AFAIK a metric uniquely determines the volume element up to to sign since the volume element since a metric will determine the length of supplied vectors and angle between them, but I do not see a way to derive a metric from the volume element. The volume form must heavily constrain the the set of compatible metrics though. So is there a nice statement of what the volume element tells you about a metric that is compatible with it? If you have a volume element what is the least additional information you need to be able to derive a metric?


Is the minimum additional information uniform structure?

A volume element gives us a kind of notion of local scale but much weaker than a metric. We can can measure a volume, but we can't tell if it looks like a sphere, a pancake or string (that is the essence of @Rhys's answer). To get a metric we need to be able to tell that a volume is "spherical", or to compare distances without assigning an actual value to them, since if we can do that then we can use use the the volume to determine the radius of a hypershere in the limit of small volumes. So the last part of my question is really asking how to do that without implicitly specifying a metric. I believe that uniform structure does this. It specifies a set of entourages that are binary relations saying that points are within some unspecified distance from each other. An entourage determines a ball of that size around each point. To induce a metric, the uniform structure needs to be compatible with the volume element by assigning the same volume to all the balls induced by the same entourage in the limit of small spheres (delta epsilonics required to make this formal). The condition will not hold for macroscopic spheres if the uniform structure is inducing a curved geometry. There must also be some conditions for the uniform structure to be compatible with the manifold structure.

Does this make sense?

  • $\begingroup$ Just to clarify: Are you working in the setting of a Riemannian manifold? And your question is: If you have a smooth $n$-manifold with a choice of non-vanishing $n$-form $\omega$, what additional information do you need to produce a Riemannian metric $g$ for which $\omega$ is the volume form for $g$? Is that right? $\endgroup$ – Jesse Madnick Mar 14 '13 at 8:05
  • $\begingroup$ And if so, why is this tagged (metric-spaces) and not, say, (differential-geometry) or (riemannian-geometry)? $\endgroup$ – Jesse Madnick Mar 14 '13 at 8:07
  • $\begingroup$ @JesseMadnick I was not specifically thinking about the Riemannian setting, but it seems natural, I just did not really think that deeply about the the formal setting. Otherwise you interpretation of the question sounds right. I tagged it (metric-spaces) just because metrics are involved :) I guess (riemannian-geometry) is a good tag. $\endgroup$ – Daniel Mahler Mar 14 '13 at 8:33
  • $\begingroup$ Doesn't assumeming the manifold is Riemannian imply a metric to begin with though? That would seem to make the question meaningless. $\endgroup$ – Daniel Mahler Mar 14 '13 at 14:39
  • $\begingroup$ Sorry, I should have said "smooth manifold." The question, as I framed it, does make sense. I was really just asking what context you were working in. (As a completely tangential aside: I assume you're aware that a Riemannian metric is not the same thing as a metric in the sense of metric spaces.) $\endgroup$ – Jesse Madnick Mar 15 '13 at 0:48

The volume form tells you very little about the metric. Let $V$ be an $n$-dimensional vector space, with volume form $v_1\wedge \ldots \wedge v_n$, where $\{v_1,\ldots,v_n\}$ are linearly independent elements of $V$ (any volume form can be written this way). Now define a metric by the condition that this be an orthonormal set; this metric gives you the above volume form. But it's far from unique, as there are many choices of $n$ vectors which give the same volume form.

The same story should apply on some small enough patch of a Riemannian manifold.


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