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Bilinear Forms associated With a Quadratic Form over Z/2

  1. Jul 26, 2011 #1
    Hi, All:

    Given a quadratic form Q(x,y) over a field of characteristic different from 2, we can

    find the bilinear form B(x,y) associated with Q by using the formula:


    I know there is a whole theory about what happens when we work over fields of
    characteristic 2, with the Arf -Invariant , Artin's and other's books on Geometric
    Algebra and everything, which I am looking into.

    Still, I wonder if someone knows the quick-and-dirty on how to transform an
    actual, specific quadratic q form over Z/2 into its associated bilinear form.

  2. jcsd
  3. Jul 26, 2011 #2


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    The reason for all the theory is that there isn't a symmetric bilinear form satisfying Q(x) = B(x,x), except for special Q's. Observe that B(x,x) is a linear function of x....
  4. Jul 26, 2011 #3
    Thanks, Hurkyl:

    Would you give me some idea on the conditions under which the
    associated B(x,y) exists?
  5. Aug 12, 2011 #4
    I'm thinking specifically of the case in which the bilinear form is (x,y)_2 ; the intersection form in H_1(Sg,Z/2) ; all defined on a symplectic basis for Sg ---Sg is the orientable, genus -g surface, and a symplectic basis {x1,y1,x2,y2,...,x2g,y2g} for Sg is one in which (xi,yi)_2=1 and (xi,yj)=0 if i=/j .

    We then say that q(x) is a quadratic form associated with the given bilinear form, if :


    And then we seem to classify these forms by their arf invariant; there seem to be 8 forms with Arf invariant 1 and 8 with Arf invariant 0 (the Arf invariant when working over Z/2 is an element of Z/2); given a choice of symplectic basis as above, the Arf invariant
    is defined as : (q(x1)q(y1)+q(x2)q(y2) ).

    Still, I don't know what the issue is with the forms with Arf invariant 1 . I know that the Arf invariant classifies the quadratic forms mod2, i.e., two forms defined over F_2 are equivalent iff they have the same Arf invariant ; just like we
    classify quadratic forms over fields of characteristic different from 2 by their resolvent, i.e., all quadratic forms over fields of characteristic different from 2 can be diagonalized ( I think by symmetry) , and the sum of the square of their diagonals is an invariant , i.e., if forms Q,Q' are equivalent, then they will have the same resolvent.
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