# Tag Info

0

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 ...

0

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

1

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 ...

Top 50 recent answers are included