# Results true in a dimension and false for higher dimensions

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?

• Related: mathoverflow.net/questions/5372 – Emil Jeřábek supports Monica Sep 14 '14 at 19:12
• This question could get more interesting if extended to fractals... – Nate Eldredge Sep 15 '14 at 4:41
• Every cubic polynomial map from $\mathbb C^n$ to $\mathbb C^n$ with nowhere vanishing Jacobian is a bijection. This is true for $n \leq 17$ but fails for larger $n$. Just kidding... but perhaps someday I'll edit the answer, change 17 appropriately and will gain some upvotes. – Marty Sep 15 '14 at 12:55
• Not sure it counts as "dimension", but for the sake of completeness: 1-generated groups vs 2-generated groups... – TT_ Sep 16 '14 at 22:44
• "A positive polynomial not reaching its minimum." Better to say, not reaching its infimum, or, not having a minimum. – Gerry Myerson Mar 6 '17 at 23:18

An n-dimensional brownian motion visits every neighborhood of $\mathbb{R}^n$ infinitely often with probability 1 iff $n \leq 2$

• That is interesting. Do you have a synthetic reference for this result' – mathcounterexamples.net Sep 14 '14 at 18:13
• The poor man's version: the unbiased random walk is recurrent in $n\le 2$, transient for $n\ge 3$. – Christian Remling Sep 14 '14 at 18:53
• A drunken man can find his way home, but a drunken bird is lost forever. – Margaret Friedland Sep 14 '14 at 20:30
• Similarly: an $n$-dimensional Brownian motion visits every point of $\mathbb{R}^n$ with probability 1 iff $n \le 1$. – Nate Eldredge Sep 15 '14 at 4:40
• @cardinal: It's attributed to Kakutani by Durrett (where I know it from). I do not have any better reference. It seems to have become one of the many recognizable quotes, so much so that a generic reference is not needed. – Margaret Friedland Sep 17 '14 at 0:20

Keller's conjecture asserts that whenever one tiles $${\bf R}^n$$ by unit cubes, there must be two cubes which share a common face. True when $$n \leq 6$$, false for $$n\geq 8$$, and still open for $$n=7$$.

UPDATE, October 2019: It appears that the conjecture has now been resolved in the affirmative by computer assisted proof in $$n=7$$: https://arxiv.org/abs/1910.03740

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? .

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):

Jeff Kahn and Gil Kalai, A counterexample to Borsuk's conjecture, Bulletin of the American Mathematical Society 29 (1993), 60–62.

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:

T Jenrich and AE Brouwer, A 64-dimensional counterexample to Borsuk’s conjecture.

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

• I'm glad you were not my math teacher when I was a child. I would have understood nothing you said. – JRaccoon Sep 19 '14 at 8:17
• @JRaccoon I said that the conjecture could be 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. – Vidit Nanda Sep 19 '14 at 14:28

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.

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 $d\ge 3$.

• Do you have a handy reference for this? Thanks in advance. – Willie Wong Sep 15 '14 at 13:52
• Is this related to the accepted answer on Brownian motion? – Tobias Fritz Sep 15 '14 at 16:08
• @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 Sep 15 '14 at 17:42
• @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 – Christian Remling Sep 15 '14 at 17:47
• @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.) – Christian Remling Sep 16 '14 at 20:24

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.

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

Banach–Tarski paradoxical decomposition of the unit ball.

• To elaborate on this: in dimensions ≥3 there are finite paradoxical decompositions of the unit ball; in dimensions 1 and 2, paradoxical decompositions must be countably infinite; and in dimension 0, there are none. – Peter LeFanu Lumsdaine Sep 15 '14 at 13:15
• @PeterLeFanuLumsdaine: true for Euclidean metric, but there exists a nice finite paradoxical decomposition of a hyperbolic plane. – Michael Sep 15 '14 at 15:55

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.

There is a proper subdomain of $\mathbb{C}^n$ biholomorphically equivalent to $\mathbb{C}^n$ (a Fatou-Bieberbach domain).

False for $n=1$, true for $n \geq 2$.

Cross product version of Hurwitz's 1, 2, 4, 8 theorem

There exists a 'cross product' on $\mathbb{R}^n$ if and only if $n=1, 3, 7$.

A cross product means a bilinear map $*: \mathbb{R}^n\times \mathbb{R}^n\rightarrow \mathbb{R}^n$ with the following usual properties:
(i) The map $(x,y,z)\mapsto(x*y)\cdot z$ is trilinear and alternating.

(ii) $(x*y)*x=(x\cdot x)y-(x\cdot y)x$

Where $\cdot$ denotes the usual dot product on $\mathbb{R}^n$.

For $n=1, 3, 7$ such products are induced by multiplication $x*y:=\frac12(xy-yx)$ where $xy$ is the multiplication of pure elements $x$ and $y$ of complex, quaternion and Cayley algebra over $\mathbb{R}$.

• Massey's MAA paper is a good source for an "elementary" proof for this fact. – BigM Oct 11 at 16:18

Busemann-Petty problem (see http://en.wikipedia.org/wiki/Busemann%E2%80%93Petty_problem) has positive solution only in dimension at most 4. The problem says: assume that in the Euclidean space $\mathbb{R}^n$ one has two convex centrally symmetric bodies $K$ and $L$. Assume that for any linear hyperplane $H$ $$vol_{n-1}(K\cap H)\leq vol_{n-1}(L\cap H).$$ Is it true that $vol(K)\leq vol(L)$?

The answer is positive for arbitrary convex centrally symmetric bodies $K,L\subset \mathbb{R}^n$ as above if and only if $n\leq 4$.

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 $\geq 3$.

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 $d \geq 5$.

References

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

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

• Could you provide a reference ? Thanks. – BS. Sep 21 '14 at 12:59

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

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).

• Contraction in Hilbert space, I suppose? – Alexander Shamov Sep 15 '14 at 10:59
• @AlexanderShamov, indeed, answer updated, thank you. – J.J. Green Sep 15 '14 at 17:52

Due to Kuratowski:

For any partition $\mathbb{R}^n=\bigcup_{i=1}^nA_i$ there exists an $i \in \{1,\dots,n\}$ and a straight line $l$ parallel to the $i$-th axis such that $A_i\cap l$ is infinite.

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

• Can you explain $|\Bbb R|$? This statement is not clear to me! – Daniele Zuddas Sep 16 '14 at 14:52
• @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. – Ramiro de la Vega Sep 16 '14 at 15:32
• So, in ZFC this is true only of $n \leq 2$, isn't? – Daniele Zuddas Sep 17 '14 at 2:31
• @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 $n \geq 3$. – Ramiro de la Vega Sep 17 '14 at 2:36
• @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 Sep 17 '14 at 12:56

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.

True for dimensions $n\le 8$: the existence of left-invariant affine structures on nilpotent Lie groups of dimension $n$. All nilpotent Lie groups of dimension $n\le 8$ admit such a structure. This is probably false for all $n\ge 9$, but this is only proved for $n=10,11,12$ so far. More precisely, there exist nilpotent Lie groups of dimension $10,11,12$ which do not admit any affine structure.

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.

The Schoenflies theorem: if $\Sigma^n \subset \Bbb R^{n+1}$ is homeomorphic to $S^n$, then there is a homeomorphism $h \colon \Bbb R^{n+1} \to \Bbb R^{n+1}$ such that $h(\Sigma^n) = S^n$ (the round $n$-sphere). Trivial for $n = 0$, true (but hard) for $n = 1$, false for $n \geq 2$ (counterexample: the Alexander horned sphere).

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

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:

• A Critical Point
• A periodic orbit
• a connected set composed of a finite number of fixed points together with homoclinic and heteroclinic orbits connecting these.

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.

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).

• ... which lets you realize that being knotted is a codimension two rather than a one dimensional phenomenon. – Andreas Thom Sep 22 '14 at 21:28
• yes you're right. – Sebastien Palcoux Sep 22 '14 at 21:34
• 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. – S. Carnahan Sep 23 '14 at 1:09
• @S.Carnahan: yes you're right. Nevertheless I read that there is no non-trivial $r$-dim. smooth knot in $n$-space for $2n>3r+3$. – Sebastien Palcoux Sep 23 '14 at 2:48

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.

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

This is only true in dimension two: Let $f$ be a harmonic function on a Riemannian manifold $(M,g)$. If the metric $g$ is changed conformally, the function $f$ is still harmonic. (The analogous result is true in any dimension $n=\dim M$ if one considers $n$-harmonic functions instead of harmonic ones, but the only case corresponding to the linear 2-Laplace equation is two dimensional.)

• This is slightly less satisfying as an answer (to me) because the result is also not true in dimension one, and one < two. – Willie Wong Sep 15 '14 at 14:27
• @WillieWong, you are right. If one considers the interesting dimensions to start at two (as I did when writing my answer), then this is a result that is true for low and false for high dimensions. But of course the situation here is somewhat nontrivial for one dimensional manifolds, so I understand your dissatisfaction. – Joonas Ilmavirta Sep 15 '14 at 14:37

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)

• I am curios to see this as a dimension phenomenon… – Dirk Sep 23 '14 at 13:26
• Given a semisimple Lie group of dimension $N$, is the corresponding Weyl group solvable? True for $N<24$... – W. Cadegan-Schlieper Oct 13 '15 at 6:44

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.