12
$\begingroup$

What are some interesting and easy-to-understand (for non-differential geometers) facts about subobjects of $\mathbb{R}^4$ that are not only false in $\mathbb{R}^3$, but also specific to the structure of $\mathbb{R}^4$ and maybe do not easily or naturally generalize to higher dimensions?

$\endgroup$
  • $\begingroup$ Do knotted 2-spheres count? $\endgroup$ – N. Owad Jan 11 '16 at 18:23
  • $\begingroup$ Sounds interesting! $\endgroup$ – Damian Reding Jan 11 '16 at 20:12
  • $\begingroup$ Does it have to be true and uninteresting in $\mathbb{R}^5$, or can it be also false in $\mathbb{R}^5$? $\endgroup$ – Brian Tung Jan 13 '16 at 21:44
  • $\begingroup$ Sure, I was just trying to weaken the requirement on what happens in $\mathbb{R}^5$. $\endgroup$ – Damian Reding Jan 14 '16 at 0:30
6
$\begingroup$

Only for $n=4$ does there exist an open set $U\subseteq\mathbb{R}^n$ that is homeomorphic to $\mathbb{R}^n$ but not diffeomorphic to $\mathbb{R}^n$ (a small exotic $\mathbb{R}^4$). What this means is not too difficult to explain (no need to explain what a manifold is, only what a homeomorphism and a diffeomorphism are between open subsets of $\mathbb{R}^n$). I don't think it qualifies as "uninteresting for $\mathbb{R}^5$", though (it's definitely not a triviality in any dimension other than $1$), but you seemed to say "false" was also OK.

$\endgroup$
11
$\begingroup$

Due to the theory of quaternions, due to Hamilton, $\bf R^4$ has a structure of a of non commutative field. The only dimension for which $\bf R^n$ is a field are $n=1,2, 4$ . As an application, the special orthogonal group in dimension 4 is not simple : it is the quotient of $U\times U$ by it center $Z/2Z$ where $U$ is the unitary group in complex dimension 2, or the set of quaternions of norm 1. In other dimension (other than 2) the special orthogonal group modulo its center is simple.

$\endgroup$
  • $\begingroup$ But then this is not true in $\mathbb{R}^5$, is it? $\endgroup$ – Brian Tung Jan 13 '16 at 21:48
10
$\begingroup$

Lets bump knot theory up a dimension. In general, an $n$-sphere can be non-trivially knotted in $\mathbb{R}^{n+2}$. Obviously, an $n$-sphere can be embedded in $\mathbb{R}^{n+1}$, where it is usually defined, but no knotting can occur. In $\mathbb{R}^{n+3}$, we can use the extra dimension and unknot every $n$-sphere.

So, for $n=2$, we have that a $2$-sphere can be knotted in $\mathbb{R}^{4}$, cannot be knotted in $\mathbb{R}^{3}$, and is uninteresting in $\mathbb{R}^{5}$, since there is only one knot type.

If you want to look into this more, you should look at any number of great books: Rolfsen, Adams, and more that I am just not thinking of at the moment.

Edit: As per Mike's comment, we should be assuming PL or locally flat here.

$\endgroup$
  • 1
    $\begingroup$ You have not said precisely what kind of knots you're considering. Your statement that the only knotted spheres in $\Bbb R^{n+2}$ are $n$-spheres; this is true if you're working with locally flat or PL maps, but not smoothly: Haefliger discovered many examples of higher codimension, smoothly knotted spheres; the first is an $S^3$ in $S^6$. It is true that 2-spheres are smoothly unknotted in $S^5$ but I do not think it is uninteresting until $S^7$. $\endgroup$ – user98602 Jan 15 '16 at 15:58
  • $\begingroup$ You make a good point. I do very little above dimension 4, so I will defer to you in this. So, I suppose I should assume locally flat or PL to make my answer complete. Thank you for the correction. $\endgroup$ – N. Owad Jan 15 '16 at 17:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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