3
$\begingroup$

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.

$\endgroup$
1
  • 5
    $\begingroup$ The Clifford algebra case follows more or less directly from the matrix case (even if you didn't have Johannes Ebert's observation below) because Clifford algebras are semisimple, and so are finite products of matrix algebras (over division rings). $\endgroup$ Dec 13, 2014 at 17:34

1 Answer 1

11
$\begingroup$

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

$\endgroup$
1
  • 3
    $\begingroup$ Another argument for $f(a)$ invertible $\Rightarrow$ $a$ a unit is simply that $x\mapsto ax$ invertible means in particular $1=ax$ has a solution. $\endgroup$ Dec 13, 2014 at 15:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.