# About elements “factorization” in Clifford Algebras

the article linked below is very instructive and advanced about real Clifford Algebras, and their relationship with Lorentz group. After a general introduction of a Clifford algebra, $$\mathcal{Cl}(V,\Phi)$$, over a real, finite dimensional vector space $$V$$ with a non-degenerate quadratic form $$\Phi$$ as the quotient algebra: $$\mathcal{Cl}(V,\Phi)\sim\frac{\mathcal{T}(V)}{\mathcal{I}(V,\Phi)}$$ where $$\mathcal{T}(V)$$ is the tensor algebra of $$V$$ and $$\mathcal{I}(V,\Phi)$$ is the bilateral ideal generated by all elements of the form: $$v \otimes v - \Phi(v)·\mathbb{1}\quad,\quad v \in V$$ and after presenting all the implications of this definition, the article go straight to define an automorphism (the is also an involution) $$\alpha:\mathcal{Cl}(V,\Phi)\rightarrow\mathcal{Cl}(V,\Phi)$$ as: $$\alpha(i(v)):=-i(v)$$ where the map $$i:V\rightarrow\mathcal{Cl}(V,\Phi)$$ is an injection. Here, in the proof of $$\alpha$$ as an involution, the authors write that:

[...]. Furthermore, every $$x\in\mathcal{Cl}(V,\Phi)$$ can be written as:
$$x=x_1···x_m$$ with $$x_j \in i(V)$$ [...].

Now, my question is: if $$\mathcal{T}(V)=\bigoplus\limits_{i=0}^{n}V^{\otimes i}$$, shouldn't $$x$$ be a superposition of "polynomials" of elements of $$i(V)$$ rather than "monomials" as above? If not, how can I demonstrate this?

• $\alpha$ is linear, right? In that case studying it on monomials should be enough. – MannyC Jul 3 at 23:46
• Yes, $\alpha$ is linear and, with the automorphism properties, i thought it would be sufficient to prove that it is an involution, too. However, the statement in the article is pretty strong and the doubt that it is related with the Cartan-Dieudonné theorem is haunting me. – Andrea Mosena Jul 4 at 7:41
• I believe the author implicitly means to say "every $x$ is of the form [...] or linear combinations thereof." – MannyC Jul 4 at 11:51
• Yes, I think you are right, but my question was not about the demostration of $\alpha$ as an involution. I was worried about if the statement used in that demoatration is actually right. Maybe it can be reformulated in other words as: can the Cartan-Dieudonné theorem be extende to the whole Clifford Algebra? – Andrea Mosena Jul 4 at 16:52
• If you are referring to the statement in yellow I think it's not right but the author meant what I wrote in the previous comment. – MannyC Jul 4 at 17:32