## Relative tensor and Levi-Civita symbol ### The volume element Let $Z = |Z_{ij}|$ be the determinant of the covariant metric tensor $Z_{ij}$. The **volume element** is defined as: $ \sqrt{Z} $ Although $Z$ is **not a tensor**, it transforms according to a specific rule, making it a **[[#Relative tensors|relative tensor]] of weight 2**: $ Z' = |Z_{i'j'}| = |Z_{ij} J^i_{i'} J^j_{j'}| = Z J^2 $ where $J = \left| J^\cdot_{\cdot'} \right|$ is the determinant of the Jacobian matrix, $Z=|Z_{\cdot \cdot}|$ and $Z'=|Z_{\cdot' \cdot'}|$. Since the inverse satisfies $ J^{-1} = \left| J^{i'}_i \right|, $ we also find that: $ \sqrt{Z'} = |J| \sqrt{Z} $ In Cartesian coordinates, $Z = 1$, so we recover the classical result $\sqrt{Z'} = |J| = \text{absolute value}(J)$. > [!Notice] > This identity holds because the inner product in Cartesian coordinates is Euclidean, i.e., $Z_{ij} = \delta_{ij}$. > As a result, the determinant of the metric is identically $1$, and the volume element reduces to the Jacobian determinant. > In this sense, Cartesian coordinates are geometrically special, serving as a baseline where the volume element becomes trivial. > [!Note] > **Geometric meaning** > The volume element $\sqrt{Z}$ measures how much a coordinate system stretches or contracts space locally: > - In 3D: $\sqrt{Z}$ gives volume scaling > - In 2D: $\sqrt{Z}$ gives area > - In 1D: $\sqrt{Z}$ gives arc length > > It also determines the correct integration measure $\mathrm{d}V = \sqrt{Z} \, \mathrm{d}x^1 \dots \mathrm{d}x^n$. > [!Example]- > Examples in common coordinates > > - Cartesian coordinates: > $ > \sqrt{Z} = 1 > $ > - Polar / Cylindrical coordinates: > $ > Z_{ij} = > \begin{pmatrix} > 1 & 0 \\ > 0 & r^2 > \end{pmatrix} > \quad \Rightarrow \quad > \sqrt{Z} = \sqrt{1 \cdot r^2} = r > $ > > - Spherical coordinates: > $ > > Z_{ij} = > \begin{pmatrix} > 1 & 0 & 0 \\ > 0 & r^2 & 0 \\ > 0 & 0 & r^2 \sin^2 \theta > \end{pmatrix} > \quad \Rightarrow \quad > \sqrt{Z} = r^2 \sin \theta > $ > > > >[!Proof]- > > Verification using tensor contraction (δ-system) > > > > The determinant of the covariant metric $Z_{ij}$ in $n$ dimensions can be written using the generalized Kronecker delta: > > $ > > \det(Z_{ij}) = \frac{1}{n!} \, \delta^{i_1 i_2 \dots i_n}_{j_1 j_2 \dots j_n} \, Z_{i_1 j_1} Z_{i_2 j_2} \dots Z_{i_n j_n} > > $ > > Therefore, the volume element is: > > $ > > \sqrt{Z} = \sqrt{ \frac{1}{n!} \, \delta^{i_1 \dots i_n}_{j_1 \dots j_n} \, Z_{i_1 j_1} \cdots Z_{i_n j_n} } > > $ > > > > In particular: > > > > - In 3D: > > $ > > \sqrt{Z} = \sqrt{ \frac{1}{3!} \, \delta^{ijk}_{rst} Z_{ir} Z_{js} Z_{kt} } > > $ > > - In 2D: > > $ > > \sqrt{Z} = \sqrt{ \frac{1}{2!} \, \delta^{ij}_{rs} Z_{ir} Z_{js} } > > $ > > > > > > **Examples**: > > > > - **Polar coordinates (2D)**: > > $ > > Z_{ij} = > > \begin{pmatrix} > > 1 & 0 \\ > > 0 & r^2 > > \end{pmatrix}, \quad > > \sqrt{Z} = \sqrt{ \delta^{ij}_{rs} Z_{ir} Z_{js} / 2 } = \sqrt{1 \cdot r^2 - 0} = r > > $ > > > > - **Spherical coordinates (3D)**: > > $ > > Z_{ij} = > > \begin{pmatrix} > > 1 & 0 & 0 \\ > > 0 & r^2 & 0 \\ > > 0 & 0 & r^2 \sin^2 \theta > > \end{pmatrix}, \quad > > \sqrt{Z} = \sqrt{ \delta^{ijk}_{rst} Z_{ir} Z_{js} Z_{kt} / 6 } = r^2 \sin \theta > > $ > > > > These results match the known volume elements and confirm the tensor-based formulation. ### The Voss–Weyl formula The **Voss–Weyl formula** expresses the divergence of a contravariant vector field $T^i$ without invoking Christoffel symbols: $ \nabla_i T^i = \frac{1}{\sqrt{Z}} \frac{\partial}{\partial Z^i} \left( \sqrt{Z} T^i \right) $ This identity is valuable in general coordinates for the following reasons. First, it provides a divergence expression **without the need for Christoffel symbols**, making it algebraically simpler and often easier to compute. Second, the formula is valid in any curvilinear coordinate system, and it is constructed from invariant quantities. Finally, it plays a key role in simplifying expressions such as the Laplacian in differential geometry and physics. > [!Proof] > We begin by applying the product rule: > $ > \frac{1}{\sqrt{Z}} \frac{\partial}{\partial Z^i} \left( \sqrt{Z} T^i \right) > = \frac{\partial T^i}{\partial Z^i} + \frac{T^i}{\sqrt{Z}} \frac{\partial \sqrt{Z}}{\partial Z^i} > $ > > We compute the derivative of the volume element via the chain rule: > $ > \frac{\partial \sqrt{Z}}{\partial Z^k} > = \frac{1}{2 \sqrt{Z}} \frac{\partial Z}{\partial Z^k} > $ > > Then, > $ > \frac{\partial Z}{\partial Z^k} > = \frac{\partial Z}{\partial Z_{ij}} \cdot \frac{\partial Z_{ij}}{\partial Z^k} > = Z Z^{ij} \left( \Gamma_{i,jk} + \Gamma_{j,ik} \right) > = 2Z \Gamma^i_{ik} > $ > Therefore, > $ > \frac{\partial \sqrt{Z}}{\partial Z^k} = \sqrt{Z} \Gamma^i_{ik} > $ > > Substitute into the earlier expression: > $ > \frac{1}{\sqrt{Z}} \frac{\partial}{\partial Z^i} \left( \sqrt{Z} T^i \right) > = \frac{\partial T^i}{\partial Z^i} + T^i \Gamma^k_{ki} > = \nabla_i T^i > $ > [!note] > **Laplacian of a scalar field** > Let $F$ be an invariant scalar. The Laplacian is given by: > $ > \nabla^i \nabla_i F = \nabla_i \left( Z^{ij} \frac{\partial F}{\partial Z^j} \right) > $ > The Voss–Weyl formula avoids computing the full Christoffel expression: > $ > \nabla_i \nabla^i F = \frac{1}{\sqrt{Z}} \frac{\partial}{\partial Z^i} \left( \sqrt{Z} Z^{ij} \frac{\partial F}{\partial Z^j} \right) > $ > This form is easier to compute in non-Cartesian coordinates and ensures coordinate-invariant structure. > [!Example]- > **Spherical coordinates example: full derivation via Christoffel vs. Voss–Weyl** > > We compute the Laplacian of a scalar field $F$ in spherical coordinates using both the Christoffel-symbol-based method and the Voss–Weyl formula. > > First write metric tensor > > The coordinate system is $(r, \theta, \phi)$. The metric tensor is: > > $ > Z_{ij} = > \begin{pmatrix} > 1 & 0 & 0 \\ > 0 & r^2 & 0 \\ > 0 & 0 & r^2 \sin^2 \theta > \end{pmatrix}, \quad > Z^{ij} = > \begin{pmatrix} > 1 & 0 & 0 \\ > 0 & \frac{1}{r^2} & 0 \\ > 0 & 0 & \frac{1}{r^2 \sin^2 \theta} > \end{pmatrix}, \quad > \sqrt{Z} = r^2 \sin \theta > $ > > > **Calculate Laplacian via Christoffel symbols**: >The Laplacian is given by: > $ > \nabla^i \nabla_i F = Z^{ij} \left( \frac{\partial^2 F}{\partial x^i \partial x^j} > - \Gamma^k_{ij} \frac{\partial F}{\partial x^k} \right) > $ > > We compute the required [[The Christoffel symbol#^657e8f|Christoffel symbols]] using: > $ > \Gamma^k_{ij} = \frac{1}{2} Z^{kl} \left( > \frac{\partial Z_{jl}}{\partial x^i} > + \frac{\partial Z_{il}}{\partial x^j} > - \frac{\partial Z_{ij}}{\partial x^l} > \right) > $ > > Non-zero Christoffel symbols: > > - $\Gamma^r_{\theta\theta} = -r$ > - $\Gamma^r_{\phi\phi} = -r \sin^2 \theta$ > - $\Gamma^\theta_{r\theta} = \Gamma^\theta_{\theta r} = \frac{1}{r}$ > - $\Gamma^\theta_{\phi\phi} = -\sin \theta \cos \theta$ > - $\Gamma^\phi_{r\phi} = \Gamma^\phi_{\phi r} = \frac{1}{r}$ > - $\Gamma^\phi_{\theta\phi} = \Gamma^\phi_{\phi\theta} = \cot \theta$ > > Compute each term of the Laplacian: > > $ > \nabla^i \nabla_i F = > \frac{\partial^2 F}{\partial r^2} > + \frac{2}{r} \frac{\partial F}{\partial r} > + \frac{1}{r^2} \left( > \frac{\partial^2 F}{\partial \theta^2} > + \cot \theta \frac{\partial F}{\partial \theta} > \right) > + \frac{1}{r^2 \sin^2 \theta} \frac{\partial^2 F}{\partial \phi^2} > $ > > This is the full Laplacian in spherical coordinates derived with Christoffel symbols. > > **Alternative: Voss–Weyl formula** >The Laplacian of a scalar using the Voss–Weyl formula: > $ > \nabla^i \nabla_i F = \frac{1}{\sqrt{Z}} \frac{\partial}{\partial x^i} > \left( \sqrt{Z} Z^{ij} \frac{\partial F}{\partial x^j} \right) > $ > > With $\sqrt{Z} = r^2 \sin \theta$, the expression becomes: > $ > \nabla^i \nabla_i F = > \frac{1}{r^2 \sin \theta} \left[ > \frac{\partial}{\partial r} \left( r^2 \sin \theta \cdot \frac{\partial F}{\partial r} \right) > + \frac{\partial}{\partial \theta} \left( \sin \theta \cdot \frac{\partial F}{\partial \theta} \right) > + \frac{\partial}{\partial \phi} \left( \frac{1}{\sin \theta} \cdot \frac{\partial F}{\partial \phi} \right) > \right] > $ > > which simplifies to: > $ > \nabla^i \nabla_i F = > \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \frac{\partial F}{\partial r} \right) > + \frac{1}{r^2 \sin \theta} \frac{\partial}{\partial \theta} > \left( \sin \theta \frac{\partial F}{\partial \theta} \right) > + \frac{1}{r^2 \sin^2 \theta} \frac{\partial^2 F}{\partial \phi^2} > $ > > This matches the classical spherical Laplacian and is obtained **without** computing any Christoffel symbols. This confirms that both approaches yield the same Laplacian, but the Voss–Weyl method avoids computing $\Gamma^k_{ij}$, making it much more efficient in curved coordinates. ### Relative tensors A **relative tensor** of weight $M$ is an object that transforms under coordinate changes like a tensor, but with an additional factor involving the determinant of the Jacobian matrix. For a type $(1,1)$ relative tensor $T^i_j$, the transformation law is: $ T^{i'}_{j'} = J^M T^i_j J^i_{i'} J^{j'}_j $ where $J = \left| J^i_{i'} \right|$ is the determinant of the Jacobian matrix. Ordinary tensors are simply the special case $M = 0$ and are referred as **absolute tensor**. A relative tensor of rank zero (i.e., a scalar function) is called a **relative invariant**. For instance, the determinant of the metric tensor, $Z = \det(Z_{ij})$ satisfies: $ Z' = J^2 Z $ and is a relative invariant of weight 2. > [!important] > **Key properties of relative tensors** > > - Multiplying relative tensors of weights $M$ and $N$ gives a tensor of weight $M + N$. > - Contracting indices preserves the weight. > - The determinant of an $n \times n$ relative covariant tensor of weight $M$ is a relative scalar of weight $2 + nM$. > - The inverse of a relative contravariant tensor of weight $M$ is a relative covariant tensor of weight $-M$. > [!Proof]- > **Product rule for weights** > Let $S^i_j$ be a relative tensor of weight $M$ and $T^k_l$ of weight $N$: > $ > S^{i'}_{j'} = J^M S^i_j J^i_{i'} J^{j'}_j, \quad > T^{k'}_{l'} = J^N T^k_l J^k_{k'} J^{l'}_l > $ > Then the product $S^i_j T^k_l$ transforms as: > $ > S^{i'}_{j'} T^{k'}_{l'} = J^{M+N} S^i_j T^k_l J^i_{i'} J^{j'}_j J^k_{k'} J^{l'}_l > $ > so the product has weight $M + N$. > > **Contraction preserves weight** > Suppose $T^{ij}_{kl}$ is a relative tensor of weight $M$, and we contract over $i$ and $k$: > $ > T^{i'j'}_{i'l'} = J^M T^{ij}_{kl} J^{i'}_i J^{j'}_j J^k_{i'} J^l_{l'} > $ > The contraction involves $J^{i'}_i J^k_{i'} = \delta^k_i$, cancelling the extra Jacobian factors. Thus, the result still has weight $M$. > > **Weight of the determinant** > Suppose $a_{ij}$ is a relative covariant tensor of weight $M$: > $ > a_{i'j'} = J^M a_{ij} J^i_{i'} J^j_{j'} > $ > Then the determinant transforms as: > $ > \det(a') = \det(J^M a_{ij} J^i_{i'} J^j_{j'}) = J^{nM} \cdot J^2 \cdot \det(a) = J^{2 + nM} \det(a) > $ > Here, $J^{nM}$ accounts for the $J^M$ factor on each row, and $J^2$ comes from the determinant of the two Jacobians. As a direct consequence, the [[#The volume element|volume element]] $\sqrt{Z}$, introduced in the previous section, is a relative scalar of weight 1. > > **Inverse weight flips sign** > Suppose $A^i_j$ is a relative contravariant tensor of weight $M$: > $ > A^{i'}_{j'} = J^M A^i_j J^{i'}_i J^j_{j'} > $ > Let $B^j_k$ be its inverse, satisfying $A^i_j B^j_k = \delta^i_k$. Then: > $ > A^{i'}_{j'} B^{j'}_{k'} = J^M A^i_j J^{i'}_i J^j_{j'} \cdot B^j_k J^{j'}_j J^k_{k'} = J^M J^{-M} A^i_j B^j_k J^{i'}_i J^k_{k'} = \delta^i_k J^{i'}_i J^k_{k'} > $ > so $B^{j'}_{k'}$ must transform with weight $-M$. Hence, the inverse of a tensor of weight $M$ transforms as a tensor of weight $-M$. ### The Levi-Civita symbols The **Levi-Civita symbols** are defined by rescaling the relative tensors $e^{ijk}$ and $e_{ijk}$ using the volume element $\sqrt{Z}$: $ \varepsilon^{ijk} = \frac{e^{ijk}}{\sqrt{Z}}, \qquad \varepsilon_{ijk} = \sqrt{Z} \, e_{ijk} $ They are **absolute tensors** under orientation-preserving coordinate changes, and play a central role in describing orientation, volume, and antisymmetry. The covariant symbol $\varepsilon_{ijk}$ can also be obtained by lowering the indices of $\varepsilon^{ijk}$ using the metric tensor: $ \varepsilon_{ijk} = \varepsilon^{rst} Z_{ir} Z_{js} Z_{kt} $ This agrees with the definition $\varepsilon_{ijk} = \sqrt{Z} e_{ijk}$, confirming consistency. > [!Proof]- > Starting from the identity $\varepsilon^{ijk} = e^{ijk} / \sqrt{Z}$, substitute into the right-hand side of the index-lowered form: > $ > \varepsilon^{rst} Z_{ir} Z_{js} Z_{kt} > = \frac{e^{rst}}{\sqrt{Z}} Z_{ir} Z_{js} Z_{kt} > = \frac{Z e_{ijk}}{\sqrt{Z}} = \sqrt{Z} e_{ijk} > = \varepsilon_{ijk} > $ > > Thus, both definitions agree. As a corollary, their product gives the generalized Kronecker delta: > $ > \delta^{ijk}_{rst} = \varepsilon^{ijk} \varepsilon_{rst} > $ > The generalized Kronecker delta $\delta^{ijk}_{rst}$ acts as an identity object for antisymmetric tensors, and since it is defined entirely via absolute tensors, it is itself an absolute tensor. > [!Notice] > While the components of $e^{ijk}$ and $e_{ijk}$ are numerically equal in Cartesian coordinates, they are **not** related by [[Index juggling|index lowering]] due to their differing transformation properties: $e^{ijk}$ has weight 1, $e_{ijk}$ has weight $-1$. ### The metrinilic property of Levi-Civita symbol *This section is mainly the proof. One may skip it if not interested.* We now prove that the Levi-Civita symbol with all lowered indices satisfies the **metrinilic property**, that is: $ \nabla_i \varepsilon_{rst} =\nabla_i \varepsilon^{ijk}= 0 $ This means the Levi-Civita symbol behaves like a true metric: it is compatible with the covariant derivative and invariant under parallel transport. We begin with the definition: $ T_{irst} := \nabla_i e_{rst} $ Since $e_{rst}$ is a constant symbol (its partial derivatives vanish), we compute the covariant derivative purely from the Christoffel correction terms: $ T_{irst} = -\Gamma^m_{ir} e_{mst} - \Gamma^m_{is} e_{rmt} - \Gamma^m_{it} e_{rsm} $ This expression is manifestly skew-symmetric in the indices $r$, $s$, $t$. To simplify, consider a specific case: $ T_{i123} = -\Gamma^m_{i1} e_{m23} - \Gamma^m_{i2} e_{1m3} - \Gamma^m_{i3} e_{12m} $ Only one term contributes in each sum: $ T_{i123} = -\Gamma^1_{i1} e_{123} - \Gamma^2_{i2} e_{123} - \Gamma^3_{i3} e_{123} = -\Gamma^m_{im} e_{123} $ Generalizing: $ T_{irst} = -\Gamma^m_{im} e_{rst} $ Now consider the Levi-Civita symbol with volume element: $ \varepsilon_{rst} = \sqrt{Z} \, e_{rst} $ Apply the product rule: $ \nabla_i \varepsilon_{rst} = \nabla_i \left( \sqrt{Z} \, e_{rst} \right) = \frac{\partial \sqrt{Z}}{\partial Z^i} e_{rst} + \sqrt{Z} \, \nabla_i e_{rst} $ Using the identity: $ \frac{\partial \sqrt{Z}}{\partial Z^i} = \sqrt{Z} \Gamma^m_{im} $ and our previous result for $\nabla_i e_{rst}$: $ \nabla_i \varepsilon_{rst} = \sqrt{Z} \Gamma^m_{im} e_{rst} - \sqrt{Z} \Gamma^m_{im} e_{rst} = 0 $ > [!note] > The Levi-Civita symbol $\varepsilon_{ijk}$ satisfies the same metrinilic property as the metric tensor: > $ > \nabla_i \varepsilon_{jkl} = 0 > $ > This ensures its consistency in curved coordinates and is essential in geometric identities involving determinants, volume forms, and antisymmetric operations. > [!important] > **Consequences of the metrinilic property** (confirms [[Properties of covariant differentiation#The metrinilic property]]) > > - The covariant Levi-Civita symbol satisfies: > $ > \nabla_m \varepsilon_{rst} = 0 > $ > > - The contravariant version satisfies: > $ > \nabla_m \varepsilon^{ijk} = 0 > $ > > - As a consequence, all Kronecker delta systems built from their products are covariantly constant: > $ > \nabla_m \delta^{ijk}_{rst} = \nabla_m \delta^{ij}_{rs} = \nabla_m \delta^i_r = 0 > $