# What's new in higher dimensions?

This is a very speculative/soft question; please keep this in mind when reading it. Here "higher" means "greater than 3".

What I am wondering about is what new geometrical phenomena are there in higher dimensions. When I say new I mean phenomena which are counterintuitive or not analogous to their lower dimensional counterparts. A good example could be hypersphere packing.

My main (and sad) impression is that almost all phenomena in higher dimensions could be thought intuitively by dimensional analogy. See for example, this link:

What this implies (for me) is the boring consequence that there is no new conceptual richness in higher dimensional geometry beyond the fact than the numbers are larger (for example my field of study is string compactifications and though, at first sight, it could sound spectacular to use orientifolding which set a loci of fixed points which are O3 and O7 planes; the reasoning is pretty much the same as in lower dimensions...)

However the question of higher dimensional geometry is very related (for me) to the idea of beauty and complexity: these projections to 2-D of higher dimensional objects totally amazes me (for example this orthonormal projection of a 12-cube) and makes me think there must be interesting higher dimensional phenomena...

I would thank anyone who could give me examples of beautiful ideas implying “visualization” of higher dimensional geometry…

• math.stackexchange.com/questions/374685/… – Nate Eldredge Feb 10 '18 at 16:28
• Do you know some smooth manifolds and algebraic topology already? – edm Feb 10 '18 at 16:28
• In very broad terms, topology in dimension $3$ and below is geometric; topology in dimension $5$ and above is algebraic (e.g., surgery obstructions), and topology in dimension $4$ is weird. More specifically, the $h$-cobordism theorem and variants only kick in at dimension $5$ (since the Whitney trick fails otherwise). – anomaly Feb 10 '18 at 16:46
• This is way too broad. You should restrict to a more specific area of mathematics. – Moishe Kohan Feb 10 '18 at 16:53
• I liked your question a lot. Sure, it might be too "soft", but it opens beautiful insights in the answers, just like it should. – Oleg Lobachev Feb 10 '18 at 22:13

In high dimensions, almost all of the volume of a ball sits at its surface. More exactly, if $V_d(r)$ is the volume of the $d$-dimensional ball with radius $r$, then for any $\epsilon>0$, no matter how small, you have $$\lim_{d\to\infty} \frac{V_d(1-\epsilon)}{V_d(1)} = 0$$ Algebraically that's obvious, but geometrically I consider it highly surprising.

Edit:

Another surprising fact: In 4D and above, you can have a flat torus, that is, a torus without any intrinsic curvature (like a cylinder in 3D). Even more: You can draw such a torus (not an image of it, the flat torus itself) on the surface of a hyperball (that is, a hypersphere). Indeed, the three-dimensional hypersphere (surface of the four-dimensional hyperball) can be almost completely partitioned into such tori, with two circles remaining in two completely orthogonal planes (thanks to anon in the comments for reminding me of those two leftover circles). Note that the circles could be considered degenerate tori, as the flat tori continuously transform into them (in much the same way as the circles of latitude on the 2-sphere transform into a point at the poles).

• For those interested, a couple of nice extended treatments of this are Geometry of High-Dimensional Space and Geometry in Very High Dimensions. – Dave L. Renfro Feb 10 '18 at 19:35
• Why do you say it is surprising? Physics students learn this in their statistical mechanics course, usually paired with the remark: "That's why you shouldn't by oranges in higher dimensions. Once you peel them, nothing is left." – marmot Feb 11 '18 at 17:56
• @marmot: That you learn it in statistical mechanics doesn't make it any less surprising. BTW, guess where I learned it. ;-) – celtschk Feb 11 '18 at 18:07
• Perhaps of interest: High dimensional hypercubes exhibit much different behavior. In particular, the distance between the center of a cube and most of the volume of the cube grows like $\sqrt{\text{dimension of cube}}$. See this post: math.stackexchange.com/questions/2085082/… – Yly Feb 11 '18 at 20:35
• @Yly All this talk of shaving small surface amounts makes me wonder if hyper-inflation was much worse in 10 dimensional Rome... – Michael Feb 15 '18 at 20:39

A number of problems in discrete geometry (typically, involving arrangements of points or other objects in $\mathbb R^d$) change behavior as the number of dimensions grows past what we have intuition for.

My favorite example is the "sausage catastrophe", because of the name. The problem here is: take $n$ unit balls in $\mathbb R^d$. How can we pack them together most compactly, minimizing the volume of the convex hull of their union? (To visualize this in $\mathbb R^3$, imagine that you're wrapping the $n$ balls in plastic wrap, creating a single object, and you want the object to be as small as possible.)

There are two competing strategies here:

1. Start with a dense sphere packing in $\mathbb R^d$, and pick out some roughly-circular piece of it.
2. Arrange all the spheres in a line, so that the convex hull of their union forms the shape of a sausage.

Which strategy is best? It depends on $d$, in kind of weird ways. For $d=2$, the first strategy (using the hexagonal circle packing, and taking a large hexagonal piece of it) is best for almost any number of circles. For $d=3$, the sausage strategy is the best known configuration for $n \le 56$ (though this is not proven) and the first strategy takes over for larger $n$ than that: the point where this switch happens is called the "sausage catastrophe".

For $d=4$, the same behavior as in $d=3$ occurs, except we're even less certain when. We've managed to show that the sausage catastrophe occurs for some $n < 375,769$. On the other hand, we're not even sure if the sausage is optimal for $n=5$.

Finally, we know that there is some sufficiently large $d$ such that the sausage strategy is always the best strategy in $\mathbb R^d$, no matter how many balls there are. We think that value is $d=5$, but the best we've shown is that the sausage is always optimal for $d\ge 42$. There are many open questions about sausages.

If you're thinking about the more general problem of packing spheres in $\mathbb R^d$ as densely as possible, the exciting stuff also happens in dimensions we can't visualize. A recent result says that the $E_8$ lattice and the Leech lattice are the densest packing in $\mathbb R^8$ and $\mathbb R^{24}$ respectively, and these are much better than the best thing we know how to do in "adjacent" dimensions. In a sense, this is saying that there are $8$-dimensional and $24$-dimensional objects with no analog in $\mathbb R^d$ for arbitrary $d$: a perfect example of something that happens in many dimensions that can't be intuitively described by comparing it to ordinary $3$-dimensional space.

Results like the Hales–Jewett theorem are another source of "new behavior" in sufficiently high-dimensional space. The Hales–Jewett theorem says, roughly speaking, that for any $n$ there is a dimension $d$ such that $n$-in-a-row tic-tac-toe on an $n \times n \times \dots \times n$ board cannot be played to a draw. (For $n=3$, that dimension is $d=3$; for $n=4$, it's somewhere between $d=7$ and $d = 10^{11}$.) However, you could complain that this result is purely combinatorial; you're not doing so much visualizing of $d$-dimensional objects here.

• Maybe you could explain — or at least give an idea — what the "sausage strategy" is... – Earthliŋ Feb 11 '18 at 9:05
• @Earthliŋ: unless I’m misunderstanding, the “sausage strategy” just means the 2nd of the two strategies listed in paragraph 3: line the spheres up, so that their convex hull is a sausage. – Peter LeFanu Lumsdaine Feb 11 '18 at 11:32
• I've looked at your paper - nice work on Hales-Jewett numbers – qwr Feb 11 '18 at 21:24
• Wow, so many open problems! Sausage problems seem like the wurst. – Mehrdad Feb 12 '18 at 10:04

In dimensions $d > 4$ there are just the obvious three regular polytopes: the simplex, the hypercube and its dual, the cross polytope. The fourth dimension stars three more, the 24-cell, the 120-cell and the 600-cell.

Pack spheres of diameter $1/2$ in the corners of a unit hypercube in dimension $d$. Then inscribe a sphere $S$ in the center of that cube tangent to the corner spheres. The long diagonal of the hypercube has length $\sqrt{d}$. It follows that the diameter of $S$ is $(\sqrt{d} -1)/2$. When $d=9$, $S$ is tangent to the facets of the hypercube. When $d> 9$ it sticks out past the facets.

$4x + 1 = \sqrt{d}$ so $2x = \frac{\sqrt{d} -1}{2}$.

My thesis advisor Andy Gleason told me once that he'd "give a lot for one good look at the fourth dimension".

• I find that hard to believe... (not that doubt you, it just seems wrong...). good one. – Aganju Feb 10 '18 at 22:44
• @peterh See my edit. – Ethan Bolker Feb 11 '18 at 14:07
• @EthanBolker Nice to see, I knew only about the 600-cell :-) – peterh Feb 11 '18 at 15:22
• Ethan, after seeing your last sentence just now (which I missed yesterday), I thought I'd mention a comment I made here here about Gleason that you might find interesting. – Dave L. Renfro Feb 11 '18 at 18:51
• > My thesis advisor Andy Gleason told me once that he'd "give a lot for one good look at the fourth dimension". He's not alone. – Hyperplane Feb 12 '18 at 15:33

Exotic spheres are a feature only of dimensions higher than 3. These are topological spaces which are homeomorphic to a sphere, but with different differential structure. Informally, you can paraphrase this result as "there are multiple distinct ways to do calculus on high dimensional spheres, but only one way on low dimensional spheres."

A related concept is exotic $\mathbb{R}^4$, but this is not characteristic of other high dimensional spaces--only $\mathbb{R}^4$.

Concentration of measure is a phenomenon that is common to many high dimensional geometric objects. The basic idea here is that most of the mass of many typical high dimensional objects (for example, the sphere) is concentrated near relatively small subsets. For instance, on the sphere, most of the mass is near the equator (or any other high dimensional analog of a "great circle").

As perhaps the best example of a phenomenon that "looks" very different in high dimensions, I suggest knot theory. The first hint that there are surprises in store is the observation that knots in the usual sense (embeddings of $S^1$ whose complement has non-trivial topology) do not exist in dimensions other than 3. However, there are analogous objects in higher dimensions: A $k$ sphere can be embedded in a $k+2$ sphere to form a sort of "high dimensional knot". So the important thing is that the "knot" has codimension 2 in the ambient space. Try thinking about what these high dimensional knots look like :)

If you are willing to consider infinite dimensions, then all bets are off. Finite intuition can be downright dangerous in this case. For example:

• For finite dimensional vector spaces, all norms are equivalent, but this is not true in infinite dimensions.
• There is no translation invariant measure in infinite dimensions. This is due to the geometric fact that you can put infinitely many balls of radius $1/2$ inside a ball of radius $2$ in infinite dimensions.
• In infinite dimensions I like, that the $S^\infty$ is contractible while the $S^n$ for $n$ finite is never contractible. – Rolf Sievers Feb 11 '18 at 13:17
• The Concentration of Measure has interesting practical consequences. A high-dimensional Gaussian distribution has nearly all its measure in a spherical shell around the center. What happens is that near the center, there is little volume and further out there is not much Gaussian. A set of points (like data entries) in such a space has all the points on this spherical shell. – ttw Feb 11 '18 at 22:51
• Either one. The spherical case is just normalized with the same variance for each variable. The probability of finding a point in a region is the product of the density times the volume of the region. In many dimensions, this product is more highly peaked than in small dimensional cases. – ttw Feb 12 '18 at 14:52
• Oooh, this is making me wonder what knot theory in infinite dimensions would look like. (A codimension 2 embedding of $S^\infty$ in $S^\infty$? O.o) – Sir Jective Feb 15 '18 at 23:33
• I think you can get codimension larger than two in some dimensions. – Akiva Weinberger Mar 16 '18 at 13:33

It is tempting to think that higher dimensions only bring "new geometrical phenomena", but they can also take away familiar properties often considered as elementary in lower dimensions.

Something we take for granted in 3D, for example, is the vector cross-product as a vector-valued, bilinear, anti-commutative, product of two 3D vectors. It may come as a surprise that there is no direct equivalent of the cross-product in $\,\mathbb{R}^n\,$ for $\,n>3\,$, except for $\,n=7\,$.

That "the seven-dimensional cross product has the same relationship to the octonions as the three-dimensional product does to the quaternions" is not immediately obvious, and the reason why the familiar cross-product only exists in $\,3\,$ or $\,7\,$ dimensions goes to the deeper Hurwitz' theorem which implies the only finite-dimensional normed division algebras must have dimension $1$, $2$, $4$, or $8$.

[ EDIT ]  Further insights can be gleaned from the related question Cross product in $\mathbb R^n$, some answers under Is the vector cross product only defined for 3D?, the quora question Why does the cross product exist only in three and seven dimensions? and J.M. Massey's article Cross Products of Vectors in Higher Dimensional Euclidean Spaces in The American Mathematical Monthly Vol. 90, No. 10 (Dec., 1983), pp. 697-701.

• Can you add a source for further reading? What exactly breaks down for $n=4$? – nbubis Feb 14 '18 at 16:17
• @nbubis I edited some references into the post. Short answer is that similar vector products can be defined in $\mathbb{R}^n$, but in dimensions other than $3$ or $7$ they lose some properties of the 3D cross product,, see in particular these 1, 2, 3. – dxiv Feb 14 '18 at 17:28

A sort of trivial thing that I haven't seen mentioned in any of the other answers yet is that rotations get weird when $d \ge 4$, messing with our three-dimensional intuition and requiring the development of completely new intuitions.

Specifically, in our three-dimensional world, we're used to things rotating around an invariant axis. The fact that in two dimensions things rotate around a point instead just seems like a special case of this — the invariant axis is still there, if we imagine the 2D space as being embedded in normal 3D space; it's just pointing out of the plane.

But in four dimensions, things don't rotate around an invariant axis; they rotate around an invariant plane. And what's worse, that invariant plane can accommodate another independent rotation at the same time! So things in four dimensions can rotate at two different speeds in two different, orthogonal planes at the same time, with only the single intersection point of the two planes staying in place.

This makes imagining elementary physics in four dimensions pretty weird.

For example, if the Earth was four-dimensional, it wouldn't have two poles and an equator; instead, it might have two equators, both orthogonal and equidistant to each other. And the 4D Moon's orbit around the 4D Earth (never mind the stability of orbits in four dimensions in the first place) could be at a constant 90° angle to the Earth's orbit around the Sun. And I won't even try to imagine what a 4D galaxy would look like, although it definitely wouldn't be a flattened disc like our own Milky Way.

Also, people living on the 4D Earth might have a hard time inventing the wheel. Certainly their wheels would look nothing like ours, what with having no axle (or, alternatively, perhaps having two orthogonal axles, or even an entire flat plane for an axle).

In higher dimensions, things get a bit more complicated yet. A general 5D rotation has two independent planes of rotation and an invariant axis, while six dimensions are enough to accommodate three independent planes of rotation. More generally, $2n$ dimensions are just enough to let an object rotate in $n$ orthogonal planes at the same time, while in $2n+1$ dimensions there will always be an extra invariant axis left over.

• +1, but to be honest this doesn't really seem counterintuitive to me, at least not after having seen linear algebra. Rotation is just any linear transformation that preserves lengths and angles. As long as you believe that, that it should make sense that you might be able to rotate "around" lots of things besides lines. – Mehrdad Feb 12 '18 at 10:16
• "the 4D Moon wouldn't be orbiting the 4D Earth in a simple circle or an ellipse" Actually, I'm pretty sure the Moon would remain in the 2-dimensional subspace determined by its initial position relative to the Earth and its initial velocity. What you say could apply to the rotation of a rigid body in 4 dimensions, but that's plenty complicated in 3D already. – Rahul Feb 12 '18 at 13:35
• Assuming the 4D moon was attracted to the 4D Earth by a 4D generalization of Newton's Laws, it would actually spiral into (or away from) the Earth at the slightest provocation. This is because orbits in a central-force motion under a potential $U \propto r^{-n}$ are unstable when $n \geq 2$, and a gravitational potential satisfying Laplace's equation in $D$ dimensions will fall off as $U \propto r^{-(D-2)}$. – Michael Seifert Feb 13 '18 at 15:24
• @Rahul: Hmm, yeah, I think you're right. It still feels kind of funny that there's a whole two-dimensional subspace that's orthogonal to the orbital plane. So, for example, if the 4D Moon had a satellite of its own, that satellite could have a circular orbit around the Moon that was perpendicular to the Moon's own orbit around the Earth. Anyway, I've edited my answer to remove the incorrect claim. – Ilmari Karonen Feb 13 '18 at 16:07
• @MichaelSeifert: Yes, that's the part I didn't want to go into, and instead linked to a Physics.SE post about it. :) – Ilmari Karonen Feb 13 '18 at 16:08

In addition to the good answers you have got so far, I think the tiling problem also fits in the scope of this question. In higher dimensions (and even in the 3D space), the tiling problem becomes highly complex and some strange results appear.

One seemingly obvious observation that becomes invalid in higher dimensions is Keller's conjecture, which states that

In any tiling of Euclidean space by identical hypercubes there are two cubes that meet face to face. For instance, as shown in figure below, in any tiling of the plane by identical squares, some two squares must meet edge to edge.

And I quote the rest from wikipedia:

It was shown to be true in dimensions at most 6 by Perron (1940). However, for higher dimensions it is false, as was shown in dimensions at least 10 by Lagarias and Shor (1992) and in dimensions at least 8 by Mackey (2002), using a reformulation of the problem in terms of the clique number of certain graphs now known as Keller graphs. Although this graph-theoretic version of the conjecture is now resolved for all dimensions, Keller's original cube-tiling conjecture remains open in dimension 7.

More precisely, the reduction to Keller graphs is equivalent to making the assumption that all hypercubes have integer or half-integer coordinates. (For example, this may be true in the image above if we assume that each vertical stripe is shifted exactly half a square's height from the stripe to its left.) So the progress on the problem to date can be summarized as:

1. Lagarias and Shor showed that even under this restriction, counterexamples to Keller's conjecture can be found in $d \ge 10$ dimensions.
2. Mackey extended this result to $d = 8$ dimensions (which implies $d=9$, too, since we can stack a bunch of 8-dimensional tilings on top of each other, shifted so none of the hypercubes in adjacent layers share faces).
3. However, Debroni et al. (2010) showed that all tilings with half-integer coordinates in $d=7$ do satisfy this conjecture.

This leaves the $d=7$ case open, but implies that if a counterexample does exist in $7$ dimensions, it must look weird and have less structure than any of the higher-dimensional counterexamples we've found.

• I'd add on to this that (based on my reading of the Debroni et al. paper) the reduction to Keller graphs can be interpreted as making the assumption that all cubes have integer or half-integer coordinates. So what we know is that (1) in dimensions 6 or less, all tilings satisfy Keller's conjecture; (2) in dimensions 8 or more, there is always a counterexample with half-integer coordinates; (3) in dimension 7, all tilings with half-integer coordinates satisfy Keller's conjecture, so any counterexample must be weirder than that. – Misha Lavrov Feb 18 '18 at 20:38
• @MishaLavrov : Thank you. That was nice and informative. Please feel free to edit the answer (and add your comment to it maybe?) – polfosol Feb 18 '18 at 21:02
• Okay, edited :) – Misha Lavrov Feb 18 '18 at 21:37

With regard to your question

"what new geometrical phenomena are there in higher dimensions. When I say new I mean phenomena which are counterintuitive or not analogous to their lower dimensional counterparts"

a good example is Borsuk's conjecture, true in dimensions 2 and 3 but false in higher dimensions. Given a compact pointset $X\subseteq \mathbb R^n$ one defines its diameter to be the maximal distance among a pair of points of $X$. Borsuk conjectured that every such $X$ can be partitioned into at most $(n+1)$ sets of diameter strictly smaller than that of $X$ itself. For example, the set of vertices of an equilateral triangle in the plane cannot be partitioned into fewer than 3 such sets, and similarly for the set of vertices of a regular $n$-simplex in $\mathbb R^n$.

By playing with figures in the plane it is easy to convince oneself that one can always partition them into subsets of smaller diameter. In 3-space the problem is trickier but can still be proved.

It came as a surprise when in 1993 by Jeff Kahn and Gil Kalai proved that the conjecture is false in sufficiently high dimension. The current best counterexample seems to be in dimension 64.

• I'm a bit confused by the 'at most $(n+1)$'. Surely if it can be partitioned into $n+1$ pieces we can get a partition in to $n+2$ pieces by further partitioning one piece into two? – Vincent Apr 4 at 14:10

The fundamental group of (connected) $n$-manifolds has a change around $n=3$. For $n=2$, the fundamental groups are cartesian products of various combinations of $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Z}$. For $n=3$, these are automatic groups. For $n=4$, any finitely presented group can be the fundamental group. So, determining triviality of the fundamental group, for the three values of $n$ mentioned, is trivial, decidable, and undecidable, respectively. If we let the dimension go to infinity, we can't even promise to have a finite presentation.

Triangulations also change with dimension. Differentiable manifolds (any dimension) have piecewise-linear (PL) triangulations. In 2 and 3 dimensions, any topological manifold has a triangulation, and the triangulations of a given manifold are piecewise-linear equivalent. In fact, in 2 and 3 dimensions, each manifold has a smooth structure, unique up to diffeomorphism. (For (complete, finite volume) hyperbolic manifolds of any dimension $\geq 3$, this is the (generalized) Mostow rigidity theorem -- every isotopy is an isometry.) In dimension 4, the $E_8$ manifold has no PL triangulation and many manifolds have infinitely many PL-inequivalent triangulations. In dimensions $\geq 4$, only some manifolds have PL triangulations (ibid.). In dimensions $\geq 5$, only some manifolds are homeomorphic to a simplicial complex -- most have no form of triangulation. (This paragraph summarizes the last paragraph of Triangulation (topology): Piecewise linear structures in the English language Wikipedia.)

And, one that surprised me the first time I ran across it in an application ($n$-dimensional diffusion, many years ago), the distribution of angles between random (nonzero) vectors in $n$-dimensional Euclidean space is concentrated (and becomes more concentrated with higher $n$) around $\pi/2$ -- everything is nearly orthogonal to everything else. This has an analog in $n=3$ -- the spherical zone between latitudes $+\pi/4$ and $-\pi/4$ has more area than the sum of the spherical caps above and below the zone. This becomes rapidly moreso as the dimension increases: the $\pm$ latitude where the measure of the zone equals the measure of the two caps goes to zero like $1/n$.

Is the $n$-sphere $S^n\subset \mathbb R^{n+1}$ the underlying real manifold of a complex holomorphic manifold?
Of course this is only possible if $n$ is even, but that is not sufficient.
The real tangent bundle of a complex manifold has the structure of a complex vector bundle and real manifolds whose tangent bundle is endowed with such a complex-linear sructure are called (unsurprisingly!) almost complex manifolds.
The complex-linear tangent bundle structure of an almost-complex manifolds comes from a complex structure of that manifold if and only an easy to check "integrability condition" holds: this is a remarkable theorem of Newlander-Nirenberg.
Now an almost complex structure on $S^n$ only exists for $n=2$ or $6$.
Of course $S^2$ has a complex structure: that of $\mathbb P^1(\mathbb C)$
The other possible sphere $S^6$ is known to have an almost-complex structure, but that structure is not integrable.
So the only remining problem is: does $S^6$ have another almost complex structure which is integrable, which would imply that $S^6$ has a holomorphic structure.
That problem is generally considered to be open, despite purported "solutions", about which the consensus seems to be that they are insufficiently convincing.

• I think Atiyah recently solved this problem (in the negative, IIRC). – Matemáticos Chibchas Feb 12 '18 at 6:26
• @MatemáticosChibchas It seems Atiyah's proof is not considered complete in the community, see here, here for example. – user99914 Feb 12 '18 at 6:36
• @John Ma is right: that's what my last senrtence alluded to (but I didn't want to be more explicit out of respect for anybody involved.) – Georges Elencwajg Feb 12 '18 at 9:16

The sphere-packing problem, already mentioned in other answers, has some bizarre behavior as the number of dimensions changes. Results in the Leech lattice lead to "monstrous moonshine", and the bizarreness of the sporadic groups. Meanwhile, in string theory, its known that certain things can only happen in 26 or 10 dimensions. Hopf fibrations show you how to deconstruct 3-D space into 2+1 dimensional space, or 11 dimensions into 7+4. But why can't this work in general N, you might wonder? The "homotopy groups of spheres" indicate that even the common-sense notion of a sphere changes from dimension to dimension.

The Laplacian, or rather, the Green's functions which provide the solutions to it, are bizarrely, counter-intuitively different in even and odd-dimensional space, and has the "simplest" form only in 3+1 dimensional space-time (in which we live. Hmmmm. Why would that be?)

• What do you mean by solutions to the Laplacian? Its eigenfunctions? – Ruslan Feb 15 '18 at 15:14
• I meant "Green's functions for the laplacian". Will edit. In 3+1 dimensions, the Green's function is $\delta(x-ct)$ with $c$ the speed of light. In higher dimensions, there are derivatives of the delta. In even spatial dimensions, there is a step-function, so that (for example) if a light blinked on and then off, we would "see" it glowing for long afterwards. Classic example is throwing a rock into water -- the ripples expand long after the rock sank. It wasn't just one single sharp wave, the same size as the rock. In 3+1 dimensions, there is just one wave, exactly the size of the rock. – Linas Mar 9 '18 at 20:59

There is already a great gap between 2 and 3 dimensions.

Here is a mildly interesting but very simple one I came up with, when thinking about this very question some time ago:

There is no partition of the 2d-plane into a finite number of path-connected regions that are translations of one another. But there is a partition of the 3d-space into just two path-connected regions that are translations of one another!

Another one is:

The mean value theorem holds for 1 and 2 dimensions, but not in 3, because of a spiral.

The following is very low-level difference, but a surprising one nonetheless.

A plane is a $2$-dimensional affine subspace (like the $xy$-plane) of $\mathbb{R}^n$. In three dimensions, it is obvious that two planes either are parallel, either intersect in a line or are exactly the same. It is impossible for two planes to intersect in only one point.

Going to $\mathbb{R}^4$ this fails. Indeed, the $xy$-plane and $zt$-plane only intersect in the origin. Any visual intuition you have becomes useless in higher dimensions as we can only visualize $3$-dimensional phenomena. On the other hand, using coordinates (as I did) it is trivial to understand that two planes in higher dimensional spaces can intersect in only one point.

I almost feel bad for posting such an obvious thing, but this difference can't be visually intuitive.

• On the other hand surfaces of $n$-dimensional objects in $\mathbb R ^ n$ are $( n - 1 )$-dimensional and such subspaces do behave pretty much like planes in $R ^ 3$. Admittedly, it is not very visually intuitive. Incidentally, a plane is not a (vector) subspace if it does not pass through the origin. – PJTraill Feb 15 '18 at 16:58
• @PJTraill: I meant affine subspace, updated the answer. Also your comment is valid, it turns out that the proper analogue of a plane is a hyperplane but who visualizes hyperplanes and stuff like that? Intuition doesn't tell you how these things generalize, algebra on the other definitely does exactly that. – Mathematician 42 Feb 15 '18 at 17:10
• "but who visualizes hyperplanes and stuff like that?" — everyone who visualizes planes in 4D ;-) – celtschk Feb 17 '18 at 6:37
• Isn't it intuitively obvious that 2d-planes in 4d generally only intersect in a 1d set? – user21820 Mar 16 '18 at 9:12
• @user21820: I don't think that's very obvious. Using linear algebra these things become very clear but before you are doing mathematics it's not clear how you can think of $4$-dimensional problems. At least, that's my opinion. – Mathematician 42 Mar 16 '18 at 9:29

If a 2-D plane is perpendicular to the diagonal of a 3-D cube and passes through the midpoint of this diagonal, its intersection with the cube is a regular hexagon.

If a 3-D hyperplane (i.e. given by an equation of the form ax + by + cz + dw = e) is perpendicular to the diagonal of a 4-D hypercube (or tesseract) and passes through the diagonal's midpont, its intersection with the hypercube is a regular octahedron.

I don't think this is very intuitive, but it makes sense after a little thought. For the 3-D cube has six 2-D faces (squares) and that plane intersects all of them. So, it makes sense that the intersection is a hexagon. The hypercube has eight 3-D hyperfaces (cubes), so that a hyperplane that intersects them all should do so in an octahedron. And symmetry suggests the intersections should be regular (although that needs to be proved).

I would say 3 dimensions is one of the most interesting dimensions as you have things like knots and 5 platonic solids. The Poincare conjecture for 3-spheres is hard to prove.

Whereas in higher dimensions tend to get simpler (and more boring/beautiful depending on your point of view.)

Things like the $E_8$ lattice and Leech lattice which solve the sphere packing problems in 8D and 24D are very interesting and are related to many symmetries.

So certain dimensions have these incredible structures but most don't.

Exotic spheres in 4D which I don't even understand!

It's lucky most of the interesting (hard) stuff happens in 3D! Because that's where we live.