Sign up ×
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 have read something about generalized geometries from Gualtieri's thesis. At section 2.3 he discussed spinors. Here I have the question: Suppose that $ x+\xi \in V \oplus V^* $ we can define a bilinear form on $ V \oplus V^* $ by $(x+ \xi , y+ \nu)=1/2( \xi (y)+ \nu (x))$ Consider the Clifford algebra generated by $V$ and the bilinear form defined as above. Question: if ${e_i}$ is an orthonormal basis for $V$ and ${e^i}$ is its dual, what is the Clifford multiplication $e_i.e^j$?

share|cite|improve this question
Er, wait, that should be $e_i.e^j+e^j.e_i=q(e_i+e^j)-q(e_i)-q(e^j)=e^j(e_i)-0(e_i)-e^j(0)=\delta^j_i.$ Which we also could anticipate by just saying that $e_i$ and $e_j$ are orthogonal for $i\neq j$ – ziggurism Jul 9 '12 at 23:55
Did you use the formula $u.v + v.u = 2g(u,v)$ for clifford multiplication? But for this formula I found $e_i.e^j=-e^j.e_i$ only. Do I miss some point? – muratguner Jul 10 '12 at 0:04
No, you didn't miss anything. So because these two vectors are independent, so is their Clifford product. And because they are orthogonal, they anticommute. – ziggurism Jul 12 '12 at 1:07

Your Answer


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

Browse other questions tagged or ask your own question.