关于笛卡尔张量协变与逆变指标等价性的证明方法及示例的技术问询

Hey there, let's break down this key proposition about Cartesian tensors and tackle your questions with clear, practical explanations.

First, let's restate the core proposition for clarity:

Proposition: While the distinction between covariant and contravariant indices must be made for general tensors, the two are equivalent for tensors in $n$-dimensional Euclidean space, and such tensors are known as Cartesian tensors.

1. Proof of the Proposition #

The equivalence boils down to the metric tensor in Euclidean Cartesian coordinates—here's a step-by-step breakdown:

• Recall the fundamental relationship between covariant and contravariant tensor components: For any tensor $T$, covariant components $T_{i_1i_2...i_k}$ and逆变 components $T^{i_1i_2...i_k}$ are linked by the metric tensor $g_{ij}$ and its inverse $g^{ij}$:
  $$T_{i_1...i_k} = g_{i_1j_1}g_{i_2j_2}...g_{i_kj_k}T^{j_1...j_k}$$
  $$T^{i_1...i_k} = g^{i_1j_1}g{i_2j_2}...g^{i_kj_k}T_{j_1...j_k}$$
• In $n$-dimensional Euclidean space with a Cartesian coordinate system, the metric tensor components simplify to the Kronecker delta:
  $$g_{ij} = \delta_{ij} = \begin{cases} 1 & \text{if } i=j \ 0 & \text{if } i\neq j \end{cases}$$
  Since the metric tensor is the identity matrix here, its inverse $g^{ij}$ is identical to $g_{ij}$ (the identity matrix's inverse is itself).
• Substitute the Kronecker delta into the index-lowering/raising formulas. Using the property $\delta_{ij}T^j = T_i$, we get:
  $$T_{i_1...i_k} = \delta_{i_1j_1}...\delta_{i_kj_k}T^{j_1...j_k} = T^{i_1...i_k}$$
  This directly shows that covariant and contravariant components are identical. Thus, the distinction between the two index types becomes irrelevant for tensors in this context—these are Cartesian tensors.

2. Worked Example: Second-Order Tensor in 2D Euclidean Space #

Let's use a concrete 2D example to make this tangible:

• Suppose we have a second-order逆变 tensor with components:
  $$T^{ij} = \begin{pmatrix} 2 & 1 \ -1 & 3 \end{pmatrix}$$
• Calculate its covariant components using the metric tensor (Kronecker delta):
  • $T_{11} = g_{1k}g_{1l}T^{kl} = \delta_{1k}\delta_{1l}T^{kl} = T^{11} = 2$
  • $T_{12} = g_{1k}g_{2l}T^{kl} = \delta_{1k}\delta_{2l}T^{kl} = T^{12} = 1$
  • $T_{21} = g_{2k}g_{1l}T^{kl} = \delta_{2k}\delta_{1l}T^{kl} = T^{21} = -1$
  • $T_{22} = g_{2k}g_{2l}T^{kl} = \delta_{2k}\delta_{2l}T^{kl} = T^{22} = 3$
• The resulting covariant tensor matrix is identical to the逆变 one:
  $$T_{ij} = \begin{pmatrix} 2 & 1 \ -1 & 3 \end{pmatrix}$$

For a simpler first-order tensor (vector) example: An逆变 vector $v^i = (3, 4)$ will have covariant components $v_i = g_{ij}v^j = \delta_{ij}v^j = (3, 4)$—exactly the same as the逆变 components.

备注：内容来源于stack exchange，提问作者Michael Levy