Let $(A,H,D)$ be a spectral triple and let $B$ be an algebra which is Morita equivalent to $A$. Then there exists a finitely generated, projective $A$ module $E$ such that $B=End_A(E)$. Endow $E$ with $A$-valued inner product $(\cdot,\cdot)$ and form a Hilbert space $H':=E \otimes_A H$ with the inner product given by $$\langle x \otimes \xi, y \otimes \eta \rangle_{H'}:=\langle (x,y)\xi,\eta \rangle_H$$ One would like impose the structure of a spectral triple on $B$: one naive choice is to put $D'(x \otimes \xi)=x \otimes D\xi$ however this map is not well defined since $H'$ is a tensor product over $A$ In order to fix this one uses a connection $\nabla$ i.e. a map $\nabla:E \to E \otimes_A \Omega_D^1$ where $\Omega^1_d=\{\sum_{j=1}^N a_jdb_j: a_j,b_j \in A, N \in \mathbb{N}\}$ and $db:=[D,b]$. This map has to satisfy $\nabla(xa)=\nabla(x)a+x\otimes da$ (note that due to Jacobi idenity for commutators the space $\Omega_D^1$ is in fact $A-A$-bimodule thus $E \otimes_A \Omega^1_D$ is a right $A$-module). Having fixed a connection one defines $$D'(x \otimes \xi)=x\otimes D\xi + \nabla(x)\xi$$ (where $\nabla(x)\xi$ is understood as follows: if $\nabla(x)=\sum_j x_j \otimes \omega_j$ then $\nabla(x)\xi=\sum_j x_j \otimes \omega_j(\xi)$ which is fine since each $\omega_j \in \Omega^1_D$ still acts on $H$). The natural action of $B$ on $H'$ is as follows $b(x \otimes \xi)bx \otimes \xi$ which is fine since $b \in B=End_A(E)$.

How to show that for each $b \in B$ the commutator $[D',b]$ (extends to a/) is a bounded operator on $H'$?

Simple computation shows that this commutator evaluated on $x \otimes \xi$ is equal to $\nabla(bx)\xi-b(\nabla x(\xi))$ and I don't see how to obtain that this operator is bounded-do we have to assume something on $\nabla$ (some sort of compatibility with the $B$-action)?

  • 1
    $\begingroup$ This absolutely deserves a detailed canonical answer, but in the meantime, the heart of the matter can be found in Section 3.4 of Blecher–Kaad–Mesland arxiv.org/abs/1703.10063 $\endgroup$ – Branimir Ćaćić May 3 '19 at 18:52
  • $\begingroup$ Actually, to be honest, everything you need is in Section 2 of Chakraborty–Mathai. arxiv.org/abs/0804.3232 $\endgroup$ – Branimir Ćaćić May 4 '19 at 21:50

Your question is entirely covered by Section 2 of Brain–Mesland–Van Suijlekom, but the fgp case is simple enough to ultimately boil down to folklore proved by Chakraborty–Mathai. Let me summarise what happens, while incorporating some technical simplifications from Blecher–Kaad–Mesland. For convenience, let me take $A$ and $B$ to be unital $C^\ast$-algebras and let $\Omega_D$ be the unital $C^\ast$-subalgebra of $B(H)$ generated by $A$ and $[D,\mathcal{A}]$.

  1. Let $\mathcal{A} := \{a \in A \mid a \operatorname{Dom}(D) \subset \operatorname{Dom}(D),\, [D,a] \in B(H)\}$ be given the Lipschitz norm $\|a\| := \|a\| + \|[D,a]\|$. Then $\mathcal{A}$ defines an involutive operator algebra and the inclusion $\mathcal{A} \hookrightarrow A$ is completely bounded with dense range that is closed under the holomorphic functional calculus; in particular, it follows that $M_N(\mathcal{A})$ is dense in $M_N(A)$ and closed under the holomorphic functional calculus for any $N \in \mathbb{N}$. You should think of $\mathcal{A}$ as defining a Lipschitz structure on the NC space $A$.
  2. Let $\mathcal{E}$ be a dense subspace of $E$ satisfying $\mathcal{E} \cdot \mathcal{A} \subset \mathcal{E}$ and $(\mathcal{E},\mathcal{E})_A \subset \mathcal{A}$. By part 1 and the properties of $\mathcal{E}$, we can find $A$-module generators $\{\xi_1,\dotsc,\xi_N\} \subset \mathcal{E}$ for $E$, such that $$ \forall e \in E, \quad e = \sum_{i=1}^N \xi_i \cdot (\xi_i,e)_A. $$ Thus, if $p := \left((\xi_i,\xi_j)_A\right)_{i,j=1}^N \in M_N(\mathcal{A})$, then $e \mapsto \left((\xi_i,e)_A\right)_{i=1}^N$ defines an isomorphism $E \cong pA^N$ of Hilbert $A$-modules that restricts to an isomorphism $\mathcal{E} \cong p\mathcal{A}^N$ of pre-Hilbert $\mathcal{A}$-modules; this now makes $\mathcal{E}$ into a (finitely generated) projective operator module over $\mathcal{A}$ in a manner that depends on the choice of $\{\xi_1,\dotsc,\xi_N\}$ only up to completely bounded isomorphism. You should think of $\mathcal{E}$ as defining a Lipschitz structure on the NC vector bundle $E$; since $E$ is fgp, this Lipschitz structure is canonically induced by the choice of Lipschitz structure on $A$. Note that such $\mathcal{E}$ exists: given an isomorphism $E \cong p_0 A^N$ of Hilbert $A$-modules for $p_0 \in M_N(A)$ an orthogonal projection, you can use the properties of $\mathcal{A}$ to find an orthogonal projection $p \in M_N(\mathcal{A})$, such that $E \cong p A^N$, in which case, you can take $\mathcal{E}$ to be the pre-image of $p\mathcal{A}^N$ and $\{\xi_1,\dotsc,\xi_N\}$ to be given by the pre-images in $E$ of the columns of $p$.
  3. Let $\mathcal{B} := \{b \in B \mid b \cdot \mathcal{E} \subset \mathcal{E}\}.$ Since $$ \mathcal{B} = \left\{b \in B \mid \left((\xi_i,b\xi_j)_A\right)_{i,j=1}^N \in M_N(\mathcal{A})\right\}, $$ it follows that $\mathcal{B}$ is $\ast$-closed and that the $\ast$-isomorphism $B \cong p M_N(A) p$ induced by the Hilbert $A$-module isomorphism $E \cong p A^N$ restricts to a $\ast$-isomorphism $\mathcal{B} \cong p M_N(\mathcal{A}) p$. Thus, $\mathcal{B}$ can be topologised as a closed $\ast$-subalgebra of the involutive operator algebra $M_N(\mathcal{A})$, thereby making $\mathcal{E}$ into a Lipschitz $(\mathcal{B},\mathcal{A})$-correspondence. In particular, you can view $\mathcal{B}$ as defining a Lipschitz structure on the NC space $B$ compatible with the Lipschitz structure $\mathcal{A}$ on the NC space $A$.
  4. Let $\nabla : \mathcal{E} \to E \hat\otimes_A \Omega_D$ be a Hermitian connection, i.e., a $\mathbb{C}$-linear map satisfying $$ \forall e \in \mathcal{E},\, \forall a \in \mathcal{A}, \quad \nabla(ea) = \nabla(e)a + e \hat\otimes [D,a],\\ \forall e_1,e_2 \in \mathcal{E}, \quad (\nabla(e_1),e_2 \hat\otimes 1)_{\Omega_D} +(e_1 \hat\otimes 1,\nabla(e_2))_{\Omega_D} = [D,(e_1,e_2)_A]; $$ for example, the Graßmann connection $\nabla_0$ induced by the frame $\{\xi_1,\dotsc,\xi_N\}$ is defined by $$ \forall e \in \mathcal{E}, \quad \nabla_0(e) := \sum_{i=1}^N \xi_i \hat\otimes [D,(\xi_i,e)_A], $$ and indeed, if $\nabla$ is any other Hermitian connection, then $\nabla = \nabla_0 + \omega$ for $\omega \in B(E,E \hat\otimes_A \Omega_D)$ defined by $$ \forall e \in \mathcal{E}, \quad \omega(e) := \sum_{i=1}^N \nabla(\xi_i)(\xi_i,e)_A. $$ Given a Hermitian connection $\nabla$, define $1 \hat\otimes_\nabla D : \mathcal{E} \otimes^{\mathrm{alg}}_{\mathcal{A}} \operatorname{Dom}(D) \to E \hat\otimes_A H$ by $$ \forall e \in \mathcal{E}, \, \forall h \in H, \quad 1 \hat\otimes_\nabla D(e \hat\otimes h) := \nabla(e)h + e \hat\otimes Dh = \omega(e)h + 1 \hat\otimes_{\nabla_0}D(e \hat\otimes h). $$ Then, by Section 2 of Chakraborty–Mathai, the operator $1 \hat\otimes_\nabla D$ is essentially self-adjoint and defines a spectral triple $(B,E \hat\otimes_A H,1 \hat\otimes_\nabla D)$ with Lipschitz algebra $\mathcal{B}$. In particular, by a direct computation, $$ \forall b \in \mathcal{B}, \, \forall e \in E, \, \forall h \in H, \quad [1 \hat\otimes_{\nabla_0} D,b](e \hat\otimes h) = \sum_{i,j=1}^N \xi_i \hat\otimes [D,(\xi_i,b\xi_j)_A](\xi_j,e)_A h; $$ so that $1 \hat\otimes_\nabla D$, which is a bounded perturbation of $1 \hat\otimes_{\nabla_0} D$, also has bounded commutators with $\mathcal{B}$.

To conclude, the missing ingredient was a Lipschitz structure on the fgp $A$-module $E$, i.e., a dense subspace $\mathcal{E}$ of $E$, such that $\mathcal{E} \cdot \mathcal{A} \subset \mathcal{E}$ and $(\mathcal{E},\mathcal{E})_A \subset \mathcal{A}$. Because $E$ is fgp, this suffices to pick out a dense $\ast$-subalgebra $\mathcal{B}$ of Lipschitz elements of $B$ and to pin down any Hermitian connection $\nabla$ on $E$ with domain $\mathcal{E}$ up to bounded perturbation, so that $\mathcal{B}$ will necessarily have bounded commutators with $1 \hat\otimes_\nabla D$; indeed, by using a normalised tight frame for $E$ consisting of vectors in $\mathcal{E}$, you can boil everything down to the compatibility of $\mathcal{A}$ with $D$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.