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Proposition 19.16 (Local Structure of Integral Manifolds). Let \( D \) be an involutive distribution of rank \( k \) on a smooth manifold \( M \), and let \( \left( {U,\left( {x}^{i}\right) }\right) \) be a flat chart for \( D \). If \( H \) is any integral manifold of \( D \), then \( H \cap U \) is a union of countab... | Proof. Let \( H \) be an integral manifold of \( D \). Because the inclusion map \( \iota : H \hookrightarrow M \) is continuous, \( H \cap U = {\iota }^{-1}\left( U\right) \) is open in \( H \), and thus consists of a countable disjoint union of connected components, each of which is open in \( H \). Let \( V \) be an... | Yes |
Theorem 19.17. Every integral manifold of an involutive distribution is weakly embedded. | Proof. Let \( M \) be a smooth \( n \) -manifold, let \( H \subseteq M \) be an integral manifold of an involutive rank- \( k \) distribution \( D \) on \( M \), and suppose \( F : N \rightarrow M \) is a smooth map such that \( F\left( N\right) \subseteq H \) . Let \( p \in N \) be arbitrary, and set \( q = F\left( p\... | Yes |
Proposition 19.23. Let \( M \) be a smooth manifold and \( \Phi : M \rightarrow M \) be a diffeomorphism. Suppose \( D \) is an involutive distribution on \( M \) and \( \mathcal{F} \) is the foliation it determines. Then \( D \) is \( \Phi \) -invariant if and only if \( \mathcal{F} \) is \( \Phi \) -invariant. | Proof. Problem 19-9. | No |
Lemma 19.24. Let \( G \) be a Lie group. If \( \mathfrak{h} \) is a Lie subalgebra of \( \operatorname{Lie}\left( G\right) \), then the subset \( D = \mathop{\bigcup }\limits_{{g \in G}}{D}_{g} \subseteq {TG} \), where\n\n\[ \n{D}_{g} = \left\{ {{X}_{g} : X \in \mathfrak{h}}\right\} \subseteq {T}_{g}G \n\]\n\n(19.7)\n\... | Proof. Each \( X \in \mathfrak{h} \) is a left-invariant vector field on \( G \) . Thus, for any \( g,{g}^{\prime } \in G \) , the differential \( d\left( {L}_{{g}^{\prime }{g}^{-1}}\right) \) restricts to an isomorphism from \( {D}_{g} \) to \( {D}_{{g}^{\prime }} \) . It follows that \( {D}_{g} \) has the same dimens... | Yes |
Theorem 19.25 (Lie Subgroups Are Weakly Embedded). Every Lie subgroup is an integral manifold of an involutive distribution, and therefore is a weakly embedded submanifold. | Proof. Suppose \( G \) is a Lie group and \( H \subseteq G \) is a Lie subgroup. Theorem 8.46 shows that the Lie algebra of \( H \) is canonically isomorphic to the Lie subalgebra \( \mathfrak{h} = {\iota }_{ * }\left( {\operatorname{Lie}\left( H\right) }\right) \subseteq \operatorname{Lie}\left( G\right) \), where \( ... | Yes |
Theorem 19.26 (The Lie Subgroup Associated with a Lie Subalgebra). Suppose \( G \) is a Lie group and \( \mathfrak{g} \) is its Lie algebra. If \( \mathfrak{h} \) is any Lie subalgebra of \( \mathfrak{g} \), then there is a unique connected Lie subgroup of \( G \) whose Lie algebra is \( \mathfrak{h} \) . | Proof. Suppose \( \mathfrak{h} \) is a Lie subalgebra of \( \mathfrak{g} \) . Let \( D \subseteq {TG} \) be the involutive distribution defined by (19.7). Let \( \mathcal{H} \) denote the foliation determined by \( D \), and for any \( g \in G \), let \( {\mathcal{H}}_{g} \) denote the leaf of \( \mathcal{H} \) contain... | Yes |
Theorem 19.27. Let \( W \subseteq {\mathbb{R}}^{n} \) be an open subset and let \( m \) be an integer such that \( 1 \leq m \leq n \) . Suppose we are given an embedded codimension- \( m \) submanifold \( S \subseteq W \), a linearly independent \( m \) -tuple of smooth vector fields \( \left( {{A}_{1},\ldots ,{A}_{m}}... | Proof. Let \( D \) be the distribution on \( W \) spanned by \( {A}_{1},\ldots ,{A}_{m} \), and let \( p \in S \) be arbitrary. It follows from (19.9) that \( D \) is involutive, so by Corollary 19.13, on some neighborhood \( U \) of \( p \) there is a flat chart for \( D \) centered at \( p \) that is also a slice cha... | Yes |
Theorem 20.1 (Characterization of One-Parameter Subgroups). Let \( G \) be a Lie group. The one-parameter subgroups of \( G \) are precisely the maximal integral curves of left-invariant vector fields starting at the identity. | Proof. First suppose \( \gamma \) is the maximal integral curve of some left-invariant vector field \( X \in \operatorname{Lie}\left( G\right) \) starting at the identity. Because left-invariant vector fields are complete (Theorem 9.18), \( \gamma \) is defined on all of \( \mathbb{R} \) . Left-invariance means that \(... | Yes |
Proposition 20.3. Suppose \( G \) is a Lie group and \( H \subseteq G \) is a Lie subgroup. The one-parameter subgroups of \( H \) are precisely those one-parameter subgroups of \( G \) whose initial velocities lie in \( {T}_{e}H \) . | Proof. Let \( \gamma : \mathbb{R} \rightarrow H \) be a one-parameter subgroup. Then the composite map\n\n\[ \mathbb{R}\overset{\gamma }{ \rightarrow }H \hookrightarrow G \]\n\n is a Lie group homomorphism and thus a one-parameter subgroup of \( G \), which clearly satisfies \( {\gamma }^{\prime }\left( 0\right) \in {T... | Yes |
Proposition 20.5. Let \( G \) be a Lie group. For any \( X \in \operatorname{Lie}\left( G\right) ,\gamma \left( s\right) = \exp {sX} \) is the one-parameter subgroup of \( G \) generated by \( X \). | Proof. Let \( \gamma : \mathbb{R} \rightarrow G \) be the one-parameter subgroup generated by \( X \), which is the integral curve of \( X \) starting at \( e \) . For any fixed \( s \in \mathbb{R} \), it follows from the rescaling lemma (Lemma 9.3) that \( \widetilde{\gamma }\left( t\right) = \gamma \left( {st}\right)... | Yes |
The results of the preceding section show that the exponential map of \( \mathrm{{GL}}\left( {n,\mathbb{R}}\right) \) (or any Lie subgroup of it) is given by \( \exp A = {e}^{A} \) | This, obviously, is the reason for the term exponential map. | No |
If \( V \) is a finite-dimensional real vector space, a choice of basis for \( V \) yields isomorphisms \( \mathrm{{GL}}\left( V\right) \cong \mathrm{{GL}}\left( {n,\mathbb{R}}\right) \) and \( \mathrm{{gI}}\left( V\right) \cong \mathrm{{gI}}\left( {n,\mathbb{R}}\right) \) . The analysis of the \( \mathrm{{GL}}\left( {... | \[ \exp A = \mathop{\sum }\limits_{{k = 0}}^{\infty }\frac{1}{k!}{A}^{k} \] where we consider \( A \in \mathfrak{{gl}}\left( V\right) \) as a linear map from \( V \) to itself, and \( {A}^{k} = A \circ \cdots \circ A \) is the \( k \) -fold composition of \( A \) with itself. | Yes |
Proposition 20.8 (Properties of the Exponential Map). Let \( G \) be a Lie group and let \( \mathfrak{g} \) be its Lie algebra.\n\n(a) The exponential map is a smooth map from \( \mathfrak{g} \) to \( G \) . | Proof. In this proof, for any \( X \in \mathfrak{g} \) we let \( {\theta }_{\left( X\right) } \) denote the flow of \( X \) . To prove (a), we need to show that the expression \( {\theta }_{\left( X\right) }^{\left( e\right) }\left( 1\right) \) depends smoothly on \( X \), which amounts to showing that the flow varies ... | Yes |
Let \( G \) be a Lie group, and let \( H \subseteq G \) be a Lie subgroup. With \( \operatorname{Lie}\left( H\right) \) considered as a subalgebra of \( \operatorname{Lie}\left( G\right) \) in the usual way, the exponential map of \( H \) is the restriction to \( \operatorname{Lie}\left( H\right) \) of the exponential ... | The fact that the exponential map of \( H \) is the restriction of that of \( G \) is an immediate consequence of Proposition 20.3. To prove the second assertion, by the way we have identified \( \operatorname{Lie}\left( H\right) \) as a subalgebra of \( \operatorname{Lie}\left( G\right) \), we need to establish the fo... | Yes |
Proposition 20.10. Let \( G \) be a Lie group and let \( \mathfrak{g} \) be its Lie algebra. For any \( X, Y \in \mathfrak{g} \), there is a smooth function \( Z : \left( {-\varepsilon ,\varepsilon }\right) \rightarrow \mathfrak{g} \) for some \( \varepsilon > 0 \) such that the following identity holds for all \( t \i... | Proof. Since the exponential map is a diffeomorphism on some neighborhood of the origin in \( \mathfrak{g} \), there is some \( \varepsilon > 0 \) such that the map \( \varphi : \left( {-\varepsilon ,\varepsilon }\right) \rightarrow \mathfrak{g} \) defined by\n\n\[ \varphi \left( t\right) = {\exp }^{-1}\left( {\exp {tX... | Yes |
Corollary 20.11. Under the hypotheses of the preceding proposition, \[ \mathop{\lim }\limits_{{n \rightarrow \infty }}{\left( \left( \exp \frac{t}{n}X\right) \left( \exp \frac{t}{n}Y\right) \right) }^{n} = \exp t\left( {X + Y}\right) . \] | Proof. Formula (20.4) implies that for any \( t \in \mathbb{R} \) and any sufficiently large \( n \in \mathbb{Z} \) , \[ \left( {\exp \frac{t}{n}X}\right) \left( {\exp \frac{t}{n}Y}\right) = \exp \left( {\frac{t}{n}\left( {X + Y}\right) + \frac{{t}^{2}}{{n}^{2}}Z\left( \frac{t}{n}\right) }\right) , \] and then Proposit... | Yes |
Lemma 20.14. Suppose \( G \) is a Lie group and \( \theta \) is a smooth right action of \( G \) on a smooth manifold \( M \) . For any \( X \in \operatorname{Lie}\left( G\right) \) and \( p \in M \), the vector fields \( X \) and \( \widehat{\theta }\left( X\right) \) are \( {\theta }^{\left( p\right) } \) -related. | Proof. Let \( X \in \operatorname{Lie}\left( G\right) \) and \( p \in M \) be arbitrary, and write \( \widehat{X} = \widehat{\theta }\left( X\right) \) . Note that the group law \( p \cdot g{g}^{\prime } = \left( {p \cdot g}\right) \cdot {g}^{\prime } \) translates to\n\n\[{\theta }^{\left( p\right) } \circ {L}_{g}\lef... | Yes |
Theorem 20.15. Suppose \( G \) is a Lie group and \( \theta \) is a smooth right action of \( G \) on a smooth manifold \( M \) . Then the map \( \widehat{\theta } : \operatorname{Lie}\left( G\right) \rightarrow \mathfrak{X}\left( M\right) \) defined above is a Lie algebra homomorphism. | Proof. For each \( p \in M \), it follows from (20.8) that \( {\widehat{X}}_{p} \) depends linearly on \( X \), so \( \widehat{\theta } \) is a linear map. Given \( p \in M \), Lemma 20.14 together with the naturality of Lie brackets implies that \( \left\lbrack {X, Y}\right\rbrack \) is \( {\theta }^{\left( p\right) }... | Yes |
Theorem 20.18. Suppose \( G \) is a Lie group and \( M \) is a smooth manifold.\n\n(a) If \( \theta \) is a smooth left action of \( G \) on \( M \), the map \( \widehat{\theta } : \operatorname{Lie}\left( G\right) \rightarrow \mathfrak{X}\left( M\right) \) defined by (20.11) is an antihomomorphism (a linear map satisf... | Proof. Problem 20-15. | No |
Theorem 20.19. Suppose \( G \) and \( H \) are Lie groups with \( G \) simply connected, and let \( \mathfrak{g} \) and \( \mathfrak{h} \) be their Lie algebras. For any Lie algebra homomorphism \( \varphi : \mathfrak{g} \rightarrow \mathfrak{h} \) , there is a unique Lie group homomorphism \( \Phi : G \rightarrow H \)... | Proof. The Lie algebra homomorphism \( \varphi : \mathfrak{g} \rightarrow \mathfrak{h} \subseteq \mathfrak{X}\left( H\right) \) is, in particular, a complete g-action on \( H \) (since every left-invariant vector field is complete). Thus, by Theorem 20.16, there is a unique smooth right \( G \) -action \( \theta : H \t... | Yes |
Corollary 20.20. If \( G \) and \( H \) are simply connected Lie groups with isomorphic Lie algebras, then \( G \) and \( H \) are isomorphic. | Proof. Let \( \mathfrak{g},\mathfrak{h} \) be the Lie algebras of \( G \) and \( H \), respectively, and let \( \varphi : \mathfrak{g} \rightarrow \mathfrak{h} \) be a Lie algebra isomorphism between them. By the preceding theorem, there are Lie group homomorphisms \( \Phi : G \rightarrow H \) and \( \Psi : H \rightarr... | Yes |
Theorem 20.21 (The Lie Correspondence). There is a one-to-one correspondence between isomorphism classes of finite-dimensional Lie algebras and isomorphism classes of simply connected Lie groups, given by associating each simply connected Lie group with its Lie algebra. | Proof. We need to show that the functor that sends a simply connected Lie group to its Lie algebra is both surjective and injective up to isomorphism. Injectivity is precisely the content of Corollary 20.20.\n\nTo prove surjectivity, suppose \( g \) is any finite-dimensional Lie algebra. By Corollary 8.50 to Ado’s theo... | Yes |
Lemma 20.23. Let \( G \) be a connected Lie group, and let \( H \subseteq G \) be a connected Lie subgroup. Let \( \mathfrak{g} \) and \( \mathfrak{h} \) denote the Lie algebras of \( G \) and \( H \), respectively. Then \( H \) is normal in \( G \) if and only if \[ \left( {\exp X}\right) \left( {\exp Y}\right) \left(... | Proof. Note that \( \exp \left( {-X}\right) = {\left( \exp X\right) }^{-1} \) . Thus if \( H \) is normal, then (20.13) holds by definition. Conversely, suppose (20.13) holds, and choose open subsets \( V \subseteq \mathfrak{g} \) containing 0 and \( U \subseteq G \) containing the identity such that exp: \( V \rightar... | Yes |
Proposition 20.24 (The Adjoint Representation). If \( G \) is a Lie group with Lie algebra \( \mathfrak{g} \), the map \( \operatorname{Ad} : G \rightarrow \mathrm{{GL}}\left( \mathfrak{g}\right) \) is a Lie group representation, called the adjoint representation of \( G \) . | Proof. Because \( {C}_{{g}_{1}{g}_{2}} = {C}_{{g}_{1}} \circ {C}_{{g}_{2}} \) for any \( {g}_{1},{g}_{2} \in G \), it follows immediately that \( \operatorname{Ad}\left( {{g}_{1}{g}_{2}}\right) = \operatorname{Ad}\left( {g}_{1}\right) \circ \operatorname{Ad}\left( {g}_{2}\right) \), and \( \operatorname{Ad}\left( g\rig... | Yes |
Theorem 20.27. Let \( G \) be a Lie group, let \( \mathfrak{g} \) be its Lie algebra, and let Ad: \( G \rightarrow \) \( \mathrm{{GL}}\left( \mathrm{g}\right) \) be the adjoint representation of \( G \). The induced Lie algebra representation \( {\mathrm{{Ad}}}_{ * } : \mathfrak{g} \rightarrow \mathfrak{{gl}}\left( \ma... | Proof. Let \( X \in \mathfrak{g} \) be arbitrary. Then \( {\operatorname{Ad}}_{ * }X \) is determined by its value at the identity, which we can interpret as an element of \( \mathfrak{{gl}}\left( \mathfrak{g}\right) \), the set of all linear maps from \( \mathfrak{g} \) to itself. Because \( t \mapsto \exp {tX} \) is ... | Yes |
Theorem 20.28 (Ideals and Normal Subgroups). Let \( G \) be a connected Lie group, and suppose \( H \subseteq G \) is a connected Lie subgroup. Then \( H \) is a normal subgroup of \( G \) if and only if \( \operatorname{Lie}\left( H\right) \) is an ideal in \( \operatorname{Lie}\left( G\right) \) . | Proof. Write \( \mathfrak{g} = \operatorname{Lie}\left( G\right) \) and \( \mathfrak{h} = \operatorname{Lie}\left( H\right) \), considering \( \mathfrak{h} \) as a Lie subalgebra of \( \mathfrak{g} \) . For any \( g \in G \), the commutative diagram (20.3) applied to the Lie group homomorphism \( {C}_{g}\left( h\right)... | Yes |
Lemma 21.1. For any continuous action of a topological group \( G \) on a topological space \( M \), the quotient map \( \pi : M \rightarrow M/G \) is an open map. | Proof. For any \( g \in G \) and any subset \( U \subseteq M \), we define a set \( g \cdot U \subseteq M \) by\n\n\[ g \cdot U = \{ g \cdot x : x \in U\} . \]\n\nIf \( U \subseteq M \) is open, then \( {\pi }^{-1}\left( {\pi \left( U\right) }\right) \) is equal to the union of all sets of the form \( g \cdot U \) as \... | Yes |
Proposition 21.4. If a Lie group acts continuously and properly on a manifold, then the orbit space is Hausdorff. | Proof. Suppose \( G \) is a Lie group acting continuously and properly on a manifold \( M \) . Let \( \Theta : G \times M \rightarrow M \times M \) be the proper map \( \Theta \left( {g, p}\right) = \left( {g \cdot p, p}\right) \), and let \( \pi : M \rightarrow M/G \) be the quotient map. Define the orbit relation \( ... | Yes |
Proposition 21.5 (Characterizations of Proper Actions). Let \( M \) be a manifold, and let \( G \) be a Lie group acting continuously on \( M \) . The following are equivalent.\n\n(a) The action is proper.\n\n(b) If \( \left( {p}_{i}\right) \) is a sequence in \( M \) and \( \left( {g}_{i}\right) \) is a sequence in \(... | Proof. Throughout this proof, let \( \Theta : G \times M \rightarrow M \times M \) denote the map \( \Theta \left( {g, p}\right) = \) \( \left( {g \cdot p, p}\right) \) ; thus, the action is proper if and only if \( \Theta \) is a proper map. We will prove (a) \( \Rightarrow \) (b) \( \Rightarrow \) (c) \( \Rightarrow ... | Yes |
Corollary 21.6. Every continuous action by a compact Lie group on a manifold is proper. | Proof. If \( \left( {p}_{i}\right) \) and \( \left( {g}_{i}\right) \) are sequences satisfying the hypotheses of Proposition 21.5(b), then a subsequence of \( \left( {g}_{i}\right) \) converges, for the simple reason that every sequence in \( G \) has a convergent subsequence. | No |
Proposition 21.7 (Orbits of Proper Actions). Suppose \( \theta \) is a proper smooth action of a Lie group \( G \) on a smooth manifold \( M \). For any point \( p \in M \), the orbit map \( {\theta }^{\left( p\right) } : G \rightarrow M \) is a proper map, and thus the orbit \( G \cdot p = {\theta }^{\left( p\right) }... | Proof. If \( K \subseteq M \) is compact, then \( {\left( {\theta }^{\left( p\right) }\right) }^{-1}\left( K\right) \) is closed in \( G \) by continuity, and since it is contained in \( {G}_{K\cup \{ p\} } \), it is compact by Proposition 21.5. Therefore, \( {\theta }^{\left( p\right) } \) is a proper map, which impli... | Yes |
Corollary 21.8. If a Lie group \( G \) acts properly on a manifold \( M \), then each orbit is a closed subset of \( M \), and each isotropy group is compact. | Proof. The first statement follows immediately from Proposition 21.7, and the second from Proposition 21.5, using the fact that the isotropy group of a point \( p \in M \) is the set \( {G}_{K} \) for \( K = \{ p\} \). | Yes |
Lemma 21.11. Suppose a discrete Lie group \( \Gamma \) acts continuously and freely on a manifold \( E \) . The action is proper if and only if the following conditions both hold:\n\n(i) Every point \( p \in E \) has a neighborhood \( U \) such that for each \( g \in \Gamma ,\left( {g \cdot U}\right) \cap \) \( U = \va... | Proof. First, suppose that the action is free and proper, and let \( \pi : E \rightarrow E/\Gamma \) denote the quotient map. By Proposition 21.4, \( E/\Gamma \) is Hausdorff. If \( p,{p}^{\prime } \in E \) are not in the same orbit, we can choose disjoint neighborhoods \( W \) of \( \pi \left( p\right) \) and \( {W}^{... | Yes |
Proposition 21.12. Let \( M \) be a smooth manifold, and let \( \pi : E \rightarrow M \) be a smooth covering map. With the discrete topology, the automorphism group \( {\operatorname{Aut}}_{\pi }\left( E\right) \) acts smoothly, freely, and properly on \( E \) . | Proof. We already showed in Proposition 7.23 that the action is smooth and free. To show it is proper, we will show that it satisfies conditions (i) and (ii) of Lemma 21.11. First, if \( p \in E \) is arbitrary, choose \( W \subseteq M \) to be an evenly covered neighborhood of \( \pi \left( p\right) \) . If \( U \) is... | Yes |
Theorem 21.13. Suppose \( E \) is a connected smooth manifold and \( \Gamma \) is a discrete Lie group acting smoothly, freely, and properly on \( E \) . Then the orbit space \( E/\Gamma \) is a topological manifold and has a unique smooth structure such that \( \pi : E \rightarrow E/\Gamma \) is a smooth normal coveri... | Proof. It follows from the quotient manifold theorem that \( E/\Gamma \) has a unique smooth manifold structure such that \( \pi \) is a smooth submersion. Because a smooth covering map is in particular a smooth submersion, any other smooth manifold structure on \( E \) making \( \pi \) into a smooth covering map must ... | Yes |
Theorem 21.17 (Homogeneous Space Construction Theorem). Let \( G \) be a Lie group and let \( H \) be a closed subgroup of \( G \). The left coset space \( G/H \) is a topological manifold of dimension equal to \( \dim G - \dim H \), and has a unique smooth structure such that the quotient map \( \pi : G \rightarrow G/... | Proof. If we let \( H \) act on \( G \) by right translation, then \( {g}_{1},{g}_{2} \in G \) are in the same \( H \)-orbit if and only if \( {g}_{1}h = {g}_{2} \) for some \( h \in H \), which is the same as saying that \( {g}_{1} \) and \( {g}_{2} \) are in the same coset of \( H \). In other words, the orbit space ... | Yes |
Theorem 21.18 (Homogeneous Space Characterization Theorem). Let \( G \) be a Lie group, let \( M \) be a homogeneous \( G \)-space, and let \( p \) be any point of \( M \). The isotropy group \( {G}_{p} \) is a closed subgroup of \( G \), and the map \( F : G/{G}_{p} \rightarrow M \) defined by \( F\left( {g{G}_{p}}\ri... | Proof. For simplicity, let us write \( H = {G}_{p} \). Note that \( H \) is closed by continuity, because \( H = {\left( {\theta }^{\left( p\right) }\right) }^{-1}\left( p\right) \), where \( {\theta }^{\left( p\right) } : G \rightarrow M \) is the orbit map.\n\nTo see that \( F \) is well defined, assume that \( {g}_{... | Yes |
Theorem 21.20. Suppose \( X \) is a set, and we are given a transitive action of a Lie group \( G \) on \( X \) such that for some point \( p \in X \), the isotropy group \( {G}_{p} \) is closed in \( G \) . Then \( X \) has a unique smooth manifold structure with respect to which the given action is smooth. With this ... | Proof. Theorem 21.17 shows that \( G/{G}_{p} \) is a smooth manifold of dimension equal to \( \dim G - \dim {G}_{p} \) . The map \( F : G/{G}_{p} \rightarrow X \) defined by \( F\left( {g{G}_{p}}\right) = g \cdot p \) is an equivariant bijection by exactly the same argument as we used in the proof of the characterizati... | Yes |
Let \( {\mathrm{G}}_{k}\left( {\mathbb{R}}^{n}\right) \) denote the Grassmannian of \( k \) - dimensional subspaces of \( {\mathbb{R}}^{n} \). The general linear group \( \mathrm{{GL}}\left( {n,\mathbb{R}}\right) \) acts transitively on \( {\mathrm{G}}_{k}\left( {\mathbb{R}}^{n}\right) \): given two subspaces \( A \) a... | \[ H = \left\{ {\left( \begin{array}{ll} A & B \\ 0 & D \end{array}\right) : A \in \mathrm{{GL}}\left( {k,\mathbb{R}}\right), D \in \mathrm{{GL}}\left( {n - k,\mathbb{R}}\right), B \in \mathrm{M}\left( {k \times \left( {n - k}\right) ,\mathbb{R}}\right) }\right\} ,\] which is easily seen to be closed in \( \mathrm{{GL}... | Yes |
Theorem 21.26 (Quotient Theorem for Lie Groups). Suppose \( G \) is a Lie group and \( K \subseteq G \) is a closed normal subgroup. Then \( G/K \) is a Lie group, and the quotient map \( \pi : G \rightarrow G/K \) is a surjective Lie group homomorphism whose kernel is \( K \) . | Proof. By the homogeneous space construction theorem, \( G/K \) is a smooth manifold and \( \pi \) is a smooth submersion; and by Theorem 21.23, \( G/K \) is a group and \( \pi : G \rightarrow G/K \) is a surjective homomorphism with kernel \( K \) . Thus, the only thing that needs to be verified is that multiplication... | Yes |
Theorem 21.27 (First Isomorphism Theorem for Lie Groups). If \( F : G \rightarrow H \) is a Lie group homomorphism, then the kernel of \( F \) is a closed normal Lie subgroup of \( G \), the image of \( F \) has a unique smooth manifold structure making it into a Lie subgroup of \( H \), and \( F \) descends to a Lie g... | Proof. By Theorem 21.24, \( \operatorname{Ker}F \) is a normal subgroup, \( \operatorname{Im}F \) is a subgroup, and \( F \) descends to a group isomorphism \( \widetilde{F} : G/\operatorname{Ker}F \rightarrow \operatorname{Im}F \) . By continuity, \( \operatorname{Ker}F \) is closed in \( G \), so it follows from Theo... | Yes |
Proposition 21.28. Every discrete subgroup of a Lie group is a closed Lie subgroup of dimension zero. | Proof. Let \( G \) be a Lie group and \( \Gamma \subseteq G \) be a discrete subgroup. With the subspace topology, \( \Gamma \) is a countable discrete space and thus a zero-dimensional Lie group. By the closed subgroup theorem, \( \Gamma \) is a closed Lie subgroup of \( G \) if and only if it is a closed subset. A di... | Yes |
Theorem 21.29 (Quotients of Lie Groups by Discrete Subgroups). If \( G \) is a connected Lie group and \( \Gamma \subseteq G \) is a discrete subgroup, then \( G/\Gamma \) is a smooth manifold and the quotient map \( \pi : G \rightarrow G/\Gamma \) is a smooth normal covering map. | Proof. The proof of Theorem 21.17 showed that \( \Gamma \) acts smoothly, freely, and properly on \( G \) on the right, and its quotient—which is the coset space \( G/\Gamma \) —is a smooth manifold. The theorem is then an immediate consequence of Theorem 21.13. | Yes |
Theorem 21.31 (Homomorphisms with Discrete Kernels). Let \( G \) and \( H \) be connected Lie groups. For any Lie group homomorphism \( F : G \rightarrow H \), the following are equivalent:\n\n(a) \( F \) is surjective and has discrete kernel.\n\n(b) \( F \) is a smooth covering map.\n\n(c) \( F \) is a local diffeomor... | Proof. We will show that (a) \( \Rightarrow \) (b) \( \Rightarrow \) (c) \( \Rightarrow \) (a) and (c) \( \Leftrightarrow \) (d). First, assume that \( F \) is surjective with discrete kernel \( \Gamma \subseteq G \) . Then Theorem 21.29 implies that the quotient map \( \pi : G \rightarrow G/\Gamma \) is a smooth cover... | Yes |
Theorem 21.32. Let \( \mathfrak{g} \) be a finite-dimensional Lie algebra. The connected Lie groups whose Lie algebras are isomorphic to \( \mathfrak{g} \) are (up to isomorphism) precisely those of the form \( G/\Gamma \), where \( G \) is the simply connected Lie group with Lie algebra \( \mathfrak{g} \), and \( \Gam... | Proof. Given \( \mathfrak{g} \), by Theorem 20.21 there exists a simply connected Lie group \( G \) with Lie algebra isomorphic to \( \mathfrak{g} \) . Suppose \( H \) is any other connected Lie group whose Lie algebra is isomorphic to \( \mathfrak{g} \), and let \( \varphi : \operatorname{Lie}\left( G\right) \rightarr... | Yes |
Proposition 21.33. Suppose a topological group \( G \) acts continuously, freely, and properly on a topological space \( M \) . If \( G \) and \( M/G \) are connected, then \( M \) is connected. | Proof. Assume for the sake of contradiction that \( G \) and \( M/G \) are connected but \( M \) is not. Then there are disjoint nonempty open subsets \( U, V \subseteq M \) whose union is \( M \) . Each \( G \) -orbit in \( M \) is the image of \( G \) under an orbit map \( {\theta }^{\left( p\right) } : G \rightarrow... | Yes |
For each \( n \geq 1 \), the Lie groups \( \mathrm{{SO}}\left( n\right) ,\mathrm{U}\left( n\right) \), and \( \mathrm{{SU}}\left( n\right) \) are connected. The group \( \mathrm{O}\left( n\right) \) has exactly two components, one of which is \( \mathrm{{SO}}\left( n\right) \) . | First, we prove by induction on \( n \) that \( \mathrm{{SO}}\left( n\right) \) is connected. For \( n = 1 \) this is obvious, because \( \mathrm{{SO}}\left( 1\right) \) is the trivial group. Now suppose we have shown that \( \operatorname{SO}\left( {n - 1}\right) \) is connected for some \( n \geq 2 \) . Because the h... | Yes |
Suppose \( v = {a}^{i}{A}_{i} + {b}^{i}{B}_{i} \in V \) satisfies \( \omega \left( {v, w}\right) = 0 \) for all \( w \in V \). Then \( 0 = \) \( \omega \left( {v,{B}_{i}}\right) = {a}^{i} \) and \( 0 = \omega \left( {v,{A}_{i}}\right) = - {b}^{i} \), which implies that \( v = 0 \). Thus \( \omega \) is nondegenerate, a... | Then \( 0 = \) \( \omega \left( {v,{B}_{i}}\right) = {a}^{i} \) and \( 0 = \omega \left( {v,{A}_{i}}\right) = - {b}^{i} \), which implies that \( v = 0 \). Thus \( \omega \) is nondegenerate, and so is a symplectic tensor. | No |
Proposition 22.5. Let \( \left( {V,\omega }\right) \) be a symplectic vector space, and let \( S \subseteq V \) be a linear subspace.\n\n(a) \( S \) is symplectic if and only if \( {S}^{ \bot } \) is symplectic.\n\n(b) \( S \) is symplectic if and only if \( {\left. \omega \right| }_{S} \) is nondegenerate.\n\n(c) \( S... | Proof. Problem 22-1. | No |
Proposition 22.7 (Canonical Form for a Symplectic Tensor). Let \( \omega \) be a symplectic tensor on an m-dimensional vector space \( V \) . Then \( V \) has even dimension \( m = {2n} \), and there exists a basis for \( V \) in which \( \omega \) has the form (22.1). | Proof. The tensor \( \omega \) has the form (22.1) with respect to a basis \( \left( {{A}_{1},{B}_{1},\ldots ,{A}_{n},{B}_{n}}\right) \) if or only if its action on basis vectors is given by (22.2). We prove the theorem by induction on \( m = \dim V \) by showing that there is a basis with this property.\n\nFor \( m = ... | Yes |
Proposition 22.8. Suppose \( V \) is a 2n-dimensional vector space and \( \omega \in {\Lambda }^{2}\left( {V}^{ * }\right) \) . Then \( \omega \) is a symplectic tensor if and only if \( {\omega }^{n} \neq 0 \) . | Proof. Suppose first that \( \omega \) is a symplectic tensor. Let \( \left( {{A}_{i},{B}_{i}}\right) \) be a symplectic basis for \( V \), and write \( \omega = \mathop{\sum }\limits_{i}{\alpha }^{i} \land {\beta }^{i} \) in terms of the dual coframe. Then \( {\omega }^{n} = \) \( \mathop{\sum }\limits_{I}{\alpha }^{{... | Yes |
Proposition 22.11. Let \( M \) be a smooth manifold. The tautological 1-form \( \tau \) is smooth, and \( \omega = - {d\tau } \) is a symplectic form on the total space of \( {T}^{ * }M \) . | Proof. Let \( \left( {x}^{i}\right) \) be smooth coordinates on \( M \), and let \( \left( {{x}^{i},{\xi }_{i}}\right) \) denote the corresponding natural coordinates on \( {T}^{ * }M \) as defined on p. 277. Recall that the coordinates of \( \left( {q,\varphi }\right) \in {T}^{ * }M \) are defined to be \( \left( {{x}... | Yes |
Proposition 22.12. Let \( M \) be a smooth manifold, and let \( \sigma \) be a smooth 1 -form on \( M \) . Thought of as a smooth map from \( M \) to \( {T}^{ * }M,\sigma \) is a smooth embedding, and \( \sigma \) is closed if and only if its image \( \sigma \left( M\right) \) is a Lagrangian submanifold of \( {T}^{ * ... | Proof. Throughout this proof we need to remember that \( \sigma : M \rightarrow {T}^{ * }M \) is playing two roles: on the one hand, it is a 1 -form on \( M \), and on the other hand, it is a smooth map between manifolds. Since they are literally the same map, we do not use different notations to distinguish between th... | Yes |
Proposition 22.15. Let \( M \) be a smooth manifold and \( J \subseteq \mathbb{R} \) be an open interval. Suppose \( V : J \times M \) is a smooth time-dependent vector field on \( M,\psi : \mathcal{E} \rightarrow M \) is its time-dependent flow, and \( A : J \times M \rightarrow {T}^{k}{T}^{ * }M \) is a smooth time-d... | Proof. For sufficiently small \( \varepsilon > 0 \), consider the smooth map \( F : \left( {{t}_{1} - \varepsilon ,{t}_{1} + \varepsilon }\right) \times \left( {{t}_{1} - \varepsilon ,{t}_{1} + \varepsilon }\right) \rightarrow {T}^{k}\left( {{T}_{p}^{ * }M}\right) \) defined by \[ F\left( {u, v}\right) = {\left( {\thet... | Yes |
Proposition 22.16 (Properties of Hamiltonian Vector Fields). Let \( \left( {M,\omega }\right) \) be a symplectic manifold and let \( f \in {C}^{\infty }\left( M\right) \) .\n\n(a) \( f \) is constant along each integral curve of \( {X}_{f} \) .\n\n(b) At each regular point of \( f \), the Hamiltonian vector field \( {X... | Proof. Both assertions follow from the fact that\n\n\[ \n{X}_{f}f = {df}\left( {X}_{f}\right) = \omega \left( {{X}_{f},{X}_{f}}\right) = 0 \n\]\n\nbecause \( \omega \) is alternating. | Yes |
Proposition 22.17 (Hamiltonian and Symplectic Vector Fields). Let \( \left( {M,\omega }\right) \) be a symplectic manifold. A smooth vector field on \( M \) is symplectic if and only if it is locally Hamiltonian. Every locally Hamiltonian vector field on \( M \) is globally Hamiltonian if and only if \( {H}_{\mathrm{{d... | Proof. By Theorem 12.37, a smooth vector field \( X \) is symplectic if and only if \( {\mathcal{L}}_{X}\omega = 0 \) . Using Cartan’s magic formula, we compute\n\n\[ \n{\mathcal{L}}_{X}\omega = d\left( {X\lrcorner \omega }\right) + X\lrcorner \left( {d\omega }\right) = d\left( {X\lrcorner \omega }\right) .\n\]\n\n(22.... | Yes |
Proposition 22.19 (Properties of the Poisson Bracket). Suppose \( \left( {M,\omega }\right) \) is a symplectic manifold, and \( f, g, h \in {C}^{\infty }\left( M\right) \) .\n\n(a) BILINEARITY: \( \{ f, g\} \) is linear over \( \mathbb{R} \) in \( f \) and in \( g \) .\n\n(b) ANTISYMMETRY: \( \{ f, g\} = - \{ g, f\} \)... | Proof. Parts (a) and (b) are obvious from the characterization \( \{ f, g\} = \omega \left( {{X}_{f},{X}_{g}}\right) \) together with the fact that \( {X}_{f} = {\widehat{\omega }}^{-1}\left( {df}\right) \) depends linearly on \( f \) . Because of the nondegeneracy of \( \omega \), to prove (d), it suffices to show tha... | Yes |
Proposition 22.21. Let \( \left( {M,\omega, H}\right) \) be a Hamiltonian system.\n\n(a) A function \( f \in {C}^{\infty }\left( M\right) \) is a conserved quantity if and only if \( \{ f, H\} = 0 \) . | Proof. Problem 22-18. | No |
Theorem 22.22 (Noether’s Theorem). Let \( \left( {M,\omega, H}\right) \) be a Hamiltonian system. If \( f \) is any conserved quantity, then its Hamiltonian vector field is an infinitesimal symmetry. Conversely, if \( {H}_{\mathrm{{dR}}}^{1}\left( M\right) = 0 \), then each infinitesimal symmetry is the Hamiltonian vec... | Proof. Suppose \( f \) is a conserved quantity. Proposition 22.21 shows that \( \{ f, H\} = \) 0 . This in turn implies that \( {X}_{f}H = \{ H, f\} = 0 \), so \( H \) is constant along the flow of \( {X}_{f} \) . Since \( \omega \) is invariant along the flow of any Hamiltonian vector field by Proposition 22.17, this ... | Yes |
Theorem 22.23 (Hamiltonian Flowout Theorem). Suppose \( \left( {M,\omega }\right) \) is a symplectic manifold, \( H \in {C}^{\infty }\left( M\right) ,\Gamma \) is an embedded isotropic submanifold of \( M \) that is contained in a single level set of \( H \), and the Hamiltonian vector field \( {X}_{H} \) is nowhere ta... | Proof. Let \( \theta \) be the flow of \( {X}_{H} \) . Recall from Theorem 9.20 that the flowout is parametrized by the restriction of \( \theta \) to a neighborhood \( {\mathcal{O}}_{\delta } \) of \( \{ 0\} \times \Gamma \) in \( \mathbb{R} \times \Gamma \) . First consider a point \( p \in \Gamma \subseteq S \) . If... | Yes |
Theorem 22.28 (The Reeb Field). Let \( \left( {M, H}\right) \) be a contact manifold, and suppose \( \theta \) is a contact form for \( H \) . There is a unique vector field \( T \in \mathfrak{X}\left( M\right) \), called the Reeb field of \( \mathbf{\theta } \), that satisfies the following two conditions:\n\n\[ T\lrc... | Proof. Define a smooth bundle homomorphism \( \Phi : {TM} \rightarrow {T}^{ * }M \) by \( \Phi \left( X\right) = \) \( X\lrcorner {d\theta } \), and for each \( p \in M \), let \( {\Phi }_{p} \) denote the linear map \( {\left. \Phi \right| }_{{T}_{p}M} : {T}_{p}M \rightarrow {T}_{p}^{ * }M \) . The fact that \( d{\the... | Yes |
Theorem 22.31 (Contact Darboux Theorem). Suppose \( \theta \) is a contact form on a \( \left( {{2n} + 1}\right) \) -dimensional manifold \( M \) . For each \( p \in M \), there are smooth coordinates \( \left( {{x}^{1},\ldots ,{x}^{n},{y}^{1},\ldots ,{y}^{n}, z}\right) \) centered at \( p \) in which \( \theta \) has ... | Proof. Let \( p \in M \) be arbitrary. Let \( \left( {U,\left( {u}^{i}\right) }\right) \) be a smooth coordinate cube centered at \( p \) in which the Reeb field of \( \theta \) has the form \( T = \partial /\partial {u}^{1} \), and let \( Y \subseteq U \) be the slice defined by \( {u}^{1} = 0 \) . Because \( T \) is ... | Yes |
Proposition 22.32. Suppose \( \left( {M, H}\right) \) is a contact manifold and \( \theta \) is a contact form for \( H \) . For any function \( f \in {C}^{\infty }\left( M\right) \), there is a unique vector field \( {X}_{f} \) , called the contact Hamiltonian vector field of \( f \), that satisfies \( \theta \left( {... | Proof. Suppose \( f \in {C}^{\infty }\left( M\right) \) . Because the restriction of \( {d\theta } \) to \( H \) is nondegenerate, there is a unique smooth vector field \( B \in \Gamma \left( H\right) \) such that \( B\lrcorner {d\theta }{\left| {}_{H} = df\right| }_{H} \) . If we set \( {X}_{f} = {fT} - B \), where \(... | Yes |
Theorem 22.33 (Characterization of Contact Vector Fields). If (M, H) is a contact manifold and \( \theta \) is a contact form for \( H \), then a smooth vector field on \( M \) is a contact vector field if and only if it is a contact Hamiltonian vector field. | Proof. Problem 22-21. | No |
Theorem 22.34 (Contact Flowout Theorem). Suppose \( \left( {M, H}\right) \) is a contact manifold, \( F \in {C}^{\infty }\left( M\right) ,\Gamma \) is an embedded isotropic submanifold of \( M \) that is contained in the zero set of \( F \), and the contact Hamiltonian vector field \( {X}_{F} \) is nowhere tangent to \... | Proof. Problem 22-23. | No |
Theorem 22.39 (The General First-Order Cauchy Problem). Suppose \( M \) is a smooth manifold, \( W \subseteq {J}^{1}M \) is an open subset, \( F : W \rightarrow \mathbb{R} \) is a smooth function, \( S \subseteq M \) is an embedded hypersurface, and \( \varphi : S \rightarrow \mathbb{R} \) is a smooth function. If the ... | Proof. Problem 22-26. | No |
Proposition 2.19 (Local Criterion for Continuity). A map \( f : X \rightarrow Y \) between topological spaces is continuous if and only if each point of \( X \) has a neighborhood on which (the restriction of) \( f \) is continuous. | Proof. If \( f \) is continuous, we may simply take each neighborhood to be \( X \) itself. Conversely, suppose \( f \) is continuous in a neighborhood of each point, and let \( U \subseteq Y \) be any open subset; we have to show that \( {f}^{-1}\left( U\right) \) is open. Any point \( x \in {f}^{-1}\left( U\right) \)... | Yes |
Let \( {\mathbb{B}}^{n} \subseteq {\mathbb{R}}^{n} \) be the unit ball, and define a map \( F : {\mathbb{B}}^{n} \rightarrow {\mathbb{R}}^{n} \) by\n\n\[ F\left( x\right) = \frac{x}{1 - \left| x\right| }.\] | Direct computation shows that the map \( G : {\mathbb{R}}^{n} \rightarrow {\mathbb{B}}^{n} \) defined by\n\n\[ G\left( y\right) = \frac{y}{1 + \left| y\right| }\]\n\nis an inverse for \( F \) . Thus \( F \) is bijective, and since \( F \) and \( {F}^{-1} = G \) are both continuous, \( F \) is a homeomorphism. It follow... | Yes |
Show that the map \( \varphi : C \rightarrow {\mathbb{S}}^{2} \) is a homeomorphism by showing that its inverse can be written\n\n\[{\varphi }^{-1}\left( {x, y, z}\right) = \frac{\left( x, y, z\right) }{\max \{ \left| x\right| ,\left| y\right| ,\left| z\right| \} }.\] | The map \( \varphi : C \rightarrow {\mathbb{S}}^{2} \) is defined as\n\n\[ \varphi \left( {x, y, z}\right) = \frac{\left( x, y, z\right) }{\sqrt{{x}^{2} + {y}^{2} + {z}^{2}}}, \]\n\nwhich is continuous on \( C \) by the usual arguments of elementary analysis (notice that the denominator is always nonzero on \( C \) ). ... | Yes |
Proposition 2.30. Suppose \( X \) and \( Y \) are topological spaces, and \( f : X \rightarrow Y \) is any map.\n\n(a) \( f \) is continuous if and only if \( f\left( \bar{A}\right) \subseteq \overline{f\left( A\right) } \) for all \( A \subseteq X \) . | Proof. Problem 2-6. | No |
Proposition 2.37. Let \( X \) be a Hausdorff space.\n\n(a) Every finite subset of \( X \) is closed.\n\n(b) If a sequence \( \left( {p}_{i}\right) \) in \( X \) converges to a limit \( p \in X \), the limit is unique. | Proof. For part (a), consider first a set \( \left\{ {p}_{0}\right\} \) containing only one point. Given \( p \neq \) \( {p}_{0} \), the Hausdorff property says that there exist disjoint neighborhoods \( U \) of \( p \) and \( V \) of \( {p}_{0} \) . In particular, \( U \) is a neighborhood of \( p \) contained in \( X... | Yes |
Proposition 2.43. Let \( X \) and \( Y \) be topological spaces and let \( \mathcal{B} \) be a basis for \( Y \) . A map \( f : X \rightarrow Y \) is continuous if and only if for every basis subset \( B \in \mathcal{B} \), the subset \( {f}^{-1}\left( B\right) \) is open in \( X \) . | Proof. One direction is easy: if \( f \) is continuous, the preimage of every open subset, and thus certainly every basis subset, is open. Conversely, suppose \( {f}^{-1}\left( B\right) \) is open for every \( B \in \mathcal{B} \) . Let \( U \subseteq Y \) be open, and let \( x \in {f}^{-1}\left( U\right) \) . Because ... | Yes |
Proposition 2.44. Let \( X \) be a set, and suppose \( \mathcal{B} \) is a collection of subsets of \( X \) . Then \( \mathcal{B} \) is a basis for some topology on \( X \) if and only if it satisfies the following two conditions:\n\n(i) \( \mathop{\bigcup }\limits_{{B \in \mathcal{B}}}B = X \) .\n\n(ii) If \( {B}_{1},... | Proof. If \( \mathcal{B} \) is a basis for some topology, the proof that it satisfies (i) and (ii) is left as an easy exercise. Conversely, suppose that \( \mathcal{B} \) satisfies (i) and (ii), and let \( \mathcal{T} \) be the collection of all unions of elements of \( \mathcal{B} \) .\n\n\n\nBecause the identity is always continuous, the characteristic property implies that \( {\iota }_{S} \) is also continuous. | No |
Corollary 3.10. Let \( X \) and \( Y \) be topological spaces, and suppose \( f : X \rightarrow Y \) is continuous.\n\n(a) RESTRICTING THE DOMAIN: The restriction of \( f \) to any subspace \( S \subseteq X \) is continuous.\n\n(b) RESTRICTING THE CODOMAIN: If \( T \) is a subspace of \( Y \) that contains \( f\left( X... | Proof. Part (a) follows from Corollary 3.9, because \( {\left. f\right| }_{S} = f \circ {\iota }_{S} \) . Part (b) follows from the characteristic property applied to the subspace \( T \) of \( Y \), and part (c) follows by composing \( f \) with the inclusion \( Y \hookrightarrow Z \) . | Yes |
Proposition 3.11 (Other Properties of the Subspace Topology). Suppose \( S \) is a subspace of the topological space \( X \) .\n\n(a) If \( R \subseteq S \) is a subspace of \( S \), then \( R \) is a subspace of \( X \) ; in other words, the subspace topologies that \( R \) inherits from \( S \) and from \( X \) agree... | Proof. For part (a), let \( U \subseteq R \) be any subset. Assume first that \( U \) is open in the subspace topology that \( R \) inherits from \( S \) . This means \( U = R \cap V \) for some open subset \( V \subseteq S \) (Fig. 3.3). The fact that \( V \) is open in \( S \) means that \( V = W \cap S \) for some o... | Yes |
Let \( F : \mathbb{R} \rightarrow {\mathbb{R}}^{2} \) be the injective continuous map \( F\left( s\right) = \left( {s,{s}^{2}}\right) \) . Its image is the parabola \( P \) defined by the equation \( y = {x}^{2} \) . Considered as a map from \( \mathbb{R} \) to \( P, F \) is bijective, and it is continuous by Corollary... | It is easy to check that its inverse is given by \( {\left. \pi \right| }_{P} \), where \( \pi : {\mathbb{R}}^{2} \rightarrow \mathbb{R} \) is the projection \( \pi \left( {x, y}\right) = x \) ; since \( \pi \) is continuous, its restriction to \( P \) is continuous by Corollary 3.10(a). Thus \( F \) is a topological e... | Yes |
Proposition 3.16. A continuous injective map that is either open or closed is a topological embedding. | Proof. Suppose \( X \) and \( Y \) are topological spaces and \( f : X \rightarrow Y \) is a continuous injective map. Note that \( f \) defines a bijective map from \( X \) to \( f\left( X\right) \), which is continuous by Corollary 3.10(b). To show that this map is a homeomorphism, it suffices by Exercise 2.29 to sho... | No |
If \( U \subseteq {\mathbb{R}}^{n} \) is an open subset and \( f : U \rightarrow {\mathbb{R}}^{k} \) is any continuous map, the graph of \( f \) (Fig. 3.5) is the subset \( \Gamma \left( f\right) \subseteq {\mathbb{R}}^{n + k} \) defined by\n\n\[ \Gamma \left( f\right) = \left\{ {\left( {x, y}\right) = \left( {{x}_{1},... | To verify that \( \Gamma \left( f\right) \) is a manifold, we show that it is in fact homeomorphic to \( U \) . Let \( {\Phi }_{f} : U \rightarrow {\mathbb{R}}^{n + k} \) be the continuous injective map\n\n\[ {\Phi }_{f}\left( x\right) = \left( {x, f\left( x\right) }\right) . \]\n\nJust as in Example 3.14, \( {\Phi }_{... | Yes |
To see that \( {\mathbb{S}}^{n} \) is a manifold, we need to show that each point has a Euclidean neighborhood. | For each \( i = 1,\ldots, n + 1 \), let \( {U}_{i}^{ + } \) denote the open subset of \( {\mathbb{R}}^{n + 1} \) consisting of points with \( {x}_{i} > 0 \), and \( {U}_{i}^{ - } \) the set of points with \( {x}_{i} < 0 \) . If \( x \) is any point in \( {\mathbb{S}}^{n} \), some coordinate \( {x}_{i} \) must be nonzer... | Yes |
Finally, consider the doughnut surface, which is the surface of revolution \( D \subseteq {\mathbb{R}}^{3} \) obtained by starting with the circle \( {\left( x - 2\right) }^{2} + {z}^{2} = 1 \) in the \( {xz} \) - plane (called the generating circle), and revolving it around the \( z \) -axis (Fig. 3.7). It is characte... | \[ F\left( {u, v}\right) = \left( {\left( {2 + \cos {2\pi u}}\right) \cos {2\pi v},\left( {2 + \cos {2\pi u}}\right) \sin {2\pi v},\sin {2\pi u}}\right) . \] This maps the plane onto \( D \) . It is not one-to-one, because \( F\left( {u + k, v + l}\right) = F\left( {u, v}\right) \) for any pair of integers \( \left( {k... | Yes |
Lemma 3.23 (Gluing Lemma). Let \( X \) and \( Y \) be topological spaces, and let \( \left\{ {A}_{i}\right\} \) be either an arbitrary open cover of \( X \) or a finite closed cover of \( X \) . Suppose that we are given continuous maps \( {f}_{i} : {A}_{i} \rightarrow Y \) that agree on overlaps: \( {\left. {f}_{i}\ri... | Proof. In either case, it follows from elementary set theory that there exists a unique map \( f \) such that \( {\left. f\right| }_{{A}_{i}} = {f}_{i} \) for each \( i \) . If the sets \( {A}_{i} \) are open, the continuity of \( f \) follows immediately from the local criterion for continuity (Proposition 2.19). On t... | Yes |
Theorem 3.24 (Uniqueness of the Subspace Topology). Suppose \( S \) is a subset of a topological space \( X \) . The subspace topology on \( S \) is the unique topology for which the characteristic property holds. | Proof. Suppose we are given an arbitrary topology on \( S \) that is known to satisfy the characteristic property. For this proof, let \( {S}_{g} \) denote the set \( S \) with the given topology, and let \( {S}_{s} \) denote \( S \) with the subspace topology. To show that the given topology is equal to the subspace t... | Yes |
Theorem 3.27 (Characteristic Property of the Product Topology). Suppose \( {X}_{1} \times \cdots \times {X}_{n} \) is a product space. For any topological space \( Y \), a map \( f : Y \rightarrow \) \( {X}_{1} \times \cdots \times {X}_{n} \) is continuous if and only if each of its component functions \( {f}_{i} = {\p... | Proof. Suppose each \( {f}_{i} \) is continuous. To prove that \( f \) is continuous, it suffices to show that the preimage of each basis subset \( {U}_{1} \times \cdots \times {U}_{k} \) is open. A point \( y \in Y \) is in \( {f}^{-1}\left( {{U}_{1} \times \cdots \times {U}_{k}}\right) \) if and only if \( {f}_{i}\le... | Yes |
Theorem 3.30 (Uniqueness of the Product Topology). Let \( {X}_{1},\ldots ,{X}_{n} \) be topological spaces. The product topology on \( {X}_{1} \times \cdots \times {X}_{n} \) is the unique topology that satisfies the characteristic property. | Proof. Suppose that \( {X}_{1} \times \cdots \times {X}_{n} \) is endowed with some topology that satisfies the characteristic property. Since the proof of Corollary 3.28 uses only the characteristic property, it follows that the canonical projections \( {\pi }_{i} \) are continuous with respect to both the product top... | Yes |
Proposition 3.33. A product of continuous maps is continuous, and a product of homeomorphisms is a homeomorphism. | Proof. Because a map is continuous provided that the preimages of basis open subsets are open, the first claim follows from the fact that \( {\left( {f}_{1} \times \cdots \times {f}_{k}\right) }^{-1}\left( {{U}_{1} \times \cdots \times }\right. \) \( \left. {U}_{k}\right) \) is just the product of the open subsets \( {... | Yes |
Proposition 3.35. If \( {M}_{1},\ldots ,{M}_{k} \) are manifolds of dimensions \( {n}_{1},\ldots ,{n}_{k} \), respectively, the product space \( {M}_{1} \times \cdots \times {M}_{k} \) is a manifold of dimension \( {n}_{1} + \cdots + {n}_{k} \) . | Proof. Proposition 3.31 shows that the product space is Hausdorff and second countable, so only the locally Euclidean property needs to be checked. Given any point \( p = \left( {{p}_{1},\ldots ,{p}_{k}}\right) \in {M}_{1} \times \cdots \times {M}_{k} \), for each \( i \) there exists a neighborhood \( {U}_{i} \) of \(... | Yes |
Proposition 3.36. The torus \( {\mathbb{T}}^{2} \) is homeomorphic to the doughnut surface \( D \) of Example 3.22. | Proof. The key geometric idea is that both surfaces are parametrized by two angles. For \( D \), the angles are \( \varphi = {2\pi u} \) and \( \theta = {2\pi v} \) as in (3.3); for \( {\mathbb{T}}^{2} \), they are the angles in the two circles. Although one must be cautious using angle functions because they cannot be... | Yes |
Corollary 3.39. If \( {\left( {X}_{\alpha }\right) }_{\alpha \in A} \) is an indexed family of nonempty topological spaces with infinitely many indices such that \( {X}_{\alpha } \) is not a trivial space, then the box topology on \( \mathop{\prod }\limits_{{\alpha \in A}}{X}_{\alpha } \) does not satisfy the character... | Proof. As we remarked above, under the hypotheses of the corollary, the box topology is not equal to the product topology. The result follows from the uniqueness statement of Theorem 3.37. | No |
Theorem 3.41 (Characteristic Property of Disjoint Union Spaces). Suppose that \( {\left( {X}_{\alpha }\right) }_{\alpha \in A} \) is an indexed family of topological spaces, and \( Y \) is any topological space. A map \( f : \mathop{\coprod }\limits_{{\alpha \in A}}{X}_{\alpha } \rightarrow Y \) is continuous if and on... | Proof. Problem 3-10. | No |
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