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Do you know properties which distinguish four-dimensional spaces among the others?

  1. What makes four-dimensional topological manifolds special?
  2. What makes four-dimensional differentiable manifolds special?
  3. What makes four-dimensional Lorentzian manifolds special?
  4. What makes four-dimensional Riemannian manifolds special?
  5. other contexts in which four dimensions or $3+1$ dimensions play a distinguishing role.

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.

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Motivation: space-time! :) – Eivind Dahl Nov 28 '10 at 8:59
The differential geometry reason seems to be connected to the fact that $2=4/2$; $2$-forms are also mid-dimensional forms. This allows for things like (anti-)selfdual Yang-Mills equation. – Torsten Ekedahl Nov 28 '10 at 10:51
I remember attending two lectures by M. Gromov that began by the remark that not only 4 equals 2 plus 2, but this equality is true in 3 different ways (meaning that there are three equipartitions in two classes of a 4-element set). According to him, a lot of exceptional behaviors in math stem from this, especially from the fact that 3 < 4 (it definitely explains why the alternating group A(4) isn't simple, but Gromov also mentioned the gauge-theoretic oddities in dimension 4). – Maxime Bourrigan Jan 6 '11 at 23:57
Four dimensions is also the last case of regular solids besides the simplex, the hypercube, and the cross-polytope (a.k.a. the dual hypercube). In fact there are three extra regular solids in dimension 4: the 24-cell (self-dual) and the 120- and 600-cells (each other's dual). These are not entirely unrelated to more advanced features of 4D space, e.g. the reflection groups $D_4$ and $H_4$, and the unit quaternions over ${\bf Q}$ (ramified at $\{\infty,2\}$) and over ${\bf Q}(\sqrt{5})$ (ramified only at the two infinite places). – Noam D. Elkies Jul 14 '11 at 19:21
3+1 is the smallest dimensionality in which general relativity is interesting. In 2+1 dimensions gravitational fields don't propagate and the theory becomes purely topological. – Ben Crowell Jul 24 '14 at 17:00

16 Answers 16

The Whitney trick 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 topological 4-manifolds with "good" fundamental group. This partially explains the striking difference between topological and smooth manifolds in dimension 4.

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 (very recently) $S^2\times S^2$ that admit infinitely many distinct smooth structures. As far as we know, it might as well be that any closed smoothable 4-manifold has infinitely many distinct structures!

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.

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Aren't all K3 surfaces diffeomorphic, and thus have the same smooth structure? – Gunnar Þór Magnússon Nov 28 '10 at 21:32
Yes, they are all diffeomorphic, that's why topologists say "the" K3-surface. However, there are (infinitely many) smooth structures on this manifold: these smooth structures do not arise from a complex structure, they are constructed with other cut-and-paste techniques (see Fintushel and Stern in 1996, and are distinguished using Seiberg-Witten invariants. – Bruno Martelli Nov 28 '10 at 22:31
As a comment, shouldn't we say that the Whitney trick works for topological 4-manifolds with good fundamental group? – Jim Conant Aug 28 '15 at 13:49
yes, you're right, I have added this hypothesis, thanks – Bruno Martelli Oct 23 '15 at 10:22
@BrunoMartelli: I don't understand your comment that "the K3 has infinitely many diffeomorphic smooth structures". Isn't it the case that every (smoothable) manifold admits infinitely many diffeomorphic smooth structures just by "conjugating" a given one by a non differentiable homeomorphism? (e.g. the standard smooth structure on the real line is diffeomorphic with the one given by the atlas $\{x\mapsto x^{2k+1}\}$, $k\in\mathbb{N}$) – Qfwfq Feb 29 at 15:45

(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).

The impact of self-duality includes: special properties of Einstein metrics, Yang-Mills connections, and twistor theory for (anti-)self-dual Riemannian manifolds.

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Note also Torsten Ekedahl's response to the question above (which I missed when posting this): in any even dimension, middle dimensional forms are not irreducible for the complexified special orthogonal group. This accounts not only for the special features of four dimensions in Riemannian geometry, but also dimensions 2 and 6, where 1-forms and 3-forms play a special role. Further, Lorentzian geometry in four dimensions is special because the bundle of 2-forms has a natural complex structure: this underpins the Petrov Classification of spacetimes, for example. – David MJC Nov 28 '10 at 23:16

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.

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.)

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Terry, I agree with all of this, but it raises a question I've had for a long time, which is why should the Ricci flow work so well for extracting the topological structure of a 3-manifold? I confess that I have not studied Perelman's work as closely as I should, but I was wondering if there is a high level explanation like the ones you give above for why one would expect in advance that Perelman's ideas would work. I myself conjectured a long time ago a completely different approach that did involve using the scale-invariant $L_p$ norm of curvature. That obviously didn't work out as well – Deane Yang Jan 6 '11 at 19:23
Good question! Certainly Ricci flow is much easier to study in two dimensions than in any other dimension, so it was a real feat of Perelman to push the three-dimensional theory as much as he did. Much of Perelman's work is in fact valid in all dimensions, though at a few points he uses (much as Hamilton did before him) the fact that Ricci curvature controls Riemann curvature, which is only true up to three dimensions. But more importantly, perhaps, in three dimensions the only singularities look like S^3, S^2 x R, or quotients thereof, and are thus amenable to surgery. (cont) – Terry Tao Jan 6 '11 at 23:41
In four dimensions, singularities with the local structure of S^2 x R^2 begin to appear, and this is bad because it is not clear at all how one could excise them by surgery. Short of finding a non-surgical approach to dealing with Ricci flow singularities, it's not clear to me at all how to push the Ricci flow strategy beyond three dimensions, for instance to tackle the smooth 4D Poincare conjecture. (The situation may be better for 4D Ricci-Kahler flow, though.) – Terry Tao Jan 6 '11 at 23:43
I should add that one of the key innovations of Perelman's approach was to not work with quantities such as the Einstein-Hilbert functional (which were best suited to two spatial dimensions), but instead to find new scale-invariant quantities (specifically, Perelman entropy and Perelman reduced volume) which had good monotonicity behaviour wrt Ricci flow in all dimensions. Once he did this, there was no longer anything particularly special about two dimensions, and in principle one could now proceed in any dimension. (And indeed, his noncollapsing theorem holds in all dimensions.) – Terry Tao Jan 6 '11 at 23:47
Hmm. I suppose this is possible, though in practice the singularity could have a more complicated 2-surface than D^2 in it (S^2 x R^2 is only the local structure of the singularity, not the global structure). The difficulties may be on the geometric side: the surgery has to respect the geometry enough that the delicate monotonicity formulae that underlie the noncollapsing theorem are not destroyed. In the 3D case the singularities are simple enough that one can locate "horns" where the singularities look geometrically like the cylinder S^2 x R^1 which is crucial for surgery. – Terry Tao Jan 7 '11 at 22:19

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.)

There's a lot going on in four dimensions. In some sense it's right at the boundary between low and high-dimensional topology.

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Thanks! Indeed, one of the particularities come from Donaldson's results (then Seiberg and Witten). – Cristi Stoica Nov 28 '10 at 9:24
+1 for the last sentence. I would also mention the h-cobordism theorem, which is alluded to elsewhere (the Whitney trick) as a (perhaps the?) key manifestation of this. – Steve Huntsman Nov 28 '10 at 14:46
And it is the only dimension that a closed smoothable manifold (possibly all of them) can have infinite smoothe structures. – Xiaolei Wu Nov 28 '10 at 21:22

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.

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The fact that 4 is the first dimension n at which every f.p. group can be realized as the fundamental group of a smooth n-manifold is interesting - and since it implies that homeomorphism of 4-manifolds is uncomputable, that implies a certain amount of "wildness" at 4 and beyond - but does this really distinguish 4 from higher numbers? That is, why its its status as the first such dimension relevant? (By analogy: 3 is the first k such that k-SAT is NP-complete, but 3-SAT is not particularly special compared to 4-SAT, 5-SAT, etc.) – Joshua Grochow Mar 2 at 5:18

Four is the dimension where a maximum number of regular polyhedra exist.

(Apart from polygons in the plane, of course. But those are "abelian", hence boring :) )

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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):

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 Freedman (formerly at the University of California, San Diego) had provided insights on topological equivalence (homeomorphism). In fact, Freedman 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.

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$4=11-7$ 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.

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WHy is that difference significant? – Mariano Suárez-Alvarez Jan 7 '11 at 0:47
In M-theory spacetime is 11-dimensional, so the question of why 4 dimensions can instead be viewed as the question of why 7 of the dimensions are so small as to be unobserveable at present energies. String theorists certainly don't know the answer to this question, but one fashionable approach involves choosing 7 of the dimensions to be those of a compact manifold with $G_2$ holonomy. – Jeff Harvey Jan 7 '11 at 1:25
Why aren't K3 surfaces considered to have special holonomy? They don't have the same holonomy as the generic 4-manifold. – Ben McKay Feb 29 at 14:59

Four is the maximum number of dimensions for which the Busemann-Petty problem has an affirmative answer. This problem is discussed in my answer to another question.

Pinning down why 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.

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 intersection body $I L$ to be the origin-symmetric star body whose radial function is given by

$$\rho_{I L}(\xi)=Vol_{n-1}(L \cap \xi^\perp),\quad \xi\in S^{n-1}.$$

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:

$$Vol_{n-1}(K\cap \xi^\perp) = \frac{1}{n-1}R(||\cdot||_K^{-n+1})(\xi),\quad \xi\in S^{n-1}.$$

Using this formula we have that

$$\rho_{I L}(\xi)=\frac{1}{n-1} R \rho_L^{n-1}(\xi),\quad \xi\in S^{n-1}.$$

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

$$Vol_{n-1}(K\cap \xi^\perp) = \frac{1}{\pi(n-1)}(||\cdot||_K^{-n+1})^\wedge(\xi),\quad \xi\in S^{n-1}.$$

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

$$||\xi||_K^{-1} = \frac{1}{\pi(n-1)}(||\cdot||_L^{-n+1})^\wedge(\xi)$$

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 $\mathbb{R}^n$.

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:

Theorem. Let $n \ge 3$ be an integer and $2 < q \le \infty$ 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 $n \le 4$.

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$so(4)$ is the only orthogonal Lie algebra that splits as a direct product of Lie algebras: $so(4)=so(3)\times so(3)$. This special property has consequences to $so(4)$-connections; in particular, the existence of Seiberg-Witten invariants.

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That is not correct. The correct statement is that $\text{Spin}(4) \cong \text{Spin}(3) \times \text{Spin}(3)$, or if you prefer that $\mathfrak{so}(4) \cong \mathfrak{so}(3) \times \mathfrak{so}(3)$. – Qiaochu Yuan Nov 3 '14 at 18:51
Yes, of course, I meant $SO(3)$. – Michael Nov 3 '14 at 19:24
But that statement is still (slightly) incorrect; $\text{SO}(4)$ (the Lie group) is a double cover of $\text{SO}(3) \times \text{SO}(3)$ (the Lie group). – Qiaochu Yuan Nov 3 '14 at 19:29
Thanks. Edited accordingly. – Michael Nov 3 '14 at 19:44
This is related to the fact that $\wedge^2 = \wedge^2_+\oplus \wedge^2_-$, i.e. Hodge duality is an endomorphism of 2-forms in dimension 4. – Ian Agol Aug 28 '15 at 19:16

One thing from physics that might be interesting: solutions to the wave equation act differently in odd+1 dimensions (so 4=3+1), versus in even+1 dimensions. For example, when an event occurs to create a wave in odd+1, the wave occurs once (if the event is instantaneous) and then spreads to fill all space. In even+1, a instantaneous point event will continue to produce waves long after the event has occurred, though with decreasing amplitude. Think of a pebble dropped in a pond: there is not only one ripple, but a continuing series of ripples based at that central point. In odd+1 this does not occur.

It's quite convenient that we're in odd+1 dimensions, because otherwise, when we turned off the lights, it would take time for the wave to be damped enough for it to get dark.

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A.Scorpan, "The wild world of 4-manifolds",(2005), AMS

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The rows of the matrix

$\begin{pmatrix} 2I_{n-1} & 0 \\ 1_{1 \times (n-1)} & 1 \end{pmatrix}$

generate a lattice for which the $n$ shortest (w/r/t the Euclidean norm) possible linearly independent vectors are only a basis if $n < 5$: otherwise they generate $(2 \mathbb{Z})^n$.

This matrix is special in this sense: see "Low-dimensional lattice basis reduction revisited" by Nguyen and Stehlé for details.

Discursive rant: this seems like it ought to be related to the Whitney trick, first in that five dimensions are necessary "to have enough room for the phenomenon to happen", and second in that there is a notion in which both rely on canceling doubles. I would be interested to know if this feeling can be formalized, undergirded, generalized, etc. through some more fundamental (perhaps categorical) fact.

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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.

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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.

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.

I used the wikipedia article "List of regular polytopes" which is available here as a source for the above information.

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Four is the dimension of the oriented Riemannian manifolds for which we can think of gwistor space. Yes, gwistor space.

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