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Chapter 4: Problem 5

If the vector fields \(Y_{1}, \ldots, Y_{\ell}\) and \(X\) are analytic, then \(\operatorname{span}\left\\{\operatorname{Ad}_{t X} Y_{j}\left(x^{0}\right), j=1, \ldots \ell, t \in \mathcal{I}_{X, x^{0}}\right\\}\) equals $$ \operatorname{span}\left\\{\operatorname{ad}_{X}^{k} Y_{j}\left(x^{0}\right), j=1, \ldots \ell, k \geq 0\right\\} $$ for each \(x^{0} \in \mathcal{O}\). Proof. Fix \(x^{0} \in \mathcal{O}\). Let \(S_{0}\) and \(S_{1}\) be, respectively, the sets of vectors \(\nu \in \mathbb{R}^{n}\) and \(\mu \in \mathbb{R}^{n}\) such that $$ \nu^{\prime} \operatorname{ad}_{X}^{k} Y_{j}\left(x^{0}\right)=0, \quad j=1, \ldots \ell, k \geq 0 $$ and $$ \mu^{\prime} \operatorname{Ad}_{t X} Y_{j}\left(x^{0}\right)=0, \quad j=1, \ldots \ell, t \in \mathcal{I}_{X, x^{0}} $$ Take any \(\nu \in S_{0}\). For each \(j=1, \ldots \ell\), by Lemma \(4.4 .3\), $$ \left.\nu^{\prime} \frac{\partial^{k} \mathrm{Ad}_{t X} Y_{j}\left(x^{0}\right)}{\partial t^{k}}\right|_{t=0}=0, \forall k \geq 0 . $$ Since, by Lemma 4.4.4, \(\operatorname{Ad}_{t} X Y_{j}\left(x^{0}\right)\) is analytic as a function of \(t\), this means that \(\nu^{\prime} \operatorname{Ad}_{t X} Y_{j}\left(x^{0}\right) \equiv 0\), so \(\nu \in S_{1}\). Conversely, if \(\nu \in S_{1}\), then \(\nu^{\prime} \operatorname{Ad}_{\ell X} Y_{j}\left(x^{0}\right) \equiv 0\) implies that all derivatives at zero vanish, so \(\nu \in S_{0}\) (analyticity is not needed here). Thus \(S_{0}=S_{1}\), and the result is proved.

Short Answer

Expert verified
For any fixed \(x^{0} \in \mathcal{O}\), we defined two sets, \(S_{0}\) and \(S_{1}\), containing vectors satisfying certain properties related to the adjoint representation and adjoint operator of vector fields. By using Lemma 4.4.3 and Lemma 4.4.4, we showed that a vector \(\nu\) belonging to \(S_{0}\) also belongs to \(S_{1}\), and vice versa. Consequently, we proved the equivalence of the spans \(S_{0} = S_{1}\), completing the proof.

Step by step solution

01

Understand the Problem

The task is to prove the equivalence of two specific spans for each \(x^{0}\) in some open set \(\mathcal{O}\). The first span is generated by the adjoint representation \(\operatorname{Ad}_{t X} Y_{j}(x^{0})\), where \(j=1,\ldots, \ell\) and \(t \in \mathcal{I}_{X, x^{0}}\). The second span is generated by \(\operatorname{ad}_{X}^{k} Y_{j}(x^{0})\), for \(j=1,\ldots, \ell\) and \(k \geq 0\). These operators, as well as the spaces they are defined over, are assumed to be analytic.
02

Begin the Proof

We start the proof by fixing \(x^{0} \in \mathcal{O}\). Next, we define \(S_{0}\) and \(S_{1}\) as the respective sets of vectors such that: - \(\nu^{\prime} \operatorname{ad}_{X}^{k} Y_{j}(x^{0})=0\), for each \(j=1, \ldots, \ell\) and \(k \geq 0\), and - \(\mu^{\prime} \operatorname{Ad}_{t X} Y_{j}(x^{0})=0\) , for each \(j=1, \ldots, \ell\) and \(t \in \mathcal{I}_{X, x^{0}}\).
03

Utilize Given Lemmas

For any given \(\nu\) in \(S_{0}\), we need to show that \(\nu\) also belongs to \(S_{1}\), which will prove that \(S_{0} = S_{1}\). To do this, we need to use Lemma 4.4.3 which assures us that for each \(j=1, \ldots, \ell\), we have: \[ \left.\nu^{\prime} \frac{\partial^{k} \mathrm{Ad}_{t X} Y_{j}(x^{0})}{\partial t^{k}}\right|_{t=0}=0, \quad \forall k \geq 0. \] Also, by Lemma 4.4.4, we know that \(\operatorname{Ad}_{t} X Y_{j}(x^{0})\) is analytic as a function of \(t\). Therefore, by applying these lemmas, we deduce that \(\nu^{\prime} \operatorname{Ad}_{t X} Y_{j}(x^{0}) \equiv 0\), which implies that \(\nu\) belongs to \(S_{1}\).
04

Complete the Proof

For the reverse direction, we take \(\nu\) in \(S_{1}\). Based on the definition of \(S_{1}\), \(\nu^{\prime} \operatorname{Ad}_{\ell X} Y_{j}(x^{0}) \equiv 0\), from which we conclude that all derivatives at zero vanish, so \(\nu\) should belong to \(S_{0}\) as well. Hence, we have shown that \(S_{0} = S_{1}\) for any fixed \(x^{0} \in \mathcal{O}\), which completes the proof.

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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, where they represent smooth, continuous transformations in a given space. An analytic vector field is defined by its ability to be expressed as a power series that converges to a function in the neighborhood of any point within the field's domain. This property of convergence implies that analytic vector fields can be differentiated infinitely often, and each derivative is also analytic.

Analyticity is essential for various reasons, one being that it allows for a precise and robust mathematical analysis of the behavior of dynamic systems. For instance, when dealing with the stability of equilibria in a control system or when studying the system's response to perturbations, analytic vector fields provide a solid groundwork as their properties are well understood and can be manipulated with a high degree of control.

In the context of the exercise, the analytic vector fields in question, denoted as \(Y_1, \.\.\., Y_{\ell}\) and \(X\), allow us to use lemmas that relate to the differentiability and manipulation of the fields around a specific point \(x^0\). The exercise demonstrates how properties at a single point can be extended over a span of functions.
Adjoint Representation
The adjoint representation is a key concept that arises in the study of Lie groups and Lie algebras, holding a place of importance in mathematical control theory. In particular, it gives us a way to understand how the elements of a Lie algebra, which often describe infinitesimal transformations, can be represented as linear transformations that act on the very same algebra.

In more concrete terms, the adjoint representation of a vector field \(X\), denoted as \(\operatorname{Ad}_{tX}\), captures how other vector fields transform as they 'flow' along the vector field \(X\). This representation is essential when looking at the structures governing control systems, especially when analyzing symmetries and conservation laws.

The exercise utilizes the notion of the adjoint representation to equate two different spans of vector fields evaluated at point \(x^0\). By proving that the spans are the same, one can infer that the behavior of the system at that point can be completely described by either set of vector fields, which is a powerful result in the context of understanding system dynamics.
Analytic Functions
Analytic functions are smooth functions that can be represented by a convergent power series within some radius of convergence around each point in their domain. This concept is similar to that of analytic vector fields but applies to functions of complex or real variables. The principal characteristic of analytic functions is that they are 'nice' in the sense that they are infinitely differentiable and agree with their Taylor series expansion at every point where the series converges.

This property of analytic functions is used in the step-by-step solution where lemmas rely on the function being analytic to ensure that \(u' \operatorname{Ad}_{t X} Y_j(x^{0}) \equiv 0\), which effectively leverages the all-or-nothing nature of analytic functions鈥攅ither their Taylor series representation agrees on an interval, or the function is not analytic on that interval.

By invoking the analyticity of the \(\operatorname{Ad}_{t} X Y_{j}(x^{0})\) function, the exercise proceeds to derive the equivalency of the spans, underscoring the importance of analytic functions in simplifying complex systems into more manageable expressions and making them amenable to rigorous analysis.

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Most popular questions from this chapter

Suppose \(X\) and \(Y\) are analytic vector fields defined on \(\mathcal{O}\). For any \(x^{0} \in \mathcal{O}\), let \(\mathcal{I}=\mathcal{I}_{X, x^{0}}:=\left\\{t \in \mathbb{R} \mid\left(t, x^{0}\right) \in \mathcal{D}_{X}\right\\}\). Then, the function \(\gamma: \mathcal{I} \rightarrow \mathbb{R}^{n}: t \mapsto \operatorname{Ad}_{t X} Y\left(x^{0}\right)\) is analytic. Proof. Let \(\alpha(t):=\left(e^{-t X}\right),\left(e^{t X} x^{0}\right)\), seen as a function \(\mathcal{I} \rightarrow \mathbb{R}^{n \times n}\). Note that \(\alpha(0)=I\), and that, by Equation (4.28), the vector \(\left(e^{t X} x^{0}, \alpha(t)\right)\) is the solution of the differential equation $$ \begin{array}{lll} \dot{x}(t) & =X(x(t)) & x(0) & =x^{0} \\ \dot{\alpha}(t) & =-\alpha(t) \cdot X_{*}(x(t)) & & \alpha(0)=1 . \end{array} $$ This is a differential equation with analytic right-hand side, so \(x(\cdot)\) and \(\alpha(\cdot)\) are both analytic (see, for instance, Proposition C.3.12). Then, \(\gamma(t)=\alpha(t) Y(x(t))\) is also analytic.

Provide an example of a set of vector fields \(f_{1}, \ldots, f_{r}\) so that (a) the distribution \(\Delta=\Delta_{f_{1}, \ldots, f_{r}}\) has constant rank \(r\) and, locally about each point \(x^{0} \in \mathcal{O}\), transforms under a diffeomorphism into \(\Delta_{c_{1}, \ldots, e_{r}}\) (in the sense of Exercise \(4.4 .17)\), but (b) there is some point \(x^{0}\) such that, for no possible diffeomorphism II defined in a neighborhood of \(x^{0},\left(\Pi_{*} f_{i}\right)(z)=e_{i}\) for all \(i\).

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

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.)

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.
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