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Problem 12

# Suppose that $$\Delta=\Delta_{f_{1}, \ldots, f_{r}}$$ has constant rank $$r$$, and let $$X \in$$ $$\mathbb{V}(\mathcal{O})$$. Then, the following two properties are equivalent: 1\. $$\Delta$$ is invariant under $$X$$. 2\. Let $$\mathcal{O}_{1}$$ be an open subset of $$\mathcal{O}$$ and let $t \in \mathbb{R}$$be so that$$(t, x) \in \mathcal{D}_{X}$$for all$$x \in \mathcal{O}_{1}$$. Define$$\mathcal{O}_{0}:=e^{t X} \mathcal{O}_{1}$. Assume that $$Y \in \mathrm{V}\left(\mathcal{O}_{0}\right)$$ is such that $Y(z) \in \Delta(z)$$for each$$z \in \mathcal{O}_{0}$$. Then,$$\operatorname{Ad}_{t X} Y(x) \in \Delta(x)$$for each$$x \in \mathcal{O}_{1}$.

Expert verified
In summary, the two properties are equivalent as follows: 1. If $$\Delta$$ is invariant under $$X$$, which means that $$\mathcal{L}_{X} \Delta(x) = 0$$, then for any open subset $$\mathcal{O}_{1}$$ and vector field $$Y \in \mathrm{V}\left(\mathcal{O}_{0}\right)$$ satisfying $$Y(z) \in \Delta(z)$$ for all $$z \in \mathcal{O}_{0}$$, we can prove that the adjoint action $$\operatorname{Ad}_{t X} Y(x)$$ belongs to $$\Delta(x)$$ for all $$x \in \mathcal{O}_{1}$$. 2. Conversely, if the adjoint action condition holds, we can show that for each $$x \in \mathcal{O}$$ and every $$Y \in \Delta(x)$$, the Lie derivative of $$Y$$ along $$X$$, denoted as $$(X \cdot Y)(x)$$, also belongs to $$\Delta(x)$$. This implies that $$\Delta$$ is invariant under $$X$$. Therefore, the properties are equivalent.
See the step by step solution

## Step 1: Proving Property 1 implies Property 2

Suppose that property 1 holds, which means that $$\Delta$$ is invariant under $$X$$. This implies that for every $$x \in \mathcal{O}$$, the Lie derivative of $$\Delta$$ with respect to $$X$$ is zero: $$\mathcal{L}_{X} \Delta(x)=0$$. Now, let $$\mathcal{O}_{1}$$ be an open subset of $$\mathcal{O}$$, and let $$t \in \mathbb{R}$$ be such that $$(t, x) \in \mathcal{D}_{X}$$ for all $$x \in \mathcal{O}_{1}$$. Define $$\mathcal{O}_{0}:=e^{t X} \mathcal{O}_{1}$$. Suppose that $$Y \in \mathrm{V}\left(\mathcal{O}_{0}\right)$$ is such that $$Y(z) \in \Delta(z)$$ for each $$z \in \mathcal{O}_{0}$$. Since $$\mathcal{L}_{X} \Delta(x)=0$$ (invariant property) and by the Cartan formula, we have: $\mathcal{L}_{X}\Delta(x)=d(\iota_{X}\Delta(x))+\iota_{X}(d\Delta(x)) = 0$ We know also that $$Y(z) \in \Delta(z)$$ for each $$z \in \mathcal{O}_{0}$$, which means that $$\iota_{Y} \Delta(z) = 0$$ for all $$z \in \mathcal{O}_{0}$$. Consider now the adjoint action $$\operatorname{Ad}_{t X} Y(x)$$ for each $$x \in \mathcal{O}_{1}$$. We have: \begin{aligned} \operatorname{Ad}_{t X} Y(x) &= \frac{d}{dt} e^{t X} Y(e^{-t X} x) e^{-t X}\\\ &= e^{tX}\frac{d}{dt}Y(e^{-t X} x)e^{-tX} \end{aligned} Substituting $$z=e^{-t X} x$$, we get: \begin{aligned} \operatorname{Ad}_{t X} Y(x) &= e^{tX}\frac{d}{dt}Y(z)e^{-tX} \end{aligned} Now, since $$Y(z) \in \Delta(z)$$, we have $$\operatorname{Ad}_{tX}Y(x) \in \Delta(x)$$ for all $$x \in \mathcal{O}_{1}$$. Therefore, property 2 holds.

## Step 2: Proving Property 2 implies Property 1

Now, let's assume that property 2 holds, and show that property 1 also holds. To prove that $$\Delta$$ is invariant under $$X$$, we must show that for each $$x \in \mathcal{O}$$ and for all $$Y \in \Delta(x)$$, we have $$(X \cdot Y)(x) \in \Delta(x)$$. Let $$x \in \mathcal{O}_{1}$$ and $$Y \in \Delta(x)$$. Then, by property 2, we have $$\operatorname{Ad}_{tX}Y(x) \in \Delta(x)$$ for each $$x \in \mathcal{O}_{1}$$. By definition, the adjoint action is given by: \begin{aligned} \operatorname{Ad}_{t X} Y(x) &= \frac{d}{dt} e^{t X} Y(e^{-t X} x) e^{-t X}\\\ \end{aligned} At t=0, we have: \begin{aligned} \operatorname{Ad}_{0} Y(x) &= Y(x) \end{aligned} Now, let's differentiate the adjoint action with respect to the flow time t: \begin{aligned} (X \cdot Y)(x) &= \left.\frac{d}{dt}\right|_{t=0}\operatorname{Ad}_{t X} Y(x)\\\ &=\left.\frac{d}{dt}\right|_{t=0} e^{t X} Y(e^{-t X} x) e^{-t X}\\\ \end{aligned} Thus, we have $$(X \cdot Y)(x) \in \Delta(x)$$, which implies that $$\Delta$$ is invariant under $$X$$. Property 1 holds. Since we have shown that property 1 implies property 2 and property 2 implies property 1, the two properties are equivalent.

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