(x-a)\leq r\}$$ is also open. Indeed, if $b\in B(a,r)^c$, then $B(b,r)$ is contained in $B(a,r)^c$ by the fact that $$v_p(u+v)=\min (v_p(u),v_p(v))$$for all $u$, $v\in \mathbb{C}_p$ such that $v_p(u)$ is different from $v_p(v)$.

Now let $O$ be any nonempty open subset of $\mathbb{C}_p$. Choose $a\in O$ and then $r$ such that $B(a,r)$ is strictly contained in $O$. Then $O$ is the union of two disjoint nonempty open sets, i.e. $B(a,r)$ and the intersection of $O$ with $B(a,r)^c$.

This shows that your definition of a function ceasing to be analytic cannot be carried over to $\mathbb{C}_p$. The same reasoning applies if you take a finite extension of $\mathbb{Q}_p$ instead of $\mathbb{C}_p$.

There are more strange phenomena. For instance, consider the function $f$ on $\mathbb{C}_p$, given by $f(x)=1$ if $v_p(x)>0$, and $f(x)=0$ if $v_p(x)<0$. This is everywhere analytic on $\mathbb{C}_p$ in the sense that it is locally given by a power series everywhere. However, it cannot be defined by a single power series on all of $\mathbb{C}_p$.

In fact, it is because of the strange property of the $p$-adic topology that every nonempty open subset of $\mathbb{C}_p$ is disconnected, that the usual complex analysis cannot be carried over.

The problem is see in $\mathbb{C}_p$ is how to cover $|z| < 2$ by finitely many balls $|z - a| < 1$? This is how we prove in $\mathbb{C}$ that a function $f$ analytic on $|z| < 2$ and given by a power series on each of those finitely many balls is given by a power series on $|z| < 2$.

In $\mathbb{C}_p$, $|z|<2$ cannot be covered by finitely many balls $|z-a|<1$. For suppose it could, say by balls$$|z-a_1|<1,\ldots, |z-a_r|<1.$$ Each ball $|z-a_i|<1$ with $|a_i|<1$ is already contained in the ball $|z|<1$, by the ultrametric inequality. So without loss of generality we may assume that $a_1=0$ and $|a_i|\geq 1$ for $i=2,\ldots,r$. Now if $|z-a_i|<1$ for some $i=2,\ldots,r$ then $|z|=|a_i|$, again by the ultrametric inequality. This shows that every $z$ with $|z|<2$ either satisfies $|z|<1$, or $|z|$ is one of the values $|a_2|,\ldots,|a_r|$. But this is impossible, for instance all numbers $z=p^{-c}$ with $c\in \mathbb{Q}$ and $1\leq p^c<2$ lie in $\mathbb{C}_p$, $|z|<2$ and have $|z|=p^c$. So if we let $z$ run through all elements of $\mathbb{C}_p$ with $1\leq |z|<2$, then $|z|$ assumes infinitely many different values.

But the main reason why the results from complex analysis cannot be carried over from $\mathbb{C}$ to $\mathbb{C}_p$ is, as said, the disconnectedness of $\mathbb{C}_p$. Here, instead of $\mathbb{C}_p$ one could take any finite extension $k$ of $\mathbb{Q}_p$; there one encounters the same problem.

Recall that $\mathbb{C}$ is connected, meaning that a nonempty open subset of $\mathbb{C}$ cannot be the union of two disjoint, nonempty open subsets.

But $\mathbb{C}_p$ or any finite extension $k$ of $\mathbb{Q}_p$ is the union of the disk $|z|<1$ and the annulus $|z|\geq 1$ which are both open. For instance, if $z$ is any element of the annulus, then so is every $w$ with $|w-z|<1$, by the ultrametric inequality. Hence if $z$ is any element of the annulus then there is an open ball around it which is also contained in the annulus.

Now define the function $f$ by $f(z)=0$ if $|z|<1$, and $f(z)=1$ if $|z|\geq 1$. This function is locally analytic everywhere. But if $r>1$, then $f(z)$ cannot be given by a power series on $|z|<r$. For first of all, such a power series should be nonconstant, since $f$ assumes two different values on $|z|<r$; and secondly, one can show that any nonconstant power series assumes infinitely many different values, while $f$ assumes only two different values.

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    $\begingroup$ Great answer (and great user name) $\endgroup$ Aug 2, 2017 at 13:00

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