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Are there any special (nontrivial) properties of $\mathbb{R}^3$ that distinguish it from any other $\mathbb{R}^n$? If there are, what are some of the important ones?

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The number of convex, regular polyhedron? –  Isaac Solomon Jan 20 '13 at 7:04
I was wondering because I recently learned that for 1 dimensional ODE, oscillating solutions are not possible, while for $\mathbb{R}^2$ and above, oscillation is possible. I was wondering if there are results that hold true for $\mathbb{R}^3$ but for no other $\mathbb{R}^n$, including $\mathbb{R}^2$ and $\mathbb{R}$ –  tacos_tacos_tacos Jan 20 '13 at 7:07
The fact that it can be embedded in $\mathbb R^3$ but not $\mathbb R^2$? –  Alex Becker Jan 20 '13 at 7:07
@AlexBecker but $\mathbb{R}^n$ can always be embedded in $\mathbb{R}^n$, but not $\mathbb{R}^{n-1}$ –  tacos_tacos_tacos Jan 20 '13 at 7:08
@jshin47 I wasn't sure if you were aware of invariance of domain. –  Alex Becker Jan 20 '13 at 7:12
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3 Answers

up vote 6 down vote accepted
  1. The unit sphere $\{x:\|x\|=1\}$ in $\mathbb R^3$ has the property that the area of each spherical slice $\{x:a\le x_1\le b\}$, $-1\le a\le b\le 1$, depends only on $b-a$. In more technical terms, the pushforward of the surface measure on a sphere under orthogonal projection to a line is a uniform measure on some segment. This property fails in all other dimensions.

  2. Nontrivial knots (informally speaking: smooth simple closed curves that cannot be continuously deformed to a circle) exist in $\mathbb R^3$ but not in $\mathbb R^n$ for $n\ne 3$.

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For the second statement, why can't you have such a knot in an embedding of $\mathbb{R}^3$ in $\mathbb{R}^n$, for $n \gt 3$? –  tacos_tacos_tacos Jan 20 '13 at 20:00
@jshin47 Then you'd be able to unknot it (deform to a circle) in $\mathbb R^n$. –  user53153 Jan 20 '13 at 20:09
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Not quite a unique property, but close:

A non-trivial vector cross product can be defined only in $\mathbb{R}^3$ and $\mathbb{R}^7$.

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Another factoid that distinguishes $\mathbb{R}^7$ is that the hypersphere has a larger surface area than a hypersphere in any other $\mathbb{R}^n$. –  Dan Brumleve Jan 20 '13 at 8:10
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A random walk on $\mathbb{Z}$ or $\mathbb{Z}^2$ will return to the origin almost surely, but this fails for $\mathbb{Z}^3$. It is not related to the reals specifically but it is a curious difference between two dimensions and three.

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Since this doesn't answer the question, don't you think it should be a comment? –  nonpop Jan 20 '13 at 7:28
@nonpop, I think it is an answer because $\mathbb{Z}^n \subset \mathbb{R}^n$? –  Dan Brumleve Jan 20 '13 at 7:29
Hmm, well I guess the question is vague enough that it could be... –  nonpop Jan 20 '13 at 7:32
This property does not distinguish $\mathbb R^3$ from $\mathbb R^n$ with $n>3$. –  user53153 Jan 20 '13 at 7:53
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