# Diffeomorphism and isometry

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1. Jul 20, 2015

### synoe

Let $f:p\mapsto f(p)$ be a diffeomorphism on a $m$ dimensional manifold $(M,g)$. In general this map doesn't preserve the length of a vector unless $f$ is the isometry.
$$g_p(V,V)\ne g_{f(p)}(f_\ast V,f_\ast V).$$
Here, $f_\ast:T_pM\to T_{f(p)}M$ is the induced map.

In spite of this fact why $ds^2=g_{\mu\nu}dx^\mu dx^\nu$, called the invariance line element, doesn't change ?

2. Jul 20, 2015

### Staff: Mentor

Because $dx^{\mu}$ and $dx^{\nu}$ are not vectors. They are coordinate differentials, i.e., infinitesimal coordinate intervals. The diffeomorphism changes those in just the right way to keep $ds^2$ invariant even though $g$ changes.

3. Jul 20, 2015

### synoe

Is it true although $dx^\mu$ transforms as a vector under the general coordinate transformation ?
$$dx^\mu\to dx^{\prime\mu}(x)=\frac{\partial x^{\prime\mu}}{\partial x^\nu}dx^\nu$$

4. Jul 20, 2015

### Staff: Mentor

Yes. Look at it this way: a general coordinate transformation does not preserve the length of a given vector, because "a given vector" here means "a vector in the tangent space of a point labeled by a given set of coordinate values". But the coordinate transformation changes the mapping of coordinate values to physical points; so "changing the length of the vector" really means changing which vector (i.e., which point's tangent space) is being referred to.

An isometry, from this perspective, is simply a coordinate transformation that happens to "move" all points to other points at which the physical metric is the same, so the lengths of vectors labeled with a given 4-tuple of coordinates don't change, because the vector being changed to has the same length as the vector being changed from.

5. Jul 20, 2015

### samalkhaiat

The invariance of the interval, $ds^{2}$, (also called the metric) under general coordinates transformation follows from the transformation laws of $g_{ab}$ and $dx^{a}$: $$\bar{g}_{ab}(\bar{x}) d\bar{x}^{a} d\bar{x}^{b} = g_{ab}(x) dx^{a} dx^{b} .$$ This invariance, however, does not mean that the functional form of the metric is also invariant. Indeed, under general infinitesimal transformation $$x^{a} \to \bar{x}^{a} = x^{a} + \epsilon^{a}(x) , \ \ \ \ (1)$$ the functional form of the metric changes according to $$\bar{g}_{ab}(x) = g_{ab}(x) + \mathcal{L}_{\epsilon} g_{ab}(x) .$$ Thus, the vanishing Lie derivative of the metric $$\mathcal{L}_{\epsilon}g_{ab}(x) = 0 ,$$ is the necessary and sufficient condition for (1) to be an isometry.

6. Jul 20, 2015

### bcrowell

Staff Emeritus
Despite the "because," I don't see the logical link between synoe's #1 and PeterDonis's #2.

I think it's also more a matter of taste whether you say that $dx^{\mu}$ is or isn't a notation for a vector. In concrete index notation, there is no other way to write a vector. Symbols like $dx$ have been interpreted in many different ways, from Leibniz to Cartan and Robinson.

7. Jul 20, 2015

### Staff: Mentor

I didn't phrase #2 very well. #4 is a better phrasing of what I was trying to say, and samalkhaiat's #5 is even better since it gives the math.

8. Jul 24, 2015

### synoe

I may resolved the question. Could you give me your opinions?

Coordinate transformations on a manifold can be seen by two different ways, passive and active. The passive one is just a reparameterization, which maps a local coordinate to another local coordinate at a same point on the manifold. The active one is a diffeomorphism, which maps a point on the manifold to another. These two interpretations are essentially different but practically equivalent.

$ds^2$, which is a scalar, is clearly invariant under reparameterizations:$ds^2(x)\to ds^{\prime2}(x^\prime)=ds^2(x)$ because vectors and tensors can be defined through coordinate transformations as
$$V^\mu(x)\to V^{\prime\mu}(x^\prime)=\frac{\partial x^{\prime\mu}}{\partial x^\nu}V^\nu(x)\\ T_{\mu\nu}(x)\to T^\prime_{\mu\nu}(x^\prime)=\frac{\partial x^\rho}{\partial x^{\prime\mu}}\frac{\partial x^\sigma}{\partial x^{\prime\nu}}T_{\rho\sigma}(x).$$

In the active point of view, this invariance corresponds to the definition of the pullback $f^\ast$:
$$(f^\ast g)(U,V)\bigr|_p=g(f_\ast U,f_\ast V)\bigr|_{f(p)}$$
where $f:p\to f(p)$ is a diffeomorphism and $f_\ast:T_pM\to T_{f(p)}M$ is the induced map. This relation can be written in terms of the components by
$$(f^\ast g)_{\mu\nu}(x)U^\mu(x)V^\nu(x)=g_{\mu\nu}(x^\prime)U^{\prime\mu}(x^\prime)V^{\prime\nu}(x^\prime).$$
This equation seems to be nonsense (other than the definition of the pullback) in the active interpretation of coordinate transformations but is apparently same as transformation law of scalars like $ds^2$ if we regard $f^\ast g$ as the metric before transforming and $g$ as after.

9. Jul 26, 2015

### Ben Niehoff

The terminology is often misused in physics. Reparametrizations in the context of GR are the passive form of isometries, not of general diffeomorphisms. Under generic diffeomorphisms, the metric tensor need not be invariant.

Stated another way, diffeomorphisms are the morphisms in the category of differential manifolds. Isometries are the morphisms in the category of Riemannian manifolds (i.e., manifolds with metric tensor).

Note that I am not talking about 1-parameter families of isometries here. That is a different issue.

Second note: Of course, tensors generally transform to their pullbacks (or pushforwards) under diffeomorphism, so it is true that given a Riemannian manifold $(M, g)$ and a map $f : M \to N$, then taking this to mean $f : (M,g) \to (N, f_*g)$ gives us an isometry. But we could also choose a different Riemannian structure on $N$. It depends in which category $f$ acts. For example, the stereographic map sends the Riemann sphere $S^2 \setminus \infty$ to $\mathbb{R}^2$; we can interpret this as a map of Riemannian manifolds that sends

$$d \theta^2 + \sin^2 \theta \, d \phi^2 \to \frac{4 (dx^2 + dy^2)}{(1+x^2 +y^2)^2},$$
or we can interpret this as a map of differentiable manifolds and "forget" the metric structure, in which case it is just a map from the sphere (minus a point) to the plane, which is clearly differentiable, but needn't be an isometry.

You should be able to cook up other maps between manifolds which are also obviously diffeomorphisms, but fail to be isometries, such as a map between two spheres of different radius.

10. Jul 26, 2015

### martinbn

A pedantic comment: these are the isomorphisms of those categories.

11. Jul 26, 2015

### synoe

I can't believe reparameterizations in GR are isometries since spacetime does not have isometries in general. At least Lagrangian level, arbitrary manifolds can be allowed.

To be an isometry, the map $f$ must be to itself $f:M\to M$ because we can't compare tensors $T_p$ at $p$ with pullbacked tensors $(f^\ast T)_p$ unless they are on the same manifold.

12. Jul 26, 2015

### Ben Niehoff

You're thinking "continuous families of isometries" rather than merely "isometries". See my first note.

Consider this: Suppose I give you two different metric tensors, and I ask you whether they represent the same Riemannian manifold. How can you answer this question? By finding a coordinate transformation that makes the metrics identical, of course. In doing so, you will have found an isometry between the original two metrics; the existence of such an isometry means that the Riemannian manifolds are (locally) "the same".

Fair enough. As Martin pointed out above, diffeomorphisms are the isomorphisms in the category of smooth manifolds; likewise isometries are the isomorphisms in Riemannian manifolds.

Therefore, in the category of smooth manifolds, if there exists a diffeomorphism $f : M \to N$, then M and N are actually "the same", as smooth manifolds. Similarly, the existence of an isometry $f : (M, g) \to (N, h)$, such that $g = f^* h$, indicates that (M, g) and (N, h) are "the same" as Riemannian manifolds.

It is frequently useful to give manifolds different names until we have shown the existence of such maps, though.

13. Jul 26, 2015

### synoe

Im not familiar with the categories so I couldn't understand precisely what you mean. But your statements seem to be different from things written in the textbooks I read. My statement is as follows.

If a map $f:M\to M^\prime$ is a diffeomorphism, then the two diffeomorphic manifolds $(M,g)$ and $(M^\prime,f^\ast g)$ have physically identical properties. This is the symmetry GR have.