An Introduction to Differentiable Manifolds and Riemannian GeometryAcademic Press, 1986/04/21 - 429 ページ An Introduction to Differentiable Manifolds and Riemannian Geometry |
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16 ページ
... zero vector at each point. Thus T(S*) = Up.s., T.(S*). This set has a natural topology: two tangent vectors X, and Y, are “close” if their initial points p and q and their terminal points are close. Similarly, if M is any of the 2 ...
... zero vector at each point. Thus T(S*) = Up.s., T.(S*). This set has a natural topology: two tangent vectors X, and Y, are “close” if their initial points p and q and their terminal points are close. Similarly, if M is any of the 2 ...
24 ページ
... zero. Thus, for example, a determinant is an analytic function of its entries and, if we exclude n x n matrices of determinant zero (which have no inverses), then each entry in the inverse A of a matrix a is an analytic (and hence C ...
... zero. Thus, for example, a determinant is an analytic function of its entries and, if we exclude n x n matrices of determinant zero (which have no inverses), then each entry in the inverse A of a matrix a is an analytic (and hence C ...
28 ページ
... zero. This completes the proof. I (2.4) Corollary If F and G are of class C" (or smooth) on U and V, respectively, then H = Go F is of class C" (or smooth) on U. Proof We prove only the statement for C*, although we will use the general ...
... zero. This completes the proof. I (2.4) Corollary If F and G are of class C" (or smooth) on U and V, respectively, then H = Go F is of class C" (or smooth) on U. Proof We prove only the statement for C*, although we will use the general ...
34 ページ
... zero on any function fe C*(a) which is constant in a neighborhood of a. Proof Because the map D is linear, it is enough to show that if 1 denotes the constant function of value 1, then D1 = 0. However, D1 = D(1 : 1) = (D1)1 + 1(D1) = D1 ...
... zero on any function fe C*(a) which is constant in a neighborhood of a. Proof Because the map D is linear, it is enough to show that if 1 denotes the constant function of value 1, then D1 = 0. However, D1 = D(1 : 1) = (D1)1 + 1(D1) = D1 ...
39 ページ
... zero at t = 0, and since it is analytic for other values of t. We let h(e” – ||x||*) e” – ||x||*) + h(|x|* – #8°) g(x) = h( Since the denominator is never zero [at either e” – ||x||* or |x|* – #e” or at both h is positive], this is a C ...
... zero at t = 0, and since it is analytic for other values of t. We let h(e” – ||x||*) e” – ||x||*) + h(|x|* – #8°) g(x) = h( Since the denominator is never zero [at either e” – ||x||* or |x|* – #e” or at both h is positive], this is a C ...
目次
1 | |
20 | |
51 | |
Chapter IV Vector Fields on a Manifold | 106 |
Chapter V Tensors and Tensor Fields on Manifolds | 176 |
Chapter VI Integration on Manifolds | 229 |
Chapter VII Differentiation on Riemannian Manifolds | 297 |
Chapter VIII Curvature | 365 |
References | 417 |
Index | 423 |
多く使われている語句
algebra basis bi-invariant C*-vector field compact completes the proof components connected coordinate frames coordinate neighborhood Corollary corresponding countable covariant tensor covering curve p(t defined definition denote derivative diffeomorphism differentiable manifold dimension domain of integration element equations equivalent Euclidean space example Exercise exists fact finite fixed point formula functions geodesic geometry given Gl(n hence homeomorphism homotopy identity imbedding inner product integral curve isometry isomorphism Lemma Let F Lie group G linear map mapping F matrix notation obtain one-parameter subgroup one-to-one open set open subset oriented orthogonal orthonormal parameter plane properly discontinuously properties prove rank real numbers regular submanifold Remark Riemannian manifold Riemannian metric Section Show structure submanifold subspace suppose surface symmetric tangent space tangent vector tensor field Theorem Let topology uniquely determined vector field vector space zero