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Clifford induces symplectic map

Statement

For each U∈Cn, conjugation induces an invertible F2-linear map SU:Vn→Vn satisfying ω(SUa,SUb)=ω(a,b) for all a,b∈Vn. Thus SU is symplectic. The assignment respects multiplication: SUV=SU∘SV for all U,V∈Cn.

Facts & Assumptions

Given: U is a unitary normalizer of the phase-aware Pauli group and Vn=Pn/Zn is identified with its binary labels.

[F1]

Binary Pauli quotient isomorphism: the label quotient is a group isomorphism, with scalar center as kernel.

[F2]

Pauli commutator is symplectic pairing: the commutator sign of two Pauli operators is (−1)ω(a,b).

[F3]

Clifford normalizer: U and U∗ normalize Pn.

Proof

1.1givenF1F3algebra

Conjugation sends each Pauli to a Pauli by [F3]. It fixes every scalar, so representatives in the same scalar coset have images in the same coset. Therefore it defines a map SU on Vn under [F1].

2.1givenF1F3step 1.1algebra

For Pauli operators P,Q, U(PQ)U∗=(UPU∗)(UQU∗), so the quotient map [F1] shows SU(a+b)=SU(a)+SU(b). An additive map between F2 vector spaces is linear because the only scalars are 0 and 1. Conjugation by U∗ is its inverse, so SU is invertible. Applying the same identity successively gives SUV=SU∘SV for Clifford unitaries U,V.

3.1givenF2step 1.1algebra∎

The commutator identity from [F2] remains the same after conjugating both sides by U, because conjugation preserves products and scalar signs. Applying [F2] to the image Paulis yields (−1)ω(SUa,SUb)=(−1)ω(a,b). The map r↦(−1)r is injective on F2, hence ω(SUa,SUb)=ω(a,b).

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