# In general relativity, are two pseudo-Riemannian manifolds physically equivalent if they are isometric, or just diffeomorphic?

In Carroll's Appendix B, he says

You will often hear it proclaimed that GR is a "diffeomorphism invariant" theory. What this means is that, if the universe is represented by a manifold $M$ with metric $g_{\mu \nu}$ and matter fields $\psi$, and $\phi : M \to M$ is a diffeomorphism, then the sets $(M, g_{\mu \nu}, \psi)$ and $(M, \phi^* g_{\mu \nu}, \phi^* \psi)$ represent the same physical situation. ... This state of affairs forces us to be very careful; it is possible that two purportedly distinct configurations (of matter and metric) in GR are actually "the same," related by a diffeomorphism.

I completely agree that two pseudo-Riemannian manifolds $R' = (M', g')$ and $R = (M, g)$, where $M', M$ are smooth manifolds and $g', g$ are metric tensors, are physically equivalent iff there exists a diffeomorphism $\phi:M' \to M$ such that $g' = \phi^* g$ (and, if there are matter fields, $\psi' = \phi^* \psi$, but for simplicity I'll focus on the vacuum case). However, I think his use of the phrase "related by a diffeomorphism" to describe this relation is a bit misleading. In the standard mathematical usage, a "diffeomorphism" is an isomorphism between smooth manifolds (the $M$ and $M'$) and doesn't "touch" the metrics at all. One can consider two Riemannian manifolds $(M, g)$ and $(M, g')$ which are diffeomorphic but have completely independent metric structures. For example, consider the flat unit disk in the $x$-$y$ plane and the upper unit hemisphere, both embedded into $\mathbb{R}^3$ and inheriting the usual Euclidean 3-D metric via the standard pullback mechanism. These Riemannian manifolds are diffeomorphic in the standard mathematical sense, but not "related by a diffeomorphism" in the sense that Carroll describes in the quotation above, because the metrics are not related by the relevant diffeomorphism pullback.

The relation between pseudo-Riemannian manifolds that Carroll describes, in which the metrics "agree" via the relevant diffeomorphism pullback, appears to be what mathematicians call an isometry, which is a very special case of a diffeomorphism. A (mathematicians') diffeomorphism is the natural notion of isomorphism between smooth manifolds, but a (mathematicians') isometry is the natural notion of isomorphism between (pseudo-)Riemannian manifolds - not only the smooth structure but also the metric structure gets "carried over" appropriately.

Question 1: Am I correct that using standard mathematical terminology, the transformation that Carroll is describing an "isometry" rather than a general "diffeomorphism"?

Putting aside Carroll's particular choice of phrasing, I believe that isometry between pseudo-Riemannian manifolds (in the standard mathematical usage linked to above, which is not the same as the usual physicists' usage) is actually the correct notion of physical equivalence in GR, rather than general diffomorphism. As discussed here, general diffeomorphisms do not map geodesics (which are physical and coordinate-independent) to geodesics - only isometries do. Moreover, in my disk and hemisphere example above, the former manifold is flat and the latter is curved, so on the latter surface initially parallel geodesics meet, triangle corners add up to more than $180^\circ$, etc. These non-isometric manifolds clearly correspond to distinct physical states, even though they are diffeomorphic.

Question 2: Am I correct that two Riemannian manifolds correspond to the same physical state iff they are isometric, not merely diffeomorphic (again, under the standard mathematical definitions of "diffeomorphism" and "isometry", not under Carroll's definitions)?

Those are my physics questions. If the answers to both questions #1 and #2 are "yes", then I have a closely related usage question. It seems to me that Carroll's usage of the word "diffeomorphism" is not a personal quirk or sloppy language, but is standard in the physics community. Many times, I've heard physicists say that diffeomorphic Riemannian manifolds are physically equivalent, or that GR is "diffeomophism-invariant".

Question 3(a): When physicists talk about a "diffeomorphism" in the context of general relativity, are they usually using the word in the standard mathematical sense, or in Carroll's sense, which mathematicians would instead call an "isometry"?

If the answer is "in Carroll's sense", then that means that the mathematics and physics (or at least GR) communities use the word "diffeomorphism" in inequivalent ways. This wouldn't surprise me, except in that if so, I've never heard anyone mention that fact.

Question 3(b): Physicists often say that the theory of general relativity is "diffeomorphism-invariant". Am I correct that this is true under the physicists' usage, but under the mathematicians' usage, GR is not diffeomorphism-invariant but only isometry-invariant?

• – tparker Aug 20 '18 at 19:29
• Regarding your edit: Let $(M,\phi)$ and $(N,\psi)$ be manifolds with metrics and $f:M\rightarrow N$ a diffeomorphism. Then the metric $\psi$ pulls back to a tensor on $g^*\psi$ on $M$. The identity map from $(M,\phi)$ to $(M,g^*\psi)$ is certainly a diffeomorphism but might or might not be an isometry. None of this has anything to do with whether $M$ is literally "the same" as $N$ or merely diffeomorphic to $N$; as a general rule, nothing that depends on that distinction can be mathematically relevant (because nothing relevant changes if you replace a category with its skeleton). – WillO Aug 21 '18 at 16:49

## 5 Answers

Indeed, "diffeomorphism invariance" of GR in physics in this context means in proper mathematical parlance that isometric (pseudo-)Riemannian manifolds are physically equivalent. In my view, this confusion between "diffeomorphism" and "isometry" is probably due to physicists usually looking at a manifold in coordinates, and a change in coordinates can be understood as an autodiffeomorphism from the manifold to itself, where the source carries one choice of coordinate charts and the target another.

Under such a change, the metric and all fields transform naturally via pushforward, and we essentially define the diffeomorphism to be an isometry so that the fields on the target with the new coordinates are equivalent to the fields on the source. So the slogan is "change of coordinates" = "diffeomorphism invariance", but the nature of the coordinate change means that the diffeomorphism is always additionally promoted to an isomorphism in whatever category of manifolds we are currently moving in.

This "diffeomorphism invariance" is emphatically not a special property of GR: Every proper physical theory does not care for the coordinates we choose. $\phi^4$-theory and Yang-Mills theory are precisely as diffeomorphism invariant in this sense as GR, just that there the diffeomorphism pushes forward not the metric, but a scalar field and a gauge connection, respectively.

Unfortunately, the phrase "diffeomorphism invariance" is occasionally also used in a different context, namely for what is more properly the local Lorentz invariance of GR (at least in the spin connection formalism). To wit, a diffeomorphism induces a map on all tensors via its Jacobian (or, more mathematically, the (co)tangent map) and GR also exhibits a invariance under the transformations from this (co)tangent map alone, i.e. it is a $\mathrm{GL}(n-1,1)$ (or $\mathrm{SO}(n-1,1)$) gauge theory, albeit an unusual one whose principal bundle is soldered. For more on this, see this answer of mine to an earlier question of yours.

• Would you agree that what physicists call "isometries" (e.g. in the context of Killing fields), mathematicians would call "autoisometries" (i.e. isometries between a single pseudo-Riemannian manifold)? – tparker Aug 22 '18 at 0:06
• @tparker: Yes, I would. – ACuriousMind Aug 22 '18 at 9:54

Note that in your second quote from Carroll, he explicitly says that the metric and the matter fields are being transformed. Therefore a diffeomorphism, by his definition, is nothing more than a renaming of the points. It may be unnecessarily confusing that he talks about a map from manifold M to itself. If we're defining a manifold in the usual way, by point-set topology, then we can actually have manifolds that are different, in the sense that the set of points is different, but homeomorphic as manifolds. Then if we talk about a diffeomorphism from M to N, it becomes a little more clear that we have to transform the metric as well -- otherwise the metric would be a function that doesn't even have the right domain to operate on N.

Moreover, as discussed here, general diffeomorphisms do not map geodesics [...] to geodesics - only isometries do.

By Carroll's definitions, diffeomorphisms do map geodesics to geodesics. For example, suppose we do a diffeomorphism of the plane from Cartesian coordinates to polar coordinates. A line that is a geodesic under the metric $ds^2=dx^2+dy^2$ is also a geodesic under the metric $ds^2=dr^2+r^2d\theta^2$, which is what you get when you transform the metric according to the diffeomorphism. It's not a geodesic under the metric $ds^2=dr^2+d\theta^2$, which is what you get if you don't transform the metric and assume there is some natural correspondence between a point $(x,y)$ and a point $(r,\theta)$.

• The point of my question is, are the transformations that Carroll (and many other physicists) call "diffeomorphisms" actually diffeomorphisms as defined by mathematicians, or are they autodiffeomorphisms, or are they isometries? I understand the procedure you describe that gives the "physically correct" answer, but I'm trying to understand the formal mathematical details. – tparker Aug 20 '18 at 20:23
• @tparker: If your question is just a question about whether Carroll's usage is consistent with the usage of most mathematicians, then I suggest you edit the question so that it says that. As written, it has lots of language that suggests that Carroll is just wrong or mathematically ignorant. – Ben Crowell Aug 20 '18 at 23:57
• Oh no, I certain agree that Carroll is describing the correct procedure for relating physically equivalent field configurations. I'm just not sure if his referring to this transformation as a "diffeomorphism" rather than an "isometry" agrees with the standard mathematical usage. I'll edit the question to clarify. – tparker Aug 21 '18 at 1:43
• I've overhauled the question - hopefully it's finally clear. – tparker Aug 21 '18 at 20:12
• @BenCrowell: it's not just Carroll. This exact point provided me a ton of confusion when I was younger. The quantum gravity community just uses "diffeomorphism" to mean "isomorphism", from what I can tell. – Jerry Schirmer Aug 21 '18 at 20:26

1) The Godel Universe $G$ is (up to homeomorphism) of the form $S\times T$ where $S$ is homeomorphic to ${\mathbb R}^3$ and $T$ is homeomorphic to ${\mathbb R}^1$. But for any $n\neq 4$, every manifold homeomorphic to ${\mathbb R}^n$ is also diffeomorphic to ${\mathbb R}^n$. Therefore $G$ is diffeomorphic to ${\mathbb R}^3\times{\mathbb R}^1={\mathbb R}^4$, and hence diffeomorphic to Minkowski space $M$.

But $G$ and $M$ are definitely not isometric, and certainly not physically equivalent in any plausible sense. So there is a counterexample to "diffeomorphic implies isometric" and a counterexample to "diffeomorphic implies physically equivalent".

2) This is not strictly a mathematical argument but: There must be a great number of spacetimes that have been around since before about 1980 and are obviously homeomorphic to ${\bf R}^4$. Sometime in the early 1980s, Michael Freedman astonished the mathematical world by proving that there are manifolds homeomorphic to ${\mathbb R}^4$ but not diffeomorphic to ${\mathbb R}^4$. This triggered a search for examples. If any of those known spacetimes had been an example, I'm betting that somebody would have soon discovered that fact and publicized it. The fact that I've never heard of such a result leads me to believe that every one of those spacetimes is in fact diffeomorphic to ${\mathbb R}^4$, and hence every one of them provides another counterexample to "diffeomorphic implies isometric".

3) The author you are quoting does seem to be using the word "diffeomorphic" to mean "isometric". Whether this is a mistake or a deliberately cultivated idiosyncracy is, without further information, an open question.

• I certainly never suggested that "diffeomorphic (in the mathematicians' sense) implies isometric." I suggested that "diffeomorphic (in the physicists' sense) is synonymous with isometric (in the mathematicians' sense)." The entire point of my question is whether physicists are using the word "diffeomorphic" in the same sense as mathematicians. I think that a physicist would say that $G$ and $M$ are not diffeomorphic, precisely because they are non isometric. – tparker Aug 21 '18 at 16:25
• And I don't really understand your third point. Carroll is using the word "diffeomorphic" in the sense that ubiquitous among physicists, who are constantly talking about how GR is "diffeomorphism invariant" and that states related by "diffeomorphisms" are physically equivalent. I therefore get the sense that the physics community and the mathematical community simply use the word "diffeomorphism" in incompatible ways. – tparker Aug 21 '18 at 16:29
• @tparker: I understood your question to be asking (at least in part) whether this particular author is using the word "diffeomorphism" the same way a mathematician would. The answer to that question is NO. But I now understand your question bo be asking whether this author is using the word in the same way that other physicists would. I'd be extremely surprised if the answer to that question is yes, but you seem to be looking for a more definitive answer. Sorry if I misinterpreted you. – WillO Aug 21 '18 at 16:40
• I've overhauled the question - hopefully it's finally clear. – tparker Aug 21 '18 at 20:13

According to ACuriousMind, physicists and mathematicians indeed use the words "diffeomorphism" and "isometry" in incompatible ways. I'm very surprised that I've never heard this before, because it seems like an important point to mention. Moreover, it's quite confusing because in both usages an "isometry" is a special case of a "diffeomorphism", but they are "offset": a mathematician's "isometry" is a physicist's "diffeomorphism". Since ACuriousMind gave some of this information in comments, for permanence here's a table translating between the two usages: $$\begin{array}{ccc} \text{Mathematicians' usage} & & \text{Physicists' usage} \\ \hline \text{diffeomorphism} & & \\[5pt] & \text{contains} & \\[10pt] \text{isometry} & = & \text{diffeomorphism} \\[5pt] & \text{contains} & \\[10pt] \text{autoisometry} & = & \text{isometry} \end{array}$$

When a manifold comes without metric, you can do anything with it, as long as the topology is the same and no kinks or ridges are made [thanks to @Willo], like a soft but unbreakable rubber membrane. This is diffeomorphism.

However, when a manifold is equipped with a metric, it becomes somehow "rigid", like a paper shell. You can only bend it when the lengths i.e. metric on it is not changed. This is why you can bend a flat paper in one direction only; in that way, the metric is not changed (so this curvature is extrinsic, and not intrinsic, as is studied in GR). This is isometry.

• Your first paragraph is incorrect; it confuses homeomorphism with diffeomorphism. – WillO Aug 21 '18 at 6:19
• Oops, forgot differentiability... – Trebor Aug 21 '18 at 6:25
• I'm afraid I have to object a second time, to your rewritten first paragraph. You can map the $x$-axis in the plane to the graph of $y=x^3$ by taking $x$ to $(x,x^3)$. This introduces no kinks or ridges, but it's still not a diffeomorphism. – WillO Aug 21 '18 at 6:36
• You are right again @WillO. I think I'll leave that and your comment here, though. Because I think the membrane analogy is enough to intuitively imply invertibility. – Trebor Aug 21 '18 at 7:42
• How does this answer my question, which is about general relativity? – tparker Aug 21 '18 at 16:17