In Penrose graphical notation, a tensor is represented as a box with as many sticks as its order. It is possible to distinguish between covariance and contravariance. The downward stick represents covariance, and upward contravariance.

Kronecker delta is represented as a stick.

A horizontal thick stick represents antisymmetry. Odd substitution of the position of the vertical stick reverses the sign. In this notation, Levi-Civita symbol is displayed as below:

The product of covariant and contravariant Levi-Civita symbols \(\epsilon^{ij\cdots k}\epsilon_{pq\cdots r}\) is drawn as the below, according to Penrose's article.

Determinant

According to the definition, \(n\times n\) matrix \(M\) satisfies

\[ \det M = \sum_{\sigma\in\mathfrak{S}_n} \mathrm{sgn}(\sigma) M_1^{\sigma_1}\cdots M_n^{\sigma_n}, \]

with symmetry group \(\mathfrak{S}_n\). Note that the sign of permutation is equivalent to that of Levi-Civita symbol, i.e. \(\mathrm{sgn}(\sigma)=\epsilon_{\sigma_1\cdots\sigma_n}^{1\cdots n}\). Therefore

\[ \det M = \epsilon_{\sigma_1\cdots\sigma_n}^{1\cdots n} M_1^{\sigma_1}\cdots M_n^{\sigma_n}, \]

but it is difficult to manipulate the lower indices in this notation, then we use the following representation:

\[ \det M = \epsilon_{\sigma_1\cdots\sigma_n}^{1\cdots n} \epsilon^{\tau_1\cdots\tau_n}_{1\cdots n} M_1^{\sigma_1}\cdots M_n^{\sigma_n} \]

Drawing this in graphical notation, we get

Cofactor expansion

Since we get the representation of determinant in graphical notation, we anticipate that cofactor expansion can be interpreted in an intrinsic way. Using the notation of determinant with two Levi-Civita symbols,

\[ \det M = \frac{1}{n} \sum_{i,j} (-1)^{i+j} M_i^j \frac{1}{(n-1)!} \epsilon_{\sigma_1\cdots\sigma_{i-1}\sigma_{i+1}\cdots\sigma_n}^{1\cdots i-1,i+1\cdots n} \epsilon^{\tau_1\cdots\tau_{j-1}\tau_{j+1}\cdots\tau_n}_{1\cdots j-1,j+1\cdots n} M_{\tau_1}^{\sigma_1} \cdots M_{\tau_n}^{\sigma_n}. \]

\(\sigma_k, \tau_k\) runs \(\{1, \cdots, i-1, i+1, \cdots, n\}, \{1, \cdots, j-1, j+1, \cdots, n\}\), respectively. Remark that Levi-Civita part is

\begin{align} &\epsilon_{\sigma_1\cdots\sigma_{i-1}\sigma_{i+1}\cdots\sigma_n}^{1\cdots i-1,i+1\cdots n} \epsilon^{\tau_1\cdots\tau_{j-1}\tau_{j+1}\cdots\tau_n}_{1\cdots j-1,j+1\cdots n} \\&\qquad= \epsilon_{\sigma_1\cdots\sigma_{i-1}i\sigma_{i+1}\cdots\sigma_n}^{1\cdots i-1,i,i+1\cdots n} \epsilon^{\tau_1\cdots\tau_{j-1}j\tau_{j+1}\cdots\tau_n}_{1\cdots j-1,j,j+1\cdots n} \\&\qquad= \delta_{\sigma_i}^i\delta_j^{\tau_j} \epsilon_{\sigma_1\cdots\sigma_{i-1}\sigma_i\sigma_{i+1}\cdots\sigma_n}^{1\cdots i-1,i,i+1\cdots n} \epsilon^{\tau_1\cdots\tau_{j-1}\tau_j\tau_{j+1}\cdots\tau_n}_{1\cdots j-1,j,j+1\cdots n} \\&\qquad= (-1)^{i+j} \delta_{\sigma_i}^i \delta_j^{\tau_j} \epsilon_{\sigma_1\cdots\sigma_{i-1}\sigma_i\sigma_{i+1}\cdots\sigma_n}^{1\cdots i-1,i,i+1\cdots n} \epsilon^{\tau_1\cdots\tau_{j-1}\tau_j\tau_{j+1}\cdots\tau_n}_{1\cdots j-1,j,j+1\cdots n} \end{align}

thus

\[ \det M = \frac{1}{n}M_{\sigma_i}^{\tau_j} \frac{1}{(n-1)!} \epsilon_{\sigma_1\cdots\sigma_{i-1}\sigma_i\sigma_{i+1}\cdots\sigma_n}^{1\cdots i-1,i,i+1\cdots n} \epsilon^{\tau_1\cdots\tau_{j-1}\tau_j\tau_{j+1}\cdots\tau_n}_{1\cdots j-1,j,j+1\cdots n} M_{\tau_1}^{\sigma_1} \cdots M_{\tau_n}^{\sigma_n}. \]

Comparing with graphical notation of determinant, \(M_{\tau_j}^{\sigma_i}\) corresponds to the first \(M\) , and remaining \(n-1 \:M\)'s composes a cofactor.

If you do not run either index of Levi-Civita symbols, the branch of the fist \(M\) is cut off to leave the same indices:

This has to be the product of matrix \(M\) and its adjugate \(\tilde{M}\). Thus the adjugate matrix is,

Inverse matrix

According to definition, \(M^{-1}=\tilde{M}/\det M\), i.e.,

Contraction in general case

\begin{align} \epsilon^{ijk}\epsilon_{pqr} = \left| \begin{matrix} \delta^i_p & \delta^i_q & \delta^i_r \\ \delta^j_p & \delta^j_q & \delta^j_r \\ \delta^k_p & \delta^k_q & \delta^k_r \end{matrix} \right| \end{align}

This formula is represented with Kronecker delta as the belows:

Connect top three points and bottom without leaks and doubles. That is, the number of terms is permutation 3!. The signing can be determined to some extent by "how many intersections there are". Nevertheless, as shown in the figure below, this rule can be broken if the line is written in a circuitous manner, so it is safe to consider "how many substitutions will return to the even-substitution form."

Contraction formulas

When using Levi-Civita symbol, some indexes are usually the same; connect the same indexes with Kronecker delta.

In case two or more kinds of indexes are the same, you have to consider permutation, because the same indexes are reversible.

If all indexes are the same, \(\epsilon^{ijk}\epsilon_{ijk}=3!\).

Contraction in general case

\begin{align} \epsilon^{ijkl}\epsilon_{pqrs} = \left| \begin{matrix} \delta^i_p & \delta^i_q & \delta^i_r & \delta^i_s \\ \delta^j_p & \delta^j_q & \delta^j_r & \delta^j_s \\ \delta^k_p & \delta^k_q & \delta^k_r & \delta^k_s \\ \delta^l_p & \delta^l_q & \delta^l_r & \delta^l_s \end{matrix} \right| \end{align}

This formula is represented with Kronecker delta as the below figure.

Contraction formulas

You can calculate contraction formulas as of 3-dim.

It is wiser to leave Levi-Civita symbols than to calculate it. The permutation follows as well as 3-dim.

If all indexes are the same, \(\epsilon^{ijkl}\epsilon_{ijkl}=4!=24\).

Minkovski metric

The above argument only holds in Euclid metric, \(g_{ij}=g^{ij}=\delta_{ij}\). Since the three spacial or one time component of metric tensor is negative in Minkovski metric, covariant and contravariant tensor has opposite signs.

\begin{align} \epsilon^{ijkl}=g^i_pg^j_qg^k_rg^l_s\epsilon_{pqrs}=-\epsilon_{pqrs} \qquad (\{i,j,k,l\}=\{p,q,r,s\}=\{0,1,2,3\}) \end{align}

Therefore, the signs of all the result above is reversed. That is,

\begin{align} \begin{array}{rcl} \epsilon^{ijkl}\epsilon_{pqrl} &=& - \left| \begin{matrix} \delta^i_p & \delta^i_q & \delta^i_r \\ \delta^j_p & \delta^j_q & \delta^j_r \\ \delta^k_p & \Delta^k_q & \delta^k_r \end{matrix} \right|, \\ \epsilon^{ijkl}\epsilon_{pqkl} &=& -2 \left| \begin{matrix} \delta^i_p & \delta^i_q \\ \delta^j_p & \delta^j_q \end{matrix} \right|, \\ \epsilon^{ijkl}\epsilon_{pjkl} &=& -6\delta^i_p, \\ \epsilon^{ijkl}\epsilon_{ijkl} &=& -24. \end{array} \end{align}