n-sphere

The set of points in (n + 1)-space, (x₁, x₂, ..., x_n+1), that define an n-sphere, ${\displaystyle S^{n}(r)}$, is represented by the equation:

${\displaystyle r^{2}=\sum _{i=1}^{n+1}(x_{i}-c_{i})^{2},}$

where c = (c₁, c₂, ..., c_n+1) is a center point, and r is the radius.

The above n-sphere exists in (n + 1)-dimensional Euclidean space and is an example of an n-manifold. The volume form ω of an n-sphere of radius r is given by

${\displaystyle \omega ={\frac {1}{r}}\sum _{j=1}^{n+1}(-1)^{j-1}x_{j}\,dx_{1}\wedge \cdots \wedge dx_{j-1}\wedge dx_{j+1}\wedge \cdots \wedge dx_{n+1}=*dr}$

where ∗ is the Hodge star operator; see Flanders (1989, §6.1) for a discussion and proof of this formula in the case r = 1. As a result,

${\displaystyle dr\wedge \omega =dx_{1}\wedge \cdots \wedge dx_{n+1}.}$

The space enclosed by an n-sphere is called an (n + 1)-ball. An (n + 1)-ball is closed if it includes the n-sphere, and it is open if it does not include the n-sphere.

Specifically:

• A 1-ball, a line segment, is the interior of a 0-sphere.
• A 2-ball, a disk, is the interior of a circle (1-sphere).
• A 3-ball, an ordinary ball, is the interior of a sphere (2-sphere).
• A 4-ball is the interior of a 3-sphere, etc.

Topologically, an n-sphere can be constructed as a one-point compactification of n-dimensional Euclidean space. Briefly, the n-sphere can be described as Sⁿ = ℝⁿ ∪ {∞}, which is n-dimensional Euclidean space plus a single point representing infinity in all directions. In particular, if a single point is removed from an n-sphere, it becomes homeomorphic to ℝⁿ. This forms the basis for stereographic projection.[1]

The 0-ball consists of a single point. The 0-dimensional Hausdorff measure is the number of points in a set. So,

${\displaystyle V_{0}=1.}$

The 0-sphere consists of its two end-points, {−1,1}. So,

${\displaystyle S_{0}=2.}$

The unit 1-ball is the interval [−1,1] of length 2. So,

${\displaystyle V_{1}=2.}$

The unit 1-sphere is the unit circle in the Euclidean plane, and this has circumference (1-dimensional measure)

${\displaystyle S_{1}=2\pi .}$

The region enclosed by the unit 1-sphere is the 2-ball, or unit disc, and this has area (2-dimensional measure)

${\displaystyle V_{2}=\pi .}$

Analogously, in 3-dimensional Euclidean space, the surface area (2-dimensional measure) of the unit 2-sphere is given by

${\displaystyle S_{2}=4\pi .}$

and the volume enclosed is the volume (3-dimensional measure) of the unit 3-ball, given by

${\displaystyle V_{3}={\tfrac {4}{3}}\pi .}$

The surface area, or properly the n-dimensional volume, of the n-sphere at the boundary of the (n + 1)-ball of radius R is related to the volume of the ball by the differential equation

${\displaystyle S_{n}R^{n}={\frac {dV_{n+1}R^{n+1}}{dR}}={(n+1)V_{n+1}R^{n}},}$

or, equivalently, representing the unit n-ball as a union of concentric (n − 1)-sphere shells,

${\displaystyle V_{n+1}=\int _{0}^{1}S_{n}r^{n}\,dr.}$

So,

${\displaystyle V_{n+1}={\frac {S_{n}}{n+1}}.}$

We can also represent the unit (n + 2)-sphere as a union of products of a circle (1-sphere) with an n-sphere. Let r = cos θ and r² + R² = 1, so that R = sin θ and dR = cos θ dθ. Then,

${\displaystyle {\begin{aligned}S_{n+2}&=\int _{0}^{\frac {\pi }{2}}S_{1}r\cdot S_{n}R^{n}\,d\theta \\[6pt]&=\int _{0}^{\frac {\pi }{2}}S_{1}\cdot S_{n}R^{n}\cos \theta \,d\theta \\[6pt]&=\int _{0}^{1}S_{1}\cdot S_{n}R^{n}\,dR\\[6pt]&=S_{1}\int _{0}^{1}S_{n}R^{n}\,dR\\[6pt]&=2\pi V_{n+1}.\end{aligned}}}$

Since S₁ = 2π V₀, the equation

${\displaystyle S_{n+1}=2\pi V_{n}}$

holds for all n.

This completes the derivation of the recurrences:

${\displaystyle {\begin{aligned}V_{0}&=1&V_{n+1}&={\frac {S_{n}}{n+1}}\\[6pt]S_{0}&=2&S_{n+1}&=2\pi V_{n}\end{aligned}}}$

Combining the recurrences, we see that

${\displaystyle V_{n+2}=2\pi {\frac {V_{n}}{n+2}}.}$

So it is simple to show by induction on k that,

${\displaystyle {\begin{aligned}V_{2k}&={\frac {\left(2\pi \right)^{k}}{(2k)!!}}={\frac {\pi ^{k}}{k!}}\\[6pt]V_{2k+1}&={\frac {2\left(2\pi \right)^{k}}{(2k+1)!!}}={\frac {2k!\left(4\pi \right)^{k}}{(2k+1)!}}\end{aligned}}}$

where !! denotes the double factorial, defined for odd natural numbers 2k + 1 by (2k + 1)!! = 1 × 3 × 5 × ... × (2k − 1) × (2k + 1) and similarly for even numbers (2k)!! = 2 × 4 × 6 × ... × (2k − 2) × (2k).

In general, the volume, in n-dimensional Euclidean space, of the unit n-ball, is given by

${\displaystyle V_{n}={\frac {\pi ^{\frac {n}{2}}}{\Gamma \left({\frac {n}{2}}+1\right)}}={\frac {\pi ^{\frac {n}{2}}}{\left({\frac {n}{2}}\right)!}}}$

where Γ is the gamma function, which satisfies Γ(1/2) = √π, Γ(1) = 1, and Γ(x + 1) = xΓ(x), and so Γ(x + 1) = x!, and where we conversely define x! = Γ(x + 1) for every x.

By multiplying V_n by Rⁿ, differentiating with respect to R, and then setting R = 1, we get the closed form

${\displaystyle S_{n-1}={\frac {n\pi ^{\frac {n}{2}}}{\Gamma \left({\frac {n}{2}}+1\right)}}={\frac {2\pi ^{\frac {n}{2}}}{\Gamma \left({\frac {n}{2}}\right)}}.}$

for the (n − 1)-dimensional surface of the sphere Sⁿ⁻¹.

The recurrences can be combined to give a "reverse-direction" recurrence relation for surface area, as depicted in the diagram:

${\displaystyle S_{n-1}={\frac {n}{2\pi }}S_{n+1}}$

n refers to the dimension of the ambient Euclidean space, which is also the intrinsic dimension of the solid whose volume is listed here, but which is 1 more than the intrinsic dimension of the sphere whose surface area is listed here. The curved red arrows show the relationship between formulas for different n. The formula coefficient at each arrow's tip equals the formula coefficient at that arrow's tail times the factor in the arrowhead (where the n in the arrowhead refers to the n value that the arrowhead points to). If the direction of the bottom arrows were reversed, their arrowheads would say to multiply by 2π/n − 2. Alternatively said, the surface area S_n+1 of the sphere in n + 2 dimensions is exactly 2πR times the volume V_n enclosed by the sphere in n dimensions.

Index-shifting n to n − 2 then yields the recurrence relations:

${\displaystyle {\begin{aligned}V_{n}&={\frac {2\pi }{n}}V_{n-2}\\[6pt]S_{n-1}&={\frac {2\pi }{n-2}}S_{n-3}\end{aligned}}}$

where S₀ = 2, V₁ = 2, S₁ = 2π and V₂ = π.

The recurrence relation for V_n can also be proved via integration with 2-dimensional polar coordinates:

${\displaystyle {\begin{aligned}V_{n}&=\int _{0}^{1}\int _{0}^{2\pi }V_{n-2}\left({\sqrt {1-r^{2}}}\right)^{n-2}\,r\,d\theta \,dr\\[6pt]&=\int _{0}^{1}\int _{0}^{2\pi }V_{n-2}\left(1-r^{2}\right)^{{\frac {n}{2}}-1}\,r\,d\theta \,dr\\[6pt]&=2\pi V_{n-2}\int _{0}^{1}\left(1-r^{2}\right)^{{\frac {n}{2}}-1}\,r\,dr\\[6pt]&=2\pi V_{n-2}\left[-{\frac {1}{n}}\left(1-r^{2}\right)^{\frac {n}{2}}\right]_{r=0}^{r=1}\\[6pt]&=2\pi V_{n-2}{\frac {1}{n}}={\frac {2\pi }{n}}V_{n-2}.\end{aligned}}}$

To express the volume element of n-dimensional Euclidean space in terms of spherical coordinates, first observe that the Jacobian matrix of the transformation is:

${\displaystyle J_{n}={\begin{pmatrix}\cos(\varphi _{1})&-r\sin(\varphi _{1})&0&0&\cdots &0\\\sin(\varphi _{1})\cos(\varphi _{2})&r\cos(\varphi _{1})\cos(\varphi _{2})&-r\sin(\varphi _{1})\sin(\varphi _{2})&0&\cdots &0\\\vdots &\vdots &\vdots &&\ddots &\vdots \\&&&&&0\\\sin(\varphi _{1})\cdots \sin(\varphi _{n-2})\cos(\varphi _{n-1})&\cdots &\cdots &&&-r\sin(\varphi _{1})\cdots \sin(\varphi _{n-2})\sin(\varphi _{n-1})\\\sin(\varphi _{1})\cdots \sin(\varphi _{n-2})\sin(\varphi _{n-1})&r\cos(\varphi _{1})\cdots \sin(\varphi _{n-1})&\cdots &&&r\sin(\varphi _{1})\cdots \sin(\varphi _{n-2})\cos(\varphi _{n-1})\end{pmatrix}}.}$

The determinant of this matrix can be calculated by induction. When n = 2, a straightforward computation shows that the determinant is r. For larger n, observe that J_n can be constructed from J_{n − 1} as follows. Except in column n, rows n − 1 and n of J_n are the same as row n − 1 of J_{n − 1}, but multiplied by an extra factor of cos φ_{n − 1} in row n − 1 and an extra factor of sin φ_{n − 1} in row n. In column n, rows n − 1 and n of J_n are the same as column n − 1 of row n − 1 of J_{n − 1}, but multiplied by extra factors of sin φ_{n − 1} in row n − 1 and cos φ_{n − 1} in row n, respectively. The determinant of J_n can be calculated by Laplace expansion in the final column. By the recursive description of J_n, the submatrix formed by deleting the entry at (n − 1, n) and its row and column almost equals J_{n − 1}, except that its last row is multiplied by sin φ_{n − 1}. Similarly, the submatrix formed by deleting the entry at (n, n) and its row and column almost equals J_{n − 1}, except that its last row is multiplied by cos φ_{n − 1}. Therefore the determinant of J_n is

${\displaystyle {\begin{aligned}|J_{n}|&=(-1)^{(n-1)+n}(-r\sin(\varphi _{1})\dotsm \sin(\varphi _{n-2})\sin(\varphi _{n-1}))(\sin(\varphi _{n-1})|J_{n-1}|)\\&\qquad {}+(-1)^{n+n}(r\sin(\varphi _{1})\dotsm \sin(\varphi _{n-2})\cos(\varphi _{n-1}))(\cos(\varphi _{n-1})|J_{n-1}|)\\&=(r\sin(\varphi _{1})\dotsm \sin(\varphi _{n-2})|J_{n-1}|(\sin ^{2}(\varphi _{n-1})+\cos ^{2}(\varphi _{n-1}))\\&=(r\sin(\varphi _{1})\dotsm \sin(\varphi _{n-2}))|J_{n-1}|.\end{aligned}}}$

Induction then gives a closed-form expression for the volume element in spherical coordinates

${\displaystyle {\begin{aligned}d^{n}V&=\left|\det {\frac {\partial (x_{i})}{\partial \left(r,\varphi _{j}\right)}}\right|dr\,d\varphi _{1}\,d\varphi _{2}\cdots d\varphi _{n-1}\\&=r^{n-1}\sin ^{n-2}(\varphi _{1})\sin ^{n-3}(\varphi _{2})\cdots \sin(\varphi _{n-2})\,dr\,d\varphi _{1}\,d\varphi _{2}\cdots d\varphi _{n-1}.\end{aligned}}}$

The formula for the volume of the n-ball can be derived from this by integration.

Similarly the surface area element of the (n − 1)-sphere of radius R, which generalizes the area element of the 2-sphere, is given by

${\displaystyle d_{S^{n-1}}V=R^{n-1}\sin ^{n-2}(\varphi _{1})\sin ^{n-3}(\varphi _{2})\cdots \sin(\varphi _{n-2})\,d\varphi _{1}\,d\varphi _{2}\cdots d\varphi _{n-1}.}$

The natural choice of an orthogonal basis over the angular coordinates is a product of ultraspherical polynomials,

${\displaystyle {\begin{aligned}&{}\quad \int _{0}^{\pi }\sin ^{n-j-1}\left(\varphi _{j}\right)C_{s}^{\left({\frac {n-j-1}{2}}\right)}\cos \left(\varphi _{j}\right)C_{s'}^{\left({\frac {n-j-1}{2}}\right)}\cos \left(\varphi _{j}\right)\,d\varphi _{j}\\[6pt]&={\frac {2^{3-n+j}\pi \Gamma (s+n-j-1)}{s!(2s+n-j-1)\Gamma ^{2}\left({\frac {n-j-1}{2}}\right)}}\delta _{s,s'}\end{aligned}}}$

for j = 1, 2,... n − 2, and the e^isφ_j for the angle j = n − 1 in concordance with the spherical harmonics.

The standard spherical coordinate system arises from writing ℝⁿ as the product ℝ × ℝ^{n − 1}. These two factors may be related using polar coordinates. For each point x of ℝⁿ, the standard Cartesian coordinates

${\displaystyle \mathbf {x} =(x_{1},\dots ,x_{n})=(y_{1},z_{1},\dots ,z_{n-1})=(y_{1},\mathbf {z} )}$

can be transformed into a mixed polar–Cartesian coordinate system:

${\displaystyle \mathbf {x} =(r\sin \theta ,(r\cos \theta ){\hat {\mathbf {z} }}).}$

This says that points in ℝⁿ may be expressed by taking the ray starting at the origin and passing through ${\displaystyle {\hat {\mathbf {z} }}=\mathbf {z} /\lVert \mathbf {z} \rVert \in S^{n-2}}$, rotating it towards ${\displaystyle (1,0,\dots ,0)}$ by ${\displaystyle \theta =\arcsin y_{1}/r}$, and traveling a distance ${\displaystyle r=\lVert \mathbf {x} \rVert }$ along the ray. Repeating this decomposition eventually leads to the standard spherical coordinate system.

Polyspherical coordinate systems arise from a generalization of this construction.[5] The space ℝⁿ is split as the product of two Euclidean spaces of smaller dimension, but neither space is required to be a line. Specifically, suppose that p and q are positive integers such that n = p + q. Then ℝⁿ = ℝ^p × ℝ^q. Using this decomposition, a point x ∈ ℝⁿ may be written as

${\displaystyle \mathbf {x} =(x_{1},\dots ,x_{n})=(y_{1},\dots ,y_{p},z_{1},\dots ,z_{q})=(\mathbf {y} ,\mathbf {z} ).}$

This can be transformed into a mixed polar–Cartesian coordinate system by writing:

${\displaystyle \mathbf {x} =((r\sin \theta ){\hat {\mathbf {y} }},(r\cos \theta ){\hat {\mathbf {z} }}).}$

Here ${\displaystyle {\hat {\mathbf {y} }}}$ and ${\displaystyle {\hat {\mathbf {z} }}}$ are the unit vectors associated to y and z. This expresses x in terms of ${\displaystyle {\hat {\mathbf {y} }}\in S^{p-1}}$, ${\displaystyle {\hat {\mathbf {z} }}\in S^{q-1}}$, r ≥ 0, and an angle θ. It can be shown that the domain of θ is [0, 2π) if p = q = 1, [0, π] if exactly one of p and q is 1, and [0, π/2] if neither p nor q are 1. The inverse transformation is

${\displaystyle {\begin{aligned}r&=\lVert \mathbf {x} \rVert ,\\\theta &=\arcsin(\lVert \mathbf {y} \rVert /\lVert \mathbf {x} \rVert )\\&=\arccos(\lVert \mathbf {z} \rVert /\lVert \mathbf {x} \rVert )\\&=\arctan(\lVert \mathbf {y} \rVert /\lVert \mathbf {z} \rVert ).\end{aligned}}}$

These splittings may be repeated as long as one of the factors involved has dimension two or greater. A polyspherical coordinate system is the result of repeating these splittings until there are no Cartesian coordinates left. Splittings after the first do not require a radial coordinate because the domains of ${\displaystyle {\hat {\mathbf {y} }}}$ and ${\displaystyle {\hat {\mathbf {z} }}}$ are spheres, so the coordinates of a polyspherical coordinate system are a non-negative radius and n − 1 angles. The possible polyspherical coordinate systems correspond to binary trees with n leaves. Each non-leaf node in the tree corresponds to a splitting and determines an angular coordinate. For instance, the root of the tree represents ℝⁿ, and its immediate children represent the first splitting into ℝ^p and ℝ^q. Leaf nodes correspond to Cartesian coordinates for S^{n − 1}. The formulas for converting from polyspherical coordinates to Cartesian coordinates may be determined by finding the paths from the root to the leaf nodes. These formulas are products with one factor for each branch taken by the path. For a node whose corresponding angular coordinate is θ_i, taking the left branch introduces a factor of sin θ_i and taking the right branch introduces a factor of cos θ_i. The inverse transformation, from polyspherical coordinates to Cartesian coordinates, is determined by grouping nodes. Every pair of nodes having a common parent can be converted from a mixed polar–Cartesian coordinate system to a Cartesian coordinate system using the above formulas for a splitting.

Polyspherical coordinates also have an interpretation in terms of the special orthogonal group. A splitting ℝⁿ = ℝ^p × ℝ^q determines a subgroup

${\displaystyle \operatorname {SO} _{p}(\mathbb {R} )\times \operatorname {SO} _{q}(\mathbb {R} )\subseteq \operatorname {SO} _{n}(\mathbb {R} ).}$

This is the subgroup that leaves each of the two factors ${\displaystyle S^{p-1}\times S^{q-1}\subseteq S^{n-1}}$ fixed. Choosing a set of coset representatives for the quotient is the same as choosing representative angles for this step of the polyspherical coordinate decomposition.

In polyspherical coordinates, the volume measure on ℝⁿ and the area measure on S^{n − 1} are products. There is one factor for each angle, and the volume measure on ℝⁿ also has a factor for the radial coordinate. The area measure has the form:

${\displaystyle dA_{n-1}=\prod _{i=1}^{n-1}F_{i}(\theta _{i})\,d\theta _{i},}$

where the factors F_i are determined by the tree. Similarly, the volume measure is

${\displaystyle dV_{n}=r^{n-1}\,dr\,\prod _{i=1}^{n-1}F_{i}(\theta _{i})\,d\theta _{i}.}$

Suppose we have a node of the tree that corresponds to the decomposition ℝ^{n₁ + n₂} = ℝ^n₁ × ℝ^n₂ and that has angular coordinate θ. The corresponding factor F depends on the values of n₁ and n₂. When the area measure is normalized so that the area of the sphere is 1, these factors are as follows. If n₁ = n₂ = 1, then

${\displaystyle F(\theta )={\frac {d\theta }{2\pi }}.}$

If n₁ > 1 and n₂ = 1, and if B denotes the beta function, then

${\displaystyle F(\theta )={\frac {\sin ^{n_{1}-1}\theta }{\mathrm {B} ({\frac {n_{1}}{2}},{\frac {1}{2}})}}\,d\theta .}$

If n₁ = 1 and n₂ > 1, then

${\displaystyle F(\theta )={\frac {\cos ^{n_{2}-1}\theta }{\mathrm {B} ({\frac {1}{2}},{\frac {n_{2}}{2}})}}\,d\theta .}$

Finally, if both n₁ and n₂ are greater than one, then

${\displaystyle F(\theta )={\frac {(\sin ^{n_{1}-1}\theta )(\cos ^{n_{2}-1}\theta )}{{\frac {1}{2}}\mathrm {B} ({\frac {n_{1}}{2}},{\frac {n_{2}}{2}})}}\,d\theta .}$

A set of uniformly distributed points on the surface of a unit 2-sphere generated using Marsaglia's algorithm.

To generate uniformly distributed random points on the unit (n − 1)-sphere (that is, the surface of the unit n-ball), Marsaglia (1972) gives the following algorithm.

Generate an n-dimensional vector of normal deviates (it suffices to use N(0, 1), although in fact the choice of the variance is arbitrary), x = (x₁, x₂,... x_n). Now calculate the "radius" of this point:

${\displaystyle r={\sqrt {x_{1}^{2}+x_{2}^{2}+\cdots +x_{n}^{2}}}.}$

The vector 1/rx is uniformly distributed over the surface of the unit n-ball.

An alternative given by Marsaglia is to uniformly randomly select a point x = (x₁, x₂,... x_n) in the unit n-cube by sampling each x_i independently from the uniform distribution over (–1,1), computing r as above, and rejecting the point and resampling if r ≥ 1 (i.e., if the point is not in the n-ball), and when a point in the ball is obtained scaling it up to the spherical surface by the factor 1/r; then again 1/rx is uniformly distributed over the surface of the unit n-ball. This method becomes very inefficient for higher dimensions, as a vanishingly small fraction of the unit cube is contained in the sphere. In ten dimensions, less than 2% of the cube is filled by the sphere, so that typically more than 50 attempts will be needed. In seventy dimensions, less than ${\displaystyle 10^{-24}}$ of the cube is filled, meaning typically a trillion quadrillion trials will be needed, far more than a computer could ever carry out.

With a point selected uniformly at random from the surface of the unit (n − 1)-sphere (e.g., by using Marsaglia's algorithm), one needs only a radius to obtain a point uniformly at random from within the unit n-ball. If u is a number generated uniformly at random from the interval [0, 1] and x is a point selected uniformly at random from the unit (n − 1)-sphere, then u^1⁄nx is uniformly distributed within the unit n-ball.

Alternatively, points may be sampled uniformly from within the unit n-ball by a reduction from the unit (n + 1)-sphere. In particular, if (x₁,x₂,...,x_n+2) is a point selected uniformly from the unit (n + 1)-sphere, then (x₁,x₂,...,x_n) is uniformly distributed within the unit n-ball (i.e., by simply discarding two coordinates).[6]

If n is sufficiently large, most of the volume of the n-ball will be contained in the region very close to its surface, so a point selected from that volume will also probably be close to the surface. This is one of the phenomena leading to the so-called curse of dimensionality that arises in some numerical and other applications.