In the Wikipedia for the Spin group, the Clifford algebra is defined as the quotient of the tensor algebra by the ideal $v\otimes v + ||v||^2$. The reverse of an element $v$ is denoted $v^r$ and for a product of orthonormal vectors: $$(e_i\otimes e_j \otimes...\otimes e_k)^r=e_k\otimes...\otimes e_j \otimes e_i$$ They say that this operation can be extended to any element of the algebra and be made an anti-automorphism by declaring $$(a\otimes b)^r=b^r \otimes a^r$$ But doesn't this assume that $a$ and $b$ are a product of orthonormal vectors? And isn't this not true of every element of a Clifford algebra? For example, $Cl(\mathbb{R}^4)$ has $e_1e_2+e_3e_4$ as an element and, unless I'm not seeing something, can't be factored into a product of vectors.

How does the reverse act as an anti-automorphism for the algebra? Do I have a misconception about Clifford algebras or is the article inaccurate?

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    $\begingroup$ The reverse of $e_1e_2+e_3e_4$ is $e_2e_1+e_4e_3$ etc. $\endgroup$ – Lord Shark the Unknown Jan 2 '18 at 6:42
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    $\begingroup$ It’s extended to arbitrary elements of the Clifford algebra by linearity; it has to be or else it wouldn’t be an (anti)automorphism. $\endgroup$ – Qiaochu Yuan Jan 2 '18 at 10:38
  • $\begingroup$ That makes more sense $\endgroup$ – Eben Cowley Jan 2 '18 at 16:51

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