$\nabla \cdot (\mathbf{B}\mathbf{B} - \frac{1}{2}B^2 \tilde{1})=(\nabla \cdot \mathbf{B})\mathbf{B} - \mathbf{B} \times (\nabla \times \mathbf{B})$ Does someone know how to show this identity?
$\nabla \cdot (\mathbf{B}\mathbf{B} - \frac{1}{2}B^2 \tilde{1})=(\nabla \cdot \mathbf{B})\mathbf{B} - \mathbf{B} \times (\nabla \times \mathbf{B})$
 A: I will work in index notation and with the Einstein summation convention.
First, let's write the LHS in index notation and simplify:
\begin{align}
\partial_i \left(B_i B_j -\frac{1}{2} B_k B_k \delta_{i j}\right)
=(\partial_i B_i)B_j +B_i (\partial_i B_j)-B_k( \partial_j B_k).
\end{align}
Here we have applied the product rule and summed over the delta function. Note that the first term is the $j$th component of $(\nabla\cdot \mathbf{B})\mathbf{B}$. That leaves the $-\mathbf{B}\times \nabla\times \mathbf{B}$ term, which in index notation may be written as 
\begin{align}
-\epsilon_{jlm}\epsilon_{mkn}B_l \, \partial_k B_{n}
&= -(\delta_{jk}\delta_{ln}-\delta_{jn}\delta_{lk})B_l \, \partial_k B_{n} =
 -B_n(\partial_j B_n) +B_k (\partial_k B_j)
\end{align}
where we have again eliminated the delta function. Accounting for dummy indices, this is exactly the two remaining terms. Hence both sides agree and the identity is valid.
A: As an alternative to Semiclassical, the problem can be tackled entirely using the machinery of clifford algebra.  Let $a$ be an arbitrary vector, and let $\underline B(a) = \overline B(a) = (B \cdot a) B$.  Then we have
$$\nabla \cdot [\underline B(a) - \frac{1}{2} B^2 a] = a \cdot [\overline B(\nabla) - \frac{1}{2} \nabla B^2] = a \cdot [B (\nabla \cdot B) + (B \cdot \nabla) B - \frac{1}{2} \nabla B^2]$$
Expansion of the $\nabla B^2$ term requires a common identity:
$$\nabla (A \cdot B) = (\nabla \wedge B) \cdot A + (\nabla \wedge A) \cdot B + (A \cdot \nabla) B + (B \cdot \nabla) A$$
You can see that, once this is proved, for $A = B$, the required expansion falls out immediately.  The hard part, then, is merely the proof of this identity:
This is where clifford algebra becomes useful.  We can write
$$\nabla (A \cdot B) = \nabla (AB) - \nabla \cdot (A \wedge B)$$
where $AB$ is understood as using the geometric product.  The resulting expansion gives
$$\nabla (AB) = (\nabla \cdot A) B + (\nabla \wedge A) \cdot B + (A \cdot \nabla) B - (A \wedge \nabla) \cdot B$$
and
$$\nabla \cdot (A \wedge B) = (A \cdot \nabla) B + (\nabla \cdot A) B - (B \cdot \nabla) A - (\nabla \cdot B) A$$
We can see some terms canceling already.  We need a slightly better expansion of terms like $(A \wedge \nabla) \cdot B$, though.  See that
$$\langle A (\nabla B )\rangle_1 = (A \cdot \nabla) B + (A \wedge \nabla) \cdot B =A (\nabla \cdot B) + A \cdot (\nabla \wedge B) $$
Putting this into the $\nabla(AB)$ expansion gives
$$\nabla(AB) = (\nabla \cdot A) B + (\nabla \wedge A) \cdot B + 2 (A \cdot \nabla) B - A (\nabla \cdot B) - A \cdot (\nabla \wedge B)$$
This, combined with the previous expansion for $\nabla \cdot (A \wedge B)$, immediately gives the desired result.  In case it's hard to keep track of all the terms, though, the following should be better formatted to convince you of the result:
$$\begin{alignat*}{6}
\nabla(AB) &= (\nabla \cdot A) B &- (\nabla \cdot B) A &+ 2(A \cdot \nabla) B &+ 0 &+ (\nabla \wedge A) \cdot B &+ (\nabla \wedge B) \cdot A \\
\nabla \cdot (A \wedge B) &= (\nabla \cdot A) B &- (\nabla \cdot B) A &+ (A \cdot \nabla) B &- (B \cdot \nabla) A &
\end{alignat*}$$
The resulting identify for $\nabla (A \cdot B)$ is then proven, and the special case needed for $A = B$ can be taken to prove the result.  All of this was proven, without resorting to index notation, thanks to the power of clifford algebra and its associated calculus.
