why preserving norm is equivalent to preserving inner product in rigid body transformation

Question

Define rigid body motion as following transformation $g$ $$g:\mathbb R^3 \to \mathbb R^3$$ such that $$|g(v)|=|v|,\forall v \in \mathbb R^3$$ $$g(u) \times g(v) = g(u\times v)$$ according to polarization identity: $\langle u,v\rangle=\frac{1}{4}(|u+v|^2-|u-v|^2)$. Inner product preservation(i.e. $\langle u,v\rangle=\langle g(u),g(v)\rangle$) can be derived while I'm stuck in reaching it. Here is my deduction: \begin{align} \langle u,v\rangle &=\frac{1}{4}(|u+v|^2-|u-v|^2) \\ &=\frac{1}{4}(|g(u+v)|^2-|g(u-v)|^2) \end{align} if $|g(u+v)|=|g(u)+g(v)|$ and $|g(u-v)|=|g(u)-g(v)|$, I can easily arrive at $\langle u,v\rangle=\langle g(u),g(v)\rangle$. However, above two conditions are not obvious to me just in a purely algebraic view. Of course, it is otherwise obvious to me in a geometrical view: So who can help proving this in a purely algebraic view?

This problem comes from Multiple View Geomotry - Lecture 3, in rigid body motion session. Here is the slide(password:mvg-ss16).

Answer 1

This is a consequence of the Theorem of Mazur-Ulam: Every bijective isometry $f: E \to F$ between normed spaces is affine.

You have in the link I provided a proof of Mazur-Ulam theorem. In your case (rigit body motion), $g$ is indeed bijective.

Answer 2

Another way to show it without using the Mazur-Ulam theorem:

$v = V - 0$

$u = U - 0$

$g_*(v) = g(V) - g(0)$

1. Using the rigid-body motion property $||g_*(x)|| = ||x||$ we have:

$||g_*(-v)|| = ||-v|| = ||v|| = ||g_*(v)|| = ||-g_*(v)||$

2. Using the rigid-body motion property $||X - Y|| = ||g(X) - g(Y)||$ we have:

$||-g_*(-v) + g_*(v)|| = ||-(g(-V) - g(0)) + (g(V) - g(0))|| = ||g(V) - g(-V)|| = ||V - (-V)|| = 2||v||$

1 + 2 imply:

$g_*(-v) = -g_*(v)$

Then:

$||u + v|| = ||(U - 0) + (V - 0)|| = ||U - (-V)|| = ||g(U) - g(-V)|| = ||(g(U) - g(0)) - (g(-V) - g(0))|| = ||g_*(u) - g_*(-v)|| = ||g_*(u) + g_*(v)||$

Also with the rigid-body motion property $||g_*(x)|| = ||x||$ we have:

$\sqrt{||g_*(u)||^2 + ||g_*(v)||^2 + 2<u, v>} = \sqrt{||u||^2 + ||v||^2 + 2<u, v>} = ||u + v|| = ||g_*(u) + g_*(v)|| = \sqrt{||g_*(u)||^2 + ||g_*(v)||^2 + 2<g_*(u), g_*(v)>}$

That implies: $<u, v> = <g_*(u), g_*(v)>$

Answer 3

The lecturer has made a mistake. He required that a rigid-body motion preserves the norm:

$ |g(v)|=|v|$

But this does not imply that the inner product is preserved. A counterexample is a function $f$ defined as $f(z)=z$ if $|z|≠1$ and $f(z)=−z$ if $|z|=1$. This function preserves norm but not inner product (see comment to this question: Does a transformation that preserves origin and norm also preserve inner product?).

This is the reason why it is so hard to show that the preservation of norm implies the preservation of the inner product - simply because it is wrong!

Instead of requiring preservation of norm, the lecturer should have required preservation of distances. Rigid-body motions are isometries:

$|g (u) − g (v )| = |u -v|, \forall u,v \in \mathbb R^3$

With this definition, the preservation of the inner product can be shown algebraically.