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nowhereman

  • 5 years ago

Does the Kirillov-form depend on the point you choose from the coadjoint orbit?

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  1. Owlfred
    • 5 years ago
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  2. nowhereman
    • 5 years ago
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    On a Lie-group \(G\) you have the conjugation action on itself: \[g,h\in G, α_g(h) := ghg^{-1}\] and it's differential at the identity element \(e \in G\) is an isomorphism of the Lie-algebra \(\mathrm{Ad(g)} := \mathrm d_e α_g \in \mathrm{GL}(\mathfrak g)\) and the dual is called the coadjoint action: \[\mathrm{Ad}^*(g) := \mathrm{Ad}(g^{-1})^* \in \mathrm{GL}(\mathfrak g^*)\] It admits an orbit at any \(α \in \mathfrak g^*\):\[\mathcal O_α := \{\mathrm{Ad}^*(g)(α)\mid g \in G\}\simeq G/G_α\] where \[G_α := \{g \in G \mid \mathrm{Ad}^*(g)(α) = α\}\] So G acts on \(\mathcal O_α\), creating fundamental vector fields \(\tilde X\) for \(X \in \mathfrak g\). Then you can define the sympletic form \[ω(\tilde X, \tilde Y) = α([X,Y])\] So the question is: For \(β\in \mathcal O_α\) do we have \[ω(\tilde X, \tilde Y) = β([X,Y])\quad \text{?}\]

  3. amistre64
    • 5 years ago
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    my guess is; it doesnt matter ....

  4. nowhereman
    • 5 years ago
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    I would like to see a proof. If you want to use the exponential map, that would be ok too ;-)

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