By Bryant R.L.
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Extra resources for An Introduction to Lie Groups and Symplectic Geometry
20. Show that SL(2, R) is not a matrix group! In fact, show that any homomorphism φ: SL(2, R) → GL(n, R) factors through the projections SL(2, R) → SL(2, R). (Hint: Recall, from earlier exercises, that the inclusion map SL(2, R) → SL(2, C) induces the zero map on π1 since SL(2, C) is simply connected. Now, any homomorphism φ: SL(2, R) → GL(n, R) induces a Lie algebra homomorphism φ (e): sl(2, R) → gl(n, R) and this may clearly be complexiﬁed to yield a Lie algebra homomorphism φ (e)C : sl(2, C) → gl(n, C).
For notational sanity, whenever the action (left or right) can be easily inferred from context, we will usually write a · m instead of λ(a, m) or m · a instead of ρ(m, a). Thus, for example, the axioms for a left action in this abbreviated notation are simply e · m = m and a · (b · m) = ab · m. For a given a left action λ: G × M → M, it is easy to see that for each ﬁxed a ∈ G the map λa : M → M deﬁned by λa (m) = λ(a, m) is a smooth diﬀeomorphism of M onto itself. Thus, G gets represented as a group of diﬀeomorphisms, or “transformations” of a manifold M.
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 .