3
$\begingroup$

This is a follow up question to this question on Clifford Algebras. As I understand it, if the associated bilinear form $\beta$ of a Clifford Algebra $C_{\ell}$ is non-degenerate, then $ker\beta$ is equal to $0$. This then implies that rad($Cl_{\ell}$) = {0}, since ) is the only thing that acts as zero in the Clifford Algebra.

But, if the bilinear form is degenerate, then the quotient $C_{\ell}/(rad(C_{\ell}))$ is not isomorphic to $C_{\ell}$, but rather to a smaller algebra. Can we say anything more than this?

$\endgroup$
2
$\begingroup$

Denote by $C\ell(V,\beta)$ the Clifford algebra of $V$ with bilinear form $\beta$.

Also define the set $\ker(\beta):=\{v\in V\mid \forall x\in V,\beta(v,x)=0\}$. It turns out that the ideal generated by $\ker(\beta)$ is exactly the Jacobson radical $C\ell(V,\beta)$, which I guess I'll denote as $rad(C\ell(V,\beta))$.

Now the bilinear form $\beta$ induces a bilinear form $\bar{\beta}$ on $\bar{V}:=V/\ker(\beta)$ in the obvious way: $\bar{\beta}(\bar{x},\bar{y})=\beta(x,y)$ for $x,y\in V$. You can easily show that $\bar{\beta}$ is nondegenerate on $\bar{V}$.

We can conclude that $C\ell(V,\beta)/rad(C\ell(V,\beta))\cong C\ell(\bar{V},\bar{\beta})$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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