# How does intuition fail for higher dimensions?

Now, Algebraic Geometry is one of the oldest, deepest, broadest and most active subjects in Mathematics with connections to almost all other branches in either a very direct or subtle way. The main motivation started with Pierre de Fermat and René Descartes who realized that to study geometry one could work with algebraic equations instead of drawings and pictures (which is now fundamental to work with higher dimensional objects, since intuition fails there).

What are these failings? Can you give me some examples?

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Does everybody have a clear intuition of objects in fourth dimension? –  ՃՃՃ Dec 14 '12 at 1:19
I have no idea. –  Vÿska Dec 14 '12 at 1:20
...fourth dimension...let alone 5th...6th...nth dimensions? –  amWhy Dec 14 '12 at 1:20
@amWhy, is that comment for me? –  ՃՃՃ Dec 14 '12 at 1:23
@use no, no...not at all; it was simply launching from your comment. That even if we struggle and arrive at some mere inkling of what a fourth dimension "looks like", how can we even fathom 5, 6, or n dimensions! I thought your comment was great. –  amWhy Dec 14 '12 at 1:35

# Spheres in a hypercube

Here is one example.

Take a square of edge length $4$. Place $2^2=4$ circles of radius $1$ into that square, in the obvious arrangement. Then place a circle at the center of all these circles, and make that fifth circle as large as possible. To compute its radius, look at the distance between the center of the square and the center of one of the large circles. That difference vector is $1$ length step for each of your two dimensions, so the distance is $\sqrt2$. As the big circle has radius $1$, your little circle has radius $\sqrt2-1$.

Now go to dimension $n$. You have a hypercube of edge length $4$, into which you place $2^n$ spheres of radius $1$ and one central sphere of radius $\sqrt n-1$. For $n=4$, the central sphere will already be as big as the outer spheres, and for $n>4$ it will be larger. At $n=9$, the radius of the central sphere will become $2$ so the central sphere will touch the containing hypercube. For $n>9$ the central sphere will no longer fit into the hypercube, even though it still touches the insides of all the radius-1-spheres contained in the hypercube.

This seems pretty counter-intuitive to me, even though the algebra is simple enough.

# Finite sphere packings

A remotely related phenomenon: take a finite number $m$ of unit spheres, and arrange them in such a way that the volume of the convex hull becomes minimal. You may imagine that for a few spheres, placing them in a straight line will be best, but the more spheres you have the more compact a clustering would appear in comparison. For 3D, more compact packings exist for e.g. $m=56$ spheres.

For $n\geq42$ dimensions (note that number!) it has been proven that the “sausage conjecture” holds: for those dimensions, the straight line sausage arrangement will always be optimal no matter the number of spheres. It has been conjectured but not proven that the sausage is always optimal for $n\geq5$ dimensions. The German Wikipedia article on this subject has some nice illustrations, even if you don't understand the language.

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Superb example! (+10 if I could). Even better than the old example which was something along the lines of "give me enough dimensions and I can fit the sun in a matchbox". –  Old John Dec 14 '12 at 8:50
@OldJohn, I can see that even at $n=4$, the 3D sun would be a infinitely thin hyperplane which I could simply roll up and fold before putting it into the 4D matchbox. But a 4D sun into a 4D matchbox seems to fail at least if you define them in terms of radius and edge length. I would like to hear the details on that. As a note: +10 is not possible, but +50 is, although I don't mean this as a request that you should offer such a bounty, just for information. –  MvG Dec 14 '12 at 9:02
The example I was thinking of was an extrapolation from the problem asking for the longest (rigid) line segment which will fit into an n-dimensional hypercube with sides of length 1 (i.e. $\sqrt{n}$, so that with enough dimensions, any given length will fit). Similarly, with enough dimensions the matchbox will hold a rigid (folding not allowed!) 3-d sphere of any prescribed radius. –  Old John Dec 14 '12 at 9:12
+1, if only for 42. –  Did Dec 14 '12 at 11:51
The thing is, the radius of the central sphere is $\sqrt{n} - 1$, but the radius of the circumsphere of the hypercube is $2 \sqrt{n}$, so the central sphere can never grow large enough to contain the whole hypercube! –  Zhen Lin Dec 14 '12 at 12:05

Intuition already breaks down at the fact that two planes can meet in a single point. The moral is that visualizing intersections in higher dimensions is hard.

Another example: the volume of the $n$-sphere. It starts at 2, then increases until dimension 4, then it decreases again, and in fact, goes to zero. (there is in fact a very long discussion of this on mathoverflow)

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No algebraic geometry in it but an example worth knowing concerns high-dimensional spheres. Roughly speaking the result is that, when the dimension is high enough, all the mass of the sphere is concentrated on anyone of its equators. A bit as if every $\varepsilon$-neighborhood of the Equator was enough to encompass nearly all the surface of the Earth. As every $\varepsilon$-neighborhood of each pair of opposite meridians...

Rigorously speaking, one introduces the uniform probability measure $u_n$ on the $n$-dimensional sphere $S^n=\{x=(x_k)_{1\leqslant k\leqslant n+1}\in\mathbb R^{n+1}\mid x_1^2+\cdots+x_{n+1}^2=1\}$ and the $\varepsilon$-neighborhood $N^n_\varepsilon$ of the great circle $x_{1}=0$, defined by $N^n_\varepsilon=\{x\in S^n\mid |x_{1}|\leqslant\varepsilon\}$. Then, for every $\varepsilon\gt0$, $u_n(S^n\setminus N^n_\varepsilon)\to0$ when $n\to\infty$.

Note that, by symmetry, the same applies to $\bar N^n_\varepsilon=\{x\in S^n\mid |x_{42}|\leqslant\varepsilon\}$.

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