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Discrete spheres: find points at maximum distance in the Hamming metric. In $\mathbb{F}_2^n$ every element has a unique antipodal point at distance $n$. In $\mathbb{F}_3^n$ every element is distance $n$ from exactly $2^n$ points, so exponentially many.

Smooth spheres: In the $2$-sphere, the surface area between two lines of longitude depends only on the difference in height, and so is proportional to the area of the part of the enclosing cylinder. (As was known to Euclid.) In the $3$-sphere, the area is biggest at the equator, and smallest at the poles; the density function is $\sqrt{1-z^2}$ for $-1 \le z \le 1$.

Discrete spheres: in $\mathbb{F}_2^n$ every element has a unique antipodal point at distance $n$. In $\mathbb{F}_3^n$ every element is distance $n$ from exactly $2^n$ points, so exponentially many.

Smooth spheres: In the $2$-sphere, the surface area between two lines of longitude depends only on the difference in height, and so is proportional to the area of the part of the enclosing cylinder. (As was known to Euclid.) In the $3$-sphere, the area is biggest at the equator, and smallest at the poles; the density function is $\sqrt{1-z^2}$ for $-1 \le z \le 1$.

Problem: find points at maximum distance in the Hamming metric. In $\mathbb{F}_2^n$ every element has a unique antipodal point at distance $n$. In $\mathbb{F}_3^n$ every element is distance $n$ from exactly $2^n$ points, so exponentially many.

Discrete spheres: find points at maximum distance in the Hamming metric. In $\mathbb{F}_2^n$ every element has a unique antipodal point at distance $n$. In $\mathbb{F}_3^n$ every element is distance $n$ from exactly $2^n$ points, so exponentially many.

Smooth spheres: In the $2$-sphere, the surface area between two lines of longitude depends only on the difference in height, and so is proportional to the area of the part of the enclosing cylinder. (As was known to Euclid.) In the $3$-sphere, the area is biggest at the equator, and smallest at the poles; the density function is $\sqrt{1-z^2}$ for $-1 \le z \le 1$.

Problem: find points at maximum distance in the Hamming metric. In $\mathbb{F}_2^n$ every element has a unique antipodal point at distance $n$. In $\mathbb{F}_3^n$ every element is distance $n$ from exactly $2^n$ points.

Problem: find points at maximum distance in the Hamming metric. In $\mathbb{F}_2^n$ every element has a unique antipodal point at distance $n$. In $\mathbb{F}_3^n$ every element is distance $n$ from exactly $2^n$ points, so exponentially many.