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By Bryant R.L.

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Note that Gg·m = g Gm g −1 . Thus, whenever H ⊂ G is the stabilizer of a point of M, then all of the conjugate subgroups of H are also stabilizers. These results imply that GM = Gm m∈M is a closed normal subgroup of G and consists of those g ∈ G for which g · m = m for all m ∈ M. Often in practice, GM is a discrete (in fact, usually finite) subgroup of G. When this is so, we say that the action is almost effective. The following theorem says that orbits and stabilizers are particularly nice objects.

Xn of g∗ , we can write the expression for the Lie bracket as an element β ∈ g ⊗ Λ2 (g∗ ), in the form β = 12 cijk xi ⊗ xj ∧ xk . 21 1 6 m m cij cm xm ⊗ xi ∧ xj ∧ xk . k + cjk ci + cki cj 32 Exercise Set 2: Lie Groups 1. Show that for any real vector space of dimension n, the Lie group GL(V ) is isomorphic to GL(n, R). ) 2. Let G be a Lie group and let H be an abstract subgroup. Show that if there is an open neighborhood U of e in G so that H ∩ U is a smooth embedded submanifold of G, then H is a Lie subgroup of G.

Mk = Gm1 ∩ Gm2 ∩ · · · ∩ Gmk . If one can arrange that this intersection is discrete, then one can explicitly compute a fundamental solution which will then yield the general solution. Example: The Riccati equation again. Consider the Riccati equation s (t) = a0 (t) + 2a1 (t)s(t) + a2 (t) s(t) 2 and suppose that we know a particular solution s0 (t). Then let g(t) = 1 s0 (t) 0 1 , so that s0 (t) = g(t) · 0 (we are using the linear fractional action of SL(2, R) on R). The stabilizer of 0 is the subgroup G0 of matrices of the form: u 0 −1 v u .

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