Vector Calculus - Curl of Vector I'm asked to prove the following identity, using index notation:
$(\nabla\times A)\times A=A \cdot\nabla A - \nabla(A \cdot A)$ 
However, when I work it out, I find that the actual solution should be: 
$(\nabla\times A)\times A=A \cdot\nabla A - \frac{1}{2}\nabla(A \cdot A)$ 
Am I missing something, or is the book wrong?
 A: Expanding everything out componentwise, I get
$(\nabla\times A)\times A$
$$
\begin{align}
&{\lower{1pt}{\Large(}}(\partial_1,\partial_2,\partial_3)\times(A_1,A_2,A_3){\lower{1pt}{\Large)}}\times(A_1,A_2,A_3)\\
&=(\partial_2A_3-\partial_3A_2,\partial_3A_1-\partial_1A_3,\partial_1A_2-\partial_2A_1)\times(A_1,A_2,A_3)\\
&=(A_3\partial_3A_1+A_2\partial_2A_1-A_3\partial_1A_3-A_2\partial_1A_2,\\
&\hphantom{=(}A_1\partial_1A_2+A_3\partial_3A_2-A_1\partial_2A_1-A_3\partial_2A_3,\\
&\hphantom{=(}A_2\partial_2A_3+A_1\partial_1A_3-A_2\partial_3A_2-A_1\partial_3A_1)
\end{align}
$$
$(A\cdot\nabla)A$
$$
\begin{align}
&(A_1\partial_1+A_2\partial_2+A_3\partial_3)(A_1,A_2,A_3)\\
&=(A_1\partial_1A_1+A_2\partial_2A_1+A_3\partial_3A_1,\\
&\hphantom{=(}A_1\partial_1A_2+A_2\partial_2A_2+A_3\partial_3A_2,\\
&\hphantom{=(}A_1\partial_1A_3+A_2\partial_2A_3+A_3\partial_3A_3)
\end{align}
$$
$\frac12\nabla(A\cdot A)$
$$
\begin{align}
&\tfrac12(\partial_1,\partial_2,\partial_3)(A_1A_1+A_2A_2+A_3A_3)\\
&=(A_1\partial_1 A_1+A_2\partial_1 A_2+A_3\partial_1 A_3,\\
&\hphantom{=(}A_1\partial_2 A_1+A_2\partial_2 A_2+A_3\partial_2 A_3,\\
&\hphantom{=(}A_1\partial_3 A_1+A_2\partial_3 A_2+A_3\partial_3 A_3)
\end{align}
$$
And from these, I get
$$
(\nabla\times A)\times A=(A\cdot\nabla)A-\tfrac12\nabla(A\cdot A)
$$
A: A similar proof in spirit to Shuhao Cao's, using geometric calculus.
In GC it's easier not to deal with the cross product or curl, but with the more fundamental entities: the generalized dot and wedge products, and the corresponding interior and exterior derivatives.
The basic relation needed to relate curl/cross product to these other operations is $a \times b = -i (a \wedge b)$.  The quantity $i$ is called the unit pseudoscalar; it might also be called a volume form, and using it finds for you the Hodge dual of a given object.  Here, that duality is captured in the natural multiplication operation called the geometric product.  I won't worry over too many of these details; just know that $i$ commutes with all objects in 3d, and $i^2 = -1$.
Hence we see that
$$\begin{align*}(\nabla \times A) \times A &= -i [(\nabla \times A) \wedge A] \\ &= -i [(-i \{\nabla \wedge A\}) \wedge A] \\&= i^2 (\nabla \wedge A) \cdot A\end{align*}$$
where in the last step we've used the duality of the dot and wedge: $(iX) \wedge Y = i (X \cdot Y)$.
To analyze $(\nabla \wedge A) \cdot A$, we can use an analogue of the BAC-CAB rule.
$$(\nabla \wedge A) \cdot A = \dot \nabla (\dot A \cdot A) - (A \cdot \nabla) A$$
But by symmetry, $\dot \nabla (\dot A \cdot A) = \frac{1}{2} \nabla (A \cdot A)$.  This basically completes the proof; in fact, nothing really exotic has been done except changing notation and using the BAC-CAB rule!  Certainly this was a surprise to me, or else I never would've bothered to change notation in the first place.
A: There is a nice identity for smooth vector fields: $\newcommand{\b}{\boldsymbol}$
$$
\nabla (\b{u}\cdot \b{v}) = (\b{u}\cdot \nabla) \b{v} + (\b{u}\cdot \nabla) \b{v} - (\nabla \times\b{u})\times\b{v} - (\nabla \times\b{u})\times\b{v} .\tag{1}
$$
For a component wise proof like robjohn gave, you can check Balanis's book. A more physical and geometrical proof would be using Feynman subscript notation:
$$
\nabla(\b{u} \cdot \b{v})=  \nabla_{\b{u}}(\b{u}  \cdot \b{v}) +  \nabla_{\b{v}} (\b{u} \cdot \b{v}) ,
$$
where the notation $\nabla_{\b{u}}$ means the gradient is applied only on $\b{u}$ and we keep $\b{v}$ fixed. 
Geometrically, $\nabla_{\b{u}}(\b{u}  \cdot \b{v}) $ can be decomposed into the normal derivative and the tangential derivative along $\b{v}$:
$$
 \nabla_{\b{u}}(\b{u}  \cdot \b{v}) = (\b{v}\cdot\nabla)\b{u}-  (\nabla \times \b{u})\times \b{v} ,
$$
hence (1) holds. 
Now in your case $\b{u} = \b{v} = \b{A}$:
$$
\nabla (\b{A}\cdot \b{A}) = (\b{A}\cdot \nabla) \b{A} + (\b{A}\cdot \nabla) \b{A} - (\nabla \times\b{A})\times\b{A} - (\nabla \times\b{A})\times\b{A},
$$
thus
$$
(\nabla \times\b{A})\times\b{A} = (\b{A}\cdot \nabla) \b{A} - \frac{1}{2}\nabla (\b{A}\cdot \b{A}) .
$$
A: A shorthand for @robjohn's answer would give:
\begin{align}
(\nabla \times A)\times A&=(e_{ijk}\partial_iA_j)(e_{klm}A_l \mathbf{g_m})&&\text{using the contracted epsilon identity:}\\
&=\mathbf{g_m}(\delta_{li}\delta_{mj}-\delta_{jl}\delta_{im})A_l\partial_iA_j\\
&= \mathbf{g_j}A_i\partial_iA_j - \mathbf{g_i}A_j\partial_iA_j&&\text{switching $i$ and $j$ for the first term+product rule}\\
&=\mathbf{g_i}(A_j\partial_j A_i - \frac 12\partial_i(A_jA_j))\\
&=(A\cdot\nabla)A- \frac12 \nabla(A\cdot A)
\end{align}
