# Derivative of cross product

Let $$f:V_1\times\dots\times V_N \to W$$ be multilinear. Then $$f$$ s differentiable and $$df(a_1,\dots,a_n)(h_1+\dots+h_n)=f(h_1,a_2,\dots,a_n)+f(a_1,h_2,a_3,\dots,a_n)+f(a_1,\dots,a_{n-1},h_n)\tag{1}$$

Problem: Calculate the derivative of the cross product $$\mathbb R^3\times \mathbb R^3 \to \mathbb R^3$$.

Solution: Let $$f$$ be the cross product. It's easy to show that $$f$$ is multilinear by a direct calculation thus we can use (1). We calculate the derivative at $$(u,v)=((u_1,u_2,u_3),(v_1,v_2,v_3))$$. We then have $$df(u,v)(h_1+h_2)=h_1\times v + u\times h_2\tag{2}$$ Let $$e_i, i=1,2,3$$ be the unit vectors in $$\mathbb R^3$$. We calculate $$e_1\times v=\begin{pmatrix}0\\ -v_3 \\ v_2\end{pmatrix}$$ $$e_2\times v=\begin{pmatrix}v_3\\ 0 \\ -v_1\end{pmatrix}$$ $$e_3\times v=\begin{pmatrix}-v_2 \\ v_1 \\ 0\end{pmatrix}$$ For $$u$$ we use the identity $$u\times e_i=-e_i\times u$$. So we get the following derivative matrix: $$df(u,v)=\begin{pmatrix}0&v_3&-v_2&0&-u_3&u_2 \\ -v_3 & 0 & v_1 & u_3 & 0 & -u_1 \\ v_2 & -v_1 & 0 & -u_2 & u_1 & 0\end{pmatrix}$$

Question: 1. Apprently, the $$+$$ means writing the result of $$f$$ as a column, so it's not the $$+$$ from e.g. $$\mathbb R^3$$. Should I see that from the definition above without the example? How should I know the dimension of the imge of $$df$$ without them specifying it somehow?

1. What $$+$$ is used in $$f(a_i)(h_1+h_2)$$? How does $$h_1$$ and $$h_2$$ and $$h_1+h_2$$ look like?
• If $f$ is $p$-linear then $f(x_1 + h_1, \ldots, x_p + h_p) = f(x_1, \ldots, x_p) + \sum_i f(x_1, \ldots, h_i, \ldots, x_p) + E(h_1, \ldots, h_p),$ where $E$ is the sum of all terms were $f$ is evaluated at least in two different $h_i.$ It follows $E(h_1, \ldots, h_p) = o(\|h\|)$ and then, $f'(x_1, \ldots, x_p) = \sum_i f(x_1, \ldots, h_i, \ldots, x_p).$ This proof does not require any sort of finite dimension. Q.E.D. – Will M. Jan 27 '19 at 1:17

For a multilinear mapping, it suffices to consider its Frechet derivative. Let $$W$$ be an $$n$$-D vector space, and each $$V_i$$ be an $$m_i$$-D vector space with $$i=1,2,...,N$$. Let $$f:V_1\times V_2\times\cdots\times V_N\to W$$ be multilinear. Then $$\forall\left(v_1,v_2,...,v_N\right)\in V_1\times V_2\times\cdots\times V_N$$, the Frechet derivative of $$f$$ at this location, denoted by $$({\rm d}f)(v_1,v_2,...,v_N)$$, is also a multilinear mapping, i.e., $$({\rm d}f)(v_1,v_2,...,v_N):V_1\times V_2\times\cdots\times V_N\to W.$$ According to Frechet, it follows that \begin{align} &({\rm d}f)(v_1,v_2,...,v_N)(h_1,h_2,...,h_N)\\ &=f(h_1,a_2,...,a_N)\\ &+f(a_1,h_2,...,a_N)\\ &+\cdots\\ &+f(a_1,a_2,...,h_N). \end{align}
Recall that, if $$g$$ is linear, its entry-wise form reads $$g_i(v)=\sum_ja_{ij}v_j,$$ and if $$g$$ is bilinear, its entry-wise form reads $$g_i(v_1,v_2)=\sum_{j_1,j_2}a_{ij_1j_2}v_{1j_1}v_{2j_2}.$$ Inductively and formally, the above multilinear $$f$$ observes the following entry-wise form $$f_i(v_1,v_2,...,v_N)=\sum_{j_1=1}^{m_1}\sum_{j_2=1}^{m_2}\cdots\sum_{j_N=1}^{m_N}a_{ij_1j_2...j_N}v_{1j_1}v_{2j_2}...v_{Nj_N}$$ for $$i=1,2,...,m$$, where each $$v_{kj_k}$$ denotes the $$j_k$$-th entry of $$v_k\in V_k$$, while $$a_{ij_1j_2...j_N}$$'s are the coefficients of $$f$$.
Thanks to this entry-wise form, we may then write down the entry-wise form of $${\rm d}f$$ as well, which reads \begin{align} &({\rm d}f)_i(v_1,v_2,...,v_N)(h_1,h_2,...,h_N)\\ &=\sum_{j_1=1}^{m_1}\sum_{j_2=1}^{m_2}\cdots\sum_{j_N=1}^{m_N}a_{ij_1j_2...j_N}h_{1j_1}v_{2j_2}...v_{Nj_N}\\ &+\sum_{j_1=1}^{m_1}\sum_{j_2=1}^{m_2}\cdots\sum_{j_N=1}^{m_N}a_{ij_1j_2...j_N}v_{1j_1}h_{2j_2}...v_{Nj_N}\\ &+\cdots\\ &+\sum_{j_1=1}^{m_1}\sum_{j_2=1}^{m_2}\cdots\sum_{j_N=1}^{m_N}a_{ij_1j_2...j_N}v_{1j_1}v_{2j_2}...h_{Nj_N}. \end{align} In other words, as $$a_{ij_1j_2...j_N}$$'s are known, the entry-wise form of $${\rm d}f$$ could be expressed straightforwardly as above.
Finally, the "$$+$$" in OP's original post, i.e., $$(h_1+h_2+\cdots+h_N)$$, is a convention in some context, which is exactly $$(h_1,h_2,...,h_N)$$ here. When there is free of ambiguity, both expressions can be used as per ones preference.