Can you explain why the "spikiness" should be associated with the hyperspheres instead of the hypercubes?
Can you explain why the "spikiness" should be associated with the hyperspheres instead of the hypercubes?
No, it would take too long. Let me summarize.
If you look at high-dimensional cubes it's pretty obvious that the corners get further and further away from the center as the dimension goes up. People (quite rightly) feel that this can be visualized as taking an ordinary 3D cube and tugging on the corners, stretching them out and thus making them "spikey".
With spheres, though, there are no corners, so people think they remain "smooth". Which they do, in a sense, but that leads to the wrong intuition. Lopping off a high-dimensional spherical cap gives you virtually no volume at all, and in our 3D world the shape that is as symmetrical as you can make it, but which when you chop it off, has almost no volume, is a spike.
Add to that these comments: http://news.ycombinator.com/item?id=3997551 and http://news.ycombinator.com/item?id=3997681
These allude to the fact that when you've on the surface of a high-dimensional sphere, almost every step takes you out. Getting into the interior is like trying to get into the interior of a spike from its tip.
Does that help?
That's why I use the term, for that shock value.
This does, yes. Starting with how this aspect changes from a circle (1-sphere) to a 2-sphere might help to get across the "spikiness" you are talking about. It seems to be basically saying the same thing as the "lopping off the cap" description, but the latter was very hard for me to visualize. Saying that a smaller and smaller fraction of the total "hypersolid angle" intersects the hypersphere's interior as you increase the dimension makes it much easier (for me) to see what's going on.
This doesn't seem to be true.
Imagine a sphere in n-dimensions, S = {Sum(x_i^2)<1}. Top of the sphere is p=(1,0,0,...,0). Now for any random direction r, if you take a very small step from p towards r, you have exactly 50% chance to be inside the sphere.
To be mathematically precise, for any r (unit vector) chosen at random, probability that there is e>0 such that p + e*r is within S is 1/2.
So half the time the steps take you out, half the time it takes you in - considering the step is small enough compared to the radius.