Cross product-like quantity for vectors in $\mathbb{R}^n$

Question

I have the quantity: \begin{equation} I= \sum_{i>j=1}^n \left( a_i b_j -a_j b_i \right)^2, \end{equation} where $a_i, b_i$ are real numbers.

In the case of $n=3$, I can interpret this as the magnitude of the vector resulting from the cross product of the two vectors in $\mathbb{R}^3$: $A=(a_1,a_2,a_3)$ and $B=(b_1,b_2,b_3)$, which geometrically, is related to the area of the parallelogram that these two vectors span.

My question is the following: when $n>3$, is there some geometric interpretation of what does this quantity define (if any) in $\mathbb{R}^n$, and more specifically for the vectors defined by $A=(a_1,\ldots,a_n)$ and $B=(b_1,\ldots,b_n)$ in analogy with the $n=3$ case?

Thanks in advance.

Answer 1

From https://en.wikipedia.org/wiki/Cross_product#Lagrange.27s_identity:

$$ \sum_{1 \le i < j \le n} \left(a_ib_j-a_jb_i \right)^2 = \left\| \mathbf a \right\| ^2 \left\| \mathbf b \right\| ^2 - (\mathbf {a \cdot b } )^2 $$

In the $n = 3$ case this matches $\|a \times b\|^2$ as noted on the same page, and so does for $n = 7$ per https://en.wikipedia.org/wiki/Lagrange%27s_identity#Seven_dimensions.

Answer 2

$$\sum_{i < j} \left(a_ib_j-a_jb_i \right)^2 = \|\mathbf a\wedge \mathbf b\|^2$$

which is just the square of the area of the parallelogram with sides $\mathbf a$ and $\mathbf b$. That's pretty geometric, I think. ;)

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If $\{\mathbf e_1, \cdots, \mathbf e_n\}$ is an orthonormal basis of $V$, then it is easily shown from the properties of the wedge product that $\mathbf a\wedge \mathbf b \in \Lambda V$ is given by $$\mathbf a\wedge \mathbf b = \sum_{i \lt j} (a_ib_j-a_jb_i)\mathbf e_i \wedge \mathbf e_j$$

Then we define the norm of this by $$\|\mathbf a \wedge \mathbf b\|^2 = \sum_{i < j} \left(a_ib_j-a_jb_i \right)^2$$