6 ms·

The spiky view of n-spheres is due to all of their volume getting concentrate next to the axes - the proportion of volume which is less than ε away from an axis

The spiky view of n-spheres is due to all of their volume getting concentrate next to the axes - the proportion of volume which is less than ε away from an axis to the entire volume of the sphere tends to 1 exponentially with n, which leads to the spiky visualization as we imagine 3d but with more axes.

kgwgk 5y ago
But there is an alternative visualization where the sides of the unit (hyper)cube are deformed because the location of the vertices gets further and further away from the origin. Of course neither is a true representation of the hyperdimensional case.
mgraczyk 5y ago
I don't think this is the right way to process this intuitively. When you use words like "volume getting concentrated" it sounds like there is some non-uniformity in the sphere, but the non-uniformity is really in our intuition about space. What's weird isn't the sphere, it's distance, and I think that's easier to process. Going from a (1d) sidewalk to a (2d) football field to a (3d) ocean, it's easier to see our intuitions about distances slowly breaking down.
- vba616 5y ago
  >it sounds like there is some non-uniformity in the sphere How I am understanding this is that the non-uniformity is in the cube, and I think it's very helpful in visualizing it. I can imagine in 3d, the centers of the sides of the cube being pulled in, so that it's kind of hyperbolic looking.
  - jiggawatts 5y ago
    That's not accurate either. Neither the cube nor the sphere "bend" in any way. Irrespective of the number of dimensions, the cube's surfaces remain (hyper)planar, and the sphere retains its convex shape and constant curvature. The diagram on the web page is just wrong. The enclosed sphere does not poke out with "spikes". The error in the diagram is that it shows the corner cubes "filled" with spheres, but this is not what happens in the higher-dimensional analogues. If you take a cube-shaped cross section of a 4D example (through the center), you wouldn't see the corner spheres at all. Think of a 2D slice through the middle of the 3D example: You'd just see a square, no circles!
    - vba616 5y ago
      >If you take a cube-shaped cross section of a 4D example (through the center), you wouldn't see the corner spheres at all. Wait wait wait, I thought the "cube" had the same number of dimensions as the "spheres".
      - jiggawatts 5y ago
        They're the same. So square + circle, cube + sphere, 4D hypercube + 4D hypersphere, etc... But cross-sections don't look the way you think they do. So if you draw a 3D intersection through the 4D case, it looks much like the 2D intersection of the 3D case, etc... In all such "intersection" diagrams you don't see the spheres in the corners of the cubes. The conceptual diagram on that page showing the high-dimensional case shows the corner spheres and the middle sphere somehow "squished out" in a spiky way. This is not at all what happens.
sdwr 5y ago
Love this explanation, visualizing corner vs centre is intuitive in a way higher dimensions usually aren't.
tgb 5y ago
I'm not certain I know what you mean, but it doesn't seem to work. If you mean to take epsilon neighborhoods in Euclidean norm of the axes, then I'm skeptical that that contains most of the mass of the sphere, since simulating random points on a thousand dimensional sphere doesn't seem to give points near these axes. If instead you mean most of the mass of the ball is in the points which have all but one coordinate within epsilon of zero, then the intuition doesn't follow. It's equally true that you get most of the mass considering just points with all coordinates within epsilon. And for that you get that the mass of the ball is concentrated in an epsilon cube at the origin, which excludes exactly the spikes that you were basing the intuition on. The weird thing in this thought experiment is the epsilon cube not the sphere. For example the epsilon cube contains points much further than epsilon from the origin and so it's maybe not surprising that it contains most of the sphere.
- momenti 5y ago
  https://news.ycombinator.com/item?id=3995615 https://news.ycombinator.com/item?id=3995615
  - tgb 5y ago
    Perhaps you can be more specific since I don't see the claim here being discussed there.
- trhway 5y ago
  >since simulating random points on a thousand dimensional sphere doesn't seem to give points near these axes. that is kind of circular argument as "random" really depends on the density measure underlying the chosen sampling distribution.
  - tgb 5y ago
    It's no more ill defined than volume itself (and less so since it doesn't need an arbitrary scale).
    - 5y ago
6gvONxR4sf7o 5y ago
The n-cube is spiky. The n-sphere is kinda defined by its non-spikiness, because it is rotationally invariant.

kgwgk 5y ago

But there is an alternative visualization where the sides of the unit (hyper)cube are deformed because the location of the vertices gets further and further away from the origin. Of course neither is a true representation of the hyperdimensional case.

mgraczyk 5y ago

I don't think this is the right way to process this intuitively. When you use words like "volume getting concentrated" it sounds like there is some non-uniformity in the sphere, but the non-uniformity is really in our intuition about space. What's weird isn't the sphere, it's distance, and I think that's easier to process. Going from a (1d) sidewalk to a (2d) football field to a (3d) ocean, it's easier to see our intuitions about distances slowly breaking down.

vba616 5y ago
>it sounds like there is some non-uniformity in the sphere How I am understanding this is that the non-uniformity is in the cube, and I think it's very helpful in visualizing it. I can imagine in 3d, the centers of the sides of the cube being pulled in, so that it's kind of hyperbolic looking.
- jiggawatts 5y ago
  That's not accurate either. Neither the cube nor the sphere "bend" in any way. Irrespective of the number of dimensions, the cube's surfaces remain (hyper)planar, and the sphere retains its convex shape and constant curvature. The diagram on the web page is just wrong. The enclosed sphere does not poke out with "spikes". The error in the diagram is that it shows the corner cubes "filled" with spheres, but this is not what happens in the higher-dimensional analogues. If you take a cube-shaped cross section of a 4D example (through the center), you wouldn't see the corner spheres at all. Think of a 2D slice through the middle of the 3D example: You'd just see a square, no circles!
  - vba616 5y ago
    >If you take a cube-shaped cross section of a 4D example (through the center), you wouldn't see the corner spheres at all. Wait wait wait, I thought the "cube" had the same number of dimensions as the "spheres".
    - jiggawatts 5y ago
      They're the same. So square + circle, cube + sphere, 4D hypercube + 4D hypersphere, etc... But cross-sections don't look the way you think they do. So if you draw a 3D intersection through the 4D case, it looks much like the 2D intersection of the 3D case, etc... In all such "intersection" diagrams you don't see the spheres in the corners of the cubes. The conceptual diagram on that page showing the high-dimensional case shows the corner spheres and the middle sphere somehow "squished out" in a spiky way. This is not at all what happens.

sdwr 5y ago

Love this explanation, visualizing corner vs centre is intuitive in a way higher dimensions usually aren't.

tgb 5y ago

I'm not certain I know what you mean, but it doesn't seem to work. If you mean to take epsilon neighborhoods in Euclidean norm of the axes, then I'm skeptical that that contains most of the mass of the sphere, since simulating random points on a thousand dimensional sphere doesn't seem to give points near these axes. If instead you mean most of the mass of the ball is in the points which have all but one coordinate within epsilon of zero, then the intuition doesn't follow. It's equally true that you get most of the mass considering just points with all coordinates within epsilon. And for that you get that the mass of the ball is concentrated in an epsilon cube at the origin, which excludes exactly the spikes that you were basing the intuition on. The weird thing in this thought experiment is the epsilon cube not the sphere. For example the epsilon cube contains points much further than epsilon from the origin and so it's maybe not surprising that it contains most of the sphere.

momenti 5y ago
https://news.ycombinator.com/item?id=3995615 https://news.ycombinator.com/item?id=3995615
- tgb 5y ago
  Perhaps you can be more specific since I don't see the claim here being discussed there.
trhway 5y ago
>since simulating random points on a thousand dimensional sphere doesn't seem to give points near these axes. that is kind of circular argument as "random" really depends on the density measure underlying the chosen sampling distribution.
- tgb 5y ago
  It's no more ill defined than volume itself (and less so since it doesn't need an arbitrary scale).
  - 5y ago

6gvONxR4sf7o 5y ago

The n-cube is spiky. The n-sphere is kinda defined by its non-spikiness, because it is rotationally invariant.