Timeline for Sufficient conditions for $\mathrm{Der}_k(A)$ to be f.g. projective
Current License: CC BY-SA 4.0
18 events
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| Jul 22, 2020 at 6:58 | comment | added | Grisha Papayanov | @YemonChoi the existence of cut-off functions shows that all derivations are local and then Hadamard lemma (Taylor series up to order two) shows that a derivation is determined by its values on coordinate functions | |
| Jul 22, 2020 at 2:02 | comment | added | Grisha Papayanov | a silly example is that this is true if A is an algebra of continious functions on a topological space (with no restictions on the space whatsoever), because these algebras do not have any derivations at all (except for the zero one) | |
| Jul 22, 2020 at 0:31 | comment | added | Yemon Choi | Thanks Tobias and @jg1896 - I withdraw my earlier caveats | |
| Jul 22, 2020 at 0:26 | comment | added | Tobias Fritz | @jg1896: great, I was actually looking for a reference for that! BTW I had stupidly forgotten to mention that $A$ can be assumed commutative, and have just edited the question accordingly. | |
| Jul 22, 2020 at 0:25 | comment | added | Tobias Fritz | @YemonChoi: yes, for me the dual is $\mathrm{Hom}_A(M,A)$. In differential geometry, being a derivation is one of several equivalent standard definitions of a vector field, and no additional continuity condition is needed. Everything happens locally, and manifolds are finite-dimensional and hence locally compact, which I think is what makes this work. | |
| Jul 22, 2020 at 0:25 | comment | added | jg1896 | A remark here: for commutative rings $A$, if $M$ is a finitely generated projective module, then so is $M^*= Hom_A(M,A)$ by the dual basis lemma: see McConnell, Robinson, Noncommutative Noetherian Rings, revised edition, 3.5.2 | |
| Jul 22, 2020 at 0:20 | history | edited | Tobias Fritz | CC BY-SA 4.0 |
forgot assumption
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| Jul 22, 2020 at 0:19 | comment | added | Yemon Choi | BTW, why is it obvious that derivations from $C^\infty(M)$ to itself are automatically continuous and hence given by vector fields? I am not a diff geometer so maybe I am overlooking something. Certainly automatic continuity problems for Banach algebras can be much tougher, but the Frechet category is niceer I admit | |
| Jul 22, 2020 at 0:16 | comment | added | Yemon Choi | Tobias, I am sufficiently fusty that I would recommend consulting books like Weibel when it comes to homological algebra rather than the nLab. Perhaps we are using different terminology from each other; if A is a unital K-algebra then K-linear derivations from A to a symmetric A-module correspond to elements of ${\rm Hom}_A( \Omega_{A/K} , M)$. For me the dual of an $A$-module $M$ is ${\rm Hom}_K (M,K)$ but perhaps for you the dual of $M$ is ${\rm Hom}_A(M,A)$ ? | |
| Jul 22, 2020 at 0:12 | comment | added | Tobias Fritz | Already purely algebraically, derivations on $C^\infty(M)$ form a fg projective module, and I'd like to know whether there are any more general statements of which this would be a special case. | |
| Jul 22, 2020 at 0:10 | comment | added | Tobias Fritz | @YemonChoi: thanks! $\mathrm{Der}(A)$ consists of the derivations $A \to A$, which by definition of Kähler differentials $\Omega^1_K$ coincides with $A$-module maps $\Omega^1_K \to A$, i.e. with the dual module $(\Omega^1_K)^*$; this is from the nLab. What am I missing? And I don't want topological constraints, although I'm well aware of the difference between Kähler differentials $\Omega^1_K$ and the usual module of 1-forms $\Omega^1$. | |
| Jul 22, 2020 at 0:03 | comment | added | Yemon Choi | Also, as soon as you want to take $C^\infty(M)$ you probably want to be imposing topological constraints: there is an old MO question mathoverflow.net/questions/6074/… | |
| Jul 22, 2020 at 0:01 | comment | added | Yemon Choi | Tobias: regarding your final statement, the dual of the Kahler module is ${\rm Der}(A,A^*)$, which is not going to be naturally isomorphic to ${\rm Der}(A)$ although perhaps they coincide for some reason when $A$ is regular Noetherian. Indeed, while there are natural instances where the Kahler module is projective, why would that make its dual projective? Surely you'd expect the dual of something projective to be injective ? | |
| Jul 21, 2020 at 23:53 | answer | added | jg1896 | timeline score: 9 | |
| Jul 21, 2020 at 20:31 | comment | added | Tobias Fritz | @JoshuaMundinger: if it's still confusing, then just ignore my statements about $C^\infty(M)$. I'm just looking for sufficient conditions to guarantee that the module of derivations is fg projective. Thus being regular Noetherian would be one possible answer, since then (if I understood correctly) the module of Kähler differentials is already fg projective and the module of derivations is its dual. | |
| Jul 21, 2020 at 20:08 | comment | added | Tobias Fritz | @JoshuaMundinger: I'd like to have conditions which apply to some reasonably general class of algebras, which in particular includes those of the form $C^\infty(M)$ in case that the ground field is $\mathbb{R}$. Indeed, since I understand this case, a sufficient condition of the form "$A$ is isomorphic to some $C^\infty(M)$" would not be interesting. | |
| Jul 21, 2020 at 19:13 | comment | added | Joshua Mundinger | The question is confusing because it implies that you would like conditions for the case you already know about (smooth functions on a manifold)... are you saying that you would like to consider when $A/k$ is not necessarily of finite type? | |
| Jul 21, 2020 at 17:48 | history | asked | Tobias Fritz | CC BY-SA 4.0 |