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P-adic analysis question

  1. Mar 21, 2010 #1
    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. jcsd
  3. Mar 24, 2010 #2
    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.
     
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