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Even subalgebras of real and complex Clifford algebras

Statement

Use the plus-square convention v2=Q(v)1. For the real standard form Qp,q(x)=∑i=1pxi2−∑j=1qxp+j2, write Clp,q0 for the even part of the canonical Z/2 grading. If p>0, then

Clp,q0≅Clq,p−1;

if p=0 and q>0, then Cl0,q0≅Cl0,q−1. For the standard complex form Qn(z)=∑i=1nzi2 with n>0, the even subalgebra is isomorphic over C to Cl0,n−1(C), the complex Clifford algebra on n−1 generators each having square −1, and hence to the standard complex Clifford algebra Cln−1(C).

Facts & Assumptions

Given: The quotient definition of a Clifford algebra, standard orthogonal generators with the displayed squares, and positive dimension.

[F1]

The Clifford quotient has the canonical grading with each generator odd, and its even subalgebra is the degree-zero part (The Clifford quotient has its canonical parity grading).

[F2]

The canonical image vˉ satisfies vˉ2=Q(v)1 (Universal Clifford algebra of a quadratic space).

[F3]

Orthogonal generators anticommute: vˉwˉ+wˉvˉ=2B(v,w)1 (Polarized relations in a Clifford algebra).

[F4]

A linear map from a quadratic space into a unital associative algebra that satisfies the square relation extends uniquely to its Clifford algebra (Universal property of the Clifford algebra).

[F5]

In dimension n, the ordered square-free Clifford monomials form a basis; the even monomials therefore give a basis of the even subalgebra (PBW basis and exterior associated graded for a Clifford algebra).

[F7]

The zero-dimensional Clifford algebra over R is R (The zero-dimensional Clifford algebra is the base field).

Proof

technique · direct generator comparison
1.1givenF1F2F3F4construct

Suppose first that p>0. Let e1 be a positive generator, let e2,…,ep be the other positive generators, and let f1,…,fq be the negative generators. Set ui=e1ei+1 for 1≤i<p and vj=e1fj for 1≤j≤q. These elements are even by [F1]. Using e12=1 and the orthogonal anticommutation in [F3], each ui2=−1, each vj2=+1, and every two distinct elements in the combined list anticommute. Thus the vj are the q positive generators and the ui the p−1 negative generators for the defining relations of Clq,p−1; [F4] gives an algebra homomorphism Clq,p−1→Clp,q0.

1.2givenF1F2F3F4construct

If p=0, choose a negative generator f1 and set uj=f1fj+1 for 1≤j<q. Each uj is even by [F1], squares to −1, and anticommutes with the other ul by [F2]–[F3]. For distinct remaining generators fi,fj, the identity (f1fi)(f1fj)=fifj shows that these pair products generate every even ordered monomial; when q=1 there are no uj and the only even monomial is the unit. Thus [F4] gives a surjection Cl0,q−1→Cl0,q0.

2.1step 1.1algebra

The image contains every product of two distinct standard generators: e1ei and e1fj are generators of the image, while for i,j>1, (e1ei)(e1ej)=−eiej, and the same identity holds for a pair of negative generators or one positive and one negative generator. Every even ordered monomial is a product of such pairs, so the homomorphism is surjective; if p=1,q=0, the only even monomial is the unit and the scalar map is already surjective.

2.2F5F7step 1.2algebra

Both algebras in step 1.2 have dimension 2q−1 by [F5], so the surjection is an isomorphism. This includes q=1, when [F7] identifies the source with R and the even part of Cl0,1 is its scalar subalgebra.

3.1F5step 2.1algebra

By [F5], Clp,q0 has dimension 2p+q−1, equal to the dimension of Clq,p−1. The surjective map of finite-dimensional vector spaces is therefore injective, proving the first isomorphism.

4.1givenF1F2F3F4F5F6step 1.1step 2.1step 3.1algebra∎

For the complex standard form, choose the first generator e1 and set uj=e1ej+1. The construction and checks in steps 1.1, 2.1, and 3.1 show that the even subalgebra is generated by n−1 pairwise anticommuting elements of square −1 and has the PBW dimension 2n−1; hence it is Cl0,n−1(C). Multiplying each of these generators by i∈C changes its square to +1 and gives a surjective homomorphism from the standard complex Clifford algebra Cln−1(C); the equal PBW dimensions make it an isomorphism. The argument includes n=1, where the even subalgebra is C.

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