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I'm a new learner of tensor product. I have seen a question that asked to show that curl of a position vector is zero. $\nabla \times r=0$

If we write the equation using epsilon, we get, $$\nabla \times r= \epsilon_{ijk} \partial_{j}r_k $$

How it could be zero?

Is that equation a special case? We get that equal to zero only if any of the indices are equal.

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As you've said, if two of the indices are equal, then the equation vanishes. This is because the Levi-Civita symbol vanishes. However, if they are all different, then we have

$$j \neq k \implies \partial_j r_k = 0 \implies \nabla \times r = 0$$

Because the coordinates of the position vector are independent (i.e. have no partial dependence).

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You're very close to getting to the answer. Here: $$\begin{align} [\nabla \times \vec{r}]_i & = \epsilon_{ijk} \partial_j r_k \\ & = \epsilon_{ijk} \delta_{jk}\end{align}.$$ You can do the contraction manually to see that you get zero, or note that $\epsilon_{ijk}$ is antisymmetric under the interchange of any two indices. Any such tensor will be traceless for any pair of indices.

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Instead of using the Levi-Civita symbol, $\epsilon_{ijk}$, I prefer to write it more explicitly as $\hat x_i\cdot (\hat x_j \times \hat x_k)$, where $\hat x_i$ is the Cartesian unit vector along the $x_i$ axis. We proceed now to write the curl of the position vector.

The $x_i$ component of the curl of the position vector $\vec r=\sum_{k=1}^3 \hat x_k x_k$ can be written

$$\hat x_i\cdot \nabla \times \vec r =\hat x_i\cdot (\hat x_j\partial_j)\times (\hat x_k x_k) \tag 1$$

where summation over repeated indices in $(1)$ is implied. Continuing, we have

$$\begin{align} \hat x_i\cdot \nabla \times \vec r &=\hat x_i\cdot (\hat x_j \times \hat x_k)\partial_j(x_k) \tag 2\\\\ &=\hat x_i\cdot (\hat x_j \times \hat x_k)\delta_{ij} \tag 3\\\\ &=\hat x_i\cdot (\hat x_j\times \hat x_j) \tag 4\\\\ &=0 \tag 5 \end{align}$$

where $\delta_{ij}=1$ for $i=j$ and $0$ otherwise is the Kronecker Delta.

In going from $(2)$ to $(3)$, we noted that the partial $\frac{\partial x_k}{\partial x_j} =\delta_{ij}$.

In going from $(3)$ to $(4)$, we exploited the sifting property of the Kronecker Delta, which sifted our the $k$ index at $k=j$.

In going from $(4)$ to $(5)$, we simply recognized that $\hat x_j \times \hat x_j=0$ for all $j$.

And we are done!

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