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I might be too ignorant to make an actual good research on google, but I can't seem to find anything about this. I know that the dot product is well defined for complex numbers, and the same applies to the cross product. So, my assumption is that the triple product is also well defined for complex numbers. But is it really?

user15532034
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    It's perfectly well-defined but it may not have all the properties you want. – Qiaochu Yuan Jul 18 '22 at 15:27
  • Thanks for the answer. Would the formula then be the same as for a real vector? As in, assuming I'm calculating the dot product and the cross product correctly already. – user15532034 Jul 18 '22 at 15:29
  • Yes, that's what I mean by "it's perfectly well-defined," but again it may not have all the properties you want. – Qiaochu Yuan Jul 18 '22 at 15:30
  • Thanks again for the answer. As for the proprerties that's also OK, this is for a personal library project I'm writing to learn new stuff. Thank you! – user15532034 Jul 18 '22 at 15:32
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    In general, for a field $K$ take $n$ vectors in $K^n$, use them as columns of an $n \times n$ determinant. Of course it would not be called "triple product" except when $n=3$. – GEdgar Jul 18 '22 at 15:46
  • See also https://math.stackexchange.com/questions/129227/cross-product-in-complex-vector-spaces – Michael Hoppe Jul 18 '22 at 16:25

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Yes, the standard triple product formula $\vec{a} \cdot (\vec{b} \times \vec{c})$ can be applied as-is to three complex 3-vectors.

$$\vec{a} \cdot (\vec{b} \times \vec{c}) = a_1b_2c_3 - a_1b_3c_2 + a_2b_3c_1 - a_2b_1c_3 + a_3b_1c_2 - a_3b_2c_1$$

Or, breaking each complex vector component into separate real and imaginary components,

$$(a_{1x}+ia_{1y})(b_{2x}+ib_{2y})(c_{3x}+ic_{3y}) - (a_{1x}+ia_{1y})(b_{3x}+ib_{3y})(c_{2x}+ic_{2y}) + (a_{2x}+ia_{2y})(b_{3x}+ib_{3y})(c_{1x}+ic_{1y}) - (a_{2x}+ia_{2y})(b_{1x}+ib_{1y})(c_{3x}+ic_{3y}) + (a_{3x}+ia_{3y})(b_{1x}+ib_{1y})(c_{2x}+ic_{2y}) - (a_{3x}+ia_{3y})(b_{2x}+ib_{2y})(c_{1x}+ic_{1y})$$

$$= a_{1x}b_{2x}c_{3x} - a_{1y}b_{2x}c_{3x} - a_{1x}b_{2y}c_{3y} - a_{1y}b_{2x}c_{3y} - a_{1x}b_{3x}c_{2x} + a_{1y}b_{3x}c_{2x} + a_{1x}b_{3y}c_{2y} + a_{1y}b_{3x}c_{2y} + a_{2x}b_{3x}c_{1x} - a_{2y}b_{3x}c_{1x} - a_{2x}b_{3y}c_{1y} - a_{2y}b_{3x}c_{1y} - a_{2x}b_{1x}c_{3x} + a_{2y}b_{1x}c_{3x} + a_{2x}b_{1y}c_{3y} + a_{2y}b_{1x}c_{3y} + a_{3x}b_{1x}c_{2x} - a_{3y}b_{1x}c_{2x} - a_{3x}b_{1y}c_{2y} - a_{3y}b_{1x}c_{2y} - a_{3x}b_{2x}c_{1x} + a_{3y}b_{2x}c_{1x} + a_{3x}b_{2y}c_{1y} + a_{3y}b_{2x}c_{1y} + i(a_{1x}b_{2y}c_{3x} + a_{1y}b_{2x}c_{3x} + a_{1x}b_{2x}c_{3y} - a_{1y}b_{2x}c_{3y} - a_{1x}b_{3y}c_{2x} - a_{1y}b_{3x}c_{2x} - a_{1x}b_{3x}c_{2y} + a_{1y}b_{3x}c_{2y} + a_{2x}b_{3y}c_{1x} + a_{2y}b_{3x}c_{1x} + a_{2x}b_{3x}c_{1y} - a_{2y}b_{3x}c_{1y} - a_{2x}b_{1y}c_{3x} - a_{2y}b_{1x}c_{3x} - a_{2x}b_{1x}c_{3y} + a_{2y}b_{1x}c_{3y} + a_{3x}b_{1y}c_{2x} + a_{3y}b_{1x}c_{2x} + a_{3x}b_{1x}c_{2y} - a_{3y}b_{1x}c_{2y} - a_{3x}b_{2y}c_{1x} - a_{3y}b_{2x}c_{1x} - a_{3x}b_{2x}c_{1y} + a_{3y}b_{2x}c_{1y})$$

I don't know what the geometric interpretation would be, though. With $\mathbb{R}^3$ vectors, $|\vec{a} \cdot (\vec{b} \times \vec{c})|$ is the volume of the parallelepiped whose edges are defined by the three vectors. But since $\mathbb{C}$ itself is two-dimensional, $\mathbb{C}^3$ is six-dimensional.

Dan
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  • Thanks for the very detailed answer. Yeah, the geometric interpretation is something too obscure for my skill level, better leave that out for now. :-) – user15532034 Jul 18 '22 at 18:05