Timeline for The minimum codimension of Lie subalgebra of $\chi^{\infty}(M)$
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Sep 20, 2017 at 10:00 | vote | accept | Ali Taghavi | ||
| Aug 9, 2016 at 12:01 | comment | added | David E Speyer | Your space $S$ is a subalgebra: The Poisson algebra $C^{\infty}(M)$ (for $M$ symplectic, compact, connected) always splits as the direct sum $\mathbb{R} \oplus S$. But I can't think of a map $\mathrm{Vect}(M) \to C^{\infty}(M)$ which doesn't land in $S$. (If $M$ is Riemmannian, then we have $\mathrm{Vect}(M) \cong \Omega^1(M) \overset{d^{\ast}}{\longrightarrow} C^{\infty}(M)$, but, at least if the Riemannian structure gives the same volume form as the symplectic one, the image of $d^{\ast}$ lands in $S$. And I have no idea if this is a map of Lie algebras anyway.) | |
| Aug 7, 2016 at 16:59 | comment | added | mme | On a side note, if one solves the first two problems, you've actually proved not only that the minimum codimension of a subalgebra is $n$, but also that all codimension $n$ subalgebras are these $L_p$. Hurtado shows that when $m=n$, the homomorphism is of the form "Lift a diffeomorphism of $M$ to a diffeomorphism of $N$ through a covering map $p: N \to M$." By your construction, the homomorphism sends $L$ to one of the $L_p$ in $\text{Diff}(N)$ (specifically, to $L_{[G]}$.) | |
| Aug 7, 2016 at 14:57 | comment | added | Ali Taghavi | 1)Is the space $S$ of all $f$ with $\int_{M} fd\Omega =0$ a Lie algebra (with poisson structure on $C^{\infty}(M)$? \new line 2) Is there a non zero Lie algebra morphism from $\chi^{\infty}(M) $ to $C^{\infty}(M)$? Is there a Lie algebra Morphism whose image is not contained in the above $S$. If the answers to the above questions would be positive then we have a codimension one Lie subalgebra of $\chi^{\infty}(M)$. | |
| Aug 7, 2016 at 14:56 | comment | added | Ali Taghavi | @DavidSpeyer What about the following alternative opposit approach for compact symplectic manifold $(M, \omega)$ with volum form $\Omega=\omega^{n}$: | |
| Aug 7, 2016 at 14:14 | comment | added | David E Speyer | @MikeMiller Agreed. I was actually thinking of coming back today to remark on this. There are a few ways we could try to get around this: (1) restrict to cases where $\pi_1(\mathrm{Diff}(M))$ is trivial (2) replace $\exp(L)$ by its closure (but then we have to argue that its closure isn't the whole group) (3) See if Hurtado's proof adapts to the universal cover of $\mathrm{Diff}(M)$. | |
| Aug 7, 2016 at 12:16 | comment | added | mme | The second step seems fishy, even in finite dimensions; if we integrate an irrational line in $\Bbb R^2$ to a subgroup in the torus, it needn't be closed, so we don't get a manifold structure on the quotient. | |
| Aug 6, 2016 at 19:17 | history | edited | David E Speyer | CC BY-SA 3.0 |
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| Aug 6, 2016 at 19:09 | history | edited | David E Speyer | CC BY-SA 3.0 |
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| Aug 6, 2016 at 18:43 | comment | added | Ali Taghavi | Thank you very much for your elegant strategy. May be the extra assumption "L is an ideal " makes some facilities? | |
| Aug 6, 2016 at 18:07 | history | answered | David E Speyer | CC BY-SA 3.0 |