- §1. Equivalence Problem
- §2. Lagrangians
- §3. Spinors
- §4. Topology
- §5. Sphere Integration
The Equivalence Problem
It is easy to see that the line elements \begin{align} ds^2 &= dx^2 + dy^2 \\ ds^2 &= dr^2 + r^2\,d\phi^2 \end{align} are equivalent; both represent flat Euclidean space, the first in rectangular coordinates, and the second in polar coordinates. What does equivalent mean? That there exists a coordinate transformation taking one line element to another, in this case the standard transformation between rectangular and polar coordinates.
What about the line element \begin{equation} ds^2 = -dt^2 + t^2 \left(\frac{dr^2}{1+r^2} + r^2 \left( d\theta^2 + \sin^2\theta\,d\phi^2 \right) \right) \end{equation} which was studied in §Friedmann Vacuum Cosmologies? With the wisdom of hindsight, we can set \begin{align} R &= tr \\ T &= t\sqrt{1+r^2} \end{align} after which the line element becomes \begin{equation} ds^2 = -dT^2 + dR^2 + R^2 \left( d\theta^2 + \sin^2\theta\,d\phi^2 \right) \end{equation} so that this line element is nothing more than Minkowski space in funny coordinates. But what if we didn't know about this coordinate transformation? Could we still have determined that this line element is that of Minkowski space?
The answer in this case is yes: Compute the curvature 2-forms; they vanish. There is only one flat geometry in each dimension and signature, so there must be a transformation from the original coordinates to Minkowski coordinates.
However, the general question of when two metrics are equivalent is much harder. The history of general relativity is full of rediscoveries of known solutions of Einstein's equation, but in new coordinates, where the solution was not immediately recognized. Is there an algorithm for determining equivalence? It was this question that led Christoffel to introduce connections!
In fact there is such an algorithm, at least in principle, and it goes something like this: Compute the components of the (Riemann) curvature tensor for both metrics. Now compute their (tensor, i.e. “covariant”) derivatives. Repeat until you have computed all derivatives of order 20. Compare your two lists of derivatives. If they are the same, that is, if they can be made to agree after some simple algebra, then yes, the metrics are equivalent. If not, they are not. This last step is not easy, as it requires putting both lists into some sort of canonical order so that they can be compared. However, that is not the main stumbling block.
How many components does the Riemann tensor have in 4 dimensions? That is, how many independent components do the curvature 2-forms have? The Riemann tensor is a 4-index object, so it has $4^4=256$ components, although symmetry reduces that number significantly. But each derivative can be in any of 4 directions, so the number of derivatives of order 20 is something like $4^{4+20}=281,474,976,710,656$. That's 281 trillion… Again, symmetry will reduce this number significantly, and the actual number of independent components through order 20 turns out to be $79,310$. But that's still too many to make the comparison realistic other than in highly symmetric situations such as flat space. What can we do?
The first simplification is to insist that all computations be done in an orthonormal frame. This turns out to reduce the order of differentiation needed from 20 to 10. That reduces the number of highest-order terms to something like $4^{4+10}=268,435,456$, and symmetry reduces the number of independent components through order 10 to $8690$.
Further simplification turns out to be possible by using spinors, rather than vectors. The relationship between spinors and differential forms is discussed briefly in §Spinors. In a certain sense, spinors are the “square roots” of vectors, and this turns out to reduce the order of differentiation to 7, and the number of independent components through order 7 to $3156$.
A few thousand terms may still seem like a lot, but in practice things are even easier, as no solutions to Einstein's equation are known for which the algorithm requires more than 3 derivatives, reducing the number of independent components to a mere $430$. Most of the known solutions have indeed been classified, in the sense of being used to generate a (long) list of components and derivatives in canonical form that can be compared with a candidate new solution, and much of this work has been automated using purpose-designed algebraic computing software.
Further discussion of these ideas can be found in \cite{Karlhede}.