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Jun 23, 2010 at 4:20 comment added Tom Goodwillie ... in case of conformal structure I would think that the vector fields preserving that structure correspond naturally to the holomorphic sections of the complex tangent bundle, so that locally such a field corresponds to one holomorphic function -- where's the antiholomorphic one? -- and globally on the Riemann sphere they form a 3-d complex Lie algebra, that of the (finite-dimensional) global conformal symmetry group. What am I miss(understand)ing?
Jun 23, 2010 at 4:14 comment added Tom Goodwillie Can you explain further? As I understand it (as a topologist not a physicist), smooth tangent vector fields on a manifold correspond to local flows = "infinitesimal actions" of the group of real numbers. The Lie algebra of all vector fields corresponds in a sort of formal sense to the group of all diffeomorphisms. One can speak of the Lie subalgebra preserving some tensor field (meaning that the Lie derivative vanishes), and this structure corresponds in the same way to local diffeomorphisms preserving it. This fits with your discussion of the Hamiltonian setup, but ...
May 31, 2010 at 16:43 comment added José Figueroa-O'Farrill Oops. I didn't see this comment before now. A pair of holomorphic functions $f(z),g(z)$ defines an infinitesimal conformal transformation $z \mapsto z + f(z) + \overline{g(z)}$. They generate the Lie algebra of conformal transformations. It is in fact isomorphic to two copies of the algebra of diffeomorphism of the circle.
May 31, 2010 at 5:07 vote accept Qiaochu Yuan
May 17, 2010 at 21:23 comment added Qiaochu Yuan The second example looks more like what I want. What do you mean by a "Lie algebra of conformal transformations"? Something like locally invertible conformal maps?
May 17, 2010 at 11:51 history edited José Figueroa-O'Farrill CC BY-SA 2.5
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May 17, 2010 at 8:40 history edited José Figueroa-O'Farrill CC BY-SA 2.5
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May 17, 2010 at 8:15 history answered José Figueroa-O'Farrill CC BY-SA 2.5