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Lemma 3.4. Let \( \left( {X, g}\right) \) be pseudo Riemannian. Let \( \eta ,\zeta \) be Jacobi lifts of a geodesic \( \alpha \) . Then\n\n\[ \n{\left\langle {D}_{{\alpha }^{\prime }}\eta ,\zeta \right\rangle }_{g} - {\left\langle \eta ,{D}_{{\alpha }^{\prime }}\zeta \right\rangle }_{g}\;\text{is constant.} \n\] | Proof. We differentiate the above expression and expect to get 0 . From the defining property of the covariant derivative, the derivative of the above expression is equal to\n\n\[ \n\left\langle {{D}_{{\alpha }^{\prime }}^{2}\eta ,\zeta }\right\rangle + \left\langle {{D}_{{\alpha }^{\prime }}\eta ,{D}_{{\alpha }^{\prim... | Yes |
Lemma 3.5. Let \( \left( {X, g}\right) \) be pseudo Riemannian. Let \( \alpha \) (defined at least on \( \left\lbrack {0,1}\right\rbrack ) \) be the geodesic such that \( \alpha \left( 0\right) = x \) and \( {\alpha }^{\prime }\left( 0\right) = v \) . Let\n\n\[ \nz \in {T}_{\alpha \left( 1\right) },\;w \in {T}_{\alpha ... | Proof. Let \( \zeta \) be the Jacobi lift of \( \alpha \) such that \( \zeta \left( 1\right) = 0 \) and \( {D}_{{\alpha }^{\prime }}\zeta \left( 1\right) = z \) . Let \( \eta \) be the Jacobi lift as in Theorem 3.1. Then\n\n\[ \n\left\langle {T{\exp }_{x}\left( v\right) w, z}\right\rangle = \left\langle {\eta \left( 1\... | Yes |
Theorem 3.6. Let \( \left( {X, g}\right) \) be Riemannian. Assume \( \left( {X, g}\right) \) has semi-negative curvature. Then for all \( x \in X \) and \( v \in {T}_{x}, v \neq 0 \), such that \( {\exp }_{x} \) is defined on the segment \( \left\lbrack {0, v}\right\rbrack \) in \( {T}_{x} \), we have\n\n\[{\begin{Vmat... | Proof. Let \( {\eta }_{w} \) be the Jacobi lift as in Proposition 3.1, so that\n\n\[ T{\exp }_{x}\left( v\right) w = {\eta }_{w}\left( 1\right) \]\n\nThe asserted inequality is then a special case of the inequality found in Proposition 2.6. This inequality implies that \( \operatorname{Ker}T{\exp }_{x}\left( v\right) =... | Yes |
Theorem 3.7 (McAlpin [McA 65]). Let \( \\left( {X, g}\\right) \) be a Riemannian-Hilbertian manifold with seminegative curvature, and let \( x \\in X \) . Assume that \( {\\exp }_{x} \) is defined on all of \( {T}_{x} \) (what we called geodesically complete at \( x) \) . Then for all \( v \\in {T}_{x} \) the map \( T{... | Proof. We have already proved that \( T{\\exp }_{x}\\left( v\\right) \) is injective and has a continuous inverse on its image. Lemma 3.5 shows that we can apply the same reasoning to the adjoint \( {\\left( T{\\exp }_{x}\\left( v\\right) \\right) }^{ * } = T{\\exp }_{y}\\left( {v}^{ * }\\right) \) for \( y = {\\exp }_... | No |
Theorem 3.8 (Cartan-Hadamard). Let \( \\left( {X, g}\\right) \) be a Riemannian manifold, connected, and such that \( {\\exp }_{x} \) is defined on all of \( {T}_{x} \) for some \( x \\in X \) (so geodesically complete). If \( {R}_{2} \\geqq 0 \) (i.e. \( X \) has seminegative curvature), then the exponential map \( {\... | Proof. We have already proved that \( {\\exp }_{x} \) is a local isomorphism. There remains to prove that \( {\\exp }_{x} \) is surjective, and that it is a covering. But all the work has been done, because we simply apply Theorem 6.9 of Chapter VIII with \( Y = {T}_{x} \) having the given metric \( h = g\\left( x\\rig... | No |
Corollary 3.9. Let \( \left( {X, g}\right) \) be a connected Riemannian manifold with seminegative curvature. Then \( \left( {X, g}\right) \) is complete if and only if the exponential map \( {\exp }_{x} \) is defined on all of \( {T}_{x} \) for some \( x \in X \), and therefore for every \( x \in X \) . | Proof. That \( \left( {X, g}\right) \) complete implies \( {\exp }_{x} \) defined on all of \( {T}_{x} \) was proved under all circumstances in Proposition 6.5 of Chapter VIII. The converse is now immediate from Theorem 2.10 and Theorem 6.9 of Chapter VIII. | Yes |
Corollary 3.10. Let \( \left( {X, g}\right) \) be a Cartan-Hadamard manifold. Let \( x \in X \) . Then for all \( v, w \in {T}_{x}X \) we have the inequality\n\n\[{\operatorname{dist}}_{g}\left( {{\exp }_{x}\left( v\right) ,{\exp }_{x}\left( w\right) }\right) \geqq \parallel v - w{\parallel }_{g}.\] | Proof. By Theorem 3.8 the exponential map has an inverse\n\n\[ \varphi : X \rightarrow {T}_{x}X \]\n\nand by Theorem 3.6 this inverse satisfies\n\n\[ \parallel {T\varphi }\left( z\right) {\parallel }_{g} \leqq 1 \]\n\nfor all \( z \in X \), where the norm is that of a continuous linear map from \( {T}_{z}X \) to \( {T}... | Yes |
Corollary 3.11. Suppose that \( \\left( {X, g}\\right) \) is a Cartan-Hadamard manifold. Then any two points can be joined by a unique geodesic whose length is the \( g \) -distance between the two points. | Proof. Immediate from Corollary 3.10, because if \( x, y \) are the two points, then \( y = {\\exp }_{x}\\left( v\\right) \) for some \( v \\in {T}_{x}X \), and the geodesic \( \\alpha \) such that \( \\alpha \\left( t\\right) = {\\exp }_{x}\\left( {tv}\\right) \) joins the two points, is unique by the Hadamard-Cartan ... | Yes |
Let \( X \) be Riemannian complete, simply connected. Let \( {x}_{0} \in X \). (a) If \( R = 0 \), i.e. if \( X \) has 0 curvature, then the exponential map \[{\exp }_{{x}_{0}} : {T}_{{x}_{0}}X \rightarrow X\] is an isometry. | For (a), we use Theorem 3.1 and Proposition 2.10 which shows that the exponential map amounts to parallel translation, so is an isometry. | No |
Theorem 3.13. Let \( X \) be Riemannian, complete, simply connected, with sectional curvature +1 . Then \( X \) is isometric to the ordinary sphere of the same dimension in Hilbert space. | Proof. The proof is similar, except that one cannot deal with the exponential defined on the whole tangent space \( {T}_{{x}_{0}}X \) . For convenience, we let \( X \) be the unit sphere in Hilbert space of a given dimension, and we let \( Y \) be Riemannian, complete simply connected with sectional curvature +1 . We c... | Yes |
Theorem 4.1. We have\n\n(1)\n\n\[ \n{\partial }_{2}h = \frac{1}{\begin{Vmatrix}{\partial }_{1}\sigma \end{Vmatrix}}{\left\langle {D}_{2}{\partial }_{1}\sigma ,{\partial }_{1}\sigma \right\rangle }_{g}, \n\] \n\n(2)\n\n\[ \n{\partial }_{2}^{2}h = \frac{1}{{\begin{Vmatrix}{\partial }_{1}\sigma \end{Vmatrix}}^{3}}\left( {... | Proof. The first formula comes directly from the definition of the metric (Levi-Civita) derivative. The second is obtained at once by using the rule for the derivative of a product, and setting \n\n\[ \n{D}_{2}^{2}{\partial }_{1}\sigma = {R}_{2}\left( {{\partial }_{2}\sigma ,{\partial }_{1}\sigma }\right) {\partial }_{... | Yes |
Theorem 4.2. Let \( X \) be a Riemannian manifold, and let \( \sigma = \sigma \left( {s, t}\right) \) be a variation of geodesics \( \left\{ {\alpha }_{t}\right\} \) . Let \( u \) be the (varying) unit vector tangent to these geodesics, namely\n\n\[ u = {\partial }_{1}\sigma /{\begin{Vmatrix}{\partial }_{1}\sigma \end{... | Proof. Immediate from Lemma 4.1 and the definitions. | No |
Theorem 4.3. Let \( X \) be a Riemannian manifold with seminegative curvature \( \left( {{R}_{2} \geqq 0}\right) \), and \( U \) a convex open set. Let \( {\beta }_{1},{\beta }_{2} \) be disjoint geodesics in \( U \), defined on the same interval. Let \( {\alpha }_{t} : \left\lbrack {a, b}\right\rbrack \rightarrow U \)... | Proof. Immediate from Theorem 4.2 and the hypothesis that \( {R}_{2} \geqq 0 \) . | No |
Theorem 4.4. Let \( X \) have seminegative curvature. Let \( U \) be a convex open subset. Let \( \gamma \) be a geodesic in \( U \) not containing a point \( x \in U \) . For each \( t \) at which \( \gamma \) is defined, let \( {\alpha }_{t} : \left\lbrack {0,1}\right\rbrack \rightarrow U \) be the geodesic joining \... | Proof. The picture is as follows. We suppose there is a point \( c \) such that \( {\ell }^{\prime \prime }\left( c\right) = 0 \) .\n\n\n\nAs in Theorem 4.2, let \( \sigma \left( {s, t}\right) = {\alpha }_{t}\left( s... | Yes |
Corollary 4.5. Let \( X \) be a Cartan-Hadamard manifold. Then every ball in \( X \) is convex. | Proof. Let \( x \) be the center of the ball, and let \( {x}_{0},{x}_{1} \) be points in the ball. If \( x \) lies on the geodesic between \( {x}_{0} \) and \( {x}_{1} \) then the Cartan-Hadamard theorem shows that this geodesic is the ray passing through the origin of the ball, so lies in the ball. If not, then we can... | No |
Theorem 4.6. Let \( X \) be a Riemannian manifold and let \( x \in X \). Let \( U \) be a convex open set in \( X \) such that\n\n\[{\exp }_{x} : V \rightarrow U\]\n\nis an isomorphism of some convex open set \( V \) in \( {T}_{x} \) containing \( {0}_{x} \), with U. Let \( \gamma \) be a curve in \( U \) not containin... | Proof. Let us first prove the result in euclidean space. Let \( t \rightarrow v\left( t\right) \) be a curve in a euclidean space, and let \( F\left( t\right) = \parallel v\left( t\right) \parallel \), with the euclidean norm [IX, §4] denoted by the double bar. Then\n\n\[\mathop{\lim }\limits_{{h \rightarrow 0 + }}\fra... | Yes |
Corollary 4.7. Let \( X \) be a Cartan-Hadamard manifold. Let \( x \in X \) and let \( \gamma \) be a geodesic which does not contain \( x \) . Then the distance \( d\left( {x,\gamma \left( t\right) }\right) \) has a unique minimum for some value \( {t}_{0} \) . The geodesic from \( x \) to \( \gamma \left( {t}_{0}\rig... | Proof. That the distance has a minimum comes from the fact that the geodesic distance goes to infinity as \( t \rightarrow \pm \infty \) . Because the line is locally compact, there is some minimum, and the convexity Theorem 4.4 shows that this is the only minimum, with the distance being strictly decreasing for \( t \... | Yes |
Theorem 4.8. Let \( X \) be a Cartan-Hadamard manifold. Let \( {ABC} \) be a geodesic triangle whose angles are \( A, B, C \) and whose sides are geodesics of lengths \( a, b \), and \( c \) . Then:\n\n(i) \( {a}^{2} + {b}^{2} \leqq {c}^{2} + {2ab}\cos C \) ;\n\n(ii) \( A + B + C \leqq \pi \) . | Proof. Let \( x \) be the vertex of angle \( C \) . Let \( {\exp }_{x}\left( v\right) \) and \( {\exp }_{x}\left( w\right) \) with \( v \) , \( w \in {T}_{x} \) be the vertices with angles \( A, B \) respectively. Then the geodesic sides of angle \( C \) are \( \alpha ,\beta \) respectively, with\n\n\[ \alpha \left( s\... | Yes |
Proposition 5.1. Let \( \eta : J \rightarrow {TX} \) be a lift of \( \alpha \) in \( {TX} \) . Then\n\n\[ \n\eta \left( t\right) = {P}^{t}\mathop{\sum }\limits_{{k = 0}}^{m}{D}_{{\alpha }^{\prime }}^{k}\eta \left( 0\right) \frac{{t}^{k}}{k!} + O\left( {t}^{m + 1}\right) \;\text{ for }t \rightarrow 0;\n\]\n\nor alternat... | Proof. The second expression is merely a reformulation of the first, taking into account the definition of parallel translation. Since \( t \rightarrow 0 \), the formula is local, and we may prove it in a chart, so we use \( \eta ,\gamma \) to denote the vector components \( {\eta }_{U},{\gamma }_{U} \) in a chart \( U... | Yes |
Lemma 5.2. Let \( \beta : J \rightarrow \mathbf{E} \) be the vector component of a lift of \( \alpha \) . If \( {D}_{{\alpha }^{\prime }}^{j}\beta \left( 0\right) = 0 \) for \( 0 \leqq j \leqq m \) then \( {\partial }^{j}\beta \left( 0\right) = 0 \) for \( 0 \leqq j \leqq m \) . | Proof. By definition,\n\n\[ \n{D}_{{\alpha }^{\prime }}\beta = {\beta }^{\prime } - B\left( {\alpha ;{\alpha }^{\prime },\beta }\right) .\n\]\n\nHence \( {D}_{{\alpha }^{\prime }}\beta \left( 0\right) = {\beta }^{\prime }\left( 0\right) \) . We can proceed by induction. Let us carry out the case of the second derivativ... | Yes |
Proposition 5.3. Suppose that \( \\alpha \) is a geodesic. Let \( w \\in {T}_{\\alpha \\left( 0\\right) }X \) and let \( {\\eta }_{w} \) be the Jacobi lift of \( \\alpha \) such that \( {\\eta }_{w}\\left( 0\\right) = 0 \) and \( {D}_{\\alpha ^{\\prime }}{\\eta }_{w}\\left( 0\\right) = w \). Then | Proof. We plug in Proposition 5.1. Since \( {D}_{\\alpha ^{\\prime }}^{2}{\\eta }_{w} = R\\left( {\\alpha ^{\\prime },{\\eta }_{w},\\alpha ^{\\prime }}\\right) \) contains \( {\\eta }_{w} \) linearly, the evaluation of the second term of the Taylor expansion at 0 is 0. As for the third term, we have to use the chain ru... | Yes |
Proposition 5.4. Let \( \left( {X, g}\right) \) be a pseudo Riemannian manifold, and let \( x \in X, \) Fix \( v, w \in {T}_{x}X \) . Then\n\n\[ \n{\exp }_{x}^{ * }\left( {tv}\right) \left( g\right) \left( {w, w}\right) = {w}^{2} + \frac{1}{3}{R}_{2}\left( {v, w}\right) {t}^{2} + O\left( {t}^{3}\right) \;\text{ for }t ... | Proof. From the theory of Jacobi lifts, applied to \( \alpha \left( t\right) = {\exp }_{x}\left( {tv}\right) \), we have the formula\n\n\[ \n\frac{1}{t}{\eta }_{w}\left( t\right) = T{\exp }_{x}\left( {tv}\right) w\n\]\n\nTherefore modulo functions which are \( O\left( {t}^{3}\right) \) for \( t \rightarrow 0 \), we get... | Yes |
Lemma 1.1. Let \( \left( {X, g}\right) \) be a pseudo-Riemannian manifold and let \( \alpha \) be a geodesic. Let \( \eta \) be a Jacobi lift, and\n\n\[ f\left( s\right) = \eta {\left( s\right) }^{2} = \langle \eta \left( s\right) ,\eta \left( s\right) {\rangle }_{g}. \]\n\nThen\n\n\[ {f}^{\prime } = 2{\left\langle {D}... | Proof. The first derivative comes from the defining property of the Levi-Civita (metric) derivative along curves, as in Chapter VIII, Theorem 4.3. This same reference then also yields the second derivative\n\n\[ {f}^{\prime \prime } = 2{\left\langle {D}_{{\alpha }^{\prime }}^{2}\eta ,\eta \right\rangle }_{g} + 2{\left\... | Yes |
Theorem 1.2. Let \( X \) be a Riemannian manifold with \( {R}_{2} \geqq 0 \) (semi-negative curvature). Let \( \alpha \) be a geodesic and \( \eta \) a Jacobi lift with \( \eta \left( 0\right) = 0 \) but \( {D}_{{\alpha }^{\prime }}\eta \left( 0\right) \neq 0 \) . Let\n\n\[ f\left( s\right) = \eta {\left( s\right) }^{2... | Proof. Immediate from the definitions and assumption on \( {R}_{2} \), taking Lemma 1.1 into account. | No |
Lemma 1.3. Let \( \eta \) be the Jacobi lift of \( \alpha \) coming from its \( \left( {\beta ,\zeta }\right) \) -variation at the beginning point. Let \( f = {\eta }^{2} . Then\n\n\[ \n{f}^{\prime }\left( 0\right) = 2{\left\langle {D}_{\eta \left( 0\right) }\zeta ,\eta \left( 0\right) \right\rangle }_{g}.\n\]\n | Proof. Starting with the expression in Lemma 1.1, we get\n\n\[ \n{f}^{\prime }\left( 0\right) = 2{\left\langle {D}_{{\alpha }^{\prime }\left( 0\right) }\eta ,\eta \right\rangle }_{g}\left( 0\right)\n\]\n\n\[ \n= 2{\left\langle {D}_{\zeta \left( 0\right) }\eta ,\eta \left( 0\right) \right\rangle }_{g}\n\]\n\n\[ \n= 2{\l... | No |
Proposition 1.4. Let \( X \) be a Riemannian manifold and let \( Y \) be a totally geodesic submanifold. Let \( \alpha \) be a geodesic in \( X,\alpha \left( 0\right) = y \in Y \) . Let \( \sigma \) be the \( \left( {\beta ,\zeta }\right) \) -variation of \( \alpha \) defined above. We suppose that \( \beta \) is a geo... | Proof. Since \( Y \) is totally geodesic, the second fundamental form \( {h}_{12}\left( {\eta ,\zeta }\right) \left( 0\right) = 0 \) by Theorem 1.4 of Chapter XIV. Then combining Theorem 1.5 of Chapter XIV and Lemma 1.3 which was just proved, we obtain \( {f}^{\prime }\left( 0\right) = 0 \) . The other assertions are i... | Yes |
Lemma 2.1. Let \( \\beta \) be a geodesic in \( Y \) with \( \\beta \\left( 0\\right) = y \) and \( {\\beta }^{\\prime }\\left( 0\\right) = z \) . For \( v \\in {N}_{{y}_{0}}Y \\), let \( {\\beta }_{v}\\left( t\\right) = \\left( {\\beta \\left( t\\right), v}\\right) \) . Let\n\n\\[ \n{\\varphi }_{v}\\left( t\\right) = ... | Proof. This is just the chain rule. | No |
Proposition 2.2. Let \( X \) be convex and let \( Y \) be a totally geodesic submanifold. Let \( {y}_{0}, y \in Y \) . Let \( v \in {N}_{{y}_{0}}Y \) . Let \( \beta ,\zeta \) be the curves defined in (1) and (2) above, and let \( \eta \) be the Jacobi lift associated with the variation \( \sigma \) defined in (3). Then... | Proof. Putting \( s = 0 \) in the definition of \( \sigma \), we obtain\n\n\[ \sigma \left( {0, t}\right) = {\exp }_{\beta \left( t\right) }\left( 0\right) = \beta \left( t\right) \]\n\nso the value \( \eta \left( 0\right) = z \) drops out. For \( \eta \left( 1\right) \), we just apply Lemma 2.1 to conclude the proof. | No |
Theorem 2.3. Let \( X \) be a convex complete Riemannian manifold, and let \( Y \) be a totally geodesic submanifold. Then \( Y \) is also convex complete. Let \( {y}_{0} \in Y \) and let\n\n\[ \n{P}_{{y}_{0}} : Y \times {N}_{{y}_{0}}Y \rightarrow {NY} \n\]\n\nbe the map such that for each \( y \in Y \) and \( v \in {N... | Proof. This simply amount to the fact that flows of differential equations depend smoothly on parameters, and that parallel translation is invertible by parallel translation along the reverse geodesic.\n\nGiven a chart \( U \) of \( Y \) at \( {y}_{0} \), it follows that \( U \times {T}_{{y}_{0}} \) is a chart at the c... | Yes |
Theorem 2.4 (Wu). Let \( X \) be a Cartan-Hadamard manifold. Let \( Y \) be a totally geodesic submanifold. Fix a point \( {y}_{0} \in Y \) . Let\n\n\[ E : Y \times {N}_{{y}_{0}}Y \rightarrow X \]\n\nbe defined by \( E\left( {y, v}\right) = {\exp }_{y}{P}_{{y}_{0}}^{y}\left( v\right) \) for \( v \in {N}_{{y}_{0}}Y \) .... | Proof. For \( z \in {T}_{y}Y \) and \( v, w \in {N}_{{y}_{0}}Y \) we have to show that\n\n\[ \parallel {TE}\left( {y, v}\right) \left( {z, w}\right) \parallel \geqq \parallel \left( {z, w}\right) \parallel . \]\n\nThe product Hilbert space metric by definition gives\n\n\[ \parallel \left( {z, w}\right) {\parallel }^{2}... | Yes |
Lemma 2.6. Let \( \mathbf{E},\mathbf{F} \) be Banach spaces. Let \( \{ A\left( s\right) \} \left( {0 \leqq s \leqq r}\right) \) be a continuous family of bounded operators, such that \( A\left( s\right) : \mathbf{E} \rightarrow \mathbf{F} \) is invertible for \( 0 \leqq s < r \), and there is a uniform lower bound \( c... | Proof. We write\n\n\[ A{\left( s\right) }^{-1} - A{\left( {s}^{\prime }\right) }^{-1} = A{\left( {s}^{\prime }\right) }^{-1}\left( {A\left( {s}^{\prime }\right) - A\left( s\right) }\right) A{\left( s\right) }^{-1}. \]\n\nTaking the norm, we see that the family \( \left\{ {A{\left( s\right) }^{-1}}\right\} \) is Cauchy,... | Yes |
Lemma 3.1. Let the notation be as above, with \( \varphi \left( s\right) = f\left( s\right) /{s}^{2} \) . Then\n\n\[ \n{\varphi }^{\prime }\left( s\right) = \frac{2}{{s}^{2}}{\eta }^{2}\left( s\right) \left( {h\left( s\right) - \frac{1}{s}}\right) \;\text{ so }\;{\varphi }^{\prime }/\varphi \left( s\right) = 2\left( {h... | Proof. Ordinary differentiation. | No |
Lemma 3.2. Let \( \alpha = {\pi \eta } \). Then\n\n\[ \n{h}^{\prime } = \frac{{\left( {D}_{ * }\eta \right) }^{2}}{{\eta }^{2}} + \frac{\left\langle {D}_{ * }^{2}\eta ,\eta \right\rangle }{{\eta }^{2}} - 2\frac{{\left\langle {D}_{ * }\eta ,\eta \right\rangle }^{2}}{{\left( {\eta }^{2}\right) }^{2}}\n\]\n\n\[ \n= \frac{... | Proof. The first equation for \( {h}^{\prime } \) is immediate from the definition of the Levi-Civita metric derivative. The second comes from the definition of \( {R}_{2} \) and the Jacobi equation for \( \eta \), as well as the definition of the orthogonal term. This concludes the proof. | No |
Lemma 3.3. Let \( {h}_{1}, h \) be a pair of functions on some interval, satisfying\n\n\[ \n{h}_{1}^{\prime } \leqq - {h}_{1}^{2}\;\text{ and }\;{h}^{\prime } \geqq - {h}^{2}.\n\]\n\nThen\n\n\[ \n{\left( \left( {h}_{1} - h\right) {e}^{\int \left( {{h}_{1} + h}\right) }\right) }^{\prime } \leqq 0.\n\]\n\nSo if \( {h}_{1... | Proof. First note that a constant of integration added to the indefinite integral in the inequality would not affect the truth of the inequality. Next, routine differentiation yields\n\n\[ \n{\left( \left( {h}_{1} - h\right) {e}^{\int \left( {{h}_{1} + h}\right) }\right) }^{\prime } = \left( {{h}_{1}^{\prime } - {h}^{\... | Yes |
Theorem 3.4. Let \( X \) be a Riemannian manifold and \( \eta \) the Jacobi lift of a curve in \( X \) . Assume \( \eta \left( 0\right) = 0 \) but \( {D}_{ * }\eta \left( 0\right) \neq 0 \) . Suppose \( {R}_{2} \geqq 0 \) (seminegative curvature). Let \( h \) be as in (1), defined on an interval \( J = \left( {0, b}\ri... | Proof. Suppose \( {h}_{0}\left( {s}_{1}\right) > h\left( {s}_{1}\right) \) for some \( {s}_{1} \in J \) . Then for some \( \delta > 0 \) ,\n\n\[ \n{h}_{0}\left( {{s}_{1} + \delta }\right) \geqq h\left( {s}_{1}\right)\n\]\n\nLet \( {h}_{1}\left( s\right) = 1/\left( {s + \delta }\right) \) . Then \( {h}_{1}^{\prime } = -... | Yes |
Proposition 4.1. Let \( X \) be a differential manifold modeled on a Banach space \( \mathbf{E} \) . Suppose that we are given a covering of \( X \) by open sets corresponding to charts \( U, V,\ldots \), and for each \( U \) we are given a morphism\n\n\[ \n{B}_{U} : U \rightarrow {L}_{\mathrm{{sym}}}^{2}\left( {\mathb... | The proof is routine, just like Proposition 3.4 of Chapter IV. | No |
Lemma 4.2. Given a spray or covariant derivative on \( X \), there is a unique vector bundle morphism over \( {TX} \) , \n\n\[ \n{\kappa }_{2} : {TTX} \rightarrow {\pi }^{ * }{TX} \n\] \n\nsuch that over a chart \( U \), we have \n\n\( \left( {5}_{U}\right) \) \n\n\[ \n{\kappa }_{2, U}\left( {x, v, z, w}\right) = \left... | Proof. Let \( h : U \rightarrow V \) be a change of charts, i.e. a differential isomorphism. In Chapter IV, \( §3 \) we gave the change of chart \( \left( {2}_{U}\right) \) of \( {TTX} \) . Let \( H = \left( {h,{h}^{\prime }}\right) \) . Then the change of chart for \( {\left( TTX\right) }_{U} \) is given by the map \n... | Yes |
Theorem 4.3 (Tensorial Splitting Theorem). Given a spray, or covariant derivative on a differential manifold \( X \), the map\n\n\[ \kappa = \left( {{\kappa }_{1},{\kappa }_{2}}\right) : {TTX} \rightarrow {\pi }^{ * }{TX}{ \oplus }_{TX}{\pi }^{ * }{TX} \]\n\n is a vector bundle isomorphism over \( {TX} \) . In the char... | Proof. With the notation \( h, H,\left( {H,{H}^{\prime }}\right) \) as in Lemma 4.2, we conclude that\n\n\[ {\kappa }_{V} \circ \left( {H,{H}^{\prime }}\right) \left( {x, v, z, w}\right) = \left( {h\left( x\right) ,{h}^{\prime }\left( x\right) v,{h}^{\prime }\left( x\right) z,{h}^{\prime }\left( x\right) w}\right) ,\]\... | Yes |
Lemma 4.4. Let \( X \) be a manifold with a spray or covariant derivative \( D \) . There exists a unique vector bundle morphism (over \( \pi \) )\n\n\[ K : {TTX} \rightarrow {TX} \]\n\nsuch that for all vector fields \( \xi ,\zeta \) on \( X \), we have\n\n(8)\n\n\[ {D}_{\xi }\zeta = K \circ {T\zeta } \circ \xi ,\;\te... | Proof. In a chart \( U \), we let the local representation\n\n\[ {K}_{U,\left( {x, v}\right) } : \mathbf{E} \times \mathbf{E} \rightarrow \mathbf{E} \]\n\nbe given by\n\n\( \left( {8}_{U}\right) \)\n\n\[ {K}_{U,\left( {x, v}\right) }\left( {z, w}\right) = w - {B}_{U}\left( {x;v, z}\right) ,\]\n\nso \( K = {S}_{2} \) sa... | Yes |
Theorem 4.5 (Dombrowski Splitting Theorem). Let \( X \) be a manifold with a spray or a covariant derivative. Then the map \[ \left( {{\pi }_{TX},{S}_{1},{S}_{2}}\right) : {TTX} \rightarrow {TX} \oplus {TX} \oplus {TX} \] is an isomorphism of fiber bundles over \( X \) . | Proof. The map is well defined, and the previous chart formulas show that it is both a bijection and a local differential isomorphism. We let readers check this out in the charts to conclude the proof. | No |
Lemma 5.1. Let \( X \) be a manifold with a spray, or equivalently a covariant derivative. Let \( \beta \) be a curve in \( X \), and let \( \zeta \) be a lift of \( \beta \) in TX. Let\n\n\[ \varphi \left( t\right) = {\exp }_{\beta \left( t\right) }\zeta \left( t\right) \]\n\nso \( \varphi \) is a curve in \( X \) . T... | Proof. This is immediate from Theorem 4.3, the local expression (2) for the covariant derivative, and formula (1). | No |
There exists a unique vector bundle morphism over \( X \) , \[ {\mathbf{T}}_{S}\exp : {TX} \oplus {TX} \rightarrow {TX} \] such that the following diagram commutes:  The two vertical maps are vector bundle morphisms,... | Proof. Routine verification that everything makes sense. | No |
Proposition 6.2. Let \( X \) be a Riemannian manifold, and let \( \Omega \) be the canonical 2-form on the tangent bundle. Let \( v \in {TX}, Z, W \in {T}_{v}{TX} \) . Write\n\n\[ \n{SZ} = \left( {{A}_{1},{B}_{1}}\right) \;\text{ and }\;{SW} = \left( {{A}_{2},{B}_{2}}\right) .\n\]\n\nThen the canonical 2-form can be ex... | Proof. This is a routine verification, which nevertheless has to be taken seriously. We use a chart. Write \( Z = \left( {{z}_{1},{z}_{2}}\right) \) and \( W = \left( {{w}_{1},{w}_{2}}\right) \) in the chart, i.e. in \( \mathbf{E} \times \mathbf{E} \) . Put together Chapter VII, \( §7 \), formula (1) for the canonical ... | No |
Theorem 6.3. Let \( X \) be a Riemannian manifold. Let \( \Omega \) be the canonical 2-form on \( {TX} \) . Let \( v \in {TX} \) (so \( v \in {T}_{x}X \) for some \( x \) ), and let \( Z \) , \( W \in {T}_{v}{TX} \) . Let \( \Phi \) be the flow of the spray on \( {TX} \) . Let\n\n\[ \psi \left( s\right) = \Omega \left(... | Proof. We use Proposition 6.2. Let \( {\eta }_{1},{\eta }_{2} \) be the Jacobi lifts of the curve \( s \mapsto {\alpha }_{v}\left( s\right) = \exp \left( {sv}\right) \) with initial conditions\n\n\[ {\eta }_{i}\left( 0\right) = {A}_{i}\;\text{ and }\;{D}_{v}{\eta }_{i}\left( 0\right) = {B}_{i}. \]\n\nThen using Theorem... | Yes |
Proposition 1.1. Let \( \alpha : \left\lbrack {a, b}\right\rbrack \rightarrow X \) be a geodesic. The index form I on \( \operatorname{Lift}\left( \alpha \right) \) also has the expression\n\n\[ I\left( {\eta ,\gamma }\right) = - {\int }_{a}^{b}\left\lbrack {{\left\langle {D}_{{\alpha }^{\prime }}^{2}\eta ,\gamma \righ... | Proof. From the defining property of the metric derivative, we know that\n\n\[ \partial {\left\langle {D}_{{\alpha }^{\prime }}\eta ,\gamma \right\rangle }_{g} = {\left\langle {D}_{{\alpha }^{\prime }}^{2}\eta ,\gamma \right\rangle }_{g} + {\left\langle {D}_{{\alpha }^{\prime }}\eta ,{D}_{{\alpha }^{\prime }}\gamma \ri... | Yes |
Theorem 1.2. Let \( \eta \in \operatorname{Lift}\left( \alpha \right) \) . Then \( I\left( {\eta ,\gamma }\right) = 0 \) for all \( \gamma \in {\operatorname{Lift}}_{0}\left( \alpha \right) \) if and only if\n\n\[{\left( {D}_{{\alpha }^{\prime }}^{2}\eta - R\left( {\alpha }^{\prime },\eta \right) {\alpha }^{\prime }\ri... | Proof. If \( \eta \) is a Jacobi lift, then by definition\n\n\[{D}_{{\alpha }^{\prime }}^{2}\eta = R\left( {{\alpha }^{\prime },\eta }\right) {\alpha }^{\prime }\n\nso \( I\left( {\eta ,\gamma }\right) = 0 \) for all \( \gamma \in {\operatorname{Lift}}_{0}\left( \alpha \right) \) . Conversely, assume this is the case. ... | Yes |
Lemma 1.4. Let \( \left( {X, g}\right) \) be a pseudo-Riemannian manifold. Let \( \alpha \) be a geodesic (not necessarily parametrized by arc length), and let \( \sigma = \sigma \left( {s, t}\right) \) be a variation of \( \alpha \) (not necessarily by geodesics), so \( \alpha = {\alpha }_{0} \), and \( {\alpha }_{t}\... | Proof. We shall keep in mind that from the definitions,\n\n\[ {D}_{{\alpha }^{\prime }}\eta \left( s\right) = {D}_{1}{\partial }_{2}\sigma \left( {s,0}\right) . \]\nFor the first derivative, we have\n\n\[ {\partial }_{2}e = {\partial }_{2}{\left\langle {\partial }_{1}\sigma ,{\partial }_{1}\sigma \right\rangle }_{g} \]... | Yes |
Corollary 1.5. Let \( \eta \) be a Jacobi lift of \( \alpha \), and \( \sigma \) a variation of \( \alpha \) such that \( \eta \left( s\right) = {\partial }_{2}\sigma \left( {s,0}\right) \) . Assume that \( t \mapsto \sigma \left( {a, t}\right) \) and \( t \mapsto \sigma \left( {b, t}\right) \) are geodesics. Then\n\n\... | Proof. Immediate from Theorem 1.3 and the alternative expressions of Proposition 1.1. | No |
Proposition 1.6. Assumptions being as in Theorem 1.3, suppose that \( {D}_{{\alpha }^{\prime }}\eta \) is orthogonal to \( {\alpha }^{\prime } \) . Then\n\n\[ \n{\left. \frac{d}{dt}L\left( {\alpha }_{t}\right) \right| }_{t = 0} = 0 \n\] | Proof. This is immediate from Lemma 1.4 (1) and I 1. | No |
Theorem 1.7. Suppose that \( \alpha \) is a geodesic whose length is the distance between its end points. Let \( \zeta \in {\operatorname{Lift}}_{0}\left( \alpha \right) \) be orthogonal to \( {\alpha }^{\prime } . Then\n\n\[ I\left( {\zeta ,\zeta }\right) \geqq 0 \] | Proof. I owe the proof to Wu. Define\n\n\[ \sigma \left( {s, t}\right) = {\exp }_{\alpha \left( s\right) }\left( {{t\zeta }\left( s\right) }\right) \]\n\nwith \( 0 \leqq s \leqq b \) and \( 0 \leqq t \leqq \epsilon \) . For each \( t,{\sigma }_{t} \) is a curve, not necessarily a geodesic, joining the endpoints of \( \... | Yes |
Corollary 1.8. Let \( \eta \) be a Jacobi lift of \( \alpha \), and let \( \xi \) be any lift of \( \alpha \), with the same end points as \( \eta \), that is\n\n\[ \n\eta \left( 0\right) = \xi \left( 0\right) \;\text{ and }\;\eta \left( b\right) = \xi \left( b\right) .\n\]\n\nSuppose that \( \eta - \xi \) is orthogona... | Proof. Let \( \zeta = \xi - \eta \) . By Theorem 1.7 we have \( I\left( {\zeta ,\zeta }\right) \geqq 0 \), so by the bilinearity of the index,\n\n(7)\n\n\[ \nI\left( {\xi ,\xi }\right) - {2I}\left( {\eta ,\xi }\right) + I\left( {\eta ,\eta }\right) \geqq 0.\n\]\n\nBut\n\n\[ \nI\left( {\eta ,\xi }\right) = {\left. \left... | Yes |
Proposition 1.9. Let \( f \) be a \( {C}^{2} \) function of a real variable. As in Theorem 1.3, let \( \sigma \) be a variation of \( \alpha \), and let \( \eta \left( s\right) = {\partial }_{2}\sigma \left( {s,0}\right) \) . Assume \( {D}_{{\alpha }^{\prime }}\eta \) orthogonal to \( {\alpha }^{\prime } \) . Then\n\n\... | Proof. Let \( F\left( t\right) = f\left( {L\left( {\alpha }_{t}\right) }\right) \) . Then\n\n\[ \n{F}^{\prime }\left( t\right) = {f}^{\prime }\left( {L\left( {\alpha }_{t}\right) }\right) \frac{d}{dt}L\left( {\alpha }_{t}\right)\n\]\n\nand\n\n\[ \n{F}^{\prime \prime }\left( t\right) = {f}^{\prime \prime }\left( {L\left... | Yes |
Proposition 2.1. The exponential map is metric preserving on rays from the origin. | Proof. We have directly from the definitions\n\n\[ \frac{1}{2}\frac{{f}^{\prime }}{f}\left( r\right) = \left\langle {{D}_{{\alpha }^{\prime }}\zeta \left( r\right) ,\zeta \left( r\right) }\right\rangle = {\left. \left\langle {D}_{{\alpha }^{\prime }}\zeta ,\zeta \right\rangle \right| }_{0}^{r} \]\n\n\[ = {I}_{0}^{r}\le... | Yes |
Lemma 2.2. Assume that \( w \bot u \) and that \( \alpha \) is contained in a convex open set. Given \( r \) as above, there exists a lift \( \xi \) of \( \alpha \) such that on \( \left\lbrack {0, r}\right\rbrack ,\xi \neq 0 \) , \( \xi \bot {\alpha }^{\prime } \), and | Proof. We have directly from the definitions\n\n\[ \frac{1}{2}\frac{{f}^{\prime }}{f}\left( r\right) = \left\langle {{D}_{{\alpha }^{\prime }}\zeta \left( r\right) ,\zeta \left( r\right) }\right\rangle = {\left. \left\langle {D}_{{\alpha }^{\prime }}\zeta ,\zeta \right\rangle \right| }_{0}^{r} \]\n\n\[ = {I}_{0}^{r}\le... | Yes |
Lemma 2.3. Let \( X \) be any manifold with a spray, and let \( \alpha : \left\lbrack {a, b}\right\rbrack \rightarrow X \) be a curve in \( X \) . Let \( \beta : \left\lbrack {a, b}\right\rbrack \rightarrow {T}_{\alpha \left( a\right) } \) be a curve in \( {T}_{\alpha \left( a\right) } \), let \( P \) be parallel trans... | Proof. We prove the relation in a chart, where we have the formula\n\n\[ \n{D}_{{\alpha }^{\prime }}\xi = {\xi }^{\prime } - B\left( {\alpha ;{\alpha }^{\prime },\xi }\right) .\n\]\n\nLet \( \gamma \left( {t, v}\right) \) be parallel translation of \( v \in {T}_{\alpha \left( a\right) } \) . Then\n\n\[ \n{\xi }^{\prime... | Yes |
Lemma 2.4. Let \( h\left( s\right) = {s}^{2}{w}^{2} \) . Then\n\n\[ \mathop{\lim }\limits_{{s \rightarrow 0}}f\left( s\right) /h\left( s\right) = 1 \] | Proof. This is immediate from the first term of the Taylor expansion given in Chapter IX, Proposition 5.1. | No |
Theorem 2.5. Under the basic assumptions, assume that \( w \bot u \) . Let \( {U}_{x} \) be an open convex neighborhood of \( x \), and \( {V}_{x} \) an open neighborhood of \( {0}_{x} \) such that \( {\exp }_{x} : {V}_{x} \rightarrow {U}_{x} \) is an isomorphism. We suppose \( \alpha \) is contained in \( {U}_{x} \) .... | Proof. By lemma 2.2, for \( \epsilon > 0 \) we find\n\n\[ \n{\int }_{\epsilon }^{r}{f}^{\prime }/f \leqq {\int }_{\epsilon }^{r}{h}^{\prime }/h + \text{ the Riemann tensor integral. }\n\]\n\nSince by hypothesis, the Riemann tensor integrand is \( \leqq 0 \), we obtain\n\n\[ \n\log f\left( r\right) /h\left( r\right) \le... | Yes |
Theorem 2.6. Let \( \left( {X, g}\right) \) be a Riemannian manifold. Let \( x \in X \) and let \( {\exp }_{X} : {V}_{x} \rightarrow {U}_{x} \) be an isomorphism of a neighborhood of \( {0}_{x} \) with an open convex neighborhood of \( x \) . Suppose \( g \) has curvature \( \geqq 0 \) on \( {U}_{x} \) . Then \( {\exp ... | Proof. We let \( u \) be the unit vector in the direction of \( v, v = {bu} \) . If \( w \) is orthogonal to \( u \), then the inequality of Theorem 2.5 together with (1) shows that\n\n\[ {\begin{Vmatrix}T{\exp }_{x}\left( ru\right) w\end{Vmatrix}}^{2} < \parallel w{\parallel }^{2}. \]\n\nFor arbitrary \( w \), we writ... | Yes |
Theorem 3.1 (Serre). Let \( X \) be a Bruhat-Tits space. Let \( S \) be a bounded subset of \( X \) . Then there exists a unique closed ball \( {\overline{\mathbf{B}}}_{r}\left( {x}_{1}\right) \) in \( X \) of minimal radius containing \( S \) . | Proof. We first prove uniqueness. Suppose there are two balls \( {\overline{\mathbf{B}}}_{r}\left( {x}_{1}\right) \) and \( {\overline{\mathbf{B}}}_{r}\left( {x}_{2}\right) \) of minimal radius containing \( S \), but \( {x}_{2} \neq {x}_{1} \) . Let \( x \) be any point of \( S \), so \( d\left( {x,{x}_{2}}\right) \le... | Yes |
Theorem 3.2 (Bruhat-Tits). Let \( X \) be a Bruhat-Tits metric space. Let \( G \) be a group of isometries of \( X \), with the action of \( G \) denoted by \( \left( {g, x}\right) \mapsto g \cdot x \) . Suppose \( G \) has a bounded orbit (this occurs if, for instance, \( G \) is compact). Then \( G \) has a fixed poi... | Proof. Let \( p \in X \) and let \( G \cdot p \) be the orbit. Let \( {\overline{\mathbf{B}}}_{r}\left( {x}_{1}\right) \) be the unique closed ball of minimal radius containing this orbit. For any \( g \in G \), the image \( g \cdot {\overline{\mathbf{B}}}_{r}\left( {x}_{1}\right) = {\overline{\mathbf{B}}}_{r}\left( {x... | Yes |
Corollary 3.3. Let \( G \) be a topological group, \( H \) a closed subgroup. Let \( K \) be a subgroup of \( G \), so that \( K \) acts by translation on the coset space \( G/H \) . Suppose \( G/H \) has a metric (distance function) such that translation by elements of \( K \) are isometries, \( G/H \) is a Bruhat-Tit... | Proof. By Corollary 3.2, the action of \( K \) has a fixed point, i.e. there exists a coset \( {xH} \) such that \( {kxH} = {xH} \) for all \( k \in K \) . Then \( {x}^{-1}{KxH} \subset H \) , whence \( {x}^{-1}{Kx} \subset H \), as was to be shown. | Yes |
Proposition 3.4. A complete Riemannian manifold satisfying EMI is a Bruhat-Tits space. A Cartan-Hadamard manifold is a Bruhat-Tits space. | Proof. On a Hilbert space, we have equality in the parallelogram law. Using the hypothesis in EMI with \( z \) as the midpoint, we see that the left side in the parallelogram law remains the same under the exponential map, the right side only increases, and hence the semi parallelogram law falls out. | No |
Theorem 3.5. Let \( X \) be a Riemannian manifold. The following three conditions are equivalent:\n\n(a) The curvature is seminegative.\n\n(b) The exponential map is locally metric semi-increasing at every point.\n\n(c) The semi parallelogram law holds locally on \( X \) . | Proof. This is merely putting together results which have been proved individually. Theorem 3.6 of Chapter IX shows that (a) implies (b). That (b) implies (c) is a local version of Proposition 3.4. Indeed, the parallelogram law holds in the tangent space \( {T}_{z} \), and if the exponential map at \( z \) is metric se... | Yes |
Theorem 3.6. Let \( \mathbf{E} \) be a Hilbert space and \( X \) a Riemannian manifold. Let \( h : \mathbf{E} \rightarrow X \) be a differential isomorphism which is metric semi-increasing, that is\n\n\[{\left| Th\left( v\right) w\right| }_{h\left( v\right) }^{2} \geqq {\left| w\right| }_{\mathbf{E}}^{2}\;\text{ for al... | Proof. The map \( {h}^{-1} : X \rightarrow \mathbf{E} \) is distance semi-decreasing. If \( \left\{ {x}_{n}\right\} \) is Cauchy in \( X \), then \( \left\{ {{h}^{-1}\left( {x}_{n}\right) }\right\} \) is Cauchy in \( \mathbf{E} \), converging to some point \( v \) , and by continuity of \( h \), it follows that \( \lef... | Yes |
Theorem 4.1. Let \( X \) be a Cartan-Hadamard manifold. Let \( Y \) be a totally geodesic submanifold. Then:\n\n(i) \( Y \) is a Cartan-Hadamard manifold.\n\n(ii) Given two distinct points of \( Y \), the unique geodesic in \( X \) passing through these points actually lies in \( Y \) . | Proof. Note that from the definition of a totally geodesic submanifold, it follows that the exponential map on \( X \), restricted to \( {TY} \), is equal to the\nexponential map on \( Y \), or in a formula, for \( y \in Y \) ,\n\n\[{\exp }_{y, Y} = {\exp }_{y, X}\;\text{ restricted to }{T}_{y}Y.\]\n\nBy hypothesis and... | Yes |
Proposition 4.2. Let \( X \) be a complete Riemannian manifold, such that given two distinct points of \( X \), there is a unique geodesic passing through these two points. Let \( Y \) be a closed submanifold. Suppose that locally, given two distinct points in \( Y \), the unique geodesic segment in \( X \) joining the... | Proof. I owe the following simple argument to Wu. One has mostly to prove that a \( Y \) -geodesic is an \( X \) -geodesic. Let \( \alpha : \lbrack 0, c) \rightarrow X \) be a geodesic in \( X \) having initial conditions in \( Y \), that is\n\n\[ \alpha \left( 0\right) = y \in Y\;\text{ and }\;{\alpha }^{\prime }\left... | Yes |
Lemma 4.3. Let \( X \) be a Cartan-Hadamard manifold. Let \( Y \) be a totally geodesic submanifold. Then the map\n\n\[ \n{\exp }_{NY} : {NY} \rightarrow X \n\]\n\nis a bijection. | Proof. The argument will follow the same pattern that is used routinely to show that given a point not in a closed subspace of a Hilbert space, there is a line through the point perpendicular to the subspace. We first prove that given \( x \in X \) but \( x \notin Y \), there exists a point \( {y}_{0} \in Y \) such tha... | Yes |
Theorem 5.1 (Rauch Comparison Theorem). Let \( \left( {X,{g}_{X}}\right) \) and \( \left( {Y,{g}_{Y}}\right) \) be Riemannian manifolds of the same dimension, which may be infinite. Let \( {\alpha }_{X} \) (resp. \( {\alpha }_{Y} \) ) be geodesics in \( X \) (resp. \( Y \) ), parametrized by arc length, and defined on ... | Proof. We shall use the definition of the index and Proposition 1.1, that is, for a Jacobi lift \( \eta \) of \( \alpha \) such that \( \eta \left( a\right) = 0 \) we have\n\n(1)\n\n\[ \n{I}_{a}^{s}\left( {\eta ,\eta }\right) = {\int }_{a}^{s}{\left( {D}_{{\alpha }^{\prime }}\eta \right) }^{2} + {R}_{2}\left( {{\alpha ... | Yes |
Theorem 1.1. The association \( g \mapsto \left\lbrack g\right\rbrack \) is a representation of \( G \) in the group of isometries of \( {\operatorname{Pos}}_{n} \), that is each \( \left\lbrack g\right\rbrack \) is an isometry. | Proof. First we note that \( \left\lbrack g\right\rbrack \) can also be viewed as a map on the whole vector space \( {\operatorname{Sym}}_{n} \), and this map is linear as a function of such matrices.\n\nHence its derivative is given by\n\n\[ \n{\left\lbrack g\right\rbrack }^{\prime }\left( p\right) w = g{w}^{t}g\;\tex... | Yes |
Theorem 1.3. The exponential map \( \exp : {\operatorname{Sym}}_{n} \rightarrow {\operatorname{Pos}}_{n} \) is metric preserving on a line through the origin. | Proof. Such a line has the form \( t \mapsto {tv} \) with some \( v \in {\operatorname{Sym}}_{n}, v \neq 0 \) . We need to prove\n\n\[ \n{\left| v\right| }_{\mathrm{{tr}}}^{2} = {\left| {\exp }^{\prime }\left( tv\right) v\right| }_{\exp {tv}}^{2} \n\]\n\nNote that\n\n\[ \n\frac{d}{dt}\exp \left( {tv}\right) = {\exp }^{... | Yes |
Theorem 1.4. Let \( p, q \in {\operatorname{Pos}}_{n} \) . Let \( {a}_{1},\ldots ,{a}_{n} \) be the roots of \( \det \left( {{tp} - q}\right) \) .\n\nThen\n\n\[ \operatorname{dist}\left( {p, q}\right) = \sum {\left( \log {a}_{i}\right) }^{2}. \] | Proof. Suppose first \( p = e \) and \( q \) is the diagonal matrix of \( {a}_{1},\ldots ,{a}_{n} \) . Let \( v = \log q \), so \( v \) is diagonal with components \( \log {a}_{1},\ldots ,\log {a}_{n} \) . The theorem is then a consequence of Theorem 1.3, since \( {v}^{2} \) has components \( {\left( \log {a}_{i}\right... | Yes |
Lemma 2.1. The maps \( {F}_{v} \) and \( {\exp }^{\prime }\left( v\right) \) are hermitian with respect to the tr-scalar product on \( \mathcal{A} \) . If \( v \in \operatorname{Sym} \), then \( {F}_{v} \) and \( {\exp }^{\prime }\left( v\right) \) map Sym into itself. | Proof. A routine verification gives for \( u, v, w \in \mathcal{A} \) :\n\n\[ \operatorname{tr}\left( {{F}_{v}\left( w\right) u}\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }\frac{1}{n!}\mathop{\sum }\limits_{{r + s = n - 1}}\operatorname{tr}\left( {\exp \left( {-v/2}\right) {v}^{r}w{v}^{s}\exp \left( {-v/2}\right... | Yes |
Lemma 2.2. Let \( v \in \operatorname{Sym} \) . Then \( {D}_{v}^{2} \) is hermitian on Sym. | Proof. Again this is routine, namely:\n\n\[ \n{D}_{v}\left( w\right) = {vw} - {wv} \n\]\n\n\[ \n{D}_{v}^{2}\left( w\right) = {v}^{2}w - {2vwv} + w{v}^{2} \n\]\n\n\[ \n\left( {{D}_{v}^{2}w}\right) u = {v}^{2}{wu} - {2vwvu} + w{v}^{2}u \n\]\n\n\[ \nw{D}_{v}^{2}u = w{v}^{2}u - {2wvuv} + {wu}{v}^{2}. \n\]\n\nApplying tr to... | Yes |
Lemma 2.3. For any \( v \in \mathcal{A} \), we have \( {D}_{v}{F}_{v} = {D}_{v}f\left( {D}_{v}\right) \) . | Proof. Let \( t \mapsto x\left( t\right) \) be a smooth curve in \( \mathcal{A} \). Then\n\n\[ x\left( {\exp x}\right) = \left( {\exp x}\right) x. \]\n\nDifferentiating both sides gives\n\n\[ {x}^{\prime }\exp x + x{\left( \exp x\right) }^{\prime } = {\left( \exp x\right) }^{\prime }x + \left( {\exp x}\right) {x}^{\pri... | Yes |
Theorem 2.4. Let \( v \in \operatorname{Sym} \). Then \( {F}_{v} = f\left( {D}_{v}\right) \) on Sym. Hence for \( w \in \) Sym, we have\n\n\[{\exp }^{\prime }\left( v\right) w = \exp \left( {v/2}\right) \cdot f\left( {D}_{v}\right) w \cdot \exp \left( {v/2}\right) . | Proof. Let \( {h}_{v} = {F}_{v} - f\left( {D}_{v}\right) \). Then \( {h}_{v} : \operatorname{Sym} \rightarrow \) Sym is hermitian, and its image is contained in the subspace \( E = \operatorname{Ker}{D}_{v} \cap \operatorname{Sym} \). Since Sym is assumed finite dimensional, it is the direct sum of \( E \) and its orth... | Yes |
Theorem 2.6. The exponential map exp is tr-norm semi-increasing on Sym, that is for all \( v, w \in \operatorname{Sym} \), putting \( p = \exp \left( v\right) \), we have\n\n\[ \n{\left| w\right| }_{\mathrm{{tr}}}^{2} = \operatorname{tr}\left( {w}^{2}\right) \leqq \operatorname{tr}\left( {\left( {p}^{-1}{\exp }^{\prime... | Proof. The right side of the above inequality is equal to\n\n\[ \n\operatorname{tr}\left( {\left( {p}^{-1}{\exp }^{\prime }\left( v\right) w\right) }^{2}\right) = \operatorname{tr}\left( {\left( \exp \left( -v/2\right) \cdot {\exp }^{\prime }\left( v\right) w \cdot \exp \left( -v/2\right) \right) }^{2}\right.\n\]\n\n\[... | Yes |
For each \( v \in \operatorname{Sym} \), the maps\n\n\[ \n{F}_{v}\;\text{ and }\;{\exp }^{\prime }\left( v\right) : \operatorname{Sym} \rightarrow \operatorname{Sym} \n\]\n\nare linear automorphisms. | Proof. Theorem 2.6 shows that Ker \( {\exp }^{\prime }\left( v\right) = 0 \), and \( {\exp }^{\prime }\left( v\right) \) is a linear isomorphism. The statement for \( {F}_{v} \) then follows because \( {F}_{v} \) is composed of \( {\exp }^{\prime }\left( v\right) \) and multiplicative translations by invertible element... | Yes |
Theorem 3.1. The Hermitian operator \( {A}_{v} \) is invertible on Sym. Furthermore, we have the formula\n\n\[ \n{A}_{v} = {\exp }^{\prime }\left( v\right) {g}_{0}\left( {D}_{v}^{2}\right) \;\text{ on Sym. } \n\] | Proof. From the definitions and Theorem 2.4, we know that\n\n\[ \n{J}_{v} = {\exp }^{\prime }\left( v\right) = \exp \left( {{L}_{v}/2}\right) \exp \left( {{R}_{v}/2}\right) f\left( {D}_{v}\right) \;\text{ on Sym. } \n\]\n\nNote that \( \exp {L}_{v} = {L}_{p} \) and \( \exp {R}_{v} = {R}_{p} \) . Abbreviate \( L = {L}_{... | Yes |
Lemma 3.2. Suppose \( X = \exp \left( V\right) \) symmetric. Given \( p, q \in X \) there exists \( y \in X \) such that \( {ypy} = q \) . In other words, \( X \) acts transitively on itself. | Proof. The condition \( {ypy} = q \) is equivalent with\n\n\[ \n{p}^{1/2}y{p}^{1/2}{p}^{1/2}y{p}^{1/2} = {p}^{1/2}q{p}^{1/2}\; \Leftrightarrow \;{\left( {p}^{1/2}y{p}^{1/2}\right) }^{2} = {p}^{1/2}q{p}^{1/2}, \n\]\n\n\[ \n\Leftrightarrow \;{p}^{1/2}y{p}^{1/2} = {\left( {p}^{1/2}q{p}^{1/2}\right) }^{1/2}, \n\]\n\n\[ \n\... | Yes |
Theorem 3.3. Let \( V \) be a vector subspace of \( \operatorname{Sym} \), and let \( X = \exp \left( V\right) \) . Then \( X \) is a symmetric submanifold of Pos if and only if: SYM 2. The map \( {D}_{v}^{2} \) maps \( V \) into itself for all \( v \in V \) . | Proof. Suppose that \( {D}_{u}^{2} \) maps \( V \) into itself for all \( u \in V \) . Note that \( {g}_{0} \) is actually real analytic, and the above equation is an ordinary differential equation for \( \xi \left( t\right) \) in \( V \) . It has a unique solution with initial condition \( \xi \left( 0\right) = \log p... | Yes |
Lemma 3.5. Let \( L \) be a Lie algebra and \( V \) a linear subspace. Then \( V \) is stable under \( {D}_{v}^{2} \) for all \( v \in V \) if and only if \( V \) is stable under all operators \( {D}_{u}{D}_{v} \) with \( u, v \in V \) . | Proof. Applying the hypothesis that \( {D}_{v}^{2} \) leaves \( V \) stable to \( u + v \) (polarization) shows that \( {D}_{u}{D}_{v} + {D}_{v}{D}_{u} \) leaves \( V \) stable, or in other words,\n\n\( \left( *\right) \)\n\n\[ \left\lbrack {u,\left\lbrack {v, w}\right\rbrack }\right\rbrack + \left\lbrack {v,\left\lbra... | Yes |
Theorem 3.7. Let \( X = \exp \left( V\right) \) . Then \( X \) is a geodesic submanifold if and only if \( X \) satisfies the (equivalent) conditions of Theorem 3.3, e.g. \( X \) is a symmetric submanifold. | Proof. Assume \( X \) is symmetric. The image of the line through 0 and an element \( v \in V, v \neq 0 \) is a geodesic which is contained in \( X \) . Since the maps \( x \mapsto {yxy} \) (for \( y \in X \) ) leave \( X \) stable, and act transitively on \( X \), it follows that \( X \) contains the geodesic between ... | Yes |
Theorem 3.9. Let \( R \) be the Riemann tensor. Then at the unit element \( e \) , with \( u, v, w \in {T}_{e}\left( \mathrm{{Pos}}\right) = \) Sym, we have\n\n\[ R\left( {v, w}\right) u = - \left\lbrack {\left\lbrack {v, w}\right\rbrack, u}\right\rbrack \]\n\nand \( {R}_{2}\left( {v, w}\right) = \langle R\left( {v, w}... | Proof. Assume the formula for the 4-tensor \( R \) . Substituting \( u = v \) and taking the tr-scalar product immediately shows that\n\n\[ \langle R\left( {v, w}\right) v, w{\rangle }_{\mathrm{{tr}}} = - 2\operatorname{tr}\left( {{\left( vw\right) }^{2} - {v}^{2}{w}^{2}}\right) . \]\n\nHence the semipositivity of \( {... | No |
Theorem 1.1. Suppose \( X \) is pseudo Riemannian and \( D \) is the metric derivative. Let \( \operatorname{gr}\left( \varphi \right) \) denote the gradient of a function \( \varphi \right) . Then\n\n\[ \n{D}^{2}\varphi \left( {\eta ,\zeta }\right) = \left\langle {{D}_{\eta }\operatorname{gr}\left( \varphi \right) ,\z... | Proof. Let \( f = \langle \operatorname{gr}\left( \varphi \right) ,\zeta \rangle = {D}_{\zeta }\varphi \) . By the definition of the metric derivative,\n\n\[ \n{D}_{\eta }f = \left\langle {{D}_{\eta }\operatorname{gr}\left( \varphi \right) ,\zeta }\right\rangle + \left\langle {\operatorname{gr}\left( \varphi \right) ,{... | Yes |
Proposition 1.2. For each vector field \( \xi, Q\left( {\eta ,\zeta }\right) \xi \) defines a bilinear tensor as a function of \( \left( {\eta ,\zeta }\right) \) . Furthermore, just as with functions, we have\n\n\[ \left( {{D}^{2}\xi }\right) \left( {\eta ,\zeta }\right) = Q\left( {\eta ,\zeta }\right) \xi \] | Proof. The expression \( Q\left( {\eta ,\zeta }\right) \) is well defined at each point of \( X \), and the local expression shows that it is a section of the vector bundle of bilinear maps of \( {TX} \) into \( {TX} \) . The formula relating it to \( {D}^{2} \) is proved by exactly the same argument as (1). Note that ... | Yes |
Proposition 1.3. For all vector fields \( \eta ,\zeta \) we have\n\n\[ Q\left( {\eta ,\zeta }\right) - Q\left( {\zeta ,\eta }\right) = R\left( {\eta ,\zeta }\right) \] | Proof. This is a short computation, namely:\n\n\[ Q\left( {\eta ,\zeta }\right) - Q\left( {\zeta ,\eta }\right) = {D}_{\eta }{D}_{\zeta } - {D}_{\zeta }{D}_{\eta } - {D}_{{D}_{\eta }\zeta } + {D}_{{D}_{\zeta }\eta } \]\n\n\[ = {D}_{\eta }{D}_{\zeta } - {D}_{\zeta }{D}_{\eta } - {D}_{\left\lbrack \eta ,\zeta \right\rbra... | Yes |
Proposition 1.4.\n\n\[ \left( {{D}^{2}\omega }\right) \left( {\eta ,\zeta }\right) = Q\left( {\eta ,\zeta }\right) \omega \] | Proof. As before,\n\n\[ \left( {{D}^{2}\omega }\right) \left( {\eta ,\zeta }\right) = \left( {{D}_{\eta }\left( {D\omega }\right) }\right) \left( \zeta \right) \]\n\n\[ = {D}_{\eta }{D}_{\zeta }\omega - \left( {D\omega }\right) \left( {{D}_{\eta }\zeta }\right) \]\n\n\[ = {D}_{\eta }{D}_{\zeta }\omega - {D}_{{D}_{\eta ... | Yes |
Proposition 1.5. Let \( A \) be a tensor field of endomorphisms of \( {TX} \), i.e. a section of \( L\left( {{TX},{TX}}\right) \) . As a function of its \( \left( {\eta ,\zeta }\right) \) variables, \( Q\left( {\eta ,\zeta }\right) A \) is tensorial. Furthermore, \( R\left( {\eta ,\zeta }\right) \) is a derivation in t... | Proof. This follows directly from Proposition 1.3 and the fact that \( {D}_{\zeta }\left( {A\eta }\right) = \left( {{D}_{\zeta }A}\right) \eta + A{D}_{\zeta }\eta \), i.e. \( {D}_{\zeta } \) is a derivation. | Yes |
For all vector fields \( \xi ,\eta ,\zeta \) we have\n\n\[ \left\lbrack {\xi ,{D}_{\zeta }\eta }\right\rbrack = Q\left( {\zeta ,\eta }\right) \xi - R\left( {\zeta ,\xi }\right) \eta + {D}_{\left\lbrack \xi ,\zeta \right\rbrack }\eta + {D}_{\zeta }\left\lbrack {\xi ,\eta }\right\rbrack \] | Proof. This is a short computation as follows:\n\n\[ \left\lbrack {\xi ,{D}_{\zeta }\eta }\right\rbrack = {D}_{\xi }{D}_{\zeta }\eta - {D}_{{D}_{\zeta }\eta }\xi \]\n\n\[ = {D}_{\zeta }{D}_{\xi }\eta + {D}_{\left\lbrack \xi ,\zeta \right\rbrack }\eta + R\left( {\xi ,\zeta }\right) \eta - {D}_{{D}_{\zeta }\eta }\xi \]\n... | Yes |
A vector field is a Killing field if and only if its restriction to every geodesic is a Jacobi lift of the geodesic. | First let \( \alpha \) be a geodesic and let \( \xi \) be Killing. We shall give two proofs that the restriction of \( \xi \) to \( \alpha \) is a Jacobi lift of \( \alpha \) . We take \( \rho = {\rho }_{s} \) to be the flow of \( \alpha \), and use Kill 1. Put\n\n\[ \sigma \left( {s, t}\right) = \rho \left( {s,\alpha ... | Yes |
Proposition 3.1. Suppose \( X \) is pseudo Riemannian. The following conditions are equivalent to a vector field \( \xi \) being g-Killing. | Proof (Cf. [O’N 83]). Assume that \( \xi \) is \( g \) -Killing. The property Kill \( {}_{g}\mathbf{1} \) then follows essentially directly from the definition of Lie derivative, because for all \( t,{\rho }_{t}^{ * }\left( g\right) = g \), so the Lie derivative of \( g \) is 0 . The converse is also immediate, because... | No |
For all vector fields \( \xi ,\eta ,\zeta \) we have\n\n\[{\mathcal{L}}_{\xi }\langle \eta ,\zeta \rangle = \left\langle {{\mathcal{L}}_{\xi }\eta ,\zeta }\right\rangle + \left\langle {\eta ,{\mathcal{L}}_{\xi }\zeta }\right\rangle + {\mathcal{L}}_{\xi }\left( g\right) \langle \eta ,\zeta \rangle\n\]\n\[= \left\langle ... | Proof. The first identity exhibits the fact that \( {\mathcal{L}}_{\xi }\left( {g\left( {\eta ,\zeta }\right) }\right) \) satisfies the Leibniz derivation product rule, relative to the triple \( \left( {g,\eta ,\zeta }\right) \), cf. Chapter V, Proposition 5.1, which applies to all multilinear forms, not just alternati... | Yes |
Proposition 3.3. Let \( \xi \) be a g-Killing field.\n\n(i) For any curve \( \alpha \), we have\n\n\[ \left\langle {\xi \left( {{\rho }_{s} \circ \alpha }\right) ,{\left( {\rho }_{s} \circ \alpha \right) }^{\prime }}\right\rangle = \left\langle {\xi \circ \alpha ,{\alpha }^{\prime }}\right\rangle . \]\n\nEquivalently, ... | Proof. The proof of (i) is immediate from (1) and (2). As for (ii), we take the derivative of the function \( \left\langle {\xi \circ \alpha ,{\alpha }^{\prime }}\right\rangle \), and find\n\n\[ \left\langle {{D}_{{\alpha }^{\prime }}\left( {\xi \circ \alpha }\right) ,{\alpha }^{\prime }}\right\rangle \]\n\nbecause \( ... | No |
Proposition 3.4. Let \( \xi \) be a g-Killing field. As usual let \( {\xi }^{2} = \langle \xi ,\xi \rangle \). Then \[ \operatorname{grad}{\xi }^{2} = - 2{D}_{\xi }\xi \] | Proof. Again let \( \alpha \) be a curve with \( \alpha \left( 0\right) = x,{\alpha }^{\prime }\left( 0\right) = v = \xi \left( x\right) \). Consider the derivative \[ h\left( {s, t}\right) = {\partial }_{t}\left\langle {{\partial }_{s}\rho \left( {s,\alpha \left( t\right) }\right) ,{\partial }_{s}\rho \left( {s,\alpha... | Yes |
Corollary 3.5. Let \( \xi \) be a \( g \) -Killing field and \( \rho = \rho \left( {s, x}\right) \) its flow. For fixed \( x \), the curve \( s \mapsto \rho \left( {s, x}\right) \) is a non-constant geodesic if and only if \( \xi \left( x\right) \neq 0 \) and \( d{\xi }^{2}\left( x\right) = 0 \) . | Proof. A curve \( s \mapsto \beta \left( s\right) \) is a geodesic if and only if \( {D}_{{\beta }^{\prime }}{\beta }^{\prime } = 0 \) . In our context, with \( \beta \left( s\right) = \rho \left( {s, x}\right) \), this means \( {D}_{s}{\partial }_{s}\rho \left( {s, x}\right) = 0 \), and so the equivalence is clear fro... | No |
Proposition 4.1. Killing fields form a Lie subalgebra of all vector fields. | Proof. It suffices to prove that if \( \xi ,\eta \) satisfy Kill 2, then so does \( \left\lbrack {\xi ,\eta }\right\rbrack \) . This is a special case of the following lemma, formulated in an abstract context because at this point I want to emphasize the extent to which the present arguments depend only on Lie algebras... | Yes |
Lemma 4.2. Let \( V \) be a Lie algebra (over a commutative ring). Suppose given a bilinear map \( V \times V \rightarrow V \), which we denote\n\n\[ \left( {y, z}\right) \mapsto {yz} \]\n\nand call the bilinear product. Let \( W \) be the submodule of \( V \) consisting of all elements \( w \in V \) such that the map\... | Proof. We carry out the short computation in full, but note that having formulated the result, the computation is forced, and no surprise occurs. For \( v, w \in W \) we have to show that \( \left\lbrack {v, w}\right\rbrack \) acts as a derivation with respect to the bilinear product. We shall use the defining property... | Yes |
Proposition 4.3. Suppose \( D \) is the metric derivative in the pseudo Riemannian case. Then the metric Killing fields form a Lie subalgebra of the Killing fields. | Proof. Property Kill \( {}_{g} \) 2 states that \( \xi \) is Killing if and only if the Lie derivative \( {\mathcal{L}}_{\xi } \) is a derivation with respect to the metric product. As in Lemma 4.2, one proves that the set of vector fields which act as a derivation with respect to such a product is a Lie subalgebra. On... | No |
Proposition 4.4. (a) \( \;\left\lbrack {{\mathfrak{m}}_{p},{\mathfrak{m}}_{p}}\right\rbrack \subset {\mathfrak{h}}_{p} \) . | Proof. For (a), we let \( \xi ,\eta \in {\mathfrak{m}}_{p} \) and we evaluate at \( p \) to get 0, by the definition of \( {\mathfrak{m}}_{p} \) . For (b), let \( \eta ,\zeta \in {\mathfrak{h}}_{p} \) . Then \[ \left\lbrack {\eta ,\zeta }\right\rbrack \left( p\right) = {D}_{\eta }\zeta \left( p\right) - {D}_{\zeta }\et... | Yes |
Proposition 4.5. Assume that the exponential map \( {\exp }_{p} : {T}_{p} \rightarrow X \) is surjective. Then \( {\mathfrak{h}}_{p} \cap {\mathfrak{m}}_{p} = \{ 0\} \), so \( {\mathfrak{h}}_{p} + {\mathfrak{m}}_{p} \) is a direct sum. More generally, the map \[ \operatorname{Kill}\left( X\right) \rightarrow {T}_{p} \t... | Proof. The first assertion is a consequence of the second, so suppose that \( \xi \left( p\right) = 0 \) and \( {D}_{\zeta }\xi \left( p\right) = 0 \) for all vector fields \( \zeta \) . We restrict \( \xi \) to a geodesic \( \alpha \) with \( \alpha \left( 0\right) = p \) . Then by Proposition 2.2, \( \xi \circ \alpha... | Yes |
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