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247 points nabla9 | 9 comments | | HN request time: 0.814s | source | bottom
1. crazygringo ◴[] No.41837962[source]
> The volume of the unit d-sphere goes to 0 as d grows! A high dimensional unit sphere encloses almost no volume!

This feels misleading to me.

Directly comparing volumes in different dimensions doesn't make any sense because the units are different. It doesn't make sense to say that a quantity in m^3 is larger or smaller than a quantity in m^4. Because it doesn't make any sense to compare the area of a circle with the volume of a sphere.

> More accurate pictorial representations of high dimensional cubes (left) and spheres (right).

The cube one is arguably accurate -- e.g. in 100 dimensions, if the distance from the center of a cube to the center of a face is 1, then the distance from the center of the cube to a corner is 10.

But the sphere one, I don't know. Every point on a 100-dimensional sphere is still the same distance away from its center. The sphere is staying spherical in an intuitive way, it's just that the corners of the enclosing cube have gotten so much further away.

So what is accurate to say is that the proportion of volume of a sphere relative to that of its bounding cube keeps decreasing. Which, rather than being supposedly "counterintuitive", makes perfect intuitive sense -- because every time you add a dimension, you can think of it as "extruding" the previous sphere into the new dimension and then shaving it round, the way a 2D circle can be extruded into a cylinder in 3D and then shaved down to make it into a sphere. Every time you add a dimension, you shave off more.

The article suggests that a 3D sphere has greater volume than a 2D circle -- with a unit radius, the sphere is 4/3π while the circle is just π. But again, they're in different units, so it's a meaningless statement. It makes much more sense to say that a 2D circle takes up (1/4)π≈0.79 of its bounding square, a 3D sphere takes of (1/6)π≈0.52 of its bounding cube, a 4D sphere takes up (π/32)π≈=0.31, and so forth. So no, the volume doesn't go up and then down -- it just goes down every time when taken as a unitless proportion (and proportions are comparable).

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2. jvanderbot ◴[] No.41839254[source]
Your image of extruding a cylinder in higher dimensions inside its bounding box then rounding it off was really insightful as a teaching tool. I've always struggled to visualize these "counterintuitive" results, which are only counterintuitive because they are harder to visualize or seem to "change" after D=3. But now they don't. Thanks!
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3. ricksunny ◴[] No.41839299[source]
Hypercubist Math has the ambitious goal of imbuing an intuitive sense of 4-dimensions to people. Currently at Vol. 2 of a several-volume series, will be interssting whether they succeed or not if user feedback is anything to go by.

https://m.youtube.com/watch?v=SwGbHsBAcZ0&t=509s&pp=ygUQaHlw...

4. yatopifo ◴[] No.41839531[source]
I think it all depends on how you define hypervolume. If you say it’s a positive real number constructed by means of integration, then you can certainly compare them across objects of various dimensions. When you say “units” I immediately think of stuff like bivectors and trivectors where you can’t reduce one to another without losing important geometric properties. But here we are talking about just the scalar part which is as “unitless” as can be.
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5. ◴[] No.41839606[source]
6. ◴[] No.41840481[source]
7. jsenn ◴[] No.41841134[source]
I don't think this is right. If you're worried about units you can calculate the (generalized) surface area to volume ratio, which turns out to be exactly D/r. In other words, as D increases, the ratio goes to infinity.

I think this fact can fairly be interpreted to mean that a high-dimensional unit sphere encloses almost no volume. The 2D cartoon drawing of a hypersphere also helps capture this: you can imagine the "spikes" stretching out and squeezing the interior portion, until it's all outside and no inside.

EDIT: another argument I've seen involves calculating the ratio of the volume of a thin shell surrounding the n-sphere's surface to its total volume. You can prove that the limit of the ratio as the dimension goes to infinity is 1. In other words, in high dimensions almost all of the volume of the sphere is concentrated near its surface.

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8. aatd86 ◴[] No.41841789[source]
Another simplistic way to see it is that it is a ratio of contained information. In higher dimensional spaces, the space is so big that below the unit spheres contain exponentially less information.

It's just something between 0 and 1 exponentized to d where d is the dimension after all (i.e. the number of eigenvectors).

d is an exponential scale factor in a sense.

9. shwouchk ◴[] No.41853924[source]
Integration when extrapolated to many dimensions has many nuances, and be careful that you don't have a circular definition of hypervolume in terms of integration.

For a simple example of difficulties consider comparing the volume of two distinct k-unit spheres embedded in R^n where n>k.