This question is from IV.3, Exercise 6 in Boothby's Introduction to Differentiable Manifolds. Given the action $$\theta(t; x, y) = (xe^{2t}, ye^{-3t})$$ from $\mathbb{R}\times\mathbb{R}^2 \rightarrow \mathbb{R}^2$ (treating $\mathbb{R}$ as your group and $\mathbb{R}^2$ as the manifold), determine the infinitesimal generator $X_p$ (herein "inf gen") and show that it is $\theta$-invariant.

He generically defines the infinitesimal generator as $$X_p f = \sum^n_{i=1} \dot{h^i}(0,x)\left(\frac{\partial \hat{f}}{\partial x^i}\right)_{\phi(p)}.$$

Here $x^i = {h^i}(0,x)$ and $\hat{f}(x^1, \ldots, x^n)$ is a local expression for the function $f\in C^\infty$. The overdot indicates a derivative with respect to $t$.

Thank you in advance for any help.

[I originally asked for hints and geometric intuition. See below for the solution attempt.]

  • $\begingroup$ Hint: The generator should be independent of $t$. And ignore the action of vector fields on functions, just think in terms of vector calculus or ODEs (for the purpose of this problem) $\endgroup$ Mar 17 at 23:01
  • $\begingroup$ Thank you. I've found the answer via your hint and math.stackexchange.com/questions/761138. $\endgroup$
    – coproduct
    Mar 18 at 15:49
  • $\begingroup$ Now you can write an answer your own question and accept your answer, to make sure this question is no long unanswered. $\endgroup$ Mar 18 at 16:13

The answer per the hint is given as follows.

We have $\theta(t; x, y) = (xe^{2t}, ye^{-3t})$. On inspection, we see that it is an action, as $\theta(0; x, y) = (x, y)$ and $\theta(t+s; x, y) = \theta(s+t; x, y)$. Per the note, we compute

\begin{equation} \label{eq1} \begin{split} \dot{\theta_1} & = 2xe^{2t} \\ & = 2\theta_1 \\ \dot{\theta_2} & = -3ye^{-3t} \\ & = -3\theta_2. \\ \end{split} \end{equation}

This means that the infinitesimal generator $X_{(x,y)} = 2x \frac{\partial}{\partial x} -3y \frac{\partial}{\partial y}$. By Theorem 3.4 in Boothby IV.3, $X$ is $\theta$-invariant. Without citing this, we can check if $X$ is $\theta$-invariant, i.e. if $\theta_{t*}(x,y) = X_{\theta_(x,y)}$.

The left-hand side has $X_{(x,y)}$ as above, with $\theta_t(x,y) = (xe^{2t}, ye^{-3t})$, which implies that $X_{\theta_t(x,y)} = 2xe^{2t}\frac{\partial}{\partial x} - 3ye^{-3t}\frac{\partial}{\partial y}$.

The right-hand side requires the computation of the pushforward $\theta_{t*}$. We consider this $\theta_{t}$ as a function of $x$ and $y$. Generally, for any $v \in T_pM$ with $v = v_1 \frac{\partial}{\partial x} +v_2 \frac{\partial}{\partial y}$, $$\theta_{t*} (v) = \begin{pmatrix} \frac{\partial \theta_{t,1}}{\partial x} & \frac{\partial \theta_{t,1}}{\partial y} \\ \frac{\partial \theta_{t,2}}{\partial x} & \frac{\partial \theta_{t,2}}{\partial y} \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} $$

where the basis for $v$ (the matrix on the right) is implicitly $(\frac{\partial}{\partial x}, \frac{\partial}{\partial y})$. Computing the Jacobian, we have

$$\theta_{t*} = \begin{pmatrix} e^{2t} & 0 \\ 0 & e^{-3t} \end{pmatrix}.$$

So $\theta_{t*}(X_{(x,y)}) = \theta_{t*}(2x \frac{\partial}{\partial x} -3y \frac{\partial}{\partial y})$, and substituting into the above, we have that

$$\theta_{t*}(X_{(x,y)}) = \begin{pmatrix} e^{2t} & 0 \\ 0 & e^{-3t} \end{pmatrix} \begin{pmatrix} 2x \\ -3y \end{pmatrix} = \begin{pmatrix} 2xe^{2t} \\ -3ye^{-3y} \end{pmatrix} .$$

Here again the basis is understood to be $(\frac{\partial}{\partial x}, \frac{\partial}{\partial y})$. So we are done.


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