Tensor notation proof of Divergence of Curl of a vector field Prove $\nabla\cdot(\nabla\times \vec{F})=\vec{0}$ using tensor notation.
Here is my shot at it:
$$\nabla\cdot(\nabla\times \vec{F})=\vec{0}$$ becomes $$\partial_{i}(\epsilon_{ijk}\partial_{j}F_{k})$$
Using the product rule.
$$\epsilon_{ijk}[F_{k}(\partial_{i}\partial_{j})+\partial_{j}(\partial_{i}F_{k})] = \epsilon_{ijk}F_{k}(\partial_{i}\partial_{j})+\epsilon_{ijk}\partial_{j}(\partial_{i}F_{k})$$
After permutation.
$$\epsilon_{jki}F_{i}\partial_{j}\partial_{k}-\epsilon_{kji}\partial_{j}\partial_{k}F_{i}$$
So wouldn't this look like $\vec{F}\cdot(\nabla\times \nabla)-\vec{F}\cdot(\nabla\times \nabla)=\vec{0}$? I am pretty sure you are not allowed to cross the gradient operator with itself. I don't think this is right.
 A: The most simple way is by noticing that $$ \partial_i\partial_j $$ is completely symmetric under the exchange of the two indices while $\epsilon_{ijk}$ is completely anti-symmetric. Now you use the fact that 

The contraction of a symmetric quantity with an antisymmetric one is always zero

You can easily see this by computing by hand the product $$\epsilon_{ijk}\partial_i\partial_j$$
At this point it's clear that $$\partial_i(\epsilon_{ijk}\partial_jF_k) = (\epsilon_{ijk}\partial_i\partial_j)F_k = 0 $$
A: You've rewritten $\partial_i(\partial_jF_k)$ as $F_k\partial_i\partial_j+\partial_j\partial_iF_k$. That would work if $\partial_j$ were an ordinary quantity you just multiply by $F_k$, but of course it's not. Indeed, your strategy also requires acknowledging $\partial_i$ is instead a differential operator, obeying the famous product rule.
The correct treatment needs no product rule. As @DavideMorgante's answer noted, you can just use the same symmetric indices argument in the proof of $A\cdot A\times F=0$ for a "normal" (i.e. non-operator-valued) vector $A$, since $\partial_i\partial_j=\partial_j\partial_i$ is just as true as $A_iA_j=A_jA_i$.
A: Here's a solution using matrix notation, instead of index notation.
In three dimensions, each vector is associated with a skew-symmetric matrix, which makes the cross product equivalent to matrix multiplication, i.e.
$$\eqalign{
A &= \left[\begin{array}{r}
   0 & -a_3 &  a_2 \\
 a_3 &    0 & -a_1 \\
-a_2 &  a_1 &    0
\end{array}\right] \\
Af &=a\times f
}$$
This suggests that the curl operation is
$$\eqalign{
\nabla\times f  &= \left[\begin{array}{r}
   0 & -\partial_3 &  \partial_2 \\
 \partial_3 &    0 & -\partial_1 \\
-\partial_2 &  \partial_1 &    0
\end{array}\right]\cdot\pmatrix{f_1\\f_2\\f_3} \\
 &=
\pmatrix{
\partial_2f_3-\partial_3f_2 \\
\partial_3f_1-\partial_1f_3 \\
\partial_1f_2-\partial_2f_1 \\
}
}$$
And the operation of interest becomes
$$\eqalign{
\nabla\cdot\nabla\times f
 &= \pmatrix{\partial_1&\partial_2&\partial_3}
\cdot
\pmatrix{
\partial_2f_3-\partial_3f_2 \\
\partial_3f_1-\partial_1f_3 \\
\partial_1f_2-\partial_2f_1 \\
} \\
&= \partial_1(\partial_2f_3-\partial_3f_2)
 + \partial_2(\partial_3f_1-\partial_1f_3)
 + \partial_3(\partial_1f_2-\partial_2f_1) \\
&= \partial_1\partial_2(f_3-f_3) 
 + \partial_2\partial_3(f_1-f_1) 
 + \partial_3\partial_1(f_2-f_2) \\
&= 0 \\
}$$
which is zero for any $f$ vector.
A: This answer uses the rules of tensor calculus with both upper and lower indices.
Let us define the divergence of a tensor field $V^i$ by using the covariant derivative $\nabla_jV^i$; where the curl is given by $V^i = \epsilon^{ijk}\nabla_jU_k$, and $\epsilon^{ijk}$ is the Levi-Cività symbol:
$$
\nabla_iV^i = \nabla_i(\epsilon^{ijk}\nabla_jU_k) = \epsilon^{ijk}\nabla_i(\nabla_j U_k) = \epsilon^{ijk}(\nabla_i\nabla_jU_k + \nabla_j\nabla_iU_k) = \epsilon^{ijk}\nabla_i\nabla_jU_k + \epsilon^{ijk}\nabla_j\nabla_iU_k.
$$
Now, since the last term does not contain live indices, we can let index $i$ become $j$ and vice-versa to give:
$$
\nabla_iV^i = \epsilon^{ijk}\nabla_i\nabla_jU_k + \epsilon^{jik}\nabla_i\nabla_jU_k = \epsilon^{ijk}\nabla_i\nabla_jU_k - \epsilon^{ijk}\nabla_i\nabla_jU_k = 0,
$$
by a single permutation in the last Levi-Cività symbol.
A: While other answers are correct, allow me to add a detailed calculation. We can write the divergence of a curl of $\vec {F}$ as:
$$
\nabla\cdot(\nabla\times \vec{F}) = \partial_i (\epsilon_{ijk} \partial_j F_k)
$$
We would have used the product rule on terms inside the bracket if they simply were a cross-product of two vectors. But as we have a differential operator, we don't need to use the product rule. We get:
$$
\nabla\cdot(\nabla\times \vec{F}) = \epsilon_{ijk} \partial_i \partial_j F_k
$$
Differential operators do commute, so $\partial_i \partial_j$ will be a symmetric quantity, we can write it as:
$\partial_i \partial_j = \frac{1}{2}\big(\partial_i \partial_j + \partial_j \partial_i \big)$. Hence we get:
$$
\Rightarrow \nabla\cdot(\nabla\times \vec{F}) = \epsilon_{ijk} \frac{1}{2}\big(\partial_i \partial_j + \partial_j \partial_i \big) F_k
$$
$$
= \frac{1}{2}\big(\epsilon_{ijk}\partial_i \partial_j F_k + \epsilon_{ijk}\partial_j \partial_i F_k \big) = \frac{1}{2}\big(\epsilon_{ijk}\partial_i \partial_j F_k - \epsilon_{jik}\partial_j \partial_i F_k \big)
$$
$$
= \frac{1}{2}\big(\nabla \cdot (\nabla \times \vec F) - \nabla \cdot (\nabla \times \vec F \big) \big) = 0
$$
Where in the second line we used the anti-symmetric property of Levi-Civita tensor: $\epsilon_{ijk} = -\epsilon_{jik}$.
