## The fundamental properties of tensors > [!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. ### Sum of tensor The sum of two tensors of the same type is also a tensor. This follows immediately from the linearity of the transformation rule. If $A^i_{jk} = B^i_{jk} + C^i_{jk}$, and both $B^i_{jk}$ and $C^i_{jk}$ are tensors, then $A^i_{jk}$ also transforms tensorially. This property allows the construction of new tensors by **linear combination**, as long as the component tensors have the same variance type (rank and index structure). ### Product of tensors The tensor product of two tensors of type $(r, s)$ and $(p, q)$ produces a tensor of type $(r + p, s + q)$. There is no restriction on their orders or types, but both must be tensors defined on compatible vector spaces. No index contraction occurs in a tensor product. > [!Note] > Tensors must be defined on compatible vector spaces or be transformable to a common frame via coordinate or geometric mappings. 1. **Tensor product of two tensors is a tensor** If $A$ is a $(r, s)$ tensor and $B$ is a $(p, q)$ tensor, then: $ (A \otimes B)^{i_1 \dots i_r \, k_1 \dots k_p}_{j_1 \dots j_s \, l_1 \dots l_q} = A^{i_1 \dots i_r}_{j_1 \dots j_s} \cdot B^{k_1 \dots k_p}_{l_1 \dots l_q} $ is a tensor of type $(r + p, s + q)$. No index contraction occurs here. 2. **Dot product of two vector-valued tensors is a tensor** For example, let $\mathbf{A} = A^i \mathbf{Z}_i$, $\mathbf{B} = B^j \mathbf{Z}_j$ be vector-valued tensors. Their dot product: $ U = \mathbf{A} \cdot \mathbf{B} = A^i B^j (\mathbf{Z}_i \cdot \mathbf{Z}_j) = A^i B^j Z_{ij} $ is a **scalar-valued** (rank $(0,0)$) tensor (i.e., invariant). It results from contracting two vector components using the metric tensor $Z_{ij}$. 3. **Product of a scalar-valued tensor and a tensor-valued vector is a tensor** Let $f(Z)$ be a scalar-valued tensor (rank $(0, 0)$, i.e., a function), and $\mathbf{V} = V^i \mathbf{Z}_i$ a vector. Then: $ f \cdot \mathbf{V} = f(Z) V^i \mathbf{Z}_i $ is still a **vector-valued tensor**. Scalars act multiplicatively on tensor components and do not affect the transformation properties. ### The contraction theorem The contraction of a tensor is also a tensor. The proof essentially relies on the transformation property of the Kronecker delta. If $T^{i_1 \dots i_r}_{j_1 \dots j_s}$ is a tensor, then contracting over one upper and one lower index, such as $T^{i_1 \dots i_k \dots}_{j_1 \dots i_k \dots}$, produces a new tensor of type $(r - 1, s - 1)$. This contraction can be interpreted as an implicit multiplication with the Kronecker delta: $ T^{\dots i \dots}_{\dots i \dots} = T^{\dots i \dots}_{\dots j \dots} \, \delta^j_i, $ where $\delta^j_i$ transforms as a tensor. The Jacobian-based construction, $ \delta^j_i = \frac{\partial Z^{i'}}{\partial Z^i} \frac{\partial Z^j}{\partial Z^{i'}}, $ demonstrates that this operation preserves tensor transformation laws. > [!Summary] > The combination of tensorial operations—**linear combinations**, **tensor products**, and **index contractions**—enables us to build new tensors and ultimately construct **invariants**. > > This process systematically removes coordinate dependencies embedded in tensor components, revealing the underlying **coordinate-free geometric meaning**. > > In this framework, invariants are viewed as **zero-order tensors**, and tensors serve as structured intermediaries that retain full transformation information. Once contracted appropriately, they produce scalar quantities valid in all coordinate systems. > > Hence, tensor calculus provides a powerful method to extract geometric content while maintaining the computational utility of coordinates. > [!Note]- > **Invariant from the Kronecker delta (exercise 98)** > The quantity $\delta^i_j$ is a tensor. Its contraction > $ > \delta^i_i = \sum_i \delta^i_i = n > $ > is an invariant scalar equal to the **dimension of the space**. Geometrically, it counts the number of independent basis directions — a coordinate-independent property. > This result can also be understood via the Jacobian representation: > $ > \delta^i_j = J^i_{k'} \bar{J}^{k'}_j \quad \Rightarrow \quad \delta^i_i = \text{Tr}(J \cdot J^{-1}) = n > $ > indicating the preservation of dimensionality under coordinate transformations. > > **Components of a vector-valued tensor are tensors (exercise 99)** > Given a vector-valued tensor $\mathbf{V}_{ij}$ with the expansion: > $ > \mathbf{V}_{ij} = V^k_{ij} \, \mathbf{Z}_k, > $ > we want to show $V^k_{ij}$ is a rank-(1,2) tensor. > > Using tensor transformation rules and expressing $\mathbf{V}_{i'j'}$ in two ways: > - via transformed components and basis: > $ > \mathbf{V}_{i'j'} = V^{k'}_{i'j'} \, \mathbf{Z}_{k'} > $ > - via Jacobians acting on the original expression: > $ > \mathbf{V}_{i'j'} = J^i_{i'} J^j_{j'} V^k_{ij} \, \mathbf{Z}_k = V^k_{ij} J^i_{i'} J^j_{j'} J^k_{k'} \, \mathbf{Z}_{k'} > $ > > Comparing both expressions gives: > $ > V^{k'}_{i'j'} = V^k_{ij} \, J^i_{i'} J^j_{j'} J^{k'}_k, > $ > which is the transformation law for a rank-(1,2) tensor. Hence, $V^k_{ij}$ is a tensor. > > **Constructing a Tensor from a Seed (exercise 100)** > In a given coordinate system $\bar{Z}^i$, suppose a quantity $T^{\bar{i}\bar{j}}_{\bar{k}}$ is arbitrarily defined. We define a new quantity in another coordinate system $Z^i$ as > $ > T^{ij}_k := T^{\bar{i}\bar{j}}_{\bar{k}} \cdot J^i_{\bar{i}} J^j_{\bar{j}} J^{\bar{k}}_k. > $ > We wish to verify whether $T^{ij}_k$ is a tensor. > > To do so, consider a third coordinate system $Z^{i'}$ and examine how $T^{ij}_k$ transforms: > $ > T^{i'j'}_{k'} = J^{i'}_i J^{j'}_j J^k_{k'} \cdot T^{ij}_k. > $ > Substituting the constructed expression gives > $ > T^{i'j'}_{k'} = J^{i'}_i J^{j'}_j J^k_{k'} \cdot (T^{\bar{i}\bar{j}}_{\bar{k}} J^i_{\bar{i}} J^j_{\bar{j}} J^{\bar{k}}_k), > $ > which simplifies to > $ > T^{i'j'}_{k'} = T^{\bar{i}\bar{j}}_{\bar{k}} \cdot (J^{i'}_i J^i_{\bar{i}})(J^{j'}_j J^j_{\bar{j}})(J^k_{k'} J^{\bar{k}}_k) > = T^{\bar{i}\bar{j}}_{\bar{k}} \cdot J^{i'}_{\bar{i}} J^{j'}_{\bar{j}} J^{\bar{k}}_{k'}. > $ > This is exactly the transformation law for a type $(2,1)$ tensor, hence $T^{ij}_k$ is a tensor. > > > [!Example] > > For example, suppose in a coordinate system $\bar{Z}^i$, we assign arbitrary values to the components: > > $ > > T^{\bar{i}\bar{j}}_{\bar{k}} = > > \begin{cases} > > T^{11}_1 = 2,\quad T^{12}_1 = 1,\quad T^{21}_1 = -1,\quad T^{22}_1 = 0, \\ > > T^{11}_2 = 0,\quad T^{12}_2 = 3,\quad T^{21}_2 = 1,\quad T^{22}_2 = -2. > > \end{cases} > > $ > > > > These values have no inherent geometric meaning and are not assumed to form a tensor. > > Now, in another coordinate system $Z^i$, define the new components via the transformation: > > $ > > T^{ij}_k := T^{\bar{i}\bar{j}}_{\bar{k}} \cdot J^i_{\bar{i}} J^j_{\bar{j}} J^{\bar{k}}_k. > > $ > > > > Then $T^{ij}_k$ is a tensor by construction, because it satisfies the correct transformation law. > > The geometric content lies not in the original values, but entirely in the consistency of their transformation across coordinate systems. > > This result illustrates that a tensor can be defined **constructively**: even if the components in one coordinate system are assigned arbitrarily and lack geometric meaning, as long as all other coordinate expressions are defined using the tensor transformation law, the resulting object is a tensor. The key is that its transformation behavior agrees with the tensorial transformation rule. > > This construction implicitly relies on the existence of a differentiable manifold structure, invertible coordinate maps, and well-defined local tensor spaces at each point. These assumptions ensure that Jacobians and their inverses are meaningful, and that the index structure corresponds to genuine geometric entities. ### The gradient The gradient of a scalar field can be introduced from an algebraic perspective as a coordinate-dependent expansion. For a scalar field $F(Z^1, Z^2, Z^3)$, it is often written as $ \nabla F = \sum_i \frac{\partial F}{\partial Z^i} \mathbf{e}_i $ where $\mathbf{e}_i$ denotes the coordinate basis vectors, and the derivatives are taken with respect to the local coordinates $Z^i$. In orthonormal Cartesian coordinates, this expression works perfectly, since the basis vectors are constant, mutually orthogonal, and of unit length. The sum simply expresses a geometric vector as a linear combination of orthonormal bases with the directional derivatives as components. However, in general curvilinear coordinates, the basis vectors $\mathbf{e}_i$ are no longer constant or orthogonal, and this form does not preserve a meaningful geometric interpretation. Although the expression is still algebraically valid, it does not define a true vector because it does not transform correctly under coordinate transformations. In tensor formalism, one might attempt to write the gradient as $ \nabla F = \sum_{i} \frac{\partial F}{\partial Z^i} \mathbf{Z}_i $ where $\mathbf{Z}_i$ are the covariant basis vectors defined as $\mathbf{Z}_i = \partial \mathbf{R} / \partial Z^i$. While both $\partial F / \partial Z^i$ and $\mathbf{Z}_i$ are well-defined quantities, they are both covariant (with lower indices), and their product does not result in a vector. The object formed by this expression is a rank-$(0,2)$ tensor, not a rank-$(1,0)$ vector, and therefore does not serve as a proper definition of the gradient. To obtain a correct geometric object—one that is coordinate-independent and properly transforms as a vector—we must pair covariant components with contravariant bases. The correct expression is $ \nabla F = \frac{\partial F}{\partial Z^i} \mathbf{Z}^i $ where $\mathbf{Z}^i$ is the contravariant basis. In this form, the contraction between the lower index of the partial derivative and the upper index of the basis ensures that the result is a true vector. This formulation is valid in all coordinate systems and satisfies the tensor transformation rules. > [!Note] > **The basis in gradient expansion is flexible.** > The expression > $ > \nabla F = \sum_{i} \frac{\partial F}{\partial Z^i} \mathbf{Z}_i > $ > is written in terms of a chosen basis, which can be covariant, orthonormal, or otherwise. This choice does not affect the validity of the gradient as a geometric object—only its representation. Therefore, one can equivalently use the contravariant basis: > $ > \nabla F = \frac{\partial F}{\partial Z^i} \mathbf{Z}^i > $ > and satisfy the contraction rule. > [!Note]- > **Exercise 101 – Why $\Delta F = Z^{ij} \partial^2 F/\partial Z^i \partial Z^j$ is not valid** > > The expression > $ > \Delta F = Z^{ij} \frac{\partial^2 F}{\partial Z^i \partial Z^j} > $ > is **not a valid scalar** in general coordinates because $\partial^2 F/\partial Z^i \partial Z^j$ is **not a tensor**. ([[General idea on tensor properties#^78f2e3|exercise 89]]) > > Although $Z^{ij}$ is a tensor (the inverse metric), the second derivative of a scalar function does not transform as a tensor, due to the nonlinearity of coordinate transformations. Therefore, this contraction does not yield a valid invariant. > > The correct construction of the Laplacian requires the use of **[[Introduction to covariant differentiation|covariant derivatives]]**: > $ > \Delta F = \nabla_i \nabla^i F > $ > or > $ > \Delta F = Z^{ij} \nabla_i \nabla_j F, > $ > which ensures the full expression is a scalar (tensor of rank 0). ### The directional derivative Given a scalar field $F(Z^i)$ and a curve $l$ parameterized by arc length $s$, let the coordinate functions along $l$ be $Z^i = Z^i(s)$. Then $F$ becomes a function of $s$ via composition: $ F(s) = F(Z(s)). $ Differentiating both sides using the chain rule gives: $ \frac{dF}{ds} = \frac{\partial F}{\partial Z^i} \frac{dZ^i}{ds}. $ This formula contains: - $\frac{\partial F}{\partial Z^i}$: the covariant components of the gradient $\nabla F$, - $\frac{dZ^i}{ds}$: the contravariant components of the unit tangent vector $\mathbf{L}$. Hence, the directional derivative of $F$ along $l$ is: $ \frac{dF}{dl} = \nabla F \cdot \mathbf{L}. $ This identity confirms that the gradient $\nabla F$ is the vector whose dot product with a unit direction vector gives the directional rate of increase of $F$. This derivation is valid in general coordinate systems as long as the gradient is properly expressed in tensorial form.