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I asked this on MSE over a month ago, but the one answer I got doesn't seem to work.




Let $\times$ denote the cross-product. $\;$ Is it the case that

For all unit vectors $\:\mathbf{x}\hspace{.01 in},\hspace{-0.03 in}\mathbf{y}\hspace{-0.03 in},\hspace{-0.02 in}\mathbf{z}\:$ in $\mathbf{R}^3$, $\;\;\;\; \left|\left|\hspace{.03 in}\mathbf{x} \hspace{-0.03 in}\times \hspace{-0.03 in}\mathbf{z}\hspace{.02 in}\right|\right| \:\: \leq \:\: \left|\left|\hspace{.03 in}\mathbf{x} \hspace{-0.03 in}\times \hspace{-0.03 in}\mathbf{y}\hspace{.02 in}\right|\right| \hspace{.02 in}+\hspace{.02 in}\left|\left|\hspace{.03 in}\mathbf{y} \hspace{-0.03 in}\times \hspace{-0.03 in}\mathbf{z}\hspace{.02 in}\right|\right| \;\;\;\;\;$.

?


(If yes, then $\;\;\; \langle [\mathbf{x}]\hspace{.01 in},\hspace{-0.03 in}[\hspace{.02 in}\mathbf{y}] \rangle \: \mapsto \: \left|\left|\hspace{.03 in}\mathbf{x} \hspace{-0.03 in}\times \hspace{-0.03 in}\mathbf{y}\hspace{.02 in}\right|\right| \;\;\;$ defines a nice metric on the projective plane.)

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  • $\begingroup$ What about the following: Fix $x$ and $z$ and consider the right hand side as a function of $y$; a map $S^2 \to \mathbb{R}$. If we can show that its minimum is larger than the constant on the right, we're done. It's easy to see that the gradient (which for simplicity we can consider to be a vector field using the usual metric on $S^2$) of $y \to || x \times y ||$ is zero at the poles ($\pm x$) and on the equator, and everywhere else points towards the equator. We can assume the vectors $x$ and $z$ are at an angle bigger than $0$ and less than $\pi/2$ (trivial cases)... $\endgroup$ Oct 9, 2013 at 20:43
  • $\begingroup$ ... Adding the gradient vector fields of $y \to ||x \times y||$ and $y \to ||y \times z||$ together, it is easy to see that the sum is zero at exactly four points: there are maxima where the two 'equators' meet, and minima at the points equidistant from $x$ and $z$ on the great circle containing them. The total function has the same value at each, so just look at the point in between $x$ and $z$. The result follows because $sin 2\theta < 2sin \theta$. A picture would help. $\endgroup$ Oct 9, 2013 at 20:49
  • $\begingroup$ I seem to have missed the editing window. It should read "the constant on the left". Also, the situation that $x$ and $z$ are orthogonal is not trivial, but does fit in with the rest of what I said. $\endgroup$ Oct 9, 2013 at 20:58
  • $\begingroup$ wow, the $\LaTeX$ markup of this looks crazy! was it generated by machine? $\endgroup$
    – Suvrit
    Oct 9, 2013 at 21:11
  • $\begingroup$ No, although it would be nice if there was a program to automatically put in $\hspace{1.76 in}$ better spacing than LaTeX uses on its own. $\:$ $\endgroup$
    – user5810
    Oct 9, 2013 at 21:18

1 Answer 1

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The alleged inequality is true.

Using the definition of the (vector) cross product, for unit vectors $x, y \in R^3$ the original claim boils down to showing that \begin{equation*} d(x,y) = \sin(\cos^{-1}(x^Ty)) \end{equation*} is a distance. But it turns out that $d(\cdot,\cdot)$ is actually is a distance on unit vectors in $R^n$.

Showing the above function to be a distance is equivalent to showing that \begin{equation*} \sqrt{1 - (x^Ty)^2} \end{equation*} is a distance over the set of unit vectors in $R^n$. But this is a known result of Wang and Zhang, "A trace inequality for unitary matrices", AMM, 1994, pp. 453-455.

EDIT: Also worth comparing with the recent determinant based distance described in this MO question (which suggests that actually after suitable normalization, other symmetric matrix functions should also generate distances, e.g., $d(X,Y) := [1-\text{trace}(\wedge^k(X^TY))]^{1/2}$.

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    $\begingroup$ What is your vector product on $\mathbb{R}^n$? $\endgroup$ Oct 9, 2013 at 22:02
  • $\begingroup$ Ok, fixed unfortunate wording (probably the vector product on $R^n$ will be somehow related to $x \wedge y$, but I'm not using that); the claim is true as written, and yields the $R^3$ case as a special case. $\endgroup$
    – Suvrit
    Oct 10, 2013 at 1:44
  • $\begingroup$ Should "vector product" be replaced with "cross product"? $\:$ If no, what do you mean by that? $\hspace{.78 in}$ $\endgroup$
    – user5810
    Oct 10, 2013 at 2:01
  • $\begingroup$ @RickyDemer: Yes, by "vector product" I mean the "cross product" in your original question. Am editing the answer once more to avoid confusion. $\endgroup$
    – Suvrit
    Oct 10, 2013 at 2:20
  • $\begingroup$ d(.,.) is not a distance since d(x,y)=0 does not imply that x=y (think about the case x=-y). It is a distance only over the open hemisphere. $\endgroup$
    – bersou
    Dec 10, 2019 at 17:04

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