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Wedge and contraction operators satisfy the Clifford relations

Statement

Let $k$ be a field of characteristic not $2$ , let $V$ be a finite-dimensional $k$ -vector space, and let $B : V \times V \to k$ be symmetric bilinear, possibly degenerate. Set $Q (v) = B (v, v)$ and let $Λ V$ be the exterior algebra of the finite free $k$ -module $V$ (Exterior Algebra Of A Finite Free Module). Define $ϵ_{v} (ω) = v \land ω$ and, on decomposable wedges,

$ι_{v} (w_{1} \land \dots \land w_{r}) = \sum_{j = 1}^{r} (- 1)^{j - 1} B (v, w_{j}) w_{1} \land \dots \land \hat{w_{j}} \land \dots \land w_{r},$

with $ι_{v} (1) = 0$ . Then $ι_{v}$ is well defined and $k$ -linear, and the operators $c (v) = ϵ_{v} + ι_{v}$ satisfy $c (v)^{2} = Q (v) I_{Λ V}$ for every $v \in V$ , where $I_{Λ V}$ is the identity map on $Λ V$ .

Facts & Assumptions

Given: A field $k$ of characteristic not $2$ , a finite-dimensional $k$ -vector space $V$ , a symmetric bilinear form $B$ on $V$ , and its diagonal quadratic form $Q (v) = B (v, v)$ .

[F1]

$T_{k} (V)$ is the direct sum of tensor powers with concatenation multiplication (Tensor algebra of a vector space).

[F2]

$Λ V = T_{k} (V) / J_{\land}$ , where $J_{\land}$ is the two-sided ideal generated by $j (w) j (w)$ for $w \in V$ ; multiplication is induced by concatenation (Exterior Algebra Of A Finite Free Module).

[F3]

Every multilinear map on a finite list of modules factors uniquely through the corresponding tensor product (Finite iterated tensor products represent multilinear maps independently of parenthesization).

[F4]

A two-sided ideal is an additive subgroup stable under multiplication on both the left and the right (Left, right and two-sided ideals).

Proof

technique · direct

1.1F1F3construct

For each $r \geq 1$ , the displayed sum is multilinear in $(w_{1}, \dots, w_{r})$ , so [F3] defines a linear map on $V^{\otimes r}$ . Together with $ι_{v} (1) = 0$ , these maps define $δ_{v} : T_{k} (V) \to T_{k} (V)$ . On a concatenated pair of pure tensors of degrees $p, q$ , the first $p$ deletion terms give $δ_{v} (x) y$ , while each remaining deletion has the extra sign $(- 1)^{p}$ and gives $(- 1)^{p} x δ_{v} (y)$ . Thus $δ_{v} (x y) = δ_{v} (x) y + (- 1)^{p} x δ_{v} (y)$ ; bilinearity extends this to homogeneous tensors.

1.2F2algebra

Left multiplication gives $ϵ_{v} ϵ_{w} + ϵ_{w} ϵ_{v} = 0$ : in $Λ V$ , $(v + w)^{2} = v^{2} = w^{2} = 0$ , so expansion gives $v w + w v = 0$ .

2.1F2F4step 1.1algebra

Let $S$ be the $k$ -linear span of the terms $x j (u)^{2} y$ for $x, y \in T_{k} (V)$ and $u \in V$ . This span contains all the squares, and multiplying its terms on the left or right only changes $x$ or $y$ , so it is a two-sided ideal by [F4]. Every two-sided ideal containing all the squares contains $S$ , hence $S = J_{\land}$ by [F2]. For $u \in V$ , $δ_{v} (j (u) j (u)) = B (v, u) j (u) - j (u) B (v, u) = 0$ . If $x$ is homogeneous of degree $p$ , then for homogeneous $y$ the derivation rule gives $δ_{v} (x j (u)^{2} y) = δ_{v} (x) j (u)^{2} y + (- 1)^{p} x δ_{v} (j (u)^{2}) y + (- 1)^{p + 2} x j (u)^{2} δ_{v} (y)$ . Each outer term lies in $J_{\land}$ by [F4], and the middle term is zero. By linearity, $δ_{v} (J_{\land}) \subseteq J_{\land}$ , so $δ_{v}$ descends to $ι_{v}$ on $Λ V$ . Since $J_{\land}$ is a two-sided ideal, left multiplication by $j (v)$ also descends to $ϵ_{v}$ .

3.1step 2.1algebra

On a decomposable wedge, every term of $ι_{v} ι_{w} + ι_{w} ι_{v}$ deletes two distinct positions. For fixed $i < j$ and fixed choice of which contraction deletes each factor, deleting position $i$ before $j$ changes the second deletion's sign from $(- 1)^{j - 1}$ to $(- 1)^{j - 2}$ ; the two orders therefore have opposite signs and identical scalar coefficient. Pairing all such terms gives zero. If $r < 2$ , both compositions vanish.

3.2step 2.1algebra

The contraction formula separates the term deleting the first factor and the terms deleting the remaining factors, giving $ι_{v} (w \land ω) = B (v, w) ω - w \land ι_{v} (ω)$ on decomposable $ω$ and hence on all $ω$ . Thus $ι_{v} ϵ_{w} + ϵ_{w} ι_{v} = B (v, w) I$ .

4.1givenstep 1.2step 3.1step 3.2algebra∎

Adding the three anticommutator identities yields $c (v) c (w) + c (w) c (v) = (B (v, w) + B (w, v)) I = 2 B (v, w) I$ . Taking $w = v$ and cancelling $2$ in $k$ gives $c (v)^{2} = Q (v) I$ . Linearity of $ϵ_{v}$ and of the contraction formula in $v$ makes $c : V \to {End}_{k} (Λ V)$ linear.

Depends on

Used by

PBW basis and exterior associated graded for a Clifford algebra Theorem

Dependency tree · two levels

11 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Eckhard Meinrenken, Clifford algebras and Lie groups, lecture notes, Chapter 2, §1.2 (15)–(16) and §2.1 Proposition 2.2, printed pp. 24–25 (standard reference, not scraped)