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Clifford involutions and the norm of a vector product

Statement

Let $V$ be finite-dimensional over $R$ , let $B$ be symmetric, put $Q (v) = B (v, v)$ , and use the Clifford convention $v^{2} = Q (v) 1$ in $C l (V, Q)$ from Universal Clifford algebra of a quadratic space. The assignments $α (v) = - v$ and $t (v) = v$ , extended respectively as an algebra automorphism and an order-reversing anti-automorphism, descend to the quotient. Put $\overset{ˉ}{x} = t (α (x))$ and $N (x) = x \overset{ˉ}{x}$ . For anisotropic $u_{1}, \dots, u_{k}$ ,

$N (u_{1} \dots u_{k}) = \prod_{j = 1}^{k} (- Q (u_{j})) 1.$

No nondegeneracy assumption is needed for the involution and norm-of-product assertions.

Facts & Assumptions

Given: The tensor-algebra quotient and relation $v^{2} = Q (v) 1$ of Universal Clifford algebra of a quadratic space.

[F1]

The Clifford algebra is the quotient of the tensor algebra by the two-sided ideal generated by $v^{2} - Q (v) 1$ for all $v \in V$ (Universal Clifford algebra of a quadratic space).

Proof

technique · direct

1.1F1construct

On the tensor algebra, send a word $v_{1} \dots v_{m}$ to $(- 1)^{m} v_{1} \dots v_{m}$ for $α$ , and to $v_{m} \dots v_{1}$ for $t$ . The first is an algebra automorphism and the second reverses products. Each defining relation $v^{2} - Q (v) 1$ is fixed by both maps. The ideal generated by these relations is therefore invariant under each map: applying either map to a generator of the ideal gives another generator, with left and right factors exchanged for $t$ . Thus both maps descend to the quotient and square to the identity.

2.1step 1.1F1algebra

Their composite $\overset{ˉ}{x} = t (α (x))$ is an order-reversing involution and satisfies $\overset{ˉ}{u} = - u$ for $u \in V$ . Hence $N (u) = u \overset{ˉ}{u} = - u^{2} = - Q (u) 1$ .

3.1step 2.1algebra∎

For $k = 0$ , the empty product is $1$ and $N (1) = 1$ . If $N (y)$ is scalar, then $N (x y) = x y \overline{x y} = x y \overset{ˉ}{y} \overset{ˉ}{x} = x N (y) \overset{ˉ}{x} = N (y) N (x)$ . Induction on $k \geq 1$ , starting with step 2.1, therefore gives $N (u_{1} \dots u_{k}) = \prod_{j = 1}^{k} (- Q (u_{j})) 1$ .

Remarks

This is $x \overset{ˉ}{x}$ with $\overset{ˉ}{v} = - v$ . Meinrenken's text often writes the reversion norm $t (x) x$ ; for a homogeneous word of parity $k$ that norm differs by $(- 1)^{k}$ , so the conventions agree on even products only.

Depends on

Universal Clifford algebra of a quadratic space

Used by

Dependency tree · two levels

3 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

Jean Gallier, Clifford Algebras, Clifford Groups, and a Generalization of the Quaternions: The Pin and Spin Groups, §§1.2–1.3 (standard reference, not scraped)
Eckhard Meinrenken, Lie groups and Clifford algebras, Chapter 2 §4.1 (standard reference, not scraped)