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Bounded tensor maps: extension, norm estimate, adjoint and commutation

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H, H^{'}, K, K^{'}$ be complex Hilbert spaces and let $S \in B (H, H^{'})$ , $T \in B (K, K^{'})$ be bounded linear operators (A bounded linear operator between normed spaces). Write $H \hat{\otimes} K$ for the completion of the algebraic tensor product constructed in Hilbert tensor construction: inner product, completion and product bases, with inclusion $κ$ .

(a) There is a unique bounded linear operator $S \otimes T : H \hat{\otimes} K \to H^{'} \hat{\otimes} K^{'}$ with $(S \otimes T) κ (u \otimes v) = κ^{'} (S u \otimes T v)$ for all $u \in H$ , $v \in K$ , and it satisfies $∥ S \otimes T ∥ \leq ∥ S ∥ ∥ T ∥$ (The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

(b) For Hilbert spaces $H^{''}, K^{''}$ and bounded $S^{'} : H^{''} \to H$ , $T^{'} : K^{''} \to K$ , one has $(S \otimes T) (S^{'} \otimes T^{'}) = (S S^{'}) \otimes (T T^{'})$ . In particular, for endomorphisms $S : H \to H$ , $T : K \to K$ , the endomorphisms $S \otimes I_{K}$ and $I_{H} \otimes T$ commute and their composite is $S \otimes T$ .

(d) If $S, T$ are unitary, then so is $S \otimes T$ . No extension of unbounded maps, and no algebraic manipulation of domains of unbounded operators, is asserted.

Facts & Assumptions

Given: Complex Hilbert spaces $H, H^{'}, K, K^{'}$ ; bounded operators $S \in B (H, H^{'})$ , $T \in B (K, K^{'})$ ; in (b), composable $S^{'} : H^{''} \to H$ , $T^{'} : K^{''} \to K$ ; the completions $H \hat{\otimes} K$ , $H^{'} \hat{\otimes} K^{'}$ of Hilbert tensor construction: inner product, completion and product bases with inclusions $κ, κ^{'}$ ; the elementary tensors $u \otimes v$ .

[F1]

The construction of Hilbert tensor construction: inner product, completion and product bases gives: the algebraic tensor product $H ⊙ K$ with $u \otimes v$ bilinear; a positive definite inner product with $⟨ u \otimes v, u^{'} \otimes v^{'} ⟩ = ⟨ u, u^{'} ⟩_{H} ⟨ v, v^{'} ⟩_{K}$ ; the completion $H \hat{\otimes} K$ with dense inclusion $κ$ ; unique orthonormal-coordinate expansions with $∥ \sum_{a, b} c_{a b} e_{a} \otimes f_{b} ∥^{2} = \sum_{a, b} ∣ c_{a b} ∣^{2}$ for orthonormal families $(e_{a}) \subseteq H$ , $(f_{b}) \subseteq K$ spanning finite-dimensional subspaces; and $∥ κ (u \otimes v) ∥ = ∥ u ∥ ∥ v ∥$ .

[F2]

If $X$ is a normed space, $D \subseteq X$ a dense normed subspace, $Y$ a Banach space and $L : D \to Y$ bounded linear, then $L$ has a unique bounded linear extension to $X$ of the same norm (A bounded linear map from a dense normed subspace into a Banach space extends uniquely with the same norm).

[F3]

A linear map is bounded exactly when it admits a finite bound $∥ L x ∥ \leq C ∥ x ∥$ , and the operator norm is the least such bound (A bounded linear operator between normed spaces, The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[F4]

The Hilbert adjoint $L^{*}$ of a bounded operator $L \in B (X, Y)$ is the unique bounded operator with $⟨ L x, y ⟩_{Y} = ⟨ x, L^{*} y ⟩_{X}$ for all $x \in X$ , $y \in Y$ ; it satisfies $(L_{1} L_{2})^{*} = L_{2}^{*} L_{1}^{*}$ , $L^{* *} = L$ and $∥ L^{*} ∥ = ∥ L ∥$ (The Hilbert-space adjoint of a bounded operator, Hilbert-adjoint identities).

Proof

1.1givenF1algebra

[The bilinear extension on the algebraic tensor product] Put $R_{0} (u \otimes v) : = κ^{'} (S u \otimes T v)$ and extend $C$ -linearly to finite sums. This is well defined and $C$ -linear on $H ⊙ K$ because $(u, v) \mapsto S u \otimes T v$ is $C$ -bilinear, composition of the linear maps $S, T$ with the bilinear tensor map; thus $R_{0}$ is the unique linear map on $H ⊙ K$ with $R_{0} (u \otimes v) = κ^{'} (S u \otimes T v)$ .

2.1step 1.1F1F3algebra

[One-sided norm estimates] Let $z = \sum_{i = 1}^{n} u_{i} \otimes v_{i} \in H ⊙ K$ . Let $V_{0} = span {v_{i}}$ and let $f_{1}, \dots, f_{q}$ be an orthonormal basis of $V_{0}$ ; grouping the expansion of each $v_{i}$ in this basis gives $z = \sum_{b = 1}^{q} u_{b} \otimes f_{b}$ with $u_{b} \in span {u_{i}}$ . By the norm formula of [F1] and $∥ S u_{b} ∥ \leq ∥ S ∥ ∥ u_{b} ∥$ , $∥ (S \otimes I) z ∥^{2} = \sum_{b} ∥ S u_{b} ∥^{2} \leq ∥ S ∥^{2} \sum_{b} ∥ u_{b} ∥^{2} = ∥ S ∥^{2} ∥ z ∥^{2}$ , so the map $S \otimes I_{K}$ on elementary tensors is bounded with norm at most $∥ S ∥$ ; symmetrically $∥ (I_{H} \otimes T) z ∥ \leq ∥ T ∥ ∥ z ∥$ . Hence $∥ R_{0} z ∥ \leq ∥ S ∥ ∥ T ∥ ∥ z ∥$ for all $z \in H ⊙ K$ , because $R_{0}$ is the composite of the two one-sided maps on elementary tensors and the estimates compose.

3.1step 1.1step 2.1F2

[Extension to the completions] By [F2] applied to the dense subspace $H ⊙ K$ of $H \hat{\otimes} K$ and the Banach space $H^{'} \hat{\otimes} K^{'}$ , the map $R_{0}$ has a unique bounded linear extension $S \otimes T : H \hat{\otimes} K \to H^{'} \hat{\otimes} K^{'}$ of the same norm; by step 2.1 its norm is at most $∥ S ∥ ∥ T ∥$ , and uniqueness also shows that $(S \otimes T) κ = R_{0}$ . This is clause (a).

4.1step 3.1F2algebra

[Products and commutation] For $u \otimes v$ one has $(S \otimes T) (S^{'} \otimes T^{'}) (u \otimes v) = κ^{'} (S S^{'} u \otimes T T^{'} v) = (S S^{'} \otimes T T^{'}) (u \otimes v)$ ; the two maps are bounded and agree on the dense subspace $H'' ⊙ K''$ , hence everywhere by uniqueness of bounded extensions, giving the first identity of (b) for elementary tensors in $H'' ⊙ K''$ . In the endomorphism case, taking $T^{'} = I_{K}$ , $S^{'} = I_{H}$ and using $I \otimes I = I$ shows $(S \otimes I_{K}) (I_{H} \otimes T) = S \otimes T$ ; applying the same identity with the roles exchanged gives $(I_{H} \otimes T) (S \otimes I_{K}) = S \otimes T$ , since $S I_{H} = S$ and $I_{K} T = T$ .

4.2step 3.1F1F4algebra

[Adjoint] For $u \in H$ , $u^{'} \in H^{'}$ and $v \in K$ , $v^{'} \in K^{'}$ , $⟨ (S \otimes T) κ (u \otimes v), κ^{'} (u^{'} \otimes v^{'}) ⟩ = ⟨ S u, u^{'} ⟩_{H^{'}} ⟨ T v, v^{'} ⟩_{K^{'}} = ⟨ u, S^{*} u^{'} ⟩_{H} ⟨ v, T^{*} v^{'} ⟩_{K} = ⟨ κ (u \otimes v), (S^{*} \otimes T^{*}) κ^{'} (u^{'} \otimes v^{'}) ⟩$ , using [F4] and the pairing formula of [F1]. Both sides are sesquilinear in the elementary tensors and continuous in each argument, and the elementary tensors span a dense subspace, so the identity holds on all of $H \hat{\otimes} K \times H^{'} \hat{\otimes} K^{'}$ ; by the uniqueness in [F4], $(S \otimes T)^{*} = S^{*} \otimes T^{*}$ .

5.1step 4.1step 4.2algebra∎

[Unitary case] If $S, T$ are unitary then $S^{*} S = I$ , $S S^{*} = I$ , $T^{*} T = I$ , $T T^{*} = I$ , so by steps 4.1 and 4.2 $(S \otimes T)^{*} (S \otimes T) = S^{*} S \otimes T^{*} T = I \otimes I = I$ and likewise $(S \otimes T) (S \otimes T)^{*} = I$ , whence $S \otimes T$ is unitary.

Bounded tensor maps: extension, norm estimate, adjoint and commutation

Statement

Facts & Assumptions

Proof

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