Tuesday, 17 June 2014

Diff(M) vs Sym(|M|) in General Relativity

In General Relativity, whole "physical universes" are represented by spacetime models, which have the following form,
$\mathcal{M} = (M, g, T, \phi^{(i)})$
Here $M$ is some differentiable manifold, $g$ and $T$ are $(0,2)$ symmetric tensors, and the $\phi^{(i)}$ are various scalar, spinor, tensor, etc., fields representing matter, electrons, photons, and so on. The laws of physics require that the "metric" tensor $g$ and the "energy-momentum" tensor $T$ be related by a differential equation called "Einstein's field equations". The details are not important here though. (For the metric tensor, some authors write $g_{ab}$, many older works write $g_{\mu \nu}$ and some just $g$. Nothing hinges on this; just clarity.)

Suppose we consider a fixed spacetime model $\mathcal{M} = (M, g, T, \phi^{(i)})$. This is to represent some whole physical universe, or world, let us call it $w$. Let $|M|$ be the set of points in $M$. (We can call it the "carrier set".)

It is known that one may apply certain mathematical operations/transformations to the model $\mathcal{M}$ and also it is part of our understanding of General Relativity that the result is an "equivalent representation" of the same physical universe. This is all intimately related to what has come to be called "the Hole argument".

The mathematical operations are certain bijections $\pi : |M| \to |M|$ of the set of points in $M$ to itself. If $\mathcal{M}$ is our starting model, then the result is denoted $\pi_{\ast}\mathcal{M}$.

[To define $\pi_{\ast}\mathcal{M}$, the whole model is "pushforward" under $\pi$; we really just take the obvious image of every tensorial field $g, T, \dots$ under the map $\pi$: in geometry there are "pushforwards" and "pullbacks", and one has to be careful about contravariant and covariant geometric fields; but when we are dealing with mappings that are bijections, it doesn't matter.]

Which of these maps $\pi$s are allowed? That is,
for which maps $\pi: |M| \to |M|$, do $\mathcal{M}$ and $\pi_{\ast}\mathcal{M}$ represent the same $w$?
It is sometimes claimed that the relevant group of transformations, for General Relativity, is $\mathsf{Diff}(M)$. This is the set of bijections of $|M|$ to itself which leave the differential structure of $M$ invariant. I.e., the automorphisms of $M$. Since $M$ is a differentiable manifold, they are diffeomorphisms. Let me call this,
Weak Leibniz Equivalence:
if $\pi \in \mathsf{Diff}(M)$, then $\mathcal{M}$ and $\pi_{\ast}\mathcal{M}$ represent the same world.
But I say that the relevant group of transformations is much bigger, and is $\mathsf{Sym}(|M|)$, the symmetric group on $|M|$. That is, the relevant group is the group of all bijections of $|M|$ to itself:
Leibniz Equivalence:
if $\pi \in \mathsf{Sym}(|M|)$, then $\mathcal{M}$ and $\pi_{\ast}\mathcal{M}$ represent the same world.
This is the main point made in the Leibniz equivalence paper linked here. I sometimes give this as a talk, with usually some physicists, philosophers of physics and mathematicians there. At the moment, I get 50% I'm wrong and 50% I'm right.

There's a much more general formulation, which is very simple (and is essentially the content given in R.M. Wald's classic textbook on GR, p. 438), and which implies the above, and it's this:
Leibniz Equivalence:
If $\mathcal{M}_1$ and $\mathcal{M}_2$ are isomorphic spacetime models, then they represent the same physical world.
The mistake that people keep making, I say, is that they claim that the points of the manifold must be permuted smoothly. This, I claim, is not so. The points in $|M|$ can be permuted anyway one likes, so long as one applies the operation to everything - topology and differential structure included!

Sometimes this is called "gauge equivalence". Personally I don't care one way or the other about the terminology. However, note that Leibniz equivalence is analogous to the standard case of gauge equivalence - the U(1)-gauge symmetry that characterizes electromagnetism. Let $\mathbb{M}^4$ be Minkowski space, and let $A$ be the 1-form electromagnetic potential. Let $\Lambda$ be a smooth scalar field on $\mathbb{M}^4$. Let $d\Lambda$ be its derivative. Then the gauge equivalence principle for electromagnetism is that $A$ and $A + d \Lambda$ are "physically equivalent". I.e.,
$(\mathbb{M}^4, A)$ and $(\mathbb{M}^4, A + d \Lambda)$ represent the same physical world.
[I'm not really very knowledgeable of the philosophy of physics, and the various revisions and so on proposed, for example, against standard quantum theory, etc.: things like Bohmian mechanics, the GRW theory and so on. Here I'm just writing about classical General Relativity.]

[UPDATE (19 June 2014): I changed the text a teeny bit and added some links to the background maths.]

1. When I started to write the comment I wanted to agree and suggest you win the argument simply by taking a GPS with you and then point out that the talk did not change depending if you are giving it at coordinates X,Y,Z or at seminar room 101 in some university.

However I then wanted to quip that you can equip the index of a street map with a differential structure in this way, which is precisely why I disagree. So consider some physical law f(x)=0 on the manifold, then for any bijection \pi with push forward \pi_star ( and its inverse \pi^star) you have

\pi_star f ( \pi^star a) = \pi_star 0

for a \in \pi_star w. So since you push forward the topology, \pi is a homomorphism and since you push forward the differential structure of M, \pi is a diffeomorphism.

(I am sorry for the half latex notation, hope it is clear enough what I am trying to say.)

2. Yoshi, thanks - yes, I get your point, I think. (Semi-latex is ok.) Your GPS idea is right - that would be somehow represented by some matter field on the whole spacetime model. Then, when *everything* gets pushedforward under any permutation of the point set, then all the "relations" between the field quantities remain "invariant". A similar point is made by Rovelli. This is the sense in which Leibnizian relationism is true.

So, I argue that we need not restrict to $\mathsf{Diff}(M)$. We may consider any $\pi \in \mathsf{Sym}(|M|)$. I.e., any permutation of the carrier set of points, no matter how "wild". It need not be a diffeomorphism of $M$ to *itself*. Maybe it just transposes two points $p,q \in |M|$! Under $\pi$, we then pushforward everything - topology, diff structure, etc. Let's forget the metric, energy-tensor, matter fields, etc., and focus just on the base manifold $M$. So, the pushedforward image is $\pi_{\ast}M$. And then, as you're saying, $\pi : M \to \pi_{\ast}M$ is a diffeomorphism. Yes! So,

(i) $\pi : M \to M$ is not a diffeomorphism.

(ii) $\pi: M \to \pi_{\ast}M$ is a diffeomorphism.

And that's my main point! As you say, any physical law or diff equation, say $f = 0$, satisfied in $M$ will be satisfied in $\pi_{\ast}M$ too. For $M$ and $\pi_{\ast}M$ are isomorphic - even when $\pi$ is any arbitrary permutation of the point set. Many physicists (and philosophers of physics) do assert that $\pi : M \to M$ must be a diffeomorphism; but I say that $\pi$ can be any permutation whatsoever of the point set.

Jeff

3. Well, I agree. However, I wonder what we gain by this, is it just that we can construct a diffeomorphism $\pi : M \mapsto \pi_{\ast}M$ from a bijection $\pi : M \mapsto \left| M \right|$? ( A distinction I previously missed, thanks for the clarification.) Or is there some context I am missing?

Additionally, I am incredibly tempted to remember that we are talking about entire space time models, and use the same arguments to construct grandiose claims about the reality of space.

4. Yoshi, the context is that many authors on GR say that the only operations on $|M|$ allowed are diffeomorphisms of $M$ to *itself*. So, the claim is that the group involved is $\mathsf{Diff}(M)$. But, in fact, the operations allowed can be any bijection $|M| \to |M|$, and so the "gauge group" involved is *not* $\mathsf{Diff}(M)$, but rather is $\mathsf{Sym}(|M|)$, the symmetric group.

It doesn't lead merely to grandiose claims about spacetime (though it does lead to Leibnizian relationism). It tells us that, loosely speaking, from the point of view of physics, "the objects don't matter, only the relations do". Making this more precise is a separate though, and a more difficult task.

Jeff