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Source. Published p. 264, Problem 1.12 and Theorem 1.13 (PDF).

For integers $n\ge r\ge2$, let $\mu(n,r)$ be the supremum of normalized surface measures of measurable $E\subseteq S^{n-1}$ containing no $r$ pairwise orthogonal vectors from the origin.

Statement. For each fixed $r\ge2$ there is $c_r>0$ such that $\mu(n,r)\le e^{-c_r n}$ for every $n\ge r$.

Proof. Let $N=4\lfloor n/4\rfloor$ and place the normalized cube $\{-1,1\}^N/\sqrt N$ in an $N$-dimensional subspace of $\mathbb R^n$. Apply a random orthogonal transformation, using the invariant probability measure specified in the external inputs. Every transformed vertex is uniform on $S^{n-1}$, so the expected number of its $2^N$ vertices in $E$ is $2^N\mu(E)$. If this exceeds the threshold $2^Ne^{-cN}$ of the proved large-dimensional Theorem 1.11, some rotation has more than that many vertices in $E$ and therefore contains $r$ orthogonal ones. Consequently $\mu(E)\le e^{-cN}$ for all large $n$, and $N\ge n-3$ gives a bound $e^{-c'n}$.

For every $n\ge r$, average instead over a random orthonormal $r$-frame. At most $r-1$ frame vectors belong to $E$, whereas their expected number is $r\mu(E)$. Hence $\mu(E)\le1-1/r$. Decrease $c_r>0$ to make $e^{-c_rn}\ge1-1/r$ in the finitely many dimensions not covered by the cube argument. Taking suprema proves the statement.

For completeness, if $r\le n<n'$, randomly rotate an $n$-dimensional subspace in $\mathbb R^{n'}$. Its spherical section of an avoiding set still avoids $r$ orthogonal vectors, and its average normalized measure equals the original measure. Thus $\mu(n',r)\le\mu(n,r)$. $\square$

Source precision. The “obvious” dimension inequality on p. 264 is printed in the opposite direction; the averaging argument gives the inequality just proved. The domain $n\ge r$ is necessary: for $n<r$ the exclusion is vacuous and the supremum equals one. No claim about an exact optimum or its present-day status is made.

Dependencies. theorem_1_11, external_inputs.