Timeline for tangent bundle on noncommutative manifold
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 21, 2021 at 19:02 | comment | added | Ken.Wong | Thank you for your clarification! | |
| Jan 21, 2021 at 14:59 | comment | added | Branimir Ćaćić | Your fundamental problem really is this, then: you can have a noncommutative tangent bundle over $A$ (which, by the Serre–Swan philosophy, should be a finitely generated projective $A$-module) or you can have noncommutative vector fields on $A$ (which, by direct analogy with the commutative case, should be derivations $A \to A$), but you generally can't have both at the same time. | |
| Jan 21, 2021 at 14:56 | comment | added | Branimir Ćaćić | Another thing to think about: if $A$ is a noncommutative algebra, then set of all derivations $A \to A$ will generally be neither a left nor right $A$-module. | |
| Jan 21, 2021 at 14:52 | comment | added | Branimir Ćaćić | Classically, if $X$ is a vector field and $f$ is a scalar field, the action of $X$ as a $C^\infty(M)$-valued derivation on $f \in C^\infty(M)$ is equal to the the $C^\infty(M)$-valued dual pairing of $X$ with the $1$-form $\mathrm{d}\!f$: $X(f) = (\mathrm{d}\!f,X)$. | |
| Jan 21, 2021 at 9:53 | comment | added | Ken.Wong | One naive question, why do we need to put derivation on $\mathfrak{X}$? If we interpret $d_{D}$ as exterior derivative, as far as I know, exterior derivative do not act on tangent bundle. In fact I can't see why we need derivation structure on $\mathfrak{X}$. If we do not need derivation on $\mathfrak{X}$, then it seems we can define $\mathfrak{X}$ as some sort of tangent bundle. | |
| Jan 19, 2021 at 16:02 | vote | accept | Ken.Wong | ||
| Jan 19, 2021 at 15:45 | comment | added | Branimir Ćaćić | That said, you can’t automatically expect your $\mathcal{A}$-bimodule of de Rham $1$-forms $\Omega^1_D$ to necessarily be finitely-generated projective as a left or right $\mathcal{A}$-module; this is something that may or may not be true for any given example, and it may be a hypothesis you have to impose in certain circumstances. | |
| Jan 19, 2021 at 15:43 | comment | added | Branimir Ćaćić | Exactly. To put things in perspective, there’s a general notion of noncommutative differential calculus based on noncommutative differential forms. Then, a spectral triple gives rise to a canonical noncommutative differential calculus sometimes called the de Rham calculus in the literature—this is because you really do recover your usual exterior derivative and differential $1$-forms (up to the inclusion of the cotangent bundle in its complexified Clifford bundle) when your spectral triple is a commutative spectral triple. | |
| Jan 19, 2021 at 15:38 | comment | added | Ken.Wong | Thank you for your great answer, I learned a lot in this answer! So instead of tangent bundle, we can actually define some sort of cotangent bundle on spectral triple? | |
| Jan 19, 2021 at 15:11 | history | answered | Branimir Ćaćić | CC BY-SA 4.0 |