TITLE: Cross product in complex vector spaces
QUESTION [7 upvotes]: When inner product is defined in complex vector space, conjugation is performed on one of the vectors. What about is the cross product of two complex 3D vectors?
I suppose that one possible generalization is 
$A\otimes B  \rightarrow  \left ( A\times B \right )^*$  where $\times$ denotes the normal cross product. The conjugation here is to ensure that the result of the cross product is orthogonal to both vectors $A$ and $B$. Is that correct ?

REPLY [4 votes]: Physicists tend to use the (nonconjugated) "real formulae" for both dot product and cross products. So physicists' dot product equals $\vec x\cdot \vec y = \vec x^T \vec y$, although their inner product is $\langle\vec x,\vec y\rangle=\vec x^H \vec y$ (or $\left\langle \vec x|\vec y \right\rangle$), where $\vec x^H = \mathrm{conj}(\vec x)^T$ (the transpose with complex-conjugated elements).
Usually physicists write $\vec x^\dagger$ instead of $\vec x^H$, whereas mathematicians may write $\vec x^*$ for $\vec x^H$ (alternatively, for the conjugate of $\vec x$, which is why I use here the Hermitian notation $\vec x^H$, to avoid ambiguity).
Non-conjugated dot and cross products are the standard practice in, e.g., almost all books on electromagnetism. Note also that physicists, modern mathematicians and almost everyone else tend to conjugate the first element in their inner products, as above, except that old-school mathematicians more often the second element, according to Wikipedia, Note 2: https://en.wikipedia.org/wiki/Inner_product_space#Notes
However, the following Wikipedia page claims that linearity in the first coordinate is [still] the prevailing condition in mathematics (not merely old-school), and uses $\left\langle \cdot ,\cdot \right\rangle$ for that and $\left\langle \cdot |\cdot \right\rangle$ for linearity in the second argument: https://en.wikipedia.org/wiki/Riesz_representation_theorem#Mathematics_vs._physics_notations_and_definitions_of_inner_product
In either case, physicists get the following formula:
$$
  (\vec x \times \vec y)\cdot \vec z=
  \left|\begin{matrix}x_1&y_1&z_1\\x_2&y_2&z_2\\x_3&y_3&z_3\\\end{matrix} 
  \right|.
$$
Mathematicians tend to have dot product equal to inner product, i.e., $\vec x\cdot \vec y = \langle\vec x,\vec y\rangle=\vec x^H \vec y$ for modern mathematicians, and $=\vec y^H \vec x$ for old-school mathematicians. (Note the old-school jumping order that makes their dot products linear with respect to their first arguments and conjugate-linear w.r.t. the second ones.)
Therefore, mathematicians tend to define cross$(x,y)=$ conj(physicscross$(x,y))$, i.e.,
$$ \vec x \times \vec y = \left[\begin{matrix}\overline{x_2y_3-x_3y_2}\\ 
 \overline{x_3y_1 - x_1y_3}\\
 \overline{x_1y_2-x_2y_1}\\
\end{matrix} 
  \right]
$$
(the overline symbols denote complex conjugation; remove them for physicists' cross product)
to preserve the formula (if conjugating the 1st argument of dot products)
$$
  (\vec x \times \vec y)\cdot\vec z=
  \left|\begin{matrix}x_1&y_1&z_1\\x_2&y_2&z_2\\x_3&y_3&z_3\\\end{matrix} 
  \right|.
$$
Old-school mathematicians have, instead, $\vec x \cdot (\vec y\times\vec z)=|\cdots|$,
as their dot product (conjugating the 2nd argument) conjugates the latter element, thus canceling the conjugation in the cross product this way.
The physicists' (non-conjugated) formulae can be found, e.g., here (gated): https://www.tandfonline.com/doi/pdf/10.1080/002073999287815
Matlab, Maple and Mathematica
Matlab uses modern mathematicians' formulae for the dot product, i.e., it conjugates the first element: dot$(x,y)=x^Hy$ (link https://se.mathworks.com/help/matlab/ref/dot.html); but it uses the physicists' (non-conjugated) formulae for the cross product: cross$(x,y)$ doesn't involve the conjugation of the output.
Thus, in Matlab, dot(i,1) = -i,
cross([0,i,0],[0,0,1]) = (+i, 0, 0).
Therefore,  $(\vec x \times \vec y)\cdot\vec z \ne |\cdots|$ in Matlab.
Mathematica conjugates neither. Maple conjugates the dot product (though you have the option to nonconjugate it) but not the cross product.
https://www.maplesoft.com/support/help/Maple/view.aspx?path=LinearAlgebra/DotProduct
https://www.maplesoft.com/support/help/Maple/view.aspx?path=LinearAlgebra/CrossProduct
My guess is that Scilab does not conjugate. Somebody hopefully certifies/corrects this paragraph.
This answer builds on the shoulders of the answer by Marc van Leeuwen on this same page.