What makes four dimensions special? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T18:00:43Z http://mathoverflow.net/feeds/question/47569 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/47569/what-makes-four-dimensions-special What makes four dimensions special? Cristi Stoica 2010-11-28T08:51:17Z 2012-05-11T16:29:19Z <p>Do you know properties which distinguish four-dimensional spaces among the others?</p> <ol> <li>What makes four-dimensional topological manifolds special?</li> <li>What makes four-dimensional differentiable manifolds special?</li> <li>What makes four-dimensional Lorentzian manifolds special?</li> <li>What makes four-dimensional Riemannian manifolds special?</li> <li>other contexts in which four dimensions or $3+1$ dimensions play a distinguishing role.</li> </ol> <p>If you feel there are many particularities, please list the most interesting from your personal viewpoint. They may be concerned with why spacetime has four dimensions, but they should not be limited to this.</p> http://mathoverflow.net/questions/47569/what-makes-four-dimensions-special/47570#47570 Answer by Jim Conant for What makes four dimensions special? Jim Conant 2010-11-28T09:02:00Z 2010-11-28T09:02:00Z <p>It's the only dimension in which the smooth Poincare conjecture is still open. It's the only dimension in which $\mathbb R^n$ has a nonstandard smooth structure. (In fact uncountably many of them.) </p> <p>There's a lot going on in four dimensions. In some sense it's right at the boundary between low and high-dimensional topology.</p> http://mathoverflow.net/questions/47569/what-makes-four-dimensions-special/47581#47581 Answer by David MJC for What makes four dimensions special? David MJC 2010-11-28T12:36:58Z 2010-11-28T12:36:58Z <p>(Riemannian geometry) Four is the only dimension $n$ in which the adjoint representation of SO($n$) is not irreducible. Since the adjoint representation is isomorphic to the representation on 2-forms, this means that the bundle of 2-forms on an oriented Riemannian manifold decomposes into self-dual and anti-self-dual forms. 2-forms are particularly significant, since the curvature of a connection is a 2-form. In particular the curvature of the Levi-Civita connection is a 2-form with values in the adjoint bundle, so it has a 4-way decomposition into self-dual and anti-self-dual pieces. Hence there are natural curvature conditions on Riemannian 4-manifolds which have no analogue in other dimensions (without imposing additional structure).</p> <p>The impact of self-duality includes: special properties of Einstein metrics, Yang-Mills connections, and twistor theory for (anti-)self-dual Riemannian manifolds.</p> http://mathoverflow.net/questions/47569/what-makes-four-dimensions-special/47584#47584 Answer by Bruno Martelli for What makes four dimensions special? Bruno Martelli 2010-11-28T13:15:57Z 2011-02-23T11:21:44Z <p>The <a href="http://en.wikipedia.org/wiki/Whitney_embedding_theorem" rel="nofollow">Whitney trick</a> is an important step in Smale's proof of the Poincaré conjecture for smooth manifolds of dimension $n\geqslant 5$. It turns out however that such a trick does not work in dimension 4. However, as shown by Freedman (using previous work by Casson), it is possible (in a non-trivial way) to make this trick work for <i> topological </i> 4-manifolds. This partially explains the striking difference between topological and smooth manifolds in dimension 4. </p> <p>As an example of this striking and exceptional difference between these two categories, we know that in every dimension $n\neq 4$ a topological closed manifold may admit only finitely many smooth structures. In dimension $n=4$ however there are 4-manifolds like the $K3$ surface or (<a href="http://arxiv.org/abs/1005.3346" rel="nofollow">very recently</a>) $S^2\times S^2$ that admit infinitely many distinct smooth structures. As far as we know, it might as well be that <i>any</i> closed smoothable 4-manifold has infinitely many distinct structures! </p> <p>The question is open for instance for $S^4$ itself, which might have any number of distinct differentiable structures ranging from 1 to $\infty$ (extremes included). That's why we say that the Poincaré Conjecture is true for topological 4-manifolds but is still (very) open for smooth 4-manifolds.</p> http://mathoverflow.net/questions/47569/what-makes-four-dimensions-special/47638#47638 Answer by Agol for What makes four dimensions special? Agol 2010-11-29T04:01:22Z 2010-11-29T04:01:22Z <p>A comment is that 4 is the first dimension for which every finitely presented group may be realized as the fundamental group of a closed smooth 4-manifold. Other special properties are that the first pontryagin class and the Kirby-Siebenmann invariant live in 4-dimensional cohomology. </p> http://mathoverflow.net/questions/47569/what-makes-four-dimensions-special/51311#51311 Answer by vonjd for What makes four dimensions special? vonjd 2011-01-06T14:16:16Z 2011-01-06T14:16:16Z <p>In his new book "The shape of inner space" (2010) fields medalist Shing-Tung Yau cites Simon Kirwan Donaldson from Imperial College London (p. 68):</p> <blockquote> <p>No one yet knows, from a fundamental standpoint, exactly what makes four dimensions so special, Donaldson admits. Prior to his work, we knew virtually nothing about “smooth equivalence” (diffeomorphism) in four dimensions, although the mathematician Michael Freeman (formerly at the University of California, San Diego) had provided insights on topological equivalence (homeomorphism). In fact, Freeman topologically classiﬁed all four-dimensional manifolds, building on the prior work of Andrew Casson (now at Yale). Donaldson provided fresh insights that could be applied to the very difficult problem of classifying smooth (diffeomorphic) four-dimensional manifolds, thereby opening a door that had previously been closed. Before his efforts, these manifolds were almost totally impenetrable. And though the mysteries largely remain, at least we now know where to start.</p> </blockquote> http://mathoverflow.net/questions/47569/what-makes-four-dimensions-special/51331#51331 Answer by Terry Tao for What makes four dimensions special? Terry Tao 2011-01-06T18:28:27Z 2011-01-06T18:34:05Z <p>The Yang-Mills functional $\int_{{\bf R}^{1+d}} F^{\mu \nu} F_{\mu \nu}\ dx dt$ is dimensionless (scale-invariant) if and only if the spacetime dimension is four. (The integrand is a quadratic function of the curvature, which is two derivatives of the metric: 2 times 2 is equal to 4. In contrast, the Dirichlet functional, which involves a quadratic function of single derivatives rather than double derivatives, becomes critical at two dimensions rather than four, which explains why harmonic functions behave particularly nicely in two spatial dimensions. Similarly, the Einstein-Hilbert functional involves a linear function of curvature, and is thus also critical at two dimensions, explaining the nice behaviour of Ricci flow and similar equations in two dimensions.) For similar reasons, the Yang-Mills energy $\int_{{\bf R}^d} T_{00}\ dx$ is dimensionless if and only if the spatial dimension is four. As such, four spatial dimensions is "critical" for the Yang-Mills equation in the sense that for a fixed energy, one gets more or less the same nonlinear behaviour at both coarse and fine scales.</p> <p>This is also related to why Yang-Mills instantons only emerge at spatial dimensions four or higher; below this dimension, (elliptic) Yang-Mills connections are always smooth. (In general, the singularities of such connections are known to have codimension at least four, a classic result of Uhlenbeck.)</p> http://mathoverflow.net/questions/47569/what-makes-four-dimensions-special/51338#51338 Answer by Mariano Suárez-Alvarez for What makes four dimensions special? Mariano Suárez-Alvarez 2011-01-06T20:16:53Z 2011-01-06T20:58:48Z <p>Four is the dimension where a maximum number of regular polyhedra exist.</p> <p>(Apart from polygons in the plane, of course. But those are "abelian", hence boring :) )</p> http://mathoverflow.net/questions/47569/what-makes-four-dimensions-special/51347#51347 Answer by Beren Sanders for What makes four dimensions special? Beren Sanders 2011-01-06T22:29:09Z 2011-01-06T22:29:09Z <p>Four is the maximum number of dimensions for which the Busemann-Petty problem has an affirmative answer. This problem is discussed in <a href="http://mathoverflow.net/questions/35468/widely-accepted-mathematical-results-that-were-later-shown-wrong/43477#43477" rel="nofollow">my answer to another question</a>.</p> <p>Pinning down <em>why</em> there is a shift between four and five dimensions is an interesting question and is probably not completely understood. Some explanations for the shift can be given but the ones I know look more like analytic artifacts and don't seem to give any profound geometric insight.</p> <p>To give a taste, let me mention an example of an analytic result which is closely related to the switch. First some preliminary explanation. To every origin-symmetric convex body $K$ in $\mathbb{R}^n$ there is an associated norm $||\cdot||_K$ on $\mathbb{R}^n$ whose unit ball is precisely $K$. If we restrict the norm to the unit sphere $S^{n-1}$ and take its reciprocal, we obtain the naturally defined radial function $\rho_K$ of $K$. In fact, any continuous, even, positive function on the sphere will be the radial function of some (not necessarily convex) origin-symmetric star body. Next, given an origin-symmetric star body $L$, we can define its so-called <em>intersection body</em> $I L$ to be the origin-symmetric star body whose radial function is given by</p> <p>$$\rho_{I L}(\xi)=Vol_{n-1}(L \cap \xi^\perp),\quad \xi\in S^{n-1}.$$</p> <p>Now, there is a simple (almost tautological) formula for the volumes of central sections of a body in terms of the spherical Radon transform of its radial function:</p> <p>$$Vol_{n-1}(K\cap \xi^\perp) = \frac{1}{n-1}R(||\cdot||_K^{-n+1})(\xi),\quad \xi\in S^{n-1}.$$</p> <p>Using this formula we have that</p> <p>$$\rho_{I L}(\xi)=\frac{1}{n-1} R \rho_L^{n-1}(\xi),\quad \xi\in S^{n-1}.$$</p> <p>It turns out that the Busemann-Petty problem is equivalent to the question of whether every origin-symmetric convex body $K$ in $\mathbb{R}^n$ is the intersection body of some star body. By the above remarks it is not hard to be convinced that this question is then related to the positivity of the inverse spherical Radon transform. Now, by utilizing a connection between Fourier analysis and the spherical Radon transform, we get that </p> <p>$$Vol_{n-1}(K\cap \xi^\perp) = \frac{1}{\pi(n-1)}(||\cdot||_K^{-n+1})^\wedge(\xi),\quad \xi\in S^{n-1}.$$</p> <p>Here the function $||\cdot||_K^{-n+1}$ is locally integrable and the Fourier transform is taken in the sense of distributions. Thus, if $K$ is an intersection body of a star body $L$ then</p> <p>$$||\xi||_K^{-1} = \frac{1}{\pi(n-1)}(||\cdot||_L^{-n+1})^\wedge(\xi)$$</p> <p>and a small argument using this formula shows that $(||\cdot||_K^{-1})^\wedge$ is a positive distribution, and hence that $||\cdot||_K^{-1}$ is a positive definite distribution. The converse also holds and we have the general result that a star body $K$ in $\mathbb{R}^n$ is an intersection body iff $||\cdot||_K^{-1}$ represents a positive definite distribution in $\mathbb{R}^n$.</p> <p>In summary, the very geometric Busemann-Petty problem is closely related to the positive definiteness of certain distributions (coming from certain negative powers of norms on $\mathbb{R}^n$). With this vague background, perhaps we can appreciate that the following analytic result is closely related to the switch between 4 to 5 dimensions in the problem:</p> <p>Theorem. Let $n \ge 3$ be an integer and $2 &lt; q \le \infty$ a real number. Then the distribution $||\cdot||_q^{-p}$ is positive definite if $p\in (0,n-3)$ and is not positive definite if $p\in [n-3,n)$. As a consequence, the unit ball of the space $\ell_q^n$ is an intersection body iff $n \le 4$.</p> http://mathoverflow.net/questions/47569/what-makes-four-dimensions-special/51359#51359 Answer by Jeff Harvey for What makes four dimensions special? Jeff Harvey 2011-01-06T23:54:12Z 2011-01-06T23:54:12Z <p>$4=11-7$ and $11$ is the maximal dimension for supersymmetry with spins $\le 2$ while $7$ is the first dimension in which there exist compact manifolds of exceptional holonomy.</p> http://mathoverflow.net/questions/47569/what-makes-four-dimensions-special/70365#70365 Answer by Al-burcas for What makes four dimensions special? Al-burcas 2011-07-14T18:52:03Z 2011-07-14T18:52:03Z <p>Four is the dimension of the oriented Riemannian manifolds for which we can think of gwistor space. Yes, gwistor space.</p> http://mathoverflow.net/questions/47569/what-makes-four-dimensions-special/96662#96662 Answer by Camilo Sarmiento for What makes four dimensions special? Camilo Sarmiento 2012-05-11T12:30:15Z 2012-05-11T12:30:15Z <p>Also related to polytopes, there is Richter-Gebert's universality theorem for 4-polytopes (which i quote from Ziegler's book): Every elementary semialgebraic set defined over $\mathbb{Z}$ is stably equivalent to the realization space of some 4-dimensional polytope.</p> http://mathoverflow.net/questions/47569/what-makes-four-dimensions-special/96684#96684 Answer by Kristal Cantwell for What makes four dimensions special? Kristal Cantwell 2012-05-11T16:29:19Z 2012-05-11T16:29:19Z <p>Here is more about regular polytopes and four dimensions. For regular convex polytopes there are six regular polytopes and then for all dimensions higher than 4 there are three. For nonconvex regular polytopes there are 10 in four dimensions and zero in all higher dimensions. For convex euclidean tessellations there are 3 in four dimensions zero in all higher dimensions. </p> <p>However there are some cases in which the 5th and 6th dimension have different values: For convex hyperbolic tessellations 4 in the fourth dimension, 5 in the fifth and zero in all higher dimensions. For nonconvex hyperbolic tessellations there are zero in four dimensions four in the fifth dimension and zero in all higher dimensions. </p> <p>I used the wikipedia article "List of regular polytopes" which is available <a href="http://en.wikipedia.org/wiki/List_of_regular_polytopes#Five-dimensional_regular_polytopes" rel="nofollow">here</a> as a source for the above information.</p>