Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 ?

share|improve this question
2  
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

1 Answer 1

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 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.

share|improve this answer
    
The meaning of triple product (x × y)⋅ z of Euclidean 3-vectors is the volume form (SL(3, ℝ) invariant), that gets an expression through dot product (O(3) invariant) and cross product (SO(3) invariant, a subgroup of SL(3, ℝ)). We can complexify all the stuff (resulting in SO(3, ℂ)-invariant vector calculus), although we will not obtain an inner product space. But if we generalize “⋅” as a sesquilinear form, then its underlying symmetry becomes U(3), whereas proposed generalization of the triple product is still ruled by SL(3, ℂ). This seemingly leads to an SU(3) vector calculus. –  Incnis Mrsi Nov 1 at 18:33

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.