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
