## Cross product, curl, and higher-dimensional generalization With [[Relative tensor and Levi-Civita symbol#The Levi-Civita symbols|the Levi-Civita symbols]], we may now write cross product and curl. ### The cross product The **cross product** of two vectors $U^i$ and $V^i$ is defined using the Levi-Civita symbol: $ W^i = \varepsilon^{ijk} U_j V_k $ This expression gives a true contravariant vector $W^i$. It is coordinate-independent and valid in general curvilinear systems. To obtain the geometric vector $\mathbf{W}$, we write: $ \mathbf{W} = W^i Z_i = \varepsilon^{ijk} U_j V_k Z_i $ > [!Notice] > This formulation confirms that the cross product is a **tensorial operation**, and does not rely on Cartesian coordinates. > [!Proof]- > **Divergence of cross product** > Using the metrinilic property of the Levi-Civita symbol: > $ > \nabla_i W^i = \nabla_i \left( \varepsilon^{ijk} U_j V_k \right) > = \varepsilon^{ijk} \nabla_i (U_j V_k) > = \varepsilon^{ijk} \left( \nabla_i U_j \cdot V_k + U_j \cdot \nabla_i V_k \right) > $ > This is the full covariant divergence, consistent with tensor analysis. > [!important] > **Key geometric and algebraic properties** > > - **Antisymmetry**: > $ > \varepsilon^{ijk} U_j V_k = -\varepsilon^{ijk} V_j U_k > \quad \Rightarrow \quad > U \times V = -V \times U > $ > > - **Orthogonality**: > $ > W^i V_i = \varepsilon^{ijk} U_j V_k V_i = 0 > $ > The result is perpendicular to both inputs. > > - **Norm (length squared)**: > $ > W_i W^i > = \varepsilon_{ijk} U^j V^k \varepsilon^{irs} U_r V_s > = \delta^{rs}_{jk} U^j V^k U_r V_s > $ > Expanding the generalized Kronecker delta: > $ > W_i W^i = U_j V_k U^j V^k - U_j V_k U^k V^j > = \left( \mathbf{U} \cdot \mathbf{U} \right) \left( \mathbf{V} \cdot \mathbf{V} \right) - \left( \mathbf{U} \cdot \mathbf{V} \right)^2 > $ > That is: > $ > |\mathbf{W}|^2 = |\mathbf{U}|^2 |\mathbf{V}|^2 - (\mathbf{U} \cdot \mathbf{V})^2 > $ Two examples for applying tensorial form are shown below. > [!Example]- > **In Cartesian and cylindrical coordinates** > > - **In Cartesian coordinates**, the basis vectors $Z_i$ are constant and orthonormal: > $ > Z_i = \{ \mathbf{i}, \mathbf{j}, \mathbf{k} \}, \quad Z_{ij} = \delta_{ij}, \quad \sqrt{Z} = 1 > $ > So the geometric cross product becomes: > $ > \mathbf{W} = \varepsilon^{ijk} U_j V_k Z_i > = \sum_{i=1}^3 \left( \sum_{j,k=1}^3 \varepsilon^{ijk} U_j V_k \right) Z_i > = > \begin{vmatrix} > \mathbf{i} & \mathbf{j} & \mathbf{k} \\ > U^1 & U^2 & U^3 \\ > V^1 & V^2 & V^3 > \end{vmatrix} > $ > > - **In cylindrical coordinates**, the basis vectors are position-dependent, and the metric tensor is: > $ > Z_{ij} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & r^2 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \quad > Z^{ij} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \frac{1}{r^2} & 0 \\ 0 & 0 & 1 \end{pmatrix}, \quad > \sqrt{Z} = r > $ > The basis vectors $Z_i$ are: > $ > Z_1 = \hat{\mathbf{r}},\quad Z_2 = r \hat{\boldsymbol{\theta}},\quad Z_3 = \hat{\mathbf{z}} > $ > Then the geometric cross product becomes: > $ > \mathbf{W} = \varepsilon^{ijk} U_j V_k Z_i > = \sum_i \left( \varepsilon^{ijk} U_j V_k \right) Z_i > = > \begin{vmatrix} > r \hat{\mathbf{r}} & r \hat{\boldsymbol{\theta}} & r \hat{\mathbf{z}} \\ > U^1 & U^2 & U^3 \\ > V^1 & V^2 & V^3 > \end{vmatrix} > $ > or more carefully: > $ > \mathbf{W} = > \begin{vmatrix} > r Z^1 & r Z^2 & r Z^3 \\ > U^1 & U^2 & U^3 \\ > V^1 & V^2 & V^3 > \end{vmatrix} > $ > where $Z^i$ is the geometric vector associated with basis direction $i$. The factor $r$ reflects the scaling from $\sqrt{Z}$ and ensures the correct geometric result. > [!Example]- > **Tensor proof of cross product non-associativity** > > We prove that: > $ > (\mathbf{U} \times \mathbf{V}) \times \mathbf{W} > = (\mathbf{U} \cdot \mathbf{W}) \mathbf{V} - (\mathbf{V} \cdot \mathbf{W}) \mathbf{U} > $ > using the Levi-Civita tensor in index notation. > > Let: > $ > (\mathbf{U} \times \mathbf{V})^i = \varepsilon^{ijk} U_j V_k > $ > Then: > $ > \left[(\mathbf{U} \times \mathbf{V}) \times \mathbf{W} \right]^m > = \varepsilon^{mnp} (\mathbf{U} \times \mathbf{V})_n W_p > = \varepsilon^{mnp} \varepsilon_{nrs} U^r V^s W_p > $ > > Using the identity: > $ > \varepsilon^{mnp} \varepsilon_{nrs} > = \delta^m_r \delta^p_s - \delta^m_s \delta^p_r > $ > we substitute: > $ > \left[(\mathbf{U} \times \mathbf{V}) \times \mathbf{W} \right]^m > = (\delta^m_r \delta^p_s - \delta^m_s \delta^p_r) U^r V^s W_p > = U^m V^s W_s - V^m U^r W_r > $ > So: > $ > \left[(\mathbf{U} \times \mathbf{V}) \times \mathbf{W} \right]^m > = (\mathbf{U} \cdot \mathbf{W}) V^m - (\mathbf{V} \cdot \mathbf{W}) U^m > $ > which matches the vector identity: > $ > (\mathbf{U} \times \mathbf{V}) \times \mathbf{W} > = (\mathbf{U} \cdot \mathbf{W}) \mathbf{V} - (\mathbf{V} \cdot \mathbf{W}) \mathbf{U} > $ > This confirms the non-associativity of the cross product using tensor notation. ### The curl The **curl** of a vector field is an invariant differential operator defined using the Levi-Civita symbol. For a vector field $U_k$, its curl is given by: $ V^i = \varepsilon^{ijk} \nabla_j U_k $ In dyadic notation, this is simply written as: $ \mathbf{V} = \nabla \times \mathbf{U} $ > [!Note]+ > **Show expression with tensor identity: $\nabla \times (\nabla \times \mathbf{U}) = \nabla (\nabla \cdot \mathbf{U}) - \nabla^2 \mathbf{U}$** > > Consider the composition $\nabla \times (\nabla \times \mathbf{U})$ in tensor notation: > $ > V_r = \varepsilon_{rsi} \nabla^s \left( \varepsilon^{ijk} \nabla_j U_k \right) > $ > Using the **metrinilic property** $\nabla^s \varepsilon^{ijk} = 0$, we move $\varepsilon^{ijk}$ outside the derivative: > $ > V_r = \varepsilon_{rsi} \varepsilon^{ijk} \nabla^s \nabla_j U_k > $ > Now use the contraction identity for Levi-Civita symbols: > $ > \varepsilon_{rsi} \varepsilon^{ijk} = \delta^{jk}_{rs} > $ > Hence, > $ > V_r = \delta^{jk}_{rs} \nabla^s \nabla_j U_k > = \delta^j_r \delta^k_s \nabla^s \nabla_j U_k - \delta^k_r \delta^j_s \nabla^s \nabla_j U_k > $ > Therefore: > $ > V_r = \nabla^k \nabla_r U_k - \nabla^s \nabla_s U_r > $ > Relabeling dummy indices yields the final result: > $ > V_i = \nabla_i \nabla_j U^j - \nabla_j \nabla^j U_i > $ > In vector notation: > $ > \nabla \times (\nabla \times \mathbf{U}) = \nabla (\nabla \cdot \mathbf{U}) - \nabla^2 \mathbf{U} > $ > Other expression can be treated in similarly manner. ### Other dimensions generalization The [[Relative tensor and Levi-Civita symbol#The Levi-Civita symbols|Levi-Civita symbol]] and the delta system naturally generalize to arbitrary dimensions. In two dimensions, indices take values 1 or 2. The permutation symbols $e_{ij}$ and $e^{ij}$ are defined analogously to the 3D case: $ e_{ij}, \, e^{ij} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} $ From these, we define the full delta system: $ \delta^{ij}_{rs} = e^{ij} e_{rs} $ which satisfies the identity: $ \delta^{ij}_{rs} = \delta^i_r \delta^j_s - \delta^j_r \delta^i_s, \qquad \delta^i_j = \delta^i_{jr} $ The determinant of a $2 \times 2$ matrix $a^i_j$ is given by: $ A = \frac{1}{2} \delta^{ij}_{rs} a^r_i a^s_j $ The Levi-Civita symbols are made absolute using the volume element $\sqrt{Z}$: $ \varepsilon_{ij} = \sqrt{Z} e_{ij}, \qquad \varepsilon^{ij} = \frac{e^{ij}}{\sqrt{Z}} $ With this, we define a 2D cross product involving a single vector: $ V_i = \varepsilon_{ij} U^j $ It satisfies the orthogonality condition: $ V_i U^i = \varepsilon_{ij} U^i U^j = 0 $ > [!note] > The 2D cross product maps a vector to an orthogonal vector. > The Levi-Civita symbol ensures antisymmetry, and this construction is crucial in 2D vector calculus. We may also define a scalar-valued curl: $ V = \varepsilon^{ij} \nabla_i U_j $ To generalize to $n$ dimensions, we define the permutation symbols $e_{i_1 \cdots i_n}$ and $e^{i_1 \cdots i_n}$, which take values $1$ or $-1$ for even and odd permutations, and $0$ otherwise. The full delta system becomes: $ \delta^{i_1 \cdots i_n}_{j_1 \cdots j_n} = e^{i_1 \cdots i_n} e_{j_1 \cdots j_n} $ and can be expressed by the determinant: $ \delta^{i_1 \cdots i_n}_{j_1 \cdots j_n} = \begin{vmatrix} \delta^{i_1}_{j_1} & \cdots & \delta^{i_1}_{j_n} \\ \vdots & \ddots & \vdots \\ \delta^{i_n}_{j_1} & \cdots & \delta^{i_n}_{j_n} \end{vmatrix} $ Delta systems satisfy contraction identities: $ \delta^{i_1 \cdots i_{k-1}}_{j_1 \cdots j_{k-1}} = (n-k) \, \delta^{i_1 \cdots i_{k-1} i_k}_{j_1 \cdots j_{k-1} i_k} $ and in particular: $ \delta^{i_1 \cdots i_k}_{i_1 \cdots i_k} = k! $ The determinant of an $n \times n$ matrix is: $ A = \frac{1}{n!} \delta^{i_1 \cdots i_n}_{j_1 \cdots j_n} a^{j_1}_{i_1} \cdots a^{j_n}_{i_n} $ In higher dimensions, the cross product generalizes to a construction involving $(n-1)$ contravariant vectors. The result is a covariant vector: $ V_i = \varepsilon_{i j_1 \cdots j_{n-1}} U^{j_1}_{(1)} \cdots U^{j_{n-1}}_{(n-1)} $ where each $U^{(k)\,j_k}$ is a contravariant vector. This generalization is well-defined in all dimensions but does **not** produce a unique direction unless $n = 3$. > [!warning] > The cross product as a binary operation is **not well defined** in dimensions other than 3. (Although it's possible, but have completely different meanings.) > In $n > 3$, the Levi-Civita symbol still defines orientation and antisymmetry, > but the curl becomes a differential form or tensor rather than a vector.