Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm fumbling a bit in my reading on Clifford algebras. I'm hoping someone can shed some light on the following isomorphism.

Suppose you have a symmetric bilinear form $G$ over a vector space $V$, and let $\mathrm{Cl}_G(V)$ be the corresponding Clifford algebra. I'll denote it by $C_G$ for short when the vector space is clear.

Now $G$ induces a form $\hat{G}$ on $V/\ker(G)$, and apparently $C_G/\mathrm{rad}(C_G)\cong C_{\hat{G}}$, where $\mathrm{rad}(C_G)$ is the radical of $C_G$. I thought this will fall out easily from some application of the isomorphism theorems, but some epimorphism $C_G\to C_{\hat{G}}$ whose kernel conveniently happens to be $\mathrm{rad}(C_G)$.

However, I can't quite find such a map. Does anyone see what the trick is here? Thanks.

Later. I believe I now understand that there is a surjective map $p:C(\beta)\to C(\bar{\beta})$ induced by the quotient map $V\to V/\ker\beta$ described below. If $I$ is the ideal generated by $\ker\beta$, then since $p(\ker\beta)=0$ in $C(\bar{\beta})$, it follows that $I\subset\ker p$. However, I don't understand why $I$ is nilpotent. What is the explanation for this last bit?

share|improve this question

1 Answer 1

$\newcommand\rad{\operatorname{rad}}$Let $\beta$ be an arbitrary symmetric form on a vector space $V$, let $\ker\beta$ be its kernel and let $\bar\beta$ be the induced non-degenerate form on $V/\ker\beta$.

  • Check that there is an surjective algebra map $p:C(\beta)\to C(\bar\beta)$ which is induced by the quotient map $V\to V/\ker\beta$ whose kernel is generaled by $\ker\beta$.

  • Notice that since the codomain of $p$ is a semisimple algebra (it is the Clifford algebra of a non-degenerate form), its kernel contains the radical $\rad C(\beta)$ of $C(\beta)$.

  • Now, every element of $\ker\beta$ is in the center of $C(\beta)$ and squares to zero. Check that the ideal $I$ generated by $\ker\beta$ in $C(\beta)$ is nilpotent and contained in $\ker p$. It follows that it is contained in the radical $\rad C(\beta)$.

  • Rejoice.

share|improve this answer
Thanks Mariano, I will give this some thought. However, do you mind expanding a bit on the first bullet point? This was my initial thought, but I was having difficulty just picturing how the induced map would act. –  Jakucha Apr 15 '12 at 7:52
The clifford algebra $C(\beta)$ is generated by the elements of $V$ and that of $C(\bar\beta)$ is generated by those of $V/\ker\beta$. There is an obvious way of mapping a generator of $C(\beta)$ to $C(\bar\beta)$! –  Mariano Suárez-Alvarez Apr 15 '12 at 8:48
Yikes! My mistake. So looking at the third bullet point, the elements of $\ker\beta$ map to zero in $C(\bar{\beta})$ since $\bar{\beta}$ is the induced form on $V/\ker\beta$, correct? Thus $I\subset\ker p$. But I'm still having trouble seeing why $I^n=0$ for some $n$. Could you please elaborate, if it's not a burden? –  Jakucha Apr 17 '12 at 7:56
The ideal is generated by finitely many nilpotent elements which are central. That is enough to show that the ideal is nilpotent. A typical element of the ideal is of the form $u=a_1v_1+\cdots a_kv_k$ with $\{v_1,\dots,v_k\}$ a basis of $\ker\beta$ and $a_1,\dots,a_k$ arbitrary elements of $C(\beta)$. Compute $u^{k+1}$ and show that it is zero. –  Mariano Suárez-Alvarez Apr 18 '12 at 6:42
I hope this is as simple as I think it is. In the expansion of $u^{k+1}$, each summand will have at least some factor $v_i^2$ for some $i$. But $v_i^2=0$, so $u^{k+1}=0$. Thanks again, I hope I didn't bother you. –  Jakucha Apr 18 '12 at 7:16

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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