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Chapter 4: Problem 18

Suppose that \(f_{1}, \ldots, f_{r}\) are smooth vector fields on an open set \(\mathcal{O} \subseteq \mathbb{R}^{n}\), and that \(f_{1}(x), \ldots, f_{r}(x)\) are linearly independent for each \(x \in \mathcal{O}\). Show that the following two properties are equivalent: \- \(\left[f_{i}, f_{j}\right]=0\) for each \(i, j \in\\{1, \ldots, r\\} .\) \- 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 \(\left(\Pi_{*} f_{i}\right)(z)=e_{i}\) for each \(z \in \mathcal{V}\), where \(e_{i}\) is the \(i\) th canonical basis vector. (That is, the vector fields commute if and only if there is a local change of variables where they all become \(\left.f_{i} \equiv e_{i} \cdot\right)\)

Short Answer

Expert verified
The two properties are equivalent: the vector fields commute if and only if there is a local change of variables where they all become the ith canonical basis vector. This is demonstrated by showing that the vanishing Lie brackets imply the existence of a diffeomorphism that takes the vector fields to the canonical basis, and that the existence of such a diffeomorphism implies the vector fields commute.

Step by step solution

01

Show that commuting vector fields implies the existence of a local change of variables taking the vector fields to the canonical basis.

Let's assume the Lie brackets of all the vector fields equal zero, i.e., \([f_{i}, f_{j}] = 0\) for each \(i, j \in \{1, \ldots, r\}\). Since all the Lie brackets vanish, that means the vector fields commute, and for each point \(x^{0}\) in the open set \(\mathcal{O}\), there is an open neighborhood \(\mathcal{O}_{0}\) containing \(x^{0}\) where we can apply the Frobenius theorem to obtain an integral manifold of the commutative distribution generated by the vector fields. This integral manifold would be diffeomorphic to \(\mathbb{R}^r\). With this integral manifold, we could define a new coordinate system on \(\mathcal{O}_{0}\) as a diffeomorphism \(\Pi\) from \(\mathcal{O}_{0}\) to some open subset \(\mathcal{V} \subseteq \mathbb{R}^{n}\). This new coordinate system takes the vector fields to \(e_{i}\), the ith canonical basis vector.
02

Show that a local change of variables taking the vector fields to the canonical basis implies that they commute.

Now, let's assume we have such a diffeomorphism \(\Pi: \mathcal{O}_{0} \rightarrow \mathcal{V}\) such that \(\left(\Pi_{*} f_{i}\right)(z)=e_{i}\) for each \(z \in \mathcal{V}\). We need to show that the vector fields commute, i.e., \([f_{i}, f_{j}] = 0\) for each \(i, j \in \{1, \ldots, r\}\). Since the pushforward of the vector fields under the diffeomorphism \(\Pi\) are given by the canonical basis vectors, we can compute the Lie bracket of the pushforward of each pair of vector fields in the local coordinate system. In the new coordinate system, the Lie brackets vanish because the Lie bracket between canonical basis vectors equals zero, i.e., \([e_{i}, e_{j}] = 0\) for each \(i, j \in \{1, \ldots, r\}\). Now to show that the Lie brackets of the original vector fields must also vanish, we note that the Lie bracket of two vector fields is a tensorial operation, and hence the Lie bracket of the pushforward vector fields will be the pushforward of the Lie bracket of the original vector fields. Since we know \([\Pi_{*} f_{i}, \Pi_{*} f_{j}] = [\Pi_{*} (f_{i}), \Pi_{*} (f_{j})] = [\Pi_{*} e_{i}, \Pi_{*} e_{j}] = 0\), it implies that \([f_i,f_j]=0\) for each \(i, j \in \{1, \ldots, r\}\). Hence, the vector fields commute. Now that we have shown both parts of the equivalence, the proof is complete. The two properties are indeed equivalent: the vector fields commute if and only if there is a local change of variables where they all become the ith canonical basis vector.

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Key Concepts

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

Lie Bracket
A Lie bracket is a mathematical operation that plays an important role in differential geometry and Lie algebra. When dealing with vector fields, the Lie bracket measures the noncommutativity of the flows generated by those vector fields.
Let's consider two vector fields, \( f_i \) and \( f_j \). Their Lie bracket is denoted by \([f_i, f_j]\). It essentially describes another vector field, showing how much the flows along \( f_i \) and \( f_j \) fail to commute.
If the Lie bracket equals zero, \([f_i, f_j] = 0\), it suggests that the vector fields "commute". This zero condition means that the difference between first moving along \( f_i \) and then \( f_j \), compared to first \( f_j \) then \( f_i \), is negligible.
  • This concept is highly useful in studying geometrical structures and transformations.
  • The vanishing of the Lie bracket often simplifies the analysis of vector fields by implying the existence of certain symmetries or simplifications.
Vector Fields
Vector fields are a fundamental concept in differential geometry, often used to provide insights into the geometry and dynamics of a manifold. Simply put, a vector field assigns a vector to every point in a space. Imagine a landscape with wind; a vector field would be like attaching an arrow (indicating both direction and strength) at every location to show how the wind is blowing.
In mathematical terms, a vector field on an open set \( \mathcal{O} \subseteq \mathbb{R}^{n} \) is a function that assigns every point \( x \) in \( \mathcal{O} \) to a vector \( f(x) \) in \( \mathbb{R}^{n} \).
Vector fields are often used to model dynamic processes such as:
  • The flow of fluids, where the field indicates fluid velocity at each point.
  • Electromagnetic fields, describing the strength and direction of the field at different locations.
  • Trajectories of objects, showing the motion of particles along the field lines.
When dealing with vector fields, mathematicians often explore interactions between them, such as through the use of the Lie bracket.
Commutative Property
The commutative property is a fundamental principle in mathematics, applicable to various operations. When we discuss vector fields and their interactions, "commutative" often pops up in the context of operations like the Lie bracket.
Two vector fields \( f_i \) and \( f_j \) are considered commutative if their Lie bracket is zero, i.e., \([f_i, f_j] = 0\). In simpler terms, this means that changing the order of operations involving these vector fields yields the same result. Thus, moving first in the direction of \( f_i \) and then in the direction of \( f_j \) is the same as doing it in the reverse order.
This property is significant in many mathematical settings:
  • It simplifies solutions by reducing the complexity when solving differential equations.
  • It can help identify symmetries, making the mathematical model easier to understand and manage.
  • In geometry, it can pave the way for a local coordinate transformation, as described in the Frobenius theorem.
Understanding the commutative property in vector fields helps in simplifying their interactions and unraveling the underlying geometrical structure.

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Most popular questions from this chapter

Let \(\dot{x}=A(t) x+B(t) u\) be a time-varying continuous time linear system, and assume that all the entries of \(A\) and \(B\) are smooth functions of \(t\). Introduce the following system with state-space \(\mathbb{R}^{n+1}\) : $$ \begin{aligned} \dot{x}_{0} &=1 \\ \dot{x} &=A\left(x_{0}\right) x+B\left(x_{0}\right) u . \end{aligned} $$ Explain the relationship between the accessibility rank condition, applied to this system, and the Kalman-like condition for controllability of \(\dot{x}=A(t) x+B(t) u\) studied in Corollary \(3.5 .18 .\) The following easy fact is worth stating, for future reference. It says that the map \(x \mapsto \phi(T, 0, x, \omega)\) is a local homeomorphism, for each fixed control \(\omega .\)

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.

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.

Let \(G: X \rightarrow \mathbb{R}^{n \times m}\) be a matrix function of class \(C^{\infty}\), defined on an open subset \(X \subseteq \mathbb{R}^{n}\). Consider the following properties: (C) For each \(x^{0} \in X\), there is some neighborhood \(X_{0}\) of \(x^{0}\), and there are two matrix functions \(C: x_{0} \rightarrow \mathbb{R}^{n \times n}\) and \(D: x_{0} \rightarrow \mathbb{R}^{m \times m}\), both of class \(C^{\infty}\) and nonsingular for each \(x \in X_{0}\), such that $$ C(x) G(x) D(x)=\left(\begin{array}{l} I \\ 0 \end{array}\right) \quad \text { for all } x \in X_{0}, $$ where \(I\) is the \(m \times m\) identity matrix. (F) For each \(x^{0} \in X\) there are \(X_{0}, C\), and \(D\) as above, and there is some diffeomorphism II : \(x_{0} \rightarrow V_{0}\) into an open subset of \(\mathbb{R}^{n}\) so that \(C(x)=\) \(\Pi_{*}(x)\) for all \(x \in X_{0}\). (a) Show that (C) holds if and only if rank \(G(x)=m\) for all \(x \in X\). (b) Show that (F) holds if and only if the columns \(g_{1}, \ldots, g_{m}\) generate a distribution \(\Delta_{g_{1}, \ldots, g_{m}}\) which is involutive and has constant rank \(m\). (That is, rank \(G(x)=m\) for all \(x \in X\) and the \(n^{2}\) Lie brackets \(\left[g_{i}, g_{j}\right]\) are pointwise linear combinations of the \(g_{i}\) 's.) 180 (c) Give an example where (C) holds but (F) fails. (d) Interpret (b), for a system without drift \(\dot{x}=G(x) u\), in terms of a change of variables (a feedback equivalence) \((x, u) \mapsto(z, v):=\left(\Pi(x), D(x)^{-1} u\right)\). What are the equations for \(\dot{z}\) in terms of \(v\), seen as a new input? (Hint: (a) and (b) are both easy, from results already given.)

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 \(i\).
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