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Generalizing the Hairy Ball Thorem - Higher Dimensions and Higher-Order Tensors

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lpetrich
#1
Nov14-12, 12:09 AM
P: 530
Hairy ball theorem - Wikipedia is not as good or as well-referenced as I'd hoped, and it mainly discusses vector fields on the 2-sphere, the ordinary sort of sphere.

In particular, it does not mention the minimum number of zero points of a continuous vector field on a sphere. I would guess that it's 2, or more generally,
|Euler characteristic|

Sphere: 2, torus: 0, etc.

There's also the question of generalization to higher-order tensors. What's the minimum number of zero points for a 2-tensor field? A 3-tensor one? A 4-tensor one? Etc.

A torus is topologically equivalent to a rectangle with periodic boundary conditions. It's easy to show that it's possible to construct an everywhere-nonzero tensor of any order -- all one needs to do is construct a constant one.

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To simplify this problem a bit, we ought to consider irreducible tensors. The 2-tensor (a)*(metric) is reducible into the scalar a, for instance. These are connected with the irreducible representations of each surface point's neighborhood's symmetry group.

For a real-valued n-surface, this group is SO(n), and for a complex-valued one, SU(n). The surface need not have that global symmetry; only that local symmetry.

For SO(n), every irreducible tensor is traceless, because a nonzero trace would enable separating out a tensor with a 2-subtracted order. For SU(n), tensors do not have that constraint.

Strictly speaking, for SO(n), we ought to also consider spinors and tensor-spinors. The tensor parts of the latter are also traceless.

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For real 2-space and 3-space, the irreducible tensors are all symmetric. Order-m tensors are composed of these numbers of basis tensors:
2D: m = 0 -> 1, m > 0 -> 2
3D; 2m+1
For 2D, the tensors can be related to Chebyshev polynomials, while for 3D, the tensors can be related to spherical harmonics.

Irreducible tensors need not be symmetric for at least 4 real dimensions or for complex dimensions.

In Lie-algebra highest-weight notation, the highest-weight vector for a symmetric (traceless) m-tensor is m * that for a vector.

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Applications of these tensor constructs?

Electromagnetic-wave polarization can be expressed as Stokes parameters composed from outer products of the electric-field vector with itself or a phase-shifted version of itself. Since electric fields of electromagnetic waves are perpendicular to the direction of motion, they thus span a 2-space. The Stokes parameters are thus
I - overall intensity - scalar
V - circular polarization - scalar
Q, U - linear polarization - symmetric traceless 2-tensor

One can make the same construction for gravitational radiation, where one starts with the metric-distortion ST 2-tensor. Its counterparts of the Stokes parameters are
I - overall intensity - scalar
V - circular polarization - scalar
Q, U - (bi)linear polarization - ST 4-tensor

So the hairy-ball theorem extended to tensor fields may imply that the Cosmic Microwave Background may have spots with zero linear polarization, and likewise for its gravitational-radiation counterpart.
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mathwonk
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Nov14-12, 08:42 AM
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the minimum number of zero points of a tangent vector field on an ordinary 2 sphere is one, but it has "index" 2.
lavinia
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Nov14-12, 04:57 PM
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Quote Quote by mathwonk View Post
the minimum number of zero points of a tangent vector field on an ordinary 2 sphere is one, but it has "index" 2.
Intuitively i tseems that you could coalesce a finite number of isolated zeros of a vector field into a single zero just by pushing them together.

lavinia
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Nov14-12, 07:36 PM
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Generalizing the Hairy Ball Thorem - Higher Dimensions and Higher-Order Tensors

For manifolds of Euler characteristic zero the answer is zero. This includes all odd dimensional manifolds. For manifolds of non-zero Euler characteristic the answer seems intuitively to be 1.

For higher order tensors I do not know why the zeros need to be isolated. Can you prove that?
lpetrich
#5
Nov15-12, 12:23 AM
P: 530
Quote Quote by mathwonk View Post
the minimum number of zero points of a tangent vector field on an ordinary 2 sphere is one, but it has "index" 2.
Quote Quote by lavinia View Post
Intuitively i tseems that you could coalesce a finite number of isolated zeros of a vector field into a single zero just by pushing them together.
So they'd be like multiple roots of a polynomial.

Wikipedia's article has a coinciding-zero case, and I've succeeded in finding an equation for it.

For position {x,y,z} with |position| = 1, it's
(1+z)*{1,0,-x} - x*{x,y,0}

As I'd expected, the zero at z = -1 is a double zero -- the field varies quadratically there and not linearly.

Quote Quote by lavinia View Post
For higher order tensors I do not know why the zeros need to be isolated. Can you prove that?
I don't think that I implied that, but if vectors can have coinciding zeros, then higher-order tensors can also have coinciding zeros.
mathwonk
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Nov15-12, 10:38 AM
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in local coordinates, tensor fields are just smooth maps between open sets of euclidean space. in order for the zeroes to be isolated, or to be finite in number say, one needs th dimension of the target to be equal to or greater than that of the source.

e.g. on the sphere the fact that tangent vector fields can have a finite number of zeroes, is due to the fact that the tangent space to a 2-sphere is a 2-plane. If we look at say alternating 2-tensors, or 2-forms, these map from the sphere into a tensor bundle whose fibers are lines (the second wedge product of a 2-plane is a line), so the zeroes must form at least curves in the sphere.

In general the bundle of k forms on a n manifold has fiber dimension equal to binomial coefficient "n choose k". So the 1 forms are always an n dimensional bundle but the n forms are always a line bundle.
lpetrich
#7
Nov16-12, 02:00 AM
P: 530
A possible problem with an alternating 2-form in a 2-surface: it is equal to a scalar multiplying the antisymmetric symbol:
Tij = T*eij

I've constructed irreducible (symmetric traceless) m-tensors from vectors on the real 2-sphere. For a vector v, construct its dual w by the cross product (r)x(v) for position r. Since r.v = r.w = 0, and using r.r = 1, w.w = v.v and v.w = 0.

There are two possible tensors, and they are constructed with s = 0 and s = 1 using:
T(s,v) = sum of all possible terms (-1)k * outer product of (m-s-2k) v's and (s+2k) w's
where 0 <= k <= (m-s)/2

These two tensors can also be constructed with
T(0,v) + i*T(1,v) = outer product of m copies of (v + i*w)
Not surprisingly, they are connected by a duality relation.

Their magnitudes, Tr(T(s,v).T(s,v)) = 2m-1 * (v.v)m
meaning that the upper limit of the minimum number of zeros is 2m.

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Going to higher dimensions in real space, one gets for a symmetric traceless m-tensor that its upper limit of its minimum number of zeros is m * (minimum number of zeros for a vector). For mixed-symmetry and antisymmetric m-tensors, one has to choose a linearly-independent set of vectors, and does linear independence create constraints on the possible zeros?

For complex space, one can construct a complex vector with v = vr + i*vi, where vr and vi are both real vectors. Since the zeros of vr and vi need not coincide, v can be everywhere nonzero. Likewise a symmetric m-tensor can be everywhere nonzero. For mixed symmetry and antisymmetry, the same possible-zero problem also occurs, however.
lpetrich
#8
Nov21-12, 07:46 AM
P: 530
For the case of a n-sphere embedded in a flat (n+1)-space, there are some simple solutions for vector fields.

I'll denote the position on the sphere in embedding space as vector r, with |r| = 1.

For the embedding space having an even number of dimensions, 2n dimensions, there is a simple solution for an always-nonzero field v in the sphere:

v = {r2, -r1, r4, -r3, ..., r2n, -r2n-1}

or more generally, a set of n disjoint interchanges of r components with one component in each interchange getting its sign reversed. I tried to construct sets of orthonormal vectors from these vectors, but I find 1 if n is odd and 3 is even, up to n = 5. I think that I can prove it more generally, however.

n = 1
{r2, -r1}
n = 2
{r2, -r1, r4, -r3}
{r3, -r4, -r1, r2}
{r4, r3, -r2, -r1}

even n: (n/2) copies of the n=2 one with appropriate indexing.

If one can construct at least n always-nonzero, linearly independent vectors, then that will mean that one can construct a tensor with arbitrary order and symmetry that will be nonzero elsewhere. But the above construction only makes it possible for n = 1 or 2.


This construction cannot be done for the embedding space having an odd number of dimensions. However, here is a simple vector field with only 2 zeros:

v = w - r*(w.r)
where w is a constant vector in embedding space. The zeros are r = +- w/|w|

For a 2-sphere, there's a related one, v = (r)x(w) also with 2 zeros. The zeros are also r = +- w/|w|

There are (n-1) linearly independent ones, and that means that a m-tensor with arbitrary symmetry will have at most 2m zeros.


Turning to the complex case, a (n-1)-sphere embedded in a flat n-space is closely analogous to a real (2n-1)-sphere in a flat 2n-space. So if one can show that it's possible to construct an arbitrary order and symmetry tensor that's always nonzero in the real space, then this result will carry over into the complex space.
mathwonk
#9
Nov21-12, 01:19 PM
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"...so the zeroes must form at least curves in the sphere."

I should have said "if non empty."


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