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I've seen a couple of constructions of the so-called Witt group: it seems that most authors start with the commutative monoid of isometry classes of quadratic spaces under direct sum, pass to the Grothendieck group, and then quotient out the subgroup generated by the hyperbolic plane (sometimes called the split quadratic spaces).

In Bump's Automorphic Forms and Representations, however, it is claimed that one need only quotient out the submonoid generated by the hyperbolic plane and the monoid thus obtained is already a group. So we can avoid the Grothendieck construction entirely. The content of this statement is that given a quadratic space, there exists another such that the direct sum of the two is split (i.e. a sum of copies of the hyperbolic plane). But I don't see why. Does anyone have a reference or an explanation?

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up vote 5 down vote accepted

Yes, this is true. (Or almost true: one doesn't take isometry classes of all quadratic forms, but only the nonsingular ones.)

If $q = \langle a_1,\ldots,a_n \rangle$ is a nonsingular diagonal quadratic space, then take $q' = \langle -a_1,\ldots,-a_n \rangle$. Then

$q \oplus q' \cong \langle a_1, -a_1 \rangle \oplus \ldots \oplus \langle a_n, - a_n \rangle$,

and it is a standard early exercise to show that for any $a \in K^{\times}$, $\langle a, - a \rangle$ is isomorphic to the hyperbolic plane. (For instance, every nonsingular isotropic quadratic form $Q$ is of the form $Q' \oplus \langle 1,-1 \rangle$ for some quadratic form $Q'$. Or see Corollary 11 in these notes.) So $q \oplus q'$ is a totally hyperbolic space and gets identified with $0$ in your quotient monoid, which is therefore a group.

If I am remembering correctly, this is indeed much closer to Witt's original take on the matter than the more modern Grothendieck group approach.

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it's good to have you back! – lhf May 16 '11 at 1:07

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