## Permutation symbols and determinants ### The permutation symbols > [!Example]- > In a three-dimensional skew-symmetric tensor $A_{ijk}$, only the six entries that are permutations of $(1,2,3)$ can be nonzero: > $ > A_{123},\ A_{132},\ A_{213},\ A_{231},\ A_{312},\ A_{321} > $ > Due to skew-symmetry, these entries satisfy: > $ > A_{123} = A_{312} = A_{231} = \alpha,\quad > A_{213} = A_{321} = A_{132} = -\alpha > $ > So the system has only one degree of freedom. When $\alpha = 1$, this system defines the **permutation symbol** $e_{ijk}$. > >[!Note] > >Skew-symmetry: an exchange of any two indices changes the sign of the entry. We define the **permutation symbol** $e_{i_1 i_2 \dots i_n}$ in $n$ dimensions as follows: $ e_{i_1 i_2 \dots i_n} = \begin{cases} +1 & \text{if } (i_1, i_2, \dots, i_n) \text{ is an even permutation of } (1, 2, \dots, n) \\ -1 & \text{if } (i_1, i_2, \dots, i_n) \text{ is an odd permutation of } (1, 2, \dots, n) \\ \ \ 0 & \text{if any two indices are equal} \end{cases} $ In three dimensions, this means for example: - $e_{123} = e_{312} = e_{231} = +1$ - $e_{213} = e_{321} = e_{132} = -1$ - $e_{112} = e_{121} = e_{111} = 0$ The same symbol can also be written with upper indices as $e^{i_1 i_2 \dots i_n}$. > [!Notice] > At this stage, $e_{i_1 i_2 \dots i_n}$ is just a symbol that encodes permutation parity and zero redundancy. We are **not** yet assuming it has any tensorial transformation properties or connections to orientation or volume. The placement of indices does not matter until we begin discussing coordinate transformations. ### Determinants We define a compact expression for the **determinant** of a $3 \times 3$ system $a^i_j$ using the permutation symbol $e^{ijk}$. > [!Notice] > At this stage, we are not concerned with the tensorial nature of $a^i_j$ or $A$. The positions of indices are purely conventional. The determinant $A$ is defined as $ A = e^{ijk} a^1_i a^2_j a^3_k $ This reproduces the standard determinant from linear algebra, where the permutation symbol $e^{ijk}$ ensures that only six terms survive, with correct signs. To express this definition in fully contracted tensor notation: $ A = \frac{1}{3!} e^{ijk} e_{rst} a^r_i a^s_j a^t_k $ This form avoids referring to specific rows and is compatible with index manipulation in tensor calculus. Definitions of the determinant can also be written for systems with different index placements (it is not a tensor under the current context): With all lower indices: $ A = e^{ijk} a_{1i} a_{2j} a_{3k} $ With all upper indices: $ A = e_{ijk} a^{1i} a^{2j} a^{3k} $ And similarly, for these two forms: $ A = \frac{1}{3!} e^{ijk} e^{rst} a_{ir} a_{js} a_{kt} $ $ A = \frac{1}{3!} e_{ijk} e_{rst} a^{ir} a^{js} a^{kt} $ > [!Example]- > **Verification and antisymmetry** > > Consider a matrix > $ > M = > \begin{pmatrix} > a^1_1 & a^1_2 & a^1_3 \\ > a^2_1 & a^2_2 & a^2_3 \\ > a^3_1 & a^3_2 & a^3_3 > \end{pmatrix} > $ > The usual definition of the determinant of a $3 \times 3$ matrix $M$ is: > $ > A = a^1_1 a^2_2 a^3_3 + a^1_2 a^2_3 a^3_1 + a^1_3 a^2_1 a^3_2 - a^1_3 a^2_2 a^3_1 - a^1_1 a^2_3 a^3_2 - a^1_2 a^2_1 a^3_3 > $ > > The permutation symbol formulation gives: > $ > A = e^{ijk} a^1_i a^2_j a^3_k > $ > There are $3! = 6$ nonzero terms in $e^{ijk}$: > $ > \begin{aligned} > A &= +a^1_1 a^2_2 a^3_3 \quad (123) \\ > &\quad + a^1_2 a^2_3 a^3_1 \quad (231) \\ > &\quad + a^1_3 a^2_1 a^3_2 \quad (312) \\ > &\quad - a^1_3 a^2_2 a^3_1 \quad (132) \\ > &\quad - a^1_1 a^2_3 a^3_2 \quad (213) \\ > &\quad - a^1_2 a^2_1 a^3_3 \quad (321) > \end{aligned} > $ > > These match exactly with the six signed terms in the standard determinant expression. The signs are correctly assigned by $e^{ijk}$. > > Moreover, the expression is **antisymmetric** with respect to row permutations. Swapping any two of the rows (e.g., exchanging $a^1$ and $a^2$) gives: > $ > e^{ijk} a^2_i a^1_j a^3_k = - e^{ijk} a^1_i a^2_j a^3_k = -A > $ > This confirms the expected sign change under row exchange. > > From the fully contracted form: > $ > e^{ijk} a^r_i a^s_j a^t_k = A e^{rst} > $ > Contracting both sides with $e_{rst}$ and using the identity $e_{rst} e^{rst} = 3!$, we find: > $ > A = \frac{1}{3!} e^{ijk} e_{rst} a^r_i a^s_j a^t_k > $ > which matches the earlier fully contracted expression. ### The delta systems We define the **delta system** $\delta^{ijk}_{rst}$ as the product of two permutation symbols: $ \delta^{ijk}_{rst} = e^{ijk} e_{rst} $ It is a natural extension of [[Coordinate change and Jacobian#Kronecker symbol|Kronecker symbol]]This object allows us to write the determinant of a second-order system $a^i_j$ in a more compact form: $ A = \frac{1}{3!} \delta^{ijk}_{rst} a^r_i a^s_j a^t_k $ The full delta system can also be expressed as a determinant of Kronecker symbols: $ \delta^{ijk}_{rst} = \begin{vmatrix} \delta^i_r & \delta^i_s & \delta^i_t \\ \delta^j_r & \delta^j_s & \delta^j_t \\ \delta^k_r & \delta^k_s & \delta^k_t \end{vmatrix} $ > [!proof]- > **Proof of determinant expression for $\delta^{ijk}_{rst}$** *(Exercise 168)* > By definition, > $ > \delta^{ijk}_{rst} = e^{ijk} e_{rst} > $ > Each permutation symbol can be written as a determinant of Kronecker deltas: > $ > e^{ijk} = > \begin{vmatrix} > \delta^i_1 & \delta^i_2 & \delta^i_3 \\ > \delta^j_1 & \delta^j_2 & \delta^j_3 \\ > \delta^k_1 & \delta^k_2 & \delta^k_3 > \end{vmatrix}, \quad > e_{rst} = > \begin{vmatrix} > \delta^1_r & \delta^1_s & \delta^1_t \\ > \delta^2_r & \delta^2_s & \delta^2_t \\ > \delta^3_r & \delta^3_s & \delta^3_t > \end{vmatrix} > $ > Taking the product and summing over 1, 2, 3 (contracting the dummy indices) gives: > $ > \delta^{ijk}_{rst} = > \begin{vmatrix} > \delta^i_r & \delta^i_s & \delta^i_t \\ > \delta^j_r & \delta^j_s & \delta^j_t \\ > \delta^k_r & \delta^k_s & \delta^k_t > \end{vmatrix} > $ > as required. > [!Example]- > > **Tensor nature** > Since the Kronecker symbol $\delta^i_j$ is a true tensor, it follows from the determinant expression that $\delta^{ijk}_{rst}$ is also a **tensor**. > Consequently, the determinant expression > $ > A = \frac{1}{3!} \delta^{ijk}_{rst} a^r_i a^s_j a^t_k > $ > is an **invariant** when $a^i_j$ has one contravariant and one covariant index. > > **Values of $\delta^{ijk}_{rst}$** > - $+1$ if the upper and lower indices are the same set and related by an even permutation > - $-1$ if related by an odd permutation > - $0$ if the index sets differ or contain repeated elements > > Examples: > $ > \delta^{12}_{12} = 1, \quad > \delta^{23}_{32} = -1, \quad > \delta^{11}_{12} = 0, \quad > \delta^{12}_{13} = 0 > $ > > **Contractions**: > $ > \delta^{ij}_{rs} = \delta^{ijk}_{rsk}, \quad > 2\delta^i_r = \delta^{ij}_{rj} > $ > > **In 4D**: > $ > \delta^{ijk}_{rst} = \delta^{ijkl}_{rstl}, \quad > 2\delta^{ij}_{rs} = \delta^{ijk}_{rsk}, \quad > 3\delta^i_j = \delta^{ij}_{rj} > $ > > **In $N$ dimensions**: > These contraction identities generalize naturally. For example: > $ > \delta^{i_1 i_2 \dots i_{N-1}}_{r_1 r_2 \dots r_{N-1}} = \delta^{i_1 i_2 \dots i_N}_{r_1 r_2 \dots r_N} \delta^{r_N}_{i_N} > $ > $ > (N - 1)\, \delta^{i_1 i_2 \dots i_{N-2}}_{r_1 r_2 \dots r_{N-2}} = \delta^{i_1 i_2 \dots i_{N-1}}_{r_1 r_2 \dots r_{N-1}} \delta^{r_{N-1}}_{i_{N-1}} > $ > > **Rank-2 delta system**: > The second-order antisymmetric delta can be expressed as: > $ > \delta^{ij}_{rs} = \delta^i_r \delta^j_s - \delta^j_r \delta^i_s > = > \begin{vmatrix} > \delta^i_r & \delta^i_s \\ > \delta^j_r & \delta^j_s > \end{vmatrix} > $ > **Generalization to $n$ indices** (see [[Cross product, curl, and higher-dimensional generalization#Other dimensions generalization]]): > The antisymmetric delta system of rank $n$ can be expressed as the determinant of Kronecker deltas: > $ > \delta^{i_1 i_2 \dots i_n}_{r_1 r_2 \dots r_n} = > \begin{vmatrix} > \delta^{i_1}_{r_1} & \delta^{i_1}_{r_2} & \cdots & \delta^{i_1}_{r_n} \\ > \delta^{i_2}_{r_1} & \delta^{i_2}_{r_2} & \cdots & \delta^{i_2}_{r_n} \\ > \vdots & \vdots & \ddots & \vdots \\ > \delta^{i_n}_{r_1} & \delta^{i_n}_{r_2} & \cdots & \delta^{i_n}_{r_n} > \end{vmatrix} > $ > This determinant vanishes if any indices repeat or if the upper and lower sets differ. It equals $+1$ or $-1$ depending on whether the sets are related by an even or odd permutation, respectively. > [!Note]- > **Why the determinant is an invariant** *(Exercise 170)* > The expression > $ > A = \frac{1}{3!} \delta^{ijk}_{rst} a^r_i a^s_j a^t_k > $ > is a **fully contracted scalar** formed from a **tensor** $\delta^{ijk}_{rst}$ and a general second-order object $a^i_j$. > If $a^i_j$ has **one upper and one lower index**, then under coordinate transformations, it transforms like a mixed tensor. Since $\delta^{ijk}_{rst}$ is a true tensor (built from Kronecker deltas), and all indices are contracted, the result $A$ is a **scalar invariant**. > If $a_{ij}$ or $a^{ij}$ were used instead, the contraction would not pair covariant with contravariant indices, and $A$ would **not** be invariant. ### Multiplication property of determinants The multiplication property of determinants states: $ |MN| = |M||N| $ Let $c^i_j = a^i_k b^k_j$ be the matrix product of $a^i_j$ and $b^j_k$. Then the determinant of $c$ satisfies: $ \det(c) = \det(a) \cdot \det(b) $ This can be shown by: $ C = \frac{1}{3!} \delta^{ijk}_{rst} c^r_i c^s_j c^t_k = \frac{1}{3!} \delta^{ijk}_{rst} a^r_l b^l_i a^s_m b^m_j a^t_n b^n_k = \frac{1}{3!} A \delta^{lmn}_{ijk} b^l_i b^m_j b^n_k = AB $ > [!Proof]- > **Details of the derivation** > Start from the expression: > $ > C = \frac{1}{3!} \delta^{ijk}_{rst} c^r_i c^s_j c^t_k > $ > Substitute $c^r_i = a^r_l b^l_i$, giving: > $ > C = \frac{1}{3!} \delta^{ijk}_{rst} a^r_l b^l_i a^s_m b^m_j a^t_n b^n_k > $ > Group the $a$ terms: > $ > C = \frac{1}{3!} \delta^{ijk}_{rst} a^r_l a^s_m a^t_n \cdot b^l_i b^m_j b^n_k > $ > Apply the identity: > $ > \delta^{ijk}_{rst} a^r_l a^s_m a^t_n = A \delta^{lmn}_{ijk} > $ > Then: > $ > C = \frac{1}{3!} A \delta^{lmn}_{ijk} b^l_i b^m_j b^n_k = AB > $ > [!Note] > This derivation illustrates the elegance of tensor notation: > The antisymmetry and permutation structure is fully encoded in the delta system, avoiding the need to explicitly track permutations or signs. ### Determinant cofactors We define a second-order system called the **cofactor** of the determinant. In $n$ dimensions, it is given by: $ A^i_r = \frac{1}{(n-1)!} \, \delta^{i i_2 \dots i_n}_{r r_2 \dots r_n} \, a^{r_2}_{i_2} \cdots a^{r_n}_{i_n} $ This generalizes the [[Linear algebra recap#Minor and adjugate matrix|signed minor]] (adjugate entry) from linear algebra, and reduces in three dimensions to: $ A^i_r = \frac{1}{2!} \, \delta^{ijk}_{rst} \, a^s_j a^t_k $ The cofactor satisfies: - The **Laplace expansion** of the determinant: $\displaystyle A = a^r_i A^i_r$ - It naturally arises in both **derivatives of the determinant** and **matrix inversion** Two important properties: 1. The cofactor $A^i_r$ is the partial derivative of the determinant: $ \frac{\partial A}{\partial a^r_i} = A^i_r $ 2. The inverse of $a^i_r$ is proportional to $A^i_r$: $ A^i_r a^r_m = A \delta^i_m $ In **three dimensions**, the formula reduces to: $ A^i_r = \frac{1}{2!} \, \delta^{ijk}_{rst} \, a^s_j a^t_k $ > [!Proof]- > The following derivation assumes a three-dimensional system. Generalization to higher dimensions follows the same delta system structure. > > **1. The cofactor is the partial derivative of the determinant** > > Start from the determinant formula: > $ > A = \frac{1}{3!} \delta^{ijk}_{rst} a^r_i a^s_j a^t_k > $ > Now compute the derivative with respect to an arbitrary component $a^u_l$. Since $i$, $j$, and $k$ are dummy indices: > $ > \frac{\partial A}{\partial a^u_l} > = \frac{1}{3!} \delta^{ijk}_{rst} \left( > \frac{\partial a^r_i}{\partial a^u_l} a^s_j a^t_k + > a^r_i \frac{\partial a^s_j}{\partial a^u_l} a^t_k + > a^r_i a^s_j \frac{\partial a^t_k}{\partial a^u_l} > \right) > $ > Each partial derivative gives a Kronecker delta: > $ > \frac{\partial a^r_i}{\partial a^u_l} = \delta^r_u \delta^i_l, \quad > \frac{\partial a^s_j}{\partial a^u_l} = \delta^s_u \delta^j_l, \quad > \frac{\partial a^t_k}{\partial a^u_l} = \delta^t_u \delta^k_l > $ > Plug in: > $ > \frac{\partial A}{\partial a^u_l} > = \frac{1}{3!} \delta^{ijk}_{rst} \left( > \delta^r_u \delta^i_l a^s_j a^t_k + > \delta^s_u \delta^j_l a^r_i a^t_k + > \delta^t_u \delta^k_l a^r_i a^s_j > \right) > $ > Distribute and relabel dummy indices for each term: > $ > \frac{\partial A}{\partial a^u_l} > = \frac{1}{3!} \left( > \delta^{ljk}_{ust} a^s_j a^t_k + > \delta^{ilk}_{rut} a^r_i a^t_k + > \delta^{ijl}_{rsu} a^r_i a^s_j > \right) > $ > Each of the three terms gives the same expression after relabeling, so the result is: > $ > \frac{\partial A}{\partial a^u_l} = \frac{3}{3!} \delta^{ljk}_{ust} a^s_j a^t_k = \frac{1}{2!} \delta^{ljk}_{ust} a^s_j a^t_k = A^l_u > $ > which proves: > $ > \frac{\partial A}{\partial a^r_i} = A^i_r > $ > > > **2. The cofactor is the inverse matrix (up to a factor $A$)** > > Consider the product: > $ > D^i_m = A^i_r a^r_m = \frac{1}{2!} \delta^{ijk}_{rst} a^s_j a^t_k a^r_m > $ > Now look at the value of $D^i_m$ in two cases: > > - If $i \ne m$: two rows in the determinant are equal, so antisymmetry gives $D^i_m = 0$ > - If $i = m$: the result is just the determinant again, so $D^i_i = A$ > > Together: > $ > A^i_r a^r_m = A \delta^i_m > $ > which confirms that $A^i_r$ behaves like $\mathrm{adj}(A)^r_i$, the right inverse up to the scalar $A$. The identity used in the proof, $ \frac{\partial a^r_i}{\partial a^u_l} = \delta^r_u \delta^i_l, $ is a standard result in tensor calculus: the partial derivative of a matrix component with respect to another is a product of [[Coordinate change and Jacobian#Kronecker symbol|Kronecker deltas]]. It ensures that only the corresponding term survives in the product rule. > [!Note] > **Cofactors with different index configurations** (exercise 183, 186) > The cofactor construction extends to tensors with other index types: > - For a system with all-lower indices $a_{ij}$, the cofactor is written as $A^{ij}$ > - For a system with all-upper indices $a^{ij}$, the cofactor becomes $A_{ij}$ > > These generalizations preserve antisymmetry and ensure proper index structure under transformation rules, though in this section we are not yet assuming any tensorial behavior.