24
$\begingroup$

One may think of this question as a duplicate of this one. I see it more like an extension.

The "inscribed sphere paradox" discussed in the aforementioned question states that if you inscribe a sphere in an $n$-dimensional cube of side $1$, then the volume of the sphere goes to $0$ as $n \to \infty$, while the volume of the cube remains the same.

A more involving paradox is Hamming's Four Circle paradox, also described as an answer to that post.

A more straightforward paradox (also discussed earlier by R. C. Hamming) is the fact that the angle $\theta$ between the diagonal of a cube $(1,1,\ldots,1)$ and any direction $(0,\ldots,1,0,\ldots,0)$ satisfies $$\cos \theta = \frac{1}{\sqrt{n}} \to 0 \mbox{ as } n \to \infty.$$ This means that, as $n$ increases, the diagonal is almost perpendicular to all $(0,\ldots,1,0,\ldots,0)$ (almost lying in (all!) corresponding hyperplanes).

My question is: are there any other elementary examples of these so-called "paradoxes" (for instance, for other objects than sphere/cubes)? I am thinking more of elementary examples in which the intuition from simple plane geometry ($\mathbb{R}^2$) fails miserably in $\mathbb{R}^n$, particularly when $n \to \infty$.

$\endgroup$
5
  • 4
    $\begingroup$ I'm not sure that these are "paradoxes": each of them computes a certain geometrical value in $R^n$ and observes that this value depends on $n$; moreover, the observed dependencies are monotone in $n$. Let me put it another way: take a basis in $R^n$, pick any element of said basis, and observe that the fraction of the basis elements you just picked, $1$ out of $n$, is $1/n \to 0$. There is no paradox in that, is it? Notice now that this observation is similar to the one about the angle between the cube diagonal and one of the cube's sides. $\endgroup$
    – Michael
    Commented Mar 10, 2014 at 2:57
  • 1
    $\begingroup$ No, they are not really paradoxes (that is the reason for the quotation marks). However, I think of them as highly counterintuitive geometric examples. The math is correct, of course, it is the "conclusion" that is counterintuitive (e.g., the diagonal is almost perpendicular to all directions, or the inner sphere goes "out" of the cube). $\endgroup$
    – Campello
    Commented Mar 10, 2014 at 3:08
  • 6
    $\begingroup$ What counts as counterintuitive will, of course, depend on how well-developed your intuition is. $\endgroup$ Commented Mar 10, 2014 at 4:34
  • 5
    $\begingroup$ +1: I find these results wonderfully paradoxical, with no quotation marks around the adjective, since they violate my intuition. If some people have a more accurate intuition, so much the better for them. $\endgroup$ Commented Mar 10, 2014 at 9:28
  • $\begingroup$ Maybe related: SURPRISING GEOMETRIC PHENOMENA IN HIGHDIMENSIONAL CONVEXITY THEORY $\endgroup$ Commented Dec 8, 2023 at 7:48

3 Answers 3

19
$\begingroup$

The unit ball in ${\bf R}^n$ has a lot of counterintuitive properties as $n \to \infty$. Almost all of the volume of the ball is concentrated near the boundary. In fact almost all the volume is concentrated near the equator. That is, for large $n$ if you choose a point in the unit ball at random it is likely to nearly lie on the unit sphere and to have first coordinate approximately zero.

The size of an orthonormal basis in ${\bf R}^n$ is exactly $n$. However, the maximal size of a set of unit vectors $v_i$ for which $|\langle v_i, v_j\rangle| < 10^{-10}$ for all $i \neq j$ grows exponentially as $n \to \infty$. (A simple volume computation shows this.)

In my second answer here I pointed out that in ${\bf R}^n$ for large $n$ we can have convex sets $A \subset B$ such that $B$ is contained in the $\epsilon$-neighborhood of $A$ but the centroid of $B$ is far away from the centroid of $A$.

$\endgroup$
5
  • $\begingroup$ or, take the vectors $v_i$ to be $(\pm 10^{-5}/\sqrt{n}, \ldots, \pm 10^{-5}/\sqrt{n})$. $\endgroup$
    – Campello
    Commented Mar 10, 2014 at 15:09
  • 1
    $\begingroup$ Those don't look like unit vectors ... $\endgroup$
    – Nik Weaver
    Commented Mar 10, 2014 at 19:09
  • $\begingroup$ oops. sorry, forget that. I skipped the "unit" vectors part in the answer. Can you give a light in the "simple volume calulation"? $\endgroup$
    – Campello
    Commented Mar 11, 2014 at 0:34
  • $\begingroup$ Yeah, let $v$ be a unit vector and suppose $w$ is any vector in the unit ball minus the balls of radius $1-\epsilon^2$ about $\epsilon v$ and $-\epsilon v$. Then $w/\|w\|$ is a unit vector whose inner product with $v$ is at most $\frac{3}{2}\epsilon$ (to first order in $\epsilon$) in absolute value. So ... $\endgroup$
    – Nik Weaver
    Commented Mar 11, 2014 at 3:04
  • $\begingroup$ ... follow a greedy algorithm: choose unit vectors $v_1, v_2, \ldots$ such that $|\langle v_i,v_j\rangle| < \frac{3}{2}\epsilon$ for all $i \neq j$. By the above, as long as the balls of radius $1 - \epsilon^2$ about $\pm \epsilon v_j$ for $j < i$ don't cover the entire unit ball, a possible choice of $v_i$ exists. Since volumes scale like ${\rm radius}^n$, you need at least $\frac{1}{2}(1-\epsilon^2)^{-n}$ vectors before no choice is possible. $\endgroup$
    – Nik Weaver
    Commented Mar 11, 2014 at 3:07
14
$\begingroup$

For the $n$-simplex $$\Delta_n = \{x\in\mathbb{R}^{n+1}\ :\ x_i\geq 0, \sum x_i = 1\},$$ its "midpoint" $m = [1,\dots, 1]/(n+1)$, and its corners $e_k$ it holds that $$\|m - e_k\|\to \infty$$ while the distance of the midpoint to the $n-1$ dimensional faces goes to zero (both for $n\to\infty$).

I found it quite counterintuitive that for a convex body like the simplex it can happen that its corners move apart from the midpoint while its "sides" move closer to it. But actually, this is somehow that standard picture of a high dimensional convex body as I learned from these slides by Roman Vershynin (who attributes this finding to V. Milman):

enter image description here

$\endgroup$
9
  • $\begingroup$ The same happens to the cube. The midpoint (1/2,...1/2) has distance 1/2 do any of the faces, whereas the distance to the corner is $1$. Thanks for the slides. $\endgroup$
    – Campello
    Commented Mar 10, 2014 at 11:19
  • 2
    $\begingroup$ Sorry... whereas the distance to the corner is $\sqrt{n}/2$, that goes to $\infty$. $\endgroup$
    – Campello
    Commented Mar 10, 2014 at 11:28
  • 2
    $\begingroup$ Incidentally, that could be a hyperbolic picture as well... $\endgroup$ Commented Mar 13, 2014 at 6:32
  • 2
    $\begingroup$ The idea that the bulk has little volume seems to contradict Nik Weaver's point that most of the volume of the unit ball is concentrated near the boundary... It also seems to be in tension with the fact that the unit sphere takes up a vanishing fraction of the volume of the unit cube. What gives? $\endgroup$ Commented Sep 8, 2014 at 1:47
  • 2
    $\begingroup$ The unit ball is indeed a special convex body. $\endgroup$
    – Dirk
    Commented Sep 8, 2014 at 13:48
3
$\begingroup$

Given any aperture less then π, when the dimension of space is large enough, a one-sided cone with such aperture can be fitted in an orthant (hyperoctant).}

Actually, it is the contrary. The widest circular cone that can be fitted into an orthant (hyperoctant) has the aperture $$ \varphi=\arccos\sqrt{\frac{n-1}{n}}$$ which approaches zero as n grows to infinity.

This is connected with the other responces that describe how faces of the cube come closer to the middle, but is counterintuitive as one would expect that "there is more room" in higher dimensions. Maybe, it is just that the circular cones capture more of the space, so there is less leftover?

This could be judged as selfpromotion, as it is the closing remark of my article, but it is relevant to the question. If there are other sources to the fact or related articles, I would feel grateful for any comments.

$\endgroup$
3
  • $\begingroup$ Isn't it true even in $\mathbb{R}^2$? $\endgroup$
    – Campello
    Commented Mar 10, 2014 at 12:37
  • $\begingroup$ In $R^2$ the quadrant is one-sided cone with aperture π/2 and half plane is one sided cone with aperture π. In $R^3$ the octant is not a circular cone, but a circular cone with aperture of just around $70^\circ$ can be fitted inside. $\endgroup$
    – Mate Kosor
    Commented Mar 11, 2014 at 17:04
  • $\begingroup$ The name of "[@MateKosor]'s article": Orthogonal projection of an infinite round cone in real Hilbert space. $\endgroup$
    – LSpice
    Commented Jun 13, 2020 at 18:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .