Spinors in various dimensions

1. Dec 16, 2009

ismaili

Dear guys,

I wanna understand the spinors in various dimensions and Clifford algebra. I tried to read the appendix B of Polchinski's volume II of his string theory book. But it's hard for me to follow and I stuck in the very beginning. I will try to figure out the outline and post my questions later.

For now, I wanna ask for very simple, introductory articles for the construction of gamma matrices and spinors in various dimensions. (Is the appendix B of Polchinski the simplest article among all?)

Ismaili

2. Dec 16, 2009

haushofer

3. Dec 18, 2009

ismaili

Excuse me,
I searched for this book in libraries nearby and on google but I couldn't find it?

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In the following, I briefly present the content and one of my question by which I stuck.

In the appendix B of Polchinski's string book.
One starts from the Clifford algebra in SO(d-1,1)
$$\{ \gamma^{\mu} , \gamma^\nu \} = 2\eta^{\mu\nu}$$
In the even dimension, $$d = 2k+2$$, one can group the $$\gamma^\mu$$ into $$k+1$$ sets of anticommuting creation and annihilation operators,
$$\gamma^{0\pm} = \frac{1}{2} (\pm\gamma^0 + \gamma^1) \quad\quad \gamma^{a\pm} = \frac{1}{2}(\gamma^{2a} \pm i \gamma^{2a+1})$$
where $$a=1,2,\cdots, k$$.
One then found that,
$$\{ \gamma^{a+}, \gamma^{b-} \} = \delta^{ab}\quad\quad \{ \gamma^{a+} , \gamma^{b+} \} = \{ \gamma^{a-} , \gamma^{b-} \} = 0$$
That is, one finds that the gamma matrices can be grouped into the creation and annihilation operators of $$k$$ species of fermions. In particular, from
$$(\gamma^{a-})^2 = 0$$
one sees there exist a vacuum $$|\xi\rangle$$ annihilated by all $$\gamma^{a-}$$.
Thus, by this observation, one constructed the representation of Clifford algebra in the following space,
$$(\gamma^{k+})^{s_k+1/2}\cdots(\gamma^{0+})^{s_0+1/2} |xi\rangle$$
, i.e. a space of the tensor product of $$k$$ species fermions; so, the dimension of this representation is $$2^{k+1}$$.

In $$d = 2$$, one can easily work out the matrix form of the gamma matrices,
$$\gamma^0 = \left(\begin{array}{cc}0 &1\\ -1 &0\end{array}\right) = i\sigma^2$$
$$\gamma^1 = \left(\begin{array}{cc}0 &1\\ 1 &0\end{array}\right) = \sigma^1$$

One can construct the representation in higher dimensional even space recursively, by $$d \rightarrow d+2$$. But now comes my question, for $$d = 6$$
$$\gamma^0 = i\sigma^2\otimes\textcolor{red}{(-\sigma^3)}\otimes\textcolor{red}{(-\sigma^3)}$$
$$\gamma^1 = \sigma^1 \otimes \textcolor{red}{(-\sigma^3)} \otimes \textcolor{red}{(-\sigma^3)}$$
$$\quad\quad\quad\vdots$$
$$\gamma^4 = I \otimes I \otimes \sigma^1$$
$$\gamma^5 = I \otimes I \otimes \sigma^2$$
where $$I$$ is the 2 by 2 unit matrix.
My question is that, why do we use $$\textcolor{red}{\sigma^3}$$? I thought it should be the 2 by 2 identity matrix!

Anybody guides me through this?
Thank you so much for your help!

4. Dec 20, 2009

ismaili

I think I know the answer to the use of $$\sigma^3$$.
The gamma matrices in d = 2 invole only $$\sigma^1, \sigma^2$$.
When we add the spacetime dimension by 2,
in order to get the correct anti-commutation relations,
we have to tensor product the original gamma matrices by $$\sigma^3$$.

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After figuring out the construction of higher dimensional gamma matrices,
I was confused by the suddenly born conjugation matrix $$B$$ and charge conjugation matrix $$C$$
...