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If char⁡F≠2, quadratic forms and symmetric bilinear forms correspond by q(v)=B(v,v) and B(u,v)=12bq(u,v)

Statement

Let char⁡F≠2. The assignments

B⟼qB,qB(v)=B(v,v),q⟼Bq,Bq(u,v)=12bq(u,v)

are inverse bijections between symmetric bilinear forms and quadratic forms.

Facts & Assumptions

Given: A field F with char⁡F≠2 and an F-vector space V.

[L1]

A quadratic form satisfies q(av)=a2q(v) and has bilinear polar form bq(u,v)=q(u+v)−q(u)−q(v) (A quadratic form q in arbitrary characteristic and its polar form bq(u,v)=q(u+v)−q(u)−q(v)).

[L2]

A symmetric bilinear form is bilinear and satisfies B(u,v)=B(v,u) (Bilinear forms, and symmetric, skew-symmetric, and alternating bilinear forms).

[L3]

The characteristic is the least positive natural multiple of 1F that is zero, or 0 if none exists (The characteristic of a ring: the least n≥1 with n⋅1R=0 when one exists, and 0 otherwise); in a field 0F≠1F and every nonzero scalar is invertible (Field).

Proof

technique · explicit inverse
1.1

Because char⁡F≠2, [L3] makes the scalar 2=1F+1F nonzero and hence invertible. If B is symmetric bilinear, then qB(av)=a2qB(v) and bqB(u,v)=B(u+v,u+v)−B(u,u)−B(v,v)=2B(u,v). Thus qB is a quadratic form.

L1L2L3algebra
1.2

If q is quadratic, [L1] makes bq bilinear, and its defining formula is symmetric. Hence Bq=12bq is symmetric bilinear by [L2] and [L3].

L1L2L3
2.1

Step 1.1 gives BqB=B. Conversely, bq(v,v)=q(2v)−2q(v)=4q(v)−2q(v)=2q(v), so qBq(v)=12bq(v,v)=q(v).

step 1.1step 1.2L1L3algebra
3.1

The two assignments are therefore mutually inverse bijections. The use of 2−1 identifies precisely where the characteristic hypothesis enters.

step 2.1L3∎

Depends on

Used by

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Sources