tangent bundle on noncommutative manifold

Question

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). Is this the canonical way to do this?

Answer 1

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).

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.

Answer 2

Here are some thoughts which essentially complement Branimir's answer. In many naturally occurring examples, in particular the theory of quantum groups, it is actually more natural to start with a noncommutative cotangent bundle and to then construct the spectral triple from it. More explicitly, start with a pre-$C^*$-algebra $A$ whose noncommutative geometry you are trying to understand.

1. Look for a dga (differential graded algebra) $\Omega^{\bullet}$ for which $A = \Omega^0$.

2. Endow $\Omega^{\bullet}$ with some type of ``Riemannian metric'', which is to say an inner product $\langle \cdot,\cdot \rangle$.

3. Confirm that the differential d of the dga is adjointable $\langle \cdot,\cdot \rangle$. Denote by d$^*$ the adjoint of d.

4. Complete $\Omega^{\bullet}$ to a Hilbert space.

5. Check that the closure of the symmetric operator $D:=$d$+$d$^*$ is a satisfies the axioms of a spectral triple.

I should stress that this is a rough plan of attack, and will only work in situations which are, in some sense, close to the classical situation. For example, $q$-deformed $C^*$-algebras, in particular examples coming from Drinfeld-Jimbo quantum groups.

The prototypical example here is the Podles sphere $\mathcal{O}_q(S^2)$, the degree zero part of a $\mathbb{Z}$-grading on $\mathcal{O}_q(SU_2)$, which $q$-deforms the algebra of algebraic functions of the $2$-sphere. This pre-$C^*$-algebra admits an essentially unique $\mathcal{O}_q(SU_2)$-covariant dga, called the Podles calculus. The calculus carries a $q$-deformation of the Fubini-Study metric, with respect to which $d$ is adjointable, and which gives a spectral triple upon completion of $\Omega^{\bullet}$ and closure of $D = $d$+$d$^*$. The dga constructed from this spectral triple (in the sense Branimir explained above) will coincide with $\Omega^{\bullet}$. This spectral triple was first constructed by Dabrowski and Sitarz (or at least the Dolbeault-Dirac version) and later reconstructed by Majid from a more geometric point of view.

This construction can be extended to a much larger class of examples called quantum flag manifolds . . . but this is very much ongoing work.