From: "Charles H. Giffen" <chg4k@virginia.edu>
Newsgroups: sci.math
Subject: Re: Arf Invariant
Date: Thu, 23 Apr 1998 16:46:01 -0400

Robin Chapman wrote:
> 
> In article <352CFA67.7B26@univ-tln.fr>,
>   Langevin Philippe <langevin@univ-tln.fr> wrote:
> >
> > Hi !
> >
> > Let K be the finite field with two elements. Let Q be a
> > quadratic form of maximal rank. The number of solution of
> > the equation q(x) = 0 is :
> >
> >         2^{m-1} - {1/2} 2^t  ; Elliptic case
> >
> >                 OU
> >
> >         2^{m-1} + {1/2} 2^t  ; Hyperbolic case
> >
> > I know that we can decide the type of q by mean of the
> > Arf's invariant.
> >
> > My question is : How does one calculate Arf(q) ?
> 
> Consider Q as a function on a vector space V over F_2. That means
> that Q(0) = 0, and B(x,y) = Q(x+y) - Q(x) - Q(y) is a bilinear form on V.
> This form is alternating, and by assumption non-singular, so there
> is a basis x_1, x_2, ..., x_n, y_1, y_2, ..., y_n of V with
> B(x_j, y_j) =  1 = B(Y_j, x_j) and B of any other pair of basis elements
> vanishes. Then the Arf invariant is
> Q(x_1)Q(y_1) + Q(x_2)Q(y_2) + ... + Q(x_n)Q(y_n).
> 
> One can generalize this to forms over k = F_{2^r}. In this case one takes
> the Arf invariant as the image of
> Q(x_1)Q(y_1) + Q(x_2)Q(y_2) + ... + Q(x_n)Q(y_n) in the 2-element group
> k/{a^2 - a: a in k}.
> 
> It's not apparent that the Arf invariant is independent of the choice
> of the x_j and y_j. There is an alternative characterization in terms
> of the Clifford algebra of V. This is the k-algebra generated by V
> with relations x^2 = Q(x) for x in V. The elements xy with x, y in V
> generate a subalgebra C^+. The centre of C^+ is isomorphic to
> k[T]/<T^2 - T - a> where a is the Arf invariant.
> 
> Robin Chapman                           + "They did not have proper
> Department of Mathematics               -  palms at home in Exeter."
> University of Exeter, EX4 4QE, UK       +
> rjc@maths.exeter.ac.uk                  -    Peter Carey,
> http://www.maths.ex.ac.uk/~rjc/rjc.html +      Oscar and Lucinda
> 
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There is another way of calculating the Arf invariant (which I learned
from  Bill Browder) -- the so-called democratic method.

Arf(Q) = 1  if  Q(x) = 1  for more than half the elements  x  of  V
Arf(Q) = 0  if  Q(x) = 0  for more than half the elements  x  of  V.

Pretty neat, eh!

--Chuck Giffen