algebraMap R (CliffordAlgebra Q)
is not always injective. #
A formalization of Darij Grinberg's answer
to a "Clifford PBW theorem for quadratic form" post on MathOverflow, that provides a counterexample
to Function.Injective (algebraMap R (CliffordAlgebra Q))
.
The outline is that we define:
- $k$ (
Q60596.K
) as the commutative ring $𝔽₂[α, β, γ] / (α², β², γ²)$ - $L$ (
Q60596.L
) as the $k$-module $⟨x,y,z⟩ / ⟨αx + βy + γz⟩$ - $Q$ (
Q60596.Q
) as the quadratic form sending $Q(\overline{ax + by + cz}) = a² + b² + c²$
and discover that $αβγ ≠ 0$ as an element of $K$, but $αβγ = 0$ as an element of $𝒞l(Q)$.
Some Zulip discussion at https://leanprover.zulipchat.com/#narrow/stream/113489-new-members/topic/.F0.9D.94.BD.E2.82.82.5B.CE.B1.2C.20.CE.B2.2C.20.CE.B3.5D.20.2F.20.28.CE.B1.C2.B2.2C.20.CE.B2.C2.B2.2C.20.CE.B3.C2.B2.29/near/222716333.
As a bonus result, we also show BilinMap.not_forall_toQuadraticMap_surjective
: that there
are quadratic forms that cannot be expressed via even non-symmetric bilinear forms.
The monomial ideal generated by terms of the form $x_ix_i$.
Equations
- Q60596.kIdeal = Ideal.span (Set.range fun (i : Fin 3) => MvPolynomial.X i * MvPolynomial.X i)
Instances For
𝔽₂[α, β, γ] / (α², β², γ²)
Equations
- Q60596.K = (MvPolynomial (Fin 3) (ZMod 2) ⧸ Q60596.kIdeal)
Instances For
The elements above square to zero
If an element multiplied by αβγ
is zero then it squares to zero.
Though αβγ
is not itself zero
The 1-form on $K^3$, the kernel of which we will take a quotient by.
Our source uses $αx - βy - γz$, though since this is characteristic two we just use $αx + βy + γz$.
Equations
- Q60596.lFunc = Q60596.K.gen 0 • LinearMap.proj 0 + Q60596.K.gen 1 • LinearMap.proj 1 + Q60596.K.gen 2 • LinearMap.proj 2
Instances For
The quotient of K^3
by the specified relation.
Equations
- Q60596.L = ((Fin 3 → Q60596.K) ⧸ LinearMap.ker Q60596.lFunc)
Instances For
The quadratic form corresponding to squaring a single coefficient.
Equations
- Q60596.sq i = QuadraticMap.sq.comp (LinearMap.proj i)
Instances For
Basis vectors in the Clifford algebra
Equations
- Q60596.gen i = (CliffordAlgebra.ι Q60596.Q) (Submodule.Quotient.mk (Pi.single i 1))
Instances For
The basis vectors square to one
By virtue of the quotient, terms of this form are zero
The core of the proof - scaling 1
by α * β * γ
gives zero
Our final result: for the quadratic form Q60596.Q
, the algebra map to the Clifford algebra
is not injective, as it sends the non-zero α * β * γ
to zero.
Bonus counterexample: Q
is a quadratic form that has no bilinear form.
The general statement: not every Clifford algebra over a module has an injective algebra map.
The general bonus statement: not every quadratic form is the diagonal of a bilinear form.