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Chapter 4: Problem 6

Let the vector fields \(Y_{1}, \ldots, Y_{\ell}\), and \(X\) be analytic, and pick any \(\left(t, x^{0}\right) \in \mathcal{D}_{X}\). Let $$ d:=\operatorname{dim} \operatorname{span}\left\\{Y_{j}\left(e^{t X} x^{0}\right), j=1, \ldots \ell\right\\} . $$ Then, $$ d \leq \operatorname{dim} \operatorname{span}\left\\{\operatorname{ad}_{X}^{k} Y_{j}\left(x^{0}\right), j=1, \ldots \ell, k \geq 0\right\\} . $$ Proof. Since \(\operatorname{Ad}_{t X} Y_{j}\left(x^{0}\right)=Q \cdot Y_{j}\left(e^{t X} x^{0}\right)\), where \(Q:=\left(e^{-t X}\right)_{*}\left(e^{t X} x^{0}\right)\) is a nonsingular matrix, \(d=\operatorname{dim} \operatorname{span}\left\\{\operatorname{Ad}_{t X} Y_{j}\left(x^{0}\right), j=1, \ldots \ell\right\\}\). The result follows then from Proposition 4.4.5.

Short Answer

Expert verified
The dimension of the span of the translated vector fields is given by \(d\). By rewriting \(d\) using the Adjoint operator and applying Proposition 4.4.5, we prove that: $$ d \leq \operatorname{dim} \operatorname{span}\{\operatorname{ad}_{X}^{k} Y_{j}(x^{0}), j=1, \ldots, \ell, k \geq 0\}. $$

Step by step solution

01

Define notation and recall properties of Adjoint operator and Lie bracket

In this step, we'll define some notations and recall the properties of the Adjoint operator and Lie bracket, which will help us in proving the given inequality. Notation: 1. \(Y_1, \ldots, Y_\ell\): Given analytic vector fields 2. \(X\): Another given vector field 3. \((t, x^0) \in \mathcal{D}_{X}\): Given point in the domain of X 4. \(d\): The dimension of the span of translated vector fields Recall the following properties: 1. Adjoint operator: \(\operatorname{Ad}_{{tX}}Y_j(x^0) = Q \cdot Y_j(e^{tX}x^0)\), where \(Q := (e^{-tX})_{*}(e^{tX}x^0)\) is a nonsingular matrix 2. Lie bracket: \(\operatorname{ad}_X^k Y_j(x^0)\) denotes the Lie bracket between vector fields X and Y, taken k times. Now, we are ready to use these properties to prove the given inequality.
02

Rewrite the dimension d using the Adjoint operator

We are given that \(d := \operatorname{dim}\operatorname{span}\{Y_j(e^{tX}x^0), j = 1, \ldots, \ell\}\). To use the properties we just recalled, we can rewrite d using the Adjoint operator: \(d = \operatorname{dim} \operatorname{span}\{\operatorname{Ad}_{tX} Y_j(x^0), j=1, \ldots, \ell\}\), since \(\operatorname{Ad}_{tX} Y_j(x^0) = Q \cdot Y_j(e^{tX} x^0)\) and Q is a nonsingular matrix.
03

Apply Proposition 4.4.5

Now that we have rewritten the dimension d in terms of the Adjoint operator, we can apply Proposition 4.4.5 to obtain the inequality: $$ d \leq \operatorname{dim} \operatorname{span}\{\operatorname{ad}_{X}^{k} Y_{j}(x^{0}), j=1, \ldots, \ell, k \geq 0\}. $$ This is the final result and proves the given inequality for the dimensions of the spans.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Analytic Vector Fields
Analytic vector fields are fundamental objects in mathematical control theory, particularly because of their smooth nature and predictable behavior. An analytic vector field is a smooth vector field that can be locally represented by a power series that converges in a neighborhood of each point in its domain.

In simpler terms, imagine being able to zoom in infinitely close to the path described by a vector field and still finding a smooth, clear trajectory without any abrupt changes; that's the essence of analyticity. This property is vital for studying the flow of a differential equation, as the solutions are as smooth as the coefficients of the equation, thus making analytic vector fields an ideal playground for mathematical analysis.
Adjoint Operator
The adjoint operator, typically denoted as 'Ad', plays a crucial role in linking the change of coordinates in vector fields with linear transformations. Think of it as a bridge between two perspectives: where you stand and how the landscape transforms as you move.

In formal terms, for any vector field X and a time parameter t, the adjoint action of a flow generated by X on another vector field Y is given by \( \operatorname{Ad}_{tX}Y_j(x^0) \). This operation effectively describes how the vector field Y is transformed under the flow induced by X after a time t. The nonsingular matrix Q mentioned in the exercise reflects how the coordinate changes preserve the vector field's structure, indicating that dimensions are not altered by these transformations -- a key insight for proving dimension-related properties.
Lie Bracket
The Lie Bracket is a measure of the non-commutativity of the flow of two vector fields. It is expressed as \( [X, Y] \) and captures the difference between the flow along X followed by Y and the flow along Y followed by X. If you're shuffling cards, the Lie bracket describes the difference you get when changing the order of your shuffles.

More technically, it is a new vector field that encapsulates how two vector fields intertwine, and taking multiple Lie brackets (denoted by \( \operatorname{ad}_X^k Y_j(x^0) \)) indicates repeated application of this process. Studying the dimensions spanned by these brackets is like looking at the possible outcomes after several precise card shuffles; it tells us about the complexity and potential patterns that can emerge from the interplay of the vector fields.
Proposition 4.4.5
Proposition 4.4.5 might seem esoteric at first, but it underpins many results in control theory by establishing bounds on dimensions within the context of Lie algebras generated by vector fields. The proposition effectively sets a ceiling to the complexity or 'reach' of the flows generated by a combination of vector fields through their repeated brackets.

It states that the dimension of the space spanned by a vector field under the flow of another (adjoint action) is no greater than the span of the infinite sequence of Lie brackets of the vector fields. This mirrors the concept that the behavior (or the 'influence') of a dynamic system is limited by its governing rules (the vector fields and their interactions). This mathematical result is a form of guardrail, ensuring that systems do not magically develop new directions of influence that weren't already present in the foundations of their structure.

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Most popular questions from this chapter

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 \(x \in \mathcal{O}_{1}\).

Let \(X \in \mathbb{V}(\mathcal{O})\) and \(\left(t^{0}, x^{0}\right) \in \mathcal{D}_{X}\). Pick any \(Y \in \mathbb{V}\left(\mathcal{O}_{Y}\right)\) such that \(e^{t^{0} X} x^{0} \in \mathcal{O}_{Y}\). Then, $$ \left.\frac{\partial \operatorname{Ad}_{t X} Y\left(x^{0}\right)}{\partial t}\right|_{t=t^{0}}=\operatorname{Ad}_{t^{0} X}[X, Y]\left(x^{0}\right) . $$ Proof. We have, for all \((t, x) \in \mathcal{D}_{X}, e^{-t X} e^{t X} x=x\), so taking \(\partial / \partial x\) there results $$ \left(e^{-t X}\right)_{*}\left(e^{t X} x\right) \cdot\left(e^{t X}\right)_{*}(x)=I $$ and, taking \(\partial / \partial t\), \(\frac{\partial}{\partial t}\left(\left(e^{-t X}\right)_{*}\left(e^{t X} x\right)\right) \cdot\left(e^{t X}\right)_{*}(x)+\left(e^{-t X}\right)_{*}\left(e^{t X} x\right) \cdot \frac{\partial}{\partial t}\left(\left(e^{t X}\right)_{*}(x)\right)=0 .\) On the other hand, $$ \begin{aligned} \frac{\partial}{\partial t}\left(\left(e^{t X}\right)_{*}(x)\right) &=\frac{\partial}{\partial t} \frac{\partial}{\partial x}\left(e^{t X} x\right)=\frac{\partial}{\partial x} \frac{\partial}{\partial t}\left(e^{t X} x\right) \\ &=\frac{\partial}{\partial x} X\left(e^{t X} x\right)=X_{*}\left(e^{t X} x\right) \cdot\left(e^{t X}\right)_{*}(x) \end{aligned} $$ so substituting in (4.27) and postmultiplying by \(\left(\left(e^{t X}\right)_{*}(x)\right)^{-1}\) there results the identity: $$ \frac{\partial}{\partial t}\left(\left(e^{-t X}\right)_{*}\left(e^{t X} x\right)\right)=-\left(e^{-t X}\right)_{*}\left(e^{t X} x\right) \cdot X_{*}\left(e^{t X} x\right) . $$ Thus, $$ \begin{gathered} \frac{\partial}{\partial t}\left(\left(e^{-t X}\right)_{*}\left(e^{t X} x\right) \cdot Y\left(e^{t X} x\right)\right) \\ =-\left(e^{-t X}\right)_{*}\left(e^{t X} x\right) \cdot X_{*}\left(e^{t X} x\right) \cdot Y\left(e^{t X} x\right) \\ +\left(e^{-t X}\right)_{*}\left(e^{t X} x\right) \cdot Y_{*}\left(e^{t X} x\right) \cdot X\left(e^{t X} x\right) \\ =\left(e^{-t X}\right)_{*}\left(e^{t X} x\right) \cdot[X, Y]\left(e^{t X} x\right) \end{gathered} $$ as required. A generalization to higher-order derivatives is as follows; it can be interpreted, in a formal sense, as saying that \(\operatorname{Ad}_{t X}=e^{t a d x}\). For each \(k=0,1, \ldots\), and each vector fields \(X, Y\), we denote \(\operatorname{ad}_{X}^{0} Y=Y\) and \(\operatorname{ad}_{X}^{k+1} Y:=\operatorname{ad}_{X}\left(\operatorname{ad}_{X}^{k} Y\right)\).

Suppose that \(\Delta=\Delta_{f_{1}, \ldots, f_{r}}\) is a distribution of constant rank \(r\) on an open set \(\mathcal{O} \subseteq \mathbb{R}^{n}\). Show that the following two properties are equivalent: \- \(\Delta\) is involutive. 176 4\. Nonlinear Controllability \- For each \(x^{0} \in \mathcal{O}\) there is an open subset \(\mathcal{O}_{0} \subseteq \mathcal{O}\) which contains \(x^{0}\) and a diffeomorphism \(\Pi: \mathcal{O}_{0} \rightarrow \mathcal{V}\) into some open subset \(\mathcal{V} \subseteq \mathbb{R}^{n}\) such that \(\left(\Pi_{*} \Delta\right)(z)=\operatorname{span}\left\\{e_{1}, \ldots, e_{r}\right\\}\) for each \(z \in \mathcal{V}\), where \(e_{i}\) is the \(i\) th canonical basis vector. (A distribution of the form \(\Delta_{e_{1}, \ldots, e_{r}}\) is said to be a flat distribution. So, the claim is that a distribution is involutive if and only if it is locally flat in appropriate coordinates.)

Suppose that \(\Delta=\Delta_{X_{1}, \ldots, X_{r}}\) has constant rank \(r\), is involutive, and is invariant under the vector field \(f\). Pick any \(x^{0} \in \mathcal{O}\), and let \(\mathcal{O}_{0}\) and II be as in Frobenius' Lemma 4.4.16. Define, for \(z \in(-\varepsilon, \varepsilon)^{n}, g(z):=\) \(\Pi_{*}\left(\Pi^{-1}(z)\right) f\left(\Pi^{-1}(z)\right)\), and partition \(g=\left(g_{1}, g_{2}\right)^{\prime}\) and \(z=\left(z_{1}, z_{2}\right)^{\prime}\) as in the proof of Theorem 11 . Show that \(g_{2}\) does not depend on \(z_{1}\), that is to say, the differential equation \(\dot{x}=f(x)\) transforms in the new coordinates \(z=\Pi(x)\) into: $$ \begin{aligned} &\dot{z}_{1}=g_{1}\left(z_{1}, z_{2}\right) \\ &\dot{z}_{2}=g_{2}\left(z_{2}\right) \end{aligned} $$ Explain how, for linear systems \(\dot{x}=A x\), this relates to the following fact from linear algebra: if \(A\) has an invariant subspace, then there is a change of coordinates so that \(A\) is brought into upper triangular form consistent with that subspace. (Hint: (For the proof that \(\partial g_{2} / \partial z_{1}=0\).) We have that \(g_{2}(\Pi(x))=\Pi_{2 *}(x) f(x)\). On the other hand, each row of \(\Pi_{2 *}(x) f(x)\) is of the form \(L_{f} \psi_{i}\), where \(\psi_{i}\) 's are the rows of \(\Pi_{2}\). We know that \(L_{X_{j}} \psi_{i}=0\) for all \(i_{1} j\) (this is what Lemma \(4.4 .16\) gives), and also \(L_{\left[f, X_{j}\right]} \psi_{i}=0\) (because \(\Delta\) is invariant under \(f)\), so conclude that \(L_{X_{j}}\left(L_{f} \psi_{i}\right)=0\). This gives that the directional derivatives of the rows of \(g_{2}(\Pi(x))\) along the directions \(e_{j}(x):=\Pi_{*}(x) X_{j}(x)\) are all zero. Now observe that the vectors \(e_{i}(x)\) are all of the form \(\left(e_{i 1}, 0\right)^{\prime}\), and they are linearly independent.)

Suppose that \(\Delta=\Delta_{f_{1}, \ldots, f_{r}}\) is a distribution of constant rank \(r\). Then, 1\. The following two properties are equivalent, for any \(f \in \mathbb{V}(\mathcal{O})\) : (a) \(f \in_{p} \Delta\) (b) For each \(x^{0} \in \mathcal{O}\), there are a neighborhood \(\mathcal{O}_{0}\) of \(x^{0}\) and \(r\) smooth functions \(\alpha_{i}: \mathcal{O}_{0} \rightarrow \mathbb{R}_{1} i=1, \ldots, r\), so that $$ f(x)=\sum_{i=1}^{r} \alpha_{i}(x) f_{i}(x) \text { for all } x \in \mathcal{O}_{0} $$ 2\. The following two properties are equivalent, for any \(f \in \mathbb{V}(\mathcal{O})\) : (a) \(\Delta\) is invariant under \(f\). (b) \(\left[f, f_{j}\right] \in_{p} \Delta\) for each \(j \in\\{1, \ldots, r\\}\). 3\. Finally, the following two properties are equivalent: (a) \(\Delta\) is involutive. (b) \(\left[f_{i}, f_{j}\right] \in_{p} \Delta\) for all \(i, j \in\\{1, \ldots, r\\}\).
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