- #1

schieghoven

- 85

- 1

- TL;DR Summary
- Trying to understand spinors given a general (+,-,-,-) spacetime metric.

I'd like to better understand spinors on curved spacetime, but started wandering along the following tangent. I've looked at but not particularly understood the sections on spinors in the texts by Penrose and (Misner, Thorne and Wheeler).

Let ##g_{ij}## be a spacetime metric (a symmetric bilinear form with signature +,-,-,-). Denote by ##SO(g)## the symmetry group of ##g##, that is the group of linear transformations ##S## on ##\mathbb{R}^4## such that ##g(Su, Sv) = g(u,v)## for all ##u,v##. It may be shown that ##SO(g)## is generated by 6 matrices ##J_{ab}## with ##a,b = 0,1,2,3## and ##J_{ab} = -J_{ba}## and

$$

(J_{ab})^k_i = g_{ai} \delta^k_b - g_{bi} \delta^k_a .

$$

The Lie bracket of the ##J_{ab}## may then be computed as

$$

[J_{ab}, J_{cd}] = -g_{ac} J_{bd} - g_{bd} J_{ac} + g_{bc} J_{ad} + g_{ad} J_{bc} .

$$

I've found a way to construct the covering group for ##SO(g)## by introducing a bar-operation (sorry: possibly non-standard notation) on the space of 2-by-2 complex matrices

$$

\bar{u} = \epsilon^T u^T \epsilon ,

$$

where

$$

\epsilon = \begin{pmatrix}

0 & 1 \\

-1 & 0

\end{pmatrix}.

$$

Equivalently if

$$

u = \begin{pmatrix}

a & b \\

c & d

\end{pmatrix},

$$

then

$$

\bar{u} = \begin{pmatrix}

d & -b \\

-c & a

\end{pmatrix}

$$

The bar-operation has the properties that (i) ##\bar{u} = u## iff ##u## is proportional to ##I##, the 2-by-2 identity matrix, (ii) ##\bar{uv} = \bar{v}\bar{u}##, (iii) ##\bar{\bar{u}} = u##, and (iv) ##\bar{u}## is Hermitian iff ##u## is Hermitian. One can now construct a representation of ##SO(g)## as follows. Let ##e_i## (##i = 0,1,2,3##) be 2-by-2 Hermitian matrices such that

$$

g_{ij} = \frac{1}{2} (e_i \bar{e_j} + e_j \bar{e_i} ).

$$

(Note that the RHS is proportional to ##I## by properties (i) and (ii) of the bar-operation.) I believe such a set of ##e_i## can be found iff ##g## has signature +,-,-,-. For example, if ##g## is the Minkowski metric diag(1,-1,-1,-1) then we take ##e_a## equal to ##I, \sigma_1, \sigma_2, \sigma_3## where ##\sigma_i## are the Pauli matrices. Then let

$$

J_{ab} = \frac{1}{4} (e_b \bar{e_a} - e_a \bar{e_b}).

$$

It is a (somewhat delicate) exercise to show that these new ##J_{ab}## satisfy the same Lie bracket as the generators of ##SO(g)##, thus providing a representation of ##SO(g)##. As expected from the Minkowski case, the group constructed in this way acts on ##\mathbb{C}^2## and is isomorphic to ##SL(2, \mathbb{C})##.

The preceding construction is I think related to the Clifford algebra of ##g##.

This finally -- and thank you for getting this far -- brings me to my question: does this help me understand what spinors look like on curved space-time? In other words, given a spacetime manifold with metric locally equal to ##g##, can I understand a spinor to be a ##\mathbb{C}^2##-valued field which transforms according to the symmetry group generated by the ##J_{ab}##?

Thanks very much for any insight.

Let ##g_{ij}## be a spacetime metric (a symmetric bilinear form with signature +,-,-,-). Denote by ##SO(g)## the symmetry group of ##g##, that is the group of linear transformations ##S## on ##\mathbb{R}^4## such that ##g(Su, Sv) = g(u,v)## for all ##u,v##. It may be shown that ##SO(g)## is generated by 6 matrices ##J_{ab}## with ##a,b = 0,1,2,3## and ##J_{ab} = -J_{ba}## and

$$

(J_{ab})^k_i = g_{ai} \delta^k_b - g_{bi} \delta^k_a .

$$

The Lie bracket of the ##J_{ab}## may then be computed as

$$

[J_{ab}, J_{cd}] = -g_{ac} J_{bd} - g_{bd} J_{ac} + g_{bc} J_{ad} + g_{ad} J_{bc} .

$$

I've found a way to construct the covering group for ##SO(g)## by introducing a bar-operation (sorry: possibly non-standard notation) on the space of 2-by-2 complex matrices

$$

\bar{u} = \epsilon^T u^T \epsilon ,

$$

where

$$

\epsilon = \begin{pmatrix}

0 & 1 \\

-1 & 0

\end{pmatrix}.

$$

Equivalently if

$$

u = \begin{pmatrix}

a & b \\

c & d

\end{pmatrix},

$$

then

$$

\bar{u} = \begin{pmatrix}

d & -b \\

-c & a

\end{pmatrix}

$$

The bar-operation has the properties that (i) ##\bar{u} = u## iff ##u## is proportional to ##I##, the 2-by-2 identity matrix, (ii) ##\bar{uv} = \bar{v}\bar{u}##, (iii) ##\bar{\bar{u}} = u##, and (iv) ##\bar{u}## is Hermitian iff ##u## is Hermitian. One can now construct a representation of ##SO(g)## as follows. Let ##e_i## (##i = 0,1,2,3##) be 2-by-2 Hermitian matrices such that

$$

g_{ij} = \frac{1}{2} (e_i \bar{e_j} + e_j \bar{e_i} ).

$$

(Note that the RHS is proportional to ##I## by properties (i) and (ii) of the bar-operation.) I believe such a set of ##e_i## can be found iff ##g## has signature +,-,-,-. For example, if ##g## is the Minkowski metric diag(1,-1,-1,-1) then we take ##e_a## equal to ##I, \sigma_1, \sigma_2, \sigma_3## where ##\sigma_i## are the Pauli matrices. Then let

$$

J_{ab} = \frac{1}{4} (e_b \bar{e_a} - e_a \bar{e_b}).

$$

It is a (somewhat delicate) exercise to show that these new ##J_{ab}## satisfy the same Lie bracket as the generators of ##SO(g)##, thus providing a representation of ##SO(g)##. As expected from the Minkowski case, the group constructed in this way acts on ##\mathbb{C}^2## and is isomorphic to ##SL(2, \mathbb{C})##.

The preceding construction is I think related to the Clifford algebra of ##g##.

This finally -- and thank you for getting this far -- brings me to my question: does this help me understand what spinors look like on curved space-time? In other words, given a spacetime manifold with metric locally equal to ##g##, can I understand a spinor to be a ##\mathbb{C}^2##-valued field which transforms according to the symmetry group generated by the ##J_{ab}##?

Thanks very much for any insight.