Let $(M,\mu)$ be a pair of a manifold $M$ ($C^\infty$ or Riemann if you like) and a probability measure $\mu$ on $M$. Is there a sensible way to put a probability measure on the space of submanifolds of M, taking into account the measure $\mu$?
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2$\begingroup$ The space of submanifolds is not finite-dimensional, so not locally compact. This makes the question very delicate. $\endgroup$– David Roberts ♦Commented Dec 14, 2021 at 11:48
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$\begingroup$ I was thinking about sampling n times iid. from M according to $\mu$, taking as the submanifold the union of all $\epsilon$ balls around the samples. This gives a distribution on the space of submanifolds. Letting $(n,\epsilon)$ tend towards $(\infty, 0)$ in a well behaved manner we could hope to get a a limit distribution but it seems very unclear how $n$ has to depend on $\epsilon$ and how to describe this limit distribution. $\endgroup$– TakirionCommented Dec 14, 2021 at 12:51
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$\begingroup$ @Takirion: why is that a measure around submanifolds, but not a measure around subsets that are not very smooth? The essence of the submanifold is its smoothness, like a fine whiskey. $\endgroup$– Ben McKayCommented Dec 14, 2021 at 16:35
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$\begingroup$ @BenMcKay: Every (open) ball is an open subset of $M$, as is the union of these balls $\cup B_\epsilon(X_i) = U$. Open subsets of (smooth) manifolds are (smooth) manifolds. Therefore for every finite n and positive $\epsilon$ we have that $\nu(A) = \mathbb{P}(U\in A)$ is a probability measure on the set of submanifolds. These measures might however not converge to a measure, even when letting $n$ and $\epsilon$ to to their limit simultaneously. $\endgroup$– TakirionCommented Dec 14, 2021 at 17:32
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$\begingroup$ You should say open submanifolds, then $\endgroup$– David Roberts ♦Commented Dec 14, 2021 at 21:08
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