How the curl of position vector is zero 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. 
 A: 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.
A: 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).
A: 
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!
