# Infinitesimal Generator of a Group Action

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.]

• 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) Mar 17, 2021 at 23:01
• Thank you. I've found the answer via your hint and math.stackexchange.com/questions/761138. Mar 18, 2021 at 15:49
• Now you can write an answer your own question and accept your answer, to make sure this question is no long unanswered. Mar 18, 2021 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

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

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.