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Paul Siegel's user avatar
Paul Siegel's user avatar
Paul Siegel
  • Member for 14 years, 9 months
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Why is $\mathbb R^{\mathbb N}$ not high-dimensional enough?
The problem is not sequences / cardinality considerations - it's the topology on the space of random variables under consideration. I think the point here is that a sequence $S_n$ in $\mathbb{R}^\mathbb{N}$ equipped with the weak topology has a limit if and only if all $S_n$ lie in some $\mathbb{R}^N \hookrightarrow \mathbb{R}^\mathbb{N}$. To prove something resembling a CLT you need a topology in which allows convergence of sequences where the dimension gets arbitrarily large.
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Naturality of Lie bracket — alternate proof
Since MO doesn't allow emoji reacts, I have to type some characters first. 🤤
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Analogous results in geometric group theory and Riemannian geometry?
Often it is possible to prove pairs of theorems like what you describe by looking for quasi-isometry invariants. This works because the universal cover of a compact Riemannian manifold is quasi-isometric to its fundamental group, and so any quasi-isometrically invariant fact about one will automatically apply to the other. Many results about compactifications take this form.
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