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Corollary 6.4. Every proper affine subspace of \( {\mathbb{R}}^{n} \) has measure zero in \( {\mathbb{R}}^{n} \) . | Proof. Let \( S \subseteq {\mathbb{R}}^{n} \) be a proper affine subspace. Suppose first that \( \dim S = n - 1 \) . Then there is at least one coordinate axis, say the \( {x}^{i} \) -axis, that is not parallel to \( S \), and in that case \( S \) is the graph of an affine function of the form \( {x}^{i} = F\left( {{x}... | Yes |
Proposition 6.5. Suppose \( A \subseteq {\mathbb{R}}^{n} \) has measure zero and \( F : A \rightarrow {\mathbb{R}}^{n} \) is a smooth map. Then \( F\left( A\right) \) has measure zero. | Proof. By definition, for each \( p \in A, F \) has an extension to a smooth map, which we still denote by \( F \), on a neighborhood of \( p \) in \( {\mathbb{R}}^{n} \) . Shrinking this neighborhood if necessary, we may assume that there is an open ball \( U \) containing \( p \) such that \( F \) extends smoothly to... | Yes |
Lemma 6.6. Let \( M \) be a smooth \( n \) -manifold with or without boundary and \( A \subseteq M \) . Suppose that for some collection \( \left\{ \left( {{U}_{\alpha },{\varphi }_{\alpha }}\right) \right\} \) of smooth charts whose domains cover \( A,{\varphi }_{\alpha }\left( {A \cap {U}_{\alpha }}\right) \) has mea... | Proof. Let \( \left( {V,\psi }\right) \) be an arbitrary smooth chart. We need to show that \( \psi \left( {A \cap V}\right) \) has measure zero. Some countable collection of the \( {U}_{\alpha } \) ’s covers \( A \cap V \) . For each such \( {U}_{\alpha } \), we have\n\n\[ \psi \left( {A \cap V \cap {U}_{\alpha }}\rig... | Yes |
Proposition 6.8. Suppose \( M \) is a smooth manifold with or without boundary and \( A \subseteq M \) has measure zero in \( M \) . Then \( M \smallsetminus A \) is dense in \( M \) . | Proof. If \( M \smallsetminus A \) is not dense, then \( A \) contains a nonempty open subset of \( M \), which implies that there is a smooth chart \( \left( {V,\psi }\right) \) such that \( \psi \left( {A \cap V}\right) \) contains a nonempty open subset of \( {\mathbb{R}}^{n} \) (where \( n = \dim M \) ). Because \(... | Yes |
Theorem 6.9. Suppose \( M \) and \( N \) are smooth \( n \) -manifolds with or without boundary, \( F : M \rightarrow N \) is a smooth map, and \( A \subseteq M \) is a subset of measure zero. Then \( F\left( A\right) \) has measure zero in \( N \) . | Proof. Let \( \left\{ \left( {{U}_{i},{\varphi }_{i}}\right) \right\} \) be a countable cover of \( M \) by smooth charts. We need to show that for each smooth chart \( \left( {V,\psi }\right) \) for \( N \), the set \( \psi \left( {F\left( A\right) \cap V}\right) \) has measure zero in \( {\mathbb{R}}^{n} \) . Note th... | Yes |
Corollary 6.11. Suppose \( M \) and \( N \) are smooth manifolds with or without boundary, and \( F : M \rightarrow N \) is a smooth map. If \( \dim M < \dim N \), then \( F\left( M\right) \) has measure zero in \( N \) . | Proof. In this case, each point of \( M \) is a critical point for \( F \) . | No |
Corollary 6.12. Suppose \( M \) is a smooth manifold with or without boundary, and \( S \subseteq M \) is an immersed submanifold with or without boundary. If \( \dim S < \dim M \) , then \( S \) has measure zero in \( M \) . | Proof. Apply Corollary 6.11 to the inclusion map \( S \hookrightarrow M \) . | No |
Lemma 6.13. Suppose \( M \subseteq {\mathbb{R}}^{N} \) is a smooth \( n \) -dimensional submanifold with or without boundary. For any \( v \in {\mathbb{R}}^{N} \smallsetminus {\mathbb{R}}^{N - 1} \), let \( {\pi }_{v} : {\mathbb{R}}^{N} \rightarrow {\mathbb{R}}^{N - 1} \) be the projection with kernel \( \mathbb{R}v \)... | Proof. In order for \( {\left. {\pi }_{v}\right| }_{M} \) to be injective, it is necessary and sufficient that \( p - q \) never be parallel to \( v \) when \( p \) and \( q \) are distinct points in \( M \) . Similarly, in order for \( {\left. {\pi }_{v}\right| }_{M} \) to be a smooth immersion, it is necessary and su... | Yes |
Corollary 6.17. Suppose \( M \) is a compact smooth \( n \) -manifold with or without boundary. If \( N \geq {2n} + 1 \), then every smooth map from \( M \) to \( {\mathbb{R}}^{N} \) can be uniformly approximated by embeddings. | Proof. Assume \( N \geq {2n} + 1 \), and let \( f : M \rightarrow {\mathbb{R}}^{N} \) be a smooth map. By the Whitney embedding theorem, there is a smooth embedding \( F : M \rightarrow {\mathbb{R}}^{{2n} + 1} \) . The map \( G = f \times F : M \rightarrow {\mathbb{R}}^{N} \times {\mathbb{R}}^{{2n} + 1} \) is also a sm... | Yes |
Theorem 6.18 (Whitney Immersion Theorem). Every smooth n-manifold with or without boundary admits a smooth immersion into \( {\mathbb{R}}^{2n} \) . | Proof. See Problem 6-2 for the case \( \partial M = \varnothing \), and Problem 9-14 for the general case. | No |
Corollary 6.22. If \( M \) is a smooth manifold with or without boundary and \( \delta : M \rightarrow \) \( \mathbb{R} \) is a positive continuous function, there is a smooth function \( e : M \rightarrow \mathbb{R} \) such that \( 0 < e\left( x\right) < \delta \left( x\right) \) for all \( x \in M \) . | Proof. Use the Whitney approximation theorem to construct a smooth function \( e : M \rightarrow \mathbb{R} \) that satisfies \( \left| {e\left( x\right) - \frac{1}{2}\delta \left( x\right) }\right| < \frac{1}{2}\delta \left( x\right) \) for all \( x \in M \) . | Yes |
Theorem 6.23. If \( M \subseteq {\mathbb{R}}^{n} \) is an embedded \( m \) -dimensional submanifold, then \( {NM} \) is an embedded \( n \) -dimensional submanifold of \( T{\mathbb{R}}^{n} \approx {\mathbb{R}}^{n} \times {\mathbb{R}}^{n} \) . | Proof. Let \( {x}_{0} \) be any point of \( M \), and let \( \left( {U,\varphi }\right) \) be a slice chart for \( M \) in \( {\mathbb{R}}^{n} \) centered at \( {x}_{0} \) . Write \( \widehat{U} = \varphi \left( U\right) \subseteq {\mathbb{R}}^{n} \), and write the coordinate functions of \( \varphi \) as \( \left( {{u... | Yes |
Proposition 6.25. Let \( M \subseteq {\mathbb{R}}^{n} \) be an embedded submanifold. If \( U \) is any tubular neighborhood of \( M \), there exists a smooth map \( r : U \rightarrow M \) that is both a retraction and a smooth submersion. | Proof. Let \( {NM} \subseteq T{\mathbb{R}}^{n} \) be the normal bundle of \( M \), and let \( {M}_{0} \subseteq {NM} \) be the set \( {M}_{0} = \{ \left( {x,0}\right) : x \in M\} \) . By definition of a tubular neighborhood, there is an open subset \( V \subseteq {NM} \) containing \( {M}_{0} \) such that \( E : V \rig... | Yes |
Theorem 6.26 (Whitney Approximation Theorem). Suppose \( N \) is a smooth manifold with or without boundary, \( M \) is a smooth manifold (without boundary), and \( F : N \rightarrow M \) is a continuous map. Then \( F \) is homotopic to a smooth map. If \( F \) is already smooth on a closed subset \( A \subseteq N \),... | Proof. By the Whitney embedding theorem, we may as well assume that \( M \) is a properly embedded submanifold of \( {\mathbb{R}}^{n} \) . Let \( U \) be a tubular neighborhood of \( M \) in \( {\mathbb{R}}^{n} \), and let \( r : U \rightarrow M \) be the smooth retraction given by Proposition 6.25. For any \( x \in M ... | Yes |
Corollary 6.27 (Extension Lemma for Smooth Maps). Suppose \( N \) is a smooth manifold with or without boundary, \( M \) is a smooth manifold, \( A \subseteq N \) is a closed subset, and \( f : A \rightarrow M \) is a smooth map. Then \( f \) has a smooth extension to \( N \) if and only if it has a continuous extensio... | Proof. If \( F : N \rightarrow M \) is a continuous extension of \( f \) to all of \( N \), the Whitney approximation theorem guarantees the existence of a smooth map \( \widetilde{F} \) (homotopic to \( F \), in fact, though we do not need that here) that agrees with \( f \) on \( A \) ; in other words, \( \widetilde{... | Yes |
Lemma 6.28. If \( N \) and \( M \) are smooth manifolds with or without boundary, smooth homotopy is an equivalence relation on the set of all smooth maps from \( N \) to \( M \) . | Proof. Reflexivity and symmetry are proved just as for ordinary homotopy. To prove transitivity, suppose \( F, G, K : N \rightarrow M \) are smooth maps, and \( {H}_{1},{H}_{2} : N \times I \rightarrow M \) are smooth homotopies from \( F \) to \( G \) and \( G \) to \( K \), respectively. Let \( \varphi : \left\lbrack... | Yes |
Theorem 6.29. Suppose \( N \) is a smooth manifold with or without boundary, \( M \) is a smooth manifold, and \( F, G : N \rightarrow M \) are smooth maps. If \( F \) and \( G \) are homotopic, then they are smoothly homotopic. If \( F \) and \( G \) are homotopic relative to some closed subset \( A \subseteq N \), th... | Proof. Suppose \( F, G : N \rightarrow M \) are smooth, and let \( H : N \times I \rightarrow M \) be a homotopy from \( F \) to \( G \) (relative to \( A \), which may be empty). We wish to show that \( H \) can be replaced by a smooth homotopy.\n\nDefine \( \bar{H} : N \times \mathbb{R} \rightarrow M \) by\n\n\[ \bar... | Yes |
Theorem 6.30. Suppose \( N \) and \( M \) are smooth manifolds and \( S \subseteq M \) is an embedded submanifold.\n\n(a) If \( F : N \rightarrow M \) is a smooth map that is transverse to \( S \), then \( {F}^{-1}\left( S\right) \) is an embedded submanifold of \( N \) whose codimension is equal to the codimension of ... | Proof. The second statement follows easily from the first, simply by taking \( F \) to be the inclusion map \( {S}^{\prime } \hookrightarrow M \), and noting that a composition of smooth embeddings \( S \cap {S}^{\prime } \hookrightarrow S \hookrightarrow M \) is again a smooth embedding.\n\nTo prove (a), let \( m \) d... | Yes |
Theorem 6.32 (Global Characterization of Graphs). Suppose \( M \) and \( N \) are smooth manifolds and \( S \subseteq M \times N \) is an immersed submanifold. Let \( {\pi }_{M} \) and \( {\pi }_{N} \) denote the projections from \( M \times N \) onto \( M \) and \( N \), respectively. The following are equivalent.\n\n... | Proof. Problem 6-15. | No |
Corollary 6.33 (Local Characterization of Graphs). Suppose \( M \) and \( N \) are smooth manifolds, \( S \subseteq M \times N \) is an immersed submanifold, and \( \left( {p, q}\right) \in S \) . If \( S \) intersects the submanifold \( \{ p\} \times N \) transversely at \( \left( {p, q}\right) \), then there exist a ... | Proof. The hypothesis guarantees that \( d{\left( {\pi }_{M}\right) }_{\left( p, q\right) } : {T}_{\left( p, q}\right) }S \rightarrow {T}_{p}M \) is an isomorphism, so \( {\left. {\pi }_{M}\right| }_{S} \) restricts to a diffeomorphism from a neighborhood \( V \) of \( \left( {p, q}\right) \) in \( S \) to a neighborho... | Yes |
Proposition 6.34. If \( \left\{ {{F}_{s} : s \in S}\right\} \) is a smooth family of maps from \( N \) to \( M \) and \( S \) is connected, then for any \( {s}_{1},{s}_{2} \in S \), the maps \( {F}_{{s}_{1}},{F}_{{s}_{2}} : N \rightarrow M \) are homotopic. | Proof. Because \( S \) is connected, it is path-connected. If \( \gamma : \left\lbrack {0,1}\right\rbrack \rightarrow S \) is any path from \( {s}_{1} \) to \( {s}_{2} \), then \( H\left( {x, s}\right) = F\left( {x,\gamma \left( s\right) }\right) \) is a homotopy from \( {F}_{{s}_{1}} \) to \( {F}_{{s}_{2}} \) . | Yes |
Theorem 6.36 (Transversality Homotopy Theorem). Suppose \( M \) and \( N \) are smooth manifolds and \( X \subseteq M \) is an embedded submanifold. Every smooth map \( f : N \rightarrow M \) is homotopic to a smooth map \( g : N \rightarrow M \) that is transverse to \( X \) . | Proof. The crux of the proof is constructing a smooth map \( F : N \times S \rightarrow M \) that is transverse to \( X \), where \( S = {\mathbb{B}}^{k} \) for some \( k \) and \( {F}_{0} = f \) . It then follows from the parametric transversality theorem that there is some \( s \in S \) such that \( {F}_{s} : N \righ... | Yes |
Example 7.3 (Lie Groups). Each of the following manifolds is a Lie group with the indicated group operation.\n\n(a) The general linear group \( \mathrm{{GL}}\left( {n,\mathbb{R}}\right) \) is the set of invertible \( n \times n \) matrices with real entries. It is a group under matrix multiplication, and it is an open ... | Inversion is smooth by Cramer’s rule. | No |
Theorem 7.5. Every Lie group homomorphism has constant rank. | Proof. Let \( F : G \rightarrow H \) be a Lie group homomorphism, and let \( e \) and \( \widetilde{e} \) denote the identity elements of \( G \) and \( H \), respectively. Suppose \( {g}_{0} \) is an arbitrary element of \( G \) . We will show that \( d{F}_{{g}_{0}} \) has the same rank as \( d{F}_{e} \) . The fact th... | Yes |
Corollary 7.6. A Lie group homomorphism is a Lie group isomorphism if and only if it is bijective. | Proof. The global rank theorem shows that a bijective Lie group homomorphism is a diffeomorphism. | Yes |
Theorem 7.9 (Uniqueness of the Universal Covering Group). For any connected Lie group \( G \), the universal covering group is unique in the following sense: if \( \widetilde{G} \) and \( {\widetilde{G}}^{\prime } \) are simply connected Lie groups that admit smooth covering maps \( \pi : \widetilde{G} \rightarrow G \)... | Proof. See Problem 7-5. | No |
For each \( n \), the map \( {\varepsilon }^{n} : {\mathbb{R}}^{n} \rightarrow {\mathbb{T}}^{n} \) given by\n\n\[ \n{\varepsilon }^{n}\left( {{x}^{1},\ldots ,{x}^{n}}\right) = \left( {{e}^{{2\pi i}{x}^{1}},\ldots ,{e}^{{2\pi i}{x}^{n}}}\right) \n\]\n\nis a Lie group homomorphism and a smooth covering map (see Example 7... | Since \( {\mathbb{R}}^{n} \) is simply connected, this shows that the universal covering group of \( {\mathbb{T}}^{n} \) is the additive Lie group \( {\mathbb{R}}^{n} \). | No |
Proposition 7.11. Let \( G \) be a Lie group, and suppose \( H \subseteq G \) is a subgroup that is also an embedded submanifold. Then \( H \) is a Lie subgroup. | Proof. We need only check that multiplication \( H \times H \rightarrow H \) and inversion \( H \rightarrow H \) are smooth maps. Because multiplication is a smooth map from \( G \times G \) into \( G \), its restriction is clearly smooth from \( H \times H \) into \( G \) (this is true even if \( H \) is merely immers... | Yes |
Lemma 7.12. Suppose \( G \) is a Lie group and \( H \subseteq G \) is an open subgroup. Then \( H \) is an embedded Lie subgroup. In addition, \( H \) is closed, so it is a union of connected components of \( G \) . | Proof. If \( H \) is open in \( G \), it is embedded by Proposition 5.1. In addition, every left coset \( {gH} = \{ {gh} : h \in H\} \) is open in \( G \) because it is the image of the open subset \( H \) under the diffeomorphism \( {L}_{g} \) . Because \( G \smallsetminus H \) is the union of the cosets of \( H \) ot... | Yes |
Proposition 7.14. Suppose \( G \) is a Lie group, and \( W \subseteq G \) is any neighborhood of the identity.\n\n(a) \( W \) generates an open subgroup of \( G \) .\n\n(b) If \( W \) is connected, it generates a connected open subgroup of \( G \) .\n\n(c) If \( G \) is connected, then \( W \) generates \( G \) . | Proof. Let \( W \subseteq G \) be any neighborhood of the identity, and let \( H \) be the subgroup generated by \( W \) . As a matter of notation, if \( A \) and \( B \) are subsets of \( G \), let us write\n\n\[ \n{AB} = \{ {ab} : a \in A, b \in B\} ,\;{A}^{-1} = \left\{ {{a}^{-1} : a \in A}\right\} .\n\]\n\n(7.5)\n\... | Yes |
Proposition 7.15. Let \( G \) be a Lie group and let \( {G}_{0} \) be its identity component. Then \( {G}_{0} \) is a normal subgroup of \( G \), and is the only connected open subgroup. Every connected component of \( G \) is diffeomorphic to \( {G}_{0} \) . | Proof. Problem 7-7. | No |
Proposition 7.16. Let \( F : G \rightarrow H \) be a Lie group homomorphism. The kernel of \( F \) is a properly embedded Lie subgroup of \( G \), whose codimension is equal to the rank of \( F \) . | Proof. Because \( F \) has constant rank, its kernel \( {F}^{-1}\left( e\right) \) is a properly embedded submanifold of codimension equal to rank \( F \) . It is thus a Lie subgroup by Proposition 7.11. | Yes |
Proposition 7.17. If \( F : G \rightarrow H \) is an injective Lie group homomorphism, the image of \( F \) has a unique smooth manifold structure such that \( F\left( G\right) \) is a Lie subgroup of \( H \) and \( F : G \rightarrow F\left( G\right) \) is a Lie group isomorphism. | Proof. Since a Lie group homomorphism has constant rank, it follows from the global rank theorem that \( F \) is a smooth immersion. Proposition 5.18 shows that \( F\left( G\right) \) has a unique smooth manifold structure such that it is an immersed subman-ifold of \( H \) and \( F \) is a diffeomorphism onto its imag... | Yes |
Theorem 7.21. Suppose \( G \) is a Lie group and \( H \subseteq G \) is a Lie subgroup. Then \( H \) is closed in \( G \) if and only if it is embedded. | Proof. Assume first that \( H \) is embedded in \( G \) . To prove that \( H \) is closed, let \( g \) be an arbitrary point of \( \bar{H} \) . Then there is a sequence \( \left( {h}_{i}\right) \) of points in \( H \) converging to \( g \) (Fig. 7.1). Let \( U \) be the domain of a slice chart for \( H \) containing th... | Yes |
Proposition 7.23. Suppose \( E \) and \( M \) are smooth manifolds with or without boundary, and \( \pi : E \rightarrow M \) is a smooth covering map. With the discrete topology, the automorphism group \( {\operatorname{Aut}}_{\pi }\left( E\right) \) is a zero-dimensional Lie group acting smoothly and freely on \( E \)... | Proof. Suppose \( \varphi \in {\operatorname{Aut}}_{\pi }\left( E\right) \) is an automorphism that fixes a point \( p \in E \) . Simply by rotating diagram (7.8), we can consider \( \varphi \) as a lift of \( \pi \) :\n\n. Let \( M \) and \( N \) be smooth manifolds and let \( G \) be a Lie group. Suppose \( F : M \rightarrow N \) is a smooth map that is equivariant with respect to a transitive smooth \( G \) -action on \( M \) and any smooth \( G \) -action on \( N \) . Then \( F \) has constant ... | Proof. Let \( \theta \) and \( \varphi \) denote the \( G \) -actions on \( M \) and \( N \), respectively, and let \( p \) and \( q \) be arbitrary points in \( M \) . Choose \( g \in G \) such that \( {\theta }_{g}\left( p\right) = q \) . (Such a \( g \) exists because we are assuming that \( G \) acts transitively o... | Yes |
Proposition 7.26 (Properties of the Orbit Map). Suppose \( \theta \) is a smooth left action of a Lie group \( G \) on a smooth manifold \( M \). For each \( p \in M \), the orbit map \( {\theta }^{\left( p\right) } : G \rightarrow M \) is smooth and has constant rank, so the isotropy group \( {G}_{p} = {\left( {\theta... | Proof. The orbit map is smooth because it is equal to the composition\n\n\[ G \approx G \times \{ p\} \hookrightarrow G \times M\overset{\theta }{ \rightarrow }M. \]\n\nIt follows from the definition of a group action that \( {\theta }^{\left( p\right) } \) is equivariant with respect to the action of \( G \) on itself... | Yes |
A real \( n \times n \) matrix \( A \) is said to be orthogonal if as a linear map \( A : {\mathbb{R}}^{n} \rightarrow {\mathbb{R}}^{n} \) it preserves the Euclidean dot product:\n\n\[ \left( {Ax}\right) \cdot \left( {Ay}\right) = x \cdot y\;\text{ for all }x, y \in {\mathbb{R}}^{n}. \] | The set \( \mathrm{O}\left( n\right) \) of all orthogonal \( n \times n \) matrices is a subgroup of \( \mathrm{{GL}}\left( {n,\mathbb{R}}\right) \), called the orthogonal group of degree \( \mathbf{n} \). It is easy to check that a matrix \( A \) is orthogonal if and only if it takes the standard basis of \( {\mathbb{... | Yes |
The special orthogonal group of degree \( n \) is defined as \( \mathrm{{SO}}\left( n\right) = \mathrm{O}\left( n\right) \cap \mathrm{{SL}}\left( {n,\mathbb{R}}\right) \subseteq \mathrm{{GL}}\left( {n,\mathbb{R}}\right) \) . Because every matrix \( A \in \mathrm{O}\left( n\right) \) satisfies\n\n\[ 1 = \det {I}_{n} = \... | it follows that \( \det A = \pm 1 \) for all \( A \in \mathrm{O}\left( n\right) \) . Therefore, \( \mathrm{{SO}}\left( n\right) \) is the open subgroup of \( \mathrm{O}\left( n\right) \) consisting of matrices of positive determinant, and is therefore also an embedded Lie subgroup of dimension \( n\left( {n - 1}\right)... | Yes |
For any positive integer \( n \), the unitary group of degree \( \mathbf{n} \) is the subgroup \( \mathrm{U}\left( n\right) \subseteq \mathrm{{GL}}\left( {n,\mathbb{C}}\right) \) consisting of complex \( n \times n \) matrices whose columns form an orthonormal basis for \( {\mathbb{C}}^{n} \) with respect to the Hermit... | It is straightforward to check that \( \mathrm{U}\left( n\right) \) consists of those matrices \( A \) such that \( {A}^{ * }A = {I}_{n} \). | Yes |
Example 7.32 (The Euclidean Group). If we consider \( {\mathbb{R}}^{n} \) as a Lie group under addition, then the natural action of \( \mathrm{O}\left( n\right) \) on \( {\mathbb{R}}^{n} \) is an action by automorphisms. The resulting semidirect product \( \mathrm{E}\left( n\right) = {\mathbb{R}}^{n} \rtimes \mathrm{O}... | This action preserves lines, distances, and angle measures, and thus all of the relationships of Euclidean geometry. | Yes |
Theorem 7.35 (Characterization of Semidirect Products). Suppose \( G \) is a Lie group, and \( N, H \subseteq G \) are closed Lie subgroups such that \( N \) is normal, \( N \cap H = \) \( \{ e\} \), and \( {NH} = G \) . Then the map \( \left( {n, h}\right) \mapsto {nh} \) is a Lie group isomorphism between \( N{ \rtim... | Proof. Problem 7-18. | No |
Proposition 7.37. Let \( G \) be a Lie group and \( V \) be a finite-dimensional vector space. A smooth left action of \( G \) on \( V \) is linear if and only if it is of the form \( g \cdot x = \rho \left( g\right) x \) for some representation \( \rho \) of \( G \) . | Proof. Every action induced by a representation is evidently linear. To prove the converse, assume that we are given a linear action of \( G \) on \( V \) . The hypothesis implies that for each \( g \in G \) there is a linear map \( \rho \left( g\right) \in \mathrm{{GL}}\left( V\right) \) such that \( g \cdot x = \rho ... | Yes |
Proposition 8.1 (Smoothness Criterion for Vector Fields). Let \( M \) be a smooth manifold with or without boundary, and let \( X : M \rightarrow {TM} \) be a rough vector field. \( {If}\left( {U,\left( {x}^{i}\right) }\right) \) is any smooth coordinate chart on \( M \), then the restriction of \( X \) to \( U \) is s... | Proof. Let \( \left( {{x}^{i},{v}^{i}}\right) \) be the natural coordinates on \( {\pi }^{-1}\left( U\right) \subseteq {TM} \) associated with the chart \( \left( {U,\left( {x}^{i}\right) }\right) \) . By definition of natural coordinates, the coordinate representation of \( X : M \rightarrow {TM} \) on \( U \) is\n\n\... | Yes |
Example 8.2 (Coordinate Vector Fields). If \( \left( {U,\left( {x}^{i}\right) }\right) \) is any smooth chart on \( M \) , the assignment\n\n\[ p \mapsto {\left. \frac{\partial }{\partial {x}^{i}}\right| }_{p} \]\ndetermines a vector field on \( U \), called the \( i \) th coordinate vector field and denoted by \( \par... | It is smooth because its component functions are constants. | No |
Example 8.4 (The Angle Coordinate Vector Field on the Circle). Let \( \theta \) be any angle coordinate on a proper open subset \( U \subseteq {\mathbb{S}}^{1} \) (see Problem 1-8), and let \( d/{d\theta } \) denote the corresponding coordinate vector field. Because any other angle coordinate \( \widetilde{\theta } \) ... | It is a smooth vector field because its component function is constant in any such chart. We denote this global vector field by \( d/{d\theta } \), even though, strictly speaking, it cannot be considered as a coordinate vector field on the entire circle at once. | Yes |
Lemma 8.6 (Extension Lemma for Vector Fields). Let \( M \) be a smooth manifold with or without boundary, and let \( A \subseteq M \) be a closed subset. Suppose \( X \) is a smooth vector field along \( A \). Given any open subset \( U \) containing \( A \), there exists a smooth global vector field \( \widetilde{X} \... | Proof. See Problem 8-1. | No |
Proposition 8.7. Let \( M \) be a smooth manifold with or without boundary. Given \( p \in M \) and \( v \in {T}_{p}M \), there is a smooth global vector field \( X \) on \( M \) such that \( {X}_{p} = v \) . | Proof. The assignment \( p \mapsto v \) is an example of a vector field along the set \( \{ p\} \) as defined above. It is smooth because it can be extended, say, to a constant-coefficient vector field in a coordinate neighborhood of \( p \) . Thus, the proposition follows from the extension lemma with \( A = \{ p\} \)... | No |
Proposition 8.11 (Completion of Local Frames). Let \( M \) be a smooth \( n \) -manifold with or without boundary.\n\n(a) If \( \\left( {{X}_{1},\\ldots ,{X}_{k}}\\right) \) is a linearly independent \( k \) -tuple of smooth vector fields on an open subset \( U \\subseteq M \), with \( 1 \\leq k < n \), then for each \... | Proof. See Problem 8-5. | No |
The standard coordinate frame is a global orthonormal frame on \( {\mathbb{R}}^{n} \) . For a less obvious example, consider the smooth vector fields defined on \( {\mathbb{R}}^{2} \smallsetminus \{ 0\} \) by\n\n\[ \n{E}_{1} = \frac{x}{r}\frac{\partial }{\partial x} + \frac{y}{r}\frac{\partial }{\partial y},\;{E}_{2} =... | A straightforward computation shows that \( \left( {{E}_{1},{E}_{2}}\right) \) is an orthonormal frame for \( {\mathbb{R}}^{2} \) over the open subset \( {\mathbb{R}}^{2} \smallsetminus \{ 0\} \) . Geometrically, \( {E}_{1} \) and \( {E}_{2} \) are unit vector fields tangent to radial lines and circles centered at the ... | Yes |
Lemma 8.13 (Gram-Schmidt Algorithm for Frames). Suppose \( \\left( {X}_{j}\\right) \) is a smooth local frame for \( T{\\mathbb{R}}^{n} \) over an open subset \( U \\subseteq {\\mathbb{R}}^{n} \) . Then there is a smooth orthonormal frame \( \\left( {E}_{j}\\right) \) over \( U \) such that \( \\operatorname{span}\\lef... | Proof. Applying the Gram-Schmidt algorithm to the vectors \( \\left( {\\left. {X}_{j}\\right| }_{p}\\right) \) at each \( p \\in U \) , we obtain an \( n \) -tuple of rough vector fields \( \\left( {{E}_{1},\\ldots ,{E}_{n}}\\right) \) given inductively by\n\n\[ \n{E}_{j} = \\frac{{X}_{j} - \\mathop{\\sum }\\limits_{{i... | Yes |
Proposition 8.14. Let \( M \) be a smooth manifold with or without boundary, and let\n\n\( X : M \rightarrow {TM} \) be a rough vector field. The following are equivalent:\n\n(a) \( X \) is smooth.\n\n(b) For every \( f \in {C}^{\infty }\left( M\right) \), the function \( {Xf} \) is smooth on \( M \) .\n\n(c) For every... | Proof. We will prove that (a) \( \Rightarrow \) (b) \( \Rightarrow \) (c) \( \Rightarrow \) (a).\n\nTo prove (a) \( \Rightarrow \) (b), assume \( X \) is smooth, and let \( f \in {C}^{\infty }\left( M\right) \) . For any \( p \in M \) , we can choose smooth coordinates \( \left( {x}^{i}\right) \) on a neighborhood \( U... | Yes |
Proposition 8.15. Let \( M \) be a smooth manifold with or without boundary. A map \( D : {C}^{\infty }\left( M\right) \rightarrow {C}^{\infty }\left( M\right) \) is a derivation if and only if it is of the form \( {Df} = {Xf} \) for some smooth vector field \( X \in \mathfrak{X}\left( M\right) \) . | Proof. We just showed that every smooth vector field induces a derivation. Conversely, suppose \( D : {C}^{\infty }\left( M\right) \rightarrow {C}^{\infty }\left( M\right) \) is a derivation. We need to concoct a vector field \( X \) such that \( {Df} = {Xf} \) for all \( f \) . From the discussion above, it is clear t... | Yes |
Proposition 8.16. Suppose \( F : M \rightarrow N \) is a smooth map between manifolds with or without boundary, \( X \in \mathfrak{X}\left( M\right) \), and \( Y \in \mathfrak{X}\left( N\right) \) . Then \( X \) and \( Y \) are \( F \) -related if and only if for every smooth real-valued function \( f \) defined on an ... | Proof. For any \( p \in M \) and any smooth real-valued \( f \) defined in a neighborhood of \( F\left( p\right) \) , \[ X\left( {f \circ F}\right) \left( p\right) = {X}_{p}\left( {f \circ F}\right) = d{F}_{p}\left( {X}_{p}\right) f, \] while \[ \left( {Yf}\right) \circ F\left( p\right) = \left( {Yf}\right) \left( {F\l... | Yes |
Proposition 8.19. Suppose \( M \) and \( N \) are smooth manifolds with or without boundary, and \( F : M \rightarrow N \) is a diffeomorphism. For every \( X \in \mathfrak{X}\left( M\right) \), there is a unique smooth vector field on \( N \) that is \( F \) -related to \( X \) . | Proof. For \( Y \in \mathfrak{X}\left( N\right) \) to be \( F \) -related to \( X \) means that \( d{F}_{p}\left( {X}_{p}\right) = {Y}_{F\left( p\right) } \) for every \( p \in M \) . If \( F \) is a diffeomorphism, therefore, we define \( Y \) by\n\n\[ \n{Y}_{q} = d{F}_{{F}^{-1}\left( q\right) }\left( {X}_{{F}^{-1}\le... | Yes |
Example 8.20 (Computing the Pushforward of a Vector Field). Let \( M \) and \( N \) be the following open submanifolds of \( {\mathbb{R}}^{2} \) :\n\n\[ M = \{ \left( {x, y}\right) : y > 0\text{ and }x + y > 0\} ,\]\n\n\[ N = \{ \left( {u, v}\right) : u > 0\text{ and }v > 0\} \]\n\nand define \( F : M \rightarrow N \) ... | The differential of \( F \) at a point \( \left( {x, y}\right) \in M \) is represented by its Jacobian matrix,\n\n\[ {DF}\left( {x, y}\right) = \left( \begin{matrix} 1 & 1 \\ \frac{1}{y} & - \frac{x}{{y}^{2}} \end{matrix}\right) \]\n\nand thus \( d{F}_{{F}^{-1}\left( {u, v}\right) } \) is represented by the matrix\n\n\... | Yes |
Proposition 8.22. Let \( M \) be a smooth manifold, \( S \subseteq M \) be an embedded subman-ifold with or without boundary, and \( X \) be a smooth vector field on \( M \) . Then \( X \) is tangent to \( S \) if and only if \( {\left. \left( Xf\right) \right| }_{S} = 0 \) for every \( f \in {C}^{\infty }\left( M\righ... | Proof. This is an immediate consequence of Proposition 5.37. | No |
Proposition 8.23 (Restricting Vector Fields to Submanifolds). Let \( M \) be a smooth manifold, let \( S \subseteq M \) be an immersed submanifold with or without boundary, and let \( \iota : S \hookrightarrow M \) denote the inclusion map. If \( Y \in \mathfrak{X}\left( M\right) \) is tangent to \( S \), then there is... | Proof. The fact that \( Y \) is tangent to \( S \) means by definition that \( {Y}_{p} \) is in the image of \( d{\iota }_{p} \) for each \( p \) . Thus, for each \( p \) there is a vector \( {X}_{p} \in {T}_{p}S \) such that \( {Y}_{p} = d{\iota }_{p}\left( {X}_{p}\right) \) . Since \( d{\iota }_{p} \) is injective, \... | Yes |
Lemma 8.25. The Lie bracket of any pair of smooth vector fields is a smooth vector field. | Proof. By Proposition 8.15, it suffices to show that \( \left\lbrack {X, Y}\right\rbrack \) is a derivation of \( {C}^{\infty }\left( M\right) \) . For arbitrary \( f, g \in {C}^{\infty }\left( M\right) \), we compute\n\n\[ \left\lbrack {X, Y}\right\rbrack \left( {fg}\right) = X\left( {Y\left( {fg}\right) }\right) - Y\... | Yes |
Proposition 8.26 (Coordinate Formula for the Lie Bracket). Let \( X, Y \) be smooth vector fields on a smooth manifold \( M \) with or without boundary, and let \( X = \) \( {X}^{i}\partial /\partial {x}^{i} \) and \( Y = {Y}^{j}\partial /\partial {x}^{j} \) be the coordinate expressions for \( X \) and \( Y \) in term... | Proof. Because we know already that \( \left\lbrack {X, Y}\right\rbrack \) is a smooth vector field, its action on a function is determined locally: \( {\left. \left( \left\lbrack X, Y\right\rbrack f\right) \right| }_{U} = \left\lbrack {X, Y}\right\rbrack \left( {\left. f\right| }_{U}\right) \) . Thus it suffices to co... | Yes |
Define smooth vector fields \( X, Y \in \mathfrak{X}\left( {\mathbb{R}}^{3}\right) \) by\n\n\[ X = x\frac{\partial }{\partial x} + \frac{\partial }{\partial y} + x\left( {y + 1}\right) \frac{\partial }{\partial z} \]\n\n\[ Y = \frac{\partial }{\partial x} + y\frac{\partial }{\partial z} \]\n\nThen (8.9) yields\n\n\[ \l... | \[ \left\lbrack {X, Y}\right\rbrack = 0\frac{\partial }{\partial x} + 1\frac{\partial }{\partial z} - 1\frac{\partial }{\partial x} - 0\frac{\partial }{\partial y} - \left( {y + 1}\right) \frac{\partial }{\partial z} \]\n\n\[ = - \frac{\partial }{\partial x} - y\frac{\partial }{\partial z} \] | Yes |
Proposition 8.28 (Properties of the Lie Bracket). The Lie bracket satisfies the following identities for all \( X, Y, Z \in \mathfrak{X}\left( M\right) \) :\n\n(a) BILINEARITY: For \( a, b \in \mathbb{R} \) ,\n\n\[ \left\lbrack {{aX} + {bY}, Z}\right\rbrack = a\left\lbrack {X, Z}\right\rbrack + b\left\lbrack {Y, Z}\rig... | Proof. Bilinearity and antisymmetry are obvious consequences of the definition. The proof of the Jacobi identity is just a computation:\n\n\[ \left\lbrack {X,\left\lbrack {Y, Z}\right\rbrack }\right\rbrack f + \left\lbrack {Y,\left\lbrack {Z, X}\right\rbrack }\right\rbrack f + \left\lbrack {Z,\left\lbrack {X, Y}\right\... | No |
Proposition 8.30 (Naturality of the Lie Bracket). Let \( F : M \rightarrow N \) be a smooth map between manifolds with or without boundary, and let \( {X}_{1},{X}_{2} \in \mathfrak{X}\left( M\right) \) and \( {Y}_{1},{Y}_{2} \in \mathfrak{X}\left( N\right) \) be vector fields such that \( {X}_{i} \) is \( F \) -related... | Proof. Using Proposition 8.16 and the fact that \( {X}_{i} \) and \( {Y}_{i} \) are \( F \) -related,\n\n\[ \n{X}_{1}{X}_{2}\left( {f \circ F}\right) = {X}_{1}\left( {\left( {{Y}_{2}f}\right) \circ F}\right) = \left( {{Y}_{1}{Y}_{2}f}\right) \circ F.\n\]\n\nSimilarly,\n\n\[ \n{X}_{2}{X}_{1}\left( {f \circ F}\right) = \... | Yes |
Corollary 8.31 (Pushforwards of Lie Brackets). Suppose \( F : M \rightarrow N \) is a diffeomorphism and \( {X}_{1},{X}_{2} \in \mathfrak{X}\left( M\right) \) . Then \( {F}_{ * }\left\lbrack {{X}_{1},{X}_{2}}\right\rbrack = \left\lbrack {{F}_{ * }{X}_{1},{F}_{ * }{X}_{2}}\right\rbrack \) . | Proof. This is just the special case of Proposition 8.30 in which \( F \) is a diffeomorphism and \( {Y}_{i} = {F}_{ * }{X}_{i} \) . | Yes |
Corollary 8.32 (Brackets of Vector Fields Tangent to Submanifolds). Let \( M \) be a smooth manifold and let \( S \) be an immersed submanifold with or without boundary in \( M \). If \( {Y}_{1} \) and \( {Y}_{2} \) are smooth vector fields on \( M \) that are tangent to \( S \), then \( \left\lbrack {{Y}_{1},{Y}_{2}}\... | Proof. By Proposition 8.23, there exist smooth vector fields \( {X}_{1} \) and \( {X}_{2} \) on \( S \) such that \( {X}_{i} \) is \( \iota \) -related to \( {Y}_{i} \) for \( i = 1,2 \) (where \( \iota : S \rightarrow M \) is the inclusion). By Proposition \( {8.30},\left\lbrack {{X}_{1},{X}_{2}}\right\rbrack \) is \(... | Yes |
Proposition 8.33. Let \( G \) be a Lie group, and suppose \( X \) and \( Y \) are smooth left-invariant vector fields on \( G \) . Then \( \left\lbrack {X, Y}\right\rbrack \) is also left-invariant. | Proof. Let \( g \in G \) be arbitrary. Since \( {\left( {L}_{g}\right) }_{ * }X = X \) and \( {\left( {L}_{g}\right) }_{ * }Y = Y \) by definition of left-invariance, it follows from Corollary 8.31 that\n\n\[ \n{\left( {L}_{g}\right) }_{ * }\left\lbrack {X, Y}\right\rbrack = \left\lbrack {{\left( {L}_{g}\right) }_{ * }... | Yes |
Corollary 8.38. Every left-invariant rough vector field on a Lie group is smooth. | Proof. Let \( X \) be a left-invariant rough vector field on a Lie group \( G \), and let \( v = {X}_{e} \) . The fact that \( X \) is left-invariant implies that \( X = {v}^{\mathrm{L}} \), which is smooth. | Yes |
Corollary 8.39. Every Lie group admits a left-invariant smooth global frame, and therefore every Lie group is parallelizable. | Proof. If \( G \) is a Lie group, every basis for \( \operatorname{Lie}\left( G\right) \) is a left-invariant smooth global frame for \( G \) . | Yes |
Let us determine the Lie algebras of some familiar Lie groups. | (a) Euclidean space \( {\mathbb{R}}^{n} \) : If we consider \( {\mathbb{R}}^{n} \) as a Lie group under addition, left translation by an element \( b \in {\mathbb{R}}^{n} \) is given by the affine map \( {L}_{b}\left( x\right) = b + x \) , whose differential \( d\left( {L}_{b}\right) \) is represented by the identity m... | Yes |
Proposition 8.41 (Lie Algebra of the General Linear Group). The composition of the natural maps\n\n\\[ \n\\operatorname{Lie}\\left( {\\mathrm{{GL}}\\left( {n,\\mathbb{R}}\\right) }\\right) \\rightarrow {T}_{{I}_{n}}\\mathrm{{GL}}\\left( {n,\\mathbb{R}}\\right) \\rightarrow \\mathrm{g}\\mathrm{I}\\left( {n,\\mathbb{R}}\... | Proof. Using the matrix entries \\( {X}_{j}^{i} \\) as global coordinates on \\( \\mathrm{{GL}}\\left( {n,\\mathbb{R}}\\right) \\subseteq \\mathrm{{gI}}\\left( {n,\\mathbb{R}}\\right) \\), the natural isomorphism \\( {T}_{{I}_{n}}\\mathrm{{GL}}\\left( {n,\\mathbb{R}}\\right) \\leftrightarrow \\mathrm{g}\\mathrm{I}\\lef... | Yes |
Corollary 8.42. If \( V \) is any finite-dimensional real vector space, the composition of the canonical isomorphisms in (8.16) yields a Lie algebra isomorphism between \( \operatorname{Lie}\left( {\mathrm{{GL}}\left( V\right) }\right) \) and \( \mathfrak{{gl}}\left( V\right) \) . | Exercise 8.43. Prove the preceding corollary by choosing a basis for \( V \) and applying Proposition 8.41. | No |
Theorem 8.44 (Induced Lie Algebra Homomorphisms). Let \( G \) and \( H \) be Lie groups, and let \( \mathfrak{g} \) and \( \mathfrak{h} \) be their Lie algebras. Suppose \( F : G \rightarrow H \) is a Lie group homomorphism. For every \( X \in \mathfrak{g} \), there is a unique vector field in \( \mathfrak{h} \) that i... | Proof. If there is any vector field \( Y \in \mathfrak{h} \) that is \( F \) -related to \( X \), it must satisfy \( {Y}_{e} = \) \( d{F}_{e}\left( {X}_{e}\right) \), and thus it must be uniquely determined by\n\n\[ Y = {\left( d{F}_{e}\left( {X}_{e}\right) \right) }^{\mathrm{L}}. \]\n\nTo show that this \( Y \) is \( ... | Yes |
Theorem 8.46 (The Lie Algebra of a Lie Subgroup). Suppose \( H \subseteq G \) is a Lie subgroup, and \( \iota : H \hookrightarrow G \) is the inclusion map. There is a Lie subalgebra \( \mathfrak{h} \subseteq \) \( \operatorname{Lie}\left( G\right) \) that is canonically isomorphic to \( \operatorname{Lie}\left( H\righ... | Proof. Because the inclusion map \( \iota : H \hookrightarrow G \) is a Lie group homomorphism, \( {\iota }_{ * }\left( {\operatorname{Lie}\left( H\right) }\right) \) is a Lie subalgebra of \( \operatorname{Lie}\left( G\right) \) . By the way we defined the induced Lie algebra homomorphism, this subalgebra is precisely... | Yes |
Example 8.47 (The Lie Algebra of \( \mathbf{O}\left( n\right) \) ). The orthogonal group \( \mathrm{O}\left( n\right) \) is a Lie subgroup of \( \mathrm{{GL}}\left( {n,\mathbb{R}}\right) \) . By Example 7.27, it is equal to the level set \( {\Phi }^{-1}\left( {I}_{n}\right) \), where \( \Phi : \mathrm{{GL}}\left( {n,\m... | By the computation in Example 7.27, this differential is \( d{\Phi }_{{I}_{n}}\left( B\right) = {B}^{T} + B \), so\n\n\[ \n{T}_{{I}_{n}}\mathrm{O}\left( n\right) = \left\{ {B \in \mathfrak{{gl}}\left( {n,\mathbb{R}}\right) : {B}^{T} + B = 0}\right\} \n\]\n\n\[ \n= \{ \text{skew-symmetric}n \times n\text{matrices}\} \te... | Yes |
Proposition 8.48 (The Lie Algebra of \( \mathrm{{GL}}\left( {n,\mathbb{C}}\right) \) ). The composition of the maps in (8.18) yields a Lie algebra isomorphism between \( \operatorname{Lie}\left( {\mathrm{{GL}}\left( {n,\mathbb{C}}\right) }\right) \) and the matrix algebra \( \mathfrak{{gl}}\left( {n,\mathbb{C}}\right) ... | Proof. The Lie group homomorphism \( \beta : \mathrm{{GL}}\left( {n,\mathbb{C}}\right) \rightarrow \mathrm{{GL}}\left( {{2n},\mathbb{R}}\right) \) that we constructed in Example 7.18(d) induces a Lie algebra homomorphism\n\n\[ \n{\beta }_{ * } : \operatorname{Lie}\left( {\mathrm{{GL}}\left( {n,\mathbb{C}}\right) }\righ... | Yes |
Corollary 8.50. Every finite-dimensional real Lie algebra is isomorphic to a Lie subalgebra of some matrix algebra \( \mathfrak{{gl}}\left( {n,\mathbb{R}}\right) \) with the commutator bracket. | Proof. Let \( \mathfrak{g} \) be a finite-dimensional real Lie algebra. By Ado’s theorem, \( \mathfrak{g} \) has a faithful representation \( \rho : \mathfrak{g} \rightarrow \mathfrak{{gl}}\left( V\right) \) for some finite-dimensional real vector space \( V \) . Choosing a basis for \( V \) yields an isomorphism of \(... | Yes |
Proposition 9.2. Let \( V \) be a smooth vector field on a smooth manifold \( M \). For each point \( p \in M \), there exist \( \varepsilon > 0 \) and a smooth curve \( \gamma : \left( {-\varepsilon ,\varepsilon }\right) \rightarrow M \) that is an integral curve of \( V \) starting at \( p \). | Proof. This is just the existence statement of Theorem D. 1 applied to the coordinate representation of \( V \). | No |
Lemma 9.3 (Rescaling Lemma). Let \( V \) be a smooth vector field on a smooth manifold \( M \), let \( J \subseteq \mathbb{R} \) be an interval, and let \( \gamma : J \rightarrow M \) be an integral curve of \( V \). For any \( a \in \mathbb{R} \), the curve \( \widetilde{\gamma } : \widetilde{J} \rightarrow M \) defin... | Proof. One way to see this is as a straightforward application of the chain rule in local coordinates. Somewhat more invariantly, we can examine the action of \( {\widetilde{\gamma }}^{\prime }\left( t\right) \) on a smooth real-valued function \( f \) defined in a neighborhood of a point \( \widetilde{\gamma }\left( {... | Yes |
Proposition 9.6 (Naturality of Integral Curves). Suppose \( M \) and \( N \) are smooth manifolds and \( F : M \rightarrow N \) is a smooth map. Then \( X \in \mathfrak{X}\left( M\right) \) and \( Y \in \mathfrak{X}\left( N\right) \) are \( F \) -related if and only if \( F \) takes integral curves of \( X \) to integr... | Proof. Suppose first that \( X \) and \( Y \) are \( F \) -related, and \( \gamma : J \rightarrow M \) is an integral curve of \( X \) . If we define \( \sigma : J \rightarrow N \) by \( \sigma = F \circ \gamma \) (see Fig. 9.3), then\n\n\[ \n{\sigma }^{\prime }\left( t\right) = {\left( F \circ \gamma \right) }^{\prime... | Yes |
Proposition 9.7. Let \( \theta : \mathbb{R} \times M \rightarrow M \) be a smooth global flow on a smooth manifold \( M \). The infinitesimal generator \( V \) of \( \theta \) is a smooth vector field on \( M \), and each curve \( {\theta }^{\left( p\right) } \) is an integral curve of \( V \). | Proof. To show that \( V \) is smooth, it suffices by Proposition 8.14 to show that \( {Vf} \) is smooth for every smooth real-valued function \( f \) defined on an open subset \( U \subseteq M \). For any such \( f \) and any \( p \in U \), just note that\n\n\[ \n{Vf}\left( p\right) = {V}_{p}f = {\theta }^{\left( p\ri... | Yes |
The flow of \( V = \partial /\partial x \) in \( {\mathbb{R}}^{2} \) is the map \( \tau : \mathbb{R} \times {\mathbb{R}}^{2} \rightarrow {\mathbb{R}}^{2} \) given by | \[ {\tau }_{t}\left( {x, y}\right) = \left( {x + t, y}\right) . \] For each nonzero \( t \in \mathbb{R},{\tau }_{t} \) translates the plane to the right \( \left( {t > 0}\right) \) or left \( \left( {t < 0}\right) \) by a distance \( \left| t\right| \) . | Yes |
Let \( M = {\mathbb{R}}^{2} \smallsetminus \{ 0\} \) with standard coordinates \( \left( {x, y}\right) \), and let \( V \) be the vector field \( \partial /\partial x \) on \( M \) . The unique integral curve of \( V \) starting at \( \left( {-1,0}\right) \in M \) is \( \gamma \left( t\right) = \left( {t - 1,0}\right) ... | This is intuitively evident because of the \ | No |
For a more subtle example, let \( M \) be all of \( {\mathbb{R}}^{2} \) and let \( W = {x}^{2}\partial /\partial x \) . You can check easily that the unique integral curve of \( W \) starting at \( \left( {1,0}\right) \) is | \[ \gamma \left( t\right) = \left( {\frac{1}{1 - t},0}\right) \] This curve also cannot be extended past \( t = 1 \), because its \( x \) -coordinate is unbounded as \( t \nearrow 1 \) . | Yes |
Proposition 9.11. If \( \theta : \mathcal{D} \rightarrow M \) is a smooth flow, then the infinitesimal generator \( V \) of \( \theta \) is a smooth vector field, and each curve \( {\theta }^{\left( p\right) } \) is an integral curve of \( V \) . | Proof. The proof is essentially identical to the analogous proof for global flows, Proposition 9.7. In the proof that \( V \) is smooth, we need only note that for any \( {p}_{0} \in M,\theta \left( {t, p}\right) \) is defined and smooth for all \( \left( {t, p}\right) \) sufficiently close to \( \left( {0,{p}_{0}}\rig... | Yes |
Proposition 9.13 (Naturality of Flows). Suppose \( M \) and \( N \) are smooth manifolds, \( F : M \rightarrow N \) is a smooth map, \( X \in \mathfrak{X}\left( M\right) \), and \( Y \in \mathfrak{X}\left( N\right) \) . Let \( \theta \) be the flow of \( X \) and \( \eta \) the flow of \( Y \) . If \( X \) and \( Y \) ... | Proof. By Proposition 9.6, for any \( p \in M \), the curve \( F \circ {\theta }^{\left( p\right) } \) is an integral curve of \( Y \) starting at \( F \circ {\theta }^{\left( p\right) }\left( 0\right) = F\left( p\right) \) . By uniqueness of integral curves, therefore, the maximal integral curve \( {\eta }^{\left( F\l... | Yes |
Lemma 9.15 (Uniform Time Lemma). Let \( V \) be a smooth vector field on a smooth manifold \( M \), and let \( \theta \) be its flow. Suppose there is a positive number \( \varepsilon \) such that for every \( p \in M \), the domain of \( {\theta }^{\left( p\right) } \) contains \( \left( {-\varepsilon ,\varepsilon }\r... | Proof. Suppose for the sake of contradiction that for some \( p \in M \), the domain \( {\mathcal{D}}^{\left( p\right) } \) of \( {\theta }^{\left( p\right) } \) is bounded above. (A similar proof works if it is bounded below.) Let \( b = \) \( \sup {\mathcal{D}}^{\left( p\right) } \), let \( {t}_{0} \) be a positive n... | Yes |
Theorem 9.16. Every compactly supported smooth vector field on a smooth manifold is complete. | Proof. Suppose \( V \) is a compactly supported vector field on a smooth manifold \( M \) , and let \( K = \operatorname{supp}V \) . For each \( p \in K \), there is a neighborhood \( {U}_{p} \) of \( p \) and a positive number \( {\varepsilon }_{p} \) that the flow of \( V \) is defined at least on \( \left( {-{\varep... | Yes |
Theorem 9.18. Every left-invariant vector field on a Lie group is complete. | Proof. Let \( G \) be a Lie group, let \( X \in \operatorname{Lie}\left( G\right) \), and let \( \theta : \mathcal{D} \rightarrow G \) denote the flow of \( X \) . There is some \( \varepsilon > 0 \) such that \( {\theta }^{\left( e\right) } \) is defined on \( \left( {-\varepsilon ,\varepsilon }\right) \) . Let \( g \... | Yes |
Lemma 9.19 (Escape Lemma). Suppose \( M \) is a smooth manifold and \( V \in \mathfrak{X}\left( M\right) \) . If \( \gamma : J \rightarrow M \) is a maximal integral curve of \( V \) whose domain \( J \) has a finite least upper bound \( b \), then for any \( {t}_{0} \in J,\gamma \left( \left\lbrack {{t}_{0}, b}\right)... | Proof. Problem 9-6. | No |
Theorem 9.20 (Flowout Theorem). Suppose \( M \) is a smooth manifold, \( S \subseteq M \) is an embedded \( k \) -dimensional submanifold, and \( V \in \mathfrak{X}\left( M\right) \) is a smooth vector field that is nowhere tangent to \( S \) . Let \( \theta : \mathcal{D} \rightarrow M \) be the flow of \( V \), let \(... | Proof. First we prove (b). Fix \( p \in S \), and let \( \sigma : {\mathcal{D}}^{\left( p\right) } \rightarrow \mathbb{R} \times S \) be the curve \( \sigma \left( t\right) = \) \( \left( {t, p}\right) \) . Then \( \Phi \circ \sigma \left( t\right) = \theta \left( {t, p}\right) \) is an integral curve of \( V \), so fo... | No |
Proposition 9.21. Let \( V \) be a smooth vector field on a smooth manifold \( M \), and let \( \theta : \mathcal{D} \rightarrow M \) be the flow generated by \( V \). If \( p \in M \) is a singular point of \( V \), then \( {\mathcal{D}}^{\left( p\right) } = \mathbb{R} \) and \( {\theta }^{\left( p\right) } \) is the ... | Proof. If \( {V}_{p} = 0 \), then the constant curve \( \gamma : \mathbb{R} \rightarrow M \) given by \( \gamma \left( t\right) \equiv p \) is clearly an integral curve of \( V \), so by uniqueness and maximality it must be equal to \( {\theta }^{\left( p\right) } \).\n\nTo verify the second statement, we prove its con... | Yes |
Theorem 9.22 (Canonical Form Near a Regular Point). Let \( V \) be a smooth vector field on a smooth manifold \( M \), and let \( p \in M \) be a regular point of \( V \) . There exist smooth coordinates \( \left( {s}^{i}\right) \) on some neighborhood of \( p \) in which \( V \) has the coordinate representation \( \p... | Proof. If no hypersurface \( S \) is given, choose any smooth coordinates \( \left( {U,\left( {x}^{i}\right) }\right) \) centered at \( p \), and let \( S \subseteq U \) be the hypersurface defined by \( {x}^{j} = 0 \), where \( j \) is chosen so that \( {V}^{j}\left( p\right) \neq 0 \) . (Recall that \( p \) is a regu... | Yes |
Let \( W = x\partial /\partial y - y\partial /\partial x \) on \( {\mathbb{R}}^{2} \). The point \( \left( {1,0}\right) \in {\mathbb{R}}^{2} \) is a regular point of \( W \), because \( {W}_{\left( 1,0\right) } = \partial /{\left. \partial y\right| }_{\left( 1,0\right) } \neq 0 \). Because \( W \) has nonzero \( y \)-c... | \[ \Psi \left( {t, s}\right) = {\theta }_{t}\left( {s,0}\right) = \left( {s\cos t, s\sin t}\right) ,\] and then solve locally for \( \left( {t, s}\right) \) in terms of \( \left( {x, y}\right) \) to obtain the following coordinate map in a neighborhood of \( \left( {1,0}\right) \): \[ \left( {t, s}\right) = {\Psi }^{-1... | Yes |
Theorem 9.24 (Boundary Flowout Theorem). Let \( M \) be a smooth manifold with nonempty boundary, and let \( N \) be a smooth vector field on \( M \) that is inward-pointing at each point of \( \partial M \) . There exist a smooth function \( \delta : \partial M \rightarrow {\mathbb{R}}^{ + } \) and a smooth embedding ... | Proof. Problem 9-11. | No |
Theorem 9.25 (Collar Neighborhood Theorem). If \( M \) is a smooth manifold with nonempty boundary, then \( \partial M \) has a collar neighborhood. | Proof. By the result of Problem 8-4, there exists a smooth vector field \( N \in \mathfrak{X}\left( M\right) \) whose restriction to \( \partial M \) is everywhere inward-pointing. Let \( \delta : M \rightarrow {\mathbb{R}}^{ + } \) and \( \Phi : {\mathcal{P}}_{\delta } \rightarrow M \) be as in Theorem 9.24, and defin... | Yes |
Theorem 9.26. Let \( M \) be a smooth manifold with nonempty boundary, and let \( \iota : \) Int \( M \hookrightarrow M \) denote inclusion. There exists a proper smooth embedding \( R : M \rightarrow \) Int \( M \) such that both \( \iota \circ R : M \rightarrow M \) and \( R \circ \iota : \operatorname{Int}M \rightar... | Proof. Theorem 9.25 shows that \( \partial M \) has a collar neighborhood \( C \) in \( M \), which is the image of a smooth embedding \( E : \lbrack 0,1) \times \partial M \rightarrow M \) satisfying \( E\left( {0, x}\right) = x \) for all \( x \in \partial M \) . To simplify notation, we will use this embedding to id... | Yes |
Theorem 9.27 (Whitney Approximation for Manifolds with Boundary). If \( M \) and \( N \) are smooth manifolds with boundary, then every continuous map from \( M \) to \( N \) is homotopic to a smooth map. | Proof. Theorem 6.26 takes care of the case in which \( \partial N = \varnothing \), so we may assume that \( \partial N \neq \varnothing \) . Let \( F : M \rightarrow N \) be a continuous map, let \( \iota : \) Int \( N \hookrightarrow N \) be inclusion, and let \( R : N \rightarrow \) Int \( N \) be the map constructe... | Yes |
Theorem 9.28. Suppose \( M \) and \( N \) are smooth manifolds with or without boundary. If \( F, G : M \rightarrow N \) are homotopic smooth maps, then they are smoothly homotopic. | Proof. Theorem 6.29 takes care of the case \( \partial N = \varnothing \), so we may assume that \( N \) has nonempty boundary. Let \( \iota : \operatorname{Int}N \hookrightarrow N \) and \( R : N \rightarrow \operatorname{Int}N \) be as in Theorem 9.26. Then \( R \circ F \) and \( R \circ G \) are homotopic smooth map... | Yes |
Lemma 9.33. Suppose \( M \) is a smooth manifold and \( D \subseteq M \) is a regular domain. If \( V \) is a smooth vector field on \( M \) that is tangent to \( \partial D \), then every integral curve of \( V \) that starts in \( D \) remains in \( D \) as long as it is defined. | Proof. Suppose \( \gamma : J \rightarrow M \) is an integral curve of \( V \) with \( \gamma \left( 0\right) \in D \) . Define \( \mathcal{T} \subseteq J \) by \( \mathcal{T} = \{ t \in J : \gamma \left( t\right) \in D\} \) . We will show that \( \mathcal{T} \) is both open and closed in \( J \) ; since \( J \) is an i... | Yes |
Theorem 9.34 (Flows on Manifolds with Boundary). The conclusions of Theorem 9.12 remain true if \( M \) is a smooth manifold with boundary and \( V \) is a smooth vector field on \( M \) that is tangent to \( \partial M \) . | Proof. Example 9.32 shows that we can consider \( M \) as a regular domain in its double \( D\left( M\right) \) . By the extension lemma for vector fields, we can extend \( V \) to a smooth vector field \( \widetilde{V} \) on \( D\left( M\right) \) . Let \( \widetilde{\theta } : \widetilde{\mathcal{D}} \rightarrow D\le... | Yes |
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