Finding an Isometry between two hyperplanes

Question

I want to show that there is a map $P$ from $\{ x\in\mathbb R ^n \mid \sum_{i=1}^nx_i =0 \}$ to $\{ x\in\mathbb R ^n \mid x_{n}=0 \}$ that preserves relative distances between points; i.e if $ d(x,y)=d(a,b)$ then $d(P(x),P(y))=d(P(a),P(b))$.

My geometrical intuition says its possible but I cannot find how to construct $P$. Technically I also believe it should be linear - but maybe it does not the case.

Thanks a lot :)

Do you know the Gram Schmidt process, which can find an orthnormal basis for your hyperplane $P$? — Lee Mosher, Jan 15 '24 at 02:10
Finding this rotation is exactly what i’m trying to do… and I do know the gram schmidt process - should I define the map by mapping one orthonormal basis to the other? — Yarin Luhmany, Jan 15 '24 at 17:53

joriki · Answer 1 · 2024-01-15T18:10:54.887

2

On $\mathbb R^n$ we have an inner product $\langle x,y\rangle=\sum_ix_iy_i$ that’s invariant under rotations. If you express the first set as $\{x\mid\langle x,y_1\rangle=0\}$ with $y_1=(1,\ldots,1,1)$ and the second one as $\{x\mid\langle x,y_2\rangle=0\}$ with $y_2=(0,\ldots,0,1)$, this invariance shows that you can transform one into the other by rotating $y_1$ into the direction of $y_2$.

edited Jan 15 '24 at 18:10

answered Jan 15 '24 at 01:49

joriki

238,052

Thank you for your answer! You mean I should find a matrix that maps the first vector to the second (by multiplication)? – Yarin Luhmany Jan 15 '24 at 17:56
@YarinLuhmany: Yes, but not just any matrix, a rotation matrix. This answer shows how to do that. – joriki Jan 15 '24 at 18:10

copper.hat · Answer 2 · 2024-01-16T04:13:19.790

One typical tool for this sort of thing is a Householder reflection (also, see this).

Let $e=(1,...,1)$. We want to rotate ${1 \over \sqrt{n}}e$ to $e_n = (0,...,0,1)$ in the two dimensional subspace spanned by these vectors (note that both have the same length).

We can do this using a Householder reflection that reflects points across a hyperplane that 'reflects' ${1 \over \sqrt{n}}e$ to $e_n$.

Let $v={1 \over \sqrt{n}}e-e_n$, and $P=I-{2 \over v^T v} v v^T$. Check that $PP^T = I$ (hence an improper rotation, which preserves distances) and $P {1 \over \sqrt{n}}e = e_n$.

Check that $P$ maps the hyperplanes as required.

Alternative, proper rotation based answer:

Here is a derivation of a proper rotation that does the job, it is essentially a Givens rotation.

Use Gram Schmidt on $v_1={1 \over \sqrt{n}}e, e_n$ to get $v_2 = {1 \over \sqrt{1-{1 \over n}}} (e_n - (v_1^T e_n ) v_1)$, and note that $e_n = {1 \over \sqrt{n}} v_1 + \sqrt{1-{1 \over n}} v_2 $. Let $R=\begin{bmatrix} {1 \over \sqrt{n}} & -\sqrt{1-{1 \over n}} \\ \sqrt{1-{1 \over n}} & {1 \over \sqrt{n}} \end{bmatrix}$, $V= \begin{bmatrix} v_1 & v_2\end{bmatrix}$ and define $P = I - V V^T + V R V^T$, and check that $P^TP = I$, $Pv_1 = e_n$ and $\det P = \det R = +1$ (to see the latter, consider an orthonormal basis of the form $v_1, v_2, ...$).

$+1$. But reflections are usually not called (proper) rotations. They're isometries, as required in the question, but improper rotations because their determinant is $-1$, i.e. they change the orientation of the space. — joriki, Jan 15 '24 at 07:37
@joriki I updated the answer to indicate the improperness of it all. I should probably have written orthogonal. — copper.hat, Jan 15 '24 at 17:54

Finding an Isometry between two hyperplanes

2 Answers2