Using the Serre-Swan's theorem, one can do vector bundle theory on noncommutative manifold $(A,H,D)$, by replacing vector bundle by finitely generated projectve module $M$. For the construction of tangent bundle, one can use derivation, but I am not sure how to define derivation on the module. The idea seems to be define derivation on $A$ satisfy Leibniz rule(see [here](https://ncatlab.org/nlab/show/derivation)). Is this the canonical way to do this?



tangent bundle on noncommutative manifold
Noncommutative Riemannian (spin) geometry via spectral triples is grounded in an approach to noncommutative differential calculus that privileges the cotangent bundle over the tangent bundle: given a spectral triple $(\mathcal{A},H,D)$, you have an $\mathcal{A}$-bimodule
$$
 \Omega^1\_D := \operatorname{Span}\{ a \cdot [D,b] \mid a,b \in \mathcal{A}\} \subset B(H),
$$
which you interpret as the bimodule of $1$-forms, and a $\ast$-derivation
$$
 \mathrm{d}\_D : \mathcal{A} \to \Omega^1\_D, \quad \mathrm{d}\_D(a) := [D,a],
$$
which you interpret as the exterior derivative from scalar fields to $1$-forms; the pair $(\Omega^1\_D,\mathrm{d}\_D)$ therefore defines something called a *first-order differential calculus*. It is not automatic that $\Omega^1\_D$ is finitely generated projective as a left or right $\mathcal{A}$-module, but hypotheses of this kind have been explored in the literature (see, e.g., [Lord–Rennie–Várilly](https://arxiv.org/abs/1109.2196)).


Now, this may begin to sound like the situation in algebraic geometry, where you construct cotangent spaces first and then *define* tangent spaces as their linear duals; certainly, in differential geometry, the tangent bundle is naturally isomorphic to the dual of the cotangent bundle. However, if you now want to define your noncommutative tangent bundle to be something like $\mathfrak{X}\_D := \operatorname{Hom}\_{\mathcal{A}}(\Omega^1\_D,\mathcal{A})$, where $\Omega^1\_D$ is viewed as a left or right $\mathcal{A}$-module, you run into a basic problem: if $\mathcal{A}$ is noncommutative, there’s no reason why $X \circ \mathrm{d}\_D : \mathcal{A} \to \mathcal{A}$ should ever be a derivation for $X \in \mathfrak{X}\_D$. You can solve this by restricting $\mathfrak{X}\_D$ to consist of $\ast$-preserving $\mathcal{A}$-bimodule maps $\Omega^1\_D \to \mathcal{A}$, but now you’ve just replaced one problem with another: if $\mathcal{A}$ is noncommutative, then this restricted $\mathfrak{X}\_D$ will generally be neither a left nor right $\mathcal{A}$-module.


To sum up, then, it’s generally impossible to construct a “noncommutative tangent bundle” $\mathfrak{X}\_D$ that simultaneously:


* defines a right or left $\mathcal{A}$-module of left or right $\mathcal{A}$-linear maps $\Omega^1\_D \to \mathcal{A}$, and
* yields $\ast$-derivations of $\mathcal{A}$ by composing with $\mathrm{d}\_D$.


There is an approach to noncommutative differential calculus that privileges the tangent bundle over the cotangent bundle and works primarily with derivations—the work of Dubois-Violette comes particularly to mind—but for the above reason, it can be difficult to apply in the context of spectral triples.


