Given a smooth manifold U, we have a map $\wedge^2\Gamma(U,TU)\to \Gamma(U,TU)$ given by $X\wedge Y\mapsto [X,Y]$, where $TU$ denotes the tangent bundle. Is it possible to describe the map $\Gamma(U,T^*U)\to \Gamma(U,\wedge^2 T^*U)$ corresponding to this map.
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8$\begingroup$ Though Victor's answer sort of tells the moral/ideological story of it, the question is still wrong as stated. The commutator of vector fields is not a well-defined map from the global sections of the exterior square of the tangent bundle, since it is not linear over the functions on the manifold, but only over the constants. $\endgroup$– Leonid PositselskiCommented Dec 8, 2010 at 0:19
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$\begingroup$ Additionally, the spaces $\Gamma(U,TU)$ and $\Gamma(U,T^{*}U)$ aren't dual to each other over the constants. $\endgroup$– Victor ProtsakCommented Dec 8, 2010 at 1:59
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1$\begingroup$ Well, when a map between the global sections of two vector bundles is linear over the functions, it does induce a map of the global sections of the dual bundles in the opposite direction. In this case, the exterior differential is not linear over the functions, and the commutator does not even exist as a map of global sections (of the kind stated in the question). $\endgroup$– Leonid PositselskiCommented Dec 8, 2010 at 2:35
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3 Answers
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To expand on Leonid's comment, if $\omega$ is a 1-form and $X,Y$ are vector fields, then $$ d\omega(X \wedge Y) = X \omega(Y) - Y \omega(X) - \omega([X,Y]). $$ If the first two terms were not there, then one could say, as in Victor's answer, that the exterior derivative is (minus) the transpose of the Lie bracket of vector fields. The fact that the first two terms are there is symptomatic of Leonid's observation that the Lie bracket is not really a tensorial map.
answered Dec 8, 2010 at 0:50
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$\begingroup$ Thanks, I was looking for a formula like the one you have written. $\endgroup$– RexCommented Dec 8, 2010 at 18:29
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2$\begingroup$ You're welcome. Personally, though, I think that Urs's answer should be the accepted answer to the question which should have been asked. $\endgroup$ Commented Dec 8, 2010 at 22:36
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The dual of a Lie bracket is the differential in the corresponding Chevalley-Eilenberg algebra.
Background, formulas, details and examples are at
nLab: Chevalley-Eilenberg algebra.
This makes sense for "Lie bracket" understood in the general sense of Lie algebroids and $L_\infty$-algebras and fully generally for $\infty$-Lie algebroids.
In the case at hand, when regarding $T X$ as a Lie algebroid (instead of regarding $\Gamma(T X)$ as just a Lie algebra) the corresponding CE-algebra is the de Rham complex
$$ CE(T X) = (\Omega^\bullet(X), d_{dR}) $$
and the general formula for the dual of a Lie bracket on a Lie algebroid reproduces the familiar formula for the de Rham differential.
answered Dec 8, 2010 at 11:29
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Yes. The dual of the Lie bracket is the exterior differential that maps 1-forms into 2-forms. See any good textbook on differential geometry.
answered Dec 7, 2010 at 23:30
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$\begingroup$ What particular book did you have in mind? $\endgroup$ Commented Dec 8, 2010 at 2:46
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5$\begingroup$ This is morally correct but technically wrong, as pointed out in Jose's answer and the comments to the question (including yours). But, yes, I do agree at a moral level that the bracket and the differential are dual. One way to make this precise, as you probably know but I'll mention it for other readers, is the Koszul duality between Lie algebra/oids and differential graded commutative algebras. When applied to tangent bundle, this Koszul duality returns the de Rham complex. $\endgroup$ Commented Dec 8, 2010 at 6:41