vector fields - Index Notation with Del Operators

I'm having trouble with some concepts of Index Notation. (Einstein notation)

If I take the divergence of curl of a vector, $\nabla \cdot (\nabla \times \vec V)$ first I do the parenthesis:

$\nabla_iV_j\epsilon_{ijk}\hat e_k$ and then I apply the outer $\nabla$...

and get: $\nabla_l(\nabla_iV_j\epsilon_{ijk}\hat e_k)\delta_{lk}$

I am not sure if I applied the outer $\nabla$ correctly. If I did do it correctly, however, what is my next step? I guess I just don't know the rules of index notation well enough. Can I apply the index of $\delta$ to the $\hat e$ inside the parenthesis? Or is that illegal?

Answer

First some notation

$$\nabla \times \vec B \rightarrow \epsilon_{ijk}\nabla_j B_k$$ $$\nabla \cdot \vec B \rightarrow \nabla_i B_i$$ $$\nabla B \rightarrow \nabla_i B$$

Now, to your problem,

$$\nabla \cdot(\nabla \times \vec V)$$

writing it in index notation

$$\nabla_i (\epsilon_{ijk}\nabla_j V_k)$$

Now, simply compute it, (remember the Levi-Civita is a constant)

$$\epsilon_{ijk} \nabla_i \nabla_j V_k$$

Here we have an interesting thing, the Levi-Civita is completely anti-symmetric on i and j and have another term $\nabla_i \nabla_j$ which is completely symmetric: it turns out to be zero.

$$\epsilon_{ijk} \nabla_i \nabla_j V_k = 0$$

Lets make the last step more clear. We can always say that $a = \frac{a+a}{2}$, so we have

$$\epsilon_{ijk} \nabla_i \nabla_j V_k = \frac{1}{2} \left[ \epsilon_{ijk} \nabla_i \nabla_j V_k + \epsilon_{ijk} \nabla_i \nabla_j V_k \right]$$

Now lets interchange in the second Levi-Civita the index $\epsilon_{ijk} = - \epsilon_{jik}$, so that

$$\epsilon_{ijk} \nabla_i \nabla_j V_k = \frac{1}{2} \left[ \epsilon_{ijk} \nabla_i \nabla_j V_k - \epsilon_{jik} \nabla_i \nabla_j V_k \right]$$

Now we can just rename the index $\epsilon_{jik} \nabla_i \nabla_j V_k = \epsilon_{ijk} \nabla_j \nabla_i V_k$ (no interchange was done here, just renamed).

$$\epsilon_{ijk} \nabla_i \nabla_j V_k = \frac{1}{2} \left[ \epsilon_{ijk} \nabla_i \nabla_j V_k - \epsilon_{ijk} \nabla_j \nabla_i V_k \right]$$

We can than put the Levi-Civita at evidency,

$$\epsilon_{ijk} \nabla_i \nabla_j V_k = \frac{\epsilon_{ijk}}{2} \left[ \nabla_i \nabla_j V_k - \nabla_j \nabla_i V_k \right]$$

And, because V_k is a good field, there must be no problem to interchange the derivatives $\nabla_j \nabla_i V_k = \nabla_i \nabla_j V_k$

$$\epsilon_{ijk} \nabla_i \nabla_j V_k = \frac{\epsilon_{ijk}}{2} \left[ \nabla_i \nabla_j V_k - \nabla_i \nabla_j V_k \right]$$

And, as you can see, what is between the parentheses is simply zero.