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A Conformal flatness of Riemannian manifolds

  1. Mar 3, 2017 #1
    Manifolds of contant curvature are conformally flat. I'm trying to find a stronger claim related to this for manifolds of dimension >2. Does anyone knows if for instance Riemannian manifolds(of dimension >2) with non-constant curvature are necessarily not conformally flat, or maybe something weaker, can it be said just for dimension n=3?
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  3. Mar 3, 2017 #2

    Ben Niehoff

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    Conformally flat manifolds always have a metric tensor of the form

    $$ds^2 = e^{2\varphi} (\delta_{ab} \, dx^a \, dx^b)$$
    So from that you can learn all the necessary facts about the curvature. It should be easy for you to compute all the curvatures. It shouldn't be hard for you to show the following:

    1. In 2d, all manifolds are (locally) conformally flat.
    2. In 3d, a manifold is (locally) conformally flat if and only if the Cotton tensor vanishes.
    3. In 4d and higher, a manifold is (locally) conformally flat if and only if the Weyl tensor vanishes.

    (In 3d, the Weyl tensor vanishes identically).
  4. Mar 4, 2017 #3
    Thanks, I did gather this much from the WP page. I'm only considering the case where the function ##\varphi## is defined for te whole manifold, not just locally. Could you perhaps give me a counterexample either in three or more dimensions in wich a non-constant curvature Riemannian manifold is conformally flat? Perhaps there is some obvious counterexample, checking the infinite possible non-constant curvatures looks like a daunting task.
  5. Mar 4, 2017 #4

    Ben Niehoff

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    If you want it to be globally conformally flat, then your constant-curvature assumption doesn't work in the first place. A sphere has constant curvature, but is not conformally flat.

    In general, any manifold with positive curvature will probably be closed (there are theorems that involve more specific criteria, I think), so they can't be conformally flat. However, negative-curvature manifolds can generally avoid being closed.

    So, I recommend doing the following: Take the ansatz I wrote for a conformally-flat metric, and work out its curvatures in terms of the function ##\varphi##. The Ricci scalar will be related to the Laplacian of ##\varphi##, for example. Next, just choose ##\varphi## to be any smooth function such that the curvature remains negative. You will get a smooth manifold which is probably homeomorphic to ##\mathbb{R}^n## (you may need to check for any non-trivial topology, but I think generically you won't have it). And then you'll have a non-constant-curvature manifold which is conformally flat.
  6. Mar 4, 2017 #5
    Oops, my bad, you are right. But at least in the negative constant curvature case of hyperbolic space, it is not only locally conformally flat but also globally, then? (since there is no topologic obstruction, i.e. it is homeomorphic to ##\mathbb{R}^n##)
    Ok. Thanks. I take it you mean that the curvature remains nonpositive?
  7. Mar 4, 2017 #6

    Ben Niehoff

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    Yes, and in fact the curvature can be positive sometimes (just not everywhere). What you really need to do is check topological invariants such as the Euler number. As long as the manifold is homeomorphic to ##\mathbb{R}^n##, then you're fine. But the condition is going to be that the integrals of some quantities (built out of the curvature) over the whole manifold need to have certain values. This can be quite permissive in terms of what can happen locally.
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