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Show that if an element of the odd part of the Clifford Algebra anticommutes with everything in the vector space, then it is 0.

Been having a really hard time making any progress with this one.

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I don't think this is algebraic number theory. And put the question in the body, summary in the title – anon Nov 23 '11 at 1:22
algebraic number theory do use clifford algebra(like quarterions,etc) sometimes, though most of the time you assume the ring to be commutative. – Kerry Nov 23 '11 at 2:17
Algebraic number theory also uses topological spaces, yet a question about them should probably not have to be tagged number-theory... – Mariano Suárez-Alvarez Aug 6 '12 at 23:07

Let $I=e_1\dots e_n$ be a pseudoscalar of $cl(V,Q)$. Since, $we_i=-e_i w$ for all $i$, we get $wI=(-1)^nIw$. On the other hand, in general, we have $Iw=a^{n-1}(w)I$, where $a$ is the grading involution of $cl(V,Q)$. By assumption, $w$ is an odd element, hence $a(w)=-w$. Therefore, comparing the two equation, we get $2wI=0$. Suppose that $char(F)\neq2$. It is easy to see that $I$ is invertible. Hence, we have $w=0$.

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If $\operatorname{char} F \neq 2$, then we have $uv+vu=2(u,v)$. Now we have $(u,v)=0, \forall v\in V$. If the form is not degenerate we should have $u=0$. But if the form is degenerate - like constant 0 - then $u$ is NOT necessarily 0. An example is the exterior algebra $\bigwedge V$ with Clifford algebra multiplication written in an orthogonal basis (Reference, Sternberg, Chapter 10).

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Where are we using that $w\in C\ell^1(V,q)$? Also, shouldn't that be if $\mathrm{char}\,F\neq2$? – Nick Nov 23 '11 at 2:33
I really could not make out why he require $u\in Cl^{1}(V,q)$. But I remember to have seen similar remark somewhere (Lawson, Chapter 1) maybe. I shall correct the symbol mistake. – Kerry Nov 23 '11 at 3:58

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