The twofold way of super holonomy

Definition A.1 (fppf)

The fppf topology on SAlgK is defined as collection of finite sets {R→Ri}i∈I (for R,Ri∈SAlgK) such that the R-supermodule ×i∈IRi is faithfully flat and all Ri are finitely presented R-superalgebras.

The topology on P as defined in Definition 3.7 agrees with the one induced by Definition A.1.

Proof.

We note first that being finitely presented is no condition in the case of Graßmann algebras. Consider a covering consisting of morphisms φi:Pi→P as in Definition 3.7. By the following proposition, Proposition A.3, φi is a submersion if and only if Ri=𝒪Pi is flat as an R=𝒪P-module with respect to φ* (a morphism in Gr). All Ri being flat, in turn, is equivalent to the condition of Definition A.1 (by [4, Lemma I.2.2] and the implication (ii) ⇒ (i) in Proposition A.3). ∎

Let φ:R→S be a morphism in Gr, and consider S as an R-module via φ. Then the following are equivalent:

1. S is faithfully flat.

2. S is flat.

3. S is free.

4. (The associated morphism of superpoints corresponding to) φ is a submersion.

Proof.

(i) ⇔ (ii). In general, an R-module M is faithfully flat if and only if M is flat and M≠m⁢M for every maximal ideal m (see [4, Proposition I.3.1]). In the case of a Graßmann algebra R, its nilpotent part Rnil is the unique maximal ideal and, obviously, S≠φ⁢(Rnil)⁢S.

(ii) ⇔ (iii). The Jacobson radical of R is Rnil, and R/Rnil≅ℝ is a field. The equivalence follows from [4, Proposition II.3.5].

(iv) ⇒ (iii). A submersion is characterised by the existence of coordinates (θi)1≤i≤n and (ηj)1≤j≤n+m on R and S, respectively, such that θi↦ηi (see [5, Proposition 5.2.5]). Then any ℝ-basis of 〈ηn+1,…,ηn+m〉 is an R-basis of S.

(ii) ⇒ (iv). This is a special case of [27, Proposition 3.6.1 (ii)]. ∎

Let φ:⋀Rn→⋀Rm be a morphism of Graßmann algebras such that ⋀Rm is free. Then φ is injective. In particular, n≤m.

Let φ:⋀Rn→⋀Rm be a morphism of Graßmann algebras such that ⋀Rm is free. Then it is also free with respect to φ∘πj:⋀Rn-1→⋀Rm.

Assuming that every morphism ψ:⋀Rn→⋀Rm, such that ⋀Rm is a free ⋀Rn-module with respect to ψ, is a submersion, the corresponding statement holds for all morphisms φ:⋀Rn+1→⋀Rm.

Proof.

Let φ:⋀ℝn+1→⋀ℝm be such that ⋀ℝm is a free ⋀ℝn+1-module. By Lemma A.5, it is also free with respect to the map φn+1:=φ∘πn+1:⋀ℝn→⋀ℝm which, by assumption, is a submersion. Therefore, there are coordinates of ⋀ℝn and ⋀ℝm, respectively, still denoted θ1,…,θn and η1,…,ηm, such that φn+1⁢(θi)=ηi (see [5, Proposition 5.2.5]). Endowing the former coordinates with the original θn+1, φ obtains the form

We may assume that φ⁢(θn+1)∈⋀ℝm∖〈η1,…,ηn〉, for if not we can modify θn+1 by subtracting from it a suitable element of 〈θ1,…,θn〉. Denoting the associated morphism of superpoints still by φ, the differential at the single topological point 0 assumes the form

Observe that φ is a submersion if and only if the lower right submatrix is nonzero. This condition is satisfied by the following argument. The last line of (d⁢φ)0 equals the differential (d⁢φn+1)0 of the map φn+1:=φ∘π1∘…∘πn, where the πj are defined with respect to the new coordinates. But φn+1 is a submersion by Lemma A.5 and the induction hypothesis for n=1. ∎

Let φ:⋀R1→⋀RL be a morphism of Graßmann algebras. Then ⋀RL is free if and only if it has an R-basis of the form (v1,…,v2L-1,φ⁢(θ1)⋅v1,…,φ⁢(θ1)⋅v2L-1).

Proof.

This is shown analogous to the proof of Lemma A.5. ∎

Let μ∈(⋀RL)1¯ and consider the ideal (μ) in ⋀RL generated by μ. Then ⋀RL admits an R-basis of the form (v1,…,v2L-1,μ⋅v1,…,μ⋅v2L-1) if and only if dimR⁡(μ)≥2L-1. In this case, dimR⁡(μ)=2L-1.

Proof.

If a basis as stated exists, then the vectors μ⋅vi∈(μ) are all linearly independent, thus the real dimension of (μ) is greater than or equal to their number, 2L-1.

Conversely, let (μ⋅w1,…,μ⋅wd) denote a real basis of (μ) with d≥2L-1. We may endow this basis by vectors vj, 1≤j≤f, to a basis (v1,…,vf,μ⋅w1,…,μ⋅wd) of ⋀ℝL. It follows that f+d=2L and thus f≤2L-1. Multiplying all vectors with μ and using μ2=0, one sees that the vectors (μ⋅v1,…,μ⋅vf) span (μ), whence f≥dim⁡(μ)=d≥2L-1, such that f=d=2L-1. In particular, (μ⋅v1,…,μ⋅vd) is a basis of (μ). Endowed with the vectors vi, we obtain a basis of ⋀ℝL as claimed. ∎

Let μ∈(⋀RL)1¯. If dimR⁡(μ)≥2L-1, then dimR⁡(φ⁢(μ))≥2L-1 for every Graßmann automorphism φ:⋀RL→⋀RL.

Let μ∈(⋀RL)1¯. Let r∈(⋀RL)0¯ and consider μ^:=μ+ηL+1⋅r∈(⋀RL+1)1¯. If dimR⁡(μ)≥2L-1 (with (μ) as an ideal in ⋀RL), then dimR⁡(μ^)≥2L (as an ideal in ⋀RL+1).

Proof.

Writing elements of ⋀ℝL+1 in the form v+ηL+1⁢w with v,w∈⋀ℝL, we obtain

By assumption and Lemma A.8, ⋀ℝL has a real basis (v1,…,v2L-1,μ⋅v1,…,μ⋅v2L-1). Let

There is a canonical map from V⊕V to the space

sending (v,w) to μ⁢v+ηL+1⁢(r⁢v-μ⁢w), which is clearly ℝ-linear and surjective. Assume μ⁢v+ηL+1⁢(r⁢v+μ⁢w)=0. Then, in particular, μ⁢v=0. By the definition of V, it follows that v=0. Then also μ⁢w=0 and, similarly, w=0. The aforementioned map is injective, and we conclude that dimℝ⁡(μ^)≥dim⁡(V⊕V)=2L. ∎

Let μ∈(⋀RL)1¯ be such that dimR⁡(μ)≥2L-1. Then there are L′≥L and μ′∈(⋀RL′)1¯ such that the following conditions hold:

1. There is a bijective correspondence

   λ:{J∣CμJ≠0}→{J′∣Cμ′J′≠0}

   such that |J|=|λ⁢(J)|.

2. We have

   ∏J′∣Cμ′J′≠0ηJ′≠0.

3. dimℝ⁡(μ′)≥2L′-1.

Proof.

We can successively built μ′ from μ as follows. Let j0 denote the smallest integer such that the generator ηj0 is contained in at least two monomials ηJ such that CμJ≠0. Let I denote one of k≥2 such multiindices. By assumption, there is r∈(⋀ℝL)0¯ such that ηI=ηj0⋅r. Consider

By Lemma A.10, it satisfies dimℝ⁡(μ^)≥2L′-1 with L′=L+1. Consider next the automorphism

defined by ηi↦ηi (i<L′) and ηL′↦(ηL′-ηj0). Then the element φ⁢(μ^) satisfies dim⁡(φ⁢(μ^))≥2L′-1 by Lemma A.9. Moreover, the number of monomials containing ηj0 is reduced to k-1. It is also clear that the multiindex bijection required in the statement is satisfied.

Now start with μ replaced by φ⁢(μ^) from the previous step, and repeat the construction until finally there is no generator ηj0 contained in more than one monomial. Since every step of the construction satisfies the first and third items in the statement, the same holds for the final result. As no two monomials therein share a common generator, the product over all is nonzero. ∎

Proof of (iii) ⇒ (iv) in Proposition A.3.

This remaining implication is proved by induction over n in R=⋀ℝn. It remains to show the base case as the inductive step was already established in Proposition A.6. Consider thus a morphism φ:⋀ℝ1→⋀ℝL such that ℝL is free as an ℝ1-module via φ. By Lemma A.7 and Lemma A.8, this property is characterised by dimℝ⁡(μ)≥2L-1 for μ:=φ⁢(θ1). The algorithm of Proposition A.11 constructs another odd μ′ with a similar shape, but such that the product over all monomials with nonvanishing coefficients in (A.1) does not vanish. This is the case treated in [7]. In [7, Theorem 4], the dimension of (μ′) is explicitly calculated to be

Together with dim⁡(μ′)≥2L′-1, this forces at least one of the multiindices J in the product to be of length |J|=1. But then the corresponding μ-multiindex λ-1⁢(J) has also length 1. It follows that φ is a submersion. ∎