5 ms·

I don't think it's correct to see the n-sphere as poking out of the sides of the cube; the sphere has constant curvature so it has no protrusions. Rather, the n

I don't think it's correct to see the n-sphere as poking out of the sides of the cube; the sphere has constant curvature so it has no protrusions. Rather, the n-cube gets more and more spiky so the constraining spheres get (relatively) smaller and further away from the centre, so constrain the central sphere less and less. The spikiness of n-cubes is apparent when you look at the (solid) angle at each vertex; it starts constant (1/2 pi radian in 2D, 1/2 pi steradian in 3D) but reduces thereafter at an increasing rate (1/8 pi^2, 1/12 pi^2, 1/64 pi^3, etc.).

shenberg 5y ago
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.
scythe 5y ago
The radius of the sphere, assuming the cube has side length 1, is (sqrt(D)-1)/2. It seems worthy of note that this radius becomes arbitrarily large.
- kgwgk 5y ago
  And the distance from a vertex of the unit cube to the center is sqrt(D)/2, which also becomes arbitrarily large. https://www.math.ucdavis.edu/~strohmer/courses/180BigData/180lecture1.pdf https://www.math.ucdavis.edu/~strohmer/courses/180BigData/18... > The result, when projected in two dimensions no longer appears convex, however all hypercubes are convex. This is part of the strangeness of higher dimensions - hypercubes are both convex and “pointy.”
  - AnimalMuppet 5y ago
    Isn't it more accurate to say that hypercubes are convex and their projections are pointy?
    - deleted 5y ago
      [deleted]
    - kgwgk 5y ago
      Yes. Hence the quotes, I guess. But hyperspheres are convex and their projections are also convex. So it's not really appropriate either to say that they are "spiky" as in https://news.ycombinator.com/item?id=29969613 https://news.ycombinator.com/item?id=29969613
thaumasiotes 5y ago
> the (solid) angle at each vertex; it starts constant (1/2 pi radian in 2D, 1/2 pi steradian in 3D) Unstudied curiosity: steradians seem to be defined in terms of the point of a cone. But the corner of a cube is the intersection of several (for a cube, 3) planes, not a cone. How do you do the calculation of steradians?
- wlesieutre 5y ago
  Cone is one scenario to think about, but what it really comes down to is the fraction of the full sphere that your solid angle includes. To make a 2D analogy, you can think of 2D angles as a portion of a circle represented as a fraction of 2pi radians. Cut it in half and you have pi radians, cut that in half (so a quarter of the circle) and you have pi/2 radians, or 90 degrees. That 90 degrees "quarter of a circle" example is looking at it as the "point of a pie slice", but it's the same 90 degree 2D angle as you have in the corner of a square. You can look at the corner of a cube the same way. The full sphere of solid angle is 4pi, a hemisphere is 2pi. Now take that hemisphere and cut it into quarters (1/8ths of a sphere). Each of those quarters is pi/2 steradians, and the solid angle at the center is the same solid angle represented by 1/8th of the sphere is the same solid angle you have at the corner of a cube. Or to put it another way, you could pack the corners of 8 cubes around a point and it would leave no empty gaps, so the corner of each cube is occupying 1/8th of a "full" 4pi steradians.
  - thaumasiotes 5y ago
    Thanks, that was helpful in thinking about it.
  - thaumasiotes 5y ago
    Now that I've thought about it, I don't think we're on solid ground measuring pointiness in n-dimensional radians. A perfectly non-pointy angle in 2D is half the circle, or π radians. A perfectly non-pointy angle in 3D is half the sphere, or 2π steradians. π is less than 2π, but the straight-line 2D angle is not sharper than the flat-plane 3D angle. What's relevant to the pointiness is the proportion of the sphere, not the numeric value of the surface area.

shenberg 5y ago

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.

scythe 5y ago

The radius of the sphere, assuming the cube has side length 1, is (sqrt(D)-1)/2. It seems worthy of note that this radius becomes arbitrarily large.

kgwgk 5y ago
And the distance from a vertex of the unit cube to the center is sqrt(D)/2, which also becomes arbitrarily large. https://www.math.ucdavis.edu/~strohmer/courses/180BigData/180lecture1.pdf https://www.math.ucdavis.edu/~strohmer/courses/180BigData/18... > The result, when projected in two dimensions no longer appears convex, however all hypercubes are convex. This is part of the strangeness of higher dimensions - hypercubes are both convex and “pointy.”
- AnimalMuppet 5y ago
  Isn't it more accurate to say that hypercubes are convex and their projections are pointy?
  - deleted 5y ago
    [deleted]
  - kgwgk 5y ago
    Yes. Hence the quotes, I guess. But hyperspheres are convex and their projections are also convex. So it's not really appropriate either to say that they are "spiky" as in https://news.ycombinator.com/item?id=29969613 https://news.ycombinator.com/item?id=29969613

thaumasiotes 5y ago

> the (solid) angle at each vertex; it starts constant (1/2 pi radian in 2D, 1/2 pi steradian in 3D) Unstudied curiosity: steradians seem to be defined in terms of the point of a cone. But the corner of a cube is the intersection of several (for a cube, 3) planes, not a cone. How do you do the calculation of steradians?

wlesieutre 5y ago
Cone is one scenario to think about, but what it really comes down to is the fraction of the full sphere that your solid angle includes. To make a 2D analogy, you can think of 2D angles as a portion of a circle represented as a fraction of 2pi radians. Cut it in half and you have pi radians, cut that in half (so a quarter of the circle) and you have pi/2 radians, or 90 degrees. That 90 degrees "quarter of a circle" example is looking at it as the "point of a pie slice", but it's the same 90 degree 2D angle as you have in the corner of a square. You can look at the corner of a cube the same way. The full sphere of solid angle is 4pi, a hemisphere is 2pi. Now take that hemisphere and cut it into quarters (1/8ths of a sphere). Each of those quarters is pi/2 steradians, and the solid angle at the center is the same solid angle represented by 1/8th of the sphere is the same solid angle you have at the corner of a cube. Or to put it another way, you could pack the corners of 8 cubes around a point and it would leave no empty gaps, so the corner of each cube is occupying 1/8th of a "full" 4pi steradians.
- thaumasiotes 5y ago
  Thanks, that was helpful in thinking about it.
- thaumasiotes 5y ago
  Now that I've thought about it, I don't think we're on solid ground measuring pointiness in n-dimensional radians. A perfectly non-pointy angle in 2D is half the circle, or π radians. A perfectly non-pointy angle in 3D is half the sphere, or 2π steradians. π is less than 2π, but the straight-line 2D angle is not sharper than the flat-plane 3D angle. What's relevant to the pointiness is the proportion of the sphere, not the numeric value of the surface area.