Hopf fibration with bivectors

Higher-dimensional geometry (previously "Polyshapes").

Hopf fibration with bivectors

Postby mr_e_man » Thu Dec 13, 2018 7:41 pm

(The first post here is for context. The second post was moved from 4D Planet: Angle of Sun, because it seemed a little off-topic there.)

PatrickPowers wrote: "The modes of rotation in 4d, is represented as a point in 6d space, where three of the 6d space forms the L axis (left-clifford) and the other three form the R axis. Any rotation, even at different speeds, can be formed by a simple sum of an L and R axis, where the two circular rotations are L+R and L-R resp."

This I do not understand. It seems it could be quite useful. Care to offer a reference? I'm thinking there must be some buzzword that I can search on.

There's always Wikipedia: https://en.wikipedia.org/wiki/Rotations ... _SO%284%29 . (I don't understand half of it, but there is some useful information there.)

You seem to be familiar with geometric algebra, so I will discuss 4D rotations in terms of bivectors.

Any rotation can be done by the geometric sandwich product with a rotor (a product of an even number of unit vectors), and any rotor is the exponential of a bivector. The set of rotations is a "Lie group", and the set of bivectors is the corresponding "Lie algebra". (But conventionally, these things are described with matrices instead of bivectors.)

The set of all bivectors in 4D is 6-dimensional, being spanned by {e1e2, e1e3, e1e4, e2e3, e2e4, e3e4}. A rotation by angle θ in the x1x2-plane corresponds to the bivector θe1e2. An isoclinic ("Clifford") rotation would be something like θ(e1e2+e3e4), and a general double rotation θe1e2+ϕe3e4.

Instead of writing everything as a sum of these simple bivectors, we could use isoclinic bivectors as a basis:

L1 = e1e2 + e3e4, L2 = e1e3 - e2e4, L3 = e1e4 + e2e3,
R1 = e1e2 - e3e4, R2 = e1e3 + e2e4, R3 = e1e4 - e2e3

For example, the simple rotation becomes θe1e2 = (θ/2)(L1+R1).

I haven't studied this in detail, but there is a non-trivial correspondence between operations in the Lie group and the Lie algebra. The Lie group operation is just composition of transformations (like applying several rotations consecutively). The Lie algebra operations are addition and the "commutator" cross product. If two bivectors commute (meaning A×B=0), then the corresponding rotors exp(A) and exp(B) also commute, and exp(A)exp(B)=exp(A+B).

The cross product of two bivectors can be calculated as follows, using the standard basis:

(e1e2)×(e1e2) = 0
(e1e2)×(e1e3) = - e2e3
(e1e2)×(e2e3) = e1e3
(e1e2)×(e3e4) = 0

etc., and using the isoclinic basis:

L2×L1 = 2L3, R1×R2 = 2R3,
L3×L2 = 2L1, R2×R3 = 2R1,
L1×L3 = 2L2, R3×R1 = 2R2,
L1×R1 = 0, L1×R2 = 0, L1×R3 = 0

Note that the cross product of any two R's is another R, so the right-isoclinic rotations form a group by themselves. (I believe this is the "swirlprism" symmetry group, and the "special unitary group".) And if each R is divided by 2, then these equations are the same as those for bivectors in 3D, or for pure-imaginary quaternions, or the ordinary vector cross product:

i×j = k,
j×k = i,
k×i = j

This isomorphism between bivectors in 4D and 3D is, of course, related to the Hopf fibration between the spheres in 4D and 3D.
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Re: Hopf fibration with bivectors

Postby mr_e_man » Thu Dec 13, 2018 7:42 pm

I just found that it's simpler than I expected! The Hopf fibration seems to be more naturally defined with bivectors.

First note that, given a bivector in 3D,

B = B12e1e2 + B13e1e3 + B23e2e3

we can extract its coordinates by taking dot products with the basis blades:

B~.(e1e2) = B12
B~.(e1e3) = B13
B~.(e2e3) = B23

(The reverse ~ is used so we don't get a negative sign.)

Now, the left-isoclinic L1=e1e2+e3e4 determines a vector field in 4D: at any point x=x1e1+x2e2+x3e3+x4e4, the vector at that point is

v = x.L1 = - x2e1 + x1e2 - x4e3 + x3e4

We can interpret x as position and v as velocity, so we have a differential equation dx/dt=x.L1; the solution curves are the Hopf fibre circles.

And it should be clear that x is orthogonal to v (meaning x.v=0), so their geometric product is a bivector:

B = xv = (x12+x22)e1e2 + (x32+x42)e3e4 + (x2x3- x1x4)(e1e3+e2e4) + (x1x3+x2x4)(e1e4- e2e3)
= (x12+x22)(L1+R1)/2 + (x32+x42)(L1- R1)/2 + (x2x3- x1x4)R2 + (x1x3+x2x4)R3

This B determines a plane, whose intersection with the sphere is, again, a Hopf fibre circle.

Taking the dot products of B with the right-isoclinic basis (using that R1.R1= -2, R1.R2=0, R1.L1=0, R1.L2=0, etc.), we get

B~.R1 = x12+x22- x32- x42
B~.R2 = 2(x2x3- x1x4)
B~.R3 = 2(x1x3+x2x4)

These can be used as the coordinates of a vector or bivector in 3D. This agrees with Wikipedia's formula (up to re-labelling of the coordinates; they also took advantage of x.x=1 on the sphere).

This treatment of the Hopf fibration is more general, in that x doesn't need to be on the sphere. In fact, B doesn't need to be derived from x or L1; this works for any bivector B, taking it from 4D to 3D. (And it's a linear function of B, despite being a quadratic function of x.)
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Re: Hopf fibration with bivectors

Postby mr_e_man » Tue Jun 04, 2019 4:36 am

This is also related to the various types of projective space.

Real projective space

There's not much we can do here. We can project a point onto the unit sphere, x --> x/||x||, but we don't have an algebraic way to make x equivalent to -x.

Complex projective space

In an even-dimensional space R2n, we have an isoclinic bivector

L1 = e1e2 + e3e4 + e5e6 + ... + e2n-1e2n

which again determines a vector field

v = x.L1 = - x2e1 + x1e2 - x4e3 + x3e4 - x6e5 + x5e6 + ... - x2ne2n-1 + x2n-1e2n

(This dot product with L1 can also be thought of as complex multiplication by i, which rotates by 90 degrees.)

Again this gives us a way to turn any point x into a bivector:

x --> x-1(x.L1) = x(x.L1)/(x.x)

The division by x2 ensures that all scalar multiples of x are equivalent (producing the same bivector). We also have (x.L1).L1 = - x, and thus

(x.L1) ((x.L1).L1) = x(x.L1)

which says that an "imaginary multiple" x.L1 produces the same bivector as x. This fits the definition of complex projective space: all complex multiples of a vector are equivalent. So complex projective space is isomorphic to the set of all bivectors of the form x(x.L1)/x2.

(Such a bivector can be visualized as a 2D plane, or a unit square or circle in that plane.)

Quaternionic projective space

In R4n, we have three isoclinic bivectors

L1 = (e1e2 + e3e4) + (e5e6 + e7e8) + ... + (e4n-3e4n-2 + e4n-1e4n)
L2 = (e1e3 - e2e4) + (e5e7 - e6e8) + ... + (e4n-3e4n-1 - e4n-2e4n)
L3 = (e1e4 + e2e3) + (e5e8 + e6e7) + ... + (e4n-3e4n + e4n-2e4n-1)

and their dot products with x act like quaternion multiplication:

(x.L1).L1 = (x.L2).L2 = (x.L3).L3 = - x,
(x.L2).L1 = x.L3,
(x.L3).L2 = x.L1,
(x.L1).L3 = x.L2

The four vectors derived from x are orthogonal (that is, (x.L1).(x.L2)=0 etc.), so their geometric product is a quadvector:

x --> x(x.L1)(x.L2)(x.L3)/x4

The division by x4 ensures that all scalar multiples of x are equivalent (producing the same quadvector). And some tedious algebra shows that any "quaternion multiple" of x,

c0x + c1x.L1 + c2x.L2 + c3x.L3

with scalars ck, is also equivalent to x. So quaternionic projective space is isomorphic to the set of all quadvectors of the form x(x.L1)(x.L2)(x.L3)/x4.

(Such a quadvector can be thought of as a 4D subspace, or a unit (hyper)cube or sphere in that subspace.)

Octonionic projective space

I haven't looked into this yet, though I know that octonion multiplication can be described in terms of geometric algebra.
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