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