## General idea on tensor properties > [!Notice] > - $Z = (Z^1, Z^2, Z^3)$: coordinates of a point in the space > - $\mathbf{R}(Z)$: position vector as a function of coordinates > - $\mathbf{Z}_i := \partial \mathbf{R} / \partial Z^i$: covariant basis vectors > - $Z_{ij} := \mathbf{Z}_i \cdot \mathbf{Z}_j$: inner product–defined covariant components > - $Z^{ij} := \mathbf{Z}^i \cdot \mathbf{Z}^j$: inner product–defined contravariant components > - The rank $(r, s)$: $r$ contravariant (upper) and $s$ covariant (lower) indices. In this part, we begin to talk about **tensor**. We will show tensor is not the data formats/structures like this term being applied in machine learning, but satisfies some important properties, which are the **key** to invariance. ### Variants, invariants and tensors **Variant**: A variant is an object than can be constructed by a similar rule **in** various coordinate systems. Under this definition, the covariant and contravariant basis, the metric tensor and the Christoffel symbol are all variants cause they could be constructed by operations like differentiation of the position vector $\mathbf{R}$. The Jacobian $J_{i}^{j'}$ is not variant, cause it is the relationship between coordinates systems (or spaces). (They could be considered as variants with respect to the coordinates or spaces themselves.) **Invariant**: As presented in [[Covariant and contravariant basis, matric tensors]], invariants are those independent with coordinates/spaces. Invariant and geometry are interchangeable in many cases. **Tensor**: Variants that transform between coordinates/spaces according to a special rule (here Jacobian) are called tensors. The **coordinate independence** is the core concept for tensor. >[!Note] >This definition relies on the transformability, it is the ability of transforming makes an object tensor, although seemingly such transformation does not have to be Jacobian, proof could be made to show why Jacobian is the proper and unique tool for such transformation. See next *[[#^d4ccae|NOTICE]]*. ### Linear transformation is the key: Jacobians and tensors A variant $T_{i}$ is called a covariant tensor if its values $T_{i}$ and $T_{i'}$ in the coordinate systems $Z^{i}$ and $Z^{i'}$ are related by $T_{i'} = T_{i} J^{i}_{i'}$ Similarly, the contravariant tensor is defined by $T^{i'} = T^{i} J^{i'}_{i}$ where $J^{i}_{i'}$ and $J^{i'}_{i}$ are the Jacobians (see [[Coordinate change and Jacobian]]). Covariant and contravariant indicate they transform in the same and opposite way as the basis $\mathbf{Z}_{i}$. The fundamental importance of the tensor property (namely transformability) is the fact that $J^{i}_{i'}$ and $J^{i'}_{i}$ are the matrix inverses of each other. $J^{i}_{i'} J^{j'}_{j} = \delta^{i}_{j}$ $J^{i'}_{i} J^{j}_{j'} = \delta^{i'}_{j'}$ >[!Note] >Just think for fun. In the generalized definition, this transformation does not have to be Jacobian, but we do need the existence of the inverse to make it a proper definition. Things could be different if we are transforming among noncommutative structures, or in a space with extreme bad properties (like fractal), and some other tools should applied. But under such context, the mathematical object may not be called *tensor*. *See the following NOTICE*. >[!Notice]- >Here is an intuitive but not strict discussion on the definition of tensor, and why the introduction of Jacobian is natural. It is the definition gives >**1. Definition of tensor** >A tensor is defined as a multilinear map that takes $r$ covariant vectors (from the dual space $V^*$) and $s$ contravariant vectors (from the vector space $V$) and maps them to the real numbers: >$T : \underbrace{V^* \times \cdots \times V^*}_{r \text{ factors}} \times \underbrace{V \times \cdots \times V}_{s \text{ factors}} \rightarrow \mathbb{R}.$ >This definition is $\textbf{coordinate-independent}$ and only assumes multilinearity. However, to express a tensor in coordinates, we need a way to handle transformations between different coordinate systems. > >**2. The Role of Taylor Expansion and Local Linear Approximation** >In a smooth manifold, coordinate transformations $x \rightarrow x'$ are generally nonlinear. However, in a sufficiently small neighborhood around any point, a $\textbf{Taylor expansion}$ allows us to approximate the transformation by a linear map. This leads to the \textbf{local linear approximation}: $ x^{i'} \approx x^{i'}(x) + \frac{\partial x^{i'}}{\partial x^i} \Delta x^i, $ where the term $ \frac{\partial x^{i'}}{\partial x^i} $ is the $\textbf{Jacobian matrix}$, capturing the first-order, or linear, part of the coordinate transformation. This approximation is crucial because it allows us to treat the transformation as locally linear, meaning that tensor components can be consistently transformed using linear mappings in small regions. (In Euclidean space we automatically satisfies such linearity.) > >**3. Deriving the Transformation Rule for Tensor Components** >Given the requirement of multilinearity in tensor transformations, the Jacobian matrix becomes essential. For a $(r, s)$ tensor $T$ with components $T^{i_1 \dots i_r}_{j_1 \dots j_s}$, the transformation rule from coordinates $x$ to $x'$ is given by: > $T^{i'_1 \dots i'_r}_{j'_1 \dots j'_s} = \left( \frac{\partial x^{i'_1}}{\partial x^{i_1}} \cdots \frac{\partial x^{i'_r}}{\partial x^{i_r}} \right) \left( \frac{\partial x^{j_1}}{\partial x^{j'_1}} \cdots \frac{\partial x^{j_s}}{\partial x^{j'_s}} \right) T^{i_1 \dots i_r}_{j_1 \dots j_s}.$ > This rule ensures that tensor components in different coordinate systems describe the $\textbf{same geometric object}$. The Jacobian matrix is therefore the unique tool that satisfies the multilinear and local linearity requirements of tensor transformations. > > **4. Implications for Space Properties and Tensor Definition** > In $\textbf{Riemannian manifolds}$, the smooth structure guarantees that local linear approximation through Taylor expansion is valid, allowing Jacobian matrices to consistently describe coordinate transformations at each point. However, in spaces lacking such smoothness: > - If the space does not support a differentiable structure (e.g., at singular points or in fractal-like structures), the Jacobian matrix cannot be defined globally or locally. > - The traditional concept of tensors, which relies on smooth coordinate transformations and linear approximation, may not apply. > > For such spaces, alternative tools may be used: > - $\textbf{Tensor densities}$, which include additional factors to adapt to singular spaces. > - $\textbf{Distributions}$ (such as Dirac delta functions) to describe physical quantities at singularities. > - $\textbf{Non-commutative geometry}$, which provides a framework for analyzing spaces that lack a smooth structure. > >**5. Conclusion** >The Jacobian matrix emerges naturally from the tensor definition when we consider the need for local linear transformations on smooth manifolds. The Taylor expansion provides a local linear approximation, making the Jacobian matrix the unique tool for satisfying the transformation requirements of tensors. In Euclidean and Riemannian manifolds, this approach works globally or locally; however, in spaces without differentiable structure, the standard tensor concept may need to be generalized or replaced by other mathematical tools. ^d4ccae ### Construct invariants with tensors Now we may show that with the aid of Jacobian and Christoffel delta (the orthonormality of Jacobians), an invariant could be constructed. Consider $S_{i}$ is a covariant tensor and $T^{i}$ is a contravariant tensor, then $U=S_{i}T^{i}$ is invariant, which means it remains the same value in all systems. >[!Proof] >Consider a primed coordinate with a quantity $U'$. This quantity is contracted by $S_{i'}$ and $T^{i'}$. $S_{i'}$ and $T^{i'}$ could be connected to the unprimed coordinate with Jacobian. Namely, >$S_{i'}=S_{i} J_{i'}^{i}$ >and >$T^{i'}=T^{k}J_{k}^{i'}$ >Therefore, >$U' = S_i J_{i'}^{i} T^k J_{k}^{i'}$ >$U' = S_i T^k \delta_k^i = S_i T^i = U$ >So we show that the variant $U$ evaluates to the same value in all coordinate systems. >What we proved here is under the assumptions that Jacobians are valid for coordinates (or system) transformation and the orthonormality of it. We show that both system gives the same contraction result, aka $U$ is an invariant. >>[!Notice]- >>The relation $T^{i'} = T^k J_k^{i'}$ follows directly from coordinate transformation rules. It describes how components change under a change of coordinates, but writing such a relation **does not itself imply** that $T^i$ is a tensor. A quantity is **defined** to be a (contra)variant tensor **if and only if** it transforms this way in all coordinate systems. >>Similarly, the contraction $U = S_i T^i$ is a symbolic operation. It evaluates to a coordinate-independent scalar (invariant) **only if** both $S_i$ and $T^i$ obey the tensor transformation laws. The contraction notation alone does not guarantee that the result is physically or geometrically meaningful unless the inputs are tensors. ### Tensor character of basis vectors and metric As shown above ([[#Construct invariants with tensors]]), an invariant can be constructed by contracting a covariant tensor with a contravariant one. This relies on the assumption that the objects involved transform as tensors. In this section, we verify that several geometric quantities, including the covariant basis vectors and the metric tensor, do in fact possess tensorial transformation properties. #### Covariant basis $\mathbf{Z}_{i}$ is tensor From the definition ([[Covariant and contravariant basis, matric tensors#Covariant basis]]), $ \mathbf{Z}_{i'} = \frac{\partial \mathbf{R}(Z')}{\partial Z^{i'}} $ To show that $\mathbf{Z}_{i'}$ is a tensor, we need to verify that it satisfies the covariant transformation rule: $ \mathbf{Z}_{i'} = \mathbf{Z}_{i} J_{i'}^{i} $ To do this, we consider the chain rule under a coordinate transformation $Z \to Z'$, where $\mathbf{R}$ is now regarded as a function of $Z$ via $Z = Z(Z')$: $ \mathbf{Z}_{i'} = \frac{\partial \mathbf{R}(Z(Z'))}{\partial Z^{i'}} = \frac{\partial \mathbf{R}}{\partial Z^{i}} \frac{\partial Z^{i}}{\partial Z^{i'}} = \mathbf{Z}_{i} J_{i'}^{i} $ This shows that the covariant basis vectors $\mathbf{Z}_i$ transform covariantly under coordinate changes. Therefore, they form a rank-$(0,1)$ tensor field. >[!Notice] > Above proof assumes that the coordinate transformation $Z \leftrightarrow Z'$ is smooth (i.e., differentiable), so that the chain rule applies. > This is standard in differential geometry, where tensors are defined with respect to smooth coordinate systems. Given the covariant transformation rule $ T_{i'} = T_i J^i_{i'}, $ we recover the unprimed component by multiplying both sides with the inverse Jacobian: $ T_i = T_{i'} J^{i'}_i. $ This follows from the identity $ J^i_{i'} J^{i'}_j = \delta^i_j. $ Similarly, for a contravariant tensor: $ T^{i'} = T^i J^{i'}_i \quad \Rightarrow \quad T^i = T^{i'} J^i_{i'}. $ In all cases, the correct reverse transformation is obtained by matching index positions. The invertibility of the Jacobian is assumed. > [!Note] > Although coordinate transformations are assumed to be smooth and invertible by default, the inverse Jacobian appears explicitly when deriving reverse transformation rules. > This emphasizes that the tensor transformation laws rely on the invertibility of the Jacobian to guarantee a consistent mapping between coordinate systems. #### Contravariant components of a vector form a tensor We show that the components $V^{i}$ of a vector $\mathbf{V}$ with respect to the basis $\mathbf{Z}_i$ transform as a contravariant tensor. In the primed coordinate system, $ \mathbf{V} = V^{i'} \mathbf{Z}_{i'} $ Using the tensor transformation rule for the basis vectors, $ \mathbf{Z}_{i'} = \mathbf{Z}_i J^i_{i'}, $ we have $ \mathbf{V} = V^{i'} \mathbf{Z}_{i'} = V^{i'} \mathbf{Z}_i J^i_{i'}. $ Rewriting the same vector in the unprimed system, $ \mathbf{V} = V^i \mathbf{Z}_i $ and comparing both expressions, we obtain $ V^i = V^{i'} J^i_{i'}. $ This shows that the components $V^i$ transform according to the contravariant transformation rule, and therefore form a rank-$(1,0)$ tensor. > [!Important] > We have [[#Construct invariants with tensors|previously shown]] that if $S_i$ and $T^i$ are a covariant tensor and a contravariant tensor, then the contraction $S_i T^i$ is an invariant. > > The converse is also useful: > If the contraction $S_i T^i$ is invariant and one of the terms (e.g., $S_i$) is known to be a tensor, then the other term (e.g., $T^i$) must also be a tensor. > > This is a special case of the more general quotient theorem. #### Contravariant basis $\mathbf{Z}^{i}$ is tensor We now verify that the contravariant basis vectors $\mathbf{Z}^i$ also transform as tensors. From the definition of dual bases, we have $ \mathbf{Z}^i \cdot \mathbf{Z}_j = \delta^i_j. $ Multiply both sides by $J^{i'}_i J^j_{j'}$ $ \left( \mathbf{Z}^i J^{i'}_i \right) \cdot \left( \mathbf{Z}_j J^j_{j'} \right) = \delta^i_j J^{i'}_i J^j_{j'}. $ The right-hand side becomes $ \delta^i_j J^{i'}_i J^j_{j'} = \delta^{i'}_{j'}. $ So we have $ \left( \mathbf{Z}^i J^{i'}_i \right) \cdot \mathbf{Z}_{j'} = \delta^{i'}_{j'}. $ But by the defining relation of the dual basis in the primed coordinate system $ \mathbf{Z}^{i'} \cdot \mathbf{Z}_{j'} = \delta^{i'}_{j'}, $ we conclude $ \mathbf{Z}^{i'} = \mathbf{Z}^i J^{i'}_i. $ This shows that $\mathbf{Z}^i$ transforms contravariantly and thus forms a rank-$(1,0)$ tensor. #### The inner product–defined metric $Z_{ij}$ and $Z^{ij}$ are tensors For higher-order tensors, the transformation rule simply involves more Jacobians, each matching the variance (covariant or contravariant) of the corresponding index. If the component objects are tensors, then their inner products or outer products are also tensors. Since $\mathbf{Z}_i$ and $\mathbf{Z}^i$ are tensors, we now show that: - $Z_{ij} := \mathbf{Z}_i \cdot \mathbf{Z}_j$ is a rank-$(0,2)$ tensor, and - $Z^{ij} := \mathbf{Z}^i \cdot \mathbf{Z}^j$ is a rank-$(2,0)$ tensor. For the covariant case: $ Z_{i'j'} = \mathbf{Z}_{i'} \cdot \mathbf{Z}_{j'} = (\mathbf{Z}_i J^i_{i'}) \cdot (\mathbf{Z}_j J^j_{j'}) = J^i_{i'} J^j_{j'} (\mathbf{Z}_i \cdot \mathbf{Z}_j) = J^i_{i'} J^j_{j'} Z_{ij}. $ Similarly, for the contravariant case: $ Z^{i'j'} = \mathbf{Z}^{i'} \cdot \mathbf{Z}^{j'} = (\mathbf{Z}^i J^{i'}_i) \cdot (\mathbf{Z}^j J^{j'}_j) = J^{i'}_i J^{j'}_j (\mathbf{Z}^i \cdot \mathbf{Z}^j) = J^{i'}_i J^{j'}_j Z^{ij}. $ This confirms that both $Z_{ij}$ and $Z^{ij}$ are tensors, and their transformation behavior directly follows from that of their first-order components. #### The Kronecker delta is a tensor Although the Kronecker delta $\delta^j_i$ numerically looks the same in all coordinate systems, it is **not** a scalar invariant, since it carries two indices and is therefore a second-order object. We now verify that $\delta^j_i$ transforms as a rank-$(1,1)$ tensor. Under a coordinate transformation, we have: $ \delta^{j'}_{i'} = J^j_{j'} J^{i'}_i \, \delta^i_j. $ This matches the transformation rule for a tensor of type $(1,1)$: $ T^{j'}_{i'} = J^j_{j'} J^{i'}_i T^j_i. $ Therefore, the Kronecker delta is a mixed tensor of rank $(1,1)$. ### Selected exercises **Gradient of scalar field is a tensor**: Show that for a scalar field $F$, the collection of partial derivatives $\partial F / \partial Z^i$ is a covariant tensor. (exercise 88) ^08d177 >[!Proof]- > Let $F$ be a scalar field. Its gradient $\partial F / \partial Z^i$ transforms as: > $ > \frac{\partial F}{\partial Z^{i'}} =\frac{\partial F}{\partial Z^i} \frac{\partial Z^i}{\partial Z^{i'}} = J_{i'}^{i}\frac{\partial F}{\partial Z^i} , > $ > which matches the transformation rule for a covariant tensor. Therefore, the gradient of a scalar field is a rank-$(0,1)$ tensor. **Partial derivatives of tensors are not necessarily tensors**: Given a scalar field $F$, show that the collection of second-order partial derivatives $\frac{\partial^2 F}{\partial Z^i \partial Z^j}$ is not a tensor. More generally, show that for a covariant tensor field $T_i$, the variant $\frac{\partial T_i}{\partial Z^j}$ is not a tensor. (exercise 89) ^78f2e3 >[!Proof]- > To check whether $\partial^2 F / \partial Z^{i'} \partial Z^{j'}$ transforms as a tensor, start from: > $\frac{\partial^2 F}{\partial Z^{i'} \partial Z^{j'}} = \frac{\partial}{\partial Z^{j'}}\left( \frac{\partial F}{\partial Z^{i'}} \right) > = \frac{\partial}{\partial Z^{j'}}\left( J^i_{i'} \frac{\partial F}{\partial Z^i} \right). > $ > Applying the product rule gives: > $ > \frac{\partial}{\partial Z^{j'}}\left( J^i_{i'} \frac{\partial F}{\partial Z^i} \right)= \frac{\partial J^i_{i'}}{\partial Z^{j'}} \frac{\partial F}{\partial Z^i} > + J^i_{i'} \frac{\partial^2 F}{\partial Z^{j'} \partial Z^i}. > $ > The first term involves derivatives of the Jacobian and prevents this expression from transforming as a second-order tensor. Therefore, $\partial^2 F / \partial Z^{i'} \partial Z^{j'}$ is not a tensor. > Similarly, for a covariant tensor field $T_{i}$, we have > $\frac{\partial T_{i}}{\partial Z^{j'}} = \frac{\partial}{\partial Z^{j'}}\left( T_{i} J_{i'}^{i} \right) > = \frac{\partial J^i_{i'}}{\partial Z^{j'}} T_{i} > + J^i_{i'} \frac{\partial T_{i}}{\partial Z^{j'} }. > $ > This also involves derivatives of the Jacobian. It's not a tensor. **Antisymmetric combinations restore tensoriality**: Show that the skew-symmetric part $S_{ij} = \partial T_i / \partial Z^j - \partial T_j / \partial Z^i$ of the non-tensorial quantity $\partial T_i / \partial Z^j$ is a tensor. (exercise 90) >[!Proof]- > Let $T_i$ be a covariant tensor. Although $\partial T_i / \partial Z^j$ is not itself a tensor (as shown in Exercise 89), we now show that the antisymmetric combination > $ > S_{ij} = \frac{\partial T_i}{\partial Z^j} - \frac{\partial T_j}{\partial Z^i} > $ > **does** transform as a tensor. > Start by expressing $T_{i'}$ in terms of unprimed quantities: > $ > T_{i'} = T_k J^k_{i'}. > $ > Taking the partial derivative: > $ > \frac{\partial T_{i'}}{\partial Z^{j'}} = \frac{\partial}{\partial Z^{j'}} (T_k J^k_{i'}) > = \frac{\partial T_k}{\partial Z^{j'}} J^k_{i'} + T_k \frac{\partial J^k_{i'}}{\partial Z^{j'}}. > $ > Similarly, > $ > \frac{\partial T_{j'}}{\partial Z^{i'}} = \frac{\partial T_k}{\partial Z^{i'}} J^k_{j'} + T_k \frac{\partial J^k_{j'}}{\partial Z^{i'}}. > $ > Forming the antisymmetric combination: > $S_{i'j'} = \left( \frac{\partial T_k}{\partial Z^{j'}} J^k_{i'} - \frac{\partial T_k}{\partial Z^{i'}} J^k_{j'} \right) > + T_k \left( \frac{\partial J^k_{i'}}{\partial Z^{j'}} - \frac{\partial J^k_{j'}}{\partial Z^{i'}} \right)$ > The first group of terms is of tensorial form, but the second group involves derivatives of the Jacobian matrix. If these extra terms cancel, then $S_{i'j'}$ will be a tensor. > > > [!Note] > > **Antisymmetric combination cancels non-tensorial terms** > > > > The Jacobian $J^k_{i'} = \frac{\partial Z^k}{\partial Z^{i'}}$ is itself a partial derivative. Therefore, its second derivatives are symmetric: > > $ > > \frac{\partial^2 Z^k}{\partial Z^{j'} \partial Z^{i'}} = \frac{\partial^2 Z^k}{\partial Z^{i'} \partial Z^{j'}}. > > $ > > This implies that: > > $ > > \frac{\partial J^k_{i'}}{\partial Z^{j'}} = \frac{\partial J^k_{j'}}{\partial Z^{i'}}. > > $ > > So the antisymmetric combination > > $ > > \frac{\partial J^k_{i'}}{\partial Z^{j'}} - \frac{\partial J^k_{j'}}{\partial Z^{i'}} > > $ > > **vanishes identically**. This cancellation removes the non-tensorial terms and ensures that $S_{ij}$ transforms as a proper tensor. > > > > > [!Tip] > > > **Classic antisymmetric cancellations** > > > > > > Many non-tensorial expressions contain symmetric components (e.g., second derivatives, Christoffel symbols). When placed into antisymmetric combinations, these symmetric parts cancel: > > > > > > - Mixed partials: $\partial^2 f / \partial x^i \partial x^j$ are symmetric, so $\partial_i \partial_j f - \partial_j \partial_i f = 0$. > > > - Jacobian derivatives: $\frac{\partial J^k_{i'}}{\partial Z^{j'}} - \frac{\partial J^k_{j'}}{\partial Z^{i'}} = 0$. > > > - Electromagnetic tensor: $F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$ cancels non-tensorial gauge terms. > > > - Christoffel symbols: $\Gamma^k_{[ij]} = 0$ due to symmetry in lower indices. > > > > > > **General rule**: antisymmetrizing over symmetric indices eliminates non-tensorial parts. > > Therefore, $S_{ij}$ is a rank-$(0,2)$ tensor. **The Christoffel symbol is not a tensor**: Derive the transformation rule for the Christoffel symbol $\Gamma^k_{ij}$ and show that it does not transform as a tensor. (exercise 92) >[!Proof]- > When transforming the Christoffel symbol defined as > $ > \Gamma^k_{ij} = \mathbf{Z}^k \cdot \frac{\partial \mathbf{Z}_i}{\partial Z^j}, > $ > the derivative $\frac{\partial \mathbf{Z}_i}{\partial Z^j}$ transforms with a term involving the derivative of the Jacobian: > $ > \frac{\partial \mathbf{Z}_{i'}}{\partial Z^{j'}} > = \frac{\partial J^i_{i'}}{\partial Z^{j'}} \mathbf{Z}_i + J^i_{i'} \frac{\partial \mathbf{Z}_i}{\partial Z^{j'}}. > $ > This leads to the transformation rule: > $ > \Gamma^{k'}_{i'j'} = J^{k'}_k J^i_{i'} \Gamma^k_{ij} + J^{k'}_i \frac{\partial J^i_{i'}}{\partial Z^{j'}}, > $ > which contains a non-tensorial term involving the derivative of the Jacobian. Hence, $\Gamma^k_{ij}$ is not a tensor. **Vanishing of a tensor is coordinate-invariant**: Show that if a tensor vanishes in one coordinate system, then it vanishes in all coordinate systems. (exercise 94) >[!Proof]- > > >[!Note] > >Saying "a tensor vanishes in one coordinate system" means that all of its components are zero in that system. Namely, > >$T^{i_1 \dots i_r}_{j_1 \dots j_s} = 0$ > > This follows from the transformation law: > $ > T^{i'_1 \dots i'_r}_{j'_1 \dots j'_s} > = \left( \prod \frac{\partial Z^{i'_k}}{\partial Z^{i_k}} \right) > \left( \prod \frac{\partial Z^{j_l}}{\partial Z^{j'_l}} \right) > T^{i_1 \dots i_r}_{j_1 \dots j_s}. > $ > If $T^{i_1 \dots i_r}_{j_1 \dots j_s} = 0$, then all transformed components must also be zero. This expresses the coordinate-invariant nature of tensor vanishing: the zero tensor is zero in all coordinate systems. > [!Note]+ > **Tensorial behavior under restricted transformation classes** > > Some objects fail to transform as tensors under general coordinate transformations, but **do** behave like tensors under a restricted class of transformations — such as **linear (affine) transformations** of the form: > $ > Z^{i'} = A^{i'}_i Z^i + b^{i'}. > $ > In such cases: > - The Jacobian $A^{i'}_i$ is constant; > - Its derivatives vanish: $\partial_j A^{i'}_i = 0$; > - This removes the non-tensorial terms (e.g., derivatives of Jacobians) from transformation rules. > > Therefore, certain objects like $\frac{\partial T_i}{\partial Z^j}$, $\frac{\partial^2 F}{\partial Z^i \partial Z^j}$, or $\Gamma^k_{ij}$ — which are **not** tensors in the general sense — may still **transform tensorially under affine transformations**. > > This reflects a general idea: **tensoriality is defined with respect to a group of allowed coordinate transformations**. Restricting that group can lead to *conditionally tensorial* behavior.