# Spinors in d dimensions and Clifford algebra

1. Aug 2, 2008

### 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

$$e_\mu e_\nu + e_\nu e_\mu = 2 \delta_{\mu \nu} Q(e_\mu) \cdot 1 ~~~(3.1)$$

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!

Last edited: Aug 2, 2008
2. Oct 9, 2008

### ophase

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