By Bryant R.L.

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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). Since SL(2, C) is simply connected, there must be a corresponding Lie group homorphism φC : SL(2, C) → GL(n, C). ) 21. An ideal in a Lie algebra g is a linear subspace h which satisﬁes [h, g] ⊂ h.

We have already seen that if T is a connected abelian Lie group with Lie algebra t, then the exponential map exp: t → T is a surjective homomorphism. ) Thus, the Lie equation for an abelian group is “solvable by quadrature” in the classical sense. Another instance where one can at least reduce the problem somewhat is when one has a homomorphism φ: G → H and knows the fundamental solution SH to the Lie equation for ϕ ◦ A: R → h. In this case, SH is the particular solution (with initial condition SH (0) = e) of the Lie equation on H associated to A by regarding φ as deﬁning a left action on H.

3 a b 0 1 a ∈ R+ , b ∈ R 35 is trivial, so any connected Lie group with the same Lie algebra is actually isomorphic to G. (In the next Lecture, we will show that whenever K is a closed normal subgroup of a Lie group G, the quotient group G/K can be given the structure of a Lie group. ) 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.

### An Introduction to Lie Groups and Symplectic Geometry by Bryant R.L.

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