Greg Egan on Nostr: Pick N points on a sphere, equally spaced on the circle at an angle θ from the pole. ...

Pick N points on a sphere, equally spaced on the circle at an angle θ from the pole.

Rotate the sphere around each of the points in turn by the same angle, α.

What value should α be, in order for the net rotation to restore the sphere to its original orientation?

Although this problem is very symmetrical, it’s not so symmetrical as to be trivial! It might be tempting to think that the symmetry guarantees that the net rotation will always be around the pole, but that’s not true: because we rotate around each of the N points in a particular order, they are *not* all treated equally, which breaks the symmetry and means the overall axis of rotation need *not* be the pole.

But here’s a sketch of a reasonably easy way to find the solution in general.

Suppose we call a rotation around the first point T, and let R be a rotation by 2π/N.

Then the rotation around the (k+1)st point will be:

R^k T R^{–k}

because R^{–k} moves the (k+1)st point to the first point, where we perform a rotation by the same angle in each case, and then R^k moves the first point back to the (k+1)st point.

This means the net result of performing all N rotations is:

R^{N–1} T R^{–(N–1)} R^{N–2} T R^{–(N–2)} ... R^2 T R^{–2} R T R^{–1} T

If we cancel positive and negative powers of R that are multiplied together, and note that

R^{N–1} = R^{–1}

this becomes:

R^{–1} (T R^{–1})^{N–1} T

Now, suppose this is equal to the identity matrix, I. We then have:

R^{–1} (T R^{–1})^{N–1} T = I
(T R^{–1})^{N–1} = R T^{–1} = (T R^{–1})^{–1}
(T R^{–1})^N = I

So we need to make the product of the rotation by α around the first point, and a rotation around the pole by –2π/N, into a rotation by an angle that is (any integer multiple of) 2π/N.