Paper XI: Dyadic quadratic refinements

13 Dyadic quadratic refinements

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Let \((V,b)\) be a symplectic vector space over \({\mathbb F}_2\). The alternating form alone is not enough to determine the finite phase.

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Definition 13.1: Quadratic refinement

A quadratic refinement of \(b\) is a map

\[ q:V\to{\mathbb F}_2 \]

satisfying

\[ q(x+y)-q(x)-q(y)=b(x,y). \]

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Proposition 13.2: Affine space of refinements

If one quadratic refinement exists, then the set of all quadratic refinements is an affine space over \(V^*\).

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Proof of Proposition 13.2

The difference of two refinements has zero polar form, so it is additive and hence linear over \({\mathbb F}_2\). Conversely, adding a linear form does not change the polar form.

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Definition 13.3: Arf invariant

Choose a symplectic basis \((e_i,f_i)_{i=1}^{r}\). Define

\[ {\operatorname{Arf}}(q) = \sum_{i=1}^{r}q(e_i)q(f_i) \in{\mathbb F}_2. \]

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Standard fact 13.4: Arf invariance and classification

The value in Definition 13.3 is independent of the symplectic basis. Two nondegenerate quadratic refinements of one finite symplectic space are isometric exactly when their Arf invariants agree.

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This is the classical classification of nonsingular quadratic forms over \({\mathbb F}_2\) [24].

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Theorem 13.5: Dyadic Gauss sign

Let \(\dim V=2r\), and let \(q\) be nondegenerate. Then

\[ \boxed{ \sum_{x\in V}(-1)^{q(x)} = (-1)^{{\operatorname{Arf}}(q)}2^r. } \]

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Proof of Theorem 13.5

Orthogonally decompose \((V,q)\) into two dimensional nonsingular planes. On a plane with Arf invariant zero, the four values of \((-1)^q\) sum to \(2\). On a plane with Arf invariant one, they sum to \(-2\). The total sum is the product of the plane sums. The parity of the number of anisotropic planes is the Arf invariant.

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Corollary 13.6: Orthogonal dyadic multiplicativity

For an orthogonal sum

\[ (V,q)=(V_1,q_1){\mathop{\perp}\limits}(V_2,q_2), \]

we have

\[ {\operatorname{Arf}}(q)={\operatorname{Arf}}(q_1)+{\operatorname{Arf}}(q_2) \]

and the normalized Gauss sign is the product of the two normalized signs.

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Proof of Corollary 13.6

Take symplectic bases of the two summands and concatenate them. The Gauss sum factors as a product over the Cartesian decomposition.

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Proposition 13.7: The form alone does not determine the phase

On every nonzero symplectic \({\mathbb F}_2\) space, there are quadratic refinements of both Arf invariants. Their polar forms agree, but their normalized Gauss signs are opposite.

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Proof of Proposition 13.7

On a symplectic plane, the standard hyperbolic refinement has Arf invariant zero. Adding a suitable nonzero linear form produces a refinement with Arf invariant one. Orthogonally adjoining a fixed refinement on the complement gives the two choices in every positive even dimension. The signs differ by Theorem 13.5.

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Proposition 13.8: Translation of quadratic refinements

For \(v\in V\), put

\[ q_v(x)=q(x)+b(v,x). \]

Then

\[ \boxed{ {\operatorname{Arf}}(q_v)={\operatorname{Arf}}(q)+q(v). } \]

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Proof of Proposition 13.8

Using the quadratic identity,

\[ q_v(x)=q(x+v)+q(v). \]

Therefore

\[ \sum_x(-1)^{q_v(x)} = (-1)^{q(v)} \sum_x(-1)^{q(x)}. \]

Compare the signs in Theorem 13.5.

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Theorem 13.9: Number of dyadic refinements

Let \(\dim V=2r\). The number of quadratic refinements with Arf invariant zero is

\[ \boxed{ 2^{r-1}(2^r+1), } \]

and the number with Arf invariant one is

\[ \boxed{ 2^{r-1}(2^r-1). } \]

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Proof of Theorem 13.9

Every refinement is \(q_v\) for a unique \(v\in V\), because \(b\) identifies \(V\) with \(V^*\). By Proposition 13.8, the Arf invariant changes according to \(q(v)\). Let \(Z\) and \(O\) be the numbers of zeros and ones of \(q\). Then

\[ Z+O=2^{2r} \]

and Theorem 13.5 gives

\[ Z-O=(-1)^{{\operatorname{Arf}}(q)}2^r. \]

Solving gives the two displayed numbers, with their roles exchanged when the starting refinement has Arf invariant one. Since translation runs through every refinement, the final counts are independent of the starting choice.

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At residue characteristic two, the quadratic refinement and its Arf invariant carry phase information not contained in the alternating form.

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