Gradient of cross product Consider $\mathbb{R}^3 \times \mathbb{R}^3$ with standard coordinates $(q_1, q_2, q_3, p_1, p_2, p_3)$. For a fixed $v \in \mathbb{R}^3$, consider the function $f : \mathbb{R}^3 \times \mathbb{R}^3 \to \mathbb{R}$ given by $$f(q, p) = \langle v, q \times p \rangle$$ Writing everything out, it's easy to show that $\nabla f = (- v \times p, v \times q)$. Is there an easier way to see this, that doesn't involve writing out the coordinate-wise formulas for cross product and inner product?
 A: Denote the triple vector product, or volume form, by $\epsilon(\cdot,\cdot,\cdot)$. Then
$$f(q,p):=\langle v,q\times p\rangle=\epsilon(v,q,p)=\left\{\eqalign{ &\langle q,p\times v\rangle\>,\cr &\langle p,v\times q\rangle\>. \cr}\right.$$
It now follows by inspection that
$$\nabla_q f=-v\times p\>,\qquad \nabla_p f=v\times q\ .$$
A: We have
$$
f(q, p) = \det(v, q, p) = \epsilon_{ijk} v_i q_j p_k
$$
so
$$
\partial_{q_s} f 
= \epsilon_{ijk} v_i \delta_{js} p_k 
= \epsilon_{isk} v_i p_k
= \epsilon_{ski} v_i p_k
= -\epsilon_{sik} v_i p_k
= -(v\times p)_s \\
\partial_{p_t} f
= \epsilon_{ijk} v_i q_j \delta_{kt}
= \epsilon_{ijt} v_i q_j
= \epsilon_{tij} v_i q_j
= (v\times q)_t
$$
A: Indeed, one can prove this without spelling everything out. We can prove for a bilinear form $ b: \mathbb{R}^n \times \mathbb{R}^m \rightarrow \mathbb{R}$ that the differential is given by
$$ Db(p,q)[u, w] = b(p, w) + b(u, q) .$$
Indeed
$$\vert  b(p+u, q+w) - b(p, q)- (b(p, w)+b(u, q)\vert
=\vert b(u, w) \vert \leq C \Vert u \Vert \cdot \Vert w \Vert = o(\Vert (u, w) \Vert)$$
and thus the differential is given by the above formula. Thus in our case we get
$$ Df(p, q)[u, w] = \langle v, p \times w\rangle + \langle v, u \times q \rangle.$$
Using the fact that that $\langle v_1, v_2 \times v_3 \rangle $doesn't change, when we permute the vectors cyclic we get the claim. 
A: This can be done succinctly using the following two facts. First, note that if $a,b$ are functions in $\mathbb{R}^3$, then
$$
\nabla (a^T b) = J[a^T]b + a^T J[b],
$$
where $J[f]$ is the Jacobian of a function $f$.
Second, one can rewrite a cross product as
$$ 
a \times b = a^\times b
$$
where 
$$
a^\times= \begin{bmatrix}
0 & -a_3 & a_2 \\
a_3 & 0 & -a_1 \\
-a_2 & a_1 & 0 
\end{bmatrix}.
$$
Then
$$
J[a\times b] = J[a^\times b] = a^\times J[b] - b^\times J[a].
$$
Using these two facts, we can write
\begin{equation}
\begin{aligned}
< v, q \times p > &= J[v^T] \ ( q \times p) + v^T \ J[q \times p] \\
&= v^T \ ( q^\times J[p] - p^\times J[q] ) \\
& = v^T \begin{bmatrix} -p^\times & q^\times \end{bmatrix} \\
& = (- v\times p , v \times q).
\end{aligned}
\end{equation}
The first two equalities use the relations above.
The third equality uses the fact that
$$
J[q] = \begin{bmatrix} I_3 & 0_3
\end{bmatrix}
$$
where $I_3$ is the identity and $0_3$ is the zero matrix.
The last equality follows as $(q^\times)^T = - q^\times$ so that
$$
v^T p^\times = (-p^\times v )^T = -p \times v = v \times p.
$$
