By Alvin K Bettinger; John A Englund

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**Additional info for Algebra and trigonometry**

**Example text**

Show that these two actions commute, and hence together deﬁne a free action of G = R×Z on the plane. Sketch the orbits and show that, even though the G-orbits of this action are closed, and the quotient space is Hausdorﬀ, the quotient space is not a manifold. ) 5. Show that if ρ: M × G → M is a right action, then the induced map ρ∗ : g → X(M) satisﬁes ρ∗ [x, y] = ρ∗ (x), ρ∗ (y) . 6. Prove Proposition 2. (Hint: you are trying to ﬁnd an open neighborhood U of {e} × M in G × M and a smooth map λ: U → M with the requisite properties.

Consider the mapping Ψ: B d × Gm → G deﬁned by Ψ(x, g) = ψ(x)g. I claim that Ψ is a diﬀeomorphism onto its image (which is an open set), say U = Ψ(B d × Gm ) ⊂ G. ) To see this, ﬁrst I show that Ψ is one-to-one: If Ψ(x1 , g1 ) = Ψ(x2 , g2 ), then (φ ◦ ψ)(x1 ) = ψ(x1 ) · m = (ψ(x1 )g1 ) · m = (ψ(x2 )g2 ) · m = ψ(x2 ) · m = (φ ◦ ψ)(x2 ), so the injectivity of φ ◦ ψ implies x1 = x2 . Since ψ(x1 )g1 = ψ(x2 )g2 , this in turn implies that g1 = g2 . Second, I must show that the derivative Ψ (x, g): Tx Rd ⊕ Tg Gm → Tψ(x)g G is an isomorphism for all (x, g) ∈ B d × Gm .

In other words in order to ﬁnd the fundamental solution of a Lie equation for G when the particular solution with initial condition g(0) = m ∈ M is known, it suﬃces to solve a Lie equation in Gm ! This observation is known as Lie’s method of reduction. It shows how knowledge of a particular solution to a Lie equation simpliﬁes the search for the general solution. ) Of course, Lie’s method can be generalized. If one knows k particular solutions with initial values m1 , . . ,mk = Gm1 ∩ Gm2 ∩ · · · ∩ Gmk .