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.
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.
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 $w=0$.
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).