nowhereman
  • nowhereman
Does the Kirillov-form depend on the point you choose from the coadjoint orbit?
Mathematics
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

Owlfred
  • Owlfred
Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!
nowhereman
  • nowhereman
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 \[ω(\tilde X, \tilde Y) = β([X,Y])\quad \text{?}\]
amistre64
  • amistre64
my guess is; it doesnt matter ....

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.