Dimension Leaps - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T08:52:36Z http://mathoverflow.net/feeds/question/5372 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/5372/dimension-leaps Dimension Leaps Andrew Stacey 2009-11-13T15:38:25Z 2011-08-29T17:52:35Z <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/5372/dimension-leaps/5376#5376 Answer by Alex Collins for Dimension Leaps Alex Collins 2009-11-13T15:45:17Z 2009-11-13T15:45:17Z <p>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.</p> http://mathoverflow.net/questions/5372/dimension-leaps/5379#5379 Answer by Qiaochu Yuan for Dimension Leaps Qiaochu Yuan 2009-11-13T15:52:10Z 2009-11-13T15:52:10Z <p>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.</p> http://mathoverflow.net/questions/5372/dimension-leaps/5383#5383 Answer by Charles Siegel for Dimension Leaps Charles Siegel 2009-11-13T16:13:13Z 2009-11-13T16:13:13Z <p>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.</p> http://mathoverflow.net/questions/5372/dimension-leaps/5396#5396 Answer by Harald Hanche-Olsen for Dimension Leaps Harald Hanche-Olsen 2009-11-13T17:09:59Z 2009-11-13T17:09:59Z <p>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.</p> http://mathoverflow.net/questions/5372/dimension-leaps/5402#5402 Answer by Sammy Black for Dimension Leaps Sammy Black 2009-11-13T17:58:41Z 2009-11-13T17:58:41Z <p>The Euclidean ball takes up the most space in dimension 5.</p> <p>$V = \frac {8 \pi^2} {15} R^5 \approx 5.26\ldots R^5$</p> http://mathoverflow.net/questions/5372/dimension-leaps/5405#5405 Answer by Qiaochu Yuan for Dimension Leaps Qiaochu Yuan 2009-11-13T18:15:14Z 2009-11-13T18:15:14Z <p>The <a href="http://en.wikipedia.org/wiki/Leech_lattice" rel="nofollow">Leech lattice</a>. </p> <p>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, <a href="http://math.ucr.edu/home/baez/numbers/" rel="nofollow">John Baez's thoughts</a>. <a href="http://books.google.com/books?id=upYwZ6cQumoC&amp;pg=PA406&amp;lpg=PA406&amp;dq=conway+lattices&amp;source=bl&amp;ots=_K0XYTh6EW&amp;sig=ryoyCU-SR_lZuqyaEatmBWZxBYA&amp;hl=en&amp;ei=FqL9Sr2OLoa6ngeQssifCw&amp;sa=X&amp;oi=book_result&amp;ct=result&amp;resnum=2&amp;ved=0CA8Q6AEwAQ#v=onepage&amp;q=&amp;f=false" rel="nofollow">Conway and Sloane</a> is a good reference.</p> http://mathoverflow.net/questions/5372/dimension-leaps/5440#5440 Answer by Dan Piponi for Dimension Leaps Dan Piponi 2009-11-13T21:36:59Z 2009-11-13T23:12:52Z <p>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 <a href="http://en.wikipedia.org/wiki/Anomaly_(physics)" rel="nofollow">anomalies</a>. 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).</p> <p>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 <a href="http://en.wikipedia.org/wiki/Division_algebra" rel="nofollow">division algebras</a>: 1, 2, 4, 8.</p> http://mathoverflow.net/questions/5372/dimension-leaps/5444#5444 Answer by Emily Peters for Dimension Leaps Emily Peters 2009-11-13T21:52:42Z 2009-11-13T21:52:42Z <p>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.</p> http://mathoverflow.net/questions/5372/dimension-leaps/5447#5447 Answer by Emily Peters for Dimension Leaps Emily Peters 2009-11-13T22:02:55Z 2009-11-13T22:02:55Z <p>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). </p> <p>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!</p> http://mathoverflow.net/questions/5372/dimension-leaps/5458#5458 Answer by James for Dimension Leaps James 2009-11-13T23:24:07Z 2009-11-13T23:24:07Z <p>The symmetric group has an outer automorphism only in degree 6.</p> http://mathoverflow.net/questions/5372/dimension-leaps/5475#5475 Answer by Richard Kent for Dimension Leaps Richard Kent 2009-11-14T03:28:50Z 2009-11-14T03:28:50Z <p>Closed hyperbolic surfaces have deformations through hyperbolic structures [Riemann], but closed hyperbolic manifolds in higher dimensions don't [Mostow].</p> <p>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].</p> <p>So hyperbolic manifolds "harden" as the dimension grows.</p> http://mathoverflow.net/questions/5372/dimension-leaps/5483#5483 Answer by Erik Davis for Dimension Leaps Erik Davis 2009-11-14T07:17:41Z 2010-11-30T01:29:34Z <p>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.</p> <p>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 $[-2,2]^d$?</p> <p>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 $d > 9$. </p> http://mathoverflow.net/questions/5372/dimension-leaps/5494#5494 Answer by Darsh Ranjan for Dimension Leaps Darsh Ranjan 2009-11-14T10:52:42Z 2009-11-14T10:52:42Z <p>For the sake of this answer, "dimension" should be interpreted as "number of variables."</p> <p>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) $a=\sigma_x\otimes 1, b=1\otimes\sigma_x, c=\sigma_z\otimes 1, d=1\otimes\sigma_z$. </p> <p>(Basically, boolean algebras are to classical logic as <i>partial boolean algebras</i> 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.)</p> http://mathoverflow.net/questions/5372/dimension-leaps/5506#5506 Answer by Jose Brox for Dimension Leaps Jose Brox 2009-11-14T14:22:10Z 2009-11-14T15:36:11Z <p>Fermat's Last Theorem: the equation $x^n+y^n=z^n$ has only nontrivial (integer) solutions if $n\leq 2$.</p> http://mathoverflow.net/questions/5372/dimension-leaps/5507#5507 Answer by Jose Brox for Dimension Leaps Jose Brox 2009-11-14T14:28:18Z 2009-11-14T14:28:18Z <p>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 $n\leq 2$.</p> http://mathoverflow.net/questions/5372/dimension-leaps/5508#5508 Answer by Jose Brox for Dimension Leaps Jose Brox 2009-11-14T14:30:44Z 2009-11-14T14:30:44Z <p>The spaces of sequences of real or complex numbers, $(l^p,||·||_p)$, are not pre-Hilbert spaces unless $p=2$.</p> http://mathoverflow.net/questions/5372/dimension-leaps/5509#5509 Answer by Jose Brox for Dimension Leaps Jose Brox 2009-11-14T14:31:53Z 2009-11-14T14:31:53Z <p>The homotopy groups of an arbitrary topological space are abelian if $n\geq 2$, but the fundamental group may not be so.</p> http://mathoverflow.net/questions/5372/dimension-leaps/5523#5523 Answer by Peter Arndt for Dimension Leaps Peter Arndt 2009-11-14T16:03:33Z 2009-11-14T16:03:33Z <p>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.</p> http://mathoverflow.net/questions/5372/dimension-leaps/9288#9288 Answer by psihodelia for Dimension Leaps psihodelia 2009-12-18T16:57:33Z 2009-12-18T16:57:33Z <p>The max number of points interconnected (every-to-every) by lines of any curvature, such that no line crosses any other line. For $\mathbb{R}^2$ it is only 4 points (smth. like Mercedes symbol) - why 4 and not 3 or 5? How many points are possible to connect in such way in $\mathbb{R}^3$? (I suggest, infinite number, but it is interesting to look at a proof). What are some special properties of the Euclidean $\mathbb{R}^3$ such that the number of interconnected points <strong>jumps from 4</strong> in $\mathbb{R}^2$ to <strong>infinity</strong> in $\mathbb{R}^3$?</p> http://mathoverflow.net/questions/5372/dimension-leaps/10250#10250 Answer by Daniel Moskovich for Dimension Leaps Daniel Moskovich 2009-12-31T08:44:26Z 2009-12-31T08:44:26Z <p>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.<br> 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.</p> http://mathoverflow.net/questions/5372/dimension-leaps/10259#10259 Answer by Colin Tan for Dimension Leaps Colin Tan 2009-12-31T11:26:39Z 2009-12-31T11:26:39Z <p>The only spheres that are topological groups are S^0, S^1, S^3. In fact these are Lie groups.</p> <p>If we widen to h-groups, then we also have S^7.</p> <p>Nothing for higher dimensional spheres. </p> http://mathoverflow.net/questions/5372/dimension-leaps/10262#10262 Answer by Jonas Meyer for Dimension Leaps Jonas Meyer 2009-12-31T12:00:30Z 2009-12-31T12:00:30Z <p>Here is a closely related pair of examples from operator theory, <a href="http://en.wikipedia.org/wiki/Von%5FNeumann%2527s%5Finequality" rel="nofollow">von Neumann's inequality</a> and the theory of <a href="http://en.wikipedia.org/wiki/Sz.-Nagy%2527s%5Fdilation%5Ftheorem" rel="nofollow">unitary dilations of contractions on Hilbert space</a>, where things work for 1 or 2 variables but not for 3 or more.</p> <p>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 <code>$\|p(T)\|\leq\sup\{|p(z)|:|z|=1\}$</code>. 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 $n$. </p> <p>These results extend to two commuting variables, as <a href="http://www.renyi.hu/~petz/pdf/andobio.pdf" rel="nofollow">Ando</a> 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 <code>$T_1^{n_1}T_2^{n_2}=PU_1^{n_1}U_2^{n_2}|_H$</code> 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 <code>$\|p(T_1,T_2)\|\leq\sup\{|p(z_1,z_2)|:|z_1|=|z_2|=1\}$</code>.</p> <p>Things aren't so nice in 3 (or more) variables. Parrott <a href="http://projecteuclid.org/DPubS?verb=Display&amp;version=1.0&amp;service=UI&amp;handle=euclid.pjm/1102976441&amp;page=record" rel="nofollow">showed</a> 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 <code>$\|p(T_1,\ldots,T_n)\|\leq K\cdot\sup\{|p(z_1,\ldots,z_n)|:|z_1|=\cdots=|z_n|=1\}$</code>. 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 $K_n\lt\infty$.</p> <p>See <a href="http://books.google.com/books?id=VtSFHDABxMIC&amp;printsec=frontcover&amp;dq=paulsen+completely&amp;ei=go08S%5FK6O4SUManM%5FcEG&amp;client=firefox-a&amp;cd=1#v=onepage&amp;q=&amp;f=false" rel="nofollow">Paulsen's book</a> (2002) for more. On page 69 he writes: </p> <blockquote> <p>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.</p> </blockquote> http://mathoverflow.net/questions/5372/dimension-leaps/18710#18710 Answer by James for Dimension Leaps James 2010-03-19T02:54:09Z 2010-03-19T02:54:09Z <p>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 $n \geq 3$.</p> http://mathoverflow.net/questions/5372/dimension-leaps/62162#62162 Answer by Qfwfq for Dimension Leaps Qfwfq 2011-04-18T19:46:26Z 2011-04-18T19:46:26Z <p>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: </p> <blockquote> <p>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 $\beta &lt; \alpha$ and it is $\mathcal{H}^{\beta}(A)=0$ for $\beta > \alpha$.</p> </blockquote> http://mathoverflow.net/questions/5372/dimension-leaps/62164#62164 Answer by Qfwfq for Dimension Leaps Qfwfq 2011-04-18T19:55:53Z 2011-04-18T19:55:53Z <p>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...</p> http://mathoverflow.net/questions/5372/dimension-leaps/62167#62167 Answer by Richard Borcherds for Dimension Leaps Richard Borcherds 2011-04-18T20:24:44Z 2011-04-18T20:24:44Z <p>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.)</p> http://mathoverflow.net/questions/5372/dimension-leaps/62168#62168 Answer by Andreas Blass for Dimension Leaps Andreas Blass 2011-04-18T20:25:47Z 2011-04-18T20:25:47Z <p>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.</p> http://mathoverflow.net/questions/5372/dimension-leaps/62172#62172 Answer by Gil Bor for Dimension Leaps Gil Bor 2011-04-18T20:47:26Z 2011-04-18T20:47:26Z <p>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. </p> http://mathoverflow.net/questions/5372/dimension-leaps/62173#62173 Answer by Spencer for Dimension Leaps Spencer 2011-04-18T20:50:09Z 2011-04-18T20:50:09Z <p>An $n=7$ cut-off appears in the theory of minimal surfaces.</p> <p>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.</p> <p>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 </p> <p>$x_1^2 + x_2^2 + x_3^2 + x_4^2 = x_5^2 + x_6^2 + x_7^2 + x_8^2$,</p> <p>is minimal and therefore shows that this is optimal - because it has an isolated singularity at the origin. </p> <p>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. </p> http://mathoverflow.net/questions/5372/dimension-leaps/62174#62174 Answer by Michael Renardy for Dimension Leaps Michael Renardy 2011-04-18T21:18:59Z 2011-04-18T21:18:59Z <p>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.</p> http://mathoverflow.net/questions/5372/dimension-leaps/64061#64061 Answer by John Stillwell for Dimension Leaps John Stillwell 2011-05-05T22:23:22Z 2011-05-05T22:23:22Z <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/5372/dimension-leaps/73950#73950 Answer by Gil Kalai for Dimension Leaps Gil Kalai 2011-08-29T07:40:18Z 2011-08-29T07:40:18Z <p>This is a very nice question. Let me describe the very different behavior of convex polytopes in 3-dimensions compared to convex polytopes in higher dimensions. In three dimensions we have the following facts:</p> <p>1) Every triangulation of $S^2$ and, more generally every realization of $S^2$ by a polyhedral complex are combinatorially equivalent to the boundary complex of a convex polytope. </p> <p>2) Every polytope is combinatorially equivalent to a rational polytope - namely to a polytope all whose vertices have rational coordinates.</p> <p>3) Every automorphism of the face lattice of a convex polytope can be realized by a rigid motion of a combinatorially equivalent polytope.</p> <p>These statements follow or extend a well known theorem of Steinitz. They are related to the Koebe-Andreev-Thurston circle packing theorem. They all fail very strongly in dimension 4 and higher. </p> http://mathoverflow.net/questions/5372/dimension-leaps/73991#73991 Answer by Simon Lyons for Dimension Leaps Simon Lyons 2011-08-29T17:52:35Z 2011-08-29T17:52:35Z <p>Brownian motion is recurrent in dimensions 1 and 2, and transient in three or more dimensions.</p>