Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Many mathematical areas have a notion of "dimension", either rigorously or naively, and different dimensions can exhibit wildly different behaviour. Often, the behaviour is similar for "nearby" dimensions, with occasional "dimension leaps" marking the boundary from one type of behaviour to another. Sometimes there is just one dimension that has is markedly different from others. Examples of this behaviour can be good provokers of the "That's so weird, why does that happen?" reaction that can get people hooked on mathematics. I want to know examples of this behaviour.

My instinct would be that as "dimension" increases, there's more room for strange behaviour so I'm more surprised when the opposite happens. But I don't want to limit answers so jumps where things get remarkably more different at a certain point are also perfectly valid.

share|improve this question
1  
There was a relevant discussion at God Plays Dice awhile ago: godplaysdice.blogspot.com/2009/07/… –  Qiaochu Yuan Nov 13 '09 at 15:47
11  
Who reads blogs anymore? They are soooo October-2009ish. –  Loop Space Nov 13 '09 at 15:51
2  
Andrew, I don't see a question in there... –  Ben Webster Nov 13 '09 at 16:05
1  
Time to put this one to bed. (ie, time to close it, I deem.) –  Loop Space Jun 23 '10 at 18:07
1  
@Gil, I've arbitrarily reopened this question. –  Scott Morrison Jun 24 '10 at 15:40
show 4 more comments

33 Answers 33

Here's a fun little example that I thought was neat... it's quite simple but tends to go against most people's geometric instinct.

We consider the cube $[-2,2]^d$ in $\mathbb{R}^d$. At the points with all coordinates equal to 1 or -1 (e.g. in $\mathbb{R}^3$, points like (1,1,1), (1,-1,-1), etc) we put unit balls. We define the "central ball" $B_d$ to be the largest ball centered at the origin that does not intersect the interior of any of the other balls we have placed. You can easily visualize this in the case $d=2$, just think of the square $[-2,2]^2$, draw 4 unit discs, one centered in each quadrant, and then $B_d$ is the little disc in the center that is big enough to just hit the boundary of these 4 balls. The question is, what is the asymptotic relationship (as d goes to infinity) between the volume of $B_d$ and the volume of $[-2,2]^d$?

The answer is that $m(B_d)/m([-2,2]^d)$ goes to infinity! Most people will try to visualize this problem in $\mathbb{R}^2$ or $\mathbb{R}^3$ to get an intuition for the behavior, and just implicitly assume that $B_d$ is contained within $[-2,2]^d$. And it certainly is in those low dimensional cases. But when you actually compute the radius of $B_d$, you see that it's $\sqrt{d}-1$, and so $B_d$ is not even contained in in $[-2,2]^d$ for $d > 9$.

share|improve this answer
9  
The fraction exceeds 1 at d=1206. The asymptotic behavior can be explained by Stirling's formula together with the fact that e times pi is larger than 8. –  S. Carnahan Dec 22 '09 at 6:15
2  
I loved this so much I'm going to give it tomorrow as a riddle to my students, along with the deceptive handout at katlas.math.toronto.edu/drorbn/AcademicPensieve/2010-11/nb/…. They no fools, they'll find this page right away, yet it's fun. Thanks! –  Dror Bar-Natan Nov 30 '10 at 10:08
add comment

The symmetric group has an outer automorphism only in degree 6.

share|improve this answer
add comment

My favourite of these is that there is precisely one differentiable structure on $\mathbb{R}^n$ up to diffeomorphism for all $n$, except when $n=4$, when there are uncountably many.

share|improve this answer
2  
Any opinions on why this one happens? Aside from more or less restating the proofs. I bet there's a variety of opinions on this one. –  Ryan Budney Nov 13 '09 at 17:52
5  
@Ryan: Has that rainy day ever come? Anyway, closely related to this: there are topological 4-manifolds that admit no smooth structures (as opposed to smaller dimensions) and some that admit infinitely many (in sharp contrast to what happens in every other dimension). –  Marco Golla Apr 18 '11 at 20:09
show 4 more comments

My favorite example is regular polytopes. The number of regular polytopes is almost monotone decreasing, from countably many in $\mathbb{R}^2$, to five in $\mathbb{R}^3$ to 3 for $\mathbb{R}^n$ for $n>4$. But in $n=4$, we get six, which is kind of weird.

share|improve this answer
4  
According to Coxeter, Schlafli really should get the credit for this result. Though most of his work went unnoticed in his lifetime, he did solve the problem completely, and before anyone else. There doesn't seem to be any question of his answer not being considered a "proof." (Though I don't read German, so I can't say for sure). What is true, though, is that there's been a lot of debate over the meaning of "polytope" in the time since then (Schlafli apparently uses "finite region bounded by a finite number of hyperplanes.") –  Emily Peters Nov 13 '09 at 22:09
show 5 more comments

The Euclidean ball takes up the most space in dimension 5.

$V = \frac {8 \pi^2} {15} R^5 \approx 5.26\ldots R^5$

share|improve this answer
8  
This is a bit silly, I think, because the "space" that's being taken up by, say, the 4-ball and the 5-ball is not really the same kind of space. –  Michael Lugo Nov 13 '09 at 18:24
22  
That's not silly. –  Richard Kent Nov 14 '09 at 3:21
add comment

The Leech lattice.

At least, in the sense that 24 is one of the only dimensions where we know what the densest lattice packing looks like. As usual, John Baez's thoughts. Conway and Sloane is a good reference.

share|improve this answer
add comment

The only spheres that are topological groups are S^0, S^1, S^3. In fact these are Lie groups.

If we widen to h-groups, then we also have S^7.

Nothing for higher dimensional spheres.

share|improve this answer
2  
The key fact here I guess is that these spheres sit inside spaces of the right dimension so they get multiplicative structure from the multiplication in the reals, complex numbers, quaternions, and octonians. To an algebraic topologist, your answer has to be the best for relating to this amazing fact, the Hopf Invariant One Theorem, hence my +1 –  David White May 5 '11 at 20:41
add comment

Closed hyperbolic surfaces have deformations through hyperbolic structures [Riemann], but closed hyperbolic manifolds in higher dimensions don't [Mostow].

Finite volume hyperbolic manifolds (usually) have deformations through complete structures in dimension two, but not in higher dimensions [Weil, Prasad]. In dimension three, they have deformations through incomplete structures [Thurston], and in dimensions four and up you don't even have that [Wang].

So hyperbolic manifolds "harden" as the dimension grows.

share|improve this answer
show 2 more comments

All manifolds in dimension $n\leq 3$ are triangulable. Conjecturally, all manifolds in dimension $n\geq 5$ can be triangulated by a simplicial complex which is not necessarily a combinatorial manifold. But "few" 4-manifolds are triangulable.
I don't think that this has anything to do with the fact that $R^4$ admits infinitely many PL structures. So perhaps dimension 4 is weird in topology for (at least) two completely different reasons.

share|improve this answer
add comment

Polya's theorem on random walks: a simple random walk on $\mathbb{Z}^n$ is recurrent for $n = 1$ and $n = 2$, but is transient for $n \geq 3$.

share|improve this answer
1  
Looking at random walks on fractals one can even see that this transition happens right at "spectral dimension" = Hausdorff dimension over walk dimension equals 2. –  BSteinhurst Apr 18 '11 at 20:33
add comment

Quantum physics is a good source of these kinds of phenomena. Classical physics often allows us to formulate a theory uniformly in any dimension. But when we quantise systems, suddenly special dimensions pop out. Quantising often involves some kind of infinite limiting process and in the limit we end up losing a symmetry that was there in the original classical system. These are called anomalies. But in special dimensions we can arrange for these anomalies to cancel. For example the simplest string theory, bosonic string theory, only works in 26 dimensions (= 2+the 24 of the Leech Lattice mentioned in another answer, no coincidence BTW).

In each case there's an interesting mathematical story to be told. For example the dimensions in which superstring theory can be made to work are related to the dimensions picked out by the division algebras: 1, 2, 4, 8.

share|improve this answer
add comment

I'm not sure if you want an example or commentary, so I'll give both: the example Michael Lugo gave above is that the Poincare conjecture was hardest to prove in three dimensions. My commentary as far as this being a general phenomenon is that in low dimensions one expects "local" obstructions to strange behavior whereas in high dimensions one expects "global" obstructions to strange behavior.

share|improve this answer
6  
Put another way, we expect topology to be governed by algebra in dimension $n\geq 5$, but not in dimensions 3 and 4, because of the Whitney embedding theorem. –  Daniel Moskovich Dec 31 '09 at 11:36
show 1 more comment

The wave equation behaves differently in even and odd space dimensions. In odd-dimensional space, radial waves satisfy a modified version of the one-dimensional wave equation. In particular, Huygens' principle holds. This is not so in even-dimensional space. This difference is reflected in the usual existence proof for solutions of the wave equation, which is easier in odd-dimensional space. Then one handles the wave equation in even-dimensional space by adding a dimension.

share|improve this answer
5  
There's a lot of dimension parity issues like 1) when a sphere has a non-zero vector field, or 2) when the n-sphere can be turned inside-out in euclidean (n+1)-space, etc. –  Ryan Budney Nov 13 '09 at 18:08
show 1 more comment

Phase changes in matter are sort of an example of this, if one replaces "dimension" by "energy." I don't know much about statistical mechanics, but as an enthusiastic amateur, the fact that (for instance) the 2-D Ising model undergoes a phase change still blows my mind.

share|improve this answer
show 2 more comments

A function $\mu: {\mathbb P}({\mathbb R}^n)\to [0,\infty]$ which is translation, rotation and reflection invariant and such that $\mu([0,1)^n)=1$ can only be finitely additive if $n\leq 2$.

share|improve this answer
add comment

The homotopy groups of an arbitrary topological space are abelian if $n\geq 2$, but the fundamental group may not be so.

share|improve this answer
add comment

Complex analysis in dimension $n>1$ is, for non obvious reasons, very different from complex analysis in dimension $1$. Think for example of Hartogs' extension theorem...

share|improve this answer
add comment

The Smith-Minkowski-Siegel mass formula implies that the number of unimodular lattices of given dimension eventually starts to increase more than exponentially fast, so one might expect that they are easy to classify in small dimensions and gradually become harder to classify in higher dimensions as the mass of the SMS formula increases. In fact this is not what happens: there is a quite precise dimension where the behavior changes qualitatively and the lattices become much harder to classify. This is the jump from dimension 25 to 26. The reason is related to the existence of the Leech lattice in dimension 24, which controls unimodular lattices in dimension up to 25. (The 25 dimensional ones were classified by hand about 30 years ago, but the 26 dimensional case is so much harder that no-one has attempted it since then even with the help of modern petaflop computers.)

share|improve this answer
show 2 more comments

The sphere $S^n$ has a set of tangent fields which are linearly indepedent at every point if and only if $n$ is 0,1,3, or 7.

share|improve this answer
2  
I feel I have to comment that this is equivalent to Adams' Hopf Invariant One theorem and relates to the fact that spheres in these dimensions sit in the reals, complex numbers, quaternions, and octonians. Hope you get the upvotes you deserve, this is a great answer –  David White May 5 '11 at 20:43
add comment

The 6-sphere is very special. It is the only sphere, other then the 2-sphere, that admits an almost complex structure. But it is yet unkown if it admits a complex structure.

share|improve this answer
add comment

Jones' index, for subfactors, is not quite an answer to this question. The range of possible values of indices of subfactors has both a discrete part (indices less than 4 must be of the form $4 \cos^2(\frac{\pi}{n})$ for $n \geq 3$) and a continuous part (any number $\geq 4$ is attainable).

The reason this is relevant is that the index measures the dimension of the subfactor inside the larger factor -- so the phenomenon which is observed to "jump" at dimension 4, is exactly the possible dimensions!

share|improve this answer
add comment

The free modular lattice on $n$ generators is finite for $n=1,2,3$, but for $n=4$ not only is it infinite but its word problem is recursively unsolvable.

share|improve this answer
add comment

I don't know if it can count as an answer, as the "dimension" involved here doesn't range through a discrete set of values, but:

For any subset $A$ of a given metric space, there is a specific dimension $\alpha$ (the Hausdorff dimension of $A$) for which the $\beta$-dimensional Hausdorff measure of $A$ is $\mathcal{H}^{\beta}(A)=+\infty$ for $\beta < \alpha$ and it is $\mathcal{H}^{\beta}(A)=0$ for $\beta > \alpha$.

share|improve this answer
show 1 more comment

An $n=7$ cut-off appears in the theory of minimal surfaces.

Bernstein's problem on golabl minimal surfaces: A global solution to the minimal surface equation on $\mathbb{R}^n$ is necessarily an affine function for $n \leq 7$, but there are counterexamples in all greater dimensions.

Also, an $n$ dimensional minimal surface in $\mathbb{R}^{n+1}$ is regular outside a singular set whose dimension is at most $n - 7$. The Simons cone, given by the set of points $x \in \mathbb{R}^8$ such that

$x_1^2 + x_2^2 + x_3^2 + x_4^2 = x_5^2 + x_6^2 + x_7^2 + x_8^2$,

is minimal and therefore shows that this is optimal - because it has an isolated singularity at the origin.

So, curiously, the set of points $x \in \mathbb{R}^6$ such that $x_1^2 + x_2^2 + x_3^2 = x_4^2 + x_5^2 + x_6^2$, just isn't minimal.

share|improve this answer
add comment

Projective spaces provide an example (counter to the usual trend) where the jump from dimension 2 to 3 actually brings greater simplicity. In any projective space of dimension 3 the Desargues theorem holds, which implies that space can be coordinatized by a skew field.

In dimension 2 (projective planes) the Desargues theorem need not hold. As a result, projective planes cannot be founded on any familiar algebraic structure and they are very hard to classify.

share|improve this answer
add comment

For the sake of this answer, "dimension" should be interpreted as "number of variables."

In quantum logic, four is the smallest $n$ such that a classically unsatisfiable propositional formula in $n$ variables can be satisfied by substituting quantum propositions in a meaningful way. One example of such a proposition is $$((a\oplus b)\oplus(c\oplus d))\oplus((a\oplus c)\oplus(b\oplus d)),$$ where $\oplus$ is exclusive-or. Note that the grouping of expressions here is crucial: in quantum logic, two propositions can only be meaningfully combined by a logical connective if the corresponding projection operators (or equivalently, the "spin" operators) commute. (For example, the proposition "I have position X and momentum Y" is not meaningful.) One "satisfying assignment" for the formula above is given by (using the spin operator convention) $a=\sigma\_x\otimes 1, b=1\otimes\sigma\_x, c=\sigma\_z\otimes 1, d=1\otimes\sigma\_z$.

(Basically, boolean algebras are to classical logic as partial boolean algebras are to quantum logic, and every 3-generator partial boolean algebra can be embedded in a boolean algebra, so there are no such formulas with three or fewer variables.)

share|improve this answer
add comment

The spaces of sequences of real or complex numbers, $(l^p,||·||_p)$, are not pre-Hilbert spaces unless $p=2$.

share|improve this answer
2  
Is this really 'dimension'? –  Spencer Apr 18 '11 at 20:22
show 1 more comment

Fermat's Last Theorem: the equation $x^n+y^n=z^n$ has only nontrivial (integer) solutions if $n\leq 2$.

share|improve this answer
add comment

Every weak 0- 1- and 2-category is equivalent to strict one (the cases 0 and 1 being silly), this is not any more true for weak 3-categories upwards.

share|improve this answer
add comment

Here is a closely related pair of examples from operator theory, von Neumann's inequality and the theory of unitary dilations of contractions on Hilbert space, where things work for 1 or 2 variables but not for 3 or more.

In one variable, von Neumann's inequality says that if $T$ is an operator on a (complex) Hilbert space $H$ with $\|T\|\leq1$ and $p$ is in $\mathbb{C}[z]$, then $\|p(T)\|\leq\sup\{|p(z)|:|z|=1\}$. Szőkefalvi-Nagy's dilation theorem says that (with the same assumptions on $T$) there is a unitary operator $U$ on a Hilbert space $K$ containing $H$ such that if $P:K\to H$ denotes orthogonal projection of $K$ onto $H$, then $T^n=PU^n|_H$ for each positive integer $n$.

These results extend to two commuting variables, as Ando proved in 1963. If $T_1$ and $T_2$ are commuting contractions on $H$, Ando's theorem says that there are commuting unitary operators $U_1$ and $U_2$ on a Hilbert space $K$ containing $H$ such that if $P:K\to H$ denotes orthogonal projection of $K$ onto $H$, then $T_1^{n_1}T_2^{n_2}=PU_1^{n_1}U_2^{n_2}|_H$ for each pair of nonnegative integers $n_1$ and $n_2$. This extension of Sz.-Nagy's theorem has the extension of von Neumann's inequality as a corollary: If $T_1$ and $T_2$ are commuting contractions on a Hilbert space and $p$ is in $\mathbb{C}[z_1,z_2]$, then $\|p(T_1,T_2)\|\leq\sup\{|p(z_1,z_2)|:|z_1|=|z_2|=1\}$.

Things aren't so nice in 3 (or more) variables. Parrott showed in 1970 that 3 or more commuting contractions need not have commuting unitary dilations. Even worse, the analogues of von Neumann's inequality don't hold for $n$-tuples of commuting contractions when $n\geq3$. Some have considered the problem of quantifying how badly the inequalities can fail. Let $K_n$ denote the infimum of the set of those positive constants $K$ such that if $T_1,\ldots,T_n$ are commuting contractions and $p$ is in $\mathbb{C}[z_1,\ldots,z_n]$, then $\|p(T_1,\ldots,T_n)\|\leq K\cdot\sup\{|p(z_1,\ldots,z_n)|:|z_1|=\cdots=|z_n|=1\}$. So von Neumann's inequality says that $K_1=1$, and Ando's Theorem yields $K_2=1$. It is known in general that $K_n\geq\frac{\sqrt{n}}{11}$. When $n>2$, it is not known whether $K_n\lt\infty$.

See Paulsen's book (2002) for more. On page 69 he writes:

The fact that von Neumann’s inequality holds for two commuting contractions but not three or more is still the source of many surprising results and intriguing questions. Many deep results about analytic functions come from this dichotomy. For example, Agler [used] Ando’s theorem to deduce an analogue of the classical Nevanlinna–Pick interpolation formula for analytic functions on the bidisk. Because of the failure of a von Neumann inequality for three or more commuting contractions, the analogous formula for the tridisk is known to be false, and the problem of finding the correct analogue of the Nevanlinna–Pick formula for polydisks in three or more variables remains open.

share|improve this answer
show 1 more comment

Your Answer

 
discard

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.