I'm studying the Einsteinian notation at the beginning of McConnell's tensor analysis. I'm puzzled by the following exercise:
Here $r, s$, are superscripts that, at the left side of the equation, have no correspondent subscript, so, at left, they are free ranging indices, ranging from 1 to 3. All other indices are under this range. On the other hand, $i, j, k$ are, at the left side of the equation, superscripts and subscripts, so they are to be understood as the indices of a sum. It is worth noting that the author, at the beginning of this chapter, explains he will try and denote sums only by indices repeated as subscripts and as superscripts, but he makes the concession that, eventually, a sum may be represented by subscripts or superscripts repeated, respectively, as subscripts or superscripts.
That said, the right side of this equation baffles me. The vertical bars around the element $a^r_s$ denote a determinant, and the element δ$^{rst}_{mnp}$ represents a δ-system of sixth order.
Until now, I thought $|a^r_s|$ could only exist for a given square matrix of established r rows and s columns, $r$ being equal to $s$, so that the notation $a^r_s$ in $|a^r_s|$ was merely a abbreviation of the whole matrix, so that it would make no sense to think about a given value of $r$ and $s$ separated from the others, that is, it would be an utter nonsense to assume $r=1$ and $s=2$ and ask what would be the value of $|a^1_2|$, the determinant of an element of a matrix being undefined.
I have solved this exercise as follows:
This solution means that, for a given value attributed to each free ranging index, namely $r, s, t, m, n, p$, the sum implied by the repetition of the subscripts of δ as the superscripts of $a$ would be equal to the determinant of the matrix $|a^x_y|$, a constant value for any $r, s, t, m, n, p$, multiplied by the δ of the given values of $r, s, t, m, n, p$.
Yet the author proposes $|a^r_s|$ multiplied by a delta of $r$ and $s$, that is, $r$ and $s$ vary their value, which renders the element $|a^r_s|$ nonsensical.
Finally, I would like to point that the very author adopts the solution proposed above, namely, to use new indices to represent a determinant of a term when this determinant is derived from an expression where all the elements have free ranging indices, as shown below:
Here $m$ and $n$ are the new indices spoken above.
The image below should clarify what is a delta system:
The determinant $|a^h_{\,\,k}|$ or $\det(a^h_{\,\,k})$ is just a (real) number. I agree it is a bit sloppy and confusing by the author to reuse the same indices later. But you should ignore any indices inside the determinant. They are just there to remind you $a^h_{\,\,k}$ is the type (1,1)-tensor. If it bothers you, just use $|a|$ or $a$ for the determinant instead. The exercise is $$\delta^{rst}_{ijk}a^i_{\,\,m}a^j_{\,\,n}a^k_{\,\,p}= \varepsilon^{rst}\varepsilon_{ijk}a^i_{\,\,m}a^j_{\,\,n}a^k_{\,\,p}=|a|\delta^{rst}_{mnp}$$
Also, as a side note I would suggest you use a more general definition of the generalized Kronecker delta using an $r\mathrm{x}r$ determinant as follows:
$$\delta^{j_1\dots j_r}_{h_1\dots h_r}=\begin{vmatrix}\delta^{j_1}_{h_1} & \delta^{j_1}_{h_2} & \dots & \delta^{j_1}_{h_r}\\ \delta^{j_2}_{h_1} & \delta^{j_2}_{h_2} & \dots & \delta^{j_2}_{h_r}\\\dots & \dots & \dots & \dots\\ \delta^{j_r}_{h_1} & \delta^{j_r}_{h_2} & \dots & \delta^{j_r}_{h_r} \end{vmatrix}$$
For $r=2$ this is the well known identity
$$\delta^{jh}_{ik}=\begin{vmatrix}\delta^j_i & \delta^j_k\\ \delta^h_i & \delta^h_k\end{vmatrix}=\delta^j_i\delta^h_k-\delta^j_k\delta^h_i$$ The generalized Kronecker delta $\delta^{j_1\dots j_r}_{h_1\dots h_r}$ is skew-symmetric under interchange of any two of the indices $j_1\dots j_r$ and under the interchange of any two of the indices $h_1\dots h_r$. In general $\delta^{j_1\dots j_r}_{h_1\dots h_r}$ is the sum of $r!$ terms, each of which is the prodcuct of $r$ Kronecker deltas. It is easy to prove that $\delta^{j_1\dots j_r}_{h_1\dots h_r}$ is a type (r,r) tensor.
Now we can use this definition to define the Levi-Civita symbols, let $r=n$ (the dimension) and
$$\varepsilon^{j_1\dots j_n}=\delta^{j_1\dots j_n}_{12\dots n},$$ $$\varepsilon_{h_1\dots h_n}=\delta^{12\dots n}_{h_1\dots h_n}$$ It follows trivially that
$$\varepsilon_{h_1\dots h_n}\varepsilon^{j_1\dots j_n}=\delta^{j_1\dots j_n}_{h_1\dots h_n}$$