Fred, thanks for your note. (How are you doing, by the way?) Yes, it would have been better to write 0 ≤ j ≤ n initially, but isn't 1 ≤ j ≤ n+1 redeemed by "(indices are modulo n+1)"? As for the later reference to R^n, isn't that actually R^n because it's the orthogonal complement in R^(N+1) of the line in the (1,1,...,1) direction? —Dan ----- To Dan --- Multiple notational confusion alert! << For all j, 1 ≤ j ≤ n+1, T(e_j) = e_(j+1) (indices are modulo n+1). >> should read 0 <= j <= n ; also elsewhere, eg. O-P only for for n even! Then half-way thru', |R(n+1) mysteriously transmogrifies into |R(n) ... That theorem is not "Chasles' theorem"; however it might well be due to Chasles, or (more likely) Euler. Fred On 9/21/20, Dan Asimov <dasimov@earthlink.net> wrote:

I worked this out last fall and found it surprising:

Let {e_0,...,e_n} be the standard basis of Euclidean n-space R^(n+1) (e_j has a 1 as its jth coordinate and 0's elsewhere).

Define the linear mapping T : R^(n+1) —> R^(n+1) as follows by its effect on basis vectors:

For all j, 1 ≤ j ≤ n+1, T(e_j) = e_(j+1) (indices are modulo n+1).

It's easy to check that T is orientation-preserving only for n even, and that T is always an isometry of R^(n+1).

It's also clear that the vector (1,1,...,1) of all 1's is a fixed point of T. This means that its orthogonal complement (1,1,...,1)^⊥, namely

(1,1,...,1)^⊥ = {(x_0,...,x_n) in R^(n+1) | Sum x_j = 0}

(which is a copy of R^n), is carried to itself by T.

A nice theorem states that any rotation T of R^n for n even has n/2 mutually orthogonal 2-planes that are each taken into themselves by a rotation by some angle. (I don't know the name of this theorem; can anyone tell me its name or who first discovered it?)

So for n even, say n = 2k, our T as defined above, when restricted to (1,1,...,1)^⊥, is actually a collection of rotations by angles that we'll call theta_j, 1 ≤ j ≤ n/2.

Question: What are these angles theta_j ?

Suggestion: Try to guess before doing a calculation.

—Dan