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Consider

i) Find the vector fields $V_1, V_2, V_3$ which generate the following smooth one-parameter groups of transformations of $\mathbb{R}$ : $$ x \mapsto \psi_1^s x=x+s, \quad x \mapsto \psi_2^s x=e^s x, \quad x \mapsto \psi_3^s x=\frac{x}{1-s x} . $$ ii) Deduce that these vector fields generate a group of transformations of the form $$ x \mapsto \frac{a x+b}{c x+d}, \quad a d-b c=1 . $$

I don't understand what is meant by a group of transformations generated by the vector fields. I don't understand how we generate elements from these. Could someone elaborate on this?

Alp Uzman
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Maths Wizzard
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  • The functions are the elements and the group operation is composition. So you'll need to find the identity element and inverses since associativity is inherited from function composition. – CyclotomicField Nov 01 '23 at 19:59
  • Hi, I am sorry but I don't understand. How can I compose two vector fields? Maybe they mean treating vector fields as operators? – Maths Wizzard Nov 01 '23 at 20:03
  • You're not composing the vector fields, they're composing the smooth one-parameter operations. So for example $\psi_1^0(x)=x$ so $\psi_1^0$ is the identity element and $\psi_1^s(\psi_1^{-s}(x)))=\psi_1^s(x-s)=x+s-s=x$ so $\psi_1^s$ is the inverse of $\psi_1^{-1}$. – CyclotomicField Nov 01 '23 at 20:27
  • Hi, I think I didn't make my self clear. I can understand how via the transformations one can generate a group. However, how does one do it via the vector fields? The question says "deduce that these vector fields generate ...". Does it just mean that I should be working with the given transformations or does it mean I should somehow consider a group of vector fields generated by the 3 given ones. – Maths Wizzard Nov 01 '23 at 20:30
  • If we restrict ourselves to $V_1$ and $\psi_1$ can you see how the vector field is related to the operation? – CyclotomicField Nov 01 '23 at 20:46
  • I can see the correspondence between a given vector field and the associated group of transformations - that is fine. What I do not understand is how say two vector fields generate a group. For example, say we have $V_2$ and $V_1$. How do these generate a group of transformations? Maybe I have misread the question? – Maths Wizzard Nov 01 '23 at 20:49
  • Given some elements of a group you can generate a new group by considering finite combinations of their elements. For example $\mathbb{Z} \times \mathbb{Z}$ is generated by $(0,1)$ and $(1,0)$ because $(n,m)=n(1,0)+m(0,1)$ and $n(1,0)$ is a "power" of $(1,0)$ and $m(0,1)$ a "power" of $(0,1)$. – CyclotomicField Nov 01 '23 at 21:04

1 Answers1

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Hint: "Generate" in this context means to "generate infinitesimally". So one (complete) vector field generates an action of $\mathbb{R}$ by diffeomorphisms (by the Existence and Uniqueness Theorem in ODEs; see e.g. What is the relationship between the vector fields of conjugate flows?), and more generally a Lie algebra $\mathfrak{g}$ generates an action of a Lie group $G$ by diffeomorphisms. Standard Lie theory then guarantees that, there is up to isomorphism a unique such $G$ (assuming connectedness).

Reverse engineering the second part, you would need to verify that the smallest Lie subalgebra of the Lie algebra of $C^\infty$ vector fields on the real line containing $V_1, V_2, V_3$ is isomorphic to the Lie algebra $\mathfrak{sl}(2,\mathbb{R})$ of $2\times 2$ traceless matrices.

It might also be useful/exciting to note that $\psi_2$ is analogous to the geodesic flow, $\psi_1$ is analogous to the unstable horocycle flow and $\psi_3$ is analogous to the stable horocycle flow (on the unit tangent bundle of a surface of constant negative curvature).

Alp Uzman
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