Covering Group of SO(g) & Understanding Spinors on Curved Spacetime

[schieghoven]

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.

 

[fresh_42]

Your bar operation is usually called inversion: $\bar u = u^{-1}$.

 

[schieghoven]

Only if $\det(u) = 1$. In general $\bar{u} = \det(u) u^{-1}$, but the latter expression obscures the fact that $\bar{u}$ is linear in $u$.

 

[fresh_42]

Only if $\det(u) = 1$. In general $\bar{u} = \det(u) u^{-1}$, but the latter expression obscures the fact that $\bar{u}$ is linear in $u$.

Yes, but you are talking about $SO(g)$ and the "S" stands for determinant one.

 

[schieghoven]

No, I defined this operation on all complex 2-by-2 matrices, and primarily used it on the matrices $e_i$. As described in the original post, the $e_i$ are 2-by-2 Hermitian and characterised by the condition
$$
g_{ij} = \frac{1}{2}( e_i \bar{e_j} + e_j \bar{e_i} ) .
$$
For a general spacetime metric $g$, the $e_i$ will not have unit determinant.

 

[PeterDonis]

For a general spacetime metric $g$, the $e_i$ will not have unit determinant.

For a general spacetime metric $g$, your whole construction doesn't work since you can't assume that $g$ has any symmetry group at all, much less the 6-parameter one you are using, which is only the symmetry group of $g$ if $g$ is a maximally symmetric metric, of which there are very few.

 

[schieghoven]

For a general spacetime metric $g$, consider the set of transformations $S$ on $\mathbb{R}^4$ such that
$$
g(u,v) = g(Su, Sv)
$$
for all $u,v$. The set of such transformations satisfies the group axioms. Maybe I gave it the wrong name, but this group is what I meant by the symmetry group of $g$. The condition on $S$ can be written in matrix form as
$$
g = S^T g S ,
$$
which has non-trivial solutions, so it is not the trivial group.

Differentiating, the Lie algebra of this group satisfies
$$
0 = J^T g + g J
$$
which has six linearly independent solutions, the generators $J_{ij}$ given in the original post.

 

[PeterDonis]

Differentiating, the Lie algebra of this group satisfies
$$
0 = J^T g + g J
$$
which has six linearly independent solutions, the generators $J_{ij}$ given in the original post.

How do you know there even exists such a group with such a Lie algebra for a general spacetime $g$? As far as I can see, that's only true for a maximally symmetric spacetime. A general spacetime $g$ might have no symmetries at all.

 

[schieghoven]

Thanks for the good question.

For the non-trivial solutions to the defining condition
$$
g = S^T g S ,
$$
suppose first that $g$ has signature $(+,+,+,+)$. We can use the eigenvalue decomposition to write $g = UDU^T$ where $U$ is orthogonal and $D$ is diagonal with positive entries. Then let $V$ be any orthogonal matrix, that is $V \in SO(4)$. We can then take
$$
S = U D^{-1/2} V D^{1/2} U^T
$$
as a non-trivial solution to the defining condition. This proves that $SO(g)$ is non-trivial, and indeed isomorphic to $SO(4)$. As isomorphic groups their Lie algebras will have the same dimension, 6.

There is a slightly different proof for $g$ with signature $(+,-,-,-)$, showing that $SO(g)$ is non-trivial and isomorphic to $SO(1,3)$.

To compute the six generators, I gave them explicitly as the first expression for $J_{ij}$ in the original post. It may be verified by substitution that they are six non-zero solutions of
$$
0 = J^T g + g J .
$$

 

[PeterDonis]

We can use the eigenvalue decomposition

I would like to see a specific worked example. Even one in two dimensions instead of four would be useful. For example, try a generic Riemannian 2-surface with topology $R^2$ and curvature which can vary from point to point. What does the metric $g$ look like? What are its eigenvalues and eigenvectors? What does your non-trivial solution $S$ to the defining condition look like?

 

[schieghoven]

I appreciate the suggestion, but I'm not sure that specific examples of the decomposition $g = UDU^T$ are particularly illuminating, even in two dimensions. This decomposition is also known as the singular value decomposition of a symmetric real matrix and can be computed by essentially any linear algebra package. Suffice to say (i) it exists; (ii) U is orthogonal, and (iii) D is diagonal.

Then, as above, let $V$ be orthogonal, and let
$$
S = U D^{-1/2} V D^{1/2} U^T .
$$
Then
$$
\begin{align*}
S^T g S &= (U D^{1/2} V^T D^{-1/2} U^T) (U D U^T) (U D^{-1/2} V D^{1/2} U^T) \\
&= U D^{1/2} V^T D^{-1/2} D D^{-1/2} V D^{1/2} U^T \\
&= U D^{1/2} V^T V D^{1/2} U^T \\
&= UDU^T \\
&= g,
\end{align*}
$$
as required.

 

[PeterDonis]

I'm not sure that specific examples of the decomposition $g = UDU^T$ are particularly illuminating, even in two dimensions.

It would be for me, since I'm having trouble seeing how one would exist for a generic metric $g$ that does not have any symmetry properties. Such a metric in two dimensions has three independent real functions of the coordinates $x_1$ and $x_2$: $g_{00} (x_1, x_2)$, $g_{11} (x_1, x_2)$, and $g_{12} (x_1, x_2) = g_{21} (x_1, x_2)$. All of the examples I can find of a singular value decomposition are for matrices that have constant coefficients, not functions of the coordinates.

 

[schieghoven]

Ah, I see now. No, the symmetry group is understood to act locally at each point, on the tangent space at that point. Thus there is a group of transformations $S$ at each point, with the group structure set by the metric $g$ at that point. The singular value decomposition, etc., can all be done pointwise as well.

 

[PeterDonis]

the symmetry group is understood to act locally at each point, on the tangent space at that point

Then that symmetry group (for the case of 4-d spacetime) is the Lorentz group, regardless of the geometry of spacetime, and the metric $g$ that appears in the symmetry is the Minkowski metric $\eta_{\mu \nu}$ of the tangent space, not the general metric $g_{\mu \nu}$ of the global geometry. In other words, you are not learning anything about the curvature of spacetime, or how to model spinors in a curved spacetime, by the procedure you describe; all you are doing is re-deriving the fact that the Lorentz group is the symmetry group of Minkowski spacetime.

The usual meaning of the term "symmetries" as that term is applied to a curved spacetime geometry is "Killing vector fields". A generic curved spacetime has zero such fields.

 

[vanhees71]

Yep, but you can derive GR (or in most general form Einstein-Cartan theory) from making Lorentz symmetry a local (gauge) symmetry.

 

[martinbn]

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 $J_{ab}$?

This helps you understand spinnors at a point. In other words just the algebra, not the geometry and analysis.