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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 ?

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Maybe it would be natural to generalize the cross product after viewing it in a sufficiently abstract setting, for example as the Hodge dual of the wedge product? I don't know enough to say whether this would apply directly to $\mathbb C^3$ though. –  Rahul Apr 8 '12 at 7:33
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up vote 2 down vote accepted

For finding the correct definition to apply, one needs to know whether the scalar product is taken to be anti-linear in its first or its second argument. Assuming the first convention, the relation one would want to preserve for $\vec x=(x_1,x_2,x_3)$ and similarly for $\vec y, \vec z$ is either that one still has $$ (\vec x \times \vec y)\cdot\vec z= \left|\begin{matrix}x_1&y_1&z_1\\x_2&y_2&z_2\\x_3&y_3&z_3\\\end{matrix} \right|. $$ Note that the determinant is linear in all of its columns, so the left hand side needs to be an expression that is linear in the vector that appears directly as a column, which explains that one cannot use $\vec x\cdot(\vec y\times\vec z)$ instead, which is anti-linear in $\vec x$. Now it is easy to see that the coordinates of $\vec x \times \vec y$ should be taken to be the complex conjugates of the expressions in their usual defintion, for instance $\overline{x_2y_3-x_3y_2}$ for the first coordinate.

One actually arrives at the same conclusion for a scalar product that is defined to be anti-linear in its second argument. However the identity that leads to this definition is different, namely the one which equates $\vec x\cdot(\vec y\times\vec z)$ to the above determinant.

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