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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![b4513aee-5efb-4549-937a-edd348f47783_182_0.jpg](images/b4513aee-5efb-4549-937a-edd348f47783_182_...
Yes
Theorem 7.25 (Equivariant Rank Theorem). 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