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You use a non-standard definition of the Clifford algebra (up to sign), but it matters nothing with respect to Spin groups. If $x\mapsto u\,x\,u^{-1}$ is a special orthogonal operator on V and u is known to be even, then $u\in\mathrm{Spin}(V)$. But if parity of u is unknown, then we can’t be sure u belongs to Spin(V). BTW, you are also confused about ...


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The Clifford algebra is defined for any quadratic module over a commutative ring. See any algebra book, e.g. Bourbaki, Algebra IX, §9.


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I think that two things must be distinguished here. One is the "traditional"/mainstream Clifford algebras, and the other, Hestenes'. The traditional definition (which you'll find in pretty much any book) takes a vector space V, and builds an algebra out of its tensor algebra. In particular, it takes a quotient of the tensor agebra by the relation $v\otimes ...



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