# Geometric meaning of the quantity $|a|^2 |b|^2 |c|^2 - (a \cdot b)(b \cdot c)(c \cdot a)$ for non-coplanar vectors $a$, $b$, $c$

For two non-collinear vectors, $$a$$ and $$b$$, the quantity $$|a|^2 |b|^2 - (a \cdot b)(b \cdot a) = |a \times b|^2$$ is the square of the area of the parallelogram spanned by these two vectors. For three non-coplanar vectors, $$a$$, $$b$$ and $$c$$, we can form a similar expression $$B = |a|^2 |b|^2 |c|^2 - (a \cdot b) (b \cdot c) (c \cdot a)$$ which is not equal to the square of the volume of the parallelepiped spanned by these three vectors.

Does $$B$$ have any geometrical (or other) meaning?

• So when you write $a^2$, do you mean $a\cdot a$? – Arnaud D. Sep 24 '18 at 14:52
• @ArnaudD. Yes, exactly! – Fizikus Sep 24 '18 at 14:53
• I took the liberty of replacing $x^2$ with $|x|^2$ to avoid further confusion about notation. – Blue Sep 24 '18 at 15:11
• Is $B$ a definition or is the right hand side of $B$ derived from some other starting point? – DWade64 Sep 24 '18 at 22:52
• @DWade64: $B$ is just the definition. – Fizikus Sep 24 '18 at 23:26

I don't know if this hint is conclusive, but this is what I got:

\begin{align*} B & = \lVert a\rVert^{2}\lVert b\rVert^{2}\lVert c\rVert^{2} - \langle a,b\rangle\langle b,c\rangle\langle c,a\rangle\\ & = \lVert a\rVert^{2}\lVert b\rVert^{2}\lVert c\rVert^{2} - \lVert a\rVert^{2}\lVert b\rVert^{2}\lVert c\rVert^{2}\cos(\theta_{1})\cos(\theta_{2})\cos(\theta_{3})\\ & = \lVert a\rVert^{2}\lVert b\rVert^{2}\lVert c\rVert^{2}[1 - \cos(\theta_{1})\cos(\theta_{2})\cos(\theta_{3})] \end{align*}

• We can see that this quantity vanishes iff $a, b, c$ are parallel, so as in the case with two vectors, $B$ is a measure of failure of the three vectors to be parallel. – Travis Willse Sep 24 '18 at 15:18

I don't know a geometric interpretation for this quantity, but there's a natural interpretation of this quantity that makes it more transparently analogous to the other candidate you mentioned, namely the square of volume $$V$$ of the parallelepiped spanned by $${\bf a}, {\bf b}, {\bf c}$$, at least for $$n = 3$$.

Recall that that squared area can be written as \begin{align*}V^2 &= \operatorname{det}\pmatrix{{\bf a} \cdot {\bf a} & {\bf a} \cdot {\bf b} & {\bf a} \cdot {\bf c} \\ {\bf b} \cdot {\bf a} & {\bf b} \cdot {\bf b} & {\bf b} \cdot {\bf c} \\ {\bf c} \cdot {\bf a} & {\bf c} \cdot {\bf b} & {\bf c} \cdot {\bf c}} \\ &= ({\bf a} \cdot {\bf a})({\bf b} \cdot {\bf b})({\bf c} \cdot {\bf c}) - \mathfrak{S}[({\bf a} \cdot {\bf a})({\bf b} \cdot {\bf c})^2] + 2 ({\bf b} \cdot {\bf c}) ({\bf c} \cdot {\bf a})({\bf a} \cdot {\bf b}),\end{align*} where $$\mathfrak{S}[\cdot]$$ denotes the cyclic sum of $$\cdot$$ in $${\bf a}, {\bf b}, {\bf c}$$.

On the other hand, recall that the determinant of an $$n \times n$$ matrix $$M = (m_{ij})$$ can be written as $$\det M = \sum_{\sigma} (\operatorname{sign} \sigma) m_{1 \sigma(1)} \cdots m_{n \sigma(n)},$$ where the sum is over all permutations $$\sigma$$ of $$n$$. Here, $$\operatorname{sign} \sigma$$ is $$+1$$ if the permutation is even and $$-1$$ if it is odd.

Now, if you know a little representation theory, you know $$\operatorname{sign}$$ is a representation of the group $$S_n$$ of permutations of $$n$$ elements. By replacing $$\operatorname{sign}$$ in the definition of the determinant with some other representation we get a generalization of the determinant (called an immanant) and---perhaps you see where this is going---can replace $$\det$$ in the first display equation above to get a new invariant for the (unordered) triple $$({\bf a}, {\bf b}, {\bf c})$$.

Taking $$n = 3$$ and the two-dimensional representation $$\lambda$$ gives the immanant $$\operatorname{imm}_{\lambda}(M) = 2 m_{11} m_{12} m_{13} - m_{23} m_{31} m_{12} - m_{32} m_{13} m_{21} .$$ Replacing $$\det$$ in the first display equation with $$\operatorname{imm}_{\lambda}$$ gives exactly twice the quantity in question: $$\boxed{\tfrac{1}{2} \operatorname{imm}_{\lambda} \pmatrix{{\bf a} \cdot {\bf a} & {\bf a} \cdot {\bf b} & {\bf a} \cdot {\bf c} \\ {\bf b} \cdot {\bf a} & {\bf b} \cdot {\bf b} & {\bf b} \cdot {\bf c} \\ {\bf c} \cdot {\bf a} & {\bf c} \cdot {\bf b} & {\bf c} \cdot {\bf c}} = ({\bf a} \cdot {\bf a})({\bf b} \cdot {\bf b})({\bf c} \cdot {\bf c}) - ({\bf b} \cdot {\bf c})({\bf c} \cdot {\bf a})({\bf a} \cdot {\bf b})} .$$

Remark Up to isomorphism there are only three irreducible representations of $$S_3$$, and the one remaining is the trivial representation. This gives rise to the permanent, $$\operatorname{per}$$, which is defined just by removing from the definition of $$\det$$ the sign of the permutation, so $$\begin{multline*}\operatorname{per}\pmatrix{{\bf a} \cdot {\bf a} & {\bf a} \cdot {\bf b} & {\bf a} \cdot {\bf c} \\ {\bf b} \cdot {\bf a} & {\bf b} \cdot {\bf b} & {\bf b} \cdot {\bf c} \\ {\bf c} \cdot {\bf a} & {\bf c} \cdot {\bf b} & {\bf c} \cdot {\bf c}} \\ = ({\bf a} \cdot {\bf a})({\bf b} \cdot {\bf b})({\bf c} \cdot {\bf c}) + \mathfrak{S}[({\bf a} \cdot {\bf a})({\bf b} \cdot {\bf c})^2] + 2 ({\bf b} \cdot {\bf c}) ({\bf c} \cdot {\bf a})({\bf a} \cdot {\bf b}),\end{multline*}$$ where $$\mathfrak{S}[\cdot]$$ denotes the cyclic sum of $$\cdot$$ in $${\bf a}, {\bf b}, {\bf c}$$. It would be interesting to have a geometric interpretation of this quantity, too.

• Very nice, I wasn't aware of these generalizations of the determinant. Do they have some practical applications? – Fizikus Sep 24 '18 at 20:06
• See this thread on MO: mathoverflow.net/questions/66284/… . This post on Terry Tao's blog about the permanent in a particular case is an interesting read, too: terrytao.wordpress.com/2008/04/16/… I don't know offhand of any applications for the Immanent – Travis Willse Sep 24 '18 at 23:27

It occurred to me that if one seeks a geometrical interpretation of $$B$$, then one has to write it in terms of other geometrical quantities, i.e., length squared $$|a|^2$$, area squared $$|a \times b|^2$$ and volume squared $$|a \cdot (b \times c)|^2$$. Doing this, we find $$B = \frac{1}{2} \left[ |a|^2 |b \times c|^2 + |b|^2 |c \times a|^2 + |c|^2 |a \times b|^2 - |a \cdot (b \times c)|^2\right].$$ Now, $$B$$ vanishes only if all three vectors are collinear so it cannot represent any geometrical volume. On the other hand, the dimensionality of $$B$$ eliminates the possibility of describing some area. Therefore, I conclude that $$B$$ does not represent any simple geometrical quantity.