Proposition. Why the subspace constant is sharp [cn-0025]

Status:

Statement. The constant in the subspace inequality [cn-0024] cannot be decreased.

Proof. Write \(W=P\oplus Q\) and divide the \(m\) indices into groups of sizes \(a=\lfloor m/2\rfloor \) and \(b=\lceil m/2\rceil \). In the first group set \((U_i,V_i)=(P,Q)\); in the second set \((U_i,V_i)=(Q,P)\). Every designated pair is disjoint. Only indices from different groups contribute, with total

\[ab\dim P+ab\dim Q=\left \lfloor \frac {m^2}{4}\right \rfloor \dim W.\]

For example, three indices split \(1+2\) give total \(2\dim W\). This construction makes the balanced product \(e(m-e)\) in the proof less mysterious.

Proof dependencies
None.
Provenance
OpenAI, 23 September 2026 [cn-0042], Remark 2.2; reconstructed as a reading account on 9 October 2026.