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More about rotations in 4 dimensions.

It is easy to take a quaternion and recognize the rotation it actually represents. This is about doing the same thing for 4 dimensional rotations. It is easy to multiply two four dimensional rotations together. Can we extract the orthogonal planes of rotation and their angles of rotations from that. We can represent a rotation as a product of two vectors ab via w mapstobavab and this will be the rotation of the ab plane which leaves the space orthogonal to that plane pointwise fixed. It will be the rotations from the vector a to the vector b applied twice. This specifies both the direction and that amount of the rotation. Since ab=g(a,b)+ab then this rotation is cos(α/2)+sin(α/2)ω where α is the amount of rotation and ω is ab scaled to have norm one by a positive scalar constant. The product ab specifies the direction in which the rotation α is applied, since any vectors cd which equals ab specifies the direction of the rotation as it is a rotation of angle α rotating starting from c and moving toward d. We can extract the plane a,b from the product ab because it is those vectors u with u(ab)=0. This is a system of 4 homogeneous equations which has rank 2 over R4. We can extract the orthogonal plane ϕ=cd to the a,b plane from ω alone. Write ω=r12e1e2+r13e1e3+r14e1e4+r23e2e3+r24e2e4+r34e3e4 where r12r34r13r24+r14r23=0. Then the orthogonal plane is ψ=r34e1e2r24e1e3+r23e1e4+r14e2e3r13e2e4+r12e3e4. Then ωψ=(r12e1e2+r13e1e3+r14e1e4+r23e2e3+r24e2e4+r34e3e4)(r34e1e2r24e1e3+r23e1e4+r14e2e3r13e2e4+r12e3e4). The constant term is r12r34+r13r24r14r23r23r14+r24r13r34r12=0. The e1e2e3e4 term is r212+r213+r214+r223+r224+r234. The e1e2 term of ωψ is r13r14e1e3e2e3r14r13e1e4e2e4r23r24e2e3e1e3+r24r23e2e4e1e4 This simplifies to zero. The other grade 2 terms should simply to zero as well. Then (a+bω)(c+dψ)=ac+(bcω+adψ)+bdωψ. So if ω and ψ were norm one then ωψ=e1e2e3e4. Thus our product becomes ac+(bcω+adψ)+bde1e2e3e4. If a=cos(α/2) and b=sin(α/2) and if c=cos(β/2) and d=sin(β/2) then ac=cos(α/2)cos(β/2) and bd=sin(α/2)sin(β/2) and consequently acbd=cos((βα)/2) and ac+bd=cos((α+β)/2). Since ac and bd are known (being the constant and the volume coefficients) then we can solve this for α and β. Now ad+bc=sin((α+β)/2) And adbc=sin((βα)/2). Now bcω+adψ=12(ad+bc)(ψ+ω)+12(adbc)(ψω). This then equals 12sin((α+β)/2)(ψ+ω)+12sin((βα)/2)(ψω) This is useful because ac and bd are not known. But we can find sin((α+β)/2) and sin((βα)/2) up to sign (of each) using ac and bd. This then becomes easy to solve for the actual coefficients rij of ω (and ψ).

Some notes on geometric algebra in 4 dimensions

Working in R4 here with the geometric algebra corresponding to the usual norm, i.e. eiej=ejei for ij and eiei=1. The inverse of a vector v is just v/v2. A geometric algebra is an algebra with zero divisors, for if v is a vector of norm 1 so vv=1 then 12(1+v) is idempotent and is also a zero divisor (by multiplying by (1v). An element of grade k is often called a k vector. Now the product of two vectors a,b is g(a,b)+ab. The product of a and bag(a,b)/g(a,a) is therefore g(a,b)g(a,a)g(a,b)/g(a,a)+(a(bag(a,b)/g(a,a)))=ab since aa=0. So the grade 2 part of a 2 blade is also always a 2 blade. Put differently, ab is always a 2 blade. Now given a 2 blade ab then as above we can assume without loss of generality that g(a,b)=0 and we can also assume that a and b are unit vectors. Then for any scalar α the two blade a(b+αa)=g(a,b+αa)+a(b+αa)=α+ab. So the grade 0 part is irrelevant to whether something is a 2 blade. If A=uv for othogonormal vectors u,v then AA=uvuv=uuvv=1. Let A=r12e1e2+r13e1e3+r14e1e4+r23e2e3+r24e2e4+r34e3e4. Then AA=1 means (r12e1e2+r13e1e3+r14e1e4+r23e2e3+r24e2e4+r34e3e4)(r12e1e2+r13e1e3+r14e1e4+r23e2e3+r24e2e4+r34e3e4)=1 Now rijeiejrijeiej=r2ijeiejeiej=r2ijeieiejej=r2ij. For the mixed terms things are uglier. The e1e2e3e4 term is r12r34e1e2e3e4+r13r24e1e3e2e4+r14r23e1e4e2e3+r23r14e2e3e1e4+r24r13e2e4e1e3+r34r12e3e4e1e2 This is equal to (r12r34r13r24+r14r23+r23r14r24r13+r34r12). Thus r12r34r13r24+r14r23=0 for a blade. The e1e2 part of AA is r13r23e1e3e2e3+r14r24e1e4e2e4+r23r13e2e3e1e3+r24r14e2e4e1e4=(r13r32r14r24+r13r23+r24r14)e1e2=0. Let B=r12e1e2+r13e1e3+r14e1e4+r23e2e3+r24e2e4+r34e3e4 in R4 and assume that r12r34r13r24+r14r23=0. How big is the space of vectors u with uB=0? If u=x1e1+x2e2+x3e3+x4e4 then uB=0 becomes x1r23e1e2e3+x1r24e1e2e4+x1r34e1e3e4+x2r13e2e1e3+x2r14e2e1e4+x2r34e2e3e4+x3r12e3e1e2+x3r14e3e1e4+x3r24e3e2e4+x4r12e4e1e2+x4r13e4e1e3+x4r23e4e2e3=0 This then becomes the system of 4 homogeneous equations: x1r23x2r13+x3r12=0, x1r24x2r14+x4r12=0, x1r34x3r14+x4r13=0, x2r34x3r24+x4r23=0. Taking r14 times the first equation minus r13 times the second we obtain x1(r14r23r13r24)+x3r14r12x4r13r12=0 If r12r34r13r24+r14r23=0 then we can rewrite this as x1r12r34+x3r14r12x4r13r12=0 which is r12 times the third equation. Similarly if we take the linear combination of the first two equations which eliminates the x1 term we should obtain the fourth equation. Thus the system has rank at most two. If B is a wedge of two orthonormal vectors then we can obtain the span of those vectors as the space of vectors u satisfying uB=0.

Rotations in 4D

So the geometric algebra specifies a rotation on Rn in quite a nice form. Given unit vectors a,b then defining σ:vbavab we get the σ is a rotation that leaves everything perpendicular to the a,b plan pointwise fixed, and within the a,b plane σ is the rotation from a to b applied twice (so that the rotation of σ is twice the angle between a and b). Now in the plane if we choose a to be e1=(1,0) and we choose b=cos(θ/2)e1+sin(θ/2)e2=(cos(θ/2),sin(θ/2)) so that applying the rotation from a to b twice is a counterclockwise rotation of the plane by angle θ then the geometric product ab=cos(θ/2)+sin(θ/2)e1e2 and the geometric ba=cos(θ/2)sin(θ/2)e1e2. Now in 4 dimensions a rotation is made up of a plane, and its orthogonal plane, each of them being setwise invariant, and each undergoing a rotation. Thus we can see a model of a 4 dimensional rotation by taking the product (cos(α/2)+sin(α/2)e1e2)(cos(β/2)+sin(β/2)e3e4). Then vq1vq is a rotation in R4 which is a rotation in the direction from e1 to e2 by angle α in the e1,e2 plane, and is a rotation in the e3,e4 plane rotating in the direction from e3 to e4 by angle β. More interestingly, given a general unit member z=a+b12e1e2+b13e1e3+b14e1e4+b23e2e3+b24e2e4+b34e3e4+ce1e2e3e4 one wonders whether we can factor it as a product of two planar orthogonal rotations. That is, can you read off the planes of rotation and the corresponding angles of rotation from any unit even member of the geometric algebra. The answer is that yes one can. Assume that the two rotations are g=cos(α/2)+sin(α/2)(r12e1e2+r13e1e3+r14e1e4+r23e2e3+r24e2e4+r34e3e4) and by h=cos(β/2)+sin(β/2)(s12e1e2+s13e1e3+s14e1e4+s23e2e3+s24e2e4+s34e3e4). For a rotation to represent a plane it must have r12r34+r13r24+r14r23=0 For the planes of these two members of the product to be orthogonal we must have r12=s34, r13=s24, r14=s23, r23=s14, r24=s13, r34=s12. Consequently the grade 4 part of the product gh is r12s34+r13s24+r14s23+r23s14+r24s13+r34s14 which is equal to r212+r213+r214+r232+r224+r234=1. Consequently if z=gh then a=cos(α/2)cos(β/2) and c=sin(α/2)sin(β/2). We then have ac=cos((α+β)/2) and a+c=cos((αβ)/2). From this we can readily extract both α and β. Then since bij=cos(β/2)sin(α/2)rij+sin(β/2)cos(α/2)sij=cos(β/2)sin(α/2)rij+sin(β/2)cos(α/2)rkl and bkl=cos(β/2)sin(α/2)rkl+sin(β/2)cos(α/2)skl=cos(β/2)sin(α/2)rkl+sin(β/2)cos(α/2)rij where kl are the two members different than ij we have equations we can readily solve for rij and rkl in terms of bij and bkl (since β and α are known). Thus one can decompose any even unit member of the geometric algebra of R4 into two orthogonal plane rotations with known angles and directions of rotation.
So the wedge product can be defined by simply setting a basis ei and requiring that eiej=ejei for ij and that eiei=0 along with commuting with scalars and distribution and associativity. It can be shown then that for any vector a one has aa=0. The wedge product defines a bit of signed area (or volume) on an oriented plane, hyperplane, etc.... just as a vetor represents a signed quantity in a specific direction. The (typical) geometric product is defined as follows for an orthonormal basis: eiej=eiej for ij and eiei=1. If a is a vector then aa=aa the usual dot product of a with itself. In general for vectors a and b the geometric product is ab=ab+ab. If b has unit length then bb=1, so b is its own inverse. Consider the conjugation τv:ve1ve1. Then for any basis vector other than e1 the result is eie1eie1=e1e1ei=ei. But e1e1e1e1=e1. Thus τv preserves the e1 axis, but is an inversion on the perpendicular space. The same is true for τa:vava for any unit vector a. The line through a is fixed pointwise, and the perpendicular space to a undergoes an involution where everything is negated. The combination of τa followed by τb then fixes everything perpendicular to the a,b plane. However, on the a,b plane, if h is rotation from a to b then τbτa is h applied twice, giving twice the rotation of a direct rotation from a to b. Thus the map v(ba)v(ab) rotates the a,b plane as if a single rotation directly from a to b had been applied twice, whereas it leaves the subspace perpendicular to the a,b plane pointwise fixed. The scalar part of ab is the cosine of ψ where ψ is the angle from a to b. Put differently, the angle of rotation is θ and the scalar part of this rotation is cos(θ/2). Because abab=aabb=1 then if we write out the taylor series expansion for exp(tab) where t is a scalar we get cos(t)+sin(t)ab. The inverse of cos(t)+sin(t)ab is cos(t)+sin(t)ba=cos(t)sin(t)ab. Thus there is a rotation subgroup cos(t) + sin(t)ab, i.e. v(cos(t)sin(t)ab)v(cos(t)+sin(t)ab) When t=0 this is the identity map and when t=π/2 this is the rotation v(ba)v(ab) Now ab=ab+ab. If a,b are unit vectors, then multiplying [cos(t)+sin(t)ab][cos(s)+sin(s)ab] yields cos(t)cos(s)+sin(t)sin(s)abab+cos(t)sin(s)ab+sin(t)cos(s)ab=(cos(t)cos(s)sin(t)sin(s))+[cos(t)sin(s)+sin(t)cos(s)]ab but this is just cos(t+s)+sin(t+s)ab In 4 dimensions a rotation is determined by the element (cos(s/2)+sin(s/2)e1e2)(cos(t/2)+sin(t/2)e3e4) Expanding gives cos(s/2)cos(t/2)+cos(s/2)sin(t/2)e3e4+cos(t/2)sin(s/2)e1e2+sin(s/2)sin(t/2)e1e2e3e4 So there is a vector and a pseudovector and one is cos(s/2)cos(t/2) and the other is sin(s/2)sin(t/2)e1e2e3e4 There is also a bivector part which is where we get real direction (because the vector and pseudovector are directionless). So given a unit versor for R4 is there necessarily a corresponding rotation, and if so, can we read off the planes of rotations and their rotation angles as we can in 3 dimensions? That is, given a + b_{12) e_1 e_2 + b_{13} e_1 e_3 + b_{14} e_1 e_4 + b_{23} e_2 e_3 + b_{24} e_2 e_4 + b_{34} e_3 e_4 + c e_1 e_2 e_3 e_4 a unit vector where the sum of squares of the coefficients is one, does this always give a rotation, and if so, what are the planes and their rotations? Now the grade two part should be the sum of two different blades, each one the orthogonal complement of the other. A single 2 blade in R4 has grade two part satisfying u12u34+u13u24+u14u23=0. The orthogonal blade grade 2 part has v12=u34 and v13=u24 etc.... One would expect that every possible choice of bij can be represented nearly uniquely as a weighted sum of two orthogonal 2 blades, i.e. choose 1, then the other is detemined, and one can choose the coefficient of the other which gives just the right dimension.

More about rotations in 4 dimensions.

It is easy to take a quaternion and recognize the rotation it actually represents. This is about doing the same thing for 4 dimensional rota...