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(a) Let ( , ) be a Riemannian metric on \(M\), and \(A\) a tensor of type \(\left(\begin{array}{l}1 \\ 1\end{array}\right)\), so that \(A(p): M_{p} \rightarrow M_{p} .\) Define a tensor \(B\) of type \(\left(\begin{array}{l}2 \\\ 0\end{array}\right)\) by $$ B(p)\left(v_{1}, v_{2}\right)=\left(A(p)\left(v_{1}\right), v_{2}\right) $$ If the expression for \(A\) in a coordinate system is $$ A=\sum_{i, j=1}^{n} A_{i}^{j} d x^{i} \otimes \frac{\partial}{\partial x^{j}} $$ show that \(B=\sum_{i, k} B_{i k} d x^{i} \otimes d x^{k}\), where $$ B_{i k}=\sum_{j=1}^{n} A_{i}^{j} g_{j k} $$ (b) Similarly, define a tensor \(C\) of type \(\left(\begin{array}{l}0 \\\ 2\end{array}\right)\) by $$ C(p)\left(\lambda_{1}, \lambda_{2}\right)=\left(A(p)^{*}\left(\lambda_{1}\right), \lambda_{2}\right) $$ Show that if \(C\) has components \(C^{k j}\), then $$ C^{k j}=\sum_{i=1}^{n} g^{k i} A_{i}^{j} $$ The tensors \(B\) and \(C\) are said to be obtained from \(A\) by "raising and lowering indices".

Short Answer

Expert verified
B is given by \( B_{ik} = \sum_{j=1}^{n} A_i^j g_{jk} \); C is given by \( C^{kj} = \sum_{i=1}^{n} g^{ki} A_i^j \). Proved using properties of the Riemannian metric.

Step by step solution

01

Understand the Tensor Definitions

Let a Riemannian metric on manifold M be given as \( ( , ) \). Let A be a tensor of type \( \left( \begin{array}{l} 1 \ 1 \end{array}\right) \), which means that A at point p is a map from the tangent space \( M_p \) to itself.
02

Define Tensor B

Tensor B of type \( \left( \begin{array}{l} 2 \ 0 \end{array}\right) \) is given by \[ B(p)(v_1, v_2) = (A(p)(v_1), v_2) \]. This means B is constructed using the Riemannian metric and tensor A.
03

Expression for A in Coordinates

In coordinates, tensor A is expressed as \[ A = \sum_{i, j=1}^{n} A_i^j dx^i \otimes \frac{\partial}{\partial x^j} \]. This notation decomposes A into its component functions \( A_i^j \) and basis vectors.
04

Expression for B in Coordinates

Tensor B needs to be expressed as \[ B = \sum_{i, k} B_{ik} dx^i \otimes dx^k \]. Therefore, we need to find an expression for \( B_{ik} \).
05

Relate B to A

By definition of B, \[ B_{ik} = B \left( \frac{\partial}{\partial x^i}, \frac{\partial}{\partial x^k} \right) = \left( A \left( \frac{\partial}{\partial x^i} \right), \frac{\partial}{\partial x^k} \right) \].
06

Use Metric

Since the inner product is given by the metric g, we have \[ \left( A \left( \frac{\partial}{\partial x^i} \right), \frac{\partial}{\partial x^k} \right) = \left( \sum_{j=1}^{n} A_i^j \frac{\partial}{\partial x^j}, \frac{\partial}{\partial x^k} \right) \].
07

Simplify

Using the bilinearity and symmetry of the metric, this becomes \[ \sum_{j=1}^{n} A_i^j g_{jk} \]. Hence, \[ B_{ik} = \sum_{j=1}^{n} A_i^j g_{jk} \]. So, we have proved the required expression for B.
08

Define Tensor C

Tensor C of type \( \left( \begin{array}{l} 0 \ 2 \end{array}\right) \) is given by \[ C(p)(\lambda_1, \lambda_2) = (A(p)^*(\lambda_1), \lambda_2) \]. This uses the dual map of A.
09

Expression for A* in Coordinates

In coordinates, \[ A^* = \sum_{i, j=1}^{n} A_j^i dx^i \otimes \frac{\partial}{\partial x^j} \], the dual map will interchange the indices due to the change of covariant and contravariant components.
10

Expression for C in Coordinates

Tensor C is expressed as \[ C = \sum_{k, j} C^{kj} dx^k \otimes dx^j \]. We need to find \( C^{kj} \).
11

Relate C to A

By definition of C, \[ C^{kj} = C \left( dx^k, dx^j \right) = \left( A^* (dx^k), dx^j \right) \].
12

Use Metric Inverse

Since the inner product is given by the inverse metric \( g^{ki} \), we have \[ \left( A^* (dx^k), dx^j \right) = \sum_{i=1}^{n} g^{ki} A_i^j \]. Hence, \[ C^{kj} = \sum_{i=1}^{n} g^{ki} A_i^j \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Tensor Algebra
Tensor algebra forms the foundation for understanding Riemannian metrics and various tensorial operations. A tensor is a multi-linear map that operates over vectors and dual vectors from a vector space. For instance, a type \(\begin{array}{l} 1 \ 1 \end{array} \)\tensor A maps one vector to another vector, taking inputs from the tangent space and acting like a linear operator. This is critical in the step-by-step solution, as the tensor A functions within the tangent space and interacts with the Riemannian metric.One essential aspect of tensor algebra is the tensor product, denoted \(\otimes \), which allows for the construction of higher-order tensors. For example, if you have two basis elements in different spaces, their tensor product combines them into a new basis element in a combined space. This operation expands the utility of tensors beyond simple vector and matrix spaces.When we express tensorial quantities in coordinates, we use basis vectors and co-vectors to form tensor components. For example, the tensor A is written as \[ A = \sum_{i, j=1}^{n} A_{i}^{j} d x^{i} \otimes \frac{\partial}{\partial x^{j}} \]This expression decomposes the tensor into components \(\text{A}_i^j dx^i \otimes \frac{\partial}{\partial x^j} \), showing how it acts over the dual basis and basis vectors. Understanding these basics of tensor algebra is crucial for grasping how tensors behave under coordinate transformations and how different types of tensors interact.
Raising and Lowering Indices
Raising and lowering indices are operations that change the type of tensors, a procedure done using the metric tensor and its inverse. Let’s take tensor A of type \(\begin{array}{l} 1 \ 1 \end{array} \)\by an example, converting it to tensors B and C of types \(\begin{array}{l} 2 \ 0 \end{array} \) and \(\begin{array}{l} 0 \ 2 \end{array} \)\respectively.For tensor B, the indices are lowered utilizing the Riemannian metric, g. As in the step-by-step solution:\( B_{ik} = \sum_{j=1}^{n} A_i^j g_{jk} \)Here, \(\text{g}_{jk} \)\represents the components of the metric tensor that converts the contravariant index j of A to a covariant index, resulting in a tensor B with two covariant indices.Conversely, tensor C is formed by raising indices using the inverse of the metric tensor, \(\text{g}^{ki} \)\( C^{kj} = \sum_{i=1}^{n} g^{ki} A_i^j \)In this operation, the covariant index i of A is raised, producing a tensor with two contravariant indices. Raising and lowering indices facilitate transforming tensor components between different spaces (contravariant and covariant), essential in many physics and engineering applications, where stresses, strains, and other tensorial quantities transform.
Coordinate Expressions
Coordinate expressions are ways to represent tensors in a specific coordinate system, crucial for practical computations in differential geometry. In our given exercise, tensor A is written as:\[ A = \sum_{i, j=1}^{n} A_i^j dx^i \otimes \frac{\partial}{\partial x^j} \]This breaks A into its components \(\text{A}_i^j \)\relative to the basis vectors \(\frac{\partial}{\partial x^j} \)and the dual basis \(\text{dx}^i \)When deriving expressions for new tensors B and C, it is important to express them in the same coordinate basis for consistency and ease of computation. For instance, B is given by:\[ B = \sum_{i, k} B_{ik} dx^i \otimes dx^k \]Here, the components \(\text{B}_{ik} \)eed to be determined in the same coordinate system. The coordinate expression allows us to perform tensor operations in a straightforward manner, as we can individually manipulate each component based on the rules of tensor algebra. Similarly, C’s coordinate expression is:\[ C = \sum_{k, j} C^{kj} dx^k \otimes dx^j \]With the components \(\text{C}^{kj} \)\being derived in a similar fashion. The use of coordinate expressions is widespread in various fields such as physics for describing fields and forces, and in engineering for stress and strain analysis.

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Most popular questions from this chapter

Let \(g: S^{n} \rightarrow \mathbb{P}^{n}\) be the map \(g(p)=[p]=\\{p,-p\\}\). (a) Show that there is a unique Riemannian metric \(\left((, \not)\right.\) on \(\mathbb{P}^{p n}\) such that \(g^{*}((,)\), is the usual Riemannian metric on \(S^{n}\) (the one that makes the inclusion of \(S^{n}\) into \(\mathbb{R}^{n+1}\) an isometry \(\\}\). (b) Show that every geodesic \(\gamma: \mathbb{R} \rightarrow \mathbb{P}^{n}\) is closed (that is, there is a number \(a\) sucl that \(\gamma(t+a)=\gamma(t)\) for all \(t)\), and that every two geodesics intersect exactly once. (c) Show that there are isometries of \(\mathbb{P}^{n}\) onto itself taking any tangent vector at one point to any tangent vector at any other point. These results show that \(\mathbb{P}^{n}\) provides a model for "elliptical" non-Euclidean geonetry. The sum of the angles in any triangle is \(>\pi\).

(a) For \(v, w \in \mathbb{R}^{n}\) with \(w \neq 0\), show that $$ \lim _{t \rightarrow 0} \frac{\|v+t w\|-\|v\|}{t}=\frac{(v, w)}{\|v\|} $$ The same result then lolds in any vector space with a Euclidean metric (, \(\rangle .\) Hint: If \(v: \mathbb{R}^{n} \rightarrow \mathbb{R}\) is the norm, then the limit is \(D v(v)(w) .\) Alternately, one can use the equation \((u, v\rangle=\|u\| \cdot\|v\|-\cos \theta\) where \(\theta\) is the angle between \(u\) and \(v\). (b) Conclude that if \(w\) is linearly independent of \(v\), then $$ \lim _{t \rightarrow 0} \frac{\|v+t w\|-\|v\|-\|t w\|}{t} \neq 0 $$ (c) Let \(\gamma:[0,1] \rightarrow M\) be a piecewise \(C^{1}\) critical point for length, and suppose that \(\gamma^{\prime}\left(t_{0}^{+}\right) \neq \gamma^{\prime}\left(t_{0}^{-}\right)\)for some \(t_{0} \in(0,1) .\) Choose \(t_{1}

(a) Let \(\gamma:[a, b] \rightarrow M\) be a geodesic, and let \(p:[\alpha, \beta] \rightarrow[a, b]\) be a diffeomorphism. Show that \(c=\gamma \circ p\) satisfies $$ \frac{d^{2} c^{k}}{d t^{2}}+\sum_{i, j=1}^{n} \Gamma_{i j}^{k}(c(t)) \frac{d c^{i}}{d t} \frac{d c^{j}}{d t}=\frac{d c^{k}}{d t} \frac{p^{\prime \prime}(t)}{p^{\prime}(t)} $$ (b) Conversely, if \(c\) satisfies this equation, then \(\gamma\) is a geodesic. (c) If \(c\) satisfles $$ \frac{d^{2} c^{k}}{d t^{2}}+\sum_{i, j=1}^{n} \Gamma_{i j}^{k}(c(t)) \frac{d c^{i}}{d t} \frac{d c^{j}}{d t}=\frac{d c^{k}}{d t} \mu(t) \quad \text { for } \mu: \mathbb{R} \rightarrow \mathbb{R} $$ then \(c\) is a reparameterization of a geodesic. (The equation \(p^{\prime \prime}(t)=p^{\prime}(t) \mu(t)\) can be solved explicitly: \(p(t)=\int^{t} e^{M(s)} d s\), where \(\left.M^{\prime}(s)=\mu(s) .\right)\)

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