Beltrami vector fields with polyhedral symmetries

0.19 MB·

by  Giedrius Alkauskas

BELTRAMI VECTOR FIELDS WITH POLYHEDRAL SYMMETRIES
GIEDRIUS ALKAUSKAS
7 Abstract. A vector field B is said to be Beltrami vector field (force free-magnetic vector
1 fieldinphysics),ifB ( B)=0. Motivatedbyourinvestigationsonprojectivesuperflows,
0 andasanimportant×sid∇e r×esult,we constructtwounique 3-dimensionalBeltramivectorfields
2
I and Y, such that I = I, Y = Y, and that both have orientation-preserving
∇ × ∇ ×
n icosahedral symmetry (group of order 60). Analogous constructions are done for tetrahedral
a
andoctahedralgroupsoforders12and24,respectively. Thisconstructionleadstonewnotions
J
of lambent vector field and lambent flow.
6
1

]
A
1. Lambent vector fields and flows
C
. 1.1. The main result. For a 3-dimensional vector field B let, as usual, B = curlB, and
h ∇×
t let us fix φ = 1+√5. One of the main results of this paper can be stated immediately.
a 2
m
Theorem 1. Let us define the vector field V = V (x,y,z),V (y,z,x),V (z,x,y) , where
x x x
[
1 Vx = 2xsin x2 sin φ2y sin 2zφ −2φxsin 2xφ si(cid:0)n y2 sin φ2z +2φ−1xsin φ2x sin 2yφ (cid:1)sin 2z
v (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17)
x φy z x y φz
8 + ysinz+2ycos cos sin 2ycos cos sin
1 2 2 2φ − 2φ 2 2
2 (cid:16)x(cid:17) (cid:16)φy(cid:17) (cid:16) z (cid:17) (cid:16)φx(cid:17) (cid:16) y(cid:17) (cid:16) z(cid:17)
+ zsiny 2zcos sin cos +2zcos sin cos ,
4 − 2 2 2φ 2 2φ 2
0 (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17)
.
1 and the vector field W = W (x,y,z),W (y,z,x),W (z,x,y) , where
x x x
0
7 Wx = xcosy−xcosz (cid:0) (cid:1)
v:1 − √5xcos x2 cos φ2y cos 2zφ +φxcos 2xφ cos y2 cos φ2z +φ−1xcos φ2x cos 2yφ cos 2z
(cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17)
Xi φ−2ysin x sin φy cos z φ2ysin x sin y cos φz +√5ysin φx sin y cos z
− 2 2 2φ − 2φ 2 2 2 2φ 2
r (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17)
a φ2zsin x cos φy sin z φ−2zsin φx cos y sin z +√5zsin x cos y sin φz .
− 2 2 2φ − 2 2φ 2 2φ 2 2
(cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17)
Then
1) The vector field I = (I ,I ,I ) = V+W has an icosahedral symmetry: if we treat I
x y z
as a map R3 R3, for any ζ I (see Subsection 2.2) one has ζ 1 I ζ = I;
−
7→ ∈ ◦ ◦
2) it satisfies I = I;
3) the Taylor∇ser×ies for I starts with a degree 6 form ̟ , where ̟ is given by
x 768
̟=(5 √5)yz5+(5+√5)y5z 20y3z3+(10+10√5)x2yz3+(10 10√5)x2y3z 10x4yz.
− − − −
Date: 15 January, 2017.
2010 Mathematics Subject Classification. Primary 37C10, 15Q31.
Key words and phrases. Beltramivectorfield,force-freemagneticfield,regularpolyhedra,Euler’sequation,
curl, irreducible representations.
The research of the author was supported by the Research Council of Lithuania grant No. MIP-072/2015.
1
Beltrami vector fields 2
The form ̟ is the numerator of the first coordinate of the icosahedral projective superflow
[2], whence the idea comes from.
Note that vector fields V and W have one additional symmetry. Indeed, let us denote by
τ the non-trivial automorphism of the number field Q(√5). Thus, τφ = φ 1. In the above
−
−
formulas for V and W , let us expand everything in Taylor series, swap y and z, leave x
x x
intact, and apply τ summand-wise. We readily obtain
τ V (x,z,y) = V (x,y,z), τ W (x,z,y) = W (x,y,z).
x x x x
−
1.2. The tetrahedral case: motivation. Let us define
1 0 0 1 0 0 0 1 0 0 1 0
−
α 0 1 0 , β 0 1 0 , η 1 0 0 , δ 0 0 1 . (1)
7→  −  7→   7→   7→  
0 0 1 0 0 1 0 0 1 1 0 0
− −
       
These are matrices of order α2 = β2 = η2 = δ3 = I. Together they generate the full tetrahe-
dral group T of order 24, and α,β and δ generate tetrahedral group T of order 12. In fact,
already α and δ generate T, and α,η and δ generate the whole group T.
b
A vector field M = (2yz,2xz,2xy) produces the tetrahedral projectivbe superflow [1]. This is
2
the unique, up to a scalar multiple, polynomial vector field of degree 2 which has a tetrahedral
symmetry T. Here we encounter the phenomenon that the symmetry a posteriori extends to
a bigger group T (see [1], Note 5).
As we will sebe now, in another direction, this vector field gives rise to another surprising
non-rational vector field T with few unique properties via the following construction. Namely,
the vector field T has a tetrahedral symmetry T, and is equals its own curl.
Since M = 0, there exist a vector field R = (r ,r ,r ) such that R = M . In
2 x y z 2
∇ · ∇ ×
expanded terms, this reads as
∂r ∂r ∂r ∂r ∂r ∂r
z y x z y x
yz = , xz = , xz = .
∂y − ∂z ∂z − ∂x ∂x − ∂y
The general solution to these equations is given by
1 ∂Q 1 ∂Q 1 ∂Q
(r ,r ,r ) = x(z2 y2)+ , y(x2 z2)+ , z(y2 x2)+ ,
x y z
2 − ∂x 2 − ∂y 2 − ∂z
!
where Q is an arbitrary function in three variables x,y,z, with needed smoothness. We want
this vector field to satisfy the following three properties (for n = 3):
i) it is n-homogeneous;
ii) R = 0;
∇·
iii) vector field R has at least a 12-fold symmetry with respect to the group T. That is,
invariance under matrices α, β and δ.
So, Q is 4 homogeneous, and
∂2 ∂2 ∂2
2Q = + + Q = 0.
∇ ∂x2 ∂y2 ∂z2
(cid:16) (cid:17)
3 G. Alkauskas
If Q = 0, the above vector field has the needed properties. Moreover, its symmetry group is
of order 24, and is equal to T I, I . Let therefore
×{ − }
1 1 1
M = x(z2 y2), y(x2 z2), z(y2 x2) .
3
2 − 2 − 2 −
(cid:16) (cid:17)
However, this is not the unique choice. We could also choose
1 1 1
M(a) = x(z2 y2), y(x2 z2), z(y2 x2)
3 2 − 2 − 2 −
(cid:16) x3 x y3 y z3(cid:17) z
+ a (y2 +z2), (z2 +x2), (x2 +y2) , a R.
3 − 2 3 − 2 3 − 2 ∈
(cid:16) (cid:17)
Now, as the next step, we will perform the same inverse problem for M as we did for M .
3 2
Namely, we will seek for a vector field M such that
4
M = M , M = 0,
4 3 4
∇× ∇·
and R = M satisfies the above three properties i), ii) and iii) for n = 4. And we indeed find
4
1 1 1
M = (y3z +yz3), (z3x+zx3), (x3y +xy3) .
4
− 6 −6 −6
(cid:16) (cid:17)
This vector field has the same symmetry group as M . Continuing, we can take
2
( 1)ℓ ( 1)ℓ ( 1)ℓ
M = − x(y2ℓ z2ℓ), − y(z2ℓ x2ℓ), − z(x2ℓ y2ℓ) ,
2ℓ+1
(2ℓ)! − (2ℓ)! − (2ℓ)! −
!
( 1)ℓyz ( 1)ℓxz ( 1)ℓxy
M = − (y2ℓ +z2ℓ), − (z2ℓ +x2ℓ), − (x2ℓ +y2ℓ) .
2ℓ+2 (2ℓ+1)! (2ℓ+1)! (2ℓ+1)!
!
The above identities holds for ℓ = 0, with the correct value M = M = 0. Let
1 2
∇×
∞
V = M = zsiny +ysinz,xsinz +zsinx,ysinx+xsiny ,
2ℓ+2
Xℓ=0 (cid:16) (cid:17)
∞
W = M = xcosy xcosz,ycosz ycosx,zcosx zcosy ,
2ℓ+1
− − −
Xℓ=0 (cid:16) (cid:17)
T = V+W = (V ,V ,V )+(W ,W ,W ) = (T ,T ,T ).
x y z x y z x y z
Thus, we have proved the following.
Proposition 1. Vector fields T,V and W have these properties:
1s) T = T (this implies T = 0);
∇× ∇·
2s) T has a group of orientation preserving symmetries of a tetrahedron as a group of its
symmetries (invariance under conjugating with matrices α, β and δ);
3s) One has
T(xt,yt,zt)
= (2yz,2xz,2xy).
t2
t=0
(cid:12)
Further, for vector fields V and W one(cid:12)has:
(cid:12)
1vw) W = V, V = W, V = 0, W = 0.
∇× ∇× ∇· ∇·
Beltrami vector fields 4
2vw) V has the group T of order 24 as the group of its symmetries; W has the group T
×
I, I of order 24 as the group of its symmetries (this group is generated by matrices
{ − }
α, β, δ, and I).b
−
The vector field T itself is given by (2); see below. The corresponding differential system
for a flow reads, for (x,y,z) = (x(t),y(t),z(t)), as
x = T , y = T , z = T .
′ x ′ y ′ z
Let x = (x,y,z). Wecall the vector field T the tetrahedral lambent field, andthe corresponding
flow F (x;t) = (x(t),y(t),z(t)), F (x;0) = x - the tetrahedral lambent flow. Further, we call
T T
the vector field V tetrahedral evenly-lambent field, and W - tetrahedral oddly-lambent field.
The definition of all these will be given in the next Subsection.
If W is the first integral for the flow with the vector field W, then as ∂W W + ∂W W +
∂x · x ∂y · y
∂W W = 0. We see that two independent solutions are given by
∂z · z
costdt
W (1) = xyz, and W (2) = Ci(x)+Ci(y)+Ci(z), Ci(x) = ∞ .
− t
Zx
So, the orbits of this flow are space curves W (1) = ξ,W (2) = χ : ξ,χ R .
{ ∈ }
1.3. Generalization: lambent flows. Let 2 be a vector Laplace operator, which maps
∇
(r ,r ,r ) to ( 2r , 2r , 2r ). If B is any smooth vector field, then we have a standard
x y z x y z
∇ ∇ ∇
identity of vector calculus
( B) = ( B) 2B.
∇× ∇× ∇ ∇· −∇
This, applied to V and W, gives
V = 2V, W = 2W.
−∇ −∇
Thus, any coordinate c of V or W satisfies a (scalar) Helmholtz equation c = 2c, which is
−∇
immediatetocheck. Intheotherdirection: ifB = 2B, and B = 0,then ( B) = B.
−∇ ∇· ∇× ∇×
Note that the last expression is curl notation-free. Thus, we may proceed with the following
definition.
Definition 1. Let Γ ֒ GL(3,R) be an exact irreducible representation of a finite group,
→
x = (x,y,z). Suppose, a vector field F satisfies these properties.
i) F = F.
∇×
ii) All three coordinates of F are entire functions, and
∞
F(x) = M (x),
ℓ
ℓ=2
X
where M (x) is an ℓ-homogeneous polynomial component of F(x).
ℓ
iii) if Γ is the group in consideration, and if F is treated as a map R3 R3, then ǫ 1
−
7→ ◦
F ǫ = F for any ǫ Γ. Of course, this implies that ǫ 1 M ǫ = M for any ǫ Γ
− ℓ ℓ
◦ ∈ ◦ ◦ ∈
and ℓ 2, too.
≥
iv) There exists ℓ 2 such that the collection M = (p ,p ,p ), up to a multiplication by
ℓ x y z
≥
a scalar, is defined by the second half of the property iii) uniquely.
5 G. Alkauskas
Then the field F is called lambent field with symmetry Γ. Corresponding term applies to a
flow generated by this field. If we drop the condition iv), the corresponding vector field is called
weakly lambent field with symmetry Γ.
The fields G = 1(F(x)+F( x)) andG = 1(F(x) F( x)) we call evenly-lambent field with
2 − 2 − −
symmetry Γ, andoddly-lambent field with symmetry Γ, respectively. These satisfy 2G = G,
∇ −
G = 0.
∇·
Thus, as the most relaxed version of these definitions, applicable to all dimensions, we
propose the following.
Definition 2. Let Γ ֒ GL(n,R) be an exact irreducible representation of a finite group.
→
Suppose, a vector field F satisfies these properties:
i) 2F = F, F = 0;
∇ − ∇·
ii) All coordinates of F are entire functions;
iii) if Γ is the group in consideration, and if F is treated as a map Rn Rn, then ǫ 1
−
7→ ◦
F ǫ = F for any ǫ Γ.
◦ ∈
Then such a vector field is called sparkling vector field with symmetry Γ.
Now we are in a position to formulate the main result of this paper.
Theorem 2. Let us define vector fields T = (T ,T ,T ) and O = (O ,O ,O ) as follows:
x y z x y z
T = zsiny +ysinz +xcosy xcosz,
x
−
T = xsinz +zsinx+ycosz ycosx, (2)
y
 −
T = ysinx+xsiny +zcosx zcosy,
 z
−
and

O = ysinz zsiny +xcosy 2sinx+xcosz,
x
− −
O = zsinx xsinz +ycosz 2siny +ycosx, (3)
y
 − −
O = xsiny ysinx+zcosx 2sinz +zcosy.
 z
− −
For the tetrahedral group (of order 12), a family of lambent fields is given by T+aO, a R.
 ∈
For the octahedral group (of order 24) the lambent vector field is given by O.
For the icosahedral group I (of order 60), a 1-parameter family of lambent vector fields is
given by I+aY, a R. Here I is given by Theorem 1, and Y is as follows.
∈
Let us define the vector field V0 = V0(x,y,z),V0(y,z,x),V0(z,x,y) , where its even part
x x x
is defined by
(cid:0) (cid:1)
x φy z x y φz φx y z
V0 = φxsin sin sin +φ2xsin sin sin xsin sin sin
x − 2 2 2φ 2φ 2 2 − 2 2φ 2
(cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17)
1+3√5 x φy z
φ2ysinz cos cos sin
− − 2 2 2 2φ
(cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17)
x y φz 5+√5 φx y z
φ3ycos cos sin ycos cos sin
− 2φ 2 2 − 2 2 2φ 2
(cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17)
x φy z
+ zsiny+z(3+2√5)cos sin cos
2 2 2φ
(cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17)
+ φ−1zcos φx sin y cos z + 5+√5zcos x sin y cos φz ,
2 2φ 2 2 2φ 2 2
(cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17)
Beltrami vector fields 6
and its odd part W0 = W0(x,y,z),W0(y,z,x),W0(z,x,y) = W0. Then the vector field
x x x ∇×
Y = V0 +W0 has an icosahedral symmetry, and satisfies Y = Y. The Taylor series for
(cid:0) ∇(cid:1)×
Y starts from the degree 10, and is given by ̟0 , where ̟ is as in (10).
x 46448640 0
Concerning this, it is apt to pose the following problem.
Problem. Classify all lambent and sparkling vector fields. Describe dynamic and qualitative
properties of lambent flows.
In this paper we only present examples of lambent vector fields, based on the solution
to the Helhmolz equation which are of the form L(x,y,z)cos(l(x,y,z)), where L and l are
linear forms; see Lemma 1. However, for any homogeneous harmonic polynomial H(y,z) in
two variables, the function H(y,z)cos(x) also solves the Helhmolz equation. There are two
linearly independent harmonic polynomials of each degree d 1, and this is in accordance
≥
with Lemma 1, where only the case d = 1 is considered. Thus, the variety of lambent vector
field is very likely much richer than found here.
1.4. Background. A vector field B is said to be Beltrami vector field, if B ( B) = 0. So,
× ∇×
for a certain scalar function α, B = αB. In this paper we deal with a special important
∇×
case α 1.
≡
There are many solutions to the equation B = B to be found in the literature. For
∇ ×
example, it is known that if the vector field is of constant length 1, then in some Euclidean
coordinates one has
B = (sinz,cosz,0).
Another solution to B = B is the Lundquist solution given by, in cylindrical coordinates
∇×
(r,ϕ,z), (x,y,z) = (rcosϕ,rsinϕ,z), by
0,J (r),J (r) ,
1 0
(cid:16) (cid:17)
where J are the Bessel functions.
⋆
In [5], Section 7.2, the author considers related problem of finding solutions to the equation
B = B B, where B is vectors length. One of rational solutions to this is given by
∇× | | | |
8(xz y) ∂ 8(x+yz) ∂ 4(1+z2 x2 y2) ∂
− + + − − .
(1+x2 +y2 +z2)2∂x (1+x2 +y2+z2)2∂y (1+x2 +y2 +z2)2 ∂z
Vector fields
b = (Asinz +Ccosy,Bsinx+Acosz,Csiny +Bcosx)
give rise to flows which are known as Arnold-Beltrami-Childress (ABC) flows. They satisfy
b = b, and have many profound dynamical properties. Symmetry-related questions of
∇ ×
these are discussed in [11].
7 G. Alkauskas
2. Other 3-dimensional cases
2.1. The octahedral case. Let us define
0 1 0 0 0 1 1 0 0
−
α 1 0 0 , β 0 1 0 , γ 0 0 1 . (4)
7→   7→  −  7→  
0 0 1 1 0 0 0 1 0
−
     
These matrices generate the octahedral group O of order 24.
Consider the vector field
yz zx xy
M = (y2 z2), (z2 x2), (x2 y2) .
4
6 − 6 − 6 −
(cid:16) (cid:17)
This vector field has a symmetry group O, and is the unique, up to multiplication by a scalar,
vector field of degree 4 with this property (see [1], Subsection 5.1). As we will shortly see, it
gives rise to the octahedral lambent vector field O.
We have
x3 x y3 y z3 z
M = M = (y2+z2), (z2 +x2), (x2 +y2) . (5)
3 4
∇× 3 − 2 3 − 2 3 − 2
(cid:16) (cid:17)
This vector field has the full octahedral symmetry O of order 48, obtained fromO by adjoining
I. The curious fact is that M does not satisfy the condition iv) of the definition, though
3
−
M does. Indeed, all vector fields b
4
ax3 bx(y2 +z2),ay3 by(z2 +x2),az3 bz(x2 +y2)
− − −
(cid:16) (cid:17)
have the full octahedral symmetry, and this is a 2-parameter family, rather than 1-parameter,
as required by the property iv). However, only for 3a = 2b it is solenoidal.
Now, since M = 0, we may act completely analogously as in the tetrahedral case.
3
∇ ×
This, for example, gives
( 1)ℓx 2x2ℓ ( 1)ℓy 2y2ℓ ( 1)ℓz 2z2ℓ
M2ℓ+1 = −(2ℓ)! y2ℓ+z2ℓ− 2ℓ+1 , −(2ℓ)! z2ℓ+x2ℓ− 2ℓ+1 , −(2ℓ)! x2ℓ+y2ℓ− 2ℓ+1 !;
(cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17)
( 1)ℓyz ( 1)ℓzx ( 1)ℓxy
M2ℓ+2 = (2−ℓ+1)!(z2ℓ−y2ℓ),(2−ℓ+1)!(x2ℓ−z2ℓ),(2−ℓ+1)!(y2ℓ−xℓ)!.
This is also valid for ℓ = 0, since M = M = 0. So, if we define again V = ∞ M ,
2 1 2ℓ+2
ℓ=0
W = ∞ M , then P
2ℓ+1
ℓ=0
P
V = ysinz zsiny,zsinx xsinz,xsiny ysinx ,
− − −
(cid:16) (cid:17)
W = xcosy 2sinx+xcosz,ycosz 2siny +ycosx,zcosx 2sinz +zcosy ,
− − −
O = V(cid:16) +W = (V ,V ,V )+(W ,W ,W ) = (O ,O ,O ). (cid:17)
x y z x y z x y z
Thus, the octahedral lambent vector field is given by (3), which gives rise to the octahedral
lambent flow.
Beltrami vector fields 8
2.2. The icosahedral case. As already defined, let φ = 1+√5, and define
2
1 φ 1
1 0 0 0 0 1
α = −0 1 0 , β = 1 0 0 , γ = φ2 −12 2φ1 . (6)
 −     2 2φ −2
0 0 1 0 1 0 1 1 φ
2φ 2 2 
   
 
These three matrices generate the icosahedral group I of order 60. Let M = (̟,̺,σ), where
6
̟ =(5 √5)yz5+(5+√5)y5z 20y3z3+(10+10√5)x2yz3+(10 10√5)x2y3z 10x4yz,
̺=(5 −√5)zx5+(5+√5)z5x −20z3x3+(10+10√5)y2zx3+(10 −10√5)y2z3x −10y4zx, (7)
 σ =(5−√5)xy5+(5+√5)x5y−20x3y3+(10+10√5)z2xy3+(10−10√5)z2x3y−10z4xy.
− − − −
These are just numerators of a vector field for the icosahedral projective superflow [2]. One
has ζ 1 M ζ = M for any ζ I, and this is the unique, up to the scalar multiple,
− 6 6
◦ ◦ ∈
6-homogeneous vector field with this property. Calculations show that M has a full
6
∇ ×
icosahedral symmetry, and ( M ) = 0.
6
∇× ∇×
Based on the previous two examples, we will look for the icosahedral lambent vector field
I = (I ,I ,I ) in an analogous form.
x y z
First, we will find evenly-lambent icosahedral vector field. Namely, we will try to find
constants a , and linear forms L , k , such that
i i i
G = V = a cos(k )+L sin(k ) . (8)
x i i i i
i
X(cid:0) (cid:1)
Two other coordinates V and V are obtained from G by a cyclic permutation.
y z
First,weknowthatV = (V ,V ,V )isinvariantunderconjugationwithα = diag( 1, 1,1),
x y z
− −
βαβ2 = diag(1, 1, 1), and β2αβ = diag( 1,1, 1). These four matrices, together with the
− − − −
unity I, produce the Klein’s fourth group K < I. This gives
G( x, y,z) = G(x,y,z), G(x, y, z) = G(x,y,z).
− − − − −
As the first three linear forms k , we just take x,y,z. We do not need x, y and z since sin
i
− − −
and cos are, respectively, an odd and an even function. Invariance under K gives, respectively,
the following sum as a part of V in (8):
x
azsiny +bysinz.
Next, let us define the following 12 linear forms and vectors as follows.
ℓ (x,y,z) = 1x+ φy + 1 z, j = 1, φ, 1
x,0 2 2 2φ x,0 2 2 2φ
ℓ (x,y,z) = 1x+ φy + 1 z, j = (cid:16) 1, φ,(cid:17)1
x,1 −2 2 2φ x,1 − 2 2 2φ
ℓ (x,y,z) = 1x φy + 1 z, j = (cid:16)1, φ, 1 (cid:17),
x,2 2 − 2 2φ x,2 2 −2 2φ
ℓ (x,y,z) = 1x+ φy 1 z, j = (cid:16)1, φ, 1 (cid:17).
x,3 2 2 − 2φ x,3 2 2 −2φ
(cid:16) (cid:17)
Similarly we define 8 other linear functions and vectors by cyclically permuting variables: for
example, ℓ = 1y φz + 1 x, and so on. Orthogonality of the matrix γ tells, for example,
y,2 2 − 2 2φ
9 G. Alkauskas
that j ,j and i areorthonormal vectors. All these relationsamount tothe same identities
x,2 z,1 y,0
φ2+φ 2+1 = 4 (unit length), or φ φ 1 1 = 0 (orthogonality). This gives 15 linear forms
− −
− −
in total, and this is our complete collection in (8).
Next, if the coordinate of the form (8) is invariant under conjugating with K, it is of the
form
G = azsiny+bysinz (9)
+ ccosℓ ccosℓ ccosℓ +ccosℓ
x,0 x,3 x,2 x,1
− −
+ K(x,y,z)sinℓ +K( x, y,z)sinℓ +K( x,y, z)sinℓ K(x, y, z)sinℓ
x,0 x,3 x,2 x,1
− − − − − − −
+ dcosℓ dcosℓ dcosℓ +dcosℓ
y,0 y,2 y,1 y,3
− −
+ L(x,y,z)sinℓ +L( x, y,z)sinℓ +L( x,y, z)sinℓ L(x, y, z)sinℓ
y,0 y,2 y,1 y,3
− − − − − − −
+ ecosℓ ecosℓ ecosℓ +ecosℓ
z,0 z,1 z,3 z,2
− −
+ M(x,y,z)sinℓ +M( x, y,z)sinℓ +M( x,y, z)sinℓ M(x, y, z)sinℓ .
z,0 z,1 z,3 z,2
− − − − − − −
Here a,b,c,d,e are arbitrary constants, and K,L,M are arbitrary linear forms.
Now, recall that for an evenly lambent vector field, 2G = G. Next, the following Lemma
∇ −
is immediate.
Lemma 1. Let a and b be two 3-dimensional vectors-rows, and x = (x,y,z)T. The function
axsin(bx) is a solution to the Helmholtz equation if and only if a,b = 0, and b = 1. The
h i | |
same holds for the cos function.
By a direct inspection, two vectors orthogonal to j are given by j and j . So,
x,0 z,2 y,1
K(x,y,z) = fℓ +gℓ ,
z,2 y,1
L(x,y,z) = hℓ +iℓ ,
x,2 z,1
M(x,y,z) = jℓ +kℓ .
y,2 x,1
We therefore have 11 free coefficients at our disposition.
Such a function G(x,y,z) is invariant under conjugation with K and satisfies the Helmholtz
equation. We will reduce the amount of free coefficients by requiring that a vector field has
an icosahedral symmetry.
2.3. Computation. Now, we will proceed with determining 11 free coefficients.
i) First, we will require that the Taylor coefficient of G of degree 6 is exactly equal to ̟,
and Taylor coefficients of degrees 5 are absent. This gives, with computer calculations per-
≤
formed on MAPLE, 6 linear relations among 11 constants. The free constants turn out to be
a,b,c,d and f, while the remaining 6 constants e,g,h,i,j,k are linear functions in a,b,c,d,f.
This way we obtain a 5-parameter function G, such that its Taylor series starts with ̟, it is
invariant under conjugation with K, and it satisfies the Helhmholz equation.
ii) As a next step, we require that G(x,y,z),G(y,z,x),G(z,x,y) = 0. This gives two
∇ ·
more relations, leaving a,b, and d as free coefficients.
(cid:0) (cid:1)
Beltrami vector fields 10
iii) Finally, a vector field H = G(x,y,z),G(y,z,x),G(z,x,y) has the tetrahedral symmetry:
it is invariant under conjugation with α and β, as in (6). If it is invariant under conjugation
(cid:0) (cid:1)
with γ, it has the icosahedral symmetry, and the problem is solved. We have:
H = γ 1 G(x,y,z),G(y,z,x),G(z,x,y) γ(x,y,z)
−
◦ ◦
(cid:16) (cid:17)
= γ 1 G(ℓ ,ℓ ,ℓ ),G(ℓ ,ℓ ,ℓ ),G(ℓ ,ℓ ,ℓ )
− x,2 z,1 y,0 z,1 y,0 x,2 y,0 x,2 z,1
◦
:= γ 1(A(cid:16),B,C). (cid:17)
−
Therefore, we finally require that
A φB C
+ + = G(x,y,z).
2 2 2φ
This gives two more relations, and we thus obtained a 1-parameter family (discarding the ho-
mothety, which in this case gives another parameter as a scaling factor, which we discard by
the property iv) of the definition) of icosahedral lambent vector fields, with a free parameter
being a. This comes as a surprise! For any choice of a we obtain a solution.
We will get a particularly symmetric example due to the following observation. Let η be
given by (1). Note that if we conjugate the vector field (̟,̺,σ) with respect to η, and
then apply the non-trivial automorphism τ of the number field Q(√5) (which exchanges
√5 √5), we arrive at the same vector field (̟,̺,σ). Now, suppose all the constants
↔ −
a through k in G(x,y,z),G(y,z,x),G(z,x,y) , where the first coordinate is given by (9), are
elements from Q(√5), and an invariance under the just described transformation holds. This
(cid:0) (cid:1)
gives the equation a = τ b. However, the last of the 10 linear equations we obtained reads as
3 √5
b = 1920 a a+384√5.
− 2 − 2
Letuschoosea = 384 2. Thisgivesb = 384 2,andthusbothrequirements arereconciled. This
· ·
yields c = d = e = 0, f = 384φ 1, g = 384φ, h = 384φ, i = 384 1, j = 384 1, k = 384φ 1.
− −
− · − ·
Thus, if we start from a collection of the coefficients (2,2,0,0,0,φ 1, φ,φ,1, 1,φ 1), we
− −
− −
arrive at the vector field 1 (̟,̺,σ), where (̟,̺,σ) is given by (7). Now, (9) gives
384
G = 2zsiny+2ysinz
+ ( x+y z)sinℓ +(x y z)sinℓ +(x+y+z)sinℓ ( x y+z)sinℓ
x,0 x,3 x,2 x,1
− − − − − − −
+ φx y sinℓ + φx+y sinℓ + φx y sinℓ φx+y sinℓ
y,0 y,2 y,1 y,3
− − − − −
+ φ 1x+z sinℓ + φ 1x+z sinℓ + φ 1x z sinℓ φ 1x z sinℓ .
(cid:0) − (cid:1) z(cid:0),0 − (cid:1) z(cid:0),1 − (cid:1) z,(cid:0)3 (cid:1)− z,2
− − − − −
Finally, l(cid:0)et us collect(cid:1) all funct(cid:0)ions in ea(cid:1)ch row as(cid:0) factors (cid:1)of x,y,z, (cid:0)and use trig(cid:1)onometric
addition formulas. For example,
x φy z
sinℓ +sinℓ +sinℓ +sinℓ = 4xsin sin sin .
− x,0 x,3 x,2 x,1 2 2 2φ
(cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17)
Thus we get (in fact, MAPLE does it for us) the first displayed formula in Theorem 1, where
(V ,V ,V ) = 1(G(x,y,z),G(y,z,x),G(z,x,y)). For the first coordinate of oddly-lambent
x y z 2
vectorfield, wehaveW = ∂Vz ∂Vy, andthisgivestheseconddisplayedformulainTheorem1.
x ∂y − ∂z