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Problem 13

# Suppose that $$\Delta=\Delta_{f_{1}, \ldots, f_{r}}$$ has constant rank $$r$$ and is invariant under each of $$X_{1}, \ldots, X_{k} \in \mathrm{V}(\mathcal{O})$$. Let $$\mathcal{O}_{k}$$ be an open subset of $$\mathcal{O}$$ and let $s_{1}, \ldots, s_{k}$ be real numbers with the following property: $$\left(s_{i}, x\right) \in \mathcal{D}_{X_{i}}, \quad \forall x \in \mathcal{O}_{i}, i=1, \ldots, k,$$ where we define $$\mathcal{O}_{k-1}:=e^{s_{k} X_{k}} \mathcal{O}_{k}, \ldots, \mathcal{O}_{1}:=e^{s_{2} X_{2}} \mathcal{O}_{2}, \mathcal{O}_{0}:=e^{s_{1} X_{1}} \mathcal{O}_{1} .$$ Assume that $$Y \in \mathbb{V}\left(\mathcal{O}_{0}\right)$$ is such that $Y(z) \in \Delta(z)$$for each$$z \in \mathcal{O}_{0}$. Then, $\operatorname{Ad}_{s_{k} X_{k}} \ldots \operatorname{Ad}_{s_{1} X_{1}} Y(x) \in \Delta(x)$$for all$$x \in \mathcal{O}_{k}$.

Expert verified
Given a distribution $$\Delta$$ with invariant constant rank $$r$$, it is invariant under the action of a given set of vector fields $$X_1,\ldots,X_k \in \mathrm{V}(\mathcal{O})$$ on an open set $$\mathcal{O}\subset M$$. We are also given a sequence of real numbers $$s_1,\ldots,s_k$$ satisfying certain properties, such as $$\left(s_{i}, x\right) \in \mathcal{D}_{X_{i}}$$. We need to show that the adjoint action of the product of the exponentials of these vector fields on a vector field $$Y\in\mathbb{V}\left(\mathcal{O}_{0}\right)$$, that takes the values in distribution $$\Delta$$, will also take the values in distribution $$\Delta$$. To prove this, we use the invariance of $$\Delta$$ under each $$X_i$$ and apply the adjoint action of each $$X_i$$ to $$Y$$. The overall result $$\operatorname{Ad}_g Y$$ will still be in the distribution $$\Delta$$. Therefore, we have $$\operatorname{Ad}_g Y(x)\in\Delta(x)$$ for all $$x\in\mathcal{O}_k$$, which proves our result.
See the step by step solution

## Step 1: Understand the Given Information

We have a distribution $$\Delta=\Delta_{f_{1},\ldots,f_{r}}$$ with constant rank $$r$$ and invariant under each vector field $$X_i\in\mathrm{V}(\mathcal{O})$$, $$i=1,\ldots,k$$. Given an open set $$\mathcal{O}_k\subset\mathcal{O}$$, and real numbers $$s_1,\ldots,s_k$$, we define sequential open sets $$\mathcal{O}_{k-1}, \ldots, \mathcal{O}_0$$, based on the given property: $\mathcal{O}_{k-1}:=e^{s_{k} X_{k}} \mathcal{O}_{k}, \ldots, \mathcal{O}_{1}:=e^{s_{2} X_{2}} \mathcal{O}_{2}, \mathcal{O}_{0}:=e^{s_{1} X_{1}} \mathcal{O}_{1} .$

## Step 2: Define the Adjoint Action

We define the adjoint action $$\operatorname{Ad}_g$$ of an element $$g\in G$$ on an element $$Y\in\mathfrak{g}$$ as: $\operatorname{Ad}_g(Y)=g Y g^{-1},$ where $$G$$ is a Lie Group and $$\mathfrak{g}$$ is its corresponding Lie Algebra.

## Step 3: Show the Result

We are given that $$Y(z)\in\Delta(z)$$, for all $$z\in\mathcal{O}_0$$. We have to show: $\operatorname{Ad}_{s_{k} X_{k}} \ldots \operatorname{Ad}_{s_{1} X_{1}} Y(x) \in \Delta(x), \; \forall x \in \mathcal{O}_{k}.$ Let $$g=e^{s_k X_k} \ldots e^{s_1 X_1}$$. Then the adjoint action of $$g$$ can be written as: $\operatorname{Ad}_g Y = g Y g^{-1}.$ We have to show that for any point $$x\in\mathcal{O}_k$$: $\operatorname{Ad}_g Y(x) \in \Delta(x).$

## Step 4: Use Invariance of $$\Delta$$ under $$X_i$$

Since $$\Delta$$ is invariant under each of the vector fields $$X_i$$, applying the adjoint action of each $$X_i$$ to $$Y$$ will ensure that the result is still in $$\Delta$$. The adjoint action of $$g$$ is the sequential application of the adjoint actions of each $$X_i$$, and since $$\Delta$$ is invariant under each action, the overall result $$\operatorname{Ad}_g Y$$ will still be in the distribution $$\Delta$$. Therefore, we have $$\operatorname{Ad}_g Y(x)\in\Delta(x)$$ for all $$x\in\mathcal{O}_k$$, which is the result we sought to prove.

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