# Non-commutative geometry – Dirac operator on an equivalent Morita algebra

Let $$(A, H, D)$$ to be a spectral triple and let $$B$$ to be an algebra that is Morita equivalent to $$A$$. Then there is a project, finely generated generated $$A$$ module $$E$$ such as $$B = End_A (E)$$. endow $$E$$ with $$A$$valued domestic product $$( cdot, cdot)$$ and forms a Hilbert space $$H = H$$ with the inner product given by
$$langle x otimes xi, y otimes eta rangle_ {H}: = langle (x, y) xi, eta rangle_H$$
We would like to impose the structure of a spectral triple at $$B$$: a naive choice is to put $$D (x otimes xi) = x otimes D xi$$ However, this map is not well defined because $$H$$ is a tensor product more than $$A$$

In order to solve this problem, use a connection $$nabla$$ it is a map $$nabla: E to E otimes_A Omega_D ^ 1$$ or $$Omega ^ 1_d = { sum_ {j = 1} ^ N a_jdb_j: a_j, b_j in A, N in mathbb {N} }$$ and $$db: =[D,b]$$. This card must satisfy $$nabla (xa) = nabla (x) a + x otimes$$ (note that because of Jacobi's idenity for switches, the space $$Omega_D ^ 1$$ is in fact $$A-A$$-bimodule then $$E otimes_A Omega ^ 1_D$$ is a right $$A$$-module). Having fixed a connection we define
$$(x otimes xi) = x otimes D xi + nabla (x) xi$$
(or $$nabla (x) xi$$ is understood as follows: $$nabla (x) = sum_j x_j otimes omega_j$$ then $$nabla (x) xi = sum_j x_j otimes omega_j ( xi)$$ which is good since each $$omega_j in Omega ^ 1_D$$ always acts on $$H$$).
The natural action of $$B$$ sure $$H$$ 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 everyone $$b B$$ the switch $$[D’,b]$$ (extends to a /) is an operator bounded on $$H$$?

A simple calculation shows that this switch evaluated on $$x otimes xi$$ is equal to $$nabla (bx) xi-b ( nabla x ( xi))$$ and I do not see how to get this operator to be bounded – do we have to assume something about $$nabla$$ (a kind of compatibility with the $$B$$-action)?

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