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Graded C_n modules and ungraded C_(n−1) modules

Statement

For n≥1, taking the even part is an equivalence from finite-dimensional graded Cn-modules to finite-dimensional ungraded Cn−1-modules, where Cn=Cl(Rn,−∥⋅∥2). An inverse is induction from the even subalgebra. For n=0, retain both parity types and treat the restriction cokernel separately.

Facts & Assumptions

Given: n≥1 and a finite-dimensional graded left Cn-module M=M0⊕M1 or a finite-dimensional ungraded left module N over Cn−1.

[F1]

A graded Clifford module has CniMj⊆Mi+j mod 2 (Real graded Clifford-module bundles and their Grothendieck groups).

[F2]

The even subalgebra of the negative-square algebra Cn=Cl0,n is isomorphic to Cn−1=Cl0,n−1 (Even subalgebras of real and complex Clifford algebras).

[F3]

Cn is finite-dimensional and has its even/odd word grading (PBW basis and exterior associated graded for a Clifford algebra).

[F4]

M0 retains the even and odd dimension coordinates separately (Clifford-module restriction cokernels A_n).

Proof

technique · explicit induction and evaluation
1.1givenF1F2construct

Let en be the last standard generator. It is odd and invertible because en2=−1. Multiplication by en maps Cn0 bijectively onto Cn1, with inverse multiplication by −en; thus Cn1=enCn0. By [F2], the action of Cn0 on M0 makes it an ungraded Cn−1-module.

2.1F2F3step 1.1construct

Define G(N)=Cn⊗Cn0N, with parity inherited from Cn and the Cn−1-action transported along the algebra isomorphism in [F2]. Because Cn=Cn0⊕enCn0 as a right Cn0-module, its even part is canonically N and its odd part is en⊗N. Hence G(N) is finite-dimensional and taking its even part returns N.

2.2givenF1step 1.1algebra

For M, define evaluation ε:Cn⊗Cn0M0→M by a⊗m↦am. It is well-defined because ab⊗m=a⊗bm for b∈Cn0. On the even part it is the canonical isomorphism Cn0⊗Cn0M0≅M0. On the odd part it sends en⊗m to enm; this is bijective since en:M0→M1 has inverse −en. Therefore evaluation is an even graded Cn-module isomorphism.

3.1F1F4step 2.1step 2.2algebra∎

The restriction functor M↦M0 and induction N↦Cn⊗Cn0N are inverse up to the natural isomorphisms in steps 2.1 and 2.2, so they give an equivalence. At n=0, C0=R is purely even; a one-dimensional module concentrated in odd parity has even part zero, while a one-dimensional even module has even part R. Thus the same equivalence fails at n=0, and the two parity generators must remain distinct in M0≅Z⊕Z.

Depends on

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