Connes Fusion
Following Chapter 3 of Unitary Quantum Symmetries Lite from Penneys, Ferrer, and Kawagoe we want to try to first establish what Connes fusion is. Big picture: just like when we compose (fuse) group representations on Hilbert spaces, we end up in the tensor product Hilbert space; Connes fusion is the equivalent for modules over an algebra.
This text gives three equivalent definitions. For now let’s restrict ourselves to the first one.
Definition: Connes Fusion.
Given a right B-module \(H_B\) and a left B-module \({}_{B}K\). We define the Connes Fusion relative tensor product Hilbert space \(H\boxtimes_B K\) can be defined using the relative tensor product for \(C^*\)-Hilbert modules. $$ Hom(L^2B_B \rightarrow H_B) \boxtimes_B K$$ with \(\mathbb{C}\)-valued inner product $$\langle f_1 \boxtimes \eta_1 \vert f_2 \boxtimes \eta_2 \rangle := \langle \langle f_2 \vert f_1 \rangle_B \eta_1 \vert \eta_2 \rangle_K$$ where we have $$\langle f_2 \vert f_1 \rangle_B = f_2^{\dagger}f1 \in End(L^2B_B)=B$$
Given a right B-module \(H_B\) and a left B-module \({}_{B}K\). We define the Connes Fusion relative tensor product Hilbert space \(H\boxtimes_B K\) can be defined using the relative tensor product for \(C^*\)-Hilbert modules. $$ Hom(L^2B_B \rightarrow H_B) \boxtimes_B K$$ with \(\mathbb{C}\)-valued inner product $$\langle f_1 \boxtimes \eta_1 \vert f_2 \boxtimes \eta_2 \rangle := \langle \langle f_2 \vert f_1 \rangle_B \eta_1 \vert \eta_2 \rangle_K$$ where we have $$\langle f_2 \vert f_1 \rangle_B = f_2^{\dagger}f1 \in End(L^2B_B)=B$$
We aim to build up the tools to understand this definition within this post.
Modules
In general, modules can be thought of as a generalazition of a vector space, where we let the scalars come from a ring rather than a field. However, what we’re really interested in is modules of algebras. Translating to that language we have:
Definition — Left Module.
(VecSpace, BilinearMap) Suppose \(A\) is a unital algebra. A (left) module \({}_{A}M\) for \(A\) is a vector space \(M\) equipped with a bilinear map \( \triangleright: A \times M \rightarrow M \) satisfying:
1. (associativity) $$a \triangleright (b \triangleright m) = (a \cdot b) \triangleright m \text{ for } m \in M \text{ and } a,b \in A$$ 2. (unitality) $$1 \triangleright m = m \text{ for } m \in M$$
(VecSpace, BilinearMap) Suppose \(A\) is a unital algebra. A (left) module \({}_{A}M\) for \(A\) is a vector space \(M\) equipped with a bilinear map \( \triangleright: A \times M \rightarrow M \) satisfying:
1. (associativity) $$a \triangleright (b \triangleright m) = (a \cdot b) \triangleright m \text{ for } m \in M \text{ and } a,b \in A$$ 2. (unitality) $$1 \triangleright m = m \text{ for } m \in M$$
Definition — Right Module.
(VecSpace, BilinearMap) Suppose \(A\) is a unital algebra. A (right) module \(M_A\) for \(A\) is a vector space \(M\) equipped with a bilinear map \( \triangleleft: M \times A \rightarrow M \) satisfying:
1. (associativity) $$(m \triangleleft a) \triangleleft b = m \triangleleft (a \cdot b) \text{ for } m \in M \text{ and } a,b \in A$$ 2. (unitality) $$m \triangleleft 1 = m \text{ for } m \in M$$
(VecSpace, BilinearMap) Suppose \(A\) is a unital algebra. A (right) module \(M_A\) for \(A\) is a vector space \(M\) equipped with a bilinear map \( \triangleleft: M \times A \rightarrow M \) satisfying:
1. (associativity) $$(m \triangleleft a) \triangleleft b = m \triangleleft (a \cdot b) \text{ for } m \in M \text{ and } a,b \in A$$ 2. (unitality) $$m \triangleleft 1 = m \text{ for } m \in M$$
Using these let’s understand the definition of a bimodule over two algebras.
(VecSpace, UnitalMap) Definition: A–B Bimodule.
An A–B bimodule \(M\), denoted \({}_{A}M_{B}\) is a right B-module, \(M_B\) equipped with a unital algebra map \( \cdot: A \rightarrow End(M_B)\)
An A–B bimodule \(M\), denoted \({}_{A}M_{B}\) is a right B-module, \(M_B\) equipped with a unital algebra map \( \cdot: A \rightarrow End(M_B)\)
The Algebraic Relative Tensor Product
Algebra-Valued Inner Products & Hilbert C*-Modules
(InnerProduct) Definition: A-valued Inner Product.
Suppose \(X_A\) is a right module for a unitary algebra \(A\). An \(A\)-valued inner product is a map \(\langle \cdot \vert \cdot \rangle_A : X \times X \rightarrow A\) satisfying:
1. (\(A\)-linear in the second variable) $$\langle \varphi \vert \zeta_1 a + \zeta_2 \rangle_A = \langle \varphi \vert \zeta_1 \rangle_A \, a + \langle \varphi \vert \zeta_2 \rangle_A$$ 2. (conjugate-symmetric) $$\langle \varphi \vert \zeta \rangle_A^{*} = \langle \zeta \vert \varphi \rangle_A$$ 3. (positive-definite) $$\langle \varphi \vert \varphi \rangle_A \geq 0 \text{ in } A, \text{ with equality iff } \varphi = 0.$$
Suppose \(X_A\) is a right module for a unitary algebra \(A\). An \(A\)-valued inner product is a map \(\langle \cdot \vert \cdot \rangle_A : X \times X \rightarrow A\) satisfying:
1. (\(A\)-linear in the second variable) $$\langle \varphi \vert \zeta_1 a + \zeta_2 \rangle_A = \langle \varphi \vert \zeta_1 \rangle_A \, a + \langle \varphi \vert \zeta_2 \rangle_A$$ 2. (conjugate-symmetric) $$\langle \varphi \vert \zeta \rangle_A^{*} = \langle \zeta \vert \varphi \rangle_A$$ 3. (positive-definite) $$\langle \varphi \vert \varphi \rangle_A \geq 0 \text{ in } A, \text{ with equality iff } \varphi = 0.$$
(Module, InnerProduct) Definition: Right Hilbert C*-Module.
A right module equipped with an C* algebra valued inner product is a right Hilbert \(C^*\)-module.
A right module equipped with an C* algebra valued inner product is a right Hilbert \(C^*\)-module.
Condition Definition: Adjointable Operator.
Suppose \(X_A, Y_A\) are right modules equipped with \(A\)-valued inner products. A right \(A\)-linear map \(x : X \rightarrow Y\) is adjointable if there is a right \(A\)-linear map \(x^{\dagger} : Y \rightarrow X\), called an adjoint, such that
$$\langle x^{\dagger}\varphi \vert \zeta \rangle_A = \langle \varphi \vert x\zeta \rangle_A \qquad \forall \zeta \in X, \; \varphi \in Y$$ where the left inner product is on \(X\) and the right one is on \(Y\).
Suppose \(X_A, Y_A\) are right modules equipped with \(A\)-valued inner products. A right \(A\)-linear map \(x : X \rightarrow Y\) is adjointable if there is a right \(A\)-linear map \(x^{\dagger} : Y \rightarrow X\), called an adjoint, such that
$$\langle x^{\dagger}\varphi \vert \zeta \rangle_A = \langle \varphi \vert x\zeta \rangle_A \qquad \forall \zeta \in X, \; \varphi \in Y$$ where the left inner product is on \(X\) and the right one is on \(Y\).
(VecSpace, InnerProduct) Trick: Realization (Self-Enrichment).
For any right Hilbert \(C^*\)-module \(X_A\), the adjointable maps \(AdHom(A_A \rightarrow X_A)\) themselves form a right Hilbert \(C^*\)-module with \(A\)-valued inner product
$$\langle f \vert g \rangle_A := f^{\dagger} \circ g \in End(A_A) = A$$
For any right Hilbert \(C^*\)-module \(X_A\), the adjointable maps \(AdHom(A_A \rightarrow X_A)\) themselves form a right Hilbert \(C^*\)-module with \(A\)-valued inner product
$$\langle f \vert g \rangle_A := f^{\dagger} \circ g \in End(A_A) = A$$
Pf. We want to show that \(\mathcal{H} := AdHom(A_A \rightarrow X_A)\), the set of adjointableWe take the \(\dagger\) operation on one of these maps to denote it's adjoint i.e. \( f \in AdHom(A_A, X_A) \implies f^{\dagger} \in Ad(Hom(X_A,A_A) \) right A-linear maps \(A_A \rightarrow X_A\), is a right Hilbert \(C^*\)-module. We'll do this in three steps: first show \(\mathcal{H}\) is a vector space, then equip it with a right A-action, and finally define the A-valued inner product.
Step 1: \(\mathcal{H}\) is a vector space. Define \((f + g)(a) := f(a) + g(a)\) and \((\lambda f)(a) := \lambda f(a)\) for \(\lambda \in \mathbb{C}\). These preserve right A-linearity, since \[(f+g)(a \cdot b) = f(a) \cdot b + g(a) \cdot b = (f+g)(a) \cdot b\] and preserve adjointability, since \((f+g)^{\dagger} = f^{\dagger} + g^{\dagger}\) and \((\lambda f)^{\dagger} = \bar{\lambda} f^{\dagger}\)
Step 2: \( \mathcal{H} \) is a right A-module. Define \(\triangleleft : \mathcal{H} \times A \rightarrow \mathcal{H}\) by \[(f \triangleleft a)(b) := f(a \cdot b)\] We verify: (a) \(f \triangleleft a \in \mathcal{H}\), (b) \(\triangleleft\) is bilinear, and (c) \(\triangleleft\) is associative and unital.
(a) \(f \triangleleft a \in \mathcal{H}.\)
Right A-linear: \[(f \triangleleft a)(b \cdot c) = f(a \cdot b \cdot c) = f(a \cdot b) \cdot c = (f \triangleleft a)(b) \cdot c\] Adjointable:
Define \(g : X \rightarrow A\) by \(g(\xi) := a^* \cdot f^{\dagger}(\xi)\). This is right A-linear, and satisfies \[\begin{aligned} \langle g(\xi) \vert b\rangle_A &= \langle a^* f^{\dagger}(\xi) \vert b\rangle_A \\ &= (a^* f^{\dagger}(\xi))^* b \\ &= f^{\dagger}(\xi)^* a b \\ &= \langle f^{\dagger}(\xi) \vert a b\rangle_A \\ &= \langle \xi \vert f(a b)\rangle_A^X \\ &= \langle \xi \vert (f \triangleleft a)(b)\rangle_A^X \end{aligned}\] so \(g = (f \triangleleft a)^{\dagger}\).
(b) Bilinearity.
Linear in \(f\): \[((f_1 + f_2) \triangleleft a)(b) = (f_1 + f_2)(a b) = f_1(ab) + f_2(ab) = (f_1 \triangleleft a)(b) + (f_2 \triangleleft a)(b)\] Linear in \(a\): \[(f \triangleleft (a_1 + a_2))(b) = f((a_1 + a_2) b) = f(a_1 b) + f(a_2 b) = (f \triangleleft a_1)(b) + (f \triangleleft a_2)(b)\] using that \(f\) is \(\mathbb{C}\)-linear and A-multiplication distributes.
(c) Associativity and unitality.
Associativity: \[((f \triangleleft a) \triangleleft b)(c) = (f \triangleleft a)(b c) = f(a b c) = (f \triangleleft (a \cdot b))(c)\] Unitality: \[(f \triangleleft 1)(b) = f(1 \cdot b) = f(b)\] So \(\mathcal{H}\) is a right A-module. \(\checkmark\)
Step 3: \(\mathcal{H}\) has an A-valued inner product.
Define \[\langle f \vert g \rangle_A := f^{\dagger} \circ g \in End(A_A) = A\] using the identification \(End(A_A) \cong A\) via \(T \mapsto T(1)\). We check the three axioms.
A-linear in the second slot.
We have \(\langle f \vert g \triangleleft a\rangle_A = f^{\dagger} \circ (g \triangleleft a)\). Evaluated at \(b \in A\): \[f^{\dagger}(g(ab)) = (f^{\dagger} \circ g)(ab) = \langle f \vert g\rangle_A \cdot a \cdot b\] so as an element of \(A\) (evaluating at \(b = 1\)), this equals \(\langle f \vert g\rangle_A \cdot a\). \(\checkmark\)
Conjugate-symmetric. \[\langle f \vert g\rangle_A^* = (f^{\dagger} \circ g)^* = g^{\dagger} \circ (f^{\dagger})^{\dagger} = g^{\dagger} \circ f = \langle g \vert f\rangle_A \; \checkmark\]
Positive-definite.
We have \(\langle f \vert f\rangle_A = f^{\dagger} \circ f\), which is positive in \(End(A_A) = A\) as the composite of an operator with its adjoint (Corollary 3.2.21). IT's defif \(f^{\dagger} f = 0\), then for all \(b \in A\), \[\langle f(b) \vert f(b)\rangle_A^X = \langle b \vert f^{\dagger} f (b)\rangle_A = 0\] so \(f(b) = 0\) by definiteness of \(\langle \cdot \vert \cdot \rangle_A^X\), hence \(f = 0\). \(\checkmark\)
Therefore \(\mathcal{H} = AdHom(A_A \rightarrow X_A)\) is a right Hilbert \(C^*\)-module. \(\blacksquare\)
Right A-linear: \[(f \triangleleft a)(b \cdot c) = f(a \cdot b \cdot c) = f(a \cdot b) \cdot c = (f \triangleleft a)(b) \cdot c\] Adjointable:
Define \(g : X \rightarrow A\) by \(g(\xi) := a^* \cdot f^{\dagger}(\xi)\). This is right A-linear, and satisfies \[\begin{aligned} \langle g(\xi) \vert b\rangle_A &= \langle a^* f^{\dagger}(\xi) \vert b\rangle_A \\ &= (a^* f^{\dagger}(\xi))^* b \\ &= f^{\dagger}(\xi)^* a b \\ &= \langle f^{\dagger}(\xi) \vert a b\rangle_A \\ &= \langle \xi \vert f(a b)\rangle_A^X \\ &= \langle \xi \vert (f \triangleleft a)(b)\rangle_A^X \end{aligned}\] so \(g = (f \triangleleft a)^{\dagger}\).
Linear in \(f\): \[((f_1 + f_2) \triangleleft a)(b) = (f_1 + f_2)(a b) = f_1(ab) + f_2(ab) = (f_1 \triangleleft a)(b) + (f_2 \triangleleft a)(b)\] Linear in \(a\): \[(f \triangleleft (a_1 + a_2))(b) = f((a_1 + a_2) b) = f(a_1 b) + f(a_2 b) = (f \triangleleft a_1)(b) + (f \triangleleft a_2)(b)\] using that \(f\) is \(\mathbb{C}\)-linear and A-multiplication distributes.
Associativity: \[((f \triangleleft a) \triangleleft b)(c) = (f \triangleleft a)(b c) = f(a b c) = (f \triangleleft (a \cdot b))(c)\] Unitality: \[(f \triangleleft 1)(b) = f(1 \cdot b) = f(b)\] So \(\mathcal{H}\) is a right A-module. \(\checkmark\)
Step 3: \(\mathcal{H}\) has an A-valued inner product.
Define \[\langle f \vert g \rangle_A := f^{\dagger} \circ g \in End(A_A) = A\] using the identification \(End(A_A) \cong A\) via \(T \mapsto T(1)\). We check the three axioms.
We have \(\langle f \vert g \triangleleft a\rangle_A = f^{\dagger} \circ (g \triangleleft a)\). Evaluated at \(b \in A\): \[f^{\dagger}(g(ab)) = (f^{\dagger} \circ g)(ab) = \langle f \vert g\rangle_A \cdot a \cdot b\] so as an element of \(A\) (evaluating at \(b = 1\)), this equals \(\langle f \vert g\rangle_A \cdot a\). \(\checkmark\)
We have \(\langle f \vert f\rangle_A = f^{\dagger} \circ f\), which is positive in \(End(A_A) = A\) as the composite of an operator with its adjoint (Corollary 3.2.21). IT's defif \(f^{\dagger} f = 0\), then for all \(b \in A\), \[\langle f(b) \vert f(b)\rangle_A^X = \langle b \vert f^{\dagger} f (b)\rangle_A = 0\] so \(f(b) = 0\) by definiteness of \(\langle \cdot \vert \cdot \rangle_A^X\), hence \(f = 0\). \(\checkmark\)
Correspondences
(VecSpace, InnerProduct) Definition: Correspondence of Unitary Algebras.
Given A,B unitary algebras. An A–B correspondence is an A–B bimodule \({}_{A}X_B\) equipped with a right B-valued inner product \(\langle \cdot \vert \cdot \rangle_B \) such that \(\rho : A \rightarrow End(X_B)\) is a unital \(\star\)–algebra map. Thus,
$$\forall a \in A, \; \; \eta, \zeta \in X \qquad \qquad \langle a^*\eta\vert\zeta\rangle_B = \langle \eta \vert a \zeta \rangle_B$$
Given A,B unitary algebras. An A–B correspondence is an A–B bimodule \({}_{A}X_B\) equipped with a right B-valued inner product \(\langle \cdot \vert \cdot \rangle_B \) such that \(\rho : A \rightarrow End(X_B)\) is a unital \(\star\)–algebra map. Thus,
$$\forall a \in A, \; \; \eta, \zeta \in X \qquad \qquad \langle a^*\eta\vert\zeta\rangle_B = \langle \eta \vert a \zeta \rangle_B$$
From this we can introduce the correct definition of relative tensor product.
(VecSpace, InnerProduct) Definition: Relative Tensor Product of Correspondences.
Given A,B,C unitary algebras and \( {}_{A}X_{B}, {}_B{Y}_{C} \) correspondences. The relative tensor product, \({}_{A}X \boxtimes_B Y_C \), is the quotient space \(X \otimes Y/N \) where $$N = span\{ \eta b \otimes \zeta - \eta \otimes b \zeta \; \vert \; \eta \in X, \; \zeta \in Y, \; \text{ and } b \in B \} $$ We denote the image of \(\eta \otimes \zeta\) by \(\eta \boxtimes \zeta\), and equip \(X \boxtimes_B Y\) with the \(C\)-valued inner product $$\langle \eta_1 \boxtimes \zeta_1 \vert \eta_2 \boxtimes \zeta_2 \rangle_C := \langle \langle \eta_2 \vert \eta_1 \rangle_B \, \zeta_1 \vert \zeta_2 \rangle_C.$$ This is well-defined and positive-definite (this is the content of Lemma 3.2.33), so \({}_{A}X \boxtimes_B Y_C\) is again an A–C correspondence.
Given A,B,C unitary algebras and \( {}_{A}X_{B}, {}_B{Y}_{C} \) correspondences. The relative tensor product, \({}_{A}X \boxtimes_B Y_C \), is the quotient space \(X \otimes Y/N \) where $$N = span\{ \eta b \otimes \zeta - \eta \otimes b \zeta \; \vert \; \eta \in X, \; \zeta \in Y, \; \text{ and } b \in B \} $$ We denote the image of \(\eta \otimes \zeta\) by \(\eta \boxtimes \zeta\), and equip \(X \boxtimes_B Y\) with the \(C\)-valued inner product $$\langle \eta_1 \boxtimes \zeta_1 \vert \eta_2 \boxtimes \zeta_2 \rangle_C := \langle \langle \eta_2 \vert \eta_1 \rangle_B \, \zeta_1 \vert \zeta_2 \rangle_C.$$ This is well-defined and positive-definite (this is the content of Lemma 3.2.33), so \({}_{A}X \boxtimes_B Y_C\) is again an A–C correspondence.
From Hilbert Space Modules to Correspondences
HilbertSpace Definition: Hilbert Space Module.
A right Hilbert space B-module \(H_B\) is a Hilbert space \(H\) equipped with a right B-action, i.e. a unital \(\star\)-homomorphism \(B^{op} \rightarrow B(H)\). Its \(\mathbb{C}\)-valued inner product does not restrict to a canonical B-valued inner product (unless \(B = \mathbb{C}\)); this is the content of Warning 3.3.3.
A right Hilbert space B-module \(H_B\) is a Hilbert space \(H\) equipped with a right B-action, i.e. a unital \(\star\)-homomorphism \(B^{op} \rightarrow B(H)\). Its \(\mathbb{C}\)-valued inner product does not restrict to a canonical B-valued inner product (unless \(B = \mathbb{C}\)); this is the content of Warning 3.3.3.
(VecSpace, InnerProduct) Trick: Realization via \(L^2 B\).
Given a Hilbert space module \(H_B\), apply the realization trick with the standard module \(L^2 B\) in place of \(B\). The adjointable maps \(Hom(L^2 B_B \rightarrow H_B)\) form a right Hilbert \(C^*\)-module with B-valued inner product
$$\langle f \vert g \rangle_B := f^{\dagger} g \in End(L^2 B_B) = B.$$ This converts the Hilbert space module \(H_B\) — which had no canonical B-valued inner product — into a genuine B-correspondence, exactly the kind of object the relative tensor product of correspondences can act on.
Given a Hilbert space module \(H_B\), apply the realization trick with the standard module \(L^2 B\) in place of \(B\). The adjointable maps \(Hom(L^2 B_B \rightarrow H_B)\) form a right Hilbert \(C^*\)-module with B-valued inner product
$$\langle f \vert g \rangle_B := f^{\dagger} g \in End(L^2 B_B) = B.$$ This converts the Hilbert space module \(H_B\) — which had no canonical B-valued inner product — into a genuine B-correspondence, exactly the kind of object the relative tensor product of correspondences can act on.
Connes Fusion
HilbertSpace Definition: Connes Fusion (the Space).
Given a right B-module \(H_B\) and a left B-module \({}_{B}K\), the Connes fusion relative tensor product Hilbert space \(H \boxtimes_B K\) is $$ Hom(L^2B_B \rightarrow H_B) \boxtimes_B K$$ equipped with the \(\mathbb{C}\)-valued inner product $$\langle f_1 \boxtimes \eta_1 \vert f_2 \boxtimes \eta_2 \rangle := \langle \langle f_2 \vert f_1 \rangle_B \eta_1 \vert \eta_2 \rangle_K$$ where $$\langle f_2 \vert f_1 \rangle_B = f_2^{\dagger}f_1 \in End(L^2B_B)=B.$$
Given a right B-module \(H_B\) and a left B-module \({}_{B}K\), the Connes fusion relative tensor product Hilbert space \(H \boxtimes_B K\) is $$ Hom(L^2B_B \rightarrow H_B) \boxtimes_B K$$ equipped with the \(\mathbb{C}\)-valued inner product $$\langle f_1 \boxtimes \eta_1 \vert f_2 \boxtimes \eta_2 \rangle := \langle \langle f_2 \vert f_1 \rangle_B \eta_1 \vert \eta_2 \rangle_K$$ where $$\langle f_2 \vert f_1 \rangle_B = f_2^{\dagger}f_1 \in End(L^2B_B)=B.$$
Bifunctor Definition: Connes Fusion (the Operation).
Given module maps \(x : H_B \rightarrow M_B\) and \(y : {}_{B}K \rightarrow {}_{B}N\), Connes fusion produces a map $$x \boxtimes_B y : H \boxtimes_B K \rightarrow M \boxtimes_B N$$ determined by $$f \boxtimes \eta \mapsto (x \circ f) \boxtimes y\eta$$ for \(f \in Hom(L^2B_B \rightarrow H_B)\) and \(\eta \in K\). Here \(x\) acts on the realized module \(Hom(L^2B_B \rightarrow H_B)\) by post-composition \(f \mapsto x \circ f\). This makes \(\boxtimes_B\) a bifunctor: it acts not only on modules but on the maps between them, and respects composition.
Given module maps \(x : H_B \rightarrow M_B\) and \(y : {}_{B}K \rightarrow {}_{B}N\), Connes fusion produces a map $$x \boxtimes_B y : H \boxtimes_B K \rightarrow M \boxtimes_B N$$ determined by $$f \boxtimes \eta \mapsto (x \circ f) \boxtimes y\eta$$ for \(f \in Hom(L^2B_B \rightarrow H_B)\) and \(\eta \in K\). Here \(x\) acts on the realized module \(Hom(L^2B_B \rightarrow H_B)\) by post-composition \(f \mapsto x \circ f\). This makes \(\boxtimes_B\) a bifunctor: it acts not only on modules but on the maps between them, and respects composition.