Re: [math-fun] commutative algebra question, sort of from algebraic geometry
Allan writes: << I am guessing that a lot of the important struts supporting the theory of algebraic curves would be severely disrupted by the admission of non-polynomials. Most of the theory pivots on results that are in some sense generalizations of the root-counting corollaries of the Fundamental Theorem of Algebra, and root-counting is the first thing that fails when you leave the comfy confines of polynomials. I doubt if there could be any such thing as an analytic Riemann-Roch theorem.
No doubt many things fall apart when one goes beyond polynomials. I think Chris has been asking about real-valued functions, but for complex analytic functions there is indeed a Riemann-Roch theorem. (See, e.g., Theorem 16.9 in "Lectures on Riemann Surfaces" by Otto Forster, 1981). --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele
Shows how shallow my knowledge base is. I was working from my memory that the R-R theorem is counting things (the dimension of some function space "over" a curve, as I recall), and that the degree of the curve (that is, the degree of the underlying polynomial) is a crucial input to the calculation. My wife and I keep looking wistfully at Fulton's terse, pull-no-punches, execrably-typeset *Algebraic Curves* and thinking that we really, really, ought to ... On Thu, Sep 24, 2009 at 5:00 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Allan writes:
<< I am guessing that a lot of the important struts supporting the theory of algebraic curves would be severely disrupted by the admission of non-polynomials. Most of the theory pivots on results that are in some sense generalizations of the root-counting corollaries of the Fundamental Theorem of Algebra, and root-counting is the first thing that fails when you leave the comfy confines of polynomials. I doubt if there could be any such thing as an analytic Riemann-Roch theorem.
No doubt many things fall apart when one goes beyond polynomials.
I think Chris has been asking about real-valued functions, but for complex analytic functions there is indeed a Riemann-Roch theorem. (See, e.g., Theorem 16.9 in "Lectures on Riemann Surfaces" by Otto Forster, 1981).
--Dan
_____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele
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Allan Wechsler -
Dan Asimov