n维球面上的点分布：特殊情形下的闭式解探究

Great question! You’re absolutely right that there’s no universal closed-form solution for maximizing the separation of k points on an n-sphere $S^n$. But when the dimension n is a power of two ($n=2^i$), we have well-known, explicit constructions for the k values you’re interested in: $2^i, 2^{i+1}, 2^{i+2}, \dots$ Let’s break this down case by case:

1. When $k=2^i$ (k equals the dimension n) #

The cleanest construction here uses Hadamard matrices:

• Since n is a power of two, we can always build an n×n Hadamard matrix—this is a matrix where every entry is $\pm1$, and any two distinct rows (or columns) are orthogonal.
• Treat each row of the Hadamard matrix as a vector in $\mathbb{R}^n$, then normalize it by dividing each component by $\sqrt{n}$. The resulting n points lie perfectly on $S^{n-1}$.
• These points are strictly equidistant: because the rows are orthogonal, their dot product is 0, so the chord length between any pair is $\sqrt{2}$ (maximizing separation for this number of points). Even better, the construction is fully explicit—you can build the Hadamard matrix iteratively: start with the 2×2 matrix $\begin{bmatrix}1&1\1&-1\end{bmatrix}$, then for larger sizes, use $H_{2m} = \begin{bmatrix}H_m & H_m \ H_m & -H_m\end{bmatrix}$.

For example, when i=2 (n=4), the normalized rows of the 4×4 Hadamard matrix give these 4 equidistant points on $S^3$:
(1/2, 1/2, 1/2, 1/2)
(1/2, 1/2, -1/2, -1/2)
(1/2, -1/2, 1/2, -1/2)
(1/2, -1/2, -1/2, 1/2)

2. When $k=2^{i+1}$ (k=2n) #

We can extend the Hadamard matrix approach easily here:

• Take all the rows from the n×n Hadamard matrix, then add their negative counterparts. This gives us 2n points total.
• The pairs of positive/negative vectors are diametrically opposite (chord length 2, the maximum possible), while all other pairs still have a chord length of $\sqrt{2}$. The minimum separation here is still $\sqrt{2}$, which is optimal for this number of points.
• An equivalent way to think about this is taking half the vertices of an n-dimensional hypercube (e.g., all vertices where the first component is positive), normalizing them, and placing them on the sphere.

For a simple example, i=1 (n=2) gives k=4: these are just the four cardinal points on the unit circle ((1,0), (0,1), (-1,0), (0,-1)), which are obviously evenly spaced and have a closed-form description.

3. When $k=2^{i+m}$ (m≥2, k is a higher multiple of $2^i$) #

For larger k values, we can use iterative tensor product constructions or binary coordinate schemes:

• If you want k equal to $2^{2i}$ (the total number of vertices in an n-dimensional hypercube), just take all possible $\pm1$ vectors in $\mathbb{R}^n$, normalize them, and place them on $S^{n-1}$. While these points aren’t all equidistant, their minimum separation is maximized—this is an optimal spherical code for this k and n.
• For k values between $2^{i+1}$ and $2^{2i}$, you can combine smaller optimal point sets (e.g., take multiple copies of the n-point set from case 1, shifted to avoid overlap) to get a closed-form set of points with maximum possible minimum separation.

If by "uniform distribution" you mean measure-theoretic uniformity (i.e., the points spread evenly over the sphere as k grows), these constructions also satisfy that—you can keep subdividing the coordinate components to generate as many evenly spread points as you need, all with explicit coordinates.

内容的提问来源于stack exchange，提问作者gota