# Representations of spinors

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1. Feb 11, 2016

I'm currently reading a book on relativistic field theory and I'm trying to understand spinors.
After the author introduces the four parts of the Lorentz group he talks about spinors and group representations:

"....With this concept we see that the 2x2 unimodular matrices A discussed in the previous section form a two-dimensional representation of the restricted Lorentz group L_+ (and arrow up)"

The derivation is not clear to me and the author is very abstract in his explanations. But I want to know how to explicitly derive this unimodular matrices. I know a little bit about group theory, for example how to represent the group Z3 as matrices with this formula $[D(g)]_{ij}=<i|D(g)|j>$ and it's simple. I know there is a difference because the Lorentz group is a continuous group but maybe there is also such a simple way to derive the spinor representation. I want to know how to explicitly derive spinors from the Lorentz group.

I know that you can write that a four vector corresponds to a 2x2 matrix via:

$\begin{pmatrix} x^{0}+x^{3} & x^{1}-ix^{2} \\ x^{1}+ix^{2} & x^{0}-x^{3} \end{pmatrix}$
Now is this already a spinor?

2. Feb 11, 2016

### George Jones

Staff Emeritus
2x2 unimodular matrices form a repesentation of the universal cover of the restricted Lorentz, i.e, the restricted Lorentz group is not simply connected.

The space of 4-vectors is a tensor product of 2-component spinor spaces.

For a somewhat readable mathematical exposition of this, see the book "The Geometry of Minkowski Spacetime" by Gregory Naber.

3. Feb 20, 2016

Ok thank you. Ok another question.
You can map a four vector to a 2x2 complex matrix like this:

$X= \begin{pmatrix} x^{0}+x^{3} & x^{1}-ix^{2} \\ x^{1}+ix^{2} & x^{0}-x^{3} \end{pmatrix}$

while

$det(X) =(x^{0})^{2}-(x^{i})^{2}$

Is the Lorentz invariant distance, which means that every transformation which preserves this length is a Lorentz transformation. Now we can make such a transformation with 2x2 unimodular matrices like:

$X' = AXA^{\dagger}$

Alright, I get all that. But how do you come to spinors now? What is missing?

4. Feb 20, 2016