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Compatible metrics?

  1. Jan 9, 2009 #1
    What does it mean to say that, given one metric, there is a compatible Riemannian metric?

    That is, is there a clear explanation of what "compatibility" means?

    I'm just starting a DG course, so I really need a definition without too much rigor. Just looking for a way to understand it (kind of heuristic definition) until I have enough lingo down to understand the rigorous definitions floating around in papers...
  2. jcsd
  3. Jan 9, 2009 #2


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    In metric topology, two metrics are said to be compatible if they result in the same open sets- which means that they give the same limits. I don't know if that same terminology is used in Riemann spaces.
  4. Jan 9, 2009 #3
    Last edited by a moderator: Apr 24, 2017
  5. Jan 9, 2009 #4
    Just having a metric is a very weak condition--most of the topological spaces you encounter will be metrizable. Having a Reimannian metric provides a lot of structure in addition to the topology (which you get from any metric). In particular, it makes a manifold into a differentiable manifold (oddly not all manifolds allow this. those that do may allow it in to different ways.)
  6. Jan 10, 2009 #5


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    Maybe if you tell us the context in which you saw it, or the original statement
    that would help.

    Otherwise, here are some comments which I hope will help:

    A Riemannian metric is not a metric in the same sense as the
    metric in a standard topological space. The Riemannian metric is actually a
    metric tensor ( or, more accurately, a tensor field), while the standard use of
    metric in a topological space X refers to a function d:XxX-->IR_+ U{0}
    ( i.e., assigns a distance to each pair of points) that satisfies the
    axioms of a metric function ( and so that the topology generated by this metric
    is usually expected to agree with the original topology of the space, if one is
    A Riemannian metric (RM) allows you to define a geometry in each tangent space,
    since an RM assigns to each tangent space T_pM in your manifold, a positive-
    definite inner-product, (which is a bilinear map; linear in each tangent vector component,
    i.e., a Riemannian metric assigns to each point p a bilinear map f:<X_p,Y_p>-->R
    this is what a 2-tensor is: an assignment of bilinear maps. Sometimes this assignment
    is needed to be smooth --or smoot, if you're from Brooklyn) and so that the inner-product is bounded in absolute value by 1 .
    This last property is expressed as:

    -1= < <X_p,Y_p> <=1

    which allows you to define an abstract cosine function (since |cost|<=1 )
    so that you can define angles in this tangent space. Once you can define
    angles, you can talk about geometric properties in your tangent space,
    i.e., if the <X_p,Y_p>=0 , then the two vectors are perpendicular, etc.

    I hope I did not go far off on my response. Hope it helped.
  7. Jan 10, 2009 #6
    This question is kinda strange because when speaking of manifolds,they are always metrizable (since it's locally metrizable and has a countable basis (i.e. countably locally finite)). Unless your definition of manifolds is different.

    Also, when you say compatible Riemannian metric, what do you mean? compatible with what? the covariant derivative? or just demand the existence of a (smoothly varying) Riemannian metric?
  8. Feb 5, 2009 #7

    So for smooth manifolds all metrics are compatible and the question is moot.
  9. Feb 5, 2009 #8
    In Riemannian geometry one starts out with two things, a differentiable manifold and a Riemannian metric on its tangent space. This Riemannian metric defines lengths of tangent vectors to the manifold. One proves that it also generates a metric on the underlying differentiable manifold i.e. it defines a distance function for pairs of points on the manifold itself. One starts out with a length function for tangent vectors and ends up with a distance function on the manifold.

    A priori this distance function may not give the same open sets as those of the underlying manifold. It may define some other topology. But it fact it does not. It gives the same open sets and so is compatible with the topology of the manifold.

    This is a standard and key theorem of Riemannian geometry.
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