2014-09-14 16:44:44 8 Comments

Some theorems are true in vector spaces or in manifolds for a given dimension $n$ but become false in higher dimensions.

Here are two examples:

- A positive polynomial not reaching its infimum. Impossible in dimension $1$ and possible in dimension $2$ or more. See more details here.
- A compact convex set whose set of extreme points is not closed. Impossible in dimension $2$ and possible in dimension $3$ or more. See more details here.

What are other "interesting" results falling in the same category?

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## 30 comments

## @BigM 2019-01-16 20:19:12

Theorem(Samelson).The only Euclidean spheres that can be made into topological groups are $\mathbb S^0,\mathbb S^1$ and $\mathbb S^3$.## @Yossi Lonke 2017-03-06 11:54:22

Let $B_2^n$ denote the Euclidean unit ball in $\mathbb{R}^n$. Then the Minkowski-sum $B_2^{n-1}+B_2^n$ is a zonoid whose polar is also a zonoid for $n\leq 4$, but not for $n\geq 6$. (here).

P.S -- I never did get to check the case $n=5$.

## @Ali Taghavi 2015-08-08 15:52:41

$S^{2}$ and $\mathbb{R}^{2}$ satisfies the Poincare Bendixon theorem but this theorem is not satisfied by higher dimensional spheres or Euclidean spaces.

For a related MSE post see the following.

https://math.stackexchange.com/questions/861231/extension-of-poincar%C3%A9-bendixson-theorem-to-mathbbr3

As another example: For $n>8$ there is no a $n$-dimensional subvector space of $M_{n}(\mathbb{R})$ which all non zero elements are invertible matrix. The only possible $n$ are $n=1,2,4,8$. Such subvector spaces correspond to matrix representation of real numbers, complex numbers, Quaternions and Cayley numbers, respectively.

## @Vladimir Tkachev 2014-11-03 13:27:38

By Liouville's theorem, a bounded subharmonic function in $\mathbb{R}^n$ is a constant. This holds not true if $n\ge 3$.

There are many similar corresponding facts in potential theory.

## @jdfadams 2014-10-30 22:31:14

If $F=(f_\lambda)_{\lambda\in\Lambda}$, is an analytic family of quadratic-like maps (with some conditions), and $M_F$ is the set of $f_\lambda$ with connected Julia set, then there is a homeomorphism of $M_F$ with the Mandelbrot set $\{c\in\mathbb{C}:f_c(z)=z^2+c\text{ has connected Julia set}\}$. See http://www.math.cornell.edu/~hubbard/PolyLikeMaps.pdf. Douady called this the "miracle of continuity" coming from the measurable Riemann mapping theorem.

Analytic families of polynomial-like maps of degree $\geq 3$ have discontinuous straighting maps. See http://arxiv.org/pdf/0903.4289v2.pdf.

## @user35546 2014-10-30 17:07:09

The James-Stein Estimator.

Suppose $Y \in \mathbb{R}^n$ is a Gaussian vector with unknown mean $\mu$ and known spherical variance $\sigma^2I$. Given an observation of $Y$, we are interested in finding an estimator $\hat{\mu}$ of $\mu$ which minimizes the expected mean squared risk $R(\hat{\mu}) := E(||\hat{\mu} - \mu||^2)$.

For $n \le 2$, the Gauss-Markov estimator, $\hat{\mu}_{GM} = Y$ minimizes $R(\hat{\mu})$. But for $n \ge 3$ the James-Stein estimator

$$ \hat{\mu}_{JS} = \left(1 - \frac{(n-2)\sigma^2}{||Y||^2}\right)Y $$

satisfies $R(\hat{\mu}_{JS}) \le R(\hat{\mu}_{GM})$. Note however, that the James-Stein estimator itself does not minimize the mean squared risk.

## @Willie Wong 2014-09-15 13:58:57

Bernstein's Problem/Theorem: a graphical minimal hypersurface of $\mathbb{R}^n$ is necessarily a hyperplane. True only for $n\leq 8$.

Related: the De Giorgi conjecture (now known to be true) concerning the geometry of level sets of bounded solutions to a certain nonlinear elliptic PDE.

## @Paul 2014-09-24 03:58:55

Riemann mapping theorem is true for (complex) dimension $1$, but is false for dimension greater or equal to $2$.

## @R W 2014-09-23 02:41:50

It's interesting that no one has so far mentioned Pythagorean triples and Fermat's Last Theorem :)

EDIT: With sufficient imagination any integer can be interpreted as the dimension of an appropriate vector space. However, in this particular case hardly any imagination is necessary: it is in dimensions 1 and 2 only that there exists 3 hypercubes with integer sides such that the sum of volumes of the first two is equal to the volume of the third one.

## @S. Carnahan 2014-09-23 02:54:47

What does this have to do with vector spaces of increasing dimension? The Fermat curve of degree $n$ defined by $x^n + y^n = z^n$ is an algebraic plane curve.

## @Joël 2014-09-24 06:09:57

RW means that $x^n$ is the hyper-volume of an hypercube of size $x$ in an Euclidean space of dimension $n$. Fermat Last Theorem is the statement that in dimension $n \geq 3$, the sum of the volumes of two hypercubes with side of (positive) integral length is not the volume of such an hypercupe. The same statement fails in dimensions $n = 2$ and $n=1$.

## @Ian Calvert 2014-09-22 21:57:17

A set X is said to be m-convex m>=2 if for every m distinct points in X at least one of the line segments determined by those points belongs to X.

For compact sets and m=3 the decomposition into convex, aka 2-convex ,sets has been known for many years. In the plane F.A.Valentine of UCLA showed in 1957 that every 3-convex set was the union of 3 convex sets , see the 5 pointed star on the USA flag. H.G.Eggleston in 1976 gave an example of a compact 3-convex set in R4 which was not the union of finitely many convex sets. In R3 some years ago I outlined an easy proof in sci.math.research that in R3 a compact 3 convex set was the union of 4 convex sets, using the 4 colour theorem of graph theory.

For information but off topic : Many years ago much was published in the Israel Journal Of Mathematics on planar decomposition, into convex sets , bounds as a function of m for compact m-convex sets. There was also a comprehensive treatment by M. Breen of planar non-closed 3-convex sets in 1977. I know of no associated higher dimension work.

## @Selim G 2014-09-23 12:35:16

Compact manifolds in dimension $2$ and $3$ which carry a metric of negative sectional curvature also carry an hyperbolic structure.

False in dimensions $\geq 4$, where counterexamples have been found by Mostow and Siu, and later by Gromov and Thurston.

## @Sebastien Palcoux 2014-09-16 22:06:37

For all $n > 3$, there is no non-trivial $1$-dim. knot included in $\mathbb{R}^n$.

Edit(sept. 22, 2014)More generally, $\forall n > r+2$, there is no non-trivial

piecewise-linear$r$-dim. knot included in $\mathbb{R}^n$,and $\forall n > (3r+3)/2$, there is no non-trivial

smooth$r$-dim. knot included in $\mathbb{R}^n$ (see this wiki page).## @Andreas Thom 2014-09-22 21:28:41

... which lets you realize that being knotted is a codimension two rather than a one dimensional phenomenon.

## @Sebastien Palcoux 2014-09-22 21:34:33

yes you're right.

## @S. Carnahan 2014-09-23 01:09:02

Once you move to higher dimensions, you are considering connected components in a moduli space of embeddings, and the obstructions are intersection-theoretic. In particular, the codimension of higher-dimensional knots is unbounded - see the wikipedia article.

## @Sebastien Palcoux 2014-09-23 02:48:11

@S.Carnahan: yes you're right. Nevertheless I read that there is no non-trivial $r$-dim. smooth knot in $n$-space for $2n>3r+3$.

## @Paul 2014-09-22 21:14:34

All two-dimensional manifolds are locally conformally flat thanks to the existence of isothermal coordinates. But this is not true in dimension greater or equal to 3. There exists manifolds which are not locally conformally flat in dimension greater or equal to 3. See here.

## @Pablo Lessa 2014-09-16 22:58:04

Reccurence of the random walk on $\mathbb{Z}^2$ implies the following: If two random walkers (say two lovers) are walking on a $2$-dimensional grid then they will eventually meet.

In dimension $3$ this isn't true, however the following property (let's call it the perfume property) is true: Eventually one of the lovers will be at a place that the other one visited before him (and hence be able to smell their perfume). In fact this will happen infinitely many times.

The perfume property is satisfied by $\mathbb{Z}^d$ if and only if $d = 1,2,3,\text{ or }4$. On $\mathbb{Z}^5$ there's a positive probability that the lovers will never even catch each other's scent.

## @Fedor Petrov 2014-09-22 16:59:18

A series of essentially equivalent statements on zonoids which hold for $d=2$ but fail for $d\geq 3$:

any convex symmetric polytope in $\mathbb{R}^d$ is a Minkowski sum of segments;

any convex symmetric body in $\mathbb{R}^d$ is a section of a unit ball in $L^1$-type space (for polytope finite-dimensional hyperoctahedron is enough);

any convex symmetric body in $\mathbb{R}^d$ is a projection of a unit ball in $L^{\infty}$-type space (for polytope finite-dimensional cube is enough);

any Banach norm in $\mathbb{R}^d$ may be expressed as $\|x\|=\int |(x,y)| d\mu(y)$ for some Borel measure $\mu$ on $\mathbb{R}^d$, where $(x,y)$ is a scalar product;

for any norm $\|\cdot\|$ on $\mathbb{R}^d$ and any vectors $v_1,u_1,\dots,v_n,u_n$ we have $\sum \|u_i\|\geq \sum \|v_i\|$ provided that $\sum |(u_i,y)|\geq \sum |(v_i,y)|$ for any vector $y$.

## @Yury 2014-09-15 15:57:06

For every set of $m$ vectors in ${\mathbb R}^d$ such that the angle between every two of them is at most $\pi/2$, there is an orthogonal transformation that maps all vectors to the positive orthant ${\mathbb R}^d_{\geq 0}$.

The statement is true for $d\leq 2$ and false for $d \geq 3$. If we require additionally that $m = d$, the statement is true for $d\leq 4$ and false for $d \geq 5$.

ReferencesGray, L. J. and Wilson, D. G. Nonnegative Factorization of Positive Semidefinite Nonnegative Matrices. Linear Algebra Appl. Appl. 31, 119-127, 1980.

Xu, Changqing. Completely Positive Matrix. From MathWorld—A Wolfram Web Resource, created by Eric W. Weisstein. http://mathworld.wolfram.com/CompletelyPositiveMatrix.html

## @BS. 2014-09-21 12:59:32

Could you provide a reference ? Thanks.

## @Lior Bary-Soroker 2014-09-21 18:48:56

What about the symmetric group $S_n$ being solvable if and only if $n<5$ (it can be stated as a dimension phenomenon, if one really really wants to)

## @Dirk 2014-09-23 13:26:43

I am curios to see this as a dimension phenomenon…

## @W. Cadegan-Schlieper 2015-10-13 06:44:15

Given a semisimple Lie group of dimension $N$, is the corresponding Weyl group solvable? True for $N<24$...

## @Matthias Wendt 2014-09-16 07:55:30

Hilbert's third problem: are two polyhedra (in Euclidean $n$-space) of the same volume scissors congruent? This is true in dimensions $\leq 2$. In dimensions $\geq 3$, an additional invariant of scissors congruence classes of polyhedra is needed: the Dehn invariant (which Dehn used in his solution of Hilbert's third problem). Volume and Dehn invariant fully characterize scissors congruence of polyhedra in dimensions 3 (Sydler) and 4 (Jessen). Unfortunately, it is not known in higher dimensions if these are the only invariants...

As a concrete consequence: in dimension $2$, we can give an elementary definition of area for polygons, based on cutting and pasting to get to rectangular shape. The negative solution to Hilbert's problem implies that such an elementary definition of volume of polyhedra is impossible in higher dimensions.

See also this MO-question for a related phenomenon where the Dehn invariant constrains tilings of euclidean space in dimensions $\geq 3$.

## @Daniel 2014-09-20 22:27:52

The Theorem by Poincare and Benedixson: Given an autonomous differential equation $\dot{x}=f(x)$ on some $U\subset\mathbb{R}^2$ with initial conditions, such that the solution $u(x)$ exists for all $t>0$, then any compact $\omega$-limit set with finitely many critical points is one of the following:

Polemically simplified: 2-dimensional autonomous systems are not too chaotic.

The Theorem does not hold in 3 or more dimensions because the Jordan curve theorem does not hold there.

## @Vidit Nanda 2014-09-15 14:24:03

My favorite example (by some distance) is the problem in discrete geometry often called Borsuk's conjecture. The basic question dates back to the 1930s and could be explained to a child:

is every bounded subset of $\mathbb{R}^d$ decomposable into $d+1$ subsets of strictly smaller diameter?You could draw figures all day and convince yourself that the answer is yes in dimensions $2$ and maybe even $3$. But then you would read the following paper (a good candidate for maximizing the ratio of importance to length):

As far as I can recall, the best lower bound on $d$ for which the conjecture is known to be false right now is $64$ or $65$.

Update (9/19/14): The current bound is $64$, if a preprint from January of this year is to be believed:

The $65$-dimensional example which they use as a starting point is due to A Bondarenko.

## @JRaccoon 2014-09-19 08:17:59

I'm glad you were not my math teacher when I was a child. I would have understood nothing you said.

## @Vidit Nanda 2014-09-19 14:28:26

@JRaccoon I said that the conjecture

couldbe explained to a child, not that I was explaining it in that fashion on a site for research-level mathematics. If it makes you feel any better, I am also glad that I was not your math teacher when you were a child.## @Oscar Cunningham 2014-09-17 13:01:33

The sausage conjecture:

Which way of arranging $M$ unit balls in $\mathbb{R}^n$ minimises the content of their convex hull? For $M$ small the answer is always to arrange them along a line, so that their convex hull is a "sausage". When $n\geq42$ this continues to be true for arbitarily large $M$.

But when $n\leq4$ it eventually becomes better to put the balls in a big round "meatball".

The cases $n=5,\dots,41$ are (as far as I know) open. The conjecture is that in these cases the sausage remains optimal.

## @Franki 2014-09-16 06:00:51

From graph theory:

Embed a fully-connected graph without multiple edges into a 2-dimensional euclidean space (the plane), some edges will intersect each-other if the number of vertices is at least a small constant (6), however, the edges will not intersect each-other for any number of vertices if the same graph is embedded into an euclidean space of dimension $n \geq 3$.

There is a weaker version of the above statement: Embed an arbitrary graph without multiple edges with minimum degree of 4 into a 2-dimensional euclidean space, some edges will intersect each-other if the number of vertices is at least a small constant (5), however, the edges will not intersect each-other if the same graph is embedded into an euclidean space of dimension $n \geq 3$.

There are similar results for spherical, cylindrical, toroidal and other topological spaces.

## @orbits 2014-09-15 14:19:51

True only in dimension 2: the 2-sphere with any smooth Riemannian metric with positive sectional curvature can be isometrically realized as convex hypersurface in the Euclidean space $\mathbb{R}^3$. For higher dimensional spheres analogous statement is not true.

## @Christian Remling 2014-09-14 18:59:42

The Schrödinger operator $$ -\Delta + gV(x) $$ in $L^2(\mathbb R^d)$ with $V\le 0$, $V\not\equiv 0$ (and, let's say, $V\in C_0^{\infty}$) has a negative eigenvalue for any $g>0$: true in $d=1, 2$, false for $d\ge 3$. (In $d\leq 2$, consider the scaling $\phi_\lambda(x) = \phi(x/\lambda)$ with $\lambda \nearrow \infty$. In $d \geq 3$, consider, for example, Hardy's inequality.)

In more physical terms: in $d=1, 2$, any attractive force, no matter how weak, can bind a particle; this is not true in $d\ge 3$.

## @Willie Wong 2014-09-15 13:52:41

Do you have a handy reference for this? Thanks in advance.

## @Tobias Fritz 2014-09-15 16:08:02

Is this related to the accepted answer on Brownian motion?

## @Christian Remling 2014-09-15 17:40:01

@TobiasFritz: I'd say not very directly, although you could say it's properties of the Laplacian that matter in both cases.

## @Christian Remling 2014-09-15 17:42:18

@WillieWong: For $d=1,2$ use quadratic forms: we need to find a test function $\varphi$ with $\int(|\nabla \varphi|^2+gV|\varphi|^2)<0$. This boils down to checking that the cost of the Laplacian (the first term) can be kept arbitrarily small, relative to $\|\varphi\|_2$.

## @Christian Remling 2014-09-15 17:47:09

@WillieWong: $d\ge 3$ follows from general eigenvalue bounds, see for example here for a survey (eq. (10) of the paper, for instance): math.uiuc.edu/~dirk/preprints/simonfest8.pdf

## @Willie Wong 2014-09-16 07:44:42

@ChristianRemling: I hope you don't mind me editing in the gist of the proofs into your answer.

## @Christian Remling 2014-09-16 20:24:31

@WillieWong: Sure, that's fine, thanks. (Actually, $d=2$ is slightly trickier: $\nabla \varphi_{\lambda}\sim 1/\lambda$ on roughly a ball of area $\lambda^2$, so $\int |\nabla \varphi_{\lambda}|^2 \sim 1$; one needs appropriate non-linear decay over a large region.)

## @Gerry Myerson 2014-09-16 03:51:52

Here's one where the only exceptional dimension is 11.

Let $A$ be an $n\times n$ matrix with non-negative entries.

The smallest positive integer $r$ such that $A^r$ has only positive entries (if there is such an $r$) is denoted $\gamma(A)$.

Lewin and Vitek conjectured that for every $n$ and every $r<1+(n^2-2n+2)/2$ there is an $n\times n$ matrix $A$ with $\gamma(A)=r$.

Zhang proved that $n=11$ is the only exception to this conjecture.

More history, and complete bibliographic references, at https://math.stackexchange.com/questions/450090/if-p-is-a-regular-transition-probability-matrix-then-pn2-has-no-zero-ele

## @Delio Mugnolo 2014-09-15 22:02:54

$\mathbb R^n$ endowed with the $1$-norm has the binary intersection property if and only if $n=1$ or $n=2$, cf here for details.

## @J.J. Green 2014-09-14 22:09:19

Von Neumann's inequality, that $\|p(A)\| \leq \|p\|_\infty$ for any polynomial $p$ and Hilbert-space contraction $A$ ($\|p\|_\infty$ being the maximum modulus of $p$ on the unit disk) holds in dimensions one (von Neumann) and two (Ando), but not for three and higher (Varopoulos).

## @Alexander Shamov 2014-09-15 10:59:28

Contraction in Hilbert space, I suppose?

## @J.J. Green 2014-09-15 17:52:09

@AlexanderShamov, indeed, answer updated, thank you.

## @Ramiro de la Vega 2014-09-15 16:52:50

Due to Kuratowski:

This is true for $n \leq m$ and false for $n>m$ where $|\mathbb{R}|=\aleph_{m-1}$.

## @Daniele Zuddas 2014-09-16 14:52:58

Can you explain $|\Bbb R|$? This statement is not clear to me!

## @Ramiro de la Vega 2014-09-16 15:32:41

@DanieleZuddas: Here $|\mathbb{R}|$ stands for the cardinality of the set of real numbers. So the specific dimension where the statement becomes false depends on your axioms of set theory. By the way, if $|\mathbb{R}|>\aleph_\omega$ then the statement is true in all finite dimensions.

## @Daniele Zuddas 2014-09-17 02:31:38

So, in ZFC this is true only of $n \leq 2$, isn't?

## @Ramiro de la Vega 2014-09-17 02:36:48

@DanieleZuddas: What does that mean? I would say $ZFC$ proves the statement for $n \leq 2$ and $ZFC$ doesn't prove either the statement or its negation for any $n \geq 3$.

## @Daniele Zuddas 2014-09-17 03:21:08

I mean, I'm not an expert of set theory, so for me the only possibility is $|\Bbb R| = \aleph_1$... What kind of axioms you use for having, for example, $|\Bbb R| = \aleph_2 = 2^{\aleph_1}$?

## @Ramiro de la Vega 2014-09-17 12:56:51

@DanieleZuddas: For each natural number $n\geq1$, $|\mathbb{R}|=\aleph_n$ is an axiom consistent with $ZFC$; the same is true for $|\mathbb{R}|>\aleph_\omega$, for $|\mathbb{R}|=\aleph_2=2^{\aleph_1}$ and for $|\mathbb{R}|=\aleph_2<2^{\aleph_1}$.

## @Ramiro de la Vega 2014-09-17 13:05:37

Let us continue this discussion in chat.

## @Terry Tao 2014-09-15 16:22:58

Claim: Every smooth $n$-dimensional manifold homeomorphic to a sphere is also diffeomorphic to a sphere. In other words, there are no exotic spheres of dimension $n$.

True for $n=1,2,3,5,6,12,61$. Open for $n=4$. False for all other $n < 126$, and for all odd $n \geq 126$ (according to forthcoming work of Behren, Hill, Hopkins, and Mahowald, plus earlier results; there are some additional results for large even $n$ also, but I don't know the precise statements). The $n=7$ case was a famous counterexample of Milnor. The problem is closely connected to that of determining the (stable) homotopy groups of spheres, see e.g. the previous MathOverflow question Exotic spheres and stable homotopy in all large dimensions? .

## @Dag Oskar Madsen 2014-09-15 15:24:20

The three altitudes of a triangle intersect in a common point, the

orthocenterof the triangle.This does not generalize to tetrahedra and higher-dimensional simplices. The altitudes of a general $d$-simplex, $d \geq 3$, are not necessarily concurrent.

Source:

Hajja, Mowaffaq; Martini, Horst. Orthocentric simplices as the true generalizations of triangles. Math. Intelligencer 35 (2013), no. 2, 16--28