[math-fun] A symmetrical rotation of Euclidean space
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
I don't exactly know what you're asking for, but maybe you are trying to make a jump off Berlekamp's complex transformations by restricting to reals? I think the answer to your questions just falls out of basic representation theory of a finite cyclic group. In some choice of coordinates, your T commutes with a cyclic permutation of basis vectors, call that transformation T'. T' is a regular representation of the generator of a cyclic group, so we know its eigenbasis, which is shared with T. Eigenvectors are classified according to their transformation properties starting with (in physics notation) isotropic A_g, g as in gerade, with components 1,1,1, etc. When n is even, there is also an A_u, u as in ungerade, with components 1,-1,1,-1, etc. When n is odd, there is no A_u, and all remaining 2*k eigenvectors can be written as Cos(2*Pi/(2*k+1)*m*x) or Sin(2*Pi/(2*k+1)*m*x), with label index m=1,...,k, and component index x=0,...2*k. The label index always determines the irreducible representation E_m, compare for example with: http://symmetry.jacobs-university.de/cgi-bin/group.cgi?group=205&option=4 Representations E_m can be thought of as planes. My guess is that your angles theta_j are just calculated by taking the angle of T modulo the translation symmetry of the E_m basis vectors. Sometimes we do want to think about E_m representations of an electronic wavefunction, because the Born Oppenheimer approximation is expected to break down due to degeneracy (the Jahn Teller effect). Other times, you want to just ignore the planes, and focus on a line where the B.O. potential describes an integrable A vibrational mode, as is the case with the problem of the Ammonia maser (see Feynman Vol. III, Ch. 9). More on this shortly... --Brad On Mon, Sep 21, 2020 at 2:31 AM Dan Asimov <dasimov@earthlink.net> wrote:
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
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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
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Brad Klee -
Dan Asimov -
Fred Lunnon