Timeline for Is it possible to construct a formal group law from a Lie group without choosing coordinates?
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| Oct 28, 2013 at 18:04 | comment | added | Vladimir Dotsenko | @Qiaochu Yuan: Isn't it the whole point in dealing with topological groups that they are uniform spaces, so knowing things close to the identity must correspond to something global? :) I do not insist on my version of the formulation of that theorem, but I am not buying into your explanation ;) | |
| Oct 23, 2013 at 20:54 | comment | added | Marguax | @Qiaochu Yuan: OK, but if the formal group is coming conceptually prior to the Lie algebra, then a "construction" of the formal group in terms of the Lie algebra is somewhat circular in flavor. I do agree with your statement about completions, but since the map to the completion is neither flat nor injective it seems a rather "bad" concept to contemplate, or rather since a canonical analytic manifold structure is available then it seems "better" to think in terms of that for any discussion involving completion; just a matter of taste. | |
| Oct 23, 2013 at 19:51 | comment | added | Qiaochu Yuan | @Marguax: actually I am happy to think of the formal group as conceptually prior to the Lie algebra. Can you expand on what the difficulties with completing the local ring are in the smooth setting? It seems like we get the same completion either way but perhaps I am missing a subtlety. | |
| Oct 23, 2013 at 13:36 | comment | added | Marguax | @Qiaochu Yuan: But how do you put the Lie algebra structure on this tangent space without either invoking the global group (through left-invariant vector fields) or the formal group which are you trying to put logically afterwards? It feels circular from the purely local point of view. (I also still think that you should regard $G$ as an analytic manifold and not just as a smooth manifold for consideration of completion of the local ring at the identity.) | |
| Oct 22, 2013 at 19:22 | comment | added | Qiaochu Yuan | @Vladimir: that's a fine theorem, but I really want a result which agrees with my intuition that what we are trying to study is a formal neighborhood of the identity, and invariant differential operators are a priori a global notion. (This is the same reason I don't want to define $\mathfrak{g}$ as left invariant vector fields and would honestly prefer to define it as the tangent space to the identity.) | |
| Oct 22, 2013 at 18:25 | comment | added | Vladimir Dotsenko | The last theorem admits a variant that I prefer a little bit since it is a bit less analytic, and also a bit more to the point of what universal enveloping algebras are: $U(\mathfrak{g})$ is the algebra of differential operators on $G$ that are invariant w.r.t. left shifts. (Similarly to how $\mathfrak{g}$ is the Lie algebra of left invariant vector fields). | |
| Oct 22, 2013 at 17:27 | history | edited | Qiaochu Yuan | CC BY-SA 3.0 |
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| Oct 22, 2013 at 17:26 | comment | added | Qiaochu Yuan | @Marguax: yes, you are right. I'll edit. | |
| Oct 22, 2013 at 9:22 | vote | accept | Richard Garner | ||
| Oct 22, 2013 at 9:21 | comment | added | Richard Garner | Thanks, this is just the sort of thing I was looking for. I had the feeling that one should use something else than an actual formal group law. I was aware of the two dual Hopf algebras arising from O(G) and U(G) but not how to construct them directly from G. I had a few questions which I've appended to Marguax's comments above. | |
| Oct 22, 2013 at 9:18 | comment | added | Marguax | It is totally non-canonical to lift $V$ back into the ring. To do that lifting is almost like choosing coordinates (not quite, but rather close). The completed local ring is canonically filtered (by powers of the maximal ideal), not canonically graded. | |
| Oct 22, 2013 at 7:42 | comment | added | Qiaochu Yuan | @Marguax: once you have the counit you have the augmentation ideal $I$ and then $V$ is canonically isomorphic to $I/I^2$. Or am I mistaken? | |
| Oct 22, 2013 at 7:33 | comment | added | Marguax | For working with completions of local rings it is "safer" to restrict to the analytic case, not the $C^{\infty}$ case, simply because local rings in the C$^{\infty}$ case are not noetherian and hence completions have to be treated with rather more care than in the noetherian case (e.g., the map from the local ring to its completion isn't even injective). But what is the intrinsic significance of the choice of $V$ in terms of your discussion with symmetric algebras? | |
| Oct 22, 2013 at 7:18 | history | edited | Qiaochu Yuan | CC BY-SA 3.0 |
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| Oct 22, 2013 at 7:12 | history | answered | Qiaochu Yuan | CC BY-SA 3.0 |