1. Mar 21, 2010

lark

The p-adic numbers Qp don't have a square root of -1, if p=3 mod 4.
So would differentiable functions from Qp -> Qp satisfy the
Cauchy-Riemann equations? I don't know why not.

To what extent would analysis in Qp have the familiar complex analysis
theorems??? You couldn't prove that Qp is algebraically complete, I
wonder what would block the complex analysis proof of that, that 1/p(x)
would be a bounded entire function if it had no roots.

Laura

2. Mar 24, 2010

lark

I looked into it some more -

Derivatives are used in the p-adics, although "strictly differentiable" is
a more useful concept than "differentiable". Strictly differentiable at a
point p means that as x and y approach p, then f(x)-f(y)/(x-y) approaches
f'(p), the derivative of f at p. If the function is strictly differentiable
and its derivative at a point is nonzero, it's locally injective, so you
get an inverse function theorem.

There's a definition of analytic functions for p-adics: a function f is analytic
in an open set D if f can be expressed as a power series around a point u in D.
Unlike with complex analysis, if a function is infinitely differentiable, that
doesn't mean it's analytic.

There are antiderivatives but no fundamental theorem of calculus in Qp, so the
antiderivatives aren't related to integration.

You can define an integral, for functions from the p-adic integers to Qp.

So, if a function is defined on a bounded subset of Qp, you could define
integration just by multiplying the argument by some power of p. Perhaps you
could use this to define an integral on an unbounded subset of Qp by a limit
process - but probably, the integral as defined by multiplying by one power of
p can't be made the same as the integral defined by multiplying by another
power of p, so that limit process might not work.

There IS a version of the Cauchy integral formula, residue theorem, and
maximum modulus principle for p-adics (see p. 129 of Koblitz "P-adic analysis:
a short course on recent work" online at http://books.google.com ).
It uses the Shnirelman integral. The Shnirelman integral is ingenious: for
z in Cp, the metric completion of the algebraic closure of Qp, and f: Cp -> Cp,
and a point a in Cp, you define integral (f, a, z) as the limit n -> oo of
1/n (sum over e: e^n=1 of f(a+ez)). In the limit, you skip n's that are
divisible by p. So, it's taking the limit of the average of f on these
points sprinkled over a circle centered at a.

Just like in complex analysis, you can use this integral to show that
a bounded analytic function is a constant.

But that's as far as it goes. In
the complex analysis proof that a polynomial p(x) has a root in C, you show
that 1/p(x) is a bounded analytic function if p(x) has no roots. But for
p(x) in Qp[x], 1/p(x) isn't analytic just because it's differentiable.