Chiral rotations in 4d

[PatrickPowers]

Rotations in 4d are chiral. Usually this is apparent. But what about a clifford rotation? Can the two chiralities be distinguished then? I don't see how.

[wendy]

The chirality of a rotation can be distinguished, even for clifford ones. The only non-chiral rotation are the great-circle ones.

If you follow two points rotating on a clifford rotation, there is a corkscrew of each around the other. Relative to either, the other progresses clockwise or anticlockwise in the direction of motion. This is the chirality of rotations. Chiral rotations exist in every even dimension, the corresponding odd dimensions results in a vector perpendicular to the rotation.

What happens is that in an even dimension, the rotation is chiral. In odd dimensions, the odd direction turns into an arrow up or down. In 3d, you have the right-hand-rule: if your right-hand fingures point towards the direction of rotation (ie appear under the fingertips), then the thumb points in the vector. Clockwise points into the clock.

In even dimensions this vector turns into a vector, which follows a circle in the even dimension.

[PatrickPowers]

Mathematically it is easy to distinguish the two chiralities. Either the signs of the two rotations are the same or they differ. The sign of each rotation is arbitrary, so there needs to be some convention to assign the sign.

On the rotating hypersphere i believe there must be some local way to ditinguish the two chiralities. The most likely prospect could be a Coriolus/Eetvos effect. These two effects have the same cause. The Corilus effect is the horizontal componenf and the Eetvos is the vertical component. For our purposes it is better to consider this to be one effect we'll call EC. It affects the path of a moving free body.

The first thing is to project the velocity vector of the free body onto the plane of rotation. Next, rotate the vector pi/2 radians in that plane. Ths rotation has the opposite sign of the rotation of the planet. The result will be the acceleration vector of the fictitious EC force acting on the free body. If you like, decompose this vector into horizontal Coriolus and vertical Eetvos components.

In 4D the only difference is that there are two planes of rotation. Compute the EC vector for each plane. The net result is the sum of the two vectors.

What was hanging me up was that with a Clifford rotation one cannot uniquely identify the planes of rotation. But this should not matter. Simply choose two arbitrary planes that meet the requirements. They should be perpendicular and intersect at the center of the planet.

We shall choose the most convenient planes. For a moment step out of the rotating frame of reference and into a stationary Cartesian frame. In this frame each point on the surface of the planet has a velocity vector. The most convenient choice is a plane aligned with that motion while still passing through the center of the plane. If the motion of the free body is purely horizontal then the acceleration vector due to the first plane will be a purely vertical Eetvos force. The second plane of rotation is then perpendicular to first. Their point of intersection is the center of the planet. If the motion of the free body is entirely horizontal then this second plane contains the horizontal Coriolus force.

This force would cause a Foucault-style pendulum to precess. The sign of the circular precession would be opposite of the sign of the rotation of the planet in that plane. So the chirality of the planet's rotation would have been detected.

[PatrickPowers]

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