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Associativity and coefficient calculus for the finite superspace Grassmann algebra

Statement

For the basis-indexed vector space and product $e_{S} e_{T} = 0$ when $S \cap T \neq \emptyset$ and $e_{S} e_{T} = (- 1)^{ν (S, T)} e_{S \cup T}$ otherwise, with $ν (S, T) = ∣ {(s, t) \in S \times T : s > t} ∣$ , multiplication is associative, unital and supercommutative. The graded tensor product with a supercommutative coefficient algebra is also associative and supercommutative. The basis formulas for left and right odd differentiation extend to derivations with the stated Leibniz rules. The ordered four-variable Berezin integral is a well-defined linear functional extracting the coefficient of $θ^{1} θ^{2} {\overset{ˉ}{θ}}^{\dot{1}} {\overset{ˉ}{θ}}^{\dot{2}}$ , and the stated conjugation is an involutive anti-automorphism.

Statement references

Martin, §4.1.

Gates et al., §3.7.

Four-dimensional $N = 1$ superspace Grassmann algebra and Berezin integration: Four-dimensional $N = 1$ superspace Grassmann algebra and Berezin integration.

Facts & Assumptions

Given: The finite basis $e_{S}$ , multiplication sign $ν (S, T)$ , coefficient superalgebra, derivative actions and ordered measures in the Statement.

[F1]

Four-dimensional $N = 1$ superspace Grassmann algebra and Berezin integration: the basis is indexed by all subsets of ${1, 2, 3, 4}$ , $e_{S} e_{T} = 0$ on overlap and otherwise equals $(- 1)^{ν (S, T)} e_{S \cup T}$ ; coefficient multiplication is the graded tensor product, and the left/right derivative actions on homogeneous coefficient-left tensors are $(- 1)^{∣ a ∣} a \otimes \partial^{L} e_{S}$ and $a \otimes \partial^{R} e_{S}$ , respectively.

[F2]

Four-dimensional $N = 1$ superspace Grassmann algebra and Berezin integration: for $i \in S$ , $\partial_{ξ_{i}}^{L} e_{S} = (- 1)^{∣ {s \in S : s < i} ∣} e_{S ∖ {i}}$ and $\partial_{ξ_{i}}^{R} e_{S} = (- 1)^{∣ {s \in S : s > i} ∣} e_{S ∖ {i}}$ , and both vanish when $i \notin S$ .

[F3]

Four-dimensional $N = 1$ superspace Grassmann algebra and Berezin integration: the one-variable integral extracts the coefficient with that variable ordered first; the four-variable measure is $\int d {\overset{ˉ}{θ}}^{\dot{2}} d {\overset{ˉ}{θ}}^{\dot{1}} d θ^{2} d θ^{1}$ .

[F4]

Four-dimensional $N = 1$ superspace Grassmann algebra and Berezin integration: conjugation reverses product order and exchanges $θ^{α}$ with ${\overset{ˉ}{θ}}^{\dot{α}}$ .

Proof

Given: The finite basis and signed product in [F1], and the specified derivative, integral and conjugation operations in [F2]–[F4].

1.1F1algebra

The element $e_{\emptyset}$ is a two-sided identity because $ν (\emptyset, S) = ν (S, \emptyset) = 0$ . For $S, T, U \subseteq {1, 2, 3, 4}$ , if any two sets overlap, both parenthesizations $(e_{S} e_{T}) e_{U}$ and $e_{S} (e_{T} e_{U})$ vanish: if the first pair overlaps one side vanishes immediately and the other retains that common generator, and the same argument applies to the other pairs. If $S, T, U$ are pairwise disjoint, the left sign exponent is $ν (S, T) + ν (S \cup T, U) = ν (S, T) + ν (S, U) + ν (T, U)$ , while the right exponent is $ν (T, U) + ν (S, T \cup U) = ν (T, U) + ν (S, T) + ν (S, U)$ . The signs agree, proving associativity on basis elements and hence by trilinearity on all elements.

2.1F1step 1.1algebra

If $S \cap T \neq \emptyset$ , both products vanish. If the sets are disjoint, every ordered pair in $S \times T$ has exactly one of $s > t$ or $t > s$ , so $ν (S, T) + ν (T, S) = ∣ S ∣ ∣ T ∣$ . Thus $e_{S} e_{T} = (- 1)^{∣ S ∣ ∣ T ∣} e_{T} e_{S}$ . This proves supercommutativity.

2.2F1F2step 1.1algebra

Put $a = ∣ {s \in S : s < i} ∣$ , $c = ∣ {s \in S : s > i} ∣$ , $b = ∣ {t \in T : t < i} ∣$ , and $d = ∣ {t \in T : t > i} ∣$ . If $S$ and $T$ are disjoint and $i \in S$ , then $ν (S, T) = ν (S ∖ {i}, T) + b$ ; the left-hand derivative exponent is $a + b + ν (S, T)$ , equal modulo two to the right-hand side exponent $a + ν (S ∖ {i}, T)$ . If $i \in T$ , then $ν (S, T) = ν (S, T ∖ {i}) + c$ ; the left-hand exponent $a + b + ν (S, T)$ equals the right-hand exponent $∣ S ∣ + b + ν (S, T ∖ {i})$ , since $i \notin S$ and $∣ S ∣ = a + c$ . For the right derivative, if $i \in S$ the identity $ν (S, T) = ν (S ∖ {i}, T) + b$ makes the left exponent $c + d + ν (S, T)$ equal to the right exponent $∣ T ∣ + c + ν (S ∖ {i}, T)$ ; if $i \in T$ , the identity $ν (S, T) = ν (S, T ∖ {i}) + c$ makes both exponents $d + ν (S, T ∖ {i})$ modulo two. If $S \cap T$ contains a common element other than $i$ , all Leibniz terms vanish because an overlap remains. If $S \cap T = {i}$ , both terms can be nonzero, but their exponents differ by one: for the left rule they are $a + ν (S ∖ {i}, T)$ and $∣ S ∣ + b + ν (S, T ∖ {i})$ ; for the right rule they are $∣ T ∣ + c + ν (S ∖ {i}, T)$ and $d + ν (S, T ∖ {i})$ . Thus they cancel in both rules, matching the derivative of $e_{S} e_{T} = 0$ . For homogeneous coefficient-left tensors $a \otimes e_{S}$ and $b \otimes e_{T}$ , the left derivative of their product has the common sign $(- 1)^{∣ S ∣ ∣ b ∣ + ∣ a ∣ + ∣ b ∣}$ multiplying the exterior left Leibniz expansion; the first tensor Leibniz term has sign $(- 1)^{∣ a ∣ + (∣ S ∣ - 1) ∣ b ∣}$ , which is equal, and the second has the additional $(- 1)^{∣ S ∣}$ required by the exterior rule. For the right derivative, the first tensor Leibniz term has sign $(- 1)^{∣ b ∣ + ∣ T ∣ + (∣ S ∣ - 1) ∣ b ∣} = (- 1)^{∣ S ∣ ∣ b ∣ + ∣ T ∣}$ , while the second carries $(- 1)^{∣ S ∣ ∣ b ∣}$ ; these are exactly the two exterior right-rule terms. Thus the stated coefficient-left derivative extensions satisfy the graded-tensor left and right Leibniz identities.

3.1F1step 1.1step 2.1algebra

For homogeneous tensors $a \otimes e_{S}$ , $b \otimes e_{T}$ and $c \otimes e_{U}$ , the sign in $((a \otimes e_{S}) (b \otimes e_{T})) (c \otimes e_{U})$ is $∣ S ∣ ∣ b ∣ + (∣ S ∣ + ∣ T ∣) ∣ c ∣$ , while the sign in $(a \otimes e_{S}) ((b \otimes e_{T}) (c \otimes e_{U}))$ is $∣ T ∣ ∣ c ∣ + ∣ S ∣ (∣ b ∣ + ∣ c ∣)$ . These exponents agree modulo two; associativity of $A$ and step 1.1 then gives associativity of the graded tensor product. Combining its sign $∣ S ∣ ∣ b ∣$ with supercommutativity of both factors gives the total sign $(- 1)^{(∣ a ∣ + ∣ S ∣) (∣ b ∣ + ∣ T ∣)}$ when the tensors are exchanged, so the tensor product is supercommutative.

3.2F1F3step 1.1step 2.1

The vectors $e_{S}$ are a specified basis, so each coefficient polynomial has a unique finite expansion. For a fixed $i$ , grouping terms with and without $i$ and moving $ξ_{i}$ to the left by the signed product rule gives a unique expression $f = f_{0} + ξ_{i} f_{1}$ with $f_{0}, f_{1}$ independent of $ξ_{i}$ ; hence the linear coefficient extraction in [F3] is well-defined. Applying the four ordered coefficient maps successively to $e_{S}$ returns $1$ exactly when $S = {1, 2, 3, 4}$ and $0$ otherwise. By linearity, the full measure extracts precisely the stated top coefficient.

4.1F1F4step 1.1step 2.1∎

Reversing a word in the four generators and exchanging each undotted generator with its dotted conjugate preserves the square-zero and anticommutation relations. The coefficient operation preserves parity and reverses products by [F4]. For homogeneous $a \in A$ and $e_{S}$ , moving $e_{S}^{*}$ back to the right of $a^{*}$ contributes $(- 1)^{∣ a ∣ ∣ S ∣}$ , so the induced tensor formula is $(a \otimes e_{S})^{*} = (- 1)^{∣ a ∣ ∣ S ∣} a^{*} \otimes e_{S}^{*}$ , as stated in [F1]. This formula respects the graded-tensor relation: applying it to $(1 \otimes e_{S}) (a \otimes 1) = (- 1)^{∣ a ∣ ∣ S ∣} (a \otimes e_{S})$ gives the same identity after reversing the factors. Applying the operation twice restores every coefficient and every generator, so it descends to an involutive anti-automorphism.

Depends on

Four-dimensional $N = 1$ superspace Grassmann algebra and Berezin integration