Complex tensor product Let $x, y, z$ be three vectors in $\mathbb{R}^m$, $\mathbb{R}^n$, $\mathbb{R}^p$, respectively. From them we can form the tensor product $x \otimes y \otimes z$, which can be interpreted as a trilinear map $\mathbb{R}^m \times \mathbb{R}^n \times \mathbb{R}^p \mapsto \mathbb{R}$ defined as
$$x \otimes y \otimes z (u, v, w) = \langle u, x \rangle \cdot \langle v, y \rangle \cdot \langle w, z \rangle.$$
$\langle u, x \rangle = \sum_{i=1}^m u_i x_i$ is the usual Euclidean inner product, with analogous definition for $\langle v, y \rangle$ and $\langle w, z \rangle$. 
I start to get confused when the field is not $\mathbb{R}$ but $\mathbb{C}$ instead. Consider the same vectors as before, but with complex numbers. Let $\langle u, x \rangle = \sum_{i=1}^m u_i \overline{x}_i$ be the Hermitian inner product and $\langle u, x \rangle_\mathbb{R} = \sum_{i=1}^m u_i x_i$. Now, what is the correct interpretation of $x \otimes y \otimes z$ as a trilinear product?
1) $x \otimes y \otimes z (u, v, w) = \langle u, x \rangle \cdot \langle v, y \rangle \cdot \langle w, z \rangle$
2) $x \otimes y \otimes z (u, v, w) = \langle x, u \rangle \cdot \langle y, v \rangle \cdot \langle z, w \rangle$
3) $x \otimes y \otimes z (u, v, w) = \langle u, x \rangle_\mathbb{R} \cdot \langle v, y \rangle_\mathbb{R} \cdot \langle w, z \rangle_\mathbb{R}$
Each choice seems to be reasonable but I don't want to guess and go for it. If possible, could you explain the reasoning behind the correct choice?
Thank you.
 A: That depends on what the correct interpretation of a vector is as a linear form. The ambiguity stems from the fact that you start off with vectors (contravariant tensors) and then switch over to linear forms (covariant tensors). 
It seems natural that the tensor obey:
$x \otimes y \otimes z (u, v, w) = x(u) \cdot  y (v)   \cdot z (w)$
And there is no ambiguity  if $x$, $y$ and $z$ are linear forms. 
$3)$ has the advantage of being fully linear: $\forall a,b, \in \mathbb{C}$, $(aX) (bY)= (ab)X(y)$ . Both $1$ and $2$ pass conjugation to one of their arguments.This leads to either  $2) (aX) (bY)= (a \overline{b})X(Y)$  or  $1)(aX) (bY)= (\overline{a} b)X(Y)$ and the main advantage is that $X(X)$ is always real. A "trilinear" form should be fully linear in its arguments, so that kind of excludes $2$ (which has the "sesquilinear" property that $X(bY) =  \overline{b} * X(Y)$). The choice would then be between $1)$ and $3)$. I would pick $3$, as it's "most linear", but there really is no correct answer. It all depends on how you chose to define the transformation between $X$ as a vector and $X$ as a linear form.      
