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Spinors in d dimensions and Clifford algebra

  1. Aug 2, 2008 #1


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    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...


    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:

    [tex] Q:x \in V \rightarrow Q(x) \in F [/tex]

    This defines a symmetric scalar product

    [tex] \phi(x,y) \equiv xy + yx = Q(x+y) - Q(x) - Q(y) [/tex]

    In particular, for [tex] e_\mu , \mu = 1 \ldots d,[/tex] 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)[/tex]

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



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

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


    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

    [tex] \delta_{\mu \nu} (Q(2 e_\mu) - Q(e_\mu) - Q(e_\mu)) [/tex]

    since the form is quadratic, this gives

    [tex] \delta_{\mu \nu} (4 Q( e_\mu) - Q(e_\mu) - Q(e_\mu)) = 2 \delta_{\mu \nu} Q(e_\mu)[/tex]

    For some reason it just clicked after I posted my question!
    Last edited: Aug 2, 2008
  2. jcsd
  3. Oct 9, 2008 #2
    I think i like the book...Who is the author of that book?? How can i buy it too??
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