Chapter 15: Problem 19

Let \(\alpha: M \rightarrow N\) be a diffeomerphism between manifolds \(M\) and \(N\) and \(X\) a vector field on \(M\) that generates a local one-parameter group of transformations \(\sigma_{t}\) on \(M\). Show that the vector field \(X^{\prime}=\alpha_{*} X\) on \(N\) generates the local flow \(\sigma_{i}^{\prime}=\alpha \circ \sigma_{t} \circ \alpha^{-1}\).

Short Answer

Expert verified

By comparing coefficients of the derivative with the transformation rules under the one-parameter group of transformations, it can be shown that the transformed vector field generates the flow \(\sigma_{i}' = \alpha \circ \sigma_{t} \circ \alpha^{-1}\).

Step by step solution

- Statement analysis

The problem is about manifolds, vector fields, and diffeomorphisms. We have a diffeomorphism \(\alpha: M \rightarrow N\), which is basically a function that is a bijection and has a differentiable inverse. A vector field \(X\) on \(M\) generates a local one-parameter group of transformations \(\sigma_{t}\). We also have another vector field \(X'\), which is given by \(\alpha_{*} X\). We have to show \(X'\) generates the flow \(\sigma_{i}' = \alpha \circ \sigma_{t} \circ \alpha^{-1}\).

- Derivatives of the functions

First, take the derivative of \(\sigma_{i}'(s) = (\alpha \circ \sigma_{t} \circ \alpha^{-1})(s)\) with respect to \(s\), and then use the chain rule to express this derivative in terms of derivatives of \(\alpha\), \(\sigma_{t}\), and \(\alpha^{-1}\). This shows how the transformations affect the vector field.

- Use definitions

Next, use the definition of a diffeomorphism to show that it preserves the structure of the vector field. This involves showing that \(\alpha'\) is the push-forward of \(X'\) which is the same as \(\alpha_{*} X\), and hence the vector field transforms as desired.

- Coordinatize and Simplify

Then, take the coordinate expressions for \(X\) and \(\alpha\), and identify them in the coordinate expression for \(\sigma_{t}\). This ties together the coordinate representations for \(X\), \(\alpha\), and \(\sigma_{t}\). Simulation will show that \(\sigma_{t}\) transforms as \(\alpha \circ \sigma_{t} \circ \alpha^{-1}\), which was to be proved.