# Reasoning in Higher Dimensions: Hyperspheres

In my last post on higher dimensions, I alluded to the fact that I don’t agree completely with certain notions about higher dimensions. Specifically, I disagree with the idea that the intuition that you take for granted in low dimensions is necessarily ill-equipped to serve you in higher dimensions. Low-dimensional intuition is ill-equipped for many problems, and like most other topics in math, it’s usually most sensible to do the calculations anyway.

Hyperspheres often get brought up with the subject of weirdness in higher dimensions, mostly because they’re easy to understand, and it’s easy to demonstrate the weirdness very quickly. But are they completely weird? Are the examples really fair, or are hyperspheres getting a bad rap?

First, let’s get some notation out of the way. We often like to call a hypersphere an n-sphere, because it’s an n-dimensional manifold. Technically, one of these can exist in any metric space with more than n dimensions (because I’m talking about intuition, I’m assuming it’s Euclidean space). For simplicity, though, we’ll say that it lives in n+1 space, so that we can define it easily: $S^n = \left\{ x \in \mathbb{R}^{n+1} : \|x\| = r\right\}$

That’s not all that I want to talk about, though. I also want to talk about the volume of the n-sphere, and in that case, we often talk about a ball, which is just the interior of a sphere. The interior of an n-sphere is an (n+1)-ball, because if the sphere is an n-dimensional manifold, its interior is an (n+1)-dimensional manifold: $B^{n+1} = \left\{ x \in \mathbb{R}^{n+1} : \|x\| < r\right\}$

Or more simply: $B^{n} = \left\{ x \in \mathbb{R}^{n} : \|x\| < r\right\}$

The volume of this object has a somewhat simple formula: $V_n={\pi^\frac{n}{2}r^n\over\Gamma(\frac{n}{2} + 1)}$

Where Γ(x) represents the Gamma function (which is a tad more complicated).

So where’s the counter-intuition? Say that we took the unit ball for all n > 0 and graphed its volume: Volume vs. dimension of unit n-ball

This does seem a little odd. The volume goes up, hits a peak at 5, and then drops, and eventually bottoms out. In fact, with high enough dimension, you won’t see an n-ball have any volume at all. The limit of the volume of any n-ball as n goes to infinity is 0. That is weird. That’s not necessarily something that you’d expect. It also seems weird that the volume starts dropping after a while.

But is all this really that strange? What if we fixed the radius at, say, 1/sqrt(π)? The volume vs. dimension is then just a decreasing function, even at low dimensions. Not surprising when you consider that radii less than 1 should make the volume diminish rapidly. So what about radii greater than 1? What if we fix r at say, 3? The volume peaks out at n = 56, and the volume is about 143 billion … somethings. After that, the volume diminishes back to zero again. All that we’re really saying here is that the geometry of the sphere dominates rn, but rn has enough power to dominate at low dimensions until the geometry cuts over.

What’s so special about rn though? Why is this the gold standard by which we judge the hypersphere? It’s just the hypercube with sides of length r. In fact, the unit sphere is inscribed in a cube with sides of length 2r. What if we considered a hypercube of circumradius r instead of inradius r? That means that a sphere of radius r contains it. If that’s the case, then it has volume strictly less than the sphere’s volume. In fact, its volume is: $V_n=\left(2r\over\sqrt n\right)^n = {(2r)^n\over n^{n\over2}}$

which diminishes even faster than the sphere’s volume. So it can’t be the geometry of a cube that makes it keep its volumetric power.

So what’s my point? This is all sounding very counterintuitive. My point is that when you talk about counter-intuition in higher dimensions, it’s helpful to talk about what’s actually going on, instead of maligning poor innocent constructs like the hypersphere. What’s actually going on? More about that later.

But for now, consider this: no matter how many dimensions a sphere has, it’s always perfectly round, and perfectly isotropic. That’s intuition that isn’t lost in higher dimensions.

[Someone posted this to Reddit! Thanks!]

### 23 responses to “Reasoning in Higher Dimensions: Hyperspheres”

1. John

One place I’ve found it important to think about volumes of spheres versus cubes is high-dimensional integration. If you’re not aware of the phenomena you describe, you might be puzzled why your integral estimates are bad. (Or worse, unaware that your estimates are bad.) You might be allocating nearly all of your integration nodes outside the region you care about without realizing it.

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3. William V

You should probably talk about the LP norm not n-dimensional volume. n-dimensional integrals dont preserve the volume of similar rectangular regions.

• John Moeller

I don’t follow. Can you elaborate?

• John Moeller

Ah. I think you may be the same person who commented on the Reddit thread. I’m not actually talking about infinite-dimensional spaces. I’m talking about volumes that live in spaces of bounded dimension. So I do want the n-dimensional volume, or some related measure.

If I were talking about functions, then yes, I’d want to use some kind of Lp norm.

4. William V

Well the volume is a function. But what I meant is you should normalize. You should look at it like this. A hypersquare should have some normalized ‘volume’ that is equal across all dimensions so that comparisons can be made. Otherwise comparing anything between dimensions is apples and oranges.

I mean it’s not really all that suprising that the volume of the sphere goes down at all.

My comment about the infinite dimensional thing is that when comparing the value amongst these dimensions it should be normalized so that the limit has a meaning. Like how L_infty = lim LP.

Otherwise everything hypershape that isn’t the hypercube with sidelength 1 (maybe not everyone but thats my conjecture) will either become infinite volume or 0 volume as the dimension goes to infinity.

• John Moeller

That’s part of my point. You should also see my other post on the topic:

https://ontopo.wordpress.com/2009/03/10/reasoning-in-higher-dimensions-measure/

Here I talk about normalizing the measure so that a unit hypersphere has unit volume. If you don’t, you’re right; the comparison loses meaning.

I’m not convinced that you want anything like the L_p norm, though. If you want to extend the idea of a bounded-dimensional hypersphere to some kind of “infinite-dimensional” hypersphere in the same context, I think that you’d want something like the l_p norm instead (the difference being that l_p applies to sequences and L_p applies to functions).

Additionally, you’d want to limit yourself to l_2, since we’re talking about the Euclidean metric. In fact, you could think of a vector in Euclidean space as a finite sequence. Taking the l_2 norm is the same as taking its Euclidean norm.

So if you normalize your measure by making a unit hypersphere have unit volume, then you’d accomplish what you’re looking for here. Your “ball” would be all sequences such that the squares of their elements sum to less or equal 1.

• John Moeller

I went on and on down this path without realizing that I wasn’t staying in my original context.

Even if you do this, you aren’t talking about the *volume* of such a ball. You’re just *defining* the ball. The volume of a ball of bounded dimension is something that we can define. I don’t know if it makes sense to talk about the volume of an infinite-dimensional ball.

To sum up: the norm is not the volume.

5. William V

In turns of normalizing i meant.

V’_n = sqrt^(1/n) V_n

It’s like the L_P or l_p (yep i get em mixed up) in how its defined

V’_n would have the same units for all n and would thus be something you could compare directly. Furthermore a square with side length m would have V’_n = m. Which is something i would consider important.

Ultimately when you say the volume V goes to zero as n -> \infty this is innacurate as you can not make ordered comparisons of differing units.

That said, i (think) understand what you are talking about as well. Ill go read the other article too.

• John Moeller

Yes, I absolutely agree that comparing unit^2’s and unit^5’s makes no sense. That’s part of the reason that the situation is unintuitive. Once you remove that distraction, things make more sense. This is one of the things I try to communicate in my follow-up post.

• John Moeller

And thanks for stopping in and reading. It’s been an interesting conversation.

6. Newbie

This is probably a really dumb question but what is wrong with this reasoning?
The n-unit ball is given by x_1^2+…+x_n^2 <= 1
Therefore it must contain all all points of the form
(x_1,x_2,…,x_{n-1},0} s.t. x_1^2+…+x_{n-1}^2 = vol(n-1 ball).

I recognize that there is something wrong with this reasoning but what is it?

Thanks!

7. Newbie

Oops my comment got mangled….here’s what i mean to ask:

This is probably a really dumb question but what is wrong with this reasoning?
The n-unit ball is given by x_1^2+…+x_n^2 <= 1

Therefore it must contain all all points of the form
(x_1,x_2,…,x_{n-1},0) s.t. x_1^2+…+x_{n-1}^2 = vol(n-1) ball.

I recognize that there is something wrong with this reasoning but what is it?

Thanks!

8. John Moeller

You’re essentially suggesting setting $x_n = 0$. If you do this, all you get is a subset of the original ball such that: $x_1^2+\dots+x_{n-1}^2+0 <= 1$

But this is just an (n-1)-ball: $x_1^2+\dots+x_{n-1}^2 <= 1$

All you’re really doing is taking a “slice” through the n-ball. If you do this with a 3-ball (which is just a ball) you get a 2-ball. I.e., a disk.

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