Take the tour ×  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.  share|improve this question 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
add comment  2 Answers  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
$a(w)=-w$. Therefo