[math-fun] Hurwitz integers --- and it sez 'ere ...
https://en.wikipedia.org/wiki/Hurwitz_quaternion << The Lipschitz quaternions L form an index 2 sublattice of H. >> "index 3" , I think ?! See eg. Conway & Smith sect. 5.5 . WFL
Definitely index 2: H = L union (L + (1/2, 1/2, 1/2, 1/2)) You might be getting confused with the fact that the Lipschitz quaternions of unit norm (abstractly Q_8) form an index-3 normal subgroup of the Hurwitz quaternions of unit norm (abstractly 2T): http://en.wikipedia.org/wiki/Binary_tetrahedral_group This is in turn an index-5 abnormal subgroup of the icosians of unit norm (abstractly 2I): http://en.wikipedia.org/wiki/Icosian APG
Sent: Tuesday, March 10, 2015 at 3:38 PM From: "Fred Lunnon" <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: [math-fun] Hurwitz integers --- and it sez 'ere ...
https://en.wikipedia.org/wiki/Hurwitz_quaternion
<< The Lipschitz quaternions L form an index 2 sublattice of H. >>
"index 3" , I think ?! See eg. Conway & Smith sect. 5.5 .
WFL
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Oops --- thanks for clarification! WFL On 3/10/15, Adam P. Goucher <apgoucher@gmx.com> wrote:
Definitely index 2: H = L union (L + (1/2, 1/2, 1/2, 1/2))
You might be getting confused with the fact that the Lipschitz quaternions of unit norm (abstractly Q_8) form an index-3 normal subgroup of the Hurwitz quaternions of unit norm (abstractly 2T):
http://en.wikipedia.org/wiki/Binary_tetrahedral_group
This is in turn an index-5 abnormal subgroup of the icosians of unit norm (abstractly 2I):
http://en.wikipedia.org/wiki/Icosian
APG
Sent: Tuesday, March 10, 2015 at 3:38 PM From: "Fred Lunnon" <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: [math-fun] Hurwitz integers --- and it sez 'ere ...
https://en.wikipedia.org/wiki/Hurwitz_quaternion
<< The Lipschitz quaternions L form an index 2 sublattice of H. >>
"index 3" , I think ?! See eg. Conway & Smith sect. 5.5 .
WFL
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It appears to me that the 24 points on the unit sphere S^3 in R^4, of form (±√½, ±√½, 0, 0) (and permutations) form a multiplicative system with inverses. QUESTION: How many such 24-element multiplicative systems with inverses are there in S^3 ? --Dan
On Mar 10, 2015, at 9:11 AM, Adam P. Goucher <apgoucher@gmx.com> wrote:
. . . the fact that the Lipschitz quaternions of unit norm (abstractly Q_8) form an index-3 normal subgroup of the Hurwitz quaternions of unit norm (abstractly 2T):
I'm not sure how you're multiplying 4-vectors. If you're multiplying them componentwise, (root2/2, root2/2,0,0) ^2 = (1/2, 1/2, 0, 0). If you're multiplying them as quaternions, roott2/2 + root2/2 * i = i So I can't tell in what sense the set of points you give is a multiplicative system. Andy On Wed, Mar 11, 2015 at 4:32 PM, Dan Asimov <asimov@msri.org> wrote:
It appears to me that the 24 points on the unit sphere S^3 in R^4, of form
(±√½, ±√½, 0, 0) (and permutations)
form a multiplicative system with inverses.
QUESTION:
How many such 24-element multiplicative systems with inverses are there in S^3 ?
--Dan
On Mar 10, 2015, at 9:11 AM, Adam P. Goucher <apgoucher@gmx.com> wrote:
. . . the fact that the Lipschitz quaternions of unit norm (abstractly Q_8) form an index-3 normal subgroup of the Hurwitz quaternions of unit norm (abstractly 2T):
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The thread started out talking about quaternions and Hurwitz integers, so I assume we are still multiplying quadruplets with the quaternion rule. On Wed, Mar 11, 2015 at 4:41 PM, Andy Latto <andy.latto@pobox.com> wrote:
I'm not sure how you're multiplying 4-vectors. If you're multiplying them componentwise,
(root2/2, root2/2,0,0) ^2 = (1/2, 1/2, 0, 0).
If you're multiplying them as quaternions,
roott2/2 + root2/2 * i = i
So I can't tell in what sense the set of points you give is a multiplicative system.
Andy
On Wed, Mar 11, 2015 at 4:32 PM, Dan Asimov <asimov@msri.org> wrote:
It appears to me that the 24 points on the unit sphere S^3 in R^4, of form
(±√½, ±√½, 0, 0) (and permutations)
form a multiplicative system with inverses.
QUESTION:
How many such 24-element multiplicative systems with inverses are there in S^3 ?
--Dan
On Mar 10, 2015, at 9:11 AM, Adam P. Goucher <apgoucher@gmx.com> wrote:
. . . the fact that the Lipschitz quaternions of unit norm (abstractly Q_8) form an index-3 normal subgroup of the Hurwitz quaternions of unit norm (abstractly 2T):
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-- Andy.Latto@pobox.com
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On Mar 11, 2015, at 1:52 PM, Allan Wechsler <acwacw@gmail.com> wrote:
The thread started out talking about quaternions and Hurwitz integers, so I assume we are still multiplying quadruplets with the quaternion rule.
Right, all I meant was quaternion multiplication, which I guess I'm so used to on S^3 = unit quaternions that I neglected to mention it.
On Wed, Mar 11, 2015 at 4:41 PM, Andy Latto <andy.latto@pobox.com> wrote:
I'm not sure how you're multiplying 4-vectors. If you're multiplying them componentwise,
(root2/2, root2/2,0,0) ^2 = (1/2, 1/2, 0, 0).
If you're multiplying them as quaternions,
roott2/2 + root2/2 * i = i
So I can't tell in what sense the set of points you give is a multiplicative system.
In S^3, (√½ + √½ i)^2 = i indeed, I screwed up yet again. (I was going by the observations that the integer quaternions Z + iZ + jZ + kZ with an even sum of coefficients (generated by ±1±i, ±i±j, ±j±k) is clearly closed under multiplication, and same for arbitrary quaternions of norm = 1.) Oh, well. To make up for wasting everyone's time, here is a research announcement: (with Joseph Gerver): There are analogues in the quaternions and octonions of the map-of-France fractal and others. The map-of-France analogue in the quaternions is based on taking a cluster of one 24-cell and its 24 neighboring ones in the 24-cell honeycomb of R^4, rotating, rescaling, and iterating, much as one does for the hex tiling of R^2 to get the map-of-France. --Dan P.S. So far we have not been able to get this to work in some other dimensions, like R^3: taking a cluster of one truncated octahedron surrounded by its 14 neighbors in the truncated octahedron honeycomb and trying to proceed mutatis mutandis does not seem to work as one might hope.
On Wed, Mar 11, 2015 at 4:32 PM, Dan Asimov <asimov@msri.org> wrote:
It appears to me that the 24 points on the unit sphere S^3 in R^4, of form
(±√½, ±√½, 0, 0) (and permutations)
form a multiplicative system with inverses.
QUESTION:
How many such 24-element multiplicative systems with inverses are there in S^3 ?
--Dan
On Mar 10, 2015, at 9:11 AM, Adam P. Goucher <apgoucher@gmx.com> wrote:
. . . the fact that the Lipschitz quaternions of unit norm (abstractly Q_8) form an index-3 normal subgroup of the Hurwitz quaternions of unit norm (abstractly 2T):
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-- Andy.Latto@pobox.com
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Is there a flowsnake for the regular octagonal tiling of the hyperbolic plane? Due to the intrinsic length scale, I think it would have to behave somewhat like the relationship between the odd numbers of Pascal's triangle and Sierpinski's triangle. It would be something like an IFS where the contraction coefficient is 1, or an L-system where the length of a segment is fixed. It wouldn't be spacefilling in quite the same sense as the Euclidean one. On Wed, Mar 11, 2015 at 2:37 PM, Dan Asimov <asimov@msri.org> wrote:
On Mar 11, 2015, at 1:52 PM, Allan Wechsler <acwacw@gmail.com> wrote:
The thread started out talking about quaternions and Hurwitz integers, so I assume we are still multiplying quadruplets with the quaternion rule.
Right, all I meant was quaternion multiplication, which I guess I'm so used to on S^3 = unit quaternions that I neglected to mention it.
On Wed, Mar 11, 2015 at 4:41 PM, Andy Latto <andy.latto@pobox.com> wrote:
I'm not sure how you're multiplying 4-vectors. If you're multiplying them componentwise,
(root2/2, root2/2,0,0) ^2 = (1/2, 1/2, 0, 0).
If you're multiplying them as quaternions,
roott2/2 + root2/2 * i = i
So I can't tell in what sense the set of points you give is a multiplicative system.
In S^3, (√½ + √½ i)^2 = i indeed, I screwed up yet again.
(I was going by the observations that the integer quaternions Z + iZ + jZ + kZ with an even sum of coefficients (generated by ±1±i, ±i±j, ±j±k) is clearly closed under multiplication, and same for arbitrary quaternions of norm = 1.)
Oh, well. To make up for wasting everyone's time, here is a research announcement:
(with Joseph Gerver): There are analogues in the quaternions and octonions of the map-of-France fractal and others. The map-of-France analogue in the quaternions is based on taking a cluster of one 24-cell and its 24 neighboring ones in the 24-cell honeycomb of R^4, rotating, rescaling, and iterating, much as one does for the hex tiling of R^2 to get the map-of-France.
--Dan
P.S. So far we have not been able to get this to work in some other dimensions, like R^3: taking a cluster of one truncated octahedron surrounded by its 14 neighbors in the truncated octahedron honeycomb and trying to proceed mutatis mutandis does not seem to work as one might hope.
On Wed, Mar 11, 2015 at 4:32 PM, Dan Asimov <asimov@msri.org> wrote:
It appears to me that the 24 points on the unit sphere S^3 in R^4, of form
(±√½, ±√½, 0, 0) (and permutations)
form a multiplicative system with inverses.
QUESTION:
How many such 24-element multiplicative systems with inverses are there in S^3 ?
--Dan
On Mar 10, 2015, at 9:11 AM, Adam P. Goucher <apgoucher@gmx.com> wrote:
. . . the fact that the Lipschitz quaternions of unit norm (abstractly Q_8) form an index-3 normal subgroup of the Hurwitz quaternions of unit norm (abstractly 2T):
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-- Andy.Latto@pobox.com
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-- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
See Conway & Sloane, pp.52--55 and 118--119. R. On Wed, 11 Mar 2015, Andy Latto wrote:
I'm not sure how you're multiplying 4-vectors. If you're multiplying them componentwise,
(root2/2, root2/2,0,0) ^2 = (1/2, 1/2, 0, 0).
If you're multiplying them as quaternions,
roott2/2 + root2/2 * i = i
So I can't tell in what sense the set of points you give is a multiplicative system.
Andy
On Wed, Mar 11, 2015 at 4:32 PM, Dan Asimov <asimov@msri.org> wrote:
It appears to me that the 24 points on the unit sphere S^3 in R^4, of form
(±√½, ±√½, 0, 0) (and permutations)
form a multiplicative system with inverses.
QUESTION:
How many such 24-element multiplicative systems with inverses are there in S^3 ?
--Dan
On Mar 10, 2015, at 9:11 AM, Adam P. Goucher <apgoucher@gmx.com> wrote:
. . . the fact that the Lipschitz quaternions of unit norm (abstractly Q_8) form an index-3 normal subgroup of the Hurwitz quaternions of unit norm (abstractly 2T):
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-- Andy.Latto@pobox.com
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Think of R^4 as chopped into cubes of side = 1/2, each corresponding to the quaternion at its (+,+,+,+) vertex. Every unit cube in 4-space is 16 of these side-(1/2) cubes, so has 16 of these (+,+,+,+) vertices. The Hurwitz quaternions correspond to 2 of these vertices, and the Lipschitz quaternions to just 1 of them. So it would appear that the index is 2. --Dan
On Mar 10, 2015, at 8:38 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
https://en.wikipedia.org/wiki/Hurwitz_quaternion
<< The Lipschitz quaternions L form an index 2 sublattice of H. >>
"index 3" , I think ?! See eg. Conway & Smith sect. 5.5 .
And it's easy to see that a nice fundamental domain for the Hurwitz integers arises from the Voronoi region of any one of them, like 0, and is a 24-cell. So this gives the tiling of R^4 by 24-cells. The Standard Mistake is to assume that, since the 24-cell is self-dual, that its tiling of R^4 is also self-dual (guilty). Hmm, let's see, what exactly is the dual tiling? --Dan
On Mar 10, 2015, at 9:15 AM, Dan Asimov <dasimov@earthlink.net> wrote:
Think of R^4 as chopped into cubes of side = 1/2, each corresponding to the quaternion at its (+,+,+,+) vertex. Every unit cube in 4-space is 16 of these side-(1/2) cubes, so has 16 of these (+,+,+,+) vertices. The Hurwitz quaternions correspond to 2 of these vertices, and the Lipschitz quaternions to just 1 of them. So it would appear that the index is 2.
--Dan
On Mar 10, 2015, at 8:38 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
https://en.wikipedia.org/wiki/Hurwitz_quaternion
<< The Lipschitz quaternions L form an index 2 sublattice of H. >>
"index 3" , I think ?! See eg. Conway & Smith sect. 5.5 .
The vertex figure of the tiling is a tesseract (with 8 cubical facets); so the dual tiling fills 4-space with their duals, which are cross-polytopes (with 16 tetrahedral facets). WFL On 3/10/15, Dan Asimov <dasimov@earthlink.net> wrote:
And it's easy to see that a nice fundamental domain for the Hurwitz integers arises from the Voronoi region of any one of them, like 0, and is a 24-cell. So this gives the tiling of R^4 by 24-cells.
The Standard Mistake is to assume that, since the 24-cell is self-dual, that its tiling of R^4 is also self-dual (guilty). Hmm, let's see, what exactly is the dual tiling?
--Dan
On Mar 10, 2015, at 9:15 AM, Dan Asimov <dasimov@earthlink.net> wrote:
Think of R^4 as chopped into cubes of side = 1/2, each corresponding to the quaternion at its (+,+,+,+) vertex. Every unit cube in 4-space is 16 of these side-(1/2) cubes, so has 16 of these (+,+,+,+) vertices. The Hurwitz quaternions correspond to 2 of these vertices, and the Lipschitz quaternions to just 1 of them. So it would appear that the index is 2.
--Dan
On Mar 10, 2015, at 8:38 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
https://en.wikipedia.org/wiki/Hurwitz_quaternion
<< The Lipschitz quaternions L form an index 2 sublattice of H. >>
"index 3" , I think ?! See eg. Conway & Smith sect. 5.5 .
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