The usual rules of index notation include the Einstein summation convention, where repeated indices imply summation, and the properties of the Kronecker delta $\delta_{ij}$ and the Levi-Civita symbol $\varepsilon_{ijk}$. Let us evaluate each expression step by step.

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(a) $\delta_{ij} \delta_{ij}$:

Using the Kronecker delta property: $\delta_{ij} \delta_{ij} = \sum_{i=1}^3 \sum_{j=1}^3 \delta_{ij} \delta_{ij}.$ The Kronecker delta satisfies $\delta_{ii} = 3$ (the trace of the identity matrix in 3D): $\delta_{ij} \delta_{ij} = \delta_{ii} = 3.$

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(b) $\varepsilon_{ijk} \varepsilon_{kji}$:

Expand the Levi-Civita symbols: $\varepsilon_{ijk} \varepsilon_{kji} = \sum_{i=1}^3 \sum_{j=1}^3 \sum_{k=1}^3 \varepsilon_{ijk} \varepsilon_{kji}.$ The property of Levi-Civita symbols states: $\varepsilon_{ijk} \varepsilon_{kji} = 3! \delta_{ij} \delta_{ji} - \delta_{ik} \delta_{jk}.$ For three dimensions, this simplifies to: $\varepsilon_{ijk} \varepsilon_{kji} = 3!(1) = 6.$

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(c) $\varepsilon_{ijk} a_i a_k$:

In this expression: $\varepsilon_{ijk} a_i a_k.$ The Levi-Civita symbol is antisymmetric in $i$ and $k$. Since $a_i a_k$ is symmetric under the interchange of $i$ and $k$, the product $\varepsilon_{ijk} a_i a_k = 0$ due to antisymmetry.

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(d) $\varepsilon_{ijk} \frac{\partial^2}{\partial x_i \partial x_j}$:

The second derivative operator $\frac{\partial^2}{\partial x_i \partial x_j}$ is symmetric under $i$ and $j$. Since $\varepsilon_{ijk}$ is antisymmetric under $i$ and $j$, their product is zero: $\varepsilon_{ijk} \frac{\partial^2}{\partial x_i \partial x_j} = 0.$

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(e) $a_i b_j \delta_{ij}$:

Using the Einstein summation convention: $a_i b_j \delta_{ij} = \sum_{i=1}^3 a_i b_i.$ This is simply the dot product of the vectors $\mathbf{a}$ and $\mathbf{b}$: $a_i b_j \delta_{ij} = \mathbf{a} \cdot \mathbf{b}.$

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Final Answers:

(a) $\delta_{ij} \delta_{ij} = 3$.
(b) $\varepsilon_{ijk} \varepsilon_{kji} = 6$.
(c) $\varepsilon_{ijk} a_i a_k = 0$.
(d) $\varepsilon_{ijk} \frac{\partial^2}{\partial x_i \partial x_j} = 0$.
(e) $a_i b_j \delta_{ij} = \mathbf{a} \cdot \mathbf{b}$.

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Do you want detailed derivations or have any questions?
Here are 5 related questions to expand your understanding:

1. What are the properties of the Levi-Civita symbol in 2D versus 3D?
2. How does the Kronecker delta simplify summations in higher dimensions?
3. What are some physical applications of these tensor operations (e.g., in fluid dynamics or electromagnetism)?
4. How does symmetry or antisymmetry of tensors impact their contraction?
5. Can you derive the determinant of a 3x3 matrix using the Levi-Civita symbol?

Tip: When working with the Levi-Civita symbol, always check the symmetry properties of the associated terms to identify when contractions vanish.