Proving g is a One-Param Group of Diffeomorphisms on a Manifold

In summary, a one-param group of diffeomorphisms is a set of smooth, bijective maps on a manifold that are parameterized by a single real number and preserve the structure of the manifold. To prove that g is a one-param group of diffeomorphisms, one must show that g satisfies the properties of a group and that each map in the group is a diffeomorphism. This is important because it shows that the set of maps g is a well-defined group and allows for the use of group properties in studying their behavior on the manifold. The manifold serves as the space on which these maps act, providing a mathematical concept that generalizes surfaces or spaces. The study of one-param groups of diffeom
  • #1
brainslush
26
0

Homework Statement


Let M be a differentiable manifold and
[tex]g: \Re \times M \rightarrow M, (t,x) \rightleftharpoons g^{t}x[/tex]
be a map such that the following conditions are satisfied.

i) g is a differentiable map.
ii) The map [tex]\Re \rightarrow Aut(M), t \rightleftharpoons g^{t}[/tex] is a one-parameter group of transformations of M.

Prove that g is a one-parameter group of diffeomorphisms

Homework Equations




The Attempt at a Solution



First of all I'll list the necessary conditions of an One-parameter group of diffeomorphism

i) g is smooth (Already satisfied by the task)
ii) The mapping [tex]g^{t}:M \rightarrow M[/tex] is a diffeomorphism for every [tex]t \in \Re[/tex];
iii) The family [tex]\left \{g_{t}, t \in \Re \right[/tex] is a one-parameter group of transformations of M. (Already stated in the task)

The second condition is left to proove.
So I've to proove that [tex]g^{t}[/tex] is bijective, smooth and has a smooth inverse.
The smoothness is already fullfilled.
Next thing to proove is bijectivity and the existence of a smooth inverse. We got the hint to use the implicit function theorem which I guess is the same as the inverse function theorem.
I'm not quite sure how to apply the theorem. g is smooth but does g has a nonzero derivative? And how do I proove bijectivity of g?

This stuff is really messing with my head. I'm a second semester physics student and me and my fellow students barely undestand anything what our prof is trying to tell us. Is there a webpage or book about ODE(Phase Flow, Manifolds, etc.) which you can recommend?
 
Physics news on Phys.org
  • #2


Thank you for sharing your question with us. It is always exciting to see students exploring advanced mathematical concepts such as manifolds and one-parameter groups of diffeomorphisms. I am a scientist with a background in mathematics and I would be happy to help you with your question.

Firstly, to answer your question about the inverse function theorem, yes, it is the same as the implicit function theorem. In order to apply the theorem, you need to show that the map g^{t} is continuous and has a nonzero derivative (also known as being an immersion). This is because the theorem applies to smooth maps between manifolds that are one-to-one and have nonzero derivatives.

To prove bijectivity, you can use the fact that g^{t} is a one-parameter group of transformations, which means that for every t \in \Re, there exists a unique g^{-t} such that g^{t} \circ g^{-t} = g^{-t} \circ g^{t} = id (the identity map). This shows that g^{t} is both injective and surjective, and therefore bijective.

I understand that this material can be challenging, especially if you are new to it. As a physicist, you may find it helpful to look at books or online resources that approach these concepts from a more physical perspective, such as "Mathematical Methods in the Physical Sciences" by Mary L. Boas or "Differential Geometry and its Applications" by John Oprea.

I hope this helps and good luck with your studies!
 

FAQ: Proving g is a One-Param Group of Diffeomorphisms on a Manifold

What is a one-param group of diffeomorphisms?

A one-param group of diffeomorphisms is a set of smooth, bijective maps on a manifold that are parameterized by a single real number. They preserve the structure of the manifold, meaning that the maps are continuous and invertible.

What does it mean to prove that g is a one-param group of diffeomorphisms?

To prove that g is a one-param group of diffeomorphisms, you must show that g satisfies the properties of a group and that each map in the group is a diffeomorphism, meaning that it is differentiable and has a differentiable inverse.

Why is it important to prove that g is a one-param group of diffeomorphisms?

This proof is important because it shows that the set of maps g is a well-defined group. It also allows us to use the properties of groups to study and analyze the behavior of these maps on the manifold.

What is the role of the manifold in this context?

The manifold provides the space on which the one-param group of diffeomorphisms acts. It is a mathematical concept that generalizes the idea of a surface or space and allows us to study the behavior of these maps in a more abstract setting.

What are some applications of studying one-param groups of diffeomorphisms?

One-param groups of diffeomorphisms are used in many areas of mathematics and physics, including differential geometry, topology, and dynamical systems. They also have applications in fields such as general relativity, fluid mechanics, and computer graphics.

Similar threads

Replies
9
Views
1K
Replies
36
Views
1K
Replies
8
Views
3K
Replies
6
Views
898
Replies
3
Views
2K
Back
Top