It is well known that the only locally finite, translation-invariant Borel measure on an infinite-dimensional, separable Frechet space is the trivial measure. I am wondering about an analogous proposition for Frechet manifolds (e.g., the manifold of Lorentz metrics over a candidate spacetime manifold, a $4$-dimensional, Hausdorff, smooth manifold). Define the notion of "locally translation-invariant measure" as follows: fix a point $p$ of the Frechet manifold, a chart $(O, \phi)$ containing the point and a measurable neighborhood $N$ of $p$ contained in $O$; then any translation of $N$ using the local Frechet linear structure that leaves $N$ entirely in $O$ preserves the measure of $N$.

Then I think the following is likely true: There is no locally finite, locally translation-invariant Borel measure on an infinite-dimensional, separable Frechet manifold. The proof would use probably use the fact that every infinite-dimensional, separable Frechet manifold is isomorphic to an open subset of the infinite-dimensional, separable Hilbert space (on which there is, of course, no such measure). Is this result, or something close to it, or a counter-example, known?


Erik Curiel

  • $\begingroup$ Isn't this notion of local invariance chart-depending? $\endgroup$ – Pietro Majer Nov 10 '14 at 8:17
  • $\begingroup$ I'm assuming it holds for all charts and all subneighborhoods of the charts, so I don't think that's an issue. $\endgroup$ – Erik Curiel Nov 10 '14 at 12:10

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