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Lemma 6.4. If \( \gamma \) is an admissible curve and \( V \) is a vector field along \( \gamma \) , then \( V \) is the variation field of some variation of \( \gamma \) . If \( V \) is proper, the variation can be taken to be proper as well.
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Proof. Set \( \Gamma \left( {s, t}\right) = \exp \left( {{sV}\left( t\right) }\right) \) (Figure 6.5). By compactness of \( \left\lbrack {a, b}\right\rbrack \), there is some positive \( \varepsilon \) such that \( \Gamma \) is defined on \( \left( {-\varepsilon ,\varepsilon }\right) \times \left\lbrack {a, b}\right\rbrack \) . Clearly \( \Gamma \) is smooth on \( \left( {-\varepsilon ,\varepsilon }\right) \times \left\lbrack {{a}_{i - 1},{a}_{i}}\right\rbrack \) for each subinterval \( \left\lbrack {{a}_{i - 1},{a}_{i}}\right\rbrack \) on which \( V \) is smooth, and is continuous on its whole domain. By the properties of the exponential map, the variation field of \( \Gamma \) is \( V \) . Moreover, if \( V\left( a\right) = V\left( b\right) = \) 0, it is immediate that \( \Gamma \left( {s, a}\right) \equiv \gamma \left( a\right) \) and \( \Gamma \left( {s, b}\right) \equiv \gamma \left( b\right) \), so \( \Gamma \) is proper.
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Yes
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Theorem 6.6. Every minimizing curve is a geodesic when it is given a unit speed parametrization.
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Proof. Suppose \( \gamma : \left\lbrack {a, b}\right\rbrack \rightarrow M \) is minimizing and unit speed, and let \( a = \) \( {a}_{0} < \cdots < {a}_{k} = b \) be a subdivision such that \( \gamma \) is smooth on \( \left\lbrack {{a}_{i - 1},{a}_{i}}\right\rbrack \) . If \( \Gamma \) is any proper variation of \( \gamma \), we conclude from elementary calculus that \( {dL}\left( {\Gamma }_{s}\right) /{ds} = 0 \) when \( s = 0 \) . Since every proper vector field along \( \gamma \) is the variation field of some proper variation, the right-hand side of (6.2) must vanish for every such \( V \) .\n\nThe first step is to show that \( {D}_{t}\dot{\gamma } = 0 \) on each subinterval \( \left\lbrack {{a}_{i - 1},{a}_{i}}\right\rbrack \), so \( \gamma \) is a \
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Yes
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Corollary 6.7. A unit speed admissible curve \( \gamma \) is a critical point for \( L \) if and only if it is a geodesic.
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Proof. If \( \gamma \) is a critical point, the proof of Theorem 6.6 goes through without modification to show that \( \gamma \) is a geodesic. Conversely, if \( \gamma \) is a geodesic, then the first term in the second variation formula vanishes by the geodesic equation, and the second term vanishes because \( \dot{\gamma } \) has no jumps.
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Yes
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Corollary 6.9. Let \( \left( {x}^{i}\right) \) be normal coordinates on a geodesic ball \( \mathcal{U} \) centered at \( p \in M \), and let \( r \) be the radial distance function as defined in (5.9). Then \( \operatorname{grad}r = \partial /\partial r \) on \( \mathcal{U} - \{ p\} \) .
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Proof. For any \( q \in \mathcal{U} - \{ p\} \) and \( Y \in {T}_{q}M \), we need to show that\n\n\[ \n{dr}\left( Y\right) = \left\langle {\frac{\partial }{\partial r}, Y}\right\rangle \n\]\n\n(6.4)\n\nThe geodesic sphere \( {\exp }_{p}\left( {\partial {B}_{R}\left( 0\right) }\right) \) through \( q \) is characterized in normal coordinates by the equation \( r = R \) . Since \( \partial /\partial r \) is transverse to this sphere, we can decompose \( Y \) as \( \alpha \partial /\partial r + X \) for some constant \( \alpha \) and some vector \( X \) tangent to the sphere (Figure 6.11). Observe that \( {dr}\left( {\partial /\partial r}\right) = 1 \) by direct computation in coordinates, and \( {dr}\left( X\right) = 0 \) since \( X \) is tangent to a level set of \( r \) . (This has nothing to do with the metric!) Therefore the left-hand side of (6.4) is\n\n\[ \n{dr}\left( {\alpha \frac{\partial }{\partial r} + X}\right) = {\alpha dr}\left( \frac{\partial }{\partial r}\right) + {dr}\left( X\right) = \alpha .\n\]\n\nOn the other hand, by Proposition 5.11(e), \( \partial /\partial r \) is a unit vector. Therefore, the right-hand side of (6.4) is\n\n\[ \n\left\langle {\frac{\partial }{\partial r},\alpha \frac{\partial }{\partial r} + X}\right\rangle = \alpha {\left| \frac{\partial }{\partial r}\right| }^{2} + \left\langle {\frac{\partial }{\partial r}, X}\right\rangle = \alpha \n\]\n\nwhere we have used the Gauss lemma to conclude that \( X \) is orthogonal to \( \partial /\partial r \) .
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Yes
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Corollary 6.11. Within any geodesic ball around \( p \in M \), the radial distance function \( r\left( x\right) \) defined by (5.9) is equal to the Riemannian distance from \( p \) to \( x \) .
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Proof. The radial geodesic \( \gamma \) from \( p \) to \( x \) is minimizing by Proposition 6.10. Since its velocity is equal to \( \partial /\partial r \), which is a unit vector in both the \( g \) norm and the Euclidean norm in normal coordinates, the \( g \) -length of \( \gamma \) is equal to its Euclidean length, which is \( r\left( x\right) \) .
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Yes
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Theorem 6.12. Every Riemannian geodesic is locally minimizing.
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Proof. Let \( \gamma : I \rightarrow M \) be a geodesic, which we may assume to be defined on an open interval, and let \( {t}_{0} \in I \) . Let \( \mathcal{W} \) be a uniformly normal neighborhood of \( \gamma \left( {t}_{0}\right) \), and let \( \mathcal{U} \subset I \) be the connected component of \( {\gamma }^{-1}\left( \mathcal{W}\right) \) containing \( {t}_{0} \) . If \( {t}_{1},{t}_{2} \in \mathcal{U} \) and \( {q}_{i} = \gamma \left( {t}_{i}\right) \), the definition of uniformly normal neighborhood implies that \( {q}_{2} \) is contained in a geodesic ball around \( {q}_{1} \) (Figure 6.13). Therefore, by Proposition 6.10, the radial geodesic from \( {q}_{1} \) to \( {q}_{2} \) is the unique minimizing curve between them. However, the restriction of \( \gamma \) is a geodesic from \( {q}_{1} \) to \( {q}_{2} \) lying in the same geodesic ball, and thus \( \gamma \) must itself be this minimizing geodesic.
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Yes
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Theorem 6.13. (Hopf-Rinow) A connected Riemannian manifold is geodesically complete if and only if it is complete as a metric space.
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Proof. Suppose first that \( M \) is complete as a metric space but not geodesically complete. Then there is some unit speed geodesic \( \gamma : \lbrack 0, b) \rightarrow M \) that extends to no interval \( \lbrack 0, b + \varepsilon ) \) for \( \varepsilon > 0 \) . Let \( \left\{ {t}_{i}\right\} \) be any increasing sequence that approaches \( b \), and set \( {q}_{i} = \gamma \left( {t}_{i}\right) \) . Since \( \gamma \) is parametrized by arc length, the length of \( {\left. \gamma \right| }_{\left\lbrack {t}_{i},{t}_{j}\right\rbrack } \) is exactly \( \left| {{t}_{j} - {t}_{i}}\right| \), so \( d\left( {{q}_{i},{q}_{j}}\right) \leq \left| {{t}_{j} - {t}_{i}}\right| \) and \( \left\{ {q}_{i}\right\} \) is a Cauchy sequence in \( M \) . By completeness, \( \left\{ {q}_{i}\right\} \) converges to some point \( q \in M \) .\n\nLet \( \mathcal{W} \) be a uniformly normal neighborhood of \( q \), and let \( \delta > 0 \) be chosen so that \( \mathcal{W} \) is contained in a geodesic \( \delta \) -ball around each of its points. For all large \( j,{q}_{j} \in \mathcal{W} \) (Figure 6.14), and by taking \( j \) large enough, we may assume \( {t}_{j} > b - \delta \) . The fact that \( {B}_{\delta }\left( {q}_{j}\right) \) is a geodesic ball means that every geodesic starting at \( {q}_{j} \) exists at least for time \( \delta \) . In particular, this is true of the geodesic \( \sigma \) with \( \sigma \left( 0\right) = {q}_{j} \) and \( \dot{\sigma }\left( 0\right) = \dot{\gamma }\left( {t}_{j}\right) \) . But by uniqueness of geodesics, this must be simply a reparametrization of \( \gamma \), so \( \widetilde{\gamma }\left( t\right) = \sigma \left( {{t}_{j} + t}\right) \) is an extension of \( \gamma \) past \( b \), which is a contradiction.
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Yes
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Corollary 6.14. If there exists one point \( p \in M \) such that the restricted exponential map \( {\exp }_{p} \) is defined on all of \( {T}_{p}M \), then \( M \) is complete.
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Null
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No
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Corollary 6.15. \( M \) is complete if and only if any two points in \( M \) can be joined by a minimizing geodesic segment.
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Null
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No
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Corollary 6.16. If \( M \) is compact, then every geodesic can be defined for all time.
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Null
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No
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Lemma 7.2. The Riemann curvature endomorphism and curvature tensor are local isometry invariants. More precisely, if \( \varphi : \left( {M, g}\right) \rightarrow \left( {\widetilde{M},\widetilde{g}}\right) \) is a local isometry, then\n\n\[{\varphi }^{ * }\widetilde{Rm} = {Rm}\]\n\n\[ \widetilde{R}\left( {{\varphi }_{ * }X,{\varphi }_{ * }Y}\right) {\varphi }_{ * }Z = {\varphi }_{ * }\left( {R\left( {X, Y}\right) Z}\right) . \]
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Exercise 7.2. Prove Lemma 7.2.
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No
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Proposition 7.4. (Symmetries of the Curvature Tensor) The curvature tensor has the following symmetries for any vector fields \( W, X, Y \) , \( Z \) :\n\n(a) \( \operatorname{Rm}\left( {W, X, Y, Z}\right) = - \operatorname{Rm}\left( {X, W, Y, Z}\right) \) .\n\n(b) \( \operatorname{Rm}\left( {W, X, Y, Z}\right) = - \operatorname{Rm}\left( {W, X, Z, Y}\right) \) .\n\n(c) \( \operatorname{Rm}\left( {W, X, Y, Z}\right) = \operatorname{Rm}\left( {Y, Z, W, X}\right) \) .\n\n(d) \( \operatorname{Rm}\left( {W, X, Y, Z}\right) + \operatorname{Rm}\left( {X, Y, W, Z}\right) + \operatorname{Rm}\left( {Y, W, X, Z}\right) = 0 \) .
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Proof of Proposition 7.4. Identity (a) is immediate from the obvious fact that \( R\left( {W, X}\right) Y = - R\left( {X, W}\right) Y \) . To prove (b), it suffices to show that \( {Rm}\left( {W, X, Y, Y}\right) = 0 \) for all \( Y \), for then (b) follows from the expansion of \( \operatorname{Rm}\left( {W, X, Y + Z, Y + Z}\right) = 0 \) . Using compatibility with the metric, we have\n\n\[ {WX}{\left| Y\right| }^{2} = W\left( {2\left\langle {{\nabla }_{X}Y, Y}\right\rangle }\right) = 2\left\langle {{\nabla }_{W}{\nabla }_{X}Y, Y}\right\rangle + 2\left\langle {{\nabla }_{X}Y,{\nabla }_{W}Y}\right\rangle \]\n\n\[ {XW}{\left| Y\right| }^{2} = X\left( {2\left\langle {{\nabla }_{W}Y, Y}\right\rangle }\right) = 2\left\langle {{\nabla }_{X}{\nabla }_{W}Y, Y}\right\rangle + 2\left\langle {{\nabla }_{W}Y,{\nabla }_{X}Y}\right\rangle \]\n\n\[ \left\lbrack {W, X}\right\rbrack {\left| Y\right| }^{2} = 2\left\langle {{\nabla }_{\left\lbrack W, X\right\rbrack }Y, Y}\right\rangle \]\n\nWhen we subtract the second and third equations from the first, the left-hand side is zero. The terms \( 2\left\langle {{\nabla }_{X}Y,{\nabla }_{W}Y}\right\rangle \) and \( 2\left\langle {{\nabla }_{W}Y,{\nabla }_{X}Y}\right\rangle \) cancel on the right-hand side, giving\n\n\[ 0 = 2\langle {\nabla }_{W}{\nabla }_{X}Y, Y\rangle - 2\langle {\nabla }_{X}{\nabla }_{W}Y, Y\rangle - 2\langle {\nabla }_{\left\lbrack W, X\right\rbrack }Y, Y\rangle \]\n\n\[ = 2\langle R\left( {W, X}\right) Y, Y\rangle \]\n\n\[ = {2Rm}\left( {W, X, Y, Y}\right) \text{.} \]\n\nNext we prove (d). From the definition of \( {Rm} \), this will follow immediately from\n\n\[ R\left( {W, X}\right) Y + R\left( {X, Y}\right) W + R\left( {Y, W}\right) X = 0. \]\n\nUsing the definition of \( R \) and the symmetry of the connection, the left-hand side expands to\n\n\[ \left( {{\nabla }_{W}{\nabla }_{X}Y - {\nabla }_{X}{\nabla }_{W}Y - {\nabla }_{\left\lbrack W, X\right\rbrack }Y}\right) \]\n\n\[ + \left( {{\nabla }_{X}{\nabla }_{Y}W - {\nabla }_{Y}{\nabla }_{X}W - {\nabla }_{\left\lbrack X, Y\right\rbrack }W}\right) \]\n\n\[ + \left( {{\nabla }_{Y}{\nabla }_{W}X - {\nabla }_{W}{\nabla }_{Y}X - {\nabla }_{\left\lbrack Y, W\right\rbrack }X}\right) \]\n\n\[ = {\nabla }_{W}\left( {{\nabla }_{X}Y - {\nabla }_{Y}X}\right) + {\nabla }_{X}\left( {{\nabla }_{Y}W - {\nabla }_{W}Y}\right) + {\nabla }_{Y}\left( {{\nabla }_{W}X - {\nabla }_{X}W}\right) \]\n\n\[ - {\nabla }_{\left\lbrack W, X\right\rbrack }Y - {\nabla }_{\left\lbrack X
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Yes
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Proposition 7.5. (Differential Bianchi Identity) The total covariant derivative of the curvature tensor satisfies the following identity:\n\n\[ \n\nabla {Rm}\left( {X, Y, Z, V, W}\right) + \nabla {Rm}\left( {X, Y, V, W, Z}\right) + \nabla {Rm}\left( {X, Y, W, Z, V}\right) = 0.\n\]
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Proof. First of all, by the symmetries of \( {Rm} \) ,(7.6) is equivalent to\n\n\[ \n\nabla {Rm}\left( {Z, V, X, Y, W}\right) + \nabla {Rm}\left( {V, W, X, Y, Z}\right) + \nabla {Rm}\left( {W, Z, X, Y, V}\right) = 0.\n\]\n\nThis can be proved by a long and tedious computation, but there is a standard shortcut for such calculations in Riemannian geometry that makes our task immeasurably easier. To prove (7.6) holds at a particular point \( p \), by multilinearity it suffices to prove the formula when \( X, Y, Z, V, W \) are basis elements with respect to some frame. The shortcut consists of choosing a special frame for each point \( p \) to simplify the computations there.\n\nLet \( \left( {x}^{i}\right) \) be normal coordinates at \( p \), and let \( X, Y, Z, V, W \) be arbitrary coordinate basis vectors \( {\partial }_{i} \) . These vectors satisfy two properties that simplify our computations enormously: (1) their commutators vanish identically, since \( \left\lbrack {{\partial }_{i},{\partial }_{j}}\right\rbrack \equiv 0 \) ; and (2) their covariant derivatives vanish at \( p \), since \( {\Gamma }_{ij}^{k}\left( p\right) = 0 \) (Proposition 5.11(f)).\n\nUsing these facts and the compatibility of the connection with the metric, the first term in (7.8) evaluated at \( p \) becomes\n\n\[ \n{\nabla }_{W}\operatorname{Rm}\left( {Z, V, X, Y}\right) = {\nabla }_{W}\langle R\left( {Z, V}\right) X, Y\rangle\n\]\n\n\[ \n= \left\langle {{\nabla }_{W}{\nabla }_{Z}{\nabla }_{V}X - {\nabla }_{W}{\nabla }_{V}{\nabla }_{Z}X, Y}\right\rangle .\n\]\n\nWrite this equation three times, with the vector fields \( W, Z, V \) cyclically permuted. Summing all three gives\n\n\[ \n\nabla \operatorname{Rm}\left( {Z, V, X, Y, W}\right) + \nabla \operatorname{Rm}\left( {V, W, X, Y, Z}\right) + \nabla \operatorname{Rm}\left( {W, Z, X, Y, V}\right)\n\]\n\n\[ \n= \langle {\nabla }_{W}{\nabla }_{Z}{\nabla }_{V}X - {\nabla }_{W}{\nabla }_{V}{\nabla }_{Z}X\n\]\n\n\[ \n+ {\nabla }_{Z}{\nabla }_{V}{\nabla }_{W}X - {\nabla }_{Z}{\nabla }_{W}{\nabla }_{V}X\n\]\n\n\[ \n\; + {\nabla }_{V}{\nabla }_{W}{\nabla }_{Z}X - {\nabla }_{V}{\nabla }_{Z}{\nabla }_{W}X, Y\rangle\n\]\n\n\[ \n= \langle R\left( {W, Z}\right) {\nabla }_{V}X + R\left( {Z, V}\right) {\nabla }_{W}X + R\left( {V, W}\right) {\nabla }_{Z}X, Y\rangle\n\]\n\n\[ \n= 0,\n\]\n\nwhere the last line follows because \( {\nabla }_{V}X = {\nabla }_{W}X = {\nabla }_{Z}X = 0 \) at \( p \) .
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Yes
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Exercise 7.5. Prove Lemma 7.6, using the symmetries of the curvature tensor.
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Null
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No
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Lemma 7.7. (Contracted Bianchi Identity) The covariant derivatives of the Ricci and scalar curvatures satisfy the following identity:\n\n\[ \operatorname{div}{Rc} = \frac{1}{2}\nabla S \]\n\nwhere div is the divergence operator (Problem 3-3). In components, this is\n\n\[ {R}_{{ij};}{}^{j} = \frac{1}{2}{S}_{;i} \]
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Proof. Formula (7.9) follows immediately by contracting the component form (7.7) of the differential Bianchi identity on the indices \( i, l \) and then again on \( j, k \), after raising one index of each pair.
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No
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Proposition 7.8. If \( g \) is an Einstein metric on a connected manifold of dimension \( n \geq 3 \), its scalar curvature is constant.
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Proof. Taking the covariant derivative of each side of (7.10) and noting that the covariant derivative of the metric is zero, we see that the Einstein condition implies\n\n\[ {R}_{{ij};k} = \frac{1}{n}{S}_{;k}{g}_{ij} \]\n\nTracing this equation on \( j \) and \( k \), and comparing with the contracted Bianchi identity (7.9), we conclude\n\n\[ \frac{1}{2}{S}_{;i} = \frac{1}{n}{S}_{;i} \]\n\nWhen \( n > 2 \), this implies \( {S}_{;i} = 0 \) . But \( {S}_{;i} \) is the component of \( \nabla S = {ds} \), so connectedness of \( M \) implies \( S \) is constant.
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Yes
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Lemma 8.1. The second fundamental form is\n\n(a) independent of the extensions of \( X \) and \( Y \) ;\n\n(b) bilinear over \( {C}^{\infty }\left( M\right) \) ; and\n\n(c) symmetric in \( X \) and \( Y \) .
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Proof. First we show that the symmetry of \( \Pi \) follows from the symmetry of the connection \( \widetilde{\nabla } \) . Let \( X \) and \( Y \) be extended arbitrarily to \( M \) . Then\n\n\[ \Pi \left( {X, Y}\right) - \Pi \left( {Y, X}\right) = {\left( {\widetilde{\nabla }}_{X}Y - {\widetilde{\nabla }}_{Y}X\right) }^{ \bot } = {\left\lbrack X, Y\right\rbrack }^{ \bot }. \]\n\nSince \( X \) and \( Y \) are tangent to \( M \) at all points of \( M \), so is their Lie bracket. (This follows easily from Exercise 2.3.) Therefore \( {\left\lbrack X, Y\right\rbrack }^{ \bot } = 0 \), so \( \Pi \) is symmetric.\n\nBecause \( {\left. {\widetilde{\nabla }}_{X}Y\right| }_{p} \) depends only on \( {X}_{p} \), it is clear that \( \Pi \left( {X, Y}\right) \) is independent of the extension chosen for \( X \), and that \( \Pi \left( {X, Y}\right) \) is linear over \( {C}^{\infty }\left( M\right) \) in \( X \) . By symmetry, the same is true for \( Y \) .
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Yes
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Theorem 8.2. (The Gauss Formula) If \( X, Y \in \mathcal{T}\left( M\right) \) are extended arbitrarily to vector fields on \( \widetilde{M} \), the following formula holds along \( M \) :
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Proof. Because of the decomposition (8.1) and the definition of the second fundamental form, it suffices to show that \( {\left( {\widetilde{\nabla }}_{X}Y\right) }^{\top } = {\nabla }_{X}Y \) at all points of \( M \) .\n\nDefine a map \( {\nabla }^{\top } : \mathcal{T}\left( M\right) \times \mathcal{T}\left( M\right) \rightarrow \mathcal{T}\left( M\right) \) by\n\n\[ {\nabla }_{X}^{\top }Y \mathrel{\text{:=}} {\left( {\widetilde{\nabla }}_{X}Y\right) }^{\top } \]\n\nwhere \( X, Y \) are extended arbitrarily to \( \widetilde{M} \) . We examined a special case of this construction, in which \( \widetilde{g} \) is the Euclidean metric, in Lemma 5.1. It follows exactly as in the proof of that lemma that \( {\nabla }^{\top } \) is a connection on \( M \) . Once we show that it is symmetric and compatible with \( g \), the uniqueness of the Riemannian connection on \( M \) shows that \( {\nabla }^{\top } = \nabla \) .\n\nTo see that \( {\nabla }^{\top } \) is symmetric, we use the symmetry of \( \widetilde{\nabla } \) and the fact that \( \left\lbrack {X, Y}\right\rbrack \) is tangent to \( M \) :\n\n\[ {\nabla }_{X}^{\top }Y - {\nabla }_{Y}^{\top }X = {\left( {\widetilde{\nabla }}_{X}Y - {\widetilde{\nabla }}_{Y}X\right) }^{\top } \]\n\n\[ = {\left\lbrack X, Y\right\rbrack }^{\top } = \left\lbrack {X, Y}\right\rbrack \]\n\nTo prove compatibility with \( g \), let \( X, Y, Z \in \mathcal{T}\left( M\right) \) be extended arbitrarily to \( \widetilde{M} \) . Using compatibility of \( \widetilde{\nabla } \) with \( \widetilde{g} \), and evaluating at points of \( M \) ,\n\n\[ X\langle Y, Z\rangle = \left\langle {{\widetilde{\nabla }}_{X}Y, Z}\right\rangle + \left\langle {Y,{\widetilde{\nabla }}_{X}Z}\right\rangle \]\n\n\[ = \left\langle {{\left( {\widetilde{\nabla }}_{X}Y\right) }^{\top }, Z}\right\rangle + \left\langle {Y,{\left( {\widetilde{\nabla }}_{X}Z\right) }^{\top }}\right\rangle \]\n\n\[ = \left\langle {{\nabla }_{X}^{\top }Y, Z}\right\rangle + \left\langle {Y,{\nabla }_{X}^{\top }Z}\right\rangle \]\n\nTherefore \( {\nabla }^{\top } \) is compatible with \( g \), so \( {\nabla }^{\top } = \nabla \) .
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Yes
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Lemma 8.3. (The Weingarten Equation) Suppose \( X, Y \in \mathfrak{T}\left( M\right) \) and \( N \in \mathcal{N}\left( M\right) \) . When \( X, Y, N \) are extended arbitrarily to \( \widetilde{M} \), the following equation holds at points of \( M \) :
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Proof. Since \( \langle N, Y\rangle \) vanishes identically along \( M \) and \( X \) is tangent to \( M \) , the following holds along \( M \) :\n\n\[ 0 = X\langle N, Y\rangle \]\n\n\[ = \left\langle {{\widetilde{\nabla }}_{X}N, Y}\right\rangle + \left\langle {N,{\widetilde{\nabla }}_{X}Y}\right\rangle \]\n\n\[ = \left\langle {{\widetilde{\nabla }}_{X}N, Y}\right\rangle + \left\langle {N,{\nabla }_{X}Y + \Pi \left( {X, Y}\right) }\right\rangle \]\n\n\[ = \left\langle {{\widetilde{\nabla }}_{X}N, Y}\right\rangle + \langle N,\Pi \left( {X, Y}\right) \rangle .\n\n
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Yes
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Theorem 8.4. (The Gauss Equation) For any \( X, Y, Z, W \in {T}_{p}M \), the following equation holds:\n\n\[ \widetilde{\operatorname{Rm}}\left( {X, Y, Z, W}\right) = \operatorname{Rm}\left( {X, Y, Z, W}\right) \]\n\n\[ - \langle \Pi \left( {X, W}\right) ,\Pi \left( {Y, Z}\right) \rangle + \langle \Pi \left( {X, Z}\right) ,\Pi \left( {Y, W}\right) \rangle . \]
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Proof. Let \( X, Y, Z, W \) be extended arbitrarily to vector fields on \( M \), and then to vector fields on \( \widetilde{M} \) that are tangent to \( M \) at points of \( M \) . Along \( M \), the Gauss formula gives\n\n\[ \widetilde{Rm}\left( {X, Y, Z, W}\right) = \left\langle {{\widetilde{\nabla }}_{X}{\widetilde{\nabla }}_{Y}Z - {\widetilde{\nabla }}_{Y}{\widetilde{\nabla }}_{X}Z - {\widetilde{\nabla }}_{\left\lbrack X, Y\right\rbrack }Z, W}\right\rangle \]\n\n\[ = \left\langle {{\widetilde{\nabla }}_{X}\left( {{\nabla }_{Y}Z + \Pi \left( {Y, Z}\right) }\right) - {\widetilde{\nabla }}_{Y}\left( {{\nabla }_{X}Z + \Pi \left( {X, Z}\right) }\right) }\right. \]\n\n\[ - \left. {\left( {{\nabla }_{\left\lbrack X, Y\right\rbrack }Z + \Pi \left( {\left\lbrack {X, Y}\right\rbrack, Z}\right) }\right), W}\right\rangle . \]\n\nSince the second fundamental form takes its values in the normal bundle and \( W \) is tangent to \( M \), the last \( \Pi \) term is zero. Apply the Weingarten equation to the other two terms involving \( \Pi \) (with \( \Pi \left( {Y, Z}\right) \) or \( \Pi \left( {X, Z}\right) \) playing the role of \( N \) ) to get\n\n\[ \widetilde{\operatorname{Rm}}\left( {X, Y, Z, W}\right) = \left\langle {{\widetilde{\nabla }}_{X}{\nabla }_{Y}Z, W}\right\rangle - \langle \Pi \left( {Y, Z}\right) ,\Pi \left( {X, W}\right) \rangle \]\n\n\[ - \left\langle {{\widetilde{\nabla }}_{Y}{\nabla }_{X}Z, W}\right\rangle + \langle \Pi \left( {X, Z}\right) ,\Pi \left( {Y, W}\right) \rangle \]\n\n\[ - \left\langle {{\nabla }_{\left\lbrack X, Y\right\rbrack }Z, W}\right\rangle \]\n\nDecomposing each term involving \( \widetilde{\nabla } \) into its tangential and normal components, we see that only the tangential component survives. Using the Gauss formula allows each to be rewritten in terms of \( \nabla \), giving\n\n\[ \widetilde{Rm}\left( {X, Y, Z, W}\right) = \left\langle {{\nabla }_{X}{\nabla }_{Y}Z, W}\right\rangle - \left\langle {{\nabla }_{Y}{\nabla }_{X}Z, W}\right\rangle - \left\langle {{\nabla }_{\left\lbrack X, Y\right\rbrack }Z, W}\right\rangle \]\n\n\[ - \langle \Pi \left( {Y, Z}\right) ,\Pi \left( {X, W}\right) \rangle + \langle \Pi \left( {X, Z}\right) ,\Pi \left( {Y, W}\right) \rangle \]\n\n\[ = \langle R\left( {X, Y}\right) Z, W\rangle \]\n\n\[ - \langle \Pi \left( {X, W}\right) ,\Pi \left( {Y, Z}\right) \rangle + \langle \Pi \left( {X, Z}\right) ,\Pi \left( {Y, W}\right) \rangle . \]\n\nThis proves the theorem.
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Yes
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Lemma 8.5. (The Gauss Formula Along a Curve) Let \( M \) be a Riemannian submanifold of \( \widetilde{M} \), and \( \gamma \) a curve in \( M \) . For any vector field \( V \) tangent to \( M \) along \( \gamma \) ,\n\n\[{\widetilde{D}}_{t}V = {D}_{t}V + \Pi \left( {\dot{\gamma }, V}\right)\]
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Proof. In terms of an adapted orthonormal frame, \( V \) can be written \( V\left( t\right) = \) \( {V}^{i}\left( t\right) {E}_{i} \), where the sum is only over \( i = 1,\ldots, n \) . Applying the product rule and the Gauss formula, we get\n\n\[{\widetilde{D}}_{t}V = {\dot{V}}^{i}{E}_{i} + {V}^{i}{\widetilde{\nabla }}_{\dot{\gamma }}{E}_{i}\]\n\n\[= {\dot{V}}^{i}{E}_{i} + {V}^{i}{\nabla }_{\dot{\gamma }}{E}_{i} + {V}^{i}\Pi \left( {\dot{\gamma },{E}_{i}}\right)\]\n\n\[= {D}_{t}V + \Pi \left( {\dot{\gamma }, V}\right)\]
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Yes
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Theorem 8.6. (Gauss’s Theorema Egregium) Let \( M \subset {\mathbf{R}}^{3} \) be a 2-dimensional submanifold and \( g \) the induced metric on \( M \). For any \( p \in M \) and any basis \( \left( {X, Y}\right) \) for \( {T}_{p}M \), the Gaussian curvature of \( M \) at \( p \) is given \( {by} \)\n\n\[ K = \frac{\operatorname{Rm}\left( {X, Y, Y, X}\right) }{{\left| X\right| }^{2}{\left| Y\right| }^{2}-\langle X, Y{\rangle }^{2}}. \]
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Proof. We begin with the special case in which \( \left( {X, Y}\right) = \left( {{E}_{1},{E}_{2}}\right) \) is an orthonormal basis for \( {T}_{p}M \). In this case the denominator in (8.5) is equal to 1. If we write \( {h}_{ij} = \dot{h}\left( {{E}_{i},{E}_{j}}\right) \), then in this basis \( K = \det s = \det \left( {h}_{ij}\right) \), and the Gauss equation (8.4) reads\n\n\[ \operatorname{Rm}\left( {{E}_{1},{E}_{2},{E}_{2},{E}_{1}}\right) = {h}_{11}{h}_{22} - {h}_{12}{h}_{21} = \det \left( {h}_{ij}\right) = K. \]\n\nThis is equivalent to (8.5).\n\nNow let \( X, Y \) be any basis for \( {T}_{p}M \). The Gram-Schmidt algorithm yields an orthonormal basis as follows:\n\n\[ {E}_{1} = \frac{X}{\left| X\right| } \]\n\n\[ {E}_{2} = \frac{Y-\langle Y,\frac{X}{\left| X\right| }\rangle \frac{X}{\left| X\right| }}{\left| Y-\langle Y,\frac{X}{\left| X\right| }\rangle \frac{X}{\left| X\right| }\right| } = \frac{Y - \frac{\langle Y, X\rangle }{{\left| X\right| }^{2}}X}{\left| Y - \frac{\langle Y, X\rangle }{{\left| X\right| }^{2}}X\right| }. \]\n\nThen by the preceding computation, the Gaussian curvature at \( p \) is\n\n\[ K = \operatorname{Rm}\left( {{E}_{1},{E}_{2},{E}_{2},{E}_{1}}\right) \]\n\n\[ = \frac{\operatorname{Rm}\left( {X, Y - \frac{\langle Y, X\rangle }{{\left| X\right| }^{2}}X, Y - \frac{\langle Y, X\rangle }{{\left| X\right| }^{2}}X, X}\right) }{{\left| X\right| }^{2}{\left| Y - \frac{\langle Y, X\rangle }{{\left| X\right| }^{2}}X\right| }^{2}} \]\n\n\[ = \frac{\operatorname{Rm}\left( {X, Y, Y, X}\right) }{{\left| X\right| }^{2}\left( {{\left| Y\right| }^{2} - 2\frac{\langle Y, X{\rangle }^{2}}{{\left| X\right| }^{2}} + \frac{\langle Y, X{\rangle }^{2}}{{\left| X\right| }^{2}}}\right) } \]\n\n\[ = \frac{\operatorname{Rm}\left( {X, Y, Y, X}\right) }{{\left| X\right| }^{2}{\left| Y\right| }^{2}-\langle X, Y{\rangle }^{2}} \]\n\n(In the third line, we used the fact that \( \operatorname{Rm}\left( {X, X,\cdot , \cdot }\right) = \operatorname{Rm}\left( {\cdot ,\cdot, X, X}\right) = 0 \) by the symmetries of the curvature tensor.) This proves the theorem.
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Yes
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Lemma 8.7. The Gaussian curvature of a Riemannian 2-manifold is related to the curvature tensor, Ricci tensor, and scalar curvature by the formulas\n\n\[ \n{Rm}\left( {X, Y, Z, W}\right) = K\left( {\langle X, W\rangle \langle Y, Z\rangle -\langle X, Z\rangle \langle Y, W\rangle }\right) \]\n\n\[ \n{Rc}\left( {X, Y}\right) = K\langle X, Y\rangle \]\n\n(8.6)\n\n\[ \nS = {2K}\text{.} \]\n\nThus \( K \) is independent of choice of frame, and completely determines the curvature tensor.
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Proof. Since both sides of the first equation are tensors, we can compute them in terms of any basis. Let \( \left( {{E}_{1},{E}_{2}}\right) \) be any orthonormal basis for \( {T}_{p}M \) , and consider the components \( {R}_{ijkl} = {Rm}\left( {{E}_{i},{E}_{j},{E}_{k},{E}_{l}}\right) \) of the curvature tensor. In terms of this basis,(8.5) gives \( K = {R}_{1221} \) . By antisymmetry, \( {R}_{ijkl} \) vanishes whenever \( i = j \) or \( k = l \), so the only nonzero components of \( {Rm} \) are\n\n\[ \n{R}_{1221} = {R}_{2112} = - {R}_{1212} = - {R}_{2121} = K. \]\n\nComparing \( \operatorname{Rm}\left( {X, Y, Z, W}\right) \) with \( K\left( {\langle X, W\rangle \langle Y, Z\rangle -\langle X, Z\rangle \langle Y, W\rangle }\right) \) when each of \( X, Y, Z, W \) is either \( {E}_{1} \) or \( {E}_{2} \) proves the first equation of (8.6).\n\nThe components of the Ricci tensor in this basis are\n\n\[ \n{R}_{ij} = {R}_{1ij1} + {R}_{2ij2} \]\n\nfrom which it follows easily that\n\n\[ \n{R}_{12} = {R}_{21} = 0;\;{R}_{11} = {R}_{22} = K, \]\n\nwhich is equivalent to the second equation. Finally, the scalar curvature is\n\n\[ \nS = {\operatorname{tr}}_{g}{Rc} = {R}_{11} + {R}_{22} = {2K}. \]\n\nBecause the scalar curvature is independent of choice of frame, so is \( K \) .
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Yes
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Proposition 8.8. If \( \\left( {X, Y}\\right) \) is any basis for a 2-plane \( \\Pi \\subset {T}_{p}M \), then\n\n\[ K\\left( {X, Y}\\right) = \\frac{\\operatorname{Rm}\\left( {X, Y, Y, X}\\right) }{{\\left| X\\right| }^{2}{\\left| Y\\right| }^{2}-\\langle X, Y{\\rangle }^{2}}.\]
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Proof. For this proof, we denote the induced metric on \( {S}_{\\Pi } \) by \( \\widetilde{g} \), and continue to denote the metric on \( M \) by \( g \) . As in the first part of this chapter, we use tildes to denote geometric quantities associated with \( \\widetilde{g} \), but note that now the roles of \( g \) and \( \\widetilde{g} \) are reversed.\n\nWe claim first that the second fundamental form of \( {S}_{\\Pi } \) vanishes at \( p \) . To see why, let \( V \\in \\Pi \\subset {T}_{p}M \), and let \( \\gamma = {\\gamma }_{V} \) be the \( M \) -geodesic with initial velocity \( V \), which lies in \( {S}_{\\Pi } \) by definition. By the Gauss formula for vector fields along curves,\n\n\[ 0 = {D}_{t}\\dot{\\gamma } = {\\widetilde{D}}_{t}\\dot{\\gamma } + \\Pi \\left( {\\dot{\\gamma },\\dot{\\gamma }}\\right) \]\n\nSince the two terms in this sum are orthogonal, each must vanish identically. Evaluating at \( t = 0 \) gives \( \\Pi \\left( {V, V}\\right) = 0 \) . Since \( V \) was an arbitrary element of \( {T}_{p}M \) and \( {II} \) is symmetric, this shows that \( {II} = 0 \) at \( p \) . (We cannot in general expect \( \\Pi \) to vanish at other points of \( {S}_{\\mathrm{{II}}} \) -it is only at \( p \) that all geodesics starting tangent to \( S \) remain in \( S \) .)\n\nNow the Gauss equation tells us that the curvature tensors of \( {S}_{\\Pi } \) and \( M \) are related at \( p \) by\n\n\[ \\widetilde{\\operatorname{Rm}}\\left( {X, Y, Z, W}\\right) = \\operatorname{Rm}\\left( {X, Y, Z, W}\\right) \]\n\nwhenever \( X, Y, Z, W \\in \\Pi \) . In particular, the Gaussian curvature of \( {S}_{\\Pi } \) at \( p \) is\n\n\[ K\\left( \\Pi \\right) = \\frac{\\widetilde{Rm}\\left( {X, Y, Y, X}\\right) }{{\\left| X\\right| }^{2}{\\left| Y\\right| }^{2} - \\langle X, Y{\\rangle }^{2}} = \\frac{{Rm}\\left( {X, Y, Y, X}\\right) }{{\\left| X\\right| }^{2}{\\left| Y\\right| }^{2}-\\langle X, Y{\\rangle }^{2}}.\]\n\nThis is what was to be proved.
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Yes
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Lemma 8.9. Suppose \( {\mathcal{R}}_{1} \) and \( {\mathcal{R}}_{2} \) are covariant 4-tensors on a vector space \( V \) with an inner product, and both have the symmetries of the curvature tensor (as described in Proposition 7.4). If for every pair of independent vectors \( X, Y \in V \), \[ \frac{{\mathcal{R}}_{1}\left( {X, Y, Y, X}\right) }{{\left| X\right| }^{2}{\left| Y\right| }^{2}-\langle X, Y{\rangle }^{2}} = \frac{{\mathcal{R}}_{2}\left( {X, Y, Y, X}\right) }{{\left| X\right| }^{2}{\left| Y\right| }^{2}-\langle X, Y{\rangle }^{2}}, \] then \( {\mathcal{R}}_{1} = {\mathcal{R}}_{2} \) .
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Proof. Setting \( \mathcal{R} = {\mathcal{R}}_{1} - {\mathcal{R}}_{2} \), it suffices to show \( \mathcal{R} = 0 \) under the assumption that \( \mathcal{R}\left( {X, Y, Y, X}\right) = 0 \) for all \( X, Y \). For any vectors \( X, Y, Z \), since \( \mathcal{R} \) also has the symmetries of the curvature tensor, \[ 0 = \mathcal{R}\left( {X + Y, Z, Z, X + Y}\right) \] \[ = \mathcal{R}\left( {X, Z, Z, X}\right) + \mathcal{R}\left( {X, Z, Z, Y}\right) + \mathcal{R}\left( {Y, Z, Z, X}\right) + \mathcal{R}\left( {Y, Z, Z, Y}\right) \] \[ = 2\mathcal{R}\left( {X, Z, Z, Y}\right) \] From this it follows that \[ 0 = \mathcal{R}\left( {X, Z + W, Z + W, Y}\right) \] \[ = \mathcal{R}\left( {X, Z, Z, Y}\right) + \mathcal{R}\left( {X, Z, W, Y}\right) + \mathcal{R}\left( {X, W, Z, Y}\right) + \mathcal{R}\left( {X, W, W, Y}\right) \] \[ = \mathcal{R}\left( {X, Z, W, Y}\right) + \mathcal{R}\left( {X, W, Z, Y}\right) . \] Therefore \( \mathcal{R} \) is antisymmetric in any adjacent pair of arguments. Now the algebraic Bianchi identity yields \[ 0 = \mathcal{R}\left( {X, Y, Z, W}\right) + \mathcal{R}\left( {Y, Z, X, W}\right) + \mathcal{R}\left( {Z, X, Y, W}\right) \] \[ = \mathcal{R}\left( {X, Y, Z, W}\right) - \mathcal{R}\left( {Y, X, Z, W}\right) - \mathcal{R}\left( {X, Z, Y, W}\right) \] \[ = 3\mathcal{R}\left( {X, Y, Z, W}\right) \text{.} \]
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Yes
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Lemma 8.10. Suppose \( \left( {M, g}\right) \) is any Riemannian \( n \) -manifold with constant sectional curvature \( C \) . The curvature endomorphism, curvature tensor, Ricci tensor, and scalar curvature of \( g \) are given by the formulas\n\n\[ R\left( {X, Y}\right) Z = C\left( {\\langle Y, Z\\rangle X-\\langle X, Z\\rangle Y}\right) \]\n\n\[ \\operatorname{Rm}\left( {X, Y, Z, W}\right) = C\left( {\\langle X, W\\rangle \\langle Y, Z\\rangle -\\langle X, Z\\rangle \\langle Y, W\\rangle }\right) \]\n\n\[ {Rc} = \\left( {n - 1}\\right) {Cg} \]\n\n\[ S = n\\left( {n - 1}\\right) C\\text{.} \]\n\nIn terms of any basis,\n\n\[ {R}_{ijkl} = C\\left( {{g}_{il}{g}_{jk} - {g}_{ik}{g}_{jl}}\\right) \]\n\n\[ {R}_{ij} = \\left( {n - 1}\\right) C{g}_{ij} \]
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Exercise 8.8. Prove Lemma 8.10.
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No
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Lemma 9.2. If \( \gamma \) is a positively oriented curved polygon in \( M \), the rotation angle of \( \gamma \) is \( {2\pi } \) .
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Proof. If we use the given coordinate chart to consider \( \gamma \) as a curved polygon in the plane, we can compute its tangent angle function either with respect to \( g \) or with respect to the Euclidean metric \( \bar{g} \) . In either case, \( \operatorname{Rot}\left( \gamma \right) \) is an integral multiple of \( {2\pi } \) because \( \theta \left( a\right) \) and \( \theta \left( b\right) \) both represent the same angle. Now for \( 0 \leq s \leq 1 \), let \( {g}_{s} = {sg} + \left( {1 - s}\right) \bar{g} \) . By the same reasoning, the rotation angle \( {\operatorname{Rot}}_{{g}_{s}}\left( \gamma \right) \) with respect to \( {g}_{s} \) is also a multiple of \( {2\pi } \) . The function \( f\left( s\right) = \left( {1/{2\pi }}\right) {\operatorname{Rot}}_{{g}_{s}}\left( \gamma \right) \) is therefore integer-valued, and is easily seen to be continuous in \( s \), so it must be constant.
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Yes
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Corollary 9.4. (Angle-Sum Theorem) The sum of the interior angles of a Euclidean triangle is \( \pi \) .
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Null
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No
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Corollary 9.5. (Circumference Theorem) The circumference of a Euclidean circle of radius \( R \) is \( {2\pi R} \) .
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Null
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No
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Corollary 9.6. (Total Curvature Theorem) If \( \gamma : \left\lbrack {a, b}\right\rbrack \rightarrow {\mathbf{R}}^{2} \) is a unit speed simple closed curve such that \( \dot{\gamma }\left( a\right) = \dot{\gamma }\left( b\right) \), and \( N \) is the inward-pointing normal, then\n\n\[{\int }_{a}^{b}{\kappa }_{N}\left( t\right) {dt} = {2\pi }\]
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Null
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No
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Theorem 9.7. (The Gauss-Bonnet Theorem) If \( M \) is a triangulated, compact, oriented, Riemannian 2-manifold, then\n\n\[{\int }_{M}{KdA} = {2\pi \chi }\left( M\right)\]
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Proof. Let \( \left\{ {{\Omega }_{i} : i = 1,\ldots ,{N}_{f}}\right\} \) denote the faces of the triangulation, and for each \( i \) let \( \left\{ {{\gamma }_{ij} : j = 1,2,3}\right\} \) be the edges of \( {\Omega }_{i} \) and \( \left\{ {{\theta }_{ij} : j = 1,2,3}\right\} \) its interior angles. Since each exterior angle is \( \pi \) minus the corresponding interior angle, applying the Gauss-Bonnet formula to each triangle and summing over \( i \) gives\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{{N}_{f}}{\int }_{{\Omega }_{i}}{KdA} + \mathop{\sum }\limits_{{i = 1}}^{{N}_{f}}\mathop{\sum }\limits_{{j = 1}}^{3}{\int }_{{\gamma }_{ij}}{\kappa }_{N}{ds} + \mathop{\sum }\limits_{{i = 1}}^{{N}_{f}}\mathop{\sum }\limits_{{j = 1}}^{3}\left( {\pi - {\theta }_{ij}}\right) = \mathop{\sum }\limits_{{i = 1}}^{{N}_{f}}{2\pi }.\]\n\n(9.6)\n\nNote that each edge integral appears exactly twice in the above sum, with opposite orientations, so the integrals of \( {\kappa }_{N} \) all cancel out. Thus (9.6) becomes\n\n\[ {\int }_{M}{KdA} + {3\pi }{N}_{f} - \mathop{\sum }\limits_{{i = 1}}^{{N}_{f}}\mathop{\sum }\limits_{{j = 1}}^{3}{\theta }_{ij} = {2\pi }{N}_{f} \]\n\n(9.7)\n\nNote also that each interior angle \( {\theta }_{ij} \) appears exactly once. At each vertex, the angles that touch that vertex must add up to \( {2\pi } \) (Figure 9.16); thus the angle sum can be rearranged to give exactly \( {2\pi }{N}_{v} \) . Equation (9.7) thus can be written\n\n\[ {\int }_{M}{KdA} = {2\pi }{N}_{v} - \pi {N}_{f} \]\n\n(9.8)\n\nFinally, since each edge appears in exactly two triangles, and each triangle has exactly three edges, the total number of edges counted with multiplicity is \( 2{N}_{e} = 3{N}_{f} \), where we count each edge once for each triangle in which it appears. This means that \( {N}_{f} = 2{N}_{e} - 2{N}_{f} \), so (9.8) finally becomes\n\n\[ {\int }_{M}{KdA} = {2\pi }{N}_{v} - {2\pi }{N}_{e} + {2\pi }{N}_{f} = {2\pi \chi }\left( M\right) \]
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Yes
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Corollary 9.8. Let \( M \) be a compact Riemannian 2-manifold and \( K \) its Gaussian curvature.\n\n(a) If \( M \) is homeomorphic to the sphere or the projective plane, then \( K > 0 \) somewhere.\n\n(b) If \( M \) is homeomorphic to the torus or the Klein bottle, then either \( K \equiv 0 \) or \( K \) takes on both positive and negative values.\n\n(c) If \( M \) is any other compact surface, then \( K < 0 \) somewhere.
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Proof. If \( M \) is orientable, the result follows immediately from the Gauss-Bonnet theorem, because a function whose integral is positive, negative, or zero must satisfy the claimed sign condition. If \( M \) is nonorientable, the result follows by applying the Gauss-Bonnet theorem to the orientable double cover \( \pi : \widetilde{M} \rightarrow M \) with the lifted metric \( \widetilde{g} = {\pi }^{ * }g \), using the fact that \( \widetilde{M} \) is the sphere if \( M = {\mathbf{P}}^{2} \), the torus if \( M \) is the Klein bottle (which is homeomorphic to the connected sum of two copies of \( {\mathbf{P}}^{2} \) ), and otherwise has \( \chi \left( \widetilde{M}\right) < 0 \) .
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Yes
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Exercise 9.2. Prove Corollary 9.9.
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Null
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No
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Lemma 10.1. If \( \\Gamma \) is any smooth admissible family of curves, and \( V \) is a smooth vector field along \( \\Gamma \), then\n\n\[ \n{D}_{s}{D}_{t}V - {D}_{t}{D}_{s}V = R\\left( {S, T}\\right) V.\n\]
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Proof. This is a local issue, so we can compute in any local coordinates.\n\nWriting \( V\\left( {s, t}\\right) = {V}^{i}\\left( {s, t}\\right) {\\partial }_{i} \), we compute\n\n\[ \n{D}_{t}V = \\frac{\\partial {V}^{i}}{\\partial t}{\\partial }_{i} + {V}^{i}{D}_{t}{\\partial }_{i}\n\]\n\nTherefore,\n\n\[ \n{D}_{s}{D}_{t}V = \\frac{{\\partial }^{2}{V}^{i}}{\\partial s\\partial t}{\\partial }_{i} + \\frac{\\partial {V}^{i}}{\\partial t}{D}_{s}{\\partial }_{i} + \\frac{\\partial {V}^{i}}{\\partial s}{D}_{t}{\\partial }_{i} + {V}^{i}{D}_{s}{D}_{t}{\\partial }_{i}.\n\]\n\nInterchanging \( {D}_{s} \) and \( {D}_{t} \) and subtracting, we see that all the terms except the last cancel:\n\n\[ \n{D}_{s}{D}_{t}V - {D}_{t}{D}_{s}V = {V}^{i}\\left( {{D}_{s}{D}_{t}{\\partial }_{i} - {D}_{t}{D}_{s}{\\partial }_{i}}\\right) .\n\]\n\n(10.1)\n\nNow we need to compute the commutator in parentheses. If we write the coordinate functions of \( \\Gamma \) as \( {x}^{j}\\left( {s, t}\\right) \), then\n\n\[ \nS = \\frac{\\partial {x}^{k}}{\\partial s}{\\partial }_{k};\\;T = \\frac{\\partial {x}^{j}}{\\partial t}{\\partial }_{j}.\n\]\n\nBecause \( {\\partial }_{i} \) is extendible,\n\n\[ \n{D}_{t}{\\partial }_{i} = {\\nabla }_{T}{\\partial }_{i} = \\frac{\\partial {x}^{j}}{\\partial t}{\\nabla }_{{\\partial }_{j}}{\\partial }_{i}\n\]\n\nand therefore, because \( {\\nabla }_{{\\partial }_{j}}{\\partial }_{i} \) is also extendible,\n\n\[ \n{D}_{s}{D}_{t}{\\partial }_{i} = {D}_{s}\\left( {\\frac{\\partial {x}^{j}}{\\partial t}{\\nabla }_{{\\partial }_{j}}{\\partial }_{i}}\\right)\n\]\n\n\[ \n= \\frac{{\\partial }^{2}{x}^{j}}{\\partial s\\partial t}{\\nabla }_{{\\partial }_{j}}{\\partial }_{i} + \\frac{\\partial {x}^{j}}{\\partial t}{\\nabla }_{S}\\left( {{\\nabla }_{{\\partial }_{j}}{\\partial }_{i}}\\right)\n\]\n\n\[ \n= \\frac{{\\partial }^{2}{x}^{j}}{\\partial s\\partial t}{\\nabla }_{{\\partial }_{j}}{\\partial }_{i} + \\frac{\\partial {x}^{j}}{\\partial t}\\frac{\\partial {x}^{k}}{\\partial s}{\\nabla }_{{\\partial }_{k}}{\\nabla }_{{\\partial }_{j}}{\\partial }_{i}\n\]\n\nInterchanging \( s \\leftrightarrow t \) and \( j \\leftrightarrow k \) and subtracting, we find that the first terms cancel out, and we get\n\n\[ \n{D}_{s}{D}_{t}{\\partial }_{i} - {D}_{t}{D}_{s}{\\partial }_{i} = \\frac{\\partial {x}^{j}}{\\partial t}\\frac{\\partial {x}^{k}}{\\partial s}\\left( {{\\nabla }_{{\\partial }_{k}}{\\nabla }_{{\\partial }_{j}}{\\partial }_{i} - {\\nabla }_{{\\partial }_{j}}{\\nabla }_{{\\partial }_{k}}{\\partial }_{i}}\\right)\n\]\n\n\[ \n= \\frac{\\partial {x}^{j}}{\\partial t}\\frac{\\partial {x}^{k}}{\\partial s}R\\left( {{\\partial }_{k},{\\partial }_{j}}\\right) {\\partial }_{i}\n\]\n\n\[ \n= R\\left( {S, T}\\right) {\\partial }_{i}\n\]\n\nFinally, inserting this into (10.1) yields the result.
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Yes
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Theorem 10.2. (The Jacobi Equation) Let \( \gamma \) be a geodesic and \( V \) a vector field along \( \gamma \) . If \( V \) is the variation field of a variation through geodesics, then \( V \) satisfies\n\n\[ \n{D}_{t}^{2}V + R\left( {V,\dot{\gamma }}\right) \dot{\gamma } = 0 \n\]
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Proof. With \( S \) and \( T \) as before, the preceding lemma implies\n\n\[ \n0 = {D}_{s}{D}_{t}T \n\]\n\n\[ \n= {D}_{t}{D}_{s}T + R\left( {S, T}\right) T \n\]\n\n\[ \n= {D}_{t}{D}_{t}S + R\left( {S, T}\right) T \n\]\n\nwhere the last step follows from the symmetry lemma. Evaluating at \( s = 0 \) , where \( S\left( {0, t}\right) = V\left( t\right) \) and \( T\left( {0, t}\right) = \dot{\gamma }\left( t\right) \), we get (10.2).
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Yes
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Lemma 10.3. Every Jacobi field along a geodesic \( \gamma \) is the variation field of some variation of \( \gamma \) through geodesics.
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Exercise 10.1. Prove Lemma 10.3. [Hint: Let \( \Gamma \left( {s, t}\right) = {\exp }_{\sigma \left( s\right) }{tW}\left( s\right) \) for a suitable curve \( \sigma \) and vector field \( W \) along \( \sigma \) .]
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No
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Proposition 10.4. (Existence and Uniqueness of Jacobi Fields) Let \( \gamma : I \rightarrow M \) be a geodesic, \( a \in I \), and \( p = \gamma \left( a\right) \) . For any pair of vectors \( X, Y \in {T}_{p}M \), there is a unique Jacobi field \( J \) along \( \gamma \) satisfying the initial conditions\n\n\[ J\left( a\right) = X;\;{D}_{t}J\left( a\right) = Y. \]
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Proof. Choose an orthonormal basis \( \left\{ {E}_{i}\right\} \) for \( {T}_{p}M \), and extend it to a parallel orthonormal frame along all of \( \gamma \) . Writing \( J\left( t\right) = {J}^{i}\left( t\right) {E}_{i} \), we can express the Jacobi equation as\n\n\[ {\ddot{J}}^{i} + {R}_{jkl}{}^{i}{J}^{j}{\dot{\gamma }}^{k}{\dot{\gamma }}^{l} = 0. \]\n\nThis is a linear system of second-order ODEs for the \( n \) functions \( {J}^{i} \) . Making the usual substitution \( {V}^{i} = {\dot{J}}^{i} \) converts it to an equivalent first-order linear system for the \( {2n} \) unknowns \( \left\{ {{J}^{i},{V}^{i}}\right\} \) . Then Theorem 4.12 guarantees the existence and uniqueness of a solution on the whole interval \( I \) with any initial conditions \( {J}^{i}\left( a\right) = {X}^{i},{V}^{i}\left( a\right) = {Y}^{i} \) .
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Yes
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Corollary 10.5. Along any geodesic \( \gamma \), the set of Jacobi fields is a \( {2n} \) - dimensional linear subspace of \( \mathcal{T}\left( \gamma \right) \) .
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Proof. Let \( p = \gamma \left( a\right) \) be any point on \( \gamma \), and consider the map from the set of Jacobi fields along \( \gamma \) to \( {T}_{p}M \oplus {T}_{p}M \) by sending \( J \) to \( \left( {J\left( a\right) ,{D}_{t}J\left( a\right) }\right) \) . The preceding proposition says precisely that this map is bijective.
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Yes
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Lemma 10.6. Let \( \gamma : I \rightarrow M \) be a geodesic, and \( a \in I \) .\n\n(a) A Jacobi field J along \( \gamma \) is normal if and only if\n\n\[ J\left( a\right) \bot \dot{\gamma }\left( a\right) \text{ and }{D}_{t}J\left( a\right) \bot \dot{\gamma }\left( a\right) . \]
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Proof. Using compatibility with the metric and the fact that \( {D}_{t}\dot{\gamma } \equiv 0 \), we compute\n\n\[ \frac{{d}^{2}}{d{t}^{2}}\langle J,\dot{\gamma }\rangle = \left\langle {{D}_{t}^{2}J,\dot{\gamma }}\right\rangle \]\n\n\[ = - \langle R\left( {J,\dot{\gamma }}\right) \dot{\gamma },\dot{\gamma }\rangle \]\n\n\[ = - \operatorname{Rm}\left( {J,\dot{\gamma },\dot{\gamma },\dot{\gamma }}\right) = 0 \]\n\nby the symmetries of the curvature tensor. Thus, by elementary calculus, \( f\left( t\right) \mathrel{\text{:=}} \langle J\left( t\right) ,\dot{\gamma }\left( t\right) \rangle \) is a linear function of \( t \) . Note that \( f\left( a\right) = \langle J\left( a\right) ,\dot{\gamma }\left( a\right) \rangle \) and \( \dot{f}\left( a\right) = \left\langle {{D}_{t}J\left( a\right) ,\dot{\gamma }\left( a\right) }\right\rangle \) . Thus \( J\left( a\right) \) and \( {D}_{t}J\left( a\right) \) are orthogonal to \( \dot{\gamma }\left( a\right) \) if and only if \( f \) and its first derivative vanish at \( a \), which happens if and only if \( f \equiv 0 \) . Similarly, if \( J \) is orthogonal to \( \dot{\gamma } \) at two points, then \( f \) vanishes at two points and is therefore identically zero.
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Yes
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Lemma 10.7. Let \( p \in M \), let \( \left( {x}^{i}\right) \) be normal coordinates on a neighborhood \( \mathcal{U} \) of \( p \), and let \( \gamma \) be a radial geodesic starting at \( p \) . For any \( W = {W}^{i}{\partial }_{i} \in {T}_{p}M \), the Jacobi field \( J \) along \( \gamma \) such that \( J\left( 0\right) = 0 \) and \( {D}_{t}J\left( 0\right) = W \) (see Figure 10.4) is given in normal coordinates by the formula\n\n\[ J\left( t\right) = t{W}^{i}{\partial }_{i} \]
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Proof. An easy computation using formula (4.10) for covariant derivatives in coordinates shows that \( J \) satisfies the specified initial conditions, so it suffices to show that \( J \) is a Jacobi field. If we set \( V = \dot{\gamma }\left( 0\right) \in {T}_{p}M \), then we know from Lemma 5.11 that \( \gamma \) is given in coordinates by the formula \( \gamma \left( t\right) = \left( {t{V}^{1},\ldots, t{V}^{n}}\right) \) . Now consider the variation \( \Gamma \) given in coordinates by\n\n\[ \Gamma \left( {s, t}\right) = \left( {t\left( {{V}^{1} + s{W}^{1}}\right) ,\ldots, t\left( {{V}^{n} + s{W}^{n}}\right) }\right) .\n\nAgain using Lemma 5.11, we see that \( \Gamma \) is a variation through geodesics. Therefore its variation field \( {\partial }_{s}\Gamma \left( {0, t}\right) \) is a Jacobi field. Differentiating \( \Gamma \left( {s, t}\right) \) with respect to \( s \) shows that its variation field is \( J\left( t\right) \).
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Yes
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Lemma 10.8. Suppose \( \\left( {M, g}\\right) \) is a Riemannian manifold with constant sectional curvature \( C \), and \( \\gamma \) is a unit speed geodesic in \( M \). The normal Jacobi fields along \( \\gamma \) vanishing at \( t = 0 \) are precisely the vector fields\n\n\[ J\\left( t\\right) = u\\left( t\\right) E\\left( t\\right) \]\n\nwhere \( E \) is any parallel normal vector field along \( \\gamma \), and \( u\\left( t\\right) \) is given by\n\n\[ u\\left( t\\right) = \\left\\{ \\begin{array}{ll} t, & C = 0 \\\\ R\\sin \\frac{t}{R}, & C = \\frac{1}{{R}^{2}} > 0 \\\\ R\\sinh \\frac{t}{R}, & C = - \\frac{1}{{R}^{2}} < 0 \\end{array}\\right. \]
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Proof. Since \( g \) has constant curvature, its curvature endomorphism is given by the formula of Lemma 8.10:\n\n\[ R\\left( {X, Y}\\right) Z = C\\left( {\\langle Y, Z\\rangle X-\\langle X, Z\\rangle Y}\\right) \]\n\nSubstituting this into the Jacobi equation, we find that a normal Jacobi field \( J \) satisfies\n\n\[ 0 = {D}_{t}^{2}J + C\\left( {\\langle \\dot{\\gamma },\\dot{\\gamma }\\rangle J-\\langle J,\\dot{\\gamma }\\rangle \\dot{\\gamma }}\\right) \]\n\n\[ = {D}_{t}^{2}J + {CJ} \]\n\nwhere we have used the facts that \( {\\left| \\dot{\\gamma }\\right| }^{2} = 1 \) and \( \\langle J,\\dot{\\gamma }\\rangle = 0 \).\n\nSince (10.7) says that the second covariant derivative of \( J \) is a multiple of \( J \) itself, it is reasonable to try to construct a solution by choosing a parallel normal vector field \( E \) along \( \\gamma \) and setting \( J\\left( t\\right) = u\\left( t\\right) E\\left( t\\right) \) for some function \( u \) to be determined. Plugging this into (10.7), we find that \( J \) is a Jacobi field provided \( u \) is a solution to the differential equation\n\n\[ \\ddot{u}\\left( t\\right) + {Cu}\\left( t\\right) = 0 \]\n\nIt is an easy matter to solve this ODE explicitly. In particular, the solutions satisfying \( u\\left( 0\\right) = 0 \) are constant multiples of the functions given in (10.6). This construction yields all the normal Jacobi fields vanishing at 0, since there is an \( \\left( {n - 1}\\right) \)-dimensional space of them, and the space of parallel normal vector fields has the same dimension.
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Yes
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Proposition 10.9. Suppose \( \left( {M, g}\right) \) is a Riemannian manifold with constant sectional curvature \( C \) . Let \( \left( {x}^{i}\right) \) be Riemannian normal coordinates on a normal neighborhood \( \mathcal{U} \) of \( p \in M \), let \( {\left| \cdot \right| }_{\bar{q}} \) be the Euclidean norm in these coordinates, and let \( r \) be the radial distance function. For any \( q \in \mathcal{U} - \{ p\} \) and \( V \in {T}_{q}M \), write \( V = {V}^{\top } + {V}^{ \bot } \), where \( {V}^{\top } \) is tangent to the sphere \( \{ r = \) constant \( \} \) through \( q \) and \( {V}^{ \bot } \) is a multiple of \( \partial /\partial r \) . The metric \( g \) can be written\n\n\[ g\left( {V, V}\right) = \left\{ \begin{array}{ll} {\left| {V}^{ \bot }\right| }_{\bar{g}}^{2} + {\left| {V}^{\top }\right| }_{\bar{g}}^{2}, & K = 0; \\ {\left| {V}^{ \bot }\right| }_{\bar{g}}^{2} + \frac{{R}^{2}}{{r}^{2}}\left( {{\sin }^{2}\frac{r}{R}}\right) {\left| {V}^{\top }\right| }_{\bar{g}}^{2}, & C = \frac{1}{{R}^{2}} > 0; \\ {\left| {V}^{ \bot }\right| }_{\bar{g}}^{2} + \frac{{R}^{2}}{{r}^{2}}\left( {{\sinh }^{2}\frac{r}{R}}\right) {\left| {V}^{\top }\right| }_{\bar{g}}^{2}, & C = - \frac{1}{{R}^{2}} < 0. \end{array}\right. \]
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Proof. By the Gauss lemma, the decomposition \( V = {V}^{\top } + {V}^{ \bot } \) is orthogonal, so \( {\left| V\right| }_{g}^{2} = {\left| {V}^{ \bot }\right| }_{g}^{2} + {\left| {V}^{\top }\right| }_{g}^{2} \) . Since \( \partial /\partial r \) is a unit vector in both the \( g \) and \( \bar{g} \) norms, it is immediate that \( {\left| {V}^{ \bot }\right| }_{g} = {\left| {V}^{ \bot }\right| }_{\bar{g}} \) . Thus we need only compute \( {\left| {V}^{\top }\right| }_{g} \).\n\nSet \( X = {V}^{\top } \), and let \( \gamma \) denote the unit speed radial geodesic from \( p \) to \( q \) . By Lemma 10.7, \( X \) is the value of a Jacobi field \( J \) along \( \gamma \) that vanishes at \( p \) (Figure 10.5), namely \( X = J\left( r\right) \), where \( r = d\left( {p, q}\right) \) and\n\n\[ J\left( t\right) = \frac{t}{r}{X}^{i}{\partial }_{i} \]\n\n(10.9)\n\nBecause \( J \) is orthogonal to \( \dot{\gamma } \) at \( p \) and \( q \), it is normal by Lemma 10.6.\n\nNow \( J \) can also be written in the form \( J\left( t\right) = u\left( t\right) E\left( t\right) \) as in Lemma 10.8. In this representation,\n\n\[ {D}_{t}J\left( 0\right) = \dot{u}\left( 0\right) E\left( 0\right) = E\left( 0\right) \]\nsince \( \dot{u}\left( 0\right) = 1 \) in each of the cases of (10.6). Therefore, since \( E \) is parallel and thus of constant length,\n\n\[ {\left| X\right| }^{2} = {\left| J\left( r\right) \right| }^{2} = {\left| u\left( r\right) \right| }^{2}{\left| E\left( r\right) \right| }^{2} = {\left| u\left( r\right) \right| }^{2}{\left| E\left( 0\right) \right| }^{2} = {\left| u\left( r\right) \right| }^{2}{\left| {D}_{t}J\left( 0\right) \right| }^{2}. \]\n\n(10.10)\n\nObserve that \( {D}_{t}J\left( 0\right) = \left( {1/r}\right) {\left. {X}^{i}{\partial }_{i}\right| }_{p} \) by (10.9). Since \( g \) agrees with \( \bar{g} \) at \( p \) , we have\n\n\[ \left| {{D}_{t}J\left( 0\right) }\right| = {\left. \frac{1}{r}{\left| {X}^{i}{\partial }_{i}\right| }_{p}\right| }_{g} = \frac{1}{r}{\left| X\right| }_{\bar{g}} \]\n\nInserting this into (10.10) and using formula (10.6) for \( u\left( r\right) \) completes the proof.
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Yes
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Proposition 10.10. (Local Uniqueness of Constant Curvature Metrics) Let \( \left( {M, g}\right) \) and \( \left( {\widetilde{M},\widetilde{g}}\right) \) be Riemannian manifolds with constant sectional curvature \( C \) . For any points \( p \in M,\widetilde{p} \in \widetilde{M} \), there exist neighborhoods \( \mathcal{U} \) of \( p \) and \( \widetilde{\mathcal{U}} \) of \( \widetilde{p} \) and an isometry \( F : \mathcal{U} \rightarrow \widetilde{\mathcal{U}} \) .
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Proof. Choose \( p \in M \) and \( \widetilde{p} \in \widetilde{M} \), and let \( \mathcal{U} \) and \( \widetilde{\mathcal{U}} \) be geodesic balls of small radius \( \varepsilon \) around \( p \) and \( \widetilde{p} \), respectively. Riemannian normal coordinates give maps \( \varphi : \mathcal{U} \rightarrow {B}_{\varepsilon }\left( 0\right) \subset {\mathbf{R}}^{n} \) and \( \widetilde{\varphi } : \widetilde{\mathcal{U}} \rightarrow {B}_{\varepsilon }\left( 0\right) \subset {\mathbf{R}}^{n} \), under which both metrics are given by (10.8) (Figure 10.6). Therefore \( {\widetilde{\varphi }}^{-1} \circ \varphi \) is the required local isometry.
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Yes
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Proposition 10.11. Suppose \( p \in M, V \in {T}_{p}M \), and \( q = {\exp }_{p}V \) . Then \( {\exp }_{p} \) is a local diffeomorphism in a neighborhood of \( V \) if and only if \( q \) is not conjugate to \( p \) along the geodesic \( \gamma \left( t\right) = {\exp }_{p}{tV},\;t \in \left\lbrack {0,1}\right\rbrack \) .
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Proof. By the inverse function theorem, \( {\exp }_{p} \) is a local diffeomorphism near \( V \) if and only if \( {\left( {\exp }_{p}\right) }_{ * } \) is an isomorphism at \( V \), and by dimensional considerations, this occurs if and only if \( {\left( {\exp }_{p}\right) }_{ * } \) is injective at \( V \) .\n\nIdentifying \( {T}_{V}\left( {{T}_{p}M}\right) \) with \( {T}_{p}M \) as usual, we can compute the push-forward \( {\left( {\exp }_{p}\right) }_{ * } \) at \( V \) as follows:\n\n\[ \n{\left. {\left( {\exp }_{p}\right) }_{ * }W = \frac{d}{ds}\right| }_{s = 0}{\exp }_{p}\left( {V + {sW}}\right) \n\]\n\nTo compute this, we define a variation of \( \gamma \) through geodesics (Figure 10.9) by\n\n\[ \n{\Gamma }_{W}\left( {s, t}\right) = {\exp }_{p}t\left( {V + {sW}}\right) \n\]\n\nThen the variation field \( {J}_{W}\left( t\right) = {\partial }_{s}{\Gamma }_{W}\left( {0, t}\right) \) is a Jacobi field along \( \gamma \), and\n\n\[ \n{J}_{W}\left( 1\right) = {\left( {\exp }_{p}\right) }_{ * }W \n\]\n\nSince \( W \in {T}_{p}M \) is arbitrary, there is an \( n \) -dimensional space of such Jacobi fields, and so these are all the Jacobi fields along \( \gamma \) that vanish at \( p \) . (If \( \gamma \) is contained in a normal neighborhood, these are just the Jacobi fields of the form (10.4) in normal coordinates.)\n\nTherefore, \( {\left( {\exp }_{p}\right) }_{ * } \) fails to be an isomorphism at \( V \) when there is a vector \( W \) such that \( {\left( ex{p}_{p}\right) }_{ * }W = 0 \), which occurs precisely when there is a Jacobi field \( {J}_{W} \) along \( \gamma \) with \( {J}_{W}\left( 0\right) = {J}_{W}\left( q\right) = 0 \) .
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Yes
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Corollary 10.13. If \( \Gamma \) is a proper variation of a unit speed geodesic \( \gamma \) whose variation field is a proper normal vector field \( V \), the second variation of \( L\left( {\Gamma }_{s}\right) \) is \( I\left( {V, V}\right) \) . In particular, if \( \gamma \) is minimizing, then \( I\left( {V, V}\right) \geq 0 \) for any proper normal vector field along \( \gamma \) .
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Null
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No
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Proposition 10.14. For any pair of proper normal vector fields \( V, W \) along a geodesic segment \( \gamma \) ,
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Proof. On any subinterval \( \left\lbrack {{a}_{i - 1},{a}_{i}}\right\rbrack \) where \( V \) and \( W \) are smooth,\n\n\[ \frac{d}{dt}\left\langle {{D}_{t}V, W}\right\rangle = \left\langle {{D}_{t}^{2}V, W}\right\rangle + \left\langle {{D}_{t}V,{D}_{t}W}\right\rangle . \]\n\nThus, by the fundamental theorem of calculus,\n\n\[ {\int }_{{a}_{i - 1}}^{{a}_{i}}\left\langle {{D}_{t}V,{D}_{t}W}\right\rangle {dt} = - {\int }_{{a}_{i - 1}}^{{a}_{i}}\left\langle {{D}_{t}^{2}V, W}\right\rangle + {\left. \left\langle {D}_{t}V, W\right\rangle \right| }_{{a}_{i - 1}}^{{a}_{i}}. \]\n\nSumming over \( i \), and noting that \( W \) is continuous at \( t = {a}_{i} \) and \( W\left( a\right) = \) \( W\left( b\right) = 0 \), we get (10.16).
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Yes
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If \( \gamma \) is a geodesic segment from \( p \) to \( q \) that has an interior conjugate point to \( p \), then there exists a proper normal vector field \( X \) along \( \gamma \) such that \( I\left( {X, X}\right) < 0 \) . In particular, \( \gamma \) is not minimizing.
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Proof. Suppose \( \gamma : \left\lbrack {0, b}\right\rbrack \rightarrow M \) is a unit speed parametrization of \( \gamma \), and \( \gamma \left( a\right) \) is conjugate to \( \gamma \left( 0\right) \) for some \( 0 < a < b \) . This means there is a nontrivial normal Jacobi field \( J \) along \( {\left. \gamma \right| }_{\left\lbrack 0, a\right\rbrack } \) that vanishes at \( t = 0 \) and \( t = a \) . Define a vector field \( V \) along all of \( \gamma \) by\n\n\[ V\left( t\right) = \left\{ \begin{array}{ll} J\left( t\right) , & t \in \left\lbrack {0, a}\right\rbrack \\ 0, & t \in \left\lbrack {a, b}\right\rbrack \end{array}\right. \]\n\nThis is a proper, normal, piecewise smooth vector field along \( \gamma \) .\n\nLet \( W \) be a smooth proper normal vector field along \( \gamma \) such that \( W\left( b\right) \) is equal to the jump \( \Delta {D}_{t}V \) at \( t = b \) (Figure 10.10). Such a vector field is easily constructed in local coordinates and extended to all of \( \gamma \) by a bump function. Note that \( \Delta {D}_{t}V = - {D}_{t}J\left( b\right) \) is not zero, because otherwise \( J \) would be a Jacobi field satisfying \( J\left( b\right) = {D}_{t}J\left( b\right) = 0 \), and thus would be identically zero.\n\nFor small positive \( \varepsilon \), let \( {X}_{\varepsilon } = V + {\varepsilon W} \) . Then\n\n\[ I\left( {{X}_{\varepsilon },{X}_{\varepsilon }}\right) = I\left( {V + {\varepsilon W}, V + {\varepsilon W}}\right) \]\n\n\[ = I\left( {V, V}\right) + {2\varepsilon I}\left( {V, W}\right) + {\varepsilon }^{2}I\left( {W, W}\right) . \]\n\nSince \( V \) satisfies the Jacobi equation on each subinterval \( \left\lbrack {0, a}\right\rbrack \) and \( \left\lbrack {a, b}\right\rbrack \) , and \( V\left( a\right) = 0,\left( {10.16}\right) \) gives\n\n\[ I\left( {V, V}\right) = - \left\langle {\Delta {D}_{t}V, V\left( a\right) }\right\rangle = 0. \]\n\nSimilarly,\n\n\[ I\left( {V, W}\right) = - \left\langle {\Delta {D}_{t}V, W\left( b\right) }\right\rangle = - {\left| W\left( b\right) \right| }^{2}. \]\n\nThus\n\n\[ I\left( {{X}_{\varepsilon },{X}_{\varepsilon }}\right) = - {2\varepsilon }{\left| W\left( b\right) \right| }^{2} + {\varepsilon }^{2}I\left( {W, W}\right) . \]\n\nIf we choose \( \varepsilon \) small enough, this is strictly negative.
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Yes
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Theorem 11.1. (Sturm Comparison Theorem) Suppose \( u \) and \( v \) are differentiable real-valued functions on \( \left\lbrack {0, T}\right\rbrack \), twice differentiable on \( \left( {0, T}\right) \) , and \( u > 0 \) on \( \left( {0, T}\right) \) . Suppose further that \( u \) and \( v \) satisfy\n\n\[ \ddot{u}\left( t\right) + a\left( t\right) u\left( t\right) = 0 \]\n\n\[ \ddot{v}\left( t\right) + a\left( t\right) v\left( t\right) \geq 0 \]\n\n\[ u\left( 0\right) = v\left( 0\right) = 0,\;\dot{u}\left( 0\right) = \dot{v}\left( 0\right) > 0 \]\n\nfor some function \( a : \left\lbrack {0, T}\right\rbrack \rightarrow \mathbf{R} \) . Then \( v\left( t\right) \geq u\left( t\right) \) on \( \left\lbrack {0, T}\right\rbrack \) .
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Proof. Consider the function \( f\left( t\right) = v\left( t\right) /u\left( t\right) \) defined on \( \left( {0, T}\right) \) . It follows from l’Hôpital’s rule that \( \mathop{\lim }\limits_{{t \rightarrow 0}}f\left( t\right) = \dot{v}\left( 0\right) /\dot{u}\left( 0\right) = 1 \) . Since \( f \) is differentiable on \( \left( {0, T}\right) \), if we could show that \( \dot{f} \geq 0 \) there it would follow from elementary calculus that \( f \geq 1 \) and therefore \( v \geq u \) on \( \left( {0, T}\right) \), and by continuity also on \( \left\lbrack {0, T}\right\rbrack \) . Differentiating,\n\n\[ \frac{d}{dt}\left( \frac{v}{u}\right) = \frac{\dot{v}u - v\dot{u}}{{u}^{2}} \]\n\nThus to show \( \dot{f} \geq 0 \) it would suffice to show \( \dot{v}u - v\dot{u} \geq 0 \) . Since \( \dot{v}\left( 0\right) u\left( 0\right) - \) \( v\left( 0\right) \dot{u}\left( 0\right) = 0 \), we need only show this expression has nonnegative derivative. Differentiating again and substituting the ODE for \( u \) ,\n\n\[ \frac{d}{dt}\left( {\dot{v}u - v\dot{u}}\right) = \ddot{v}u + \dot{v}\dot{u} - \dot{v}\dot{u} - v\ddot{u} = \ddot{v}u + {avu} \geq 0. \]\n\nThis proves the theorem.
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Yes
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Corollary 11.3. (Conjugate Point Comparison Theorem) Suppose all sectional curvatures of \( \left( {M, g}\right) \) are bounded above by a constant \( C \) . If\n\n\( C \leq 0 \), then no point of \( M \) has conjugate points along any geodesic. If \( C = 1/{R}^{2} > 0 \), then the first conjugate point along any geodesic occurs at a distance of at least \( {\pi R} \) .
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Proof. If \( C \leq 0 \), the Jacobi field comparison theorem implies that any nontrivial normal Jacobi field vanishing at \( t = 0 \) satisfies \( \left| {J\left( t\right) }\right| > 0 \) for all \( t > 0 \) . Similarly, if \( C > 0 \), then \( \left| {J\left( t\right) }\right| \geq \left( \text{constant}\right) \sin \left( {t/R}\right) > 0 \) for \( 0 < t < {\pi R} \) .
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Yes
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Corollary 11.4. (Metric Comparison Theorem) Suppose all sectional curvatures of \( \left( {M, g}\right) \) are bounded above by a constant \( C \) . In any normal coordinate chart, \( g\left( {V, V}\right) \geq {g}_{C}\left( {V, V}\right) \), where \( {g}_{C} \) is the constant curvature metric given by formula (10.8).
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Proof. Decomposing a vector \( V \) into components \( {V}^{\top } \) tangent to the geodesic sphere and \( {V}^{ \bot } \) tangent to the radial geodesics as in the proof of Proposition 10.9 gives\n\n\[ g\left( {V, V}\right) = g\left( {{V}^{ \bot },{V}^{ \bot }}\right) + g\left( {{V}^{\top },{V}^{\top }}\right) .\n\]\n\nJust as in that proof, \( g\left( {{V}^{ \bot },{V}^{ \bot }}\right) = \bar{g}\left( {{V}^{ \bot },{V}^{ \bot }}\right) = {g}_{C}\left( {{V}^{ \bot },{V}^{ \bot }}\right) \) . Also, \( {V}^{\top } \) is the value of some normal Jacobi field vanishing at \( t = 0 \), so the Jacobi field comparison theorem gives \( g\left( {{V}^{\top },{V}^{\top }}\right) \geq {g}_{C}\left( {{V}^{\top },{V}^{\top }}\right) \) .
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Yes
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Theorem 11.5. (The Cartan-Hadamard Theorem) If \( M \) is a complete, connected manifold all of whose sectional curvatures are nonpositive, then for any point \( p \in M,{\exp }_{p} : {T}_{p}M \rightarrow M \) is a covering map. In particular, the universal covering space of \( M \) is diffeomorphic to \( {\mathbf{R}}^{n} \) . If \( M \) is simply connected, then \( M \) itself is diffeomorphic to \( {\mathbf{R}}^{n} \) .
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Proof. The assumption of nonpositive curvature guarantees that \( p \) has no conjugate points along any geodesic, which can be shown by using either the conjugate point comparison theorem above or Problem 10-2. Therefore, by Proposition 10.11, \( {\exp }_{p} \) is a local diffeomorphism on all of \( {T}_{p}M \) . Let \( \widetilde{g} \) be the (variable-coefficient) 2-tensor field \( {\exp }_{p}^{ * }g \) defined on \( {T}_{p}M \) . Because \( {\exp }_{p}^{ * } \) is everywhere nonsingular, \( \widetilde{g} \) is a Riemannian metric, and \( {\exp }_{p} : \left( {{T}_{p}M,\widetilde{g}}\right) \rightarrow \left( {M, g}\right) \) is a local isometry. It then follows from Lemma 11. \( \dot{6} \) below that \( {\exp }_{p} \) is a covering map. The remaining statements of the theorem follow immediately from uniqueness of the universal covering space.
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Yes
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Lemma 11.6. Suppose \( \widetilde{M} \) and \( M \) are connected Riemannian manifolds, with \( \widetilde{M} \) complete, and \( \pi : \widetilde{M} \rightarrow M \) is a local isometry. Then \( M \) is complete and \( \pi \) is a covering map.
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Proof. A fundamental property of covering maps is the path-lifting property: any continuous path \( \gamma \) in \( M \) lifts to a path \( \widetilde{\gamma } \) in \( \widetilde{M} \) such that \( \pi \circ \widetilde{\gamma } = \gamma \) . We begin by proving that \( \pi \) possesses the path-lifting property for geodesics: If \( p \in M,\widetilde{p} \in {\pi }^{-1}\left( p\right) \), and \( \gamma : I \rightarrow M \) is a geodesic starting at \( p \), then \( \gamma \) has a unique lift starting at \( \widetilde{p} \) (Figure 11.1). The lifted curve is necessarily also a geodesic because \( \pi \) is a local isometry.\n\nTo prove the path-lifting property for geodesics, let \( V = \dot{\gamma }\left( 0\right) \) and \( \widetilde{V} = \) \( {\pi }_{ * }^{-1}\dot{\gamma }\left( 0\right) \in {T}_{\widetilde{p}}\widetilde{M} \) (which is well defined because \( {\pi }_{ * } \) is an isomorphism at each point), and let \( \widetilde{\gamma } \) be the geodesic in \( \widetilde{M} \) with initial point \( \widetilde{p} \) and initial velocity \( \widetilde{V} \) . Because \( \widetilde{M} \) is complete, \( \widetilde{\gamma } \) is defined for all time. Since \( \pi \) is a local isometry, it takes geodesics to geodesics; and since by construction \( \pi \left( {\widetilde{\gamma }\left( 0\right) }\right) = \gamma \left( 0\right) \) and \( {\pi }_{ * }\dot{\widetilde{\gamma }}\left( 0\right) = \dot{\gamma }\left( 0\right) \), we must have \( \pi \circ \widetilde{\gamma } = \gamma \) on \( I \) . In particular, \( \pi \circ \widetilde{\gamma } \) is a geodesic defined for all \( t \) that coincides with \( \gamma \) on \( I \) , so \( \gamma \) extends to all of \( \mathbf{R} \) and thus \( M \) is complete.\n\nNext we show that \( \pi \) is surjective. Choose some point \( \widetilde{p} \in \widetilde{M} \), write \( p = \) \( \pi \left( \widetilde{p}\right) \), and let \( q \in M \) be arbitrary. Because \( M \) is connected and complete, there is a minimizing geodesic segment \( \gamma \) from \( p \) to \( q \) . Letting \( \widetilde{\gamma } \) be the lift of \( \gamma \) starting at \( \widetilde{p} \) and \( r = d\left( {p, q}\right) \), we have \( \pi \left( {\widetilde{\gamma }\left( r\right) }\right) = \gamma \left( r\right) = q \), so \( q \) is in the image of \( \pi \) .
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Yes
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Theorem 11.8. (Myers’s Theorem) Suppose \( M \) is a complete, connected Riemannian n-manifold whose Ricci tensor satisfies the following inequality for all \( V \in {TM} \) :\n\n\[ \operatorname{Rc}\left( {V, V}\right) \geq \frac{n - 1}{{R}^{2}}{\left| V\right| }^{2} \]\n\nThen \( M \) is compact, with a finite fundamental group, and diameter at most \( {\pi R} \) .
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Proof. As in the proof of Bonnet's theorem, it suffices to prove the diameter estimate. As before, let \( \gamma \) be a minimizing unit speed geodesic segment of\n\nlength \( L > {\pi R} \) . Let \( \left( {{E}_{1},\ldots ,{E}_{n}}\right) \) be a parallel orthonormal frame along \( \gamma \) such that \( {E}_{n} = \dot{\gamma } \), and for each \( i = 1,\ldots, n - 1 \) let \( {V}_{i} \) be the proper normal vector field\n\n\[ {V}_{i}\left( t\right) = \left( {\sin \frac{\pi t}{L}}\right) {E}_{i}\left( t\right) \]\n\nBy the same computation as before,\n\n\[ I\left( {{V}_{i},{V}_{i}}\right) = {\int }_{0}^{L}\left( {{\sin }^{2}\frac{\pi t}{L}}\right) \left( {\frac{{\pi }^{2}}{{L}^{2}} - \operatorname{Rm}\left( {{E}_{i},\dot{\gamma },\dot{\gamma },{E}_{i}}\right) }\right) {dt}. \]\n\n(11.1)\n\nIn this case, we cannot conclude that each of these terms is negative. However, because \( \left\{ {E}_{i}\right\} \) is an orthonormal frame, the Ricci tensor at points along \( \gamma \) is given by\n\n\[ \operatorname{Rc}\left( {\dot{\gamma },\dot{\gamma }}\right) = \mathop{\sum }\limits_{{i = 1}}^{n}\operatorname{Rm}\left( {{E}_{i},\dot{\gamma },\dot{\gamma },{E}_{i}}\right) = \mathop{\sum }\limits_{{i = 1}}^{{n - 1}}\operatorname{Rm}\left( {{E}_{i},\dot{\gamma },\dot{\gamma },{E}_{i}}\right) \]\n\n(because \( {Rm}\left( {{E}_{n},\dot{\gamma },\dot{\gamma },{E}_{n}}\right) = {Rm}\left( {\dot{\gamma },\dot{\gamma },\dot{\gamma },\dot{\gamma }}\right) = 0 \) ). Therefore, summing (11.1) over \( i \) gives\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{{n - 1}}I\left( {{V}_{i},{V}_{i}}\right) = {\int }_{0}^{L}\left( {{\sin }^{2}\frac{\pi t}{L}}\right) \left( {\left( {n - 1}\right) \frac{{\pi }^{2}}{{L}^{2}} - \operatorname{Rc}\left( {\dot{\gamma },\dot{\gamma }}\right) }\right) {dt} \]\n\n\[ \leq {\int }_{0}^{L}\left( {{\sin }^{2}\frac{\pi t}{L}}\right) \left( {\frac{\left( {n - 1}\right) {\pi }^{2}}{{L}^{2}} - \frac{n - 1}{{R}^{2}}}\right) {dt} < 0. \]\n\nThis means at least one of the terms \( I\left( {{V}_{i},{V}_{i}}\right) \) must be negative, and again we have a contradiction to \( \gamma \) being minimizing.
|
Yes
|
Theorem 11.9. (Rauch Comparison Theorem) Let \( M \) and \( \widetilde{M} \) be Riemannian manifolds, let \( \gamma : \left\lbrack {0, T}\right\rbrack \rightarrow M \) and \( \widetilde{\gamma } : \left\lbrack {0, T}\right\rbrack \rightarrow \widetilde{M} \) be unit speed geodesic segments such that \( \widetilde{\gamma }\left( 0\right) \) has no conjugate points along \( \widetilde{\gamma } \), and let \( J,\widetilde{J} \) be normal Jacobi fields along \( \gamma \) and \( \widetilde{\gamma } \) such that \( J\left( 0\right) = \widetilde{J}\left( 0\right) = 0 \) and \( \left| {{D}_{t}J\left( 0\right) }\right| = \left| {{\widetilde{D}}_{t}\widetilde{J}\left( 0\right) }\right| \) (Figure 11.5). Suppose that the sectional curvatures of \( M \) and \( \widetilde{M} \) satisfy \( K\left( \Pi \right) \leq \widetilde{K}\left( \widetilde{\Pi }\right) \) whenever \( \Pi \subset {T}_{\gamma \left( t\right) }M \) is a 2- plane containing \( \dot{\gamma }\left( t\right) \) and \( \widetilde{\Pi } \subset {T}_{\widetilde{\gamma }\left( t\right) }\widetilde{M} \) is a 2-plane containing \( \dot{\widetilde{\gamma }}\left( t\right) \) . Then \( \left| {J\left( t\right) }\right| \geq \left| {\widetilde{J}\left( t\right) }\right| \) for all \( t \in \left\lbrack {0, T}\right\rbrack \) .
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You can find proofs in [dC92], [CE75], and [Spi79, volume 4]. Letting \( \widetilde{M} \) be one of our constant curvature model spaces, we recover the Jacobi field comparison theorem above. On the other hand, if instead we take \( M \) to have constant curvature, we get the same result with the inequalities reversed.
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No
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Theorem 11.10. (The Sphere Theorem) Suppose \( M \) is a complete, simply-connected, Riemannian n-manifold that is strictly \( \frac{1}{4} \) -pinched. Then \( M \) is homeomorphic to \( {\mathbf{S}}^{n} \) .
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The proof, which can be found in [CE75] or [dC92], is an elaborate application of the Rauch comparison theorem together with the Morse index theorem mentioned in Chapter 10. This result is sharp, at least in even dimensions, because the Fubini-Study metrics on complex projective spaces are \( \frac{1}{4} \) -pinched (Problem 8-12).
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Yes
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Theorem 11.11. (Hamilton) Suppose \( M \) is a simply-connected compact Riemannian 3-manifold with strictly positive Ricci curvature. Then \( M \) is diffeomorphic to \( {\mathbf{S}}^{3} \) .
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Null
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No
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Corollary 11.13. (Classification of Constant Curvature Metrics) Suppose \( M \) is a complete, connected Riemannian manifold with constant sectional curvature. Then \( M \) is isometric to \( \widetilde{M}/\Gamma \), where \( \widetilde{M} \) is one of the constant curvature model spaces \( {\mathbf{R}}^{n},{\mathbf{S}}_{R}^{n} \), or \( {\mathbf{H}}_{R}^{n} \), and \( \Gamma \) is a discrete subgroup of \( \mathcal{J}\left( \widetilde{M}\right) \), isomorphic to \( {\pi }_{1}\left( M\right) \), and acting freely and properly discontinuously on \( \widetilde{M} \) .
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Proof. If \( \pi : \widetilde{M} \rightarrow M \) is the universal covering space of \( M \) with the lifted metric \( \widetilde{g} = {\pi }^{ * }g \), the preceding theorem shows that \( \left( {\widetilde{M},\widetilde{g}}\right) \) is isometric to one of the model spaces. From covering space theory [Sie92, Mas67] it follows that the group \( \Gamma \) of covering transformations is isomorphic to \( {\pi }_{1}\left( M\right) \) and acts freely and properly discontinuously on \( \widetilde{M} \), and \( M \) is diffeomorphic to the quotient \( \widetilde{M}/\Gamma \) . Moreover, if \( \varphi \) is any covering transformation, \( \pi \circ \varphi = \pi \) , and so \( {\varphi }^{ * }\widetilde{g} = {\varphi }^{ * }{\pi }^{ * }g = {\pi }^{ * }g = \widetilde{g} \), so \( \Gamma \) acts by isometries. Finally, suppose \( \left\{ {\varphi }_{i}\right\} \subset \Gamma \) is an infinite set with an accumulation point in \( \mathcal{J}\left( \widetilde{M}\right) \) . Since the action of \( \Gamma \) is fixed-point free, for any point \( \widetilde{p} \in \widetilde{M} \) the set \( \left\{ {{\varphi }_{i}\left( \widetilde{p}\right) }\right\} \) is infinite, and by continuity of the action it has an accumulation point in \( \widetilde{M} \) . But this is impossible, since the points \( \left\{ {{\varphi }_{i}\left( \widetilde{p}\right) }\right\} \) all project to the same point in \( M \), and so form a discrete set. Thus \( \Gamma \) is discrete in \( \mathcal{J}\left( \widetilde{M}\right) \) .
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Yes
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Lemma 1. Let \( S \) have density \( \alpha \) and \( 0 \in S \) . Then \( S \oplus S \) has density at least \( {2\alpha } - {\alpha }^{2} \) .
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Proof. All the gaps in the set \( S \) are covered in part by the translation of \( S \) by the term of \( S \) just before this gap. Hence, at least the fraction \( \alpha \) of this gap gets covered. So from this covering we have density \( \alpha \) from \( S \) itself and \( \alpha \) times the gaps. Altogether, then, we indeed have \( \alpha + \alpha \left( {1 - \alpha }\right) = {2\alpha } - {\alpha }^{2} \), as claimed.
|
Yes
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Lemma 2. If \( S \) has density \( \alpha > \frac{1}{2} \), then \( S \oplus S \) contains all the positive integers.
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Proof. Fix an integer \( n \) which is arbitrary, let \( A \) be the subset of \( S \) which lies \( \leq n \), and let \( B \) be the set of all \( n \) minus elements of \( S \) . Since \( A \) contains more than \( n/2 \) elements and \( B \) contains at least \( n/2 \) elements, the Pigeonhole principle guarantees that they overlap. So suppose they overlap at \( k \) . Since \( k \in A \), we get \( k \in S \), and since \( k \in B \), we get \( n - k \in S \) . These are the two elements of \( S \) which sum to \( n \).
|
Yes
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Lemma 3. Let \( k > 1 \) be a fixed integer. There exists a \( {C}_{1} \) such that, for any positive integers \( N, a, b \) with \( \left( {a, b}\right) = 1 \) , \[ \left| {\mathop{\sum }\limits_{{n = 1}}^{N}e\left( {\frac{a}{b}{n}^{k}}\right) }\right| \leq {C}_{1}{N}^{1 + o\left( 1\right) }{b}^{-{2}^{1 - k}}. \]
|
Null
|
No
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Lemma 4. There exists \( \epsilon > 0 \) and \( {C}_{2} \) such that, throughout any interval \( {I}_{a, b, N} \) , \[ \left| {\mathop{\sum }\limits_{{n = 1}}^{N}e\left( {x{n}^{k}}\right) }\right| \leq \frac{{C}_{2}N}{{\left( b + j\right) }^{\epsilon }} \]
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Proof. This is almost trivial if \( b > {N}^{2/3} \), for, since the derivative of \( \left| {\mathop{\sum }\limits_{{n = 1}}^{N}e\left( {x{n}^{k}}\right) }\right| \) is bounded by \( {2\pi }{N}^{k + 1} \) , \[ \left| {\mathop{\sum }\limits_{{n = 1}}^{N}e\left( {x{n}^{k}}\right) }\right| \leq \left| {\mathop{\sum }\limits_{{n = 1}}^{N}e\left( {\frac{a}{b}{n}^{k}}\right) }\right| + \left| {x - \frac{a}{b}}\right| {2\pi }{N}^{k + 1} \] \[ \leq \frac{{N}^{1 + o\left( 1\right) }}{{b}^{\frac{1}{{2}^{k - 1}}}} + \frac{{2\pi }{N}^{3/2}}{b} \leq \frac{{N}^{1 + o\left( 1\right) }}{{b}^{\frac{1}{{2}^{k - 1}}}} + \frac{2\pi N}{{b}^{1/4}}, \] by \( \mathrm{C} \), which gives the result, since \( j = 0 \) automatically. Assume therefore that \( b \leq {N}^{2/3} \), and note the following two simple facts (A) and (B). For details see [K. Knopp, Theory and Application of Infinite Series, Blackie & Sons, Glasgow, 1946.] and [G. Pólya und G. Szegö, Aufgaben und Lehrsätze aus der Analysis, Dover Publications, New York 1945, Vol. 1, Part II, p. 37]. Q.E.D.
|
Yes
|
Theorem 1.1.4 Let \( {A}_{0},{A}_{1},{A}_{2},\ldots \) be countable sets. Then their union \( A = \mathop{\bigcup }\limits_{0}^{\infty }{A}_{n} \) is countable.
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Proof. For each \( n \), choose an enumeration \( {a}_{n0},{a}_{n1},{a}_{n2},\ldots \) of \( {A}_{n} \) . We enumerate \( A = \mathop{\bigcup }\limits_{n}{A}_{n} \) following the above diagonal method.
|
Yes
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Theorem 1.1.4 Let \( {A}_{0},{A}_{1},{A}_{2},\ldots \) be countable sets. Then their union \( A = \mathop{\bigcup }\limits_{0}^{\infty }{A}_{n} \) is countable.
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Proof. For each \( n \), choose an enumeration \( {a}_{n0},{a}_{n1},{a}_{n2},\ldots \) of \( {A}_{n} \) . We enumerate \( A = \mathop{\bigcup }\limits_{n}{A}_{n} \) following the above diagonal method.
|
Yes
|
Theorem 1.1.8 (Cantor) For any two real numbers \( a, b \) with \( a < b \), the interval \( \left\lbrack {a, b}\right\rbrack \) is uncountable.
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Proof. (Cantor) Let \( \left( {a}_{n}\right) \) be a sequence in \( \left\lbrack {a, b}\right\rbrack \) . Define an increasing sequence \( \left( {b}_{n}\right) \) and a decreasing sequence \( \left( {c}_{n}\right) \) in \( \left\lbrack {a, b}\right\rbrack \) inductively as follows: Put \( {b}_{0} = a \) and \( {c}_{0} = b \) . For some \( n \in \mathbb{N} \), suppose\n\n\[ \n{b}_{0} < {b}_{1} < \cdots < {b}_{n} < {c}_{n} < \cdots < {c}_{1} < {c}_{0} \n\] \n\nhave been defined. Let \( {i}_{n} \) be the first integer \( i \) such that \( {b}_{n} < {a}_{i} < {c}_{n} \) and \( {j}_{n} \) the first integer \( j \) such that \( {a}_{{i}_{n}} < {a}_{j} < {c}_{n} \) . Since \( \left\lbrack {a, b}\right\rbrack \) is infinite \( {i}_{n},{j}_{n} \) exist. Put \( {b}_{n + 1} = {a}_{{i}_{n}} \) and \( {c}_{n + 1} = {a}_{{j}_{n}} \) .\n\nLet \( x = \sup \left\{ {{b}_{n} : n \in \mathbb{N}}\right\} \) . Clearly, \( x \in \left\lbrack {a, b}\right\rbrack \) . Suppose \( x = {a}_{k} \) for some \( k \) . Clearly, \( x \leq {c}_{m} \) for all \( m \) . So, by the definition of the sequence \( \left( {b}_{n}\right) \) there is an integer \( i \) such that \( {b}_{i} > {a}_{k} = x \) . This contradiction shows that the range of the sequence \( \left( {a}_{n}\right) \) is not the whole of \( \left\lbrack {a, b}\right\rbrack \) . Since \( \left( {a}_{n}\right) \) was an arbitrary sequence, the result follows.
|
Yes
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Theorem 1.1.8 (Cantor) For any two real numbers \( a, b \) with \( a < b \), the interval \( \left\lbrack {a, b}\right\rbrack \) is uncountable.
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Proof. (Cantor) Let \( \left( {a}_{n}\right) \) be a sequence in \( \left\lbrack {a, b}\right\rbrack \) . Define an increasing sequence \( \left( {b}_{n}\right) \) and a decreasing sequence \( \left( {c}_{n}\right) \) in \( \left\lbrack {a, b}\right\rbrack \) inductively as follows: Put \( {b}_{0} = a \) and \( {c}_{0} = b \) . For some \( n \in \mathbb{N} \), suppose\n\n\[ \n{b}_{0} < {b}_{1} < \cdots < {b}_{n} < {c}_{n} < \cdots < {c}_{1} < {c}_{0} \n\] \n\nhave been defined. Let \( {i}_{n} \) be the first integer \( i \) such that \( {b}_{n} < {a}_{i} < {c}_{n} \) and \( {j}_{n} \) the first integer \( j \) such that \( {a}_{{i}_{n}} < {a}_{j} < {c}_{n} \) . Since \( \left\lbrack {a, b}\right\rbrack \) is infinite \( {i}_{n},{j}_{n} \) exist. Put \( {b}_{n + 1} = {a}_{{i}_{n}} \) and \( {c}_{n + 1} = {a}_{{j}_{n}} \) .\n\nLet \( x = \sup \left\{ {{b}_{n} : n \in \mathbb{N}}\right\} \) . Clearly, \( x \in \left\lbrack {a, b}\right\rbrack \) . Suppose \( x = {a}_{k} \) for some \( k \) . Clearly, \( x \leq {c}_{m} \) for all \( m \) . So, by the definition of the sequence \( \left( {b}_{n}\right) \) there is an integer \( i \) such that \( {b}_{i} > {a}_{k} = x \) . This contradiction shows that the range of the sequence \( \left( {a}_{n}\right) \) is not the whole of \( \left\lbrack {a, b}\right\rbrack \) . Since \( \left( {a}_{n}\right) \) was an arbitrary sequence, the result follows.
|
Yes
|
Theorem 1.1.9 The set \( \{ 0,1{\} }^{\mathbb{N}} \), consisting of all sequences of 0’s and 1’s, is uncountable.
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Proof. Let \( \left( {\alpha }_{n}\right) \) be a sequence in \( \{ 0,1{\} }^{\mathbb{N}} \) . Define \( \alpha \in \{ 0,1{\} }^{\mathbb{N}} \) by\n\n\[ \alpha \left( n\right) = 1 - {\alpha }_{n}\left( n\right), n \in \mathbb{N}. \]\n\nThen \( \alpha \neq {\alpha }_{i} \) for all \( i \) . Since \( \left( {\alpha }_{n}\right) \) was arbitrary, our result is proved.
|
Yes
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Theorem 1.2.1 (Cantor) For any set \( X, X{ < }_{c}\mathcal{P}\left( X\right) \) .
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Proof. First assume that \( X = \varnothing \) . Then \( \mathcal{P}\left( X\right) = \{ \varnothing \} \) . The only function on \( X \) is the empty function \( \varnothing \), which is not onto \( \{ \varnothing \} \) . This observation proves the result when \( X = \varnothing \) .\n\nNow assume that \( X \) is nonempty. The map \( x \rightarrow \{ x\} \) from \( X \) to \( \mathcal{P}\left( X\right) \) is one-to-one. Therefore, \( X{ \leq }_{c}\mathcal{P}\left( X\right) \) . Let \( f : X \rightarrow \mathcal{P}\left( X\right) \) be any map. We show that \( f \) cannot be onto \( \mathcal{P}\left( X\right) \) . This will complete the proof.\n\nConsider the set\n\n\[ A = \{ x \in X \mid x \notin f\left( x\right) \} .\n\]\n\nSuppose \( A = f\left( {x}_{0}\right) \) for some \( {x}_{0} \in X \) . Then\n\n\[ {x}_{0} \in A \Leftrightarrow {x}_{0} \notin A.\n\]\n\nThis contradiction proves our claim.
|
Yes
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Theorem 1.2.3 (Schröder - Bernstein Theorem) For any two sets \( X \) and \( Y \) , \[ \left( {X{ \leq }_{c}Y\& Y{ \leq }_{c}X}\right) \Rightarrow X \equiv Y. \]
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Proof. (Dedekind) Let \( X{ \leq }_{c}Y \) and \( Y{ \leq }_{c}X \) . Fix one-to-one maps \( f : X \rightarrow Y \) and \( g : Y \rightarrow X \) . We have to show that \( X \) and \( Y \) have the same cardinality; i.e., that there is a bijection \( h \) from \( X \) onto \( Y \) .\n\nWe first show that there is a set \( E \subseteq X \) such that \[ {g}^{-1}\left( {X \smallsetminus E}\right) = Y \smallsetminus f\left( E\right) \] \( \left( *\right) \)\n\n(See Figure 1.1.) Assuming that such a set \( E \) exists, we complete the proof as follows. Define \( h : X \rightarrow Y \) by \[ h\left( x\right) = \left\{ \begin{array}{ll} f\left( x\right) & \text{ if }x \in E \\ {g}^{-1}\left( x\right) & \text{ otherwise. } \end{array}\right. \] The map \( h : X \rightarrow Y \) is clearly seen to be one-to-one and onto.\n\nWe now show the existence of a set \( E \subseteq X \) satisfying \( \left( \star \right) \) . Consider the map \( \mathcal{H} : \mathcal{P}\left( X\right) \rightarrow \mathcal{P}\left( X\right) \) defined by \[ \mathcal{H}\left( A\right) = X \smallsetminus g\left( {Y \smallsetminus f\left( A\right) }\right) ,\;A \subseteq X. \] It is easy to check that (i) \( A \subseteq B \subseteq X \Rightarrow \mathcal{H}\left( A\right) \subseteq \mathcal{H}\left( B\right) \), and (ii) \( \mathcal{H}\left( {\mathop{\bigcup }\limits_{n}{A}_{n}}\right) = \mathop{\bigcup }\limits_{n}\mathcal{H}\left( {A}_{n}\right) \).\n\nNow define a sequence \( \left( {A}_{n}\right) \) of subsets of \( X \) inductively as follows: \( {A}_{0} = \varnothing \), and \( {A}_{n + 1} = \mathcal{H}\left( {A}_{n}\right), n = 0,1,2,\ldots \) Let \( E = \mathop{\bigcup }\limits_{n}{A}_{n} \) . Then, \( \mathcal{H}\left( E\right) = E \) . The set \( E \) clearly satisfies \( \left( \star \right) \).
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Yes
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Corollary 1.2.4 For sets \( A \) and \( B \) ,
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\[ A{ < }_{c}B \Leftrightarrow A{ \leq }_{c}B\& B{ \nleq }_{c}A. \]
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Yes
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Example 1.2.5 Define \( f : \mathcal{P}\left( \mathbb{N}\right) \rightarrow \mathbb{R} \), the set of all real numbers, by\n\n\[ f\left( A\right) = \mathop{\sum }\limits_{{n \in A}}\frac{2}{{3}^{n + 1}}, A \subseteq \mathbb{N}. \]\n\nThen \( f \) is one-to-one. Therefore, \( \mathcal{P}\left( \mathbb{N}\right) { \leq }_{c}\mathbb{R} \) .
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Now consider the map \( g \) :\n\n\( \mathbb{R} \rightarrow \mathcal{P}\left( \mathbb{Q}\right) \) by\n\n\[ g\left( x\right) = \{ r \in \mathbb{Q} \mid r < x\}, x \in \mathbb{R}. \]\n\nClearly, \( g \) is one-to-one and so \( \mathbb{R}{ \leq }_{c}\mathcal{P}\left( \mathbb{Q}\right) \) . As \( \mathbb{Q} \equiv \mathbb{N},\mathcal{P}\left( \mathbb{Q}\right) \equiv \mathcal{P}\left( \mathbb{N}\right) \) . Therefore, \( \mathbb{R}{ \leq }_{c}\mathcal{P}\left( \mathbb{N}\right) \) . By the Schröder - Bernstein theorem, \( \mathbb{R} \equiv \mathcal{P}\left( \mathbb{N}\right) \) . Since \( \mathcal{P}\left( \mathbb{N}\right) \equiv \{ 0,1{\} }^{\mathbb{N}},\mathbb{R} \equiv \{ 0,1{\} }^{\mathbb{N}} \) .
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Yes
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Example 1.2.6 Fix a one-to-one map \( x \rightarrow \left( {{x}_{0},{x}_{1},{x}_{2},\ldots }\right) \) from \( \mathbb{R} \) onto \( \{ 0,1{\} }^{\mathbb{N}} \), the set of sequences of 0 ’s and 1’s. Then the function \( \left( {x, y}\right) \rightarrow \) \( \left( {{x}_{0},{y}_{0},{x}_{1},{y}_{1},\ldots }\right) \) from \( {\mathbb{R}}^{2} \) to \( \{ 0,1{\} }^{\mathbb{N}} \) is one-to-one and onto. So, \( {\mathbb{R}}^{2} \equiv \) \( \{ 0,1{\} }^{\mathbb{N}} \equiv \mathbb{R} \) . By induction on the positive integers \( k \), we can now show that \( {\mathbb{R}}^{k} \) and \( \mathbb{R} \) are equinumerous.
|
Null
|
No
|
Theorem 1.3.1 If \( X \) is infinite and \( A \subseteq X \) finite, then \( X \smallsetminus A \) and \( X \) have the same cardinality.
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Proof. Let \( A = \left\{ {{a}_{0},{a}_{1},\ldots ,{a}_{n}}\right\} \) with the \( {a}_{i} \) ’s distinct. By \( \mathbf{{AC}} \), there exist distinct elements \( {a}_{n + 1},{a}_{n + 2},\ldots \) in \( X \smallsetminus A \) . To see this, fix a choice function \( f : \mathcal{P}\left( X\right) \smallsetminus \{ \varnothing \} \rightarrow X \) such that \( f\left( E\right) \in E \) for every nonempty subset \( E \) of \( X \) . Such a function exists by \( \mathbf{{AC}} \) . Now inductively define \( {a}_{n + 1},{a}_{n + 2},\ldots \) such that\n\n\[ \n{a}_{n + k + 1} = f\left( {X \smallsetminus \left\{ {{a}_{0},{a}_{1},\ldots ,{a}_{n + k}}\right\} }\right)\n\]\n\n\( k = 0,1,\ldots \) Define \( h : X \rightarrow X \smallsetminus A \) by\n\n\[ \nh\left( x\right) = \left\{ \begin{array}{ll} {a}_{n + k + 1} & \text{ if }x = {a}_{k} \\ x & \text{ otherwise. } \end{array}\right.\n\]\n\nClearly, \( h : X \rightarrow X \smallsetminus A \) is one-to-one and onto.
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Yes
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Corollary 1.3.2 Show that for any infinite set \( X,\mathbb{N}{ \leq }_{c}X \) ; i.e., every infinite set \( X \) has a countable infinite subset.
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Null
|
No
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Example 1.3.4 Let \( X \) and \( Y \) be any two sets. A partial function \( f \) : \( X \rightarrow Y \) is a function with domain a subset of \( X \) and range contained in \( Y \) . Let \( f : X \rightarrow Y \) and \( g : X \rightarrow Y \) be partial functions. We say that \( g \) extends \( f \), or \( f \) is a restriction of \( g \), written \( g \succcurlyeq f \) or \( f \preccurlyeq g \), if domain \( \left( f\right) \) is contained in \( \operatorname{domain}\left( g\right) \) and \( f\left( x\right) = g\left( x\right) \) for all \( x \in \operatorname{domain}\left( f\right) \) . If \( f \) is a restriction of \( g \) and \( \operatorname{domain}\left( f\right) = A \), we write \( f = g \mid A \) . Let\n\n\[ \n{Fn}\left( {X, Y}\right) = \{ f : f\text{ a one-to-one partial function from }X\text{ to }Y\} .\n\]\n\nSuppose \( Y \) has more than one element and \( X \neq \varnothing \) . Then \( \left( {{Fn}\left( {X, Y}\right) , \preccurlyeq }\right) \) is a poset that is not linearly ordered.
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Null
|
No
|
Example 1.3.5 Let \( V \) be a vector space over any field \( F \) and \( P \) the set of all independent subsets of \( V \) ordered by the inclusion \( \subseteq \) . Then \( P \) is a poset that is not a linearly ordered set.
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In 1.3.5, Let \( C \) be a chain in \( P \) . Then for any two elements \( E \) and \( F \) of \( P \), either \( E \subseteq F \) or \( F \subseteq E \) . It follows that \( \bigcup C \) itself is an independent set and so is an upper bound of \( C \) .
|
No
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Proposition 1.3.7 Every vector space \( V \) has a basis.
|
Proof. Let \( P \) be the poset defined in 1.3.5; i.e., \( P \) is the set of all independent subsets of \( V \) . Since every singleton set \( \{ v\}, v \neq 0 \), is an independent set, \( P \neq \varnothing \) . As shown earlier, every chain in \( P \) has an upper bound. Therefore, by Zorn’s lemma, \( P \) has a maximal element, say \( B \) . Suppose \( B \) does not span \( V \) . Take \( v \in V \smallsetminus \operatorname{span}\left( B\right) \) . Then \( B\bigcup \{ v\} \) is an independent set properly containing \( B \) . This contradicts the maximality of \( B \) . Thus \( B \) is a basis of \( V \) .
|
Yes
|
Theorem 1.4.1 For any two sets \( X \) and \( Y \), at least one of\n\n\[ X{ \leq }_{c}Y\text{or}Y{ \leq }_{c}X \]\n\nholds.
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Proof. Without loss of generality we can assume that both \( X \) and \( Y \) are nonempty. We need to show that either there exists a one-to-one map \( f : X \rightarrow Y \) or there exists a one-to-one map \( g : Y \rightarrow X \) . To show this, consider the poset \( {Fn}\left( {X, Y}\right) \) of all one-to-one partial functions from \( X \) to \( Y \) as defined in 1.3.4. It is clearly nonempty. As shown earlier, every chain in \( {Fn}\left( {X, Y}\right) \) has an upper bound. Therefore, by Zorn’s lemma, \( P \) has a maximal element, say \( {f}_{0} \) . Then, either \( \operatorname{domain}\left( {f}_{0}\right) = X \) or range \( \left( {f}_{0}\right) = Y \) . If \( \operatorname{domain}\left( {f}_{0}\right) = X \), then \( {f}_{0} \) is a one-to-one map from \( X \) to \( Y \) . So, in this case, \( X{ \leq }_{c}Y \) . If \( \operatorname{range}\left( {f}_{0}\right) = Y \), then \( {f}_{0}^{-1} \) is a one-to-one map from \( Y \) to \( X \), and so \( Y{ \leq }_{c}X \) .
|
Yes
|
Corollary 1.4.2 Let \( A \) and \( B \) be any two sets. Then exactly one of\n\n\[ A{ < }_{c}B, A \equiv B,\text{ and }B{ < }_{c}A \] \n\nholds.
|
Null
|
No
|
Theorem 1.4.3 For every infinite set \( X \) ,\n\n\[ X \times \{ 0,1\} \equiv X \]
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Proof. Let\n\n\[ P = \{ \left( {A, f}\right) : A \subseteq X\\text{ and }f : A \times \{ 0,1\} \rightarrow A\\text{ a bijection }\} .\n\]\n\nSince \( X \) is infinite, it contains a countably infinite set, say \( D \) . By 1.1.3, \( D \times \{ 0,1\} \equiv D \) . Therefore, \( P \) is nonempty. Consider the partial order \( \\propto \) on \( P \) defined by\n\n\[ \\left( {A, f}\right) \\propto \\left( {B, g}\right) \\Leftrightarrow A \\subseteq B\\& f \\preccurlyeq g.\n\]\n\nFollowing the argument contained in the proof of 1.4.1, we see that the hypothesis of Zorn’s lemma is satisfied by \( P \) . So, \( P \) has a maximal element, say \( \\left( {A, f}\right) \) .\n\nTo complete the proof we show that \( A \\equiv X \) . Since \( X \) is infinite, by 1.3.1, it will be sufficient to show that \( X \\smallsetminus A \) is finite. Suppose not. By 1.3.2, there is a \( B \\subseteq X \\smallsetminus A \) such that \( B \\equiv \\mathbb{N} \) . So there is a one-to-one map \( g \) from \( B \times \{ 0,1\} \) onto \( B \) . Combining \( f \) and \( g \) we get a bijection\n\n\[ h : \\left( {A\\bigcup B}\right) \times \{ 0,1\} \\rightarrow A\\bigcup B\n\]\n\nthat extends \( f \) . This contradicts the maximality of \( \\left( {A, f}\right) \) . Hence, \( X \\smallsetminus A \) is finite. Therefore, \( A \\equiv X \) . The proof is complete.
|
Yes
|
Theorem 1.4.5 For every infinite set \( X \) , \[ X \times X \equiv X. \]
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Proof. Let \[ P = \{ \left( {A, f}\right) : A \subseteq X\text{ and }f : A \times A \rightarrow A\text{ a bijection }\} . \] Note that \( P \) is nonempty. Consider the partial order \( \propto \) on \( P \) defined by \[ \left( {A, f}\right) \propto \left( {B, g}\right) \Leftrightarrow A \subseteq B\& f \preccurlyeq g. \] By Zorn’s lemma, take a maximal element \( \left( {A, f}\right) \) of \( P \) as in the proof of 1.4.3. Note that \( A \) must be infinite. To complete the proof, we shall show that \( A \equiv X \) . Suppose not. Then \( A{ < }_{c}X \) . We first show that \( X \smallsetminus A \equiv X \) . Suppose \( X \smallsetminus A{ < }_{c}X \) . By 1.4.1, either \( A{ \leq }_{c}X \smallsetminus A \) or \( X \smallsetminus A{ \leq }_{c}A \) . Assume first \( X \smallsetminus A{ \leq }_{c}A \) . Using 1.4.3, take two disjoint sets \( {A}_{1},{A}_{2} \) of the same cardinality as \( A \) and \( {A}_{1}\bigcup {A}_{2} = A \) . Now, \[ X = A\bigcup \left( {X \smallsetminus A}\right) { \leq }_{c}{A}_{1}\bigcup {A}_{2} \equiv A{ < }_{c}X. \] This is a contradiction. Similarly we arrive at a contradiction from the other inequality. Thus, by 1.4.2, \( X \smallsetminus A \equiv X \) . Now choose \( B \subseteq X \smallsetminus A \) such that \( B \equiv A \) . By 1.4.4, write \( B \) as the union of three disjoint sets, say \( {B}_{1},{B}_{2} \), and \( {B}_{3} \), each of the same cardinality as \( A \) . Since there is a one-to-one map from \( A \times A \) onto \( A \), there exist bijections \( {f}_{1} : B \times A \rightarrow {B}_{1},{f}_{2} : B \times B \rightarrow {B}_{2} \), and \( {f}_{3} : A \times B \rightarrow {B}_{3} \) . Let \( C = \) \( A\bigcup B \) . Combining these four bijections, we get a bijection \( g : C \times C \rightarrow C \) that is a proper extension of \( f \) . This contradicts the maximality of \( \left( {A, f}\right) \) . Thus, \( A \equiv X \) . The proof is now complete.
|
Yes
|
Theorem 1.4.5 For every infinite set \( X \) , \[ X \times X \equiv X. \]
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Proof. Let \[ P = \{ \left( {A, f}\right) : A \subseteq X\text{ and }f : A \times A \rightarrow A\text{ a bijection }\} . \] Note that \( P \) is nonempty. Consider the partial order \( \propto \) on \( P \) defined by \[ \left( {A, f}\right) \propto \left( {B, g}\right) \Leftrightarrow A \subseteq B\& f \preccurlyeq g. \] By Zorn’s lemma, take a maximal element \( \left( {A, f}\right) \) of \( P \) as in the proof of 1.4.3. Note that \( A \) must be infinite. To complete the proof, we shall show that \( A \equiv X \) . Suppose not. Then \( A{ < }_{c}X \) . We first show that \( X \smallsetminus A \equiv X \) . Suppose \( X \smallsetminus A{ < }_{c}X \) . By 1.4.1, either \( A{ \leq }_{c}X \smallsetminus A \) or \( X \smallsetminus A{ \leq }_{c}A \) . Assume first \( X \smallsetminus A{ \leq }_{c}A \) . Using 1.4.3, take two disjoint sets \( {A}_{1},{A}_{2} \) of the same cardinality as \( A \) and \( {A}_{1}\bigcup {A}_{2} = A \) . Now, \[ X = A\bigcup \left( {X \smallsetminus A}\right) { \leq }_{c}{A}_{1}\bigcup {A}_{2} \equiv A{ < }_{c}X. \] This is a contradiction. Similarly we arrive at a contradiction from the other inequality. Thus, by 1.4.2, \( X \smallsetminus A \equiv X \) . Now choose \( B \subseteq X \smallsetminus A \) such that \( B \equiv A \) . By 1.4.4, write \( B \) as the union of three disjoint sets, say \( {B}_{1},{B}_{2} \), and \( {B}_{3} \), each of the same cardinality as \( A \) . Since there is a one-to-one map from \( A \times A \) onto \( A \), there exist bijections \( {f}_{1} : B \times A \rightarrow {B}_{1},{f}_{2} : B \times B \rightarrow {B}_{2} \), and \( {f}_{3} : A \times B \rightarrow {B}_{3} \) . Let \( C = \) \( A\bigcup B \) . Combining these four bijections, we get a bijection \( g : C \times C \rightarrow C \) that is a proper extension of \( f \) . This contradicts the maximality of \( \left( {A, f}\right) \) . Thus, \( A \equiv X \) . The proof is now complete.
|
Yes
|
Proposition 1.4.8 (J. König,[58]) Let \( \\left\\{ {{X}_{i} : i \\in I}\\right\\} \) and \( \\left\\{ {{Y}_{i} : i \\in I}\\right\\} \) be families of sets such that \( {X}_{i}{ < }_{c}{Y}_{i} \) for each \( i \\in I \) . Then there is no map \( f \) from \( \\mathop{\\bigcup }\\limits_{i}{X}_{i} \) onto \( {\\Pi }_{i}{Y}_{i} \) .
|
Proof. Let \( f : \\mathop{\\bigcup }\\limits_{i}{X}_{i} \\rightarrow {\\Pi }_{i}{Y}_{i} \) be any map. For any \( i \\in I \\), let\n\n\[ \n{A}_{i} = {Y}_{i} \\smallsetminus {\\pi }_{i}\\left( {f\\left( {X}_{i}\\right) }\\right)\n\]\n\nwhere \( {\\pi }_{i} : \\mathop{\\prod }\\limits_{j}{Y}_{j} \\rightarrow {Y}_{i} \) is the projection map. Since for evry \( i,{X}_{i}{ < }_{c}{Y}_{i} \) , each \( {A}_{i} \) is nonempty. By \( \\mathbf{{AC}},{\\Pi }_{i}{A}_{i} \\neq \\varnothing \) . But\n\n\[ \n{\\Pi }_{i}{A}_{i}\\bigcap \\operatorname{range}\\left( f\\right) = \\varnothing .\n\]\n\nIt follows that \( f \) is not onto.
|
Yes
|
\[ {2}^{\mathfrak{c}} \leq {\aleph }_{0}^{\mathfrak{c}} \]
|
\[ {2}^{\mathfrak{c}} \leq {\aleph }_{0}^{\mathfrak{c}}\;\text{ (since }2 \leq {\aleph }_{0}\text{ ) } \] \[ \leq {\mathfrak{c}}^{\mathfrak{c}}\;\text{ (since }{\aleph }_{0} \leq \mathfrak{c}\text{ ) } \] \[ = {\left( {2}^{{\aleph }_{0}}\right) }^{\mathfrak{c}}\;\left( {\text{since}\mathfrak{c} = {2}^{{\aleph }_{0}}}\right) \] \[ = {2}^{{\aleph }_{0} \cdot \mathfrak{c}}\;\text{(since for nonempty sets}X, Y, Z,{\left( {X}^{Y}\right) }^{Z} \equiv {X}^{Y \times Z}\text{)} \] \[ \leq {2}^{\mathfrak{c} \cdot \mathfrak{c}}\;\text{ (since }{\aleph }_{0} < \mathfrak{c}\text{ ) } \] \[ = {2}^{\mathfrak{c}}\;\text{(since}\mathfrak{c} \cdot \mathfrak{c} = \mathfrak{c}\text{).} \] So, by the Schröder - Bernstein theorem, \( {2}^{\mathfrak{c}} = {\aleph }_{0}^{\mathfrak{c}} = {\mathfrak{c}}^{\mathfrak{c}} \) . It follows that \[ \{ 0,1{\} }^{\mathbb{R}} \equiv {\mathbb{N}}^{\mathbb{R}} \equiv {\mathbb{R}}^{\mathbb{R}} \]
|
Yes
|
Proposition 1.6.2 A linearly ordered set \( \\left( {W, \\leq }\\right) \) is well-ordered if and only if there is no descending sequence \( {w}_{0} > {w}_{1} > {w}_{2} > \\cdots \) in \( W \) .
|
Proof. Let \( W \) be not well-ordered. Then there is a nonempty subset \( A \) of \( W \) not having a least element. Choose any \( {w}_{0} \\in A \) . Since \( {w}_{0} \) is not the first element of \( A \), there is a \( {w}_{1} \\in A \) such that \( {w}_{1} < {w}_{0} \) . Since \( {w}_{1} \) is not the first element of \( A \), we get \( {w}_{2} < {w}_{1} \) in \( A \) . Proceeding similarly, we get a descending sequence \( \\left\\{ {{w}_{n} : n \\geq 0}\\right\\} \) in \( W \) . This completes the proof of the \
|
Yes
|
Example 1.6.3 Let \( W = \mathbb{N}\bigcup \{ \infty \} \). Let \( \leq \) be defined in the usual way on \( \mathbb{N} \) and let \( i < \infty \) for \( i \in \mathbb{N} \). Clearly, \( W \) is a well-ordered set. Since \( W \) has a last element and \( {\omega }_{0} \) does not, \( \left( {W, \leq }\right) \) is not isomorphic to \( {\omega }_{0} \). Thus there exist nonisomorphic well-ordered sets of the same cardinality.
|
Null
|
No
|
Proposition 1.6.5 No well-ordered set \( W \) is order isomorphic to an initial segment \( W\left( u\right) \) of itself.
|
Proof. Let \( W \) be a well-ordered set and \( u \in W \) . Suppose \( W \) and \( W\left( u\right) \) are isomorphic. Let \( f : W \rightarrow W\left( u\right) \) be an order isomorphism. For \( n \in \mathbb{N} \) , let \( {w}_{n} = {f}^{n}\left( u\right) \) . Note that\n\n\[ \n{w}_{0} = {f}^{0}\left( u\right) = u > {f}^{1}\left( u\right) = f\left( u\right) = {w}_{1}.\n\]\n\nBy induction on \( n \), we see that \( {w}_{n} > {w}_{n + 1} \) for all \( n \), i.e., \( \left( {w}_{n}\right) \) is a descending sequence in \( W \) . By 1.6.2, \( W \) is not well-ordered. This contradiction proves our result.
|
Yes
|
Proposition 1.7.1 (Proof by induction) For each \( n \in \mathbb{N} \), let \( {P}_{n} \) be a mathematical proposition. Suppose \( {P}_{0} \) is true and for every \( n,{P}_{n + 1} \) is true whenever \( {P}_{n} \) is true. Then for every \( n,{P}_{n} \) is true. Symbolically, we can express this as follows.
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The proof of this proposition uses two basic properties of the set of natural numbers. First, it is well-ordered by the usual order, and second, every nonzero element in it is a successor. A repeated application of 1.7.1 gives us the following.
|
No
|
Proposition 1.7.2 (Definition by induction) Let \( X \) be any nonempty set. Suppose \( {x}_{0} \) is a fixed point of \( X \) and \( g : X \rightarrow X \) any map. Then there is a unique map \( f : \mathbb{N} \rightarrow X \) such that \( f\left( 0\right) = {x}_{0} \) and \( f\left( {n + 1}\right) = g\left( {f\left( n\right) }\right) \) for all \( n \) .
|
Null
|
No
|
Theorem 1.7.3 (Proof by transfinite induction) Let \( \\left( {W, \\leq }\\right) \) be a well-ordered set, and for every \( w \\in W \), let \( {P}_{w} \) be a mathematical proposition. Suppose that for each \( w \\in W \), if \( {P}_{v} \) is true for each \( v < w \), then \( {P}_{w} \) is true. Then for every \( w \\in W,{P}_{w} \) is true. Symbolically, we express this as\n\n\[ \n\\left( {\\forall w \\in W}\\right) \\left( {\\left( {\\left( {\\forall v < w}\\right) {P}_{v}}\\right) \\Rightarrow {P}_{w}}\\right) \\Rightarrow \\left( {\\forall w \\in W}\\right) {P}_{w}.\n\]
|
Proof. Let\n\n\[ \n\\left( {\\forall w \\in W}\\right) \\left( {\\left( {\\left( {\\forall v < w}\\right) {P}_{v}}\\right) \\Rightarrow {P}_{w}}\\right) .\n\]\n\n\( \\left( *\\right) \)\n\nSuppose \( {P}_{w} \) is false for some \( w \\in W \) . Consider\n\n\[ \nA = \\left\{ {w \\in W : {P}_{w}\\text{ does not hold }}\\right\} .\n\]\n\nBy our assumptions, \( A \\neq \\varnothing \) . Let \( {w}_{0} \) be the least element of \( A \) . Then for every \( v < {w}_{0},{P}_{v} \) holds. However, \( {P}_{{w}_{0}} \) does not hold. This contradicts \( \\left( \\star \\right) \). Therefore, for every \( w \\in W,{P}_{w} \) holds.
|
Yes
|
Theorem 1.7.4 (Definition by transfinite induction) Let \( \\left( {W, \\leq }\\right) \) be a well-ordered set, \( X \) a set, and \( \\mathcal{F} \) the set of all maps with domain an initial segment of \( W \) and range contained in \( X \) . If \( G : \\mathcal{F} \\rightarrow X \) is any map, then there is a unique map \( f : W \\rightarrow X \) such that for every \( u \\in W \) ,\n\n\[ f\\left( u\\right) = G\\left( {f \\mid W\\left( u\\right) }\\right) . \]
|
Proof. For each \( w \\in W \), let \( {P}_{w} \) be the proposition \
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No
|
Theorem 1.7.5 (Trichotomy theorem for well-ordered sets) For any two well-ordered sets \( W \) and \( {W}^{\prime } \), exactly one of\n\n\[ W \prec {W}^{\prime }, W \sim {W}^{\prime },\text{ and }{W}^{\prime } \prec W \]\n\nholds.
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Proof. It is easy to see that no two of these can hold simultaneously. For example, if \( W \sim {W}^{\prime } \) and \( {W}^{\prime } \prec W \), then \( W \) is isomorphic to an initial segment of itself. This is impossible by 1.6.5.\n\nTo show that at least one of these holds, take \( X = {W}^{\prime }\bigcup \{ \infty \} \), where \( \infty \) is a point outside \( {W}^{\prime } \) . Now define a map \( f : W \rightarrow X \) by transfinite induction as follows. Let \( w \in W \) and assume that \( f \) has been defined on \( W\left( w\right) \) . If \( {W}^{\prime } \smallsetminus f\left( {W\left( w\right) }\right) \neq \varnothing \), then we take \( f\left( w\right) \) to be the least element of \( {W}^{\prime } \smallsetminus f\left( {W\left( w\right) }\right) \) ; otherwise, \( f\left( w\right) = \infty \) . By 1.7.4, such a function exists.\n\nLet us assume that \( \infty \notin f\left( W\right) \) . Then\n\n(i) the map \( f \) is one-to-one and order preserving, and\n\n(ii) the range of \( f \) is either whole of \( {W}^{\prime } \) or an initial segment of \( {W}^{\prime } \) .\n\nSo, in this case at least one of \( W \sim {W}^{\prime } \) or \( W \prec {W}^{\prime } \) holds.\n\nIf \( \infty \in f\left( W\right) \), then let \( w \) be the first element of \( W \) such that \( f\left( w\right) = \infty \) . Then \( f \mid W\left( w\right) \) is an order isomorphism from \( W\left( w\right) \) onto \( {W}^{\prime } \) . Thus in this case \( {W}^{\prime } \prec W \) .
|
Yes
|
Corollary 1.7.6 Let \( \\left( {W, \\leq }\\right) ,\\left( {{W}^{\\prime },{ \\leq }^{\\prime }}\\right) \) be well-ordered sets. Then \( W \\preccurlyeq {W}^{\\prime } \) if and only if there is a one-to-one order-preserving map from \( W \) into \( {W}^{\\prime } \) .
|
Proof. Suppose there is a one-to-one order-preserving map \( g \) from \( W \) into \( {W}^{\\prime } \) . Let \( X \) and \( f : W \\rightarrow X \) be as in the proof of 1.7.5. Then, by induction on \( w \), we easily show that for every \( w \\in W, f\\left( w\\right) { \\leq }^{\\prime }g\\left( w\\right) \) . Therefore, \( \\infty \\notin f\\left( W\\right) \) . Hence, \( W \\preccurlyeq {W}^{\\prime } \) . The converse is clear.
|
Yes
|
Theorem 1.7.7 Let \( \mathcal{W} = \left\{ {\left( {{W}_{i},{ \leq }_{i}}\right) : i \in I}\right\} \) be a family of pairwise non-isomorphic well-ordered sets. Then there is a \( W \in \mathcal{W} \) such that \( W \prec {W}^{\prime } \) for every \( {W}^{\prime } \in \mathcal{W} \) different from \( W \) .
|
Proof. Suppose no such \( W \) exists. Then there is a descending sequence\n\n\[ \cdots \prec {W}_{n} \prec \cdots \prec {W}_{1} \prec {W}_{0} \]\n\nin \( \mathcal{W} \) . For \( n \in \mathbb{N} \), choose a \( {w}_{n}^{\prime } \in {W}_{n} \) such that \( {W}_{n + 1} \sim {W}_{n}\left( {w}_{n}^{\prime }\right) \) . Fix an order isomorphism \( {f}_{n} : {W}_{n + 1} \rightarrow {W}_{n}\left( {w}_{n}^{\prime }\right) \) . Let \( {w}_{0} = {w}_{0}^{\prime } \), and for \( n > 0 \), \n\n\[ {w}_{n} = {f}_{0}\left( {{f}_{1}\left( {\cdots {f}_{n - 1}\left( {w}_{n}^{\prime }\right) }\right) }\right) . \]\n\n(See Figure 1.2.) Then \( \left( {w}_{n}\right) \) is a descending sequence in \( {W}_{0} \) . This is a contradiction. The result follows.
|
Yes
|
Theorem 1.8.3 Every ordinal \( \alpha \) can be uniquely written as \[ \alpha = \beta + n \] where \( \beta \) is a limit ordinal and \( n \) finite.
|
Proof. Let \( \alpha \) be an ordinal number. We first show that there exists a limit ordinal \( \beta \) and an \( n \in \omega \) such that \( \alpha = \beta + n \) . Choose a well-ordered set \( W \) such that \( t\left( W\right) = \alpha \) . If \( W \) has no last element, then we take \( \beta = \alpha \) and \( n = 0 \) . Suppose \( W \) has a last element, say \( {w}_{0} \) . If \( {w}_{0} \) has no immediate predecessor, then take \( \beta = t\left( {W\left( {w}_{0}\right) }\right) \) and \( n = 1 \) . Now suppose that \( {w}_{0} \) does have an immediate predecessor, say \( {w}_{1} \) . If \( {w}_{1} \) has no immediate predecessor, then we take \( \beta = t\left( {W\left( {w}_{1}\right) }\right) \) and \( n = 2 \) . Since \( W \) has no descending sequence, this process ends after finitely many steps. Thus we get \( {w}_{0},{w}_{1},\ldots ,{w}_{k - 1} \) such that \( {w}_{i} = {w}_{i - 1}^{ - } \) for all \( i > 0 \), and \( {w}_{k - 1} \) has no immediate predecessor. We take \( \beta = t\left( {W\left( {w}_{k - 1}\right) }\right) \) and \( n = k \) . We now show that \( \alpha \) has a unique representation of the type mentioned above. Let \( W,{W}^{\prime } \) be well-ordered sets with no last element, and \( {A}_{n},{B}_{m} \) finite well-ordered sets of cardinality \( n \) and \( m \) respectively such that \[ {A}_{n} \cap W = {B}_{m} \cap {W}^{\prime } = \varnothing . \] Let \( f : W + {A}_{n} \rightarrow {W}^{\prime } + {B}_{m} \) be an order isomorphism. It is easy to check that \( f\left( W\right) = {W}^{\prime } \) and \( f\left( {A}_{n}\right) = {B}_{m} \) . Uniqueness now follows.
|
Yes
|
Theorem 1.8.5 The set \( \Omega \) of all countable ordinals is uncountable.
|
Proof. Suppose \( \Omega \) is countable. Fix an enumeration \( {\alpha }_{0},{\alpha }_{1},\ldots \) of \( \Omega \) . Then\n\n\[ \alpha = \mathop{\sum }\limits_{n}{\alpha }_{n} + 1 \]\n\n is a countable ordinal strictly larger than each \( {\alpha }_{n} \) . This is a contradiction. So, \( \Omega \) is uncountable.
|
Yes
|
Proposition 1.8.6 Let \( \alpha \) be a countable limit ordinal. Then there exist \( {\alpha }_{0} < {\alpha }_{1} < \cdots \) such that \( \sup \left\{ {{\alpha }_{n} : n \in \mathbb{N}}\right\} = \alpha \) .
|
Proof. Since \( \alpha \) is countable, \( \{ \beta \in \mathbf{{ON}} : \beta < \alpha \} \) is countable. Fix an enumeration \( \left\{ {{\beta }_{n} : n \in \mathbb{N}}\right\} \) of all ordinals less than \( \alpha \) . We now define a sequence of ordinals \( \left( {\alpha }_{n}\right) \) by induction on \( n \) . Choose \( {\alpha }_{0} \) such that \( {\beta }_{0} < \) \( {\alpha }_{0} < \alpha \) . Since \( \alpha \) is a limit ordinal, such an ordinal exists. Suppose \( {\alpha }_{n} \) has been defined. Choose \( {\alpha }_{n + 1} \) greater than \( {\alpha }_{n} \) such that \( {\beta }_{n + 1} < {\alpha }_{n + 1} < \alpha \) . Clearly,\n\n\[ \n\alpha = \sup \left\{ {{\beta }_{n} : n \in \mathbb{N}}\right\} \leq \sup \left\{ {{\alpha }_{n} : n \in \mathbb{N}}\right\} \leq \alpha .\n\]\n\nSo, \( \sup \left\{ {{\alpha }_{n} : n \in \mathbb{N}}\right\} = \alpha \) .
|
Yes
|
Proposition 1.10.7 (König’s infinity lemma, [57]) Let \( T \) be a finitely splitting, infinite tree on \( A \) . Then \( T \) is ill-founded.
|
Proof. Let \( T \) be a finitely splitting, infinite tree on \( A \) . Let \( \left( {a}_{0}\right) \) be a node of \( T \) with infinitely many extensions in \( T \) . Since \( T \) is finitely splitting (and \( e \in T),\{ a \in A : \left( a\right) \in T\} \) is finite. Further, \( T \) is infinite. So, \( \left( {a}_{0}\right) \) exists. By the same argument we get \( {a}_{1} \in A \) such that \( {s}_{1} = \left( {{a}_{0},{a}_{1}}\right) \) has infinitely many extensions in \( T \) . Proceeding similarly we get an \( \alpha = \left( {{a}_{0},{a}_{1},\ldots }\right) \) such that for all \( k,\alpha \mid k \) has infinitely many extensions in \( T \) . In particular, \( \alpha \in \left\lbrack T\right\rbrack \) , and the result is proved.
|
Yes
|
Example 1.10.3 The tree\n\n\\[ T = \\{ e\\} \\bigcup \\left\\{ {i{0}^{j} : j \\leq i, i \\in \\mathbb{N}}\\right\\} \\]\n\nis infinite and well-founded.
|
Null
|
No
|
Example 1.10.4 Let \( T \) be a tree and \( u \) a node of \( T \) . The set\n\n\[ \n{T}_{u} = \\left\\{ {v \\in {A}^{ < \\mathbb{N}} : u \\hat{} v \\in T}\\right\\} \n\]\n\nforms a tree. (See Figure 1.4.)
|
Null
|
No
|
Proposition 1.10.7 (König’s infinity lemma, [57]) Let \( T \) be a finitely splitting, infinite tree on \( A \) . Then \( T \) is ill-founded.
|
Proof. Let \( T \) be a finitely splitting, infinite tree on \( A \) . Let \( \left( {a}_{0}\right) \) be a node of \( T \) with infinitely many extensions in \( T \) . Since \( T \) is finitely splitting (and \( e \in T),\{ a \in A : \left( a\right) \in T\} \) is finite. Further, \( T \) is infinite. So, \( \left( {a}_{0}\right) \) exists. By the same argument we get \( {a}_{1} \in A \) such that \( {s}_{1} = \left( {{a}_{0},{a}_{1}}\right) \) has infinitely many extensions in \( T \) . Proceeding similarly we get an \( \alpha = \left( {{a}_{0},{a}_{1},\ldots }\right) \) such that for all \( k,\alpha \mid k \) has infinitely many extensions in \( T \) . In particular, \( \alpha \in \left\lbrack T\right\rbrack \) , and the result is proved.
|
Yes
|
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