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Meaning of diffeomorphism invariance?

  1. Mar 17, 2008 #1

    kdv

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    I initially posted this question in the Beyond the Standard Model forum since diffeomorphism invariance is a key ingredient of loop quantum gravity but it was suggested that I post the question here.

    Why is Einstein's theory diffeomorphism invariant? A diffeomorphism is basically a map of the points of the manifold into other points of the manifold so the points are actually moved and the manifold gets distorted (which is totally different than a coordinate transformation in which the points are not changed but simply relabelled).

    At first, it seems impossible for GR to be invariant under an arbitrary mapping of the points. Consider a region of space very far from any mass/energy distribution. GR would say that this region is near flat. But if we can perform an arbitrary mapping of the points, what prevents us from transforming this region into a region where there is a nobn-zero curvature? (which would clearly violate Einstein's equations).

    So does it mean that there is some well-defined restriction on the allowed mappings? But this never seems to be mentioned...

    Or am I completely misunderstanding what diffeomorphism invariance means?
     
  2. jcsd
  3. Mar 17, 2008 #2
    When you move points around you also have to 'bring' the geometry along with you, so everything is ok. It just means that there is no absolute location (or the absolute coordinate system) in spacetime. Suppose the spacetime is a plane, and suppose you have chosen some coordinate system on this plane. And suppose that a massive object is near the origin, and another massive object is at (0,1). Then, by diffeomorphism you can move these objects to, say, points (10,10) and (100,23), of course you also have to transform the geometry of the plane appropriately. Then the underlying physics will not change, the physics is only in the relative position (or geometry) of these points, it is not whether we have the point (0,1) or (100,23). Remember the coordinate system is not changed, only the geometry changed, so these coordinate values can identify the points on the mathematical manifold.
     
  4. Mar 17, 2008 #3

    kdv

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    Thank you. I am still a bit confused so let me ask more questions.

    First, what do you mean by "you also have to bring the geometry along with you"? This is the part that confuses me. So we are not free to map the points to anything we want. For example, I can't map part of the plane to a curved surface, right? (I mean, I can't generate a "bump" in the surface through the diffeomorphism). I am not sure how this type of restriction is implemented mathematically. what is a definition of a mapping of points that "brings along the geometry"?

    A secodn question is about the fact that the coordinate system is not changed so as you say the coordinates of the points changes. So this means that the metric must change too when a diffeomorphism is applied, right? Otherwise it seems that physical resultys would be changed. SO my question is: what happens to the metric during a diffeomorphism...am I right that it gets transformed? And if so what is the definition of the transformation law of the metric through a diffeomorphism?

    Thank you very much.
     
  5. Mar 19, 2008 #4

    George Jones

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    Diiffeomorphisms are the active versions of passive coordinate transformations.

    If [itex]\phi : M \rightarrow M[/itex] is a diffeomorphism and [itex]p \in M[/itex], then [itex]\phi[/itex] can be used to shift all geometric objects, including the metric tensor, at [itex]p[/itex] to corresponding geometric objects at [itex]\phi\left(p\right)[/itex], either by pushing forward with [itex]\phi[/itex], or by pulling back with [itex]\phi^{-1}[/itex].

    In particular, [itex]\left( M , g \right)[/itex] goes to [itex]\left( M , \phi_* g \right)[/itex], where the new metric tensor [itex]\phi_* g[/itex] "acts" at [itex]\phi\left(p\right)[/itex] like [itex]g[/itex] does at [itex]p[/itex].

    Everything actively shifts, vector fields, metric, curvature, etc. This shift is represented by new objects at, for example, [itex]\phi\left(p\right)[/itex] that "act like" the old objects at [itex]p[/itex].

    Now let's start adding a bit of detail. To do this, generalize to [itex]\phi : M \rightarrow N[/itex], where [itex]\phi[/itex] is smooth but not necessarily a diffeomorphism. For example, [itex]\phi[/itex] might not even be invertible.

    Consider a scalar field [itex]f : N \maps \mathgbb{R}[/itex] on manifold [itex]N[/itex]. Define the scalar [itex]\phi^* f : M \maps \mathgbb{R}[/itex] (think of it a as all one symbol for a function) on [itex]M[/itex] by [itex]\phi^* f := f \circ \phi[/itex]. The function [itex]\phi^* f[/itex] has the same value at [itex]p \in M[/itex] that the function [itex]f[/itex] has at [itex]\phi\left(p\right) \in N[/itex], i.e., [itex]\left( \phi^* f \right) \left(p\right) = f \left( \phi\left(p\right) \right)[/itex].

    The smooth mapping [itex]\phi[/itex] has been used to pullback the function [itex]f[/itex] on [itex]N[/itex] to a new function [itex]\phi^* f[/itex] on [itex]M[/itex]. Note that because the compositions don't work out correctly, [itex]\phi[/itex] cannot be used to push forward a function [itex]h : M \rightarrow \mathbb{R}[/itex] to a new function on [itex]N[/itex].

    Now, on to vectors, covectors, and tensors.
     
  6. Mar 20, 2008 #5

    kdv

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    Thanks George. I did not reply earlier because I needed time to mull this over (and was busy with midterm tests). I think that very slowly things are making more sense. But I still have ways to go.

    The above sentence makes sense. I used to think that "mapping of the manifold to itself" meant mapping of, say, a 3-manifold to any 3-manifold. But now (if I understand correctly!) one is actually mapping points of the manifold to other points already there in the manifold, right? Then that makes much more sense to me.

    So, to check my understanding of what you said let me ask this: one way to generate a diffeomoprhism is to do the following: one may relabel the points of the manifold with new coordinates (at this point the manifold has not been changed in any way, just the relabelling). A second step is now to map the points of the manifold (with their new coordinates) back to the points which had the same coordinates before we relabeled the points. Is this correct?


    Does this have any physical content? I am curious because this diffeomorphism invariance is a bid deal in LQG. I guess that the point is that if one sums over manifolds, one must treat the manifolds related by diffeomorphism as gauge equivalent.
     
  7. Mar 22, 2008 #6

    George Jones

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    This works beautifully when it can work, but circumstances don't always allow it to work.

    For example, consider the plane [itex]\mathbb{R}^2[/itex] with Cartesian and polar coordinates. Given a point p, it's not always possible to find a point q such that polar coordinates of q equal the Cartesian coordinates of p.

    More abstractly, a coordinate system on an n-dimensional manifold [itex]M[/itex] is a bijective smooth mapping from a "bit" of [itex]M[/itex] to a bit of [itex]\mathbb{R}^n[/itex] such that the mapping's inverse is also smooth. In general these bits of [itex]M[/itex] and [itex]\mathbb{R}^n[/itex] aren't all of [itex]M[/itex] and [itex]\mathbb{R}^n[/itex]; a number of coordinate systems may be required to cover [itex]M[/itex].

    More formally, a coordinate system on an n-dimensional differentiable manifold [itex]M[/itex] is a diffeomorphism [itex]\psi: U \rightarrow \psi\left[U\right][/itex] between an open subset [itex]U[/itex] of [itex]M[/itex] and an open subset [itex]\psi\left[U\right][/itex] of [itex]\mathbb{R}^n[/itex]. So, for [itex]p \in M[/itex], [itex]\psi\left(p\right)[/itex] is an n-tuple of real numbers, the coordinates of p.

    Now suppose we have two coordinate systems between all of [itex]M[/itex] and all of [itex]\mathbb{R}^n[/itex], [itex]\psi: M \rightarrow \mathbb{R}^n[/itex] and [itex]\psi': M \rightarrow \mathbb{R}^n[/itex]. Then [itex]\psi^{-1} \circ \psi' : M \rightarrow M[/itex] and [itex]\psi'^{-1} \circ \psi : M \rightarrow M[/itex] are diffeomorphisms of [itex]M[/itex]. For example, let [itex]\phi = \psi^{-1} \circ \psi'[/itex] and [itex]p \in M[/itex]. The [itex]q = \phi\left(p\right)[/itex] is also in [itex]M[/itex], and [itex]\psi\left(q\right) = \psi'\left(p\right)[/itex], i.e., the coordinates of q in one coordinate system equal the coordinates of q in the other coordinate system.

    This formalizes (maybe overly so!)your nice idea.

    If my points are particularly obtuse or compressed, let me know. I don't think things would seem as abstract if we could stand together with appropriate tools in our hands (chalk!) in front of a chalkboard.

    I don't know. I've heard about this sort of thing, but I know nothing about it.
     
  8. Mar 23, 2008 #7

    kdv

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    Thanks for your reply. It is starting to make a whole lot more sense to me now.
    I am glad to see that my (admittedly very rough and unsophisticated) idea was in th eright ballpark
    This makes a lot of sense!! I guess I always need a very concrete picture of something before understanding the more formal (and complete) description. Even if the concrete picture is not completely general, it gives me an "anchor" to build the more general formalism around, you see what I mean?

    So, if I understand correctly, the maps [itex] \psi [/itex] and [itex] \psi' [/itex] you are talking about must absolutely map the whole manifold to [itex]\mathbb{R}^n[/itex], right?
    We can't work with patches only? This seems to me a big problem because then it would sound as if it is too restrictive. For example, consider the surface of a sphere (a plain, two dimensional euclidian surface of a sphere). There is no map that covers the entire manifold, right? So does that mean that we can't have any diffeomorphism in that case?


    That would be great but this is already much clearer in my mind, thansk to you. It's just that I am slow and I need to have a concrete picture in my mind before grasping the more formal approach. Thanks a lot, again. I hope we can soon discuss general coordinate transformations vs Lorentz symmetry in GR as well.
     
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