Spinors in d dimensions and Clifford algebra

[nrqed]

I bought a book on susy and there is a chapter on spinors in d-dimensions.
Now, maybe I am extremely dumb but I just can't understand the first few lines!

EDIT: I was being very dumb except that I think there is a typo...See below...


BEGINNING OF QUOTE

Consider a d-dimensional vector space V over the field F for which we shall choose two alternatives F=R or F=C. Let Q be a quadratic form on V:

Q:x \in V \rightarrow Q(x) \in F

This defines a symmetric scalar product

\phi(x,y) \equiv xy + yx = Q(x+y) - Q(x) - Q(y)

In particular, for e_\mu , \mu = 1 \ldots d, a basis of V, orthogonal with respect to Q, we then have

<br /> e_\mu e_\nu + e_\nu e_\mu = 2 \delta_{\mu \nu} Q(e_\mu) \cdot 1 ~~~(3.1)

The associative algebra with unit element generated by the e_\mu with the defining relation (3.1) is called the Clifford algebra C(Q) of the quadratic form Q. The dimension of C(Q) is 2^d


END OF QUOTE

Questions:



1) What does it mean to say that the basis is orthogonal with respect to Q? Does that mean Q(e_\mu + e_\nu) = 0 unless mu = nu? Or is that a typo an dhe meant to write orthogonal with respect to \phi, not Q?

2) No matter what I try I don't see how to get from the definition of phi to equation 3.1!


EDIT: I THINK I GOT IT

Just as I posted I think that I finally understood.

First, I think the author really meant that the basis is orthogonal with respect to phi and not Q. In that case, we get

\delta_{\mu \nu} (Q(2 e_\mu) - Q(e_\mu) - Q(e_\mu))

since the form is quadratic, this gives

\delta_{\mu \nu} (4 Q( e_\mu) - Q(e_\mu) - Q(e_\mu)) = 2 \delta_{\mu \nu} Q(e_\mu)


For some reason it just clicked after I posted my question!

 

[ophase]

I think i like the book...Who is the author of that book?? How can i buy it too??