Q
stringlengths 4
3.96k
| A
stringlengths 1
3k
| Result
stringclasses 4
values |
|---|---|---|
Lemma 3.15 For all integers \( 0 < a < b \) we have\n\n(i) \( \mathop{\sum }\limits_{{n = a}}^{b}\frac{\chi \left( n\right) }{{n}^{1/2}} = O\left( {a}^{-1/2}\right) \), \n\n(ii) \( \mathop{\sum }\limits_{{n = a}}^{b}\frac{\chi \left( n\right) }{n} = O\left( {a}^{-1}\right) \).
|
Proof. This argument is similar to the proof of Proposition 3.4; we use summation by parts. Let \( {s}_{n} = \mathop{\sum }\limits_{{1 \leq k \leq n}}\chi \left( k\right) \), and remember that \( \left| {s}_{n}\right| \leq q \) for all \( n \) . Then\n\n\[ \mathop{\sum }\limits_{{n = a}}^{b}\frac{\chi \left( n\right) }{{n}^{1/2}} = \mathop{\sum }\limits_{{n = a}}^{{b - 1}}{s}_{n}\left\lbrack {{n}^{-1/2} - {\left( n + 1\right) }^{-1/2}}\right\rbrack + O\left( {a}^{-1/2}\right) \]\n\n\[ = O\left( {\mathop{\sum }\limits_{{n = a}}^{\infty }{n}^{-3/2}}\right) + O\left( {a}^{-1/2}\right) . \]\n\nBy comparing the sum \( \mathop{\sum }\limits_{{n = a}}^{\infty }{n}^{-3/2} \) with the integral of \( f\left( x\right) = {x}^{-3/2} \), we find that the former is also \( O\left( {a}^{-1/2}\right) \). \n\nA similar argument establishes (ii).
|
Yes
|
Lemma 1.2 If \( f \) is real-valued integrable on \( \left\lbrack {a, b}\right\rbrack \) and \( \varphi \) is a real-valued continuous function on \( \mathbb{R} \), then \( \varphi \circ f \) is also integrable on \( \left\lbrack {a, b}\right\rbrack \) .
|
Proof. Let \( \epsilon > 0 \) and remember that \( f \) is bounded, say \( \left| f\right| \leq M \) . Since \( \varphi \) is uniformly continuous on \( \left\lbrack {-M, M}\right\rbrack \) we may choose \( \delta > 0 \) so that if \( s, t \in \left\lbrack {-M, M}\right\rbrack \) and \( \left| {s - t}\right| < \delta \), then \( \left| {\varphi \left( s\right) - \varphi \left( t\right) }\right| < \epsilon \) . Now choose a partition \( P = \left\{ {{x}_{0},\ldots ,{x}_{N}}\right\} \) of \( \left\lbrack {a, b}\right\rbrack \) with \( \mathcal{U}\left( {P, f}\right) - \mathcal{L}\left( {P, f}\right) < {\delta }^{2} \) . Let \( {I}_{j} = \) \( \left\lbrack {{x}_{j - 1},{x}_{j}}\right\rbrack \) and distinguish two classes: we write \( j \in \Lambda \) if \( \mathop{\sup }\limits_{{x \in {I}_{j}}}f\left( x\right) - \) \( \mathop{\inf }\limits_{{x \in {I}_{j}}}f\left( x\right) < \delta \) so that by construction\n\n\[ \mathop{\sup }\limits_{{x \in {I}_{j}}}\varphi \circ f\left( x\right) - \mathop{\inf }\limits_{{x \in {I}_{j}}}\varphi \circ f\left( x\right) < \epsilon . \]\n\nOtherwise, we write \( j \in {\Lambda }^{\prime } \) and note that\n\n\[ \delta \mathop{\sum }\limits_{{j \in {\Lambda }^{\prime }}}\left| {I}_{j}\right| \leq \mathop{\sum }\limits_{{j \in {\Lambda }^{\prime }}}\left\lbrack {\mathop{\sup }\limits_{{x \in {I}_{j}}}f\left( x\right) - \mathop{\inf }\limits_{{x \in {I}_{j}}}f\left( x\right) }\right\rbrack \left| {I}_{j}\right| \leq {\delta }^{2} \]\n\nso \( \mathop{\sum }\limits_{{j \in {\Lambda }^{\prime }}}\left| {I}_{j}\right| < \delta \) . Therefore, separating the cases \( j \in \Lambda \) and \( j \in {\Lambda }^{\prime } \) we find that\n\n\[ \mathcal{U}\left( {P,\varphi \circ f}\right) - \mathcal{L}\left( {P,\varphi \circ f}\right) \leq \epsilon \left( {b - a}\right) + 2\mathcal{B}\delta , \]\n\nwhere \( \mathcal{B} \) is a bound for \( \varphi \) on \( \left\lbrack {-M, M}\right\rbrack \) . Since we can also choose \( \delta < \epsilon \) , we see that the proposition is proved.
|
Yes
|
Proposition 1.3 A bounded monotonic function \( f \) on an interval \( \left\lbrack {a, b}\right\rbrack \) is integrable.
|
Proof. We may assume without loss of generality that \( a = 0, b = 1 \) , and \( f \) is monotonically increasing. Then, for each \( N \), we choose the uniform partition \( {P}_{N} \) given by \( {x}_{j} = j/N \) for all \( j = 0,\ldots, N \) . If \( {\alpha }_{j} = \) \( f\left( {x}_{j}\right) \), then we have\n\n\[ \mathcal{U}\left( {{P}_{N}, f}\right) = \frac{1}{N}\mathop{\sum }\limits_{{j = 1}}^{N}{\alpha }_{j}\;\text{ and }\;\mathcal{L}\left( {{P}_{N}, f}\right) = \frac{1}{N}\mathop{\sum }\limits_{{j = 1}}^{N}{\alpha }_{j - 1}. \]\n\nTherefore, if \( \left| {f\left( x\right) }\right| \leq B \) for all \( x \) we have\n\n\[ \mathcal{U}\left( {{P}_{N}, f}\right) - \mathcal{L}\left( {{P}_{N}, f}\right) = \frac{{\alpha }_{N} - {\alpha }_{0}}{N} \leq \frac{2B}{N}, \]\n\nand the proposition is proved.
|
Yes
|
Proposition 1.4 Let \( f \) be a bounded function on the compact interval \( \left\lbrack {a, b}\right\rbrack \) . If \( c \in \left( {a, b}\right) \), and if for all small \( \delta > 0 \) the function \( f \) is integrable on the intervals \( \left\lbrack {a, c - \delta }\right\rbrack \) and \( \left\lbrack {c + \delta, b}\right\rbrack \), then \( f \) is integrable on \( \left\lbrack {a, b}\right\rbrack \) .
|
Proof. Suppose \( \left| f\right| \leq M \) and let \( \epsilon > 0 \) . Choose \( \delta > 0 \) (small) so that \( {4\delta M} \leq \epsilon /3 \) . Now let \( {P}_{1} \) and \( {P}_{2} \) be partitions of \( \left\lbrack {a, c - \delta }\right\rbrack \) and \( \lbrack c + \) \( \delta, b\rbrack \) so that for each \( i = 1,2 \) we have \( \mathcal{U}\left( {{P}_{i}, f}\right) - \mathcal{L}\left( {{P}_{i}, f}\right) < \epsilon /3 \) . This is possible since \( f \) is integrable on each one of the intervals. Then by taking as a partition \( P = {P}_{1} \cup \{ c - \delta \} \cup \{ c + \delta \} \cup {P}_{2} \) we immediately see that \( \mathcal{U}\left( {P, f}\right) - \mathcal{L}\left( {P, f}\right) < \epsilon . \
|
Yes
|
Lemma 1.6 The union of countably many sets of measure 0 has measure 0.
|
Proof. Say \( {E}_{1},{E}_{2},\ldots \) are sets of measure 0, and let \( E = { \cup }_{i = 1}^{\infty }{E}_{i} \) . Let \( \epsilon > 0 \), and for each \( i \) choose open interval \( {I}_{i,1},{I}_{i,2},\ldots \) so that\n\n\[ \n{E}_{i} \subset \mathop{\bigcup }\limits_{{k = 1}}^{\infty }{I}_{i, k}\;\text{ and }\;\mathop{\sum }\limits_{{k = 1}}^{\infty }\left| {I}_{i, k}\right| < \epsilon /{2}^{i}.\n\]\n\nNow clearly we have \( E \subset \mathop{\bigcup }\limits_{{i, k = 1}}^{\infty }{I}_{i, k} \), and\n\n\[ \n\mathop{\sum }\limits_{{i = 1}}^{\infty }\mathop{\sum }\limits_{{k = 1}}^{\infty }\left| {I}_{i, k}\right| \leq \mathop{\sum }\limits_{{i = 1}}^{\infty }\frac{\epsilon }{{2}^{i}} \leq \epsilon\n\]\n\nas was to be shown.
|
Yes
|
Lemma 1.8 If \( \epsilon > 0 \), then the set \( {A}_{\epsilon } \) is closed and therefore compact.
|
Proof. The argument is simple. Suppose \( {c}_{n} \in {A}_{\epsilon } \) converges to \( c \) and assume that \( c \notin {A}_{\epsilon } \) . Write \( \operatorname{osc}\left( {f, c}\right) = \epsilon - \delta \) where \( \delta > 0 \) . Select \( r \) so that \( \operatorname{osc}\left( {f, c, r}\right) < \epsilon - \delta /2 \), and choose \( n \) with \( \left| {{c}_{n} - c}\right| < r/2 \) . Then \( \operatorname{osc}\left( {f,{c}_{n}, r/2}\right) < \epsilon \) which implies \( \operatorname{osc}\left( {f,{c}_{n}}\right) < \epsilon \), a contradiction.
|
No
|
Theorem 2.1 Let \( f \) be a continuous function defined on a closed rectangle \( R \subset {\mathbb{R}}^{d} \). Suppose \( R = {R}_{1} \times {R}_{2} \) where \( {R}_{1} \subset {\mathbb{R}}^{{d}_{1}} \) and \( {R}_{2} \subset {\mathbb{R}}^{{d}_{2}} \) with \( d = {d}_{1} + {d}_{2} \). If we write \( x = \left( {{x}_{1},{x}_{2}}\right) \) with \( {x}_{i} \in {\mathbb{R}}^{{d}_{i}} \), then \( F\left( {x}_{1}\right) = {\int }_{{R}_{2}}f\left( {{x}_{1},{x}_{2}}\right) d{x}_{2} \) is continuous on \( {R}_{1} \), and we have \[ {\int }_{R}f\left( x\right) {dx} = {\int }_{{R}_{1}}\left( {{\int }_{{R}_{2}}f\left( {{x}_{1},{x}_{2}}\right) d{x}_{2}}\right) d{x}_{1}. \]
|
Proof. The continuity of \( F \) follows from the uniform continuity of \( f \) on \( R \) and the fact that \[ \left| {F\left( {x}_{1}\right) - F\left( {x}_{1}^{\prime }\right) }\right| \leq {\int }_{{R}_{2}}\left| {f\left( {{x}_{1},{x}_{2}}\right) - f\left( {{x}_{1}^{\prime },{x}_{2}}\right) }\right| d{x}_{2}. \] To prove the identity, let \( {P}_{1} \) and \( {P}_{2} \) be partitions of \( {R}_{1} \) and \( {R}_{2} \), respectively. If \( S \) and \( T \) are subrectangles in \( {P}_{1} \) and \( {P}_{2} \), respectively, then the key observation is that \[ \mathop{\sup }\limits_{{S \times T}}f\left( {{x}_{1},{x}_{2}}\right) \geq \mathop{\sup }\limits_{{{x}_{1} \in S}}\left( {\mathop{\sup }\limits_{{{x}_{2} \in T}}f\left( {{x}_{1},{x}_{2}}\right) }\right) \] and \[ \mathop{\inf }\limits_{{S \times T}}f\left( {{x}_{1},{x}_{2}}\right) \leq \mathop{\inf }\limits_{{{x}_{1} \in S}}\left( {\mathop{\inf }\limits_{{{x}_{2} \in T}}f\left( {{x}_{1},{x}_{2}}\right) }\right) . \] Then, \[ \mathcal{U}\left( {P, f}\right) = \mathop{\sum }\limits_{{S, T}}\left\lbrack {\mathop{\sup }\limits_{{S \times T}}f\left( {{x}_{1},{x}_{2}}\right) }\right\rbrack \left| {S \times T}\right| \] \[ \geq \mathop{\sum }\limits_{S}\mathop{\sum }\limits_{T}\mathop{\sup }\limits_{{{x}_{1} \in S}}\left\lbrack {\mathop{\sup }\limits_{{{x}_{2} \in T}}f\left( {{x}_{1},{x}_{2}}\right) }\right\rbrack \left| T\right| \times \left| S\right| \] \[ \geq \mathop{\sum }\limits_{S}\mathop{\sup }\limits_{{{x}_{1} \in S}}\left( {{\int }_{{R}_{2}}f\left( {{x}_{1},{x}_{2}}\right) d{x}_{2}}\right) \left| S\right| \] \[ \geq \mathcal{U}\left( {{P}_{1},{\int }_{{R}_{2}}f\left( {{x}_{1},{x}_{2}}\right) d{x}_{2}}\right) . \] Arguing similarly for the lower sums, we find that \[ \mathcal{L}\left( {P, f}\right) \leq \mathcal{L}\left( {{P}_{1},{\int }_{{R}_{2}}f\left( {{x}_{1},{x}_{2}}\right) d{x}_{2}}\right) \leq \mathcal{U}\left( {{P}_{1},{\int }_{{R}_{2}}f\left( {{x}_{1},{x}_{2}}\right) d{x}_{2}}\right) \leq \mathcal{U}\left( {P, f}\right) , \] and the theorem follows from these inequalities.
|
Yes
|
Theorem 2.2 Suppose \( A \) and \( B \) are compact subsets of \( {\mathbb{R}}^{d} \) and \( g : A \rightarrow B \) is a diffeomorphism of class \( {C}^{1} \) . If \( f \) is continuous on \( B \) , then\n\n\[ \n{\int }_{g\left( A\right) }f\left( x\right) {dx} = {\int }_{A}f\left( {g\left( y\right) }\right) \left| {\det \left( {Dg}\right) \left( y\right) }\right| {dy}.\n\]
|
The proof of this theorem consists first of an analysis of the special situation when \( g \) is a linear transformation \( L \) . In this case, if \( R \) is a rectangle, then\n\n\[ \n\left| {g\left( R\right) }\right| = \left| {\det \left( L\right) }\right| \left| R\right|\n\]\nwhich explains the term \( \left| {\det \left( {Dg}\right) }\right| \) . Indeed, this term corresponds to the new infinitesimal element of volume after the change of variables.
|
Yes
|
Theorem 1.6. (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\[{\int }_{a}^{b}{\kappa }_{N}\left( t\right) {dt} = {2\pi }\]
|
The second will be derived as a consequence of a more general result in Chapter 9; the proof of the first is left to Problem 9-6.
|
No
|
Lemma 2.1. Let \( V \) be a finite-dimensional vector space. There is a natural (basis-independent) isomorphism between \( {T}_{l + 1}^{k}\left( V\right) \) and the space of multilinear maps\n\n\[ \underset{l}{\underbrace{{V}^{ * } \times \cdots \times {V}^{ * }}} \times \underset{k}{\underbrace{V \times \cdots \times V}} \rightarrow V. \]
|
Exercise 2.1. Prove Lemma 2.1. [Hint: In the special case \( k = 1, l = 0 \) , consider the map \( \Phi : \operatorname{End}\left( V\right) \rightarrow {T}_{1}^{1}\left( V\right) \) by letting \( {\Phi A} \) be the \( \left( \begin{array}{l} 1 \\ 1 \end{array}\right) \) -tensor defined by \( {\Phi A}\left( {\omega, X}\right) = \omega \left( {AX}\right) \) . The general case is similar.]
|
No
|
Lemma 2.2. Let \( M \) be a smooth manifold, \( E \) a set, and \( \pi : E \rightarrow M \) a surjective map. Suppose we are given an open covering \( \left\{ {U}_{\alpha }\right\} \) of \( M \) together with bijective maps \( {\varphi }_{\alpha } : {\pi }^{-1}\left( {U}_{\alpha }\right) \rightarrow {U}_{\alpha } \times {\mathbf{R}}^{k} \) satisfying \( {\pi }_{1} \circ {\varphi }_{\alpha } = \pi \), such that whenever \( {U}_{\alpha } \cap {U}_{\beta } \neq \varnothing \), the composite map \[ {\varphi }_{\alpha } \circ {\varphi }_{\beta }^{-1} : {U}_{\alpha } \cap {U}_{\beta } \times {\mathbf{R}}^{k} \rightarrow {U}_{\alpha } \cap {U}_{\beta } \times {\mathbf{R}}^{k} \] is of the form \[ {\varphi }_{\alpha } \circ {\varphi }_{\beta }^{-1}\left( {p, V}\right) = \left( {p,\tau \left( p\right) V}\right) \] for some smooth map \( \tau : {U}_{\alpha } \cap {U}_{\beta } \rightarrow {GL}\left( {k,\mathbf{R}}\right) \). Then \( E \) has a unique structure as a smooth \( k \) -dimensional vector bundle over \( M \) for which the maps \( {\varphi }_{\alpha } \) are local trivializations.
|
Proof. For each \( p \in M \), let \( {E}_{p} = {\pi }^{-1}\left( p\right) \). If \( p \in {U}_{\alpha } \), observe that the map \( {\left( {\varphi }_{\alpha }\right) }_{p} : {E}_{p} \rightarrow \{ p\} \times {\mathbf{R}}^{k} \) obtained by restricting \( {\varphi }_{\alpha } \) is a bijection. We can define a vector space structure on \( {E}_{p} \) by declaring this map to be a linear isomorphism. This structure is well defined, since for any other set \( {U}_{\beta } \) containing \( p,\left( {2.4}\right) \) guarantees that \( {\left( {\varphi }_{\alpha }\right) }_{p} \circ {\left( {\varphi }_{\beta }\right) }_{p}^{-1} = \tau \left( p\right) \) is an isomorphism. Shrinking the sets \( {U}_{\alpha } \) and taking more of them if necessary, we may assume each of them is diffeomorphic to some open set \( {\widetilde{U}}_{\alpha } \subset {\mathbf{R}}^{n} \). Following \( {\varphi }_{\alpha } \) with such a diffeomorphism, we get a bijection \( {\pi }^{-1}\left( {U}_{\alpha }\right) \rightarrow {\widetilde{U}}_{\alpha } \times {\mathbf{R}}^{k} \), which we can use as a coordinate chart for \( E \). Because (2.4) shows that the \( {\varphi }_{\alpha } \) s overlap smoothly, these charts determine a locally Euclidean topology and a smooth manifold structure on \( E \). It is immediate that each map \( {\varphi }_{\alpha } \) is a diffeomorphism with respect to this smooth structure, and the rest of the conditions for a vector bundle follow automatically.
|
Yes
|
Lemma 3.1. Let \( g \) be a Riemannian metric on a manifold \( M \) . There is a unique fiber metric on each tensor bundle \( {T}_{l}^{k}M \) with the property that if \( \left( {{E}_{1},\ldots ,{E}_{n}}\right) \) is an orthonormal basis for \( {T}_{p}M \) and \( \left( {{\varphi }^{1},\ldots ,{\varphi }^{n}}\right) \) is the corresponding dual basis, then the collection of tensors given by (2.1) forms an orthonormal basis for \( {T}_{l}^{k}\left( {{T}_{p}M}\right) \) .
|
Exercise 3.8. Prove Lemma 3.1 by showing that in any local coordinate system, the required inner product is given by\n\n\[ \langle F, G\rangle = {g}^{{i}_{1}{r}_{1}}\cdots {g}^{{i}_{k}{r}_{k}}{g}_{{j}_{1}{s}_{1}}\cdots {g}_{{j}_{l}{s}_{l}}{F}_{{i}_{1}\ldots {i}_{k}}^{{j}_{1}\ldots {j}_{l}}{G}_{{r}_{1}\ldots {r}_{k}}^{{s}_{1}\ldots {s}_{l}}. \]
|
No
|
Lemma 3.2. On any oriented Riemannian n-manifold \( \left( {M, g}\right) \), there is a unique \( n \) -form \( {dV} \) satisfying the property that \( {dV}\left( {{E}_{1},\ldots ,{E}_{n}}\right) = 1 \) whenever \( \left( {{E}_{1},\ldots ,{E}_{n}}\right) \) is an oriented orthonormal basis for some tangent space \( {T}_{p}M \) .
|
Exercise 3.9. Prove Lemma 3.2, and show that the expression for \( {dV} \) with respect to any oriented local frame \( \left\{ {E}_{i}\right\} \) is\n\n\[ \n{dV} = \sqrt{\det \left( {g}_{ij}\right) }{\varphi }^{1} \land \cdots \land {\varphi }^{n}, \n\] \n\nwhere \( {g}_{ij} = \left\langle {{E}_{i},{E}_{j}}\right\rangle \) are the coefficients of \( g \) and \( \left\{ {\varphi }^{i}\right\} \) is the dual coframe.
|
No
|
Proposition 3.3. \( O\left( {n + 1}\right) \) acts transitively on orthonormal bases on \( {\mathbf{S}}_{R}^{n} \) . More precisely, given any two points \( p,\widetilde{p} \in {\mathbf{S}}_{R}^{n} \), and orthonormal bases \( \left\{ {E}_{i}\right\} \) for \( {T}_{p}{\mathbf{S}}_{R}^{n} \) and \( \left\{ {\widetilde{E}}_{i}\right\} \) for \( {T}_{\widetilde{p}}{\mathbf{S}}_{R}^{n} \), there exists \( \varphi \in O\left( {n + 1}\right) \) such that \( \varphi \left( p\right) = \widetilde{p} \) and \( {\varphi }_{ * }{E}_{i} = {\widetilde{E}}_{i} \) . In particular, \( {\mathbf{S}}_{R}^{n} \) is homogeneous and isotropic.
|
Proof. It suffices to show that given any \( p \in {\mathbf{S}}_{R}^{n} \) and any orthonormal basis \( \left\{ {E}_{i}\right\} \) for \( {T}_{p}{\mathbf{S}}_{R}^{n} \), there is an orthogonal map that takes the \
|
No
|
Lemma 3.4. Stereographic projection is a conformal equivalence between \( {\mathbf{S}}_{R}^{n} - \{ N\} \) and \( {\mathbf{R}}^{n} \) .
|
Proof. The inverse map \( {\sigma }^{-1} \) is a local parametrization, so we will use it to compute the pullback metric. Consider an arbitrary point \( q \in {\mathbf{R}}^{n} \) and a vector \( V \in {T}_{q}{\mathbf{R}}^{n} \), and compute\n\n\[ \n{\left( {\sigma }^{-1}\right) }^{ * }{\overset{ \circ }{g}}_{R}\left( {V, V}\right) = {\overset{ \circ }{g}}_{R}\left( {{\sigma }_{ * }^{-1}V,{\sigma }_{ * }^{-1}V}\right) = \bar{g}\left( {{\sigma }_{ * }^{-1}V,{\sigma }_{ * }^{-1}V}\right) , \n\]\n\nwhere \( \bar{g} \) denotes the Euclidean metric on \( {\mathbf{R}}^{n + 1} \) . Writing \( V = {V}^{i}{\partial }_{i} \) and \( {\sigma }^{-1}\left( u\right) = \left( {\xi \left( u\right) ,\tau \left( u\right) }\right) \), the usual formula for the push-forward of a vector\ncan be written\n\n\[ \n{\sigma }_{ * }^{-1}V = {V}^{i}\frac{\partial {\xi }^{j}}{\partial {u}^{i}}\frac{\partial }{\partial {\xi }^{j}} + {V}^{i}\frac{\partial \tau }{\partial {u}^{i}}\frac{\partial }{\partial \tau } \n\]\n\n\[ \n= V{\xi }^{j}\frac{\partial }{\partial {\xi }^{j}} + {V\tau }\frac{\partial }{\partial \tau } \n\]\n\nNow\n\n\[ \nV{\xi }^{j} = V\left( \frac{2{R}^{2}{u}^{j}}{{\left| u\right| }^{2} + {R}^{2}}\right) \n\]\n\n\[ \n= \frac{2{R}^{2}{V}^{j}}{{\left| u\right| }^{2} + {R}^{2}} - \frac{4{R}^{2}{u}^{j}\langle V, u\rangle }{{\left( {\left| u\right| }^{2} + {R}^{2}\right) }^{2}} \n\]\n\n\[ \n{V\tau } = V\left( {R\frac{{\left| u\right| }^{2} - {R}^{2}}{{\left| u\right| }^{2} + {R}^{2}}}\right) \n\]\n\n\[ \n= \frac{{2R}\langle V, u\rangle }{{\left| u\right| }^{2} + {R}^{2}} - \frac{{2R}\left( {{\left| u\right| }^{2} - {R}^{2}}\right) \langle V, u\rangle }{{\left( {\left| u\right| }^{2} + {R}^{2}\right) }^{2}} \n\]\n\n\[ \n= \frac{4{R}^{3}\langle V, u\rangle }{{\left( {\left| u\right| }^{2} + {R}^{2}\right) }^{2}}, \n\]\n\nwhere we have used the notation \( V\left( {\left| u\right| }^{2}\right) = 2\mathop{\sum }\limits_{k}{V}^{k}{u}^{k} = 2\langle V, u\rangle \) . Therefore,\n\n\[ \n\bar{g}\left( {{\sigma }_{ * }^{-1}V,{\sigma }_{ * }^{-1}V}\right) = \mathop{\sum }\limits_{{j = 1}}^{n}{\left( V{\xi }^{j}\right) }^{2} + {\left( V\tau \right) }^{2} \n\]\n\n\[ \n= \frac{4{R}^{4}|V{|}^{2}}{(|u{|}^{2} + {R}^{2}{)}^{2}} - \frac{{16}{R}^{4}\langle V, u{\rangle }^{2}}{(|u{|}^{2} + {R}^{2}{)}^{3}} + \frac{{16}{R}^{4}|u{|}^{2}\langle V, u{\rangle }^{2}}{(|u{|}^{2} + {R}^{2}{)}^{4}} \n\]\n\n\[ \n\; + \frac{{16}{R}^{6}\langle V, u{\rangle }^{2}}{{\left( {\left| u\right| }^{2} + {R}^{2}\right) }^{4}} \n\]\n\n\[ \n= \frac{4{R}^{4}{\left| V\right| }^{2}}{{\left( {\left| u\right| }^{2} + {R}^{2}\right) }^{2}}. \n\]\n\nIn other words,\n\n\[ \n{\left( {\sigma }^{-1}\right) }^{ * }{\overset{ \circ }{g}}_{R} = \frac{4{R}^{4}}{{\left( {\left| u\right| }^{2} + {R}^{2}\right) }^{2}}\bar{g} \n\]\n\n(3.10)\n\nwhere now \( \bar{g} \) represents the Euclidean metric on \( {\mathbf{R}}^{n} \), and so \( \sigma \) is a conformal equivalence.
|
Yes
|
Proposition 3.6. \( {O}_{ + }\left( {n,1}\right) \) acts transitively on the set of orthonormal bases on \( {\mathbf{H}}_{R}^{n} \), and therefore \( {\mathbf{H}}_{R}^{n} \) is homogeneous and isotropic.
|
Proof. The argument is entirely analogous to the proof of Proposition 3.3, so we give only a sketch. If \( p \in {\mathbf{H}}_{R}^{n} \) and \( \left\{ {E}_{i}\right\} \) is an orthonormal basis for \( {T}_{p}{\mathbf{H}}_{R}^{n} \), an easy computation shows that \( \left\{ {{E}_{1},\ldots ,{E}_{n},{E}_{n + 1} = p/R}\right\} \) is a basis for \( {\mathbf{R}}^{n + 1} \) such that \( m \) has the following expression in terms of the dual basis:\n\n\[ m = {\left( {\varphi }^{1}\right) }^{2} + \cdots + {\left( {\varphi }^{n}\right) }^{2} - {\left( {\varphi }^{n + 1}\right) }^{2}. \]\n\nIt follows easily that the matrix whose columns are the \( {E}_{i}\mathrm{\;s} \) is an element of \( {O}_{ + }\left( {n,1}\right) \) sending \( N = \left( {0,\ldots ,0, R}\right) \) to \( p \) and \( {\partial }_{i} \) to \( {E}_{i} \) (Figure 3.5).
|
Yes
|
Lemma 4.1. If \( \nabla \) is a connection in a bundle \( E, X \in \mathcal{T}\left( M\right), Y \in \mathcal{E}\left( M\right) \) , and \( p \in M \), then \( {\left. {\nabla }_{X}Y\right| }_{p} \) depends only on the values of \( X \) and \( Y \) in an arbitrarily small neighborhood of \( p \) . More precisely, if \( X = \widetilde{X} \) and \( Y = \widetilde{Y} \) on a neighborhood of \( p \), then \( {\left. {\nabla }_{X}Y\right| }_{p} = {\left. {\nabla }_{\widetilde{X}}\widetilde{Y}\right| }_{p} \) .
|
Proof. First consider \( Y \) . Replacing \( Y \) by \( Y - \widetilde{Y} \), it clearly suffices to show that \( {\left. {\nabla }_{X}Y\right| }_{p} = 0 \) if \( Y \) vanishes on a neighborhood \( U \) of \( p \) .\n\nChoose a bump function \( \varphi \in {C}^{\infty }\left( M\right) \) with support in \( U \) such that \( \varphi \left( p\right) = \) 1. The hypothesis that \( Y \) vanishes on \( U \) implies that \( {\varphi Y} \equiv 0 \) on all of \( M \) , so \( {\nabla }_{X}\left( {\varphi Y}\right) = {\nabla }_{X}\left( {0 \cdot {\varphi Y}}\right) = 0{\nabla }_{X}\left( {\varphi Y}\right) = 0 \) . Thus for any \( X \in \mathcal{T}\left( M\right) \), the product rule gives\n\n\[ 0 = {\nabla }_{X}\left( {\varphi Y}\right) = \left( {X\varphi }\right) Y + \varphi \left( {{\nabla }_{X}Y}\right) \]\n\n(4.1)\n\nNow \( Y \equiv 0 \) on the support of \( \varphi \), so the first term on the right is identically zero. Evaluating (4.1) at \( p \) shows that \( {\left. {\nabla }_{X}Y\right| }_{p} = 0 \) . The argument for \( X \) is similar but easier.
|
Yes
|
Lemma 4.2. With notation as in Lemma 4.1, \( {\left. {\nabla }_{X}Y\right| }_{p} \) depends only on the values of \( Y \) in a neighborhood of \( p \) and the value of \( X \) at \( p \) .
|
Proof. By linearity, it suffices to show that \( {\left. {\nabla }_{X}Y\right| }_{p} = 0 \) whenever \( {X}_{p} = \) 0 . Choose a coordinate neighborhood \( U \) of \( p \), and write \( X = {X}^{i}{\partial }_{i} \) in coordinates on \( U \), with \( {X}^{i}\left( p\right) = 0 \) . Then, for any \( Y \in \mathcal{E}\left( M\right) \) ,\n\n\[ \n{\left. {\nabla }_{X}Y\right| }_{p} = {\left. {\nabla }_{{X}^{i}{\partial }_{i}}Y\right| }_{p} = {\left. {X}^{i}\left( p\right) {\nabla }_{{\partial }_{i}}Y\right| }_{p} = 0.\n\]\n\nIn the first equality, we used Lemma 4.1, which allows us to evaluate \( {\left. {\nabla }_{X}Y\right| }_{p} \) by computing locally in \( U \) ; in the second, we used linearity of \( {\nabla }_{X}Y \) over \( {C}^{\infty }\left( M\right) \) in \( X \) .
|
Yes
|
Lemma 4.3. Let \( \nabla \) be a linear connection, and let \( X, Y \in \mathcal{T}\left( U\right) \) be expressed in terms of a local frame by \( X = {X}^{i}{E}_{i}, Y = {Y}^{j}{E}_{j} \) . Then\n\n\[ \n{\nabla }_{X}Y = \left( {X{Y}^{k} + {X}^{i}{Y}^{j}{\Gamma }_{ij}^{k}}\right) {E}_{k} \n\]\n\n(4.3)
|
Proof. Just use the defining rules for a connection and compute:\n\n\[ \n{\nabla }_{X}Y = {\nabla }_{X}\left( {{Y}^{j}{E}_{j}}\right) \n\]\n\n\[ \n= \left( {X{Y}^{j}}\right) {E}_{j} + {Y}^{j}{\nabla }_{{X}^{i}{E}_{i}}{E}_{j} \n\]\n\n\[ \n= \left( {X{Y}^{j}}\right) {E}_{j} + {X}^{i}{Y}^{j}{\nabla }_{{E}_{i}}{E}_{j} \n\]\n\n\[ \n= X{Y}^{j}{E}_{j} + {X}^{i}{Y}^{j}{\Gamma }_{ij}^{k}{E}_{k}. \n\]\n\nRenaming the dummy index in the first term yields (4.3).
|
Yes
|
Lemma 4.4. Suppose \( M \) is a manifold covered by a single coordinate chart. There is a one-to-one correspondence between linear connections on \( M \) and choices of \( {n}^{3} \) smooth functions \( \left\{ {\Gamma }_{ij}^{k}\right\} \) on \( M \), by the rule\n\n\[{\nabla }_{X}Y = \left( {{X}^{i}{\partial }_{i}{Y}^{k} + {X}^{i}{Y}^{j}{\Gamma }_{ij}^{k}}\right) {\partial }_{k}\]\n\n(4.5)
|
Proof. Observe that (4.5) is equivalent to (4.3) when \( {E}_{i} = {\partial }_{i} \) is a coordinate frame, so for every connection the functions \( \left\{ {\Gamma }_{ij}^{k}\right\} \) defined by (4.2) satisfy (4.5). On the other hand, given \( \left\{ {\Gamma }_{ij}^{k}\right\} \), it is easy to see by inspection that (4.5) is smooth if \( X \) and \( Y \) are, linear over \( \mathbf{R} \) in \( Y \), and linear over \( {C}^{\infty }\left( M\right) \) in \( X \), so only the product rule requires checking; this is a straightforward computation left to the reader.
|
No
|
Proposition 4.5. Every manifold admits a linear connection.
|
Proof. Cover \( M \) with coordinate charts \( \left\{ {U}_{\alpha }\right\} \) ; the preceding lemma guarantees the existence of a connection \( {\nabla }^{\alpha } \) on each \( {U}_{\alpha } \) . Choosing a partition of unity \( \left\{ {\varphi }_{\alpha }\right\} \) subordinate to \( \left\{ {U}_{\alpha }\right\} \), we’d like to patch the \( {\nabla }^{\alpha }\mathrm{s} \) together by the formula\n\n\[ \n{\nabla }_{X}Y = \mathop{\sum }\limits_{\alpha }{\varphi }_{\alpha }{\nabla }_{X}^{\alpha }Y \n\]\n\n(4.6)\n\nAgain, it is obvious by inspection that this expression is smooth, linear over \( \mathbf{R} \) in \( Y \), and linear over \( {C}^{\infty }\left( M\right) \) in \( X \) . We have to be a bit careful with the product rule, though, since a linear combination of connections is not necessarily a connection. (You can check, for example, that if \( {\nabla }^{1} \) and \( {\nabla }^{2} \n\nare connections, neither \( \frac{1}{2}{\nabla }^{1} \) nor \( {\nabla }^{1} + {\nabla }^{2} \) satisfies the product rule.) By direct computation,\n\n\[ \n{\nabla }_{X}\left( {fY}\right) = \mathop{\sum }\limits_{\alpha }{\varphi }_{\alpha }{\nabla }_{X}^{\alpha }\left( {fY}\right) \n\]\n\n\[ \n= \mathop{\sum }\limits_{\alpha }{\varphi }_{\alpha }\left( {\left( {Xf}\right) Y + f{\nabla }_{X}^{\alpha }Y}\right) \n\]\n\n\[ \n= \left( {Xf}\right) Y + f\mathop{\sum }\limits_{\alpha }{\varphi }_{\alpha }{\nabla }_{X}^{\alpha }Y \n\]\n\n\[ \n= \left( {Xf}\right) Y + f{\nabla }_{X}Y \n\]
|
Yes
|
Lemma 4.7. If \( \nabla \) is a linear connection on \( M \), and \( F \in {\mathcal{T}}_{l}^{k}\left( M\right) \), the map \( \nabla F : {\mathcal{T}}^{1}\left( M\right) \times \cdots \times {\mathcal{T}}^{1}\left( M\right) \times \mathcal{T}\left( M\right) \times \cdots \times \mathcal{T}\left( M\right) \rightarrow {C}^{\infty }\left( M\right) \), given by\n\n\[ \nabla F\left( {{\omega }^{1},\ldots ,{\omega }^{l},{Y}_{1},\ldots ,{Y}_{k}, X}\right) = {\nabla }_{X}F\left( {{\omega }^{1},\ldots ,{\omega }^{l},{Y}_{1},\ldots ,{Y}_{k}}\right) ,\]\ndefines a \( \left( \begin{matrix} k + 1 \\ l \end{matrix}\right) \) -tensor field.
|
Proof. This follows immediately from the tensor characterization lemma: \( {\nabla }_{X}F \) is a tensor field, so it is multilinear over \( {C}^{\infty }\left( M\right) \) in its \( k + l \) arguments; and it is linear over \( {C}^{\infty }\left( M\right) \) in \( X \) by definition of a connection.
|
Yes
|
Lemma 4.8. Let \( \nabla \) be a linear connection. The components of the total covariant derivative of a \( \left( \begin{array}{l} k \\ l \end{array}\right) \) -tensor field \( F \) with respect to a coordinate system are given by \[ {F}_{{i}_{1}\ldots {i}_{k};m}^{{j}_{1}\ldots {j}_{l}} = {\partial }_{m}{F}_{{i}_{1}\ldots {i}_{k}}^{{j}_{1}\ldots {j}_{l}} + \mathop{\sum }\limits_{{s = 1}}^{l}{F}_{{i}_{1}\ldots {i}_{k}}^{{j}_{1}\ldots p\ldots {j}_{l}}{\Gamma }_{mp}^{{j}_{s}} - \mathop{\sum }\limits_{{s = 1}}^{k}{F}_{{i}_{1}\ldots p\ldots {i}_{k}}^{{j}_{1}\ldots {j}_{l}}{\Gamma }_{m{i}_{s}}^{p}. \]
|
Exercise 4.6. Prove Lemma 4.8.
|
No
|
Lemma 4.9. Let \( \nabla \) be a linear connection on \( M \) . For each curve \( \gamma : I \rightarrow \) \( M,\nabla \) determines a unique operator\n\n\[ \n{D}_{t} : \mathcal{T}\left( \gamma \right) \rightarrow \mathcal{T}\left( \gamma \right)\n\]\n\nsatisfying the following properties:\n\n(a) Linearity over \( \mathbf{R} \) :\n\n\[ \n{D}_{t}\left( {{aV} + {bW}}\right) = a{D}_{t}V + b{D}_{t}W\;\text{ for }a, b \in \mathbf{R}.\n\]\n\n(b) Product rule:\n\n\[ \n{D}_{t}\left( {fV}\right) = \dot{f}V + f{D}_{t}V\;\text{ for }f \in {C}^{\infty }\left( I\right) .\n\]\n\n(c) If \( V \) is extendible, then for any extension \( \widetilde{V} \) of \( V \) ,\n\n\[ \n{D}_{t}V\left( t\right) = {\nabla }_{\dot{\gamma }\left( t\right) }\widetilde{V}\n\]\n\nFor any \( V \in \mathcal{T}\left( \gamma \right) ,{D}_{t}V \) is called the covariant derivative of \( V \) along \( \gamma \) .
|
Proof. First we show uniqueness. Suppose \( {D}_{t} \) is such an operator, and let \( {t}_{0} \in I \) be arbitrary. An argument similar to that of Lemma 4.1 shows that the value of \( {D}_{t}V \) at \( {t}_{0} \) depends only on the values of \( V \) in any interval \( \left( {{t}_{0} - \varepsilon ,{t}_{0} + \varepsilon }\right) \) containing \( {t}_{0} \) . (If \( I \) has an endpoint, extend \( \gamma \) to a slightly bigger open interval, prove the lemma there, and then restrict back to \( I \) .)\n\nChoose coordinates near \( \gamma \left( {t}_{0}\right) \), and write\n\n\[ \nV\left( t\right) = {V}^{j}\left( t\right) {\partial }_{j}\n\]\n\nnear \( {t}_{0} \) . Then by the properties of \( {D}_{t} \), since \( {\partial }_{j} \) is extendible,\n\n\[ \n{D}_{t}V\left( {t}_{0}\right) = {\dot{V}}^{j}\left( {t}_{0}\right) {\partial }_{j} + {V}^{j}\left( {t}_{0}\right) {\nabla }_{\dot{\gamma }\left( {t}_{0}\right) }{\partial }_{j}\n\]\n\n\[ \n= \left( {{\dot{V}}^{k}\left( {t}_{0}\right) + {V}^{j}\left( {t}_{0}\right) {\dot{\gamma }}^{i}\left( {t}_{0}\right) {\Gamma }_{ij}^{k}\left( {\gamma \left( {t}_{0}\right) }\right) }\right) {\partial }_{k}.\n\]\n\n(4.10)\n\nThis shows that such an operator is unique if it exists.\n\nFor existence, if \( \gamma \left( I\right) \) is contained in a single chart, we can define \( {D}_{t}V \) by (4.10); the easy verification that it satisfies the requisite properties is left to the reader. In the general case, we can cover \( \gamma \left( I\right) \) with coordinate charts and define \( {D}_{t}V \) by this formula in each chart, and uniqueness implies the various definitions agree whenever two or more charts overlap.
|
Yes
|
Theorem 4.10. (Existence and Uniqueness of Geodesics) Let \( M \) be a manifold with a linear connection. For any \( p \in M \), any \( V \in {T}_{p}M \), and any \( {t}_{0} \in \mathbf{R} \), there exist an open interval \( I \subset \mathbf{R} \) containing \( {t}_{0} \) and a geodesic \( \gamma : I \rightarrow M \) satisfying \( \gamma \left( {t}_{0}\right) = p,\dot{\gamma }\left( {t}_{0}\right) = V \). Any two such geodesics agree on their common domain.
|
Proof. Choose coordinates \( \left( {x}^{i}\right) \) on some neighborhood \( U \) of \( p \). From (4.10), a curve \( \gamma : I \rightarrow U \) is a geodesic if and only if its component functions \( \gamma \left( t\right) = \left( {{x}^{1}\left( t\right) ,\ldots ,{x}^{n}\left( t\right) }\right) \) satisfy the geodesic equation\n\n\[{\ddot{x}}^{k}\left( t\right) + {\dot{x}}^{i}\left( t\right) {\dot{x}}^{j}\left( t\right) {\Gamma }_{ij}^{k}\left( {x\left( t\right) }\right) = 0.\]\n\n(4.11)\n\nThis is a second-order system of ordinary differential equations for the functions \( {x}^{i}\left( t\right) \). The usual trick for proving existence and uniqueness for a second-order system is to introduce auxiliary variables \( {v}^{i} = {\dot{x}}^{i} \) to convert it to the following equivalent first-order system in twice the number of variables:\n\n\[{\dot{x}}^{k}\left( t\right) = {v}^{k}\left( t\right)\]\n\n\[{\dot{v}}^{k}\left( t\right) = - {v}^{i}\left( t\right) {v}^{j}\left( t\right) {\Gamma }_{ij}^{k}\left( {x\left( t\right) }\right).\]\n\nBy the existence and uniqueness theorem for first-order ODEs (see, for example,[Boo86, Theorem IV.4.1]), for any \( \left( {p, V}\right) \in U \times {\mathbf{R}}^{n} \), there exist \( \varepsilon > 0 \) and a unique solution \( \eta : \left( {{t}_{0} - \varepsilon ,{t}_{0} + \varepsilon }\right) \rightarrow U \times {\mathbf{R}}^{n} \) to this system satisfying the initial condition \( \eta \left( {t}_{0}\right) = \left( {p, V}\right) \). If we write the component functions of \( \eta \) as \( \eta \left( t\right) = \left( {{x}^{i}\left( t\right) ,{v}^{i}\left( t\right) }\right) \), then we can easily check that the curve \( \gamma \left( t\right) = \left( {{x}^{1}\left( t\right) ,\ldots ,{x}^{n}\left( t\right) }\right) \) in \( U \) satisfies the existence claim of the lemma.\n\nTo prove the uniqueness claim, suppose \( \gamma ,\sigma : I \rightarrow M \) are geodesics defined on an open interval with \( \gamma \left( {t}_{0}\right) = \sigma \left( {t}_{0}\right) \) and \( \dot{\gamma }\left( {t}_{0}\right) = \dot{\sigma }\left( {t}_{0}\right) \). By the uniqueness part of the ODE theorem, they agree on some neighborhood of \( {t}_{0} \). Let \( \beta \) be the supremum of numbers \( b \) such that they agree on \( \left\lbrack {{t}_{0}, b}\right\rbrack \) . If \( \beta \in I \), then by continuity \( \gamma \left( \beta \right) = \sigma \left( \beta \right) \) and \( \dot{\gamma }\left( \beta \right) = \dot{\sigma }\left( \beta \right) \), and applying local uniqueness in a neighborhood of \( \beta \), we conclude that they agree on a slightly larger interval (Figure 4.6), which is a contradiction. Arguing similarly to the left of \( {t}_{0} \), we conclude that they agree on all of \( I \).
|
Yes
|
Theorem 4.11. (Parallel Translation) Given a curve \( \gamma : I \rightarrow M,{t}_{0} \in \) \( I \), and a vector \( {V}_{0} \in {T}_{\gamma \left( {t}_{0}\right) }M \), there exists a unique parallel vector field \( V \) along \( \gamma \) such that \( V\left( {t}_{0}\right) = {V}_{0} \) .
|
Proof of Theorem 4.11. First suppose \( \gamma \left( I\right) \) is contained in a single coordinate chart. Then, using formula (4.10), \( V \) is parallel along \( \gamma \) if and only if\n\n\[{\dot{V}}^{k}\left( t\right) = - {V}^{j}\left( t\right) {\dot{\gamma }}^{i}\left( t\right) {\Gamma }_{ij}^{k}\left( {\gamma \left( t\right) }\right) ,\;k = 1,\ldots, n.\]\n\n(4.13)\n\nThis is a linear system of ODEs for \( \left( {{V}^{1}\left( t\right) ,\ldots ,{V}^{n}\left( t\right) }\right) \) . Thus Theorem 4.12 guarantees the existence and uniqueness of a solution on all of \( I \) with any initial condition \( V\left( {t}_{0}\right) = {V}_{0} \) .\n\nNow suppose \( \gamma \left( I\right) \) is not covered by a single chart. Let \( \beta \) denote the supremum of all \( b > {t}_{0} \) for which there is a unique parallel translate on \( \left\lbrack {{t}_{0}, b}\right\rbrack \) . Clearly \( \beta > {t}_{0} \), since for \( b \) close enough to \( {t}_{0},\gamma \left\lbrack {{t}_{0}, b}\right\rbrack \) is contained in a single chart and the above argument applies. Then a unique parallel translate \( V \) exists on \( \left\lbrack {{t}_{0},\beta }\right) \) (Figure 4.8). If \( \beta \in I \), choose coordinates on an open set containing \( \gamma \left( {\beta - \delta ,\beta + \delta }\right) \).
|
Yes
|
Theorem 4.12. (Existence and Uniqueness for Linear ODEs) Let \( I \subset \mathbf{R} \) be an interval, and for \( 1 \leq j, k \leq n \) let \( {A}_{j}^{k} : I \rightarrow \mathbf{R} \) be arbitrary smooth functions. The linear initial-value problem\n\n\[ \n{\dot{V}}^{k}\left( t\right) = {A}_{j}^{k}\left( t\right) {V}^{j}\left( t\right) \n\]\n\n(4.12)\n\n\[ \n{V}^{k}\left( {t}_{0}\right) = {B}^{k} \n\]\n\nhas a unique solution on all of \( I \) for any \( {t}_{0} \in I \) and any initial vector \( \left( {{B}^{1},\ldots ,{B}^{n}}\right) \in {\mathbf{R}}^{n} \) .
|
Exercise 4.11. Prove Theorem 4.12, as follows. Consider the vector field \( Y \) on \( I \times {\mathbf{R}}^{n} \) given by\n\n\[ \n{Y}^{0}\left( {{x}^{0},\ldots ,{x}^{n}}\right) = 1 \n\]\n\n\[ \n{Y}^{k}\left( {{x}^{0},\ldots ,{x}^{n}}\right) = {A}_{j}^{k}\left( {x}^{0}\right) {x}^{j},\;k = 1,\ldots, n. \n\]\n\n(a) Show that any solution to (4.12) is the projection to \( {\mathbf{R}}^{n} \) of an integral curve of \( Y \) .\n\n(b) For any compact subinterval \( K \subset I \), show there exists a positive constant \( C \) such that every solution \( V\left( t\right) = \left( {{V}^{1}\left( t\right) ,\ldots ,{V}^{n}\left( t\right) }\right) \) to (4.12) on \( K \) satisfies\n\n\[ \n\frac{d}{dt}\left( {{e}^{-{Ct}}{\left| V\left( t\right) \right| }^{2}}\right) \leq 0 \n\]\n\n(Here \( \left| {V\left( t\right) }\right| \) is just the Euclidean norm.)\n\n(c) If an integral curve of \( Y \) is defined only on some proper subinterval of \( I \), use Exercise 4.10 above to derive a contradiction.
|
No
|
Lemma 5.1. The operator \( {\nabla }^{\top } \) is well defined, and is a connection on \( M \) .
|
Proof. Since the value of \( {\bar{\nabla }}_{X}Y \) at a point \( p \in M \) depends only on \( {X}_{p} \) , \( {\nabla }_{X}^{\top }Y \) is clearly independent of the choice of vector field extending \( X \) . On the other hand, because of the result of Exercise 4.7, the value of \( {\bar{\nabla }}_{X}Y \) at \( p \) depends only on the values of \( Y \) along a curve whose initial tangent vector is \( {X}_{p} \) ; taking the curve to lie entirely in \( M \) shows that \( {\nabla }_{X}^{\top }Y \) depends only on the original vector field \( Y \in \mathcal{T}\left( M\right) \) . Thus \( {\nabla }^{\top } \) is well defined. Smoothness follows easily by expressing \( {\bar{\nabla }}_{X}Y \) in terms of an adapted orthonormal frame as in Problem 3-1.\n\nIt is obvious from the definition that \( {\nabla }_{X}^{\top }Y \) is linear over \( {C}^{\infty }\left( M\right) \) in \( X \) and over \( \mathbf{R} \) in \( Y \), so to show that it is a connection, only the product rule needs checking. Let \( f \in {C}^{\infty }\left( M\right) \) be extended arbitrarily to \( {\mathbf{R}}^{n} \) . Evaluating along \( M \), we get\n\n\[{\nabla }_{X}^{\top }\left( {fY}\right) = {\pi }^{\top }\left( {{\bar{\nabla }}_{X}\left( {fY}\right) }\right)\]\n\n\[= \left( {Xf}\right) {\pi }^{\top }Y + f{\pi }^{\top }\left( {{\bar{\nabla }}_{X}Y}\right)\]\n\n\[= \left( {Xf}\right) Y + f{\nabla }_{X}^{\top }Y\]\n\nThus \( {\nabla }^{\top } \) is a connection.
|
Yes
|
Lemma 5.3. The tangential connection on an embedded submanifold \( M \subset \) \( {\mathbf{R}}^{n} \) is symmetric.
|
Exercise 5.3. Prove Lemma 5.3. [Hint: If \( X \) and \( Y \) are vector fields on \( {\mathbf{R}}^{n} \) that are tangent to \( M \) at points of \( M \), so is \( \left\lbrack {X, Y}\right\rbrack \) by Exercise 2.3.]
|
No
|
Theorem 5.4. (Fundamental Lemma of Riemannian Geometry) Let \( \left( {M, g}\right) \) be a Riemannian (or pseudo-Riemannian) manifold. There exists a unique linear connection \( \nabla \) on \( M \) that is compatible with \( g \) and symmetric.
|
Proof. We prove uniqueness first, by deriving a formula for \( \nabla \) . Suppose, therefore, that \( \nabla \) is such a connection, and let \( X, Y, Z \in \mathcal{T}\left( M\right) \) be arbitrary vector fields. Writing the compatibility equation three times with \( X, Y, Z \) cyclically permuted, we obtain\n\n\[ X\langle Y, Z\rangle = \left\langle {{\nabla }_{X}Y, Z}\right\rangle + \left\langle {Y,{\nabla }_{X}Z}\right\rangle \]\n\n\[ Y\langle Z, X\rangle = \left\langle {{\nabla }_{Y}Z, X}\right\rangle + \left\langle {Z,{\nabla }_{Y}X}\right\rangle \]\n\n\[ Z\langle X, Y\rangle = \left\langle {{\nabla }_{Z}X, Y}\right\rangle + \left\langle {X,{\nabla }_{Z}Y}\right\rangle \]\n\nUsing the symmetry condition on the last term in each line, this can be rewritten as\n\n\[ X\langle Y, Z\rangle = \left\langle {{\nabla }_{X}Y, Z}\right\rangle + \left\langle {Y,{\nabla }_{Z}X}\right\rangle + \langle Y,\left\lbrack {X, Z}\right\rbrack \rangle \]\n\n\[ Y\langle Z, X\rangle = \left\langle {{\nabla }_{Y}Z, X}\right\rangle + \left\langle {Z,{\nabla }_{X}Y}\right\rangle + \langle Z,\left\lbrack {Y, X}\right\rbrack \rangle \]\n\n\[ Z\langle X, Y\rangle = \left\langle {{\nabla }_{Z}X, Y}\right\rangle + \left\langle {X,{\nabla }_{Y}Z}\right\rangle + \langle X,\left\lbrack {Z, Y}\right\rbrack \rangle .\n\nAdding the first two of these equations and subtracting the third, we obtain\n\n\[ X\langle Y, Z\rangle + Y\langle Z, X\rangle - Z\langle X, Y\rangle = \]\n\n\[ 2\left\langle {{\nabla }_{X}Y, Z}\right\rangle + \langle Y,\left\lbrack {X, Z}\right\rbrack \rangle + \langle Z,\left\lbrack {Y, X}\right\rbrack \rangle - \langle X,\left\lbrack {Z, Y}\right\rbrack \rangle .\n\nFinally, solving for \( \left\langle {{\nabla }_{X}Y, Z}\right\rangle \), we get\n\n\[ \left\langle {{\nabla }_{X}Y, Z}\right\rangle = \frac{1}{2}(X\langle Y, Z\rangle + Y\langle Z, X\rangle - Z\langle X, Y\rangle \]\n\n\[ - \langle Y,\left\lbrack {X, Z}\right\rbrack \rangle - \langle Z,\left\lbrack {Y, X}\right\rbrack \rangle + \langle X,\left\lbrack {Z, Y}\right\rbrack \rangle ). \]\n\n(5.1)\n\nNow suppose \( {\nabla }^{1} \) and \( {\nabla }^{2} \) are two connections that are symmetric and compatible with \( g \) . Since the right-hand side of (5.1) does not depend on the connection, it follows that \( \left\langle {{\nabla }_{X}^{1}Y - {\nabla }_{X}^{2}Y, Z}\right\rangle = 0 \) for all \( X, Y, Z \) . This can only happen if \( {\nabla }_{X}^{1}Y = {\nabla }_{X}^{2}Y \) for all \( X \) and \( Y \), so \( {\nabla }^{1} = {\nabla }^{2} \) .
|
Yes
|
Lemma 5.5. All Riemannian geodesics are constant speed curves.
|
Proof. Let \( \gamma \) be a Riemannian geodesic. Since \( \dot{\gamma } \) is parallel along \( \gamma \), its length \( \left| \dot{\gamma }\right| = \langle \dot{\gamma },\dot{\gamma }{\rangle }^{1/2} \) is constant by Lemma 5.2(d).
|
Yes
|
Proposition 5.6. (Naturality of the Riemannian Connection) Suppose \( \varphi : \left( {M, g}\right) \rightarrow \left( {\widetilde{M},\widetilde{g}}\right) \) is an isometry.\n\n(a) \( \varphi \) takes the Riemannian connection \( \nabla \) of \( g \) to the Riemannian connection \( \widetilde{\nabla } \) of \( \widetilde{g} \), in the sense that\n\n\[{\varphi }_{ * }\left( {{\nabla }_{X}Y}\right) = {\widetilde{\nabla }}_{{\varphi }_{ * }X}\left( {{\varphi }_{ * }Y}\right)\]\n\n(b) If \( \gamma \) is a curve in \( M \) and \( V \) is a vector field along \( \gamma \), then\n\n\[{\varphi }_{ * }{D}_{t}V = {\widetilde{D}}_{t}\left( {{\varphi }_{ * }V}\right)\]\n\n(c) \( \varphi \) takes geodesics to geodesics: if \( \gamma \) is the geodesic in \( M \) with initial point \( p \) and initial velocity \( V \), then \( \varphi \circ \gamma \) is the geodesic in \( \widetilde{M} \) with initial point \( \varphi \left( p\right) \) and initial velocity \( {\varphi }_{ * }V \) .
|
Exercise 5.4. Prove Proposition 5.6 as follows. For part (a), define a map\n\n\[{\varphi }^{ * }\widetilde{\nabla } : \mathfrak{T}\left( M\right) \times \mathfrak{T}\left( M\right) \rightarrow \mathfrak{T}\left( M\right)\]\n\nby\n\n\[{\left( {\varphi }^{ * }\widetilde{\nabla }\right) }_{X}Y = {\varphi }_{ * }^{-1}\left( {{\widetilde{\nabla }}_{{\varphi }_{ * }X}\left( {{\varphi }_{ * }Y}\right) }\right)\]\n\nShow that \( {\varphi }^{ * }\widetilde{\nabla } \) is a connection on \( M \) (called the pullback connection), and that it is symmetric and compatible with \( g \) ; therefore \( {\varphi }^{ * }\widetilde{\nabla } = \nabla \) by uniqueness of the Riemannian connection. You will have to unwind the definition of the push-forward of a vector field very carefully. For part (b), define an operator \( {\varphi }^{ * }{\widetilde{D}}_{t} : \mathcal{T}\left( \gamma \right) \rightarrow \mathcal{T}\left( \gamma \right) \) by a similar formula and show that it is equal to \( {D}_{t} \) .
|
No
|
Proposition 5.7. (Properties of the Exponential Map)\n\n(a) \( \\mathcal{E} \) is an open subset of TM containing the zero section, and each set \( {\\mathcal{E}}_{p} \) is star-shaped with respect to 0 .\n\n(b) For each \( V \\in {TM} \), the geodesic \( {\\gamma }_{V} \) is given by\n\n\[ \n{\\gamma }_{V}\\left( t\\right) = \\exp \\left( {tV}\\right) \n\]\n\nfor all \( t \) such that either side is defined.\n\n(c) The exponential map is smooth.
|
Proof of Proposition 5.7. The rescaling lemma with \( t = 1 \) says precisely that \( \\exp \\left( {cV}\\right) = {\\gamma }_{cV}\\left( 1\\right) = {\\gamma }_{V}\\left( c\\right) \) whenever either side is defined; this is (b). Moreover, if \( V \\in {\\mathcal{E}}_{p} \), by definition \( {\\gamma }_{V} \) is defined at least on \( \\left\\lbrack {0,1}\\right\\rbrack \) . Thus for \( 0 \\leq t \\leq 1 \), the rescaling lemma says that\n\n\[ \n\\exp \\left( {tV}\\right) = {\\gamma }_{tV}\\left( 1\\right) = {\\gamma }_{V}\\left( t\\right) \n\]\n\nis defined. This shows that \( {\\mathcal{E}}_{p} \) is star-shaped.\n\nIt remains to show that \( \\mathcal{E} \) is open and exp is smooth. To do so, we revisit the proof of the existence and uniqueness theorem for geodesics (Theorem\n\n4.10) and reformulate it in a more invariant way. Let \( \\le
|
No
|
Lemma 5.8. (Rescaling Lemma) For any \( V \in {TM} \) and \( c, t \in R \) , \[ {\gamma }_{cV}\left( t\right) = {\gamma }_{V}\left( {ct}\right) \] whenever either side is defined.
|
Proof. It suffices to show that \( {\gamma }_{cV}\left( t\right) \) exists and (5.5) holds whenever the right-hand side is defined, for then the converse statement follows by replacing \( V \) by \( {cV}, t \) by \( {ct} \), and \( c \) by \( 1/c \) . Suppose the domain of \( {\gamma }_{V} \) is the open interval \( I \subset \mathbf{R} \) . For simplicity, write \( \gamma = {\gamma }_{V} \), and define a new curve \( \widetilde{\gamma } \) by \( \widetilde{\gamma }\left( t\right) = \gamma \left( {ct}\right) \), defined on \( {c}^{-1}I \mathrel{\text{:=}} \) \( \{ t : {ct} \in I\} \) . We will show that \( \widetilde{\gamma } \) is a geodesic with initial point \( p \) and initial velocity \( {cV} \) ; it then follows by uniqueness that it must be equal to \( {\gamma }_{cV} \) . It is immediate from the definition that \( \widetilde{\gamma }\left( 0\right) = \gamma \left( 0\right) = p \) . Writing \( \gamma \left( t\right) = \) \( \left( {{\gamma }^{1}\left( t\right) ,\ldots ,{\gamma }^{n}\left( t\right) }\right) \) in any local coordinates, the chain rule gives \[ {\dot{\widetilde{\gamma }}}^{i}\left( t\right) = \frac{d}{dt}{\gamma }^{i}\left( {ct}\right) \] \[ = c{\dot{\gamma }}^{i}\left( {ct}\right) \] In particular, it follows that \( \dot{\widetilde{\gamma }}\left( 0\right) = c\dot{\gamma }\left( 0\right) = {cV} \) . Now let \( {D}_{t} \) and \( {\widetilde{D}}_{t} \) denote the covariant differentiation operators along \( \gamma \) and \( \widetilde{\gamma } \), respectively. Using the chain rule again in coordinates, \[ {\widetilde{D}}_{t}\dot{\widetilde{\gamma }}\left( t\right) = \left( {\frac{d}{dt}{\dot{\widetilde{\gamma }}}^{k}\left( t\right) + {\Gamma }_{ij}^{k}\left( {\widetilde{\gamma }\left( t\right) }\right) {\dot{\widetilde{\gamma }}}^{i}\left( t\right) {\dot{\widetilde{\gamma }}}^{j}\left( t\right) }\right) {\partial }_{k} \] \[ = \left( {{c}^{2}{\ddot{\gamma }}^{k}\left( {ct}\right) + {c}^{2}{\Gamma }_{ij}^{k}\left( {\gamma \left( {ct}\right) }\right) {\dot{\gamma }}^{i}\left( {ct}\right) {\dot{\gamma }}^{j}\left( {ct}\right) }\right) {\partial }_{k} \] \[ = {c}^{2}{D}_{t}\dot{\gamma }\left( {ct}\right) = 0. \] Thus \( \widetilde{\gamma } \) is a geodesic, and so \( \widetilde{\gamma } = {\gamma }_{cV} \) as claimed.
|
Yes
|
Lemma 5.10. (Normal Neighborhood Lemma) For any \( p \in M \), there is a neighborhood \( \mathcal{V} \) of the origin in \( {T}_{p}M \) and a neighborhood \( \mathcal{U} \) of \( p \) in \( M \) such that \( {\exp }_{p} : \mathcal{V} \rightarrow \mathcal{U} \) is a diffeomorphism.
|
Proof. This follows immediately from the inverse function theorem, once we show that \( {\left( {\exp }_{p}\right) }_{ * } \) is invertible at 0 . Since \( {T}_{p}M \) is a vector space, there is a natural identification \( {T}_{0}\left( {{T}_{p}M}\right) = {T}_{p}M \) . Under this identification, we will show that \( {\left( {\exp }_{p}\right) }_{ * } : {T}_{0}\left( {{T}_{p}M}\right) = {T}_{p}M \rightarrow {T}_{p}M \) has a particularly simple expression: it is the identity map!\n\nTo compute \( {\left( {\exp }_{p}\right) }_{ * }V \) for an arbitrary vector \( V \in {T}_{p}M \), we just need to choose a curve \( \tau \) in \( {T}_{p}M \) starting at 0 whose initial tangent vector is \( V \) , and compute the initial tangent vector of the composite curve \( {\exp }_{p} \circ \tau \left( t\right) \) . An obvious such curve is \( \tau \left( t\right) = {tV} \) . Thus\n\n\[ \n{\left. {\left( {\exp }_{p}\right) }_{ * }V = \frac{d}{dt}\right| }_{t = 0}\left( {{\exp }_{p} \circ \tau }\right) \left( t\right) = {\left. \frac{d}{dt}\right| }_{t = 0}{\exp }_{p}\left( {tV}\right) = {\left. \frac{d}{dt}\right| }_{t = 0}{\gamma }_{V}\left( t\right) = V. \n\]
|
Yes
|
Lemma 6.1. For any curve segment \( \gamma : \left\lbrack {a, b}\right\rbrack \rightarrow M \), and any reparametrization \( \widetilde{\gamma } \) of \( \gamma, L\left( \gamma \right) = L\left( \widetilde{\gamma }\right) \) .
|
Exercise 6.1. Prove Lemma 6.1.
|
No
|
Lemma 6.3. (Symmetry Lemma) Let \( \Gamma : \left( {-\varepsilon ,\varepsilon }\right) \times \left\lbrack {a, b}\right\rbrack \rightarrow M \) be an admissible family of curves in a Riemannian (or pseudo-Riemannian) manifold. On any rectangle \( \left( {-\varepsilon ,\varepsilon }\right) \times \left\lbrack {{a}_{i - 1},{a}_{i}}\right\rbrack \) where \( \Gamma \) is smooth,\n\n\[ \n{D}_{s}{\partial }_{t}\Gamma = {D}_{t}{\partial }_{s}\Gamma \n\]
|
Proof. This is a local question, so we may compute in coordinates \( \left( {x}^{i}\right) \) around any point \( \Gamma \left( {{s}_{0},{t}_{0}}\right) \) . Writing the components of \( \Gamma \) as \( \Gamma \left( {s, t}\right) = \) \( \left( {{x}^{1}\left( {s, t}\right) ,\ldots ,{x}^{n}\left( {s, t}\right) }\right) \), we have\n\n\[ \n{\partial }_{t}\Gamma = \frac{\partial {x}^{k}}{\partial t}{\partial }_{k};\;{\partial }_{s}\Gamma = \frac{\partial {x}^{k}}{\partial s}{\partial }_{k}.\n\]\n\nThen, using the coordinate formula (4.10) for covariant derivatives along curves,\n\n\[ \n{D}_{s}{\partial }_{t}\Gamma = \left( {\frac{{\partial }^{2}{x}^{k}}{\partial s\partial t} + \frac{\partial {x}^{i}}{\partial t}\frac{\partial {x}^{j}}{\partial s}{\Gamma }_{ji}^{k}}\right) {\partial }_{k}\n\]\n\n\[ \n{D}_{t}{\partial }_{s}\Gamma = \left( {\frac{{\partial }^{2}{x}^{k}}{\partial t\partial s} + \frac{\partial {x}^{i}}{\partial s}\frac{\partial {x}^{j}}{\partial t}{\Gamma }_{ji}^{k}}\right) {\partial }_{k}.\n\]\n\nReversing the roles of \( i \) and \( j \) in the second line above, and using the symmetry condition \( {\Gamma }_{ji}^{k} = {\Gamma }_{ij}^{k} \), we see immediately that these two expressions are equal.
|
Yes
|
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.
|
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.
|
Yes
|
Theorem 6.6. Every minimizing curve is a geodesic when it is given a unit speed parametrization.
|
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 \
|
Yes
|
Corollary 6.7. A unit speed admissible curve \( \gamma \) is a critical point for \( L \) if and only if it is a geodesic.
|
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.
|
Yes
|
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 \( \mathcal{U} - \{ p\} \) .
|
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 \( \partial /\partial r \) .
|
Yes
|
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 \( x \) .
|
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 \( r\left( x\right) \) .
|
Yes
|
Theorem 6.12. Every Riemannian geodesic is locally minimizing.
|
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.
|
Yes
|
Theorem 6.13. (Hopf-Rinow) A connected Riemannian manifold is geodesically complete if and only if it is complete as a metric space.
|
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.
|
Yes
|
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\[ \widetilde{R}\left( {{\varphi }_{ * }X,{\varphi }_{ * }Y}\right) {\varphi }_{ * }Z = {\varphi }_{ * }\left( {R\left( {X, Y}\right) Z}\right) . \]
|
Exercise 7.2. Prove Lemma 7.2.
|
No
|
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) \( \operatorname{Rm}\left( {W, X, Y, Z}\right) + \operatorname{Rm}\left( {X, Y, W, Z}\right) + \operatorname{Rm}\left( {Y, W, X, Z}\right) = 0 \) .
|
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
|
Yes
|
Proposition 7.5. (Differential Bianchi Identity) The total covariant derivative of the curvature tensor satisfies the following identity:\n\n\[ \n\nabla {Rm}\left( {X, Y, Z, V, W}\right) + \nabla {Rm}\left( {X, Y, V, W, Z}\right) + \nabla {Rm}\left( {X, Y, W, Z, V}\right) = 0.\n\]
|
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 \( p \) .
|
Yes
|
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\[ {R}_{{ij};}{}^{j} = \frac{1}{2}{S}_{;i} \]
|
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.
|
No
|
Proposition 7.8. If \( g \) is an Einstein metric on a connected manifold of dimension \( n \geq 3 \), its scalar curvature is constant.
|
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.
|
Yes
|
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 \( Y \) .
|
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 \( Y \) .
|
No
|
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 \( M \) :
|
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 \( {\nabla }^{\top } = \nabla \) .
|
Yes
|
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 \( M \) :
|
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
|
Yes
|
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\[ - \langle \Pi \left( {X, W}\right) ,\Pi \left( {Y, Z}\right) \rangle + \langle \Pi \left( {X, Z}\right) ,\Pi \left( {Y, W}\right) \rangle . \]
|
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.
|
Yes
|
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\[{\widetilde{D}}_{t}V = {D}_{t}V + \Pi \left( {\dot{\gamma }, V}\right)\]
|
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\[= {D}_{t}V + \Pi \left( {\dot{\gamma }, V}\right)\]
|
Yes
|
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\[ K = \frac{\operatorname{Rm}\left( {X, Y, Y, X}\right) }{{\left| X\right| }^{2}{\left| Y\right| }^{2}-\langle X, Y{\rangle }^{2}}. \]
|
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.
|
Yes
|
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.
|
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 \( K \) .
|
Yes
|
Proposition 8.8. If \( \\left( {X, Y}\\right) \) is any basis for a 2-plane \( \\Pi \\subset {T}_{p}M \), then\n\n\[ K\\left( {X, Y}\\right) = \\frac{\\operatorname{Rm}\\left( {X, Y, Y, X}\\right) }{{\\left| X\\right| }^{2}{\\left| Y\\right| }^{2}-\\langle X, Y{\\rangle }^{2}}.\]
|
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.
|
Yes
|
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 \( {\mathcal{R}}_{1} = {\mathcal{R}}_{2} \) .
|
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 \[ 0 = \mathcal{R}\left( {X, Y, Z, W}\right) + \mathcal{R}\left( {Y, Z, X, W}\right) + \mathcal{R}\left( {Z, X, Y, W}\right) \] \[ = \mathcal{R}\left( {X, Y, Z, W}\right) - \mathcal{R}\left( {Y, X, Z, W}\right) - \mathcal{R}\left( {X, Z, Y, W}\right) \] \[ = 3\mathcal{R}\left( {X, Y, Z, W}\right) \text{.} \]
|
Yes
|
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\[ {R}_{ij} = \\left( {n - 1}\\right) C{g}_{ij} \]
|
Exercise 8.8. Prove Lemma 8.10.
|
No
|
Lemma 9.2. If \( \gamma \) is a positively oriented curved polygon in \( M \), the rotation angle of \( \gamma \) is \( {2\pi } \) .
|
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.
|
Yes
|
Theorem 9.7. (The Gauss-Bonnet Theorem) If \( M \) is a triangulated, compact, oriented, Riemannian 2-manifold, then\n\n\[{\int }_{M}{KdA} = {2\pi \chi }\left( M\right)\]
|
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\[ {\int }_{M}{KdA} = {2\pi }{N}_{v} - {2\pi }{N}_{e} + {2\pi }{N}_{f} = {2\pi \chi }\left( M\right) \]
|
Yes
|
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.
|
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 \( \chi \left( \widetilde{M}\right) < 0 \) .
|
Yes
|
Lemma 10.1. If \( \\Gamma \) is any smooth admissible family of curves, and \( V \) is a smooth vector field along \( \\Gamma \), then\n\n\[ \n{D}_{s}{D}_{t}V - {D}_{t}{D}_{s}V = R\\left( {S, T}\\right) V.\n\]
|
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.
|
Yes
|
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\[ \n{D}_{t}^{2}V + R\left( {V,\dot{\gamma }}\right) \dot{\gamma } = 0 \n\]
|
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).
|
Yes
|
Lemma 10.3. Every Jacobi field along a geodesic \( \gamma \) is the variation field of some variation of \( \gamma \) through geodesics.
|
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 \( \sigma \) .]
|
No
|
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\[ J\left( a\right) = X;\;{D}_{t}J\left( a\right) = Y. \]
|
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 \( {J}^{i}\left( a\right) = {X}^{i},{V}^{i}\left( a\right) = {Y}^{i} \) .
|
Yes
|
Corollary 10.5. Along any geodesic \( \gamma \), the set of Jacobi fields is a \( {2n} \) - dimensional linear subspace of \( \mathcal{T}\left( \gamma \right) \) .
|
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.
|
Yes
|
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\[ J\left( a\right) \bot \dot{\gamma }\left( a\right) \text{ and }{D}_{t}J\left( a\right) \bot \dot{\gamma }\left( a\right) . \]
|
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.
|
Yes
|
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\[ J\left( t\right) = t{W}^{i}{\partial }_{i} \]
|
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 \( J\left( t\right) \).
|
Yes
|
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\[ u\\left( t\\right) = \\left\\{ \\begin{array}{ll} t, & C = 0 \\\\ R\\sin \\frac{t}{R}, & C = \\frac{1}{{R}^{2}} > 0 \\\\ R\\sinh \\frac{t}{R}, & C = - \\frac{1}{{R}^{2}} < 0 \\end{array}\\right. \]
|
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.
|
Yes
|
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\[ g\left( {V, V}\right) = \left\{ \begin{array}{ll} {\left| {V}^{ \bot }\right| }_{\bar{g}}^{2} + {\left| {V}^{\top }\right| }_{\bar{g}}^{2}, & K = 0; \\ {\left| {V}^{ \bot }\right| }_{\bar{g}}^{2} + \frac{{R}^{2}}{{r}^{2}}\left( {{\sin }^{2}\frac{r}{R}}\right) {\left| {V}^{\top }\right| }_{\bar{g}}^{2}, & C = \frac{1}{{R}^{2}} > 0; \\ {\left| {V}^{ \bot }\right| }_{\bar{g}}^{2} + \frac{{R}^{2}}{{r}^{2}}\left( {{\sinh }^{2}\frac{r}{R}}\right) {\left| {V}^{\top }\right| }_{\bar{g}}^{2}, & C = - \frac{1}{{R}^{2}} < 0. \end{array}\right. \]
|
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.
|
Yes
|
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 \( F : \mathcal{U} \rightarrow \widetilde{\mathcal{U}} \) .
|
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.
|
Yes
|
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 \( \gamma \left( t\right) = {\exp }_{p}{tV},\;t \in \left\lbrack {0,1}\right\rbrack \) .
|
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 \( {J}_{W}\left( 0\right) = {J}_{W}\left( q\right) = 0 \) .
|
Yes
|
Proposition 10.14. For any pair of proper normal vector fields \( V, W \) along a geodesic segment \( \gamma \) ,
|
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).
|
Yes
|
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.
|
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.
|
Yes
|
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 \( \left\lbrack {0, T}\right\rbrack \) .
|
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.
|
Yes
|
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 \( {\pi R} \) .
|
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 \( 0 < t < {\pi R} \) .
|
Yes
|
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).
|
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 \( g\left( {{V}^{\top },{V}^{\top }}\right) \geq {g}_{C}\left( {{V}^{\top },{V}^{\top }}\right) \) .
|
Yes
|
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 \( {\mathbf{R}}^{n} \) .
|
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.
|
Yes
|
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.
|
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 \( \pi \) .
|
Yes
|
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 \( {\pi R} \) .
|
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 \( t \in \left\lbrack {0, T}\right\rbrack \) .
|
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.
|
No
|
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 \( {\mathbf{S}}^{n} \) .
|
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).
|
No
|
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 \( \widetilde{M} \) .
|
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 \( \mathcal{J}\left( \widetilde{M}\right) \) .
|
Yes
|
Lemma 1. Let \( S \) have density \( \alpha \) and \( 0 \in S \) . Then \( S \oplus S \) has density at least \( {2\alpha } - {\alpha }^{2} \) .
|
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
|
Lemma 2. If \( S \) has density \( \alpha > \frac{1}{2} \), then \( S \oplus S \) contains all the positive integers.
|
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
|
Lemma 4. There exists \( \epsilon > 0 \) and \( {C}_{2} \) such that, throughout any interval \( {I}_{a, b, N} \) , \[ \left| {\mathop{\sum }\limits_{{n = 1}}^{N}e\left( {x{n}^{k}}\right) }\right| \leq \frac{{C}_{2}N}{{\left( b + j\right) }^{\epsilon }} \]
|
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.
|
No
|
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.
|
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.
|
No
|
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.
|
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.
|
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 1.1.8 (Cantor) For any two real numbers \( a, b \) with \( a < b \), the interval \( \left\lbrack {a, b}\right\rbrack \) is uncountable.
|
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.
|
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
|
Theorem 1.2.1 (Cantor) For any set \( X, X{ < }_{c}\mathcal{P}\left( X\right) \) .
|
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
|
Theorem 1.2.3 (Schröder - Bernstein Theorem) For any two sets \( X \) and \( Y \) , \[ \left( {X{ \leq }_{c}Y\& Y{ \leq }_{c}X}\right) \Rightarrow X \equiv Y. \]
|
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 \( \left( \star \right) \).
|
Yes
|
Corollary 1.2.4 For sets \( A \) and \( B \) ,
|
\[ A{ < }_{c}B \Leftrightarrow A{ \leq }_{c}B\& B{ \nleq }_{c}A. \]
|
Yes
|
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, \( \mathcal{P}\left( \mathbb{N}\right) { \leq }_{c}\mathbb{R} \) .
|
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 \( \mathcal{P}\left( \mathbb{N}\right) \equiv \{ 0,1{\} }^{\mathbb{N}},\mathbb{R} \equiv \{ 0,1{\} }^{\mathbb{N}} \) .
|
Yes
|
Theorem 1.3.1 If \( X \) is infinite and \( A \subseteq X \) finite, then \( X \smallsetminus A \) and \( X \) have the same cardinality.
|
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.
|
Yes
|
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.
|
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"
|
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.
|
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
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.