TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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Universal property of the Clifford algebra

Statement

Let $k, V, B, Q$ be as in Universal Clifford algebra of a quadratic space, and let $A$ be a unital associative $k$ -algebra. For every $k$ -linear map $f : V \to A$ satisfying $f (v)^{2} = Q (v) 1_{A}$ for all $v \in V$ , there is a unique unital $k$ -algebra homomorphism $\overset{ˉ}{f} : C l (V, Q) \to A$ such that $\overset{ˉ}{f} \circ i_{Q} = f$ .

Facts & Assumptions

Given: The quadratic space $(V, Q)$ , a unital associative $k$ -algebra $A$ , and a $k$ -linear map $f : V \to A$ satisfying the stated quadratic relations.

[F1]

Every linear map from $V$ to a unital associative $k$ -algebra extends uniquely to a unital algebra homomorphism from $T_{k} (V)$ (Universal property of the tensor algebra).

[F2]

A homomorphism from $T_{k} (V)$ that kills every Clifford relation factors uniquely through $C l (V, Q)$ (Factorization through the Clifford quotient).

[F3]

The canonical map is $i_{Q} = q_{Q} \circ j$ , and the quotient map $q_{Q}$ is surjective (Universal Clifford algebra of a quadratic space).

[F4]

A unital $k$ -algebra is a unital ring with a central unital structure map from $k$ (Algebras over a commutative ring, central structure maps, and algebra homomorphisms).

Proof

technique · direct

1.1givenF1F4construct

By [F4], $A$ has the central unital $k$ -algebra structure required by [F1], so $f$ extends uniquely to a unital $k$ -algebra homomorphism $\tilde{f} : T_{k} (V) \to A$ with $\tilde{f} \circ j = f$ .

2.1givenstep 1.1F2

For every $v \in V$ , $\tilde{f} (j (v)^{2} - Q (v) 1) = f (v)^{2} - Q (v) 1_{A} = 0$ . Thus [F2] gives a unique unital homomorphism $\overset{ˉ}{f} : C l (V, Q) \to A$ satisfying $\overset{ˉ}{f} \circ q_{Q} = \tilde{f}$ .

3.1givenF1F3step 2.1∎

Since $i_{Q} = q_{Q} \circ j$ , step 2.1 gives $\overset{ˉ}{f} \circ i_{Q} = \tilde{f} \circ j = f$ . Let $g : C l (V, Q) \to A$ be any other unital $k$ -algebra homomorphism with $g \circ i_{Q} = f$ . Then $g \circ q_{Q} : T_{k} (V) \to A$ and $\tilde{f}$ are unital algebra extensions of the same map $f$ along $j$ , because $(g \circ q_{Q}) \circ j = g \circ i_{Q} = f = \tilde{f} \circ j$ . Uniqueness in [F1] gives $g \circ q_{Q} = \tilde{f} = \overset{ˉ}{f} \circ q_{Q}$ . Since $q_{Q}$ is surjective by [F3], $g = \overset{ˉ}{f}$ .

Depends on

Used by

Dependency tree · two levels

10 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 (standard reference, not scraped)
M. F. Atiyah, R. Bott, and A. Shapiro, Clifford Modules, Part I, §1 (standard reference, not scraped)