A Coxeter diagram with N nodes, and labels ni,j on edges between nodes i and j, defines an N-dimensional vector space with dot product
ei•ei = 1,
ei•ej = - cos(π/ni,j).
If this dot product is positive-definite (signature ++...++), then the symmetry group is finite, and can be used to describe (Wythoffian) polytopes in N-dimensional Euclidean space, or tilings of (N-1)-dimensional spheric space.
If the dot product is positive-semidefinite with 1 dimension of degeneracy (signature ++...+0), then the symmetry group can be used to describe tilings of (N-1)-dimensional Euclidean space.
If the dot product is indefinite with signature ++...+-, then the symmetry group is hyperbolic.
What about other signatures? One example is o5o4o3o5o5o, which has signature ++++--. This doesn't match any of the previous types. In fact a 2D negative-definite subspace can't exist in a space with signature +++...+-, regardless of the number of '+' signs. So the polytope x5o4o3o5o5o doesn't fit in hyperbolic space of any dimension! At least, not with the expected symmetry.
(Hence the term "ultra-hyperbolic". Apparently Wikipedia calls it "Lorentzian", but I'm not sure it means exactly the same thing.)
Ultra-hyperbolic Coxeter groups
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Ultra-hyperbolic Coxeter groups
by mr_e_man » Mon Sep 20, 2021 4:13 am
ΓΔΘΛΞΠΣΦΨΩ αβγδεζηθϑικλμνξοπρϱσςτυϕφχψωϖ °±∓½⅓⅔¼¾×÷†‡• ⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎
ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩
ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩
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Re: Ultra-hyperbolic Coxeter groups
by wendy » Mon Sep 27, 2021 10:29 am
The normal form of the Dynkin matrix is to double all of the values you give, so \(e_{i,i} = 2 \quad e_{i,j} = -2 \cos ( \pi / n )\). This greatly simplifies calculations. The edge for these groups, when a given value g is taken, is to calculate the determinate with and without the value of g, divide these one to another, and then by 2.
The exact kind of polytope that arises can not be completely determined from the matrix alone, but some guide helps. The values given against the W number, represent the negative square of the value used in the Dynkin matrix, so x4o corresponds to xW2o, and the value used is \(-\sqrt{2}\)
If one colours the edges of an x8o4o, in alternating colours, you get separate 'red' and 'blue' lines, which are the crossings of x4oW6.828o, in much the same way that the same done to x4o4o gives x2oUo. The vertex figure of this is xW6,828, which corresponds to the girthing polygon of say {8,4}. In hyperbolic geometry, ultraideal points are represented by their equatorial plane, and where the ordinary polygon might come to meet at a vertex, this is the effect of a complex-vertex, or paravertex.
Symmetries like {8,3,4} occur, and these produce compact (piecewise finite) groups if several of them are allowed to interact. One of the (obvious) subgroups of this is {4,8,4}, which leads to that {8,3,4} can sit on its symmetry in two different ways. This is fairly common in H-space. For example, {5,3,4} is such, that if the dodecahedra are divided into eight by the three right angles, this merges 5 cells of symmetry of {5,3} against two cells of {3,4} into a symmetry that {5,3,4} can be restored onto it in two distinct ways.
When one has several complex points in the symmetry, these result in crossing planes (which can still have an angle). The mathematicians stop at 'paracompact' or finite-content groups. These are those groups, for which the removal of any node gives rise to a matrix whose determinate is 0 or positive. When one has a complete positive run, it is 'compact' or finite content, such as {5,3,4}, its curvature in this case is given by the overall determinate.
The exact kind of polytope that arises can not be completely determined from the matrix alone, but some guide helps. The values given against the W number, represent the negative square of the value used in the Dynkin matrix, so x4o corresponds to xW2o, and the value used is \(-\sqrt{2}\)
If one colours the edges of an x8o4o, in alternating colours, you get separate 'red' and 'blue' lines, which are the crossings of x4oW6.828o, in much the same way that the same done to x4o4o gives x2oUo. The vertex figure of this is xW6,828, which corresponds to the girthing polygon of say {8,4}. In hyperbolic geometry, ultraideal points are represented by their equatorial plane, and where the ordinary polygon might come to meet at a vertex, this is the effect of a complex-vertex, or paravertex.
Symmetries like {8,3,4} occur, and these produce compact (piecewise finite) groups if several of them are allowed to interact. One of the (obvious) subgroups of this is {4,8,4}, which leads to that {8,3,4} can sit on its symmetry in two different ways. This is fairly common in H-space. For example, {5,3,4} is such, that if the dodecahedra are divided into eight by the three right angles, this merges 5 cells of symmetry of {5,3} against two cells of {3,4} into a symmetry that {5,3,4} can be restored onto it in two distinct ways.
When one has several complex points in the symmetry, these result in crossing planes (which can still have an angle). The mathematicians stop at 'paracompact' or finite-content groups. These are those groups, for which the removal of any node gives rise to a matrix whose determinate is 0 or positive. When one has a complete positive run, it is 'compact' or finite content, such as {5,3,4}, its curvature in this case is given by the overall determinate.
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Re: Ultra-hyperbolic Coxeter groups
by mr_e_man » Fri Oct 01, 2021 5:41 am
Of course, scaling the dot product by any positive number, such as 2, doesn't change its signature.
I forgot to mention that ei denotes the i'th basis vector. It is nice to have unit vectors, rather than vectors with length √2. To reflect some vector v across the i'th mirror, we can use the geometric product -eivei, or equivalently the formula
v - 2(v•ei)ei.
(These are related by the identity vei + eiv = 2v•ei, along with ei2 = 1, and the associativity of the geometric product.) So the dot product is naturally being doubled anyway.
But when it's represented directly by a matrix, the factor of 2 is not there, so you need to insert it artificially to get nice algebraic-integer values.
The signature can be written in the form (+)P(-)Q(0)R, or just (P,Q,R), where P+Q+R=N. The determinant doesn't give much information; ignoring positive scale factors, it is exactly the product (+1)P(-1)Q(0)R; it only tells whether Q is odd or even, and that only if R=0. (This is some general background for other readers. I guess Wendy is familiar with most of this.)
Yet the determinants of sub-matrices can tell the full signature. Assuming that the space spanned by the first k vectors (k<N) is non-degenerate, the next vector ek+1 can be projected away from that k-dimensional subspace; call the result fk+1. Then fk+12 is the determinant of the matrix of dot products of the first k+1 vectors, divided by that of the first k vectors. In other words, the next term in the signature is given by the ratio of determinants of two sub-matrices. (Of course the first term is e12 = +1.) So, by looking at the sign changes in the sequence of determinants of a 1×1 sub-matrix, a 2×2 sub-matrix, a 3×3 sub-matrix, and so on to the full N×N matrix, we can figure out the signature (assuming we never get a 0).
"Complex points", also called "ultra-ideal" or "beyond infinity", are still what I consider to be hyperbolic. Such a point is represented by a vector with positive square, in a space with signature ++...+- . The finite hyperbolic points have negative square. Ideal points are null vectors.
By "ultra-hyperbolic" I mean a space with signature ++...-- ; with at least 2 minus signs. (I haven't decided whether degeneracy should be allowed.)
I forgot to mention that ei denotes the i'th basis vector. It is nice to have unit vectors, rather than vectors with length √2. To reflect some vector v across the i'th mirror, we can use the geometric product -eivei, or equivalently the formula
v - 2(v•ei)ei.
(These are related by the identity vei + eiv = 2v•ei, along with ei2 = 1, and the associativity of the geometric product.) So the dot product is naturally being doubled anyway.
But when it's represented directly by a matrix, the factor of 2 is not there, so you need to insert it artificially to get nice algebraic-integer values.
The signature can be written in the form (+)P(-)Q(0)R, or just (P,Q,R), where P+Q+R=N. The determinant doesn't give much information; ignoring positive scale factors, it is exactly the product (+1)P(-1)Q(0)R; it only tells whether Q is odd or even, and that only if R=0. (This is some general background for other readers. I guess Wendy is familiar with most of this.)
Yet the determinants of sub-matrices can tell the full signature. Assuming that the space spanned by the first k vectors (k<N) is non-degenerate, the next vector ek+1 can be projected away from that k-dimensional subspace; call the result fk+1. Then fk+12 is the determinant of the matrix of dot products of the first k+1 vectors, divided by that of the first k vectors. In other words, the next term in the signature is given by the ratio of determinants of two sub-matrices. (Of course the first term is e12 = +1.) So, by looking at the sign changes in the sequence of determinants of a 1×1 sub-matrix, a 2×2 sub-matrix, a 3×3 sub-matrix, and so on to the full N×N matrix, we can figure out the signature (assuming we never get a 0).
"Complex points", also called "ultra-ideal" or "beyond infinity", are still what I consider to be hyperbolic. Such a point is represented by a vector with positive square, in a space with signature ++...+- . The finite hyperbolic points have negative square. Ideal points are null vectors.
By "ultra-hyperbolic" I mean a space with signature ++...-- ; with at least 2 minus signs. (I haven't decided whether degeneracy should be allowed.)
ΓΔΘΛΞΠΣΦΨΩ αβγδεζηθϑικλμνξοπρϱσςτυϕφχψωϖ °±∓½⅓⅔¼¾×÷†‡• ⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎
ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩
ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩
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Re: Ultra-hyperbolic Coxeter groups
by mr_e_man » Mon Nov 01, 2021 10:23 pm
- Code: Select all
o
[2]
det = +2
- Code: Select all
o5o
⌈ 2 -φ⌉
⌊-φ 2⌋
det = +1.3820
- Code: Select all
o5o4o
⌈ 2 -φ 0 ⌉
|-φ 2 -√2|
⌊ 0 -√2 2 ⌋
det = -1.2361
- Code: Select all
o5o4o3o
⌈ 2 -φ 0 0 ⌉
|-φ 2 -√2 0 |
| 0 -√2 2 -1 |
⌊ 0 0 -1 2 ⌋
det = -3.8541
- Code: Select all
o5o4o3o5o
⌈ 2 -φ 0 0 0 ⌉
|-φ 2 -√2 0 0 |
| 0 -√2 2 -1 0 |
| 0 0 -1 2 -φ |
⌊ 0 0 0 -φ 2 ⌋
det = -4.4721
- Code: Select all
o5o4o3o5o5o
⌈ 2 -φ 0 0 0 0 ⌉
|-φ 2 -√2 0 0 0 |
| 0 -√2 2 -1 0 0 |
| 0 0 -1 2 -φ 0 |
| 0 0 0 -φ 2 -φ |
⌊ 0 0 0 0 -φ 2 ⌋
det = +1.1459
ΓΔΘΛΞΠΣΦΨΩ αβγδεζηθϑικλμνξοπρϱσςτυϕφχψωϖ °±∓½⅓⅔¼¾×÷†‡• ⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎
ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩
ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩
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Re: Ultra-hyperbolic Coxeter groups
by mr_e_man » Thu Feb 10, 2022 10:52 pm
Here's something on MO: https://mathoverflow.net/questions/1652 ... eter-group
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ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩
ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩
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Re: Ultra-hyperbolic Coxeter groups
by wendy » Tue Mar 08, 2022 8:11 am
The sad thing is that most of this is wrong.
The determinate of this matrix is the value calculted by the Schläfli process, also given by Th. Gosset. You can use it to calculate the diameter of a regular, from 2S(v)/S(s) where v is the vertex figure and s is the solid, S() is the determinate.
The "signature" is little more than if the edge and radius are of the same curvature. For example, ordinary spheric polytopes are +/+. When the radius goes hyperbolic, it goes +/- or -. When the edge goes hyperbolic, it's -/- = +.
The matrix used to demonstrate this proof is of course, the Coxeter Dynkin matrix, which is really little more than the CD in matrix form. The resulting values actually represent the required dot product of the normals to the mirrors, and are thus not the ones to represent the distances. You need a matrix, for which the product of it and this matrix will give you the identity, or better the identity times the Schläfli determinate. This matrix exists, it's called the Stott matrix.
The "unit vectors" of the Stott matrix are those rays from the origin to xPoQo, oPxQo, and oPoQx. Extend for dimensions. Using these vectors allows you to determine the dot product of the stott vectors, when the input is a pair of dynkin vectors.
The determinate of this matrix is the value calculted by the Schläfli process, also given by Th. Gosset. You can use it to calculate the diameter of a regular, from 2S(v)/S(s) where v is the vertex figure and s is the solid, S() is the determinate.
The "signature" is little more than if the edge and radius are of the same curvature. For example, ordinary spheric polytopes are +/+. When the radius goes hyperbolic, it goes +/- or -. When the edge goes hyperbolic, it's -/- = +.
The matrix used to demonstrate this proof is of course, the Coxeter Dynkin matrix, which is really little more than the CD in matrix form. The resulting values actually represent the required dot product of the normals to the mirrors, and are thus not the ones to represent the distances. You need a matrix, for which the product of it and this matrix will give you the identity, or better the identity times the Schläfli determinate. This matrix exists, it's called the Stott matrix.
The "unit vectors" of the Stott matrix are those rays from the origin to xPoQo, oPxQo, and oPoQx. Extend for dimensions. Using these vectors allows you to determine the dot product of the stott vectors, when the input is a pair of dynkin vectors.
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Re: Ultra-hyperbolic Coxeter groups
by mr_e_man » Wed Mar 09, 2022 6:27 pm
wendy wrote:The "signature" is little more than if the edge and radius are of the same curvature. For example, ordinary spheric polytopes are +/+. When the radius goes hyperbolic, it goes +/- or -. When the edge goes hyperbolic, it's -/- = +.
No, it's much more. The signature is an n-tuple of ± signs, not a single ± sign.
Are you not familiar with the 6-dimensional space where the dot product is (w₁,x₁,y₁,z₁,u₁,v₁)•(w₂,x₂,y₂,z₂,u₂,v₂) = w₁w₂ + x₁x₂ + y₁y₂ + z₁z₂ - u₁u₂ - v₁v₂ ?
The space of o5o4o3o5o5o has exactly this non-Euclidean dot product, after some change of coordinates. That is what it means to have signature ++++-- . (One such change of coordinates is given by the Gram-Schmidt orthogonalization process, suitably modified to avoid division by zero or square roots of negatives.)
Equivalently, given the Dynkin matrix D =
- Code: Select all
⌈ 2 -φ 0 0 0 0 ⌉
|-φ 2 -√2 0 0 0 |
| 0 -√2 2 -1 0 0 |
| 0 0 -1 2 -φ 0 |
| 0 0 0 -φ 2 -φ |
⌊ 0 0 0 0 -φ 2 ⌋ ,
- Code: Select all
⌈ 1 0 0 0 0 0 ⌉
| 0 1 0 0 0 0 |
| 0 0 1 0 0 0 |
| 0 0 0 1 0 0 |
| 0 0 0 0 -1 0 |
⌊ 0 0 0 0 0 -1 ⌋ .
And the Stott matrix has the same signature (in this case), because (det D)D-1 = (√(det D) I)T (D-1) (√(det D) I), where I is the identity matrix.
ΓΔΘΛΞΠΣΦΨΩ αβγδεζηθϑικλμνξοπρϱσςτυϕφχψωϖ °±∓½⅓⅔¼¾×÷†‡• ⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎
ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩
ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩
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Re: Ultra-hyperbolic Coxeter groups
by wendy » Thu Mar 10, 2022 12:42 pm
This is the matrix dot for {5,4,3,5,5}. The bulk of the numbers belong to Z5, although the first row is Z5.q (ie pentagonal integers times sqrt(2).
If you are taking dot products in hyperbolic space, you should not use the dynkin matrix, but the stott matrix. The sign signature that you have supplied is the signs of the succesive vertex-figure determinates, and does not apply to any distances here.
- Code: Select all
[d:\prompt]matrix4 fqsff
fqsff 1.145898033750315455 ! 6D f. **
-7.70820393 -10.23606798 -5.65685425 3.16227766 7.40491835 5.99070478
-10.23606798 -12.65247584 -6.99225639 3.90879015 9.15298245 7.40491835
-5.65685425 -6.99225639 -3.41640786 1.90983006 4.47213595 3.61803399
3.16227766 3.90879015 1.90983006 -1.70820393 -4.00000000 -3.23606798
7.40491835 9.15298245 4.47213595 -4.00000000 -7.70820393 -6.23606798
5.99070478 7.40491835 3.61803399 -3.23606798 -6.23606798 -4.47213595
If you are taking dot products in hyperbolic space, you should not use the dynkin matrix, but the stott matrix. The sign signature that you have supplied is the signs of the succesive vertex-figure determinates, and does not apply to any distances here.
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Re: Ultra-hyperbolic Coxeter groups
by mr_e_man » Tue Mar 15, 2022 10:12 pm
wendy wrote:If you are taking dot products in hyperbolic space, you should not use the dynkin matrix, but the stott matrix.
This is not in hyperbolic space. That's my point! This thing doesn't fit in 6D Euclidean space, and it doesn't fit in 6D Lorentzian space (also called spacetime) which contains 5D hyperbolic space.
If you calculate the eigenvalues of that Dynkin matrix, or that Stott matrix you provided, you'll find that 4 of them are positive and 2 are negative. If it was hyperbolic, 5 would be positive and 1 negative (or 1 positive and 5 negative).
To put it another way: If you try to find a set of 6 vectors in 6D Euclidean or Lorentzian space, whose dot products with each other are given by either of those matrices (or any non-zero scalar multiple of them), there is no solution.
(Or are you using "hyperbolic" more generally, for what others would call "pseudo-Euclidean"?)
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ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩
ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩
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