Re: What is a "differential field"?
From: James Dolan (jdolan_at_math-cl-n03.math.ucr.edu)
Date: 08/16/04
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Date: Mon, 16 Aug 2004 05:11:23 +0000 (UTC)
in article <pan.2004.08.15.14.17.09.974947@lycos.com>,
igor khavkine <k_igor_k@lycos.com> wrote:
|On Sun, 15 Aug 2004 07:44:00 +0000, Ted Hwa wrote:
|
|Dear Ted, thanks a lot for taking the time to answer.
|
|> :> Igor Khavkine <k_igor_k@lycos.com> wrote:
|
|> : What are algebraic groups exactly? Is that some fancy way of saying
|> : matrix groups? What intrinsic property makes them algebraic?
|>
|> An algebraic group is an affine algebraic variety (set of solutions to a
|> finite number of polynomials over a finite number of variables, over some
|> field) together with a group structure, where multiplication and inversion
|> are rational maps.
|>
|> There are two very different kinds of algebraic groups:
|>
|> (1) matrix groups, which are subgroups of GL(n,F) defined by polynomials
|> in
|> the matrix entries. These are also called linear algebraic groups. The
|> differential Galois group of a Picard-Vessiot extension is always a
|> linear algebraic group (forgot to mention that in the previous post).
|
|I've only recently learned what an affine algebraic variety is. So let me
|see if I can prove that GL(n,F) satisfies the requirements to be one.
|First, n x n matrices are elements of a n^2 dimensional vector space over
|F, which is an affine space. So GL(n,F) would certainly be an affine
|variety, since it is embedded in an affine space. The condition for a
|matrix M to be in the group is GL(n,F) is that det(M) != 0. Interesting,
|this is an inequality and not an equation. Does that qualify GL(n,F) to
|be a variety at all? SL(n,F) would definitely qualify, since the only
|condition on its elements is det(M)=1.
|
|About rationality of group operations. If M=NP, then the matrix elements
|of M are certainly rational (polynomial even) functions of matrix elements
|of N and P. To get the inverse we can use the formula M^(-1) = M*/det(M),
|where M* is the matrix adjoint, hence inversion is also rational.
|Everything checks out here. However I'm a little bothered by the
|requirement of rationality here. I've been indoctrinated by modern
|differential geometry that intrinsic geometric properties/notions must be
|coordinate independent. Yet rationality is always defined with respect to
|some coordinates (variables). Granted that rationality here is invariant
|under linear transformations of the affine space in which the algebraic
|group is embedded, but I would still like to see some sort of intrinsic
|condition on the group that shows it to be an "algebraic group".
the vague impression that i have about this is that a lie group's
being algebraic (i think affine algebraic in the strict sense
actually, meaning that the group operation is not just rational but
polynomial) is secretly very closely related to the representation
theory of the group. if the representation theory of the group is
sufficiently nice, meaning basically that the group has "enough"
representations on finite-dimensional vector spaces, then the
so-called "representative" functions on the group will form an affine
coordinate ring making the group into an algebraic group.
a "representative" function on a group is a polynomial function of the
matrix by which a group element is represented under a
finite-dimensional representation of the group. the nice trick that
shows that the representative functions on a group form a
(commutative) ring is that the sum of representative functions is
representative because you can take the direct sum of group
representations, while the product of representative functions is
representative because you can take the tensor product of group
representations.
-- [e-mail address jdolan@math.ucr.edu]
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