Meaning of expression similar to cross product Given are 2 vectors $\vec{x},\vec{y}$ in $\mathbb{R}^4$. I have a function  $f(\vec{x},\vec{y})$ that looks quite symmetrically. Therefore I think that there must be a meaning or any algebraic rule. What is the meaning of $f$?
$$f(\vec{x},\vec{y})=(x_1y_2-x_2y_1)^2
+(x_2y_3-x_3y_2)^2
+(x_3y_4-x_4y_3)^2
+(x_4y_1-x_1y_4)^2
+(x_1y_3-x_3y_1)^2
+(x_2y_4-x_4y_2)^2$$
The first 4 summands look like the squared norm of a cross product. Of course such a cross product has not properties as in $\mathbb{R}^3$. By cross product I just mean the application of the construction rule of a cross product in  $\mathbb{R}^3$ that is analogically extended to  $\mathbb{R}^4$.
Another interpretation would be the squared norm of a (so far unknown) vector product $\mathbb{R}^4\times\mathbb{R}^4\to\mathbb{R}^6$.

Edit:
For simplicity above was given  only a part but the general form looks like
$$f(\vec{x},\vec{y},\vec{z})=\\
(x_1y_2-x_2y_1)^2
+(x_2y_3-x_3y_2)^2
+(x_3y_4-x_4y_3)^2
+(x_4y_1-x_1y_4)^2
+(x_1y_3-x_3y_1)^2
+(x_2y_4-x_4y_2)^2+\\
(x_1z_2-x_2z_1)^2
+(x_2z_3-x_3z_2)^2
+(x_3z_4-x_4z_3)^2
+(x_4z_1-x_1z_4)^2
+(x_1z_3-x_3z_1)^2
+(x_2z_4-x_4z_2)^2+\\
(y_1z_2-y_2z_1)^2
+(y_2z_3-y_3z_2)^2
+(y_3z_4-y_4z_3)^2
+(y_4z_1-y_1z_4)^2
+(y_1z_3-y_3z_1)^2
+(y_2z_4-y_4z_2)^2
$$
 A: As Ted says in the comments, $f(x, y)$ is the squared norm of the wedge product or exterior product
$$\mathbb{R}^4 \times \mathbb{R}^4 \ni v, w \mapsto v \wedge w \in \wedge^2(\mathbb{R}^4) \cong \mathbb{R}^6$$
which is given explicitly in coordinates by exactly the maps $x_i y_j - x_j y_i$ that you are squaring to compute $f$. This follows from writing $v = \sum x_i e_i, w = \sum y_j e_j$ where $e_i$ is the standard basis and computing that
$$v \wedge w = \sum_{i, j} x_i y_j (e_i \wedge e_j) = \sum_{i < j} (x_i y_j - x_j y_i)(e_i \wedge e_j)$$
since $e_i \wedge e_i = 0$ and $e_j \wedge e_i = - e_i \wedge e_j$. Next we invoke the existence of an inner product on $\wedge^2(V)$ for $V$ an inner product space with the property that if $\{ e_i \}$ is an orthonormal basis for $V$ then $\{ e_i \wedge e_j \}, i < j$ is an orthonormal basis for $\wedge^2(V)$; this inner product can be constructed in a few different ways, for example as a determinant
$$\langle v_1 \wedge w_1, v_2 \wedge w_2 \rangle = \det \left[ \begin{array}{cc} \langle v_1, v_2 \rangle & \langle v_1, w_2 \rangle \\ \langle w_1, v_2 \rangle & \langle w_1, w_2 \rangle \end{array} \right].$$
The definition of the wedge product in coordinates, while concrete, obscures many important features of the wedge product, such as its associativity and how it behaves under change of coordinates.
$f(x, y, z)$ is, also as Ted says, $\| x \wedge y \|^2 + \| y \wedge z \|^2 + \| z \wedge x \|^2$. It can be described more conceptually by reinterpreting $x, y, z \in \mathbb{R}^4$ as the columns of a $4 \times 3$ matrix $M$, thought of as a linear transformation $M : \mathbb{R}^3 \to \mathbb{R}^4$ such that $M(a, b, c) = ax + by + cz$. Then $M$ is the squared Frobenius norm of the wedge product
$$\wedge^2(M) : \wedge^2(\mathbb{R}^3) \to \wedge^2(\mathbb{R}^4).$$
$\wedge^2(M)$ is given explicitly in coordinates by a $6 \times 3$ matrix whose entries are the $2 \times 2$ minors of $M$, which again are exactly the maps that you are squaring to compute $f$. And again this definition, while concrete, obscures many important features of the wedge product, such as its multiplicativity: we have $\wedge^2(MN) = \wedge^2(M) \wedge^2(N)$ but working in coordinates makes this very far from obvious.
A: There is a simple meaning that can be understood without knowing anything about abstract $\wedge$ notation or exterior algebra: $$f(\vec{x},\vec{y})=4 A^2,$$ where $A$ is the area of the triangle in $\mathbb{R}^4$  that is spanned by vectors $\vec{x},\vec{y}$.
Proof
The area of a triangle in any dimension is given by
$$A=\frac{1}{2}\|\vec{x}\|\|\vec{y}\|\sin\alpha$$
where $\|\vec{x}\|,\|\vec{y}\|$ are 2 side lengths that enclose angle $\alpha$.
Using dot product $\cdot$ and relations $$\sin^2\alpha+\cos^2\alpha=1$$ $$\cos\alpha=\frac{\vec{x}\cdot\vec{y}}{\|\vec{x}\|\|\vec{y}\|}$$
we get
$$4A^2=\|\vec{x}\|^2\|\vec{y}\|^2-(\vec{x}\cdot\vec{y})^2\\=(\vec{x}\cdot\vec{x})(
\vec{y}\cdot\vec{y})-(\vec{x}\cdot\vec{y})^2\\
=(x_1^2+x_2^2+x_3^2+x_4^2)(y_1^2+y_2^2+y_3^2+y_4^2)-(x_1y_1+x_2y_2+x_3y_3+x_4y_4)^2\\
=x_1^2y_2^2-2x_1x_2y_1y_2+x_2^2y_1^2\\
+x_2^2y_3^2-2x_2x_3y_2y_3+x_3^2y_2^2\\
+x_3^2y_4^2-2x_3x_4y_3y_4+x_4^2y_3^2\\
+x_4^2y_1^2-2x_1x_4y_1y_4+x_1^2y_4^2\\
+ x_1^2y_3^2-2x_1x_3y_1y_3+ x_3^2y_1^2 \\
+x_2^2y_4^2-2x_2x_4y_2y_4+x_4^2y_2^2\\
=(x_1y_2-x_2y_1)^2
+(x_2y_3-x_3y_2)^2
+(x_3y_4-x_4y_3)^2
+(x_4y_1-x_1y_4)^2
+(x_1y_3-x_3y_1)^2
+(x_2y_4-x_4y_2)^2\\
=f(\vec{x},\vec{y})$$
