TurboQuant

Proof

Step 1: The Slicing Argument

Consider the unit sphere $\mathbb{S}^{d-1} = \{\boldsymbol{x} \in \mathbb{R}^d : \|\boldsymbol{x}\| = 1\}$. Fix a coordinate index $j$ (by symmetry, the choice of $j$ doesn’t matter — let’s take $j = 1$ for concreteness).

Slice the sphere at height $x_1 = x$ for some fixed $x \in (-1, 1)$. The slice is the set of all points on $\mathbb{S}^{d-1}$ whose first coordinate equals $x$:

The constraint $x_2^2 + \cdots + x_d^2 = 1 - x^2$ means the remaining $d-1$ coordinates lie on a sphere of radius $r(x) = \sqrt{1 - x^2}$ in $\mathbb{R}^{d-1}$. So:

Step 2: Surface Area of a Sphere

We need the surface area of $\mathbb{S}^{m-1}(r)$, the $(m-1)$-dimensional sphere of radius $r$ in $\mathbb{R}^m$. The formula is:

Step 3: From Surface Area to Density

Now we derive $f_X(x)$. The key idea: “uniform on $\mathbb{S}^{d-1}$” means probability is proportional to surface area. So the probability that $x_1 \in [x, x + dx]$ is proportional to the surface area of the thin strip of $\mathbb{S}^{d-1}$ between heights $x_1 = x$ and $x_1 = x + dx$.

The strip is NOT a flat disk. This is the subtle point. The sphere surface between $x_1 = x$ and $x_1 = x + dx$ is a thin band on a curved surface. We need to account for the tilt of the surface relative to the horizontal.

Computing the strip width. Consider a point on $\mathbb{S}^{d-1}$ at height $x_1 = x$. As we move along the sphere surface to increase $x_1$ by $dx$, we travel a distance $ds$ along the sphere. These are related by the geometry of the sphere:

At height $x_1 = x$, the sphere surface makes an angle with the horizontal. The relationship between the infinitesimal height change $dx$ and the arc length $ds$ along the sphere is:

Why? Parameterize the sphere locally. A point on $\mathbb{S}^{d-1}$ at height $x_1 = x$ satisfies $x_1^2 + \|\boldsymbol{x}_{2:d}\|^2 = 1$. If we increase $x_1$ by $dx$ while staying on the sphere, the transverse radius changes by $dr = \frac{-x\,dx}{\sqrt{1-x^2}}$, and the total displacement along the sphere surface has length:

(We used $1 + \frac{x^2}{1-x^2} = \frac{1}{1-x^2}$.)

Area of the strip. The strip at height $x$ has:

• “Circumference”: the $(d-2)$-dimensional area of the cross-section, which is $A_{d-1}(\sqrt{1-x^2})$
• Width (along the sphere surface): $ds = \frac{dx}{\sqrt{1-x^2}}$

So the surface area of the strip is:

The density is this strip area divided by the total surface area of $\mathbb{S}^{d-1}$:

Step 4: Substituting the Area Formula

Plug in $A_m(r) = \frac{2\pi^{m/2}}{\Gamma(m/2)} \cdot r^{m-1}$:

Numerator (cross-section area $\times$ Jacobian):

(We combined the exponents: $\frac{d-2}{2} - \frac{1}{2} = \frac{d-3}{2}$.)

Denominator (total sphere area):

Step 5: Simplification

The $2$‘s cancel. For the $\pi$ terms:

For the $\Gamma$ terms, the denominator’s $\Gamma(d/2)$ moves to the numerator and vice versa:

Connection to the Standard Beta Distribution

The standard Beta distribution $\text{Beta}(\alpha, \beta)$ has density on $[0, 1]$:

with mean $\frac{\alpha}{\alpha+\beta}$ and variance $\frac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)}$.

Mapping from $f_X$ to $\text{Beta}$. The density $f_X(x) \propto (1 - x^2)^{(d-3)/2}$ lives on $[-1, 1]$. To see the Beta connection, substitute $t = \frac{x + 1}{2}$ (mapping $[-1, 1] \to [0, 1]$):

Absorbing the constant $4^{(d-3)/2} \cdot 2$ (Jacobian $dx = 2\,dt$) into the normalization:

with $\alpha = \beta = \frac{d - 1}{2}$. Therefore:

This is a symmetric Beta distribution (since $\alpha = \beta$), centered at $t = 1/2$ (i.e., $x = 0$).

[!note] Why “Beta” matters The symmetric Beta form tells us that $f_X$ is unimodal, sub-Gaussian, and supported on a compact interval. These properties guarantee that the Lloyd-Max quantizer converges quickly, the codebook can be precomputed exactly, and the Panter-Dite formula applies with well-controlled error.