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Quasilocal completion of finite matrix registers

Definition

Fix $q \geq 1$ and the ordered registers $V_{N} = (C^{q})^{\otimes N}$ , with $V_{0} = C$ . Here $End (V_{N})$ is the space of linear maps $V_{N} \to V_{N}$ as in Linear map between vector spaces over the same field. For such a map define its operator norm by $∥ A ∥ : = \sup_{∥ x ∥ = 1} ∥ A x ∥$ , using the ordered tensor inner product of Ordered tensor registers, their inner product, and wire permutations. In its ordered orthonormal product basis define the adjoint by $(A^{*})_{i j} = \overline{A_{j i}}$ . For the linear-first inner product this gives $⟨ A x, y ⟩ = ⟨ x, A^{*} y ⟩$ for all $x, y \in V_{N}$ . Embed $A_{N} = End (V_{N})$ into $A_{N + 1}$ by $ι_{N} (A) = A \otimes I_{q}$ . Write a vector in $V_{N} \otimes C^{q}$ as $x = \sum_{j = 1}^{q} x_{j} \otimes ∣ j ⟩$ . Then $∥ (A \otimes I_{q}) x ∥^{2} = \sum_{j = 1}^{q} ∥ A x_{j} ∥^{2} \leq ∥ A ∥^{2} \sum_{j = 1}^{q} ∥ x_{j} ∥^{2} = ∥ A ∥^{2} ∥ x ∥^{2} .$ Testing on unit simple tensors $v \otimes ∣ 1 ⟩$ and taking the supremum over unit $v$ gives the reverse norm inequality, hence $∥ A \otimes I_{q} ∥ = ∥ A ∥$ . The embeddings are isometric unital *-homomorphisms. For $m \geq N$ , define $ι_{N, m} (A) = A \otimes I_{q}^{\otimes (m - N)}$ and identify $A \in A_{N}$ with its image at every larger register. For a two-step inclusion the maps fit into this commutative diagram:

The curved arrow tensors with two identities, and the upper path tensors with one identity twice; both send $A$ to $A \otimes I_{q} \otimes I_{q}$ . Let $A_{l o c} = ⋃_{N \geq 0} A_{N}$ with the induced norm, product and adjoint. Define $A = \overline{A_{l o c}}$ in this norm. This is the quasilocal matrix algebra for the one-dimensional ordered register chain; it is a unital C*-algebra with unit the common image of $1 \in A_{0}$ .

The sequence $(A_{N}, ι_{N})$ is the specified inductive system. This definition concerns one fixed local matrix size $q$ and makes no claim about a physical lattice geometry or dynamics.

Facts & Assumptions

Given: $q \geq 1$ , the ordered finite tensor registers, and their finite operator norms.

[F1]

Ordered tensor registers, their inner product, and wire permutations: the ordered product basis spans each finite tensor register and its inner product is the product inner product.

[F2]

C star algebra: a C*-algebra is a complete normed -algebra satisfying the C-identity.

[F3]

Finite Hermitian diagonalization: every finite Hermitian matrix has an orthonormal eigenbasis with real eigenvalues.

[F4]

Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs: the inner-product pairing obeys $∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥ ∥ v ∥$ .

Proof

1.1givenF1algebra

The decomposition $x = \sum_{j} x_{j} \otimes ∣ j ⟩$ is orthogonal in the product basis [F1], and $A \otimes I_{q}$ sends each summand to $A x_{j} \otimes ∣ j ⟩$ . The displayed sum-of-squares formula follows. It proves $∥ A \otimes I_{q} ∥ \leq ∥ A ∥$ . For every unit $v$ , $∥ (A \otimes I_{q}) (v \otimes ∣ 1 ⟩) ∥ = ∥ A v ∥$ ; taking the supremum over unit $v$ proves the opposite inequality. Thus $ι_{N}$ is isometric.

2.1givenF1F2F3F4step 1.1algebra∎

The conjugate-transpose formula gives $(α A + β B)^{*} = \overline{α} A^{*} + \overline{β} B^{*}$ , $(A^{*})^{*} = A$ and $(A B)^{*} = B^{*} A^{*}$ by conjugating and reversing the finite coordinate sums for matrix multiplication. For $A \in A_{N}$ , the finite Hermitian matrix $A^{*} A$ is nonnegative, since $x^{*} A^{*} A x = ∥ A x ∥^{2} \geq 0$ . By [F3] its eigenvalues are nonnegative, and in an orthonormal eigenbasis $∥ A ∥^{2} = \sup_{∥ x ∥ = 1} x^{*} A^{*} A x = \max spec (A^{*} A) = ∥ A^{*} A ∥ .$ Indeed, if its eigenvalues are $λ_{i} \geq 0$ , then $x^{*} A^{*} A x = \sum_{i} λ_{i} ∣ x_{i} ∣^{2}$ and $∥ A^{*} A x ∥^{2} = \sum_{i} λ_{i}^{2} ∣ x_{i} ∣^{2}$ . The unit-vector suprema are $\max_{i} λ_{i}$ and $(\max_{i} λ_{i})^{2}$ , respectively. Thus every local matrix algebra satisfies the C*-identity. Tensor multiplication and adjoint give $(A \otimes I) (B \otimes I) = (A B) \otimes I$ and $(A \otimes I)^{*} = A^{*} \otimes I$ , while $I \otimes I$ is the unit. Hence these maps are unital -homomorphisms, and their compatible inclusions make the union operations well-defined. Operator-norm submultiplicativity follows from $∥ A B x ∥ \leq ∥ A ∥ ∥ B ∥ ∥ x ∥$ . By [F4], $∥ z ∥ = \sup_{∥ x ∥ = 1} ∣ ⟨ z, x ⟩ ∣$ , with equality at $x = z / ∥ z ∥$ when $z \neq 0$ . The adjoint identity gives $∥ A^{*} y ∥ = \sup_{∥ x ∥ = 1} ∣ ⟨ y, A x ⟩ ∣ \leq ∥ A ∥ ∥ y ∥$ ; applying this to $A^{*}$ proves $∥ A^{*} ∥ = ∥ A ∥$ . Hence the union is a normed -algebra satisfying the C-identity. The norm completion preserves product, involution and that identity by continuity, and is complete by construction. By [F2] it is a C-algebra.

Quasilocal completion of finite matrix registers

Definition

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

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