- #1
lpetrich
- 988
- 180
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.
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.
-
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.
-
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.
-
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.