Dan, I don’t understand the step near the end where you find the rotation angles (“it is easy to check”), and I don’t see how it can be correct. The permutation you describe is of order n, while the rotation you describe is of order L. Am I missing something? Jim On Sunday, June 24, 2018, Dan Asimov <dasimov@earthlink.net> wrote:

Let n be a positive integer. We can think of n as a set with n elements.

An even prettier representation of n is as the cyclic permutation on n elements.

And given any permutation on n elements, we can always think of that as a permutation of the standard basis elements of R^n. Which of course defines a linear mapping

P : R^n —> R^n.

Fine, so we are imagining the mapping defined by

P(e_k) = e_(k+1)

for 0 <= k < n, i.e.,

P(x_0, x_1, ..., x_(n-1)) = (x_(n-1), x_0, x_1, ..., x_(n-2))

that just cyclically permutes the coordinates. So what?

Well, it's easy to see this is an orthogonal transformation, or in other words it preserves distances. If n is an odd number, it's in fact an *orientation-preserving* distance-preserving linear transformation of R^n, and this can only mean one thing: P is a rotation.

Since the (1,1,...,1) direction in R^n is preserved by P, P must also preserve its orthogonal complement, which is the plane given by

(R^n)_0 = {(x_0,...,x_(n-1)) | Sum x_j = 0}.

So still assuming n is odd, say:

n = 2L + 1

then the plane (R^n)^0, is a copy of R^(2L), on which P is a rotation.

But we have a lovely canonical form for rotations on even-dimensional spaces: They are determined by 2-plane rotations, by possibly distinct angles applied respectively to a series of mutually perpendicular 2-planes that together span the space, R^(2L) in the current situation.

All such decompositions of an even-dimensional space into mutually perpendicular 2-planes, so the only interesting this here is the set of angles.

It's easy to check that in this case, the 2-plane rotations are by the angles

{theta, 2*theta, ..., L*theta = pi}

for theta = 2*pi/(2L).

But why? What do these rotations have to do with the cyclic permutation of n elements?

[Please forgive my naïveté, for I have never taken a course in linear algebra.]

—Dan

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