Suppose we have a Clifford algebra $Cl(V,q)$, $V\simeq \mathbb{R}^n$ and $q$ non-degenerate bilinear form. Then every non-zero element of $V\subset Cl(V,q)$ invertible, but they are not the only ones (of course the product of invertible elements is again an invertible one). In fact, according to *Lawson, H. Blaine; Michelsohn, Marie-Louise (1989). Spin Geometry*, the set of invertible elements $Cl^{\times}(V,q)$ is a $2^n$-dimensional Lie group, while $Cl(V,q)$ is $2^n$-dimensional as a vector space. In such a situation, one wonders whether we have the same situation as in $GL(n,\mathbb{R})\subset \mathcal{M}\_n(\mathbb{R})$, where the set of excluded elements is defined by an algebraic equation, namely the determinant of the matrix (and consequently we have a zero-measure excluded subset). **Is there anything analogous for the Clifford algebra case?**


Any suggestions will be welcomed.



Equation for non-invertible elements in Clifford algebras
For any finite-dimensional associative unital $\mathbb{R}$-algebra $A$, the set $S$ of noninvertible elements is the zero-set of a polynomial. Namely, let $f:A \to Hom (A,A)$ be the map $a \mapsto ( x \mapsto ax)$ to the linear endomorphisms of $a$. $f$ is an injective algebra homomorphism, since $f(a)=0$ means $ax =0$ for all $x \in A$, in particular $a= a1 =0$. 


Next, if $f(a)$ is invertible, then $a$ is a unit. This is because if $f(a)$ is invertible, then there exists a polynomial $p$ with $ f(a)^{-1}=p(f(a)) =f(p(a))$. Therefore, $f (p(a)a)=1$ and hence, as $f$ is injective $f(a)a=1$, and so $a$ is a unit.


Therefore $S= (det \circ f)^{-1}(0)$.


