BELTRAMI VECTOR FIELDS WITH POLYHEDRAL SYMMETRIES
arXiv:1701.04218v1 [math.CA] 16 Jan 2017
GIEDRIUS ALKAUSKAS Abstract. A vector field B is said to be Beltrami vector field (force free-magnetic vector field in physics), if B × (∇ × B) = 0. Motivated by our investigations on projective superflows, and as an important side result, we construct two unique 3-dimensional Beltrami vector fields I and Y, such that ∇ × I = I, ∇ × Y = Y, and that both have orientation-preserving icosahedral symmetry (group of order 60). Analogous constructions are done for tetrahedral and octahedral groups of orders 12 and 24, respectively. This construction leads to new notions of lambent vector field and lambent flow.
1. Lambent vector fields and flows 1.1. The main √ result. For a 3-dimensional vector field B let, as usual, ∇ × B = curl B, and 1+ 5 let us fix φ = 2 . One of the main results of this paper can be stated immediately. Theorem 1. Let us define the vector field V = Vx (x, y, z), Vx (y, z, x), Vx (z, x, y) , where Vx
=
+ +
z x y φz φx y z − 2φx sin sin sin + 2φ−1 x sin sin sin 2 2 2φ 2φ 2 2 2 2φ 2 x φy z x y φz y sin z + 2y cos cos sin − 2y cos cos sin 2 2 2φ 2φ 2 2 φy z φx y z x sin cos + 2z cos sin cos , z sin y − 2z cos 2 2 2φ 2 2φ 2 2x sin
x
sin
φy
sin
and the vector field W = Wx (x, y, z), Wx (y, z, x), Wx (z, x, y) , where Wx
=
− − −
x cos y − x cos z φy z x y φz φx y z x √ cos cos + φx cos cos cos + φ−1 x cos cos cos 5x cos 2 2 2φ 2φ 2 2 2 2φ 2 φy z x y φz √ y z x φx sin cos − φ2 y sin sin cos + 5y sin sin cos φ−2 y sin 2 2 2φ 2φ 2 2 2 2φ 2 x x φy z φx y z √ y φz 2 −2 φ z sin cos sin − φ z sin cos sin + 5z sin cos sin . 2 2 2φ 2 2φ 2 2φ 2 2
Then 1) The vector field I = (Ix , Iy , Iz ) = V + W has an icosahedral symmetry: if we treat I as a map R3 7→ R3 , for any ζ ∈ I (see Subsection 2.2) one has ζ −1 ◦ I ◦ ζ = I; 2) it satisfies ∇ × I = I; ̟ , where ̟ is given by 3) the Taylor series for Ix starts with a degree 6 form 768 ̟ = (5 −
√
5)yz 5 + (5 +
√
√ √ 5)y 5 z − 20y 3 z 3 + (10 + 10 5)x2 yz 3 + (10 − 10 5)x2 y 3 z − 10x4 yz.
Date: 15 January, 2017. 2010 Mathematics Subject Classification. Primary 37C10, 15Q31. Key words and phrases. Beltrami vector field, force-free magnetic field, regular polyhedra, Euler’s equation, 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, τ φ = −φ−1 . In the above formulas for Vx and Wx , let us expand everything in Taylor series, swap y and z, leave x intact, and apply τ summand-wise. We readily obtain τ Vx (x, z, y) = Vx (x, y, z),
τ Wx (x, z, y) = −Wx (x, y, z).
1.2. The tetrahedral case: motivation. Let us define 1 0 0 α 7→ 0 −1 0 , 0 0 −1
−1 0 0 β 7→ 0 1 0 , 0 0 −1
0 1 0 η 7→ 1 0 0 , 0 0 1
0 1 0 δ→ 7 0 0 1 . 1 0 0
(1)
These are matrices of order α2 = β 2 = η 2 = δ 3 = I. Together they generate the full tetraheb of order 24, and α, β and δ generate tetrahedral group T of order 12. In fact, dral group T b already α and δ generate T, and α, η and δ generate the whole group T.
A vector field M2 = (2yz, 2xz, 2xy) produces the tetrahedral projective superflow [1]. This is 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 b (see [1], Note 5). a bigger group T As we will see 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 ∇ · M2 = 0, there exist a vector field R = (rx , ry , rz ) such that ∇ × R = M2 . In expanded terms, this reads as ∂rz ∂ry ∂rx ∂rz ∂ry ∂rx yz = − , xz = − , xz = − . ∂y ∂z ∂z ∂x ∂x ∂y The general solution to these equations is given by ! ∂Q 1 ∂Q 1 ∂Q 1 , x(z 2 − y 2 ) + , y(x2 − z 2 ) + , z(y 2 − x2 ) + (rx , ry , rz ) = 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 ∇2 Q = Q = 0. + + ∂x2 ∂y 2 ∂z 2
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 M3 = x(z 2 − y 2 ), y(x2 − z 2 ), z(y 2 − x2 ) . 2 2 2 However, this is not the unique choice. We could also choose 1 1 1 (a) M3 = x(z 2 − y 2 ), y(x2 − z 2 ), z(y 2 − x2 ) 2 2 2 x3 x 3 y y z3 z − (y 2 + z 2 ), − (z 2 + x2 ), − (x2 + y 2 ) , a ∈ R. + a 3 2 3 2 3 2 Now, as the next step, we will perform the same inverse problem for M3 as we did for M2 . Namely, we will seek for a vector field M4 such that ∇ × M4 = M3 ,
∇ · M4 = 0,
and R = M4 satisfies the above three properties i), ii) and iii) for n = 4. And we indeed find 1 1 3 1 3 3 3 3 3 M4 = − (y z + yz ), − (z x + zx ), − (x y + xy ) . 6 6 6 This vector field has the same symmetry group as M2 . Continuing, we can take ! ℓ ℓ (−1) (−1) (−1)ℓ x(y 2ℓ − z 2ℓ ), y(z 2ℓ − x2ℓ ), z(x2ℓ − y 2ℓ ) , M2ℓ+1 = (2ℓ)! (2ℓ)! (2ℓ)! ! ℓ ℓ (−1)ℓ yz 2ℓ (−1) xz (−1) xy M2ℓ+2 = (y + z 2ℓ ), (z 2ℓ + x2ℓ ), (x2ℓ + y 2ℓ ) . (2ℓ + 1)! (2ℓ + 1)! (2ℓ + 1)! The above identities holds for ℓ = 0, with the correct value M1 = ∇ × M2 = 0. Let ∞ X M2ℓ+2 = z sin y + y sin z, x sin z + z sin x, y sin x + x sin y , V = W =
ℓ=0 ∞ X ℓ=0
M2ℓ+1 = x cos y − x cos z, y cos z − y cos x, z cos x − z cos y ,
T = V + W = (Vx , Vy , Vz ) + (Wx , Wy , Wz ) = (Tx , Ty , Tz ). 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 Further, for vector fields V and W one has: 1vw) ∇ × W = V, ∇ × V = W, ∇ · V = 0, ∇ · W = 0.
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Beltrami vector fields
b of order 24 as the group of its symmetries; W has the group T × 2vw) V 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). 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′ = Tx ,
y ′ = Ty ,
z ′ = Tz .
Let x = (x, y, z). We call the vector field T the tetrahedral lambent field, and the corresponding flow FT (x; t) = (x(t), y(t), z(t)), FT (x; 0) = x - the tetrahedral lambent flow. Further, we call 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 · Wx + ∂x ∂W · Wz = 0. We see that two independent solutions are given by ∂z Z ∞ cos t dt (1) (2) W = xyz, and W = Ci(x) + Ci(y) + Ci(z), Ci(x) = − . t x
∂W ∂y
· Wy +
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 (rx , ry , rz ) to (∇2 rx , ∇2 ry , ∇2 rz ). If B is any smooth vector field, then we have a standard identity of vector calculus ∇ × (∇ × B) = ∇(∇ · B) − ∇2 B.
This, applied to V and W, gives
V = −∇2 V,
W = −∇2 W.
Thus, any coordinate c of V or W satisfies a (scalar) Helmholtz equation c = −∇2 c, which is immediate to check. In the other direction: if B = −∇2 B, 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) =
∞ X
Mℓ (x),
ℓ=2
where Mℓ (x) is an ℓ-homogeneous polynomial component of F (x). iii) if Γ is the group in consideration, and if F is treated as a map R3 7→ R3 , then ǫ−1 ◦ F ◦ ǫ = F for any ǫ ∈ Γ. Of course, this implies that ǫ−1 ◦ Mℓ ◦ ǫ = Mℓ for any ǫ ∈ Γ and ℓ ≥ 2, too. iv) There exists ℓ ≥ 2 such that the collection Mℓ = (px , py , pz ), up to a multiplication by a scalar, is defined by the second half of the property iii) uniquely.
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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 = 21 (F (x)+F (−x)) and G = 12 (F (x)−F (−x)) we call evenly-lambent field with symmetry Γ, and oddly-lambent field with symmetry Γ, respectively. These satisfy ∇2 G = −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) ∇2 F = −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 7→ Rn , then ǫ−1 ◦ 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 = (Tx , Ty , Tz ) and O = (Ox , Oy , Oz ) as follows: Tx = z sin y + y sin z + x cos y − x cos z, Ty = x sin z + z sin x + y cos z − y cos x, (2) Tz = y sin x + x sin y + z cos x − z cos y, and
Ox = y sin z − z sin y + x cos y − 2 sin x + x cos z, Oy = z sin x − x sin z + y cos z − 2 sin y + y cos x, Oz = x sin y − y sin x + z cos x − 2 sin z + z cos y.
(3)
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 = V0x (x, y, z), V0x (y, z, x), V0x (z, x, y) , where its even part is defined by V0x
= − − + +
x
z x y φz φx y z + φ2 x sin sin sin − x sin sin sin 2 2 2φ 2φ 2 2 2 2φ 2 √ φy z x 1+3 5 cos sin cos φ2 y sin z − 2 2 2 2φ y φz 5 + √5 y z x φx 3 cos sin − cos sin φ y cos y cos 2φ 2 2 2 2 2φ 2 φy z x √ sin cos z sin y + z(3 + 2 5) cos 2 2 2φ φx y z 5 + √5 y φz x φ−1 z cos sin cos + sin cos , z cos 2 2φ 2 2 2φ 2 2
−φx sin
sin
φy
sin
Beltrami vector fields
6
and its odd part W0 = W0x (x, y, z), W0x (y, z, x), W0x (z, x, y) = ∇ × W0 . Then the vector field Y = V0 + W0 has an icosahedral symmetry, and satisfies ∇ × Y = Y. The Taylor series for ̟0 , where ̟0 is as in (10). Yx starts from the degree 10, and is given by 46448640 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 = (sin z, cos z, 0). Another solution to ∇ × B = B is the Lundquist solution given by, in cylindrical coordinates (r, ϕ, z), (x, y, z) = (r cos ϕ, r sin ϕ, z), by 0, J1 (r), J0(r) ,
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(x + yz) 4(1 + z 2 − x2 − y 2 ) ∂ ∂ ∂ 8(xz − y) + + . (1 + x2 + y 2 + z 2 )2 ∂x (1 + x2 + y 2 + z 2 )2 ∂y (1 + x2 + y 2 + z 2 )2 ∂z Vector fields b = (A sin z + C cos y, B sin x + A cos z, C sin y + B cos x) 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].
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G. Alkauskas
2. Other 3-dimensional cases 2.1. The octahedral 0 α 7→ 1 0
case. Let us define 1 0 0 0 1 0 0 , β → 7 0 −1 0 , 0 −1 1 0 0
−1 0 0 γ 7→ 0 0 1 . 0 1 0
(4)
These matrices generate the octahedral group O of order 24. Consider the vector field
zx 2 2 xy 2 2 (y − z ), (z − x ), (x − y ) . M4 = 6 6 6 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
yz
2
2
x y3 y z3 z − (y 2 + z 2 ), − (z 2 + x2 ), − (x2 + y 2 ) . (5) 3 2 3 2 3 2 b of order 48, obtained from O by adjoining This vector field has the full octahedral symmetry O −I. The curious fact is that M3 does not satisfy the condition iv) of the definition, though M4 does. Indeed, all vector fields ax3 − bx(y 2 + z 2 ), ay 3 − by(z 2 + x2 ), az 3 − bz(x2 + y 2 ) M3 = ∇ × M4 =
x3
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 ∇ × M3 = 0, we may act completely analogously as in the tetrahedral case. This, for example, gives M2ℓ+1
=
M2ℓ+2
=
! (−1)ℓ x 2ℓ 2x2ℓ (−1)ℓ y 2ℓ 2y 2ℓ (−1)ℓ z 2ℓ 2z 2ℓ ; y + z 2ℓ − , z + x2ℓ − , x + y 2ℓ − (2ℓ)! 2ℓ + 1 (2ℓ)! 2ℓ + 1 (2ℓ)! 2ℓ + 1 ! (−1)ℓ yz 2ℓ (−1)ℓ zx 2ℓ (−1)ℓ xy 2ℓ 2ℓ 2ℓ ℓ (z − y ), (x − z ), (y − x ) . (2ℓ + 1)! (2ℓ + 1)! (2ℓ + 1)!
This is also valid for ℓ = 0, since M2 = M1 = 0. So, if we define again V = W=
∞ P
∞ P
M2ℓ+2 ,
ℓ=0
M2ℓ+1 , then
ℓ=0
y sin z − z sin y, z sin x − x sin z, x sin y − y sin x , W = x cos y − 2 sin x + x cos z, y cos z − 2 sin y + y cos x, z cos x − 2 sin z + z cos y , V =
O = V + W = (Vx , Vy , Vz ) + (Wx , Wy , Wz ) = (Ox , Oy , Oz ).
Thus, the octahedral lambent vector field is given by (3), which gives rise to the octahedral lambent flow.
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Beltrami vector fields 2.2. The icosahedral −1 α= 0 0
√
case. As already defined, let φ = 1+2 5 , and define φ 1 1 − 0 0 0 0 1 2 2 2φ 1 − 21 . −1 0 , β = 1 0 0 , γ = φ2 2φ φ 1 1 0 1 0 1 0 2φ
2
(6)
2
These three matrices generate the icosahedral group I of order 60. Let M6 = (̟, ̺, σ), where √ √ √ √ ̟ = (5 −√ 5)yz 5 + (5 + √ 5)y 5 z − 20y 3 z 3 + (10 + 10√ 5)x2 yz 3 + (10 − 10√ 5)x2 y 3 z − 10x4 yz, ̺ = (5 − √5)zx5 + (5 + √5)z 5 x − 20z 3x3 + (10 + 10 √5)y 2 zx3 + (10 − 10 √5)y 2 z 3 x − 10y 4zx, σ = (5 − 5)xy 5 + (5 + 5)x5 y − 20x3 y 3 + (10 + 10 5)z 2 xy 3 + (10 − 10 5)z 2 x3 y − 10z 4 xy.
(7)
These are just numerators of a vector field for the icosahedral projective superflow [2]. One has ζ −1 ◦ M6 ◦ ζ = M6 for any ζ ∈ I, and this is the unique, up to the scalar multiple, 6-homogeneous vector field with this property. Calculations show that ∇ × M6 has a full icosahedral symmetry, and ∇ × (∇ × M6 ) = 0. Based on the previous two examples, we will look for the icosahedral lambent vector field I = (Ix , Iy , Iz ) in an analogous form. First, we will find evenly-lambent icosahedral vector field. Namely, we will try to find constants ai , and linear forms Li , ki , such that X G = Vx = ai cos(ki ) + Li sin(ki ) . (8) i
Two other coordinates Vy and Vz are obtained from G by a cyclic permutation.
First, we know that V = (Vx , Vy , Vz ) is invariant under conjugation with α = diag(−1, −1, 1), βαβ 2 = diag(1, −1, −1), and β 2 αβ = 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 ki , we just take x, y, z. We do not need −x, −y and −z since sin and cos are, respectively, an odd and an even function. Invariance under K gives, respectively, the following sum as a part of Vx in (8): az sin y + by sin z. Next, let us define the following 12 linear forms and vectors as follows. φ 1 1 1 φ 1 ℓx,0(x, y, z) = 2 x + 2 y + 2φ z, jx,0 = 2 , 2 , 2φ 1 1 ℓx,1(x, y, z) = − 12 x + φ2 y + 2φ z, jx,1 = − 12 , φ2 , 2φ 1 1 , z, jx,2 = 12 , − φ2 , 2φ ℓx,2(x, y, z) = 21 x − φ2 y + 2φ 1 1 . z, jx,3 = 12 , φ2 , − 2φ ℓx,3(x, y, z) = 21 x + φ2 y − 2φ
Similarly we define 8 other linear functions and vectors by cyclically permuting variables: for 1 example, ℓy,2 = 21 y − φ2 z + 2φ x, and so on. Orthogonality of the matrix γ tells, for example,
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G. Alkauskas
that jx,2 , jz,1 and iy,0 are orthonormal vectors. All these relations amount to the same identities 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 = + + + + + +
az sin y + by sin z (9) c cos ℓx,0 − c cos ℓx,3 − c cos ℓx,2 + c cos ℓx,1 K(x, y, z) sin ℓx,0 + K(−x, −y, z) sin ℓx,3 + K(−x, y, −z) sin ℓx,2 − K(x, −y, −z) sin ℓx,1 d cos ℓy,0 − d cos ℓy,2 − d cos ℓy,1 + d cos ℓy,3 L(x, y, z) sin ℓy,0 + L(−x, −y, z) sin ℓy,2 + L(−x, y, −z) sin ℓy,1 − L(x, −y, −z) sin ℓy,3 e cos ℓz,0 − e cos ℓz,1 − e cos ℓz,3 + e cos ℓz,2 M (x, y, z) sin ℓz,0 + M (−x, −y, z) sin ℓz,1 + M (−x, y, −z) sin ℓz,3 − M (x, −y, −z) sin ℓz,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, ∇2 G = −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 ax sin(bx) is a solution to the Helmholtz equation if and only if ha, bi = 0, and |b| = 1. The same holds for the cos function. By a direct inspection, two vectors orthogonal to jx,0 are given by jz,2 and jy,1 . So, K(x, y, z) = f ℓz,2 + gℓy,1 , L(x, y, z) = hℓx,2 + iℓz,1 , M(x, y, z) = jℓy,2 + kℓx,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 performed 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.
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 with γ, it has the icosahedral symmetry, and the problem is solved. We have: −1 H = γ ◦ G(x, y, z), G(y, z, x), G(z, x, y) ◦ γ(x, y, z) = γ −1 ◦ G(ℓx,2 , ℓz,1, ℓy,0 ), G(ℓz,1 , ℓy,0 , ℓx,2), G(ℓy,0 , ℓx,2 , ℓz,1) := γ −1 (A, B, C).
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 homothety, 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 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 Let us choose a = 384·2. This gives b = 384·2, and thus both requirements are reconciled. 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 1 arrive at the vector field 384 (̟, ̺, σ), where (̟, ̺, σ) is given by (7). Now, (9) gives G = 2z sin y + 2y sin z + (−x + y − z) sin ℓx,0 + (x − y − z) sin ℓx,3 + (x + y + z) sin ℓx,2 − (−x − y + z) sin ℓx,1 + φx − y sin ℓy,0 + − φx + y sin ℓy,2 + − φx − y sin ℓy,1 − φx + y sin ℓy,3 + − φ−1 x + z sin ℓz,0 + φ−1 x + z sin ℓz,1 + φ−1 x − z sin ℓz,3 − − φ−1 x − z sin ℓz,2 .
Finally, let us collect all functions in each row as factors of x, y, z, and use trigonometric addition formulas. For example, φy z x sin sin . − sin ℓx,0 + sin ℓx,3 + sin ℓx,2 + sin ℓx,1 = 4x sin 2 2 2φ
Thus we get (in fact, MAPLE does it for us) the first displayed formula in Theorem 1, where (Vx , Vy , Vz ) = 21 (G(x, y, z), G(y, z, x), G(z, x, y)). For the first coordinate of oddly-lambent y z − ∂V , and this gives the second displayed formula in Theorem 1. vector field, we have Wx = ∂V ∂y ∂z
11
G. Alkauskas
As we now know, any icosahedral lambent vector field of the form (9) is given by I + a′ Y, a ∈ R. Calculations show that Y is obtained by specializing φ−1 φ3 φ φ2 1 φ (a, b, c, d, e, f, g, h, i, j, k) = 1, −φ2 , 0, 0, 0, , , ,− ,− ,− . 2 2 2 2 2 2 ̟0 Taylor series of Yx starts at the degree 10, and is given by 46448640 , where ′
̟0
= −
+
√ √ √ (10) (9 + 9 5)x8 yz − (252 + 84 5)x6 y 3 z + 168x6 yz 3 + (378 + 126 5)x4 y 5 z − 252x4 yz 5 √ 2 5 3 √ 2 3 5 √ 2 7 √ 2 7 (288 + 72 5)x y z + (1260 + 252 5)x y z − (1260 + 252 5)x y z + (252 + 36 5)x yz √ √ √ √ √ (8 − 2 5)y 9 z + 48 5y 7 z 3 − (126 + 126 5)y 5 z 5 + (120 + 72 5)z 7 y 3 − (17 + 7 5)z 9 y.
This gives, with the help of MAPLE, the formula in Theorem 2. References [1] G. Alkauskas, Projective superflows. I. http://arxiv.org/abs/1601.06570. [2] G. Alkauskas, Projective superflows. II. O(3) and the icosahedral group. http://arxiv.org/abs/1606.05772. [3] G. Alkauskas, Projective superflows. III. Finite subgroups of U (2). http://arxiv.org/abs/1608.02522. [4] G. Alkauskas, Projective superflows. IV. Arithmetic of the orbits (in preparation, 2017). [5] D. E. Blair, Riemannian geometry of contact and symplectic manifolds. Progress in Mathematics, 203. Birkh¨auser Boston, Inc., Boston, MA, (2002). [6] S. Chandrasekhar, P.C. Kendall, On force-free magnetic fields. Astrophys. J. 126 (1957), 457–460. [7] S. Chandrasekhar, On force-free magnetic fields. Proc. Nat. Acad. Sci. U.S.A. 42 (1956), 1–5. [8] S. Chandrasekhar, L. Woltjer, On force-free magnetic fields. Proc. Nat. Acad. Sci. U.S.A. 44, 1958, 285–289. [9] J. Etnyre, R. Ghrist, Robert, Contact topology and hydrodynamics. I. Beltrami fields and the Seifert conjecture. Nonlinearity 13(2) (2000) 441-458. [10] Q. Z. Feng, On force-free magnetic fields and Beltrami flows. Appl. Math. Mech. (English Ed.) 18(10) (1997), 997-1003. [11] P. Fre, A.S. Sorin, Classification of Arnold-Beltrami Flows and their Hidden Symmetries. https://arxiv.org/abs/1501.04604. [12] R. Hiptmair, P.R. Kotiuga, S. Tordeux, Self-adjoint curl operators. Ann. Mat. Pura Appl. (4), 191(3) (2012), 431–457. [13] R. Kaiser, M. Neudert, W. von Wahl, On the existence of force-free magnetic fields with small nonconstant α in exterior domains. Comm. Math. Phys. 211(1) (2000), 111-136. [14] V.V. Kravchenko, On force free magnetic fields. Quaternionic approach. Math. Methods Appl. Sci. 28(4) (2005), 79-386. [15] M.A. MacLeod, The spherical curl transform of a linear force-free magnetic field. J. Math. Phys. 39(3) (1998), 1642–1658. [16] E.C. Morse, Eigenfunctions of the curl in annular cylindrical and rectangular geometry. J. Math. Phys. 48(8) (2007). [17] E. R. Priest, Solar Magnetohydrodynamics. Geophysics and Astrophysics Monographs, Volume 21, Springer (1982). [18] K. Saygili, Trkalian fields and Radon transformation. J. Math. Phys. 51(3) (2010). [19] K. Saygili, Trkalian fields: ray transforms and mini-twistors. J. Math. Phys. 54(10) (2013). [20] E. Tassi, F. Pegoraro, G. Cicogna, Solutions and symmetries of force-free magnetic fields. Physics of Plasmas 15(9) (2008). [21] S.M. Mahajan, Z. Yoshida, Double Curl Beltrami Flow: Diamagnetic Structures. Phys. Rev. Lett. 81(22), 4863, November 1998. [22] S.M. Mahajan, Z. Yoshida, Simultaneous Beltrami conditions in coupled vortex dynamics. J. Math. Phys. 40(10), (1999), 5080-5091.
Beltrami vector fields
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[23] N.A. Salingaros, On solutions of the equation ∇ × a = ka. Journal of Physics A: Mathematical and General, 19(3), 1986. [24] T. Wiegelmann, T. Sakurai, Solar Force-free Magnetic Fields. Living Reviews in Solar Physics, 9(5), (2012). [25] L. Woltjer, A theorem on force-free magnetic fields. Proc. Nat. Acad. Sci. U.S.A. 44, 1958, 489-491. Vilnius University, Department of Mathematics and Informatics, Naugarduko 24, LT-03225 Vilnius, Lithuania E-mail address:
[email protected]
