Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
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 $n\ne 3$.

share|cite|improve this answer
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

Not quite a unique property, but close:

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

share|cite|improve this answer
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

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.