Thread: "Hyperbolic Honeycomb {7,3,3}"

Hi Nan,

I just wanted to address a couple of points that caught my eye
in your message (and in the part of Roice's that you quoted)...

On Sun, Jun 24, 2012 at 08:14:31AM -0000, schuma wrote:
>
>
> Hi everyone,
> I'm continuing talking about my honeycomb/polytope viewer applet. I added
> a new honeycomb, and I think it deserves a new topic. This is {7,3,3}.
> Each cell is a hyperbolic tiling {7,3}. Please check it here:
> http://people.bu.edu/nanma/InsideH3/H3.html
> I first heard of this thing together with {3,3,7} in emails with Roice
> Nelson. He had been exchanging emails with Andrey Astrelin about them. We
> have NOT seen any publication talking about these honeycombs. Even when
> Coxeter enumerate the hyperbolic honeycombs, he stopped at honeycombs like
> {6,3,3}, where each cell is at most an Euclidean tessellation like {6,3}.
> He said, "we shall restrict consideration to cases where the fundamental
> region of the symmetry group has a finite content" (content = volume?),

Right. The fundamental region is the characteristic simplex,
so (since even ideal simplices have finite volume)
this is the same as saying that all the vertices
of the characteristic simplex (i.e. the honeycomb vertex, edge center,
face center, cell center) are "accessible" (either finite, or infinite
i.e. at some definite location on the boundary of the poincare ball).
So you're examining some cases where that condition is partially
relaxed, i.e. the fundamental region contains more of the horizon than just
isolated points there... and the characteristic tetrahedron
is actually missing one or more of its vertices.

> and hence didn't consider {7,3,3}, where each cell is a hyperbolic
> tessellation {7,3}.
> We think {3,3,7} and {7,3,3} and other similar objects are constructable.

{7,3,3} yes, in the sense that the vertices/edges/faces are finite,
and there's clear local structure around them, and, as you observe,
the edge length formula works out fine
(but not the cell in-radius nor circum-radius formula)
and you can render it (as you have-- nice!)...

{3,3,7} less so... its vertices are not simply at infinity (as in {3,3,6}),
they are "beyond infinity"...
If you try to draw this one, none of the edges will meet at all (not even at
infinity)... they all diverge! You'll see each edge
emerging from somewhere on the horizon (although there's no vertex
there) and leaving somewhere else on the horizon...
so nothing meets up, which kind of makes the picture less satisfying.
If you run the formula for edge length or cell circumradius, you'll get, not infinity,
but an imaginary or complex number (although the cell in-radius is finite, of
course, being equal to the half-edge-length of the dual {7,3,3}).

It may be Coxeter refrained omitted these figures
because the "beyond infinity" parts are awkward to talk about,
and if you insist on running the formulas and completing the tables,
a lot of it will consist of imaginary and complex numbers
that aren't all that meaningful physically, and might scare some readers away
(even though, as you've noted, some of the entries
are perfectly fine finite numbers or plain old infinity).



> I derived the edge length of {n,3,3} for general n, and then computed the
> coordinates of several vertices of {7,3,3}, then I plotted them. There's
> really nothing so weird about this honeycomb. It looks just like, or, as
> weird as, {6,3,3}. The volume of the fundamental region of {7,3,3} may be
> infinite, but as long as we talk about the edge length, face area,
> everything is finite and looks normal.
> I could go and make {8,3,3}, {9,3,3} etc. I also believe {7,3,4} and
> {7,3,5} are also pretty well behaved, and looks just like {6,3,4} and
> {6,3,5} respectively. Or even {7,4,3}. As long as the vertex figure is
> finite (not like {3,4,4}), the image shouldn't be crazy. Since we are
> facing an infinite number of honeycombs here, I feel I should stop at some
> point. After all we don't understand {7,3,3} well, which is the smallest
> representative of them. I'd like to spend more energy making sense of
> {7,3,3} rather than go further.
> It's not clear for me whether we can identify some heptagons in {7,3} to
> make it Klein Quartic, in {7,3,3}. For example, in the hypercube {4,3,3},
> we can replace each cubic cell by hemi-cube by identification. The result
> is that all the vertices end up identified as only one vertex. I don't
> know what'll happen if I replace {7,3} by Klein Quartic ({7,3}_8). It will
> be awesome if we can fit three KQ around each edge to make a polytope
> based on {7,3,3}. If "three" doesn't work, maybe the one based on {7,3,4}
> or {7,3,5} works. I actually also don't know what'll happen if I replace
> the dodecahedral cells of 120-cell by hemi-dodecahedra. Does anyone know?
> I still suspect people have discussed it somewhere in literature. But I
> haven't found anything really related. Roice found the following statement
> and references. I don't haven't check them yet.
> __________
>
> I checked 'Abstract Regular Polytopes', and was not able to find
> anything on the {7,3,3}. H3 honeycombs make several appearances at
> various places in the book, but the language seems to be similar to
> Coxeter, and their charts also limited to the same ones. On page 77,
> they distinguish between "compact" and "non-compact" hyperbolic types,
> and say:
> Coxeter groups of hyperbolic type exist only in ranks 3 to 10, and there
> are only finitely many such groups in ranks 4 to 10. Groups of compact
> hyperbolic type exist only in ranks 3, 4, and 5.
> But as best I can tell, "non-compact" still only refers to the same
> infinite honeycombs Coxeter enumerated. They reference the following
> book:
> J. E. Humphreys, Reflection Groups and Coxeter Groups, Cambridge
> University Press (Cambridge, 1990).
> When researching just now on wikipedia, the page on uniform hyperbolic
> honeycombs has a short section on noncompact hyperbolic honeycombs, and
> also references the same book by Humphreys. So maybe this book could be
> a good reference to dig up, even though I suspect it will still not
> mention the {7,3,3}.
> Also: Abstract Regular Polytopes, p78:
> For the general theory of hyperbolic reflexion groups, the reader is
> referred to Vinberg [431-433]. We remark that there are examples of
> discrete groups generated by hyperplane reflexions in a hyperbolic space
> which are Coxeter groups, but do not have a simplex as a fundamental
> region.
> These honeycombs fall into that category.
> Here are those references:
> [431] E. B. Vinberg, Discrete groups in Lobachevskii spaces generated by
> reflections, Mat. Sb. 72 (1967), 471-488 (= Math. USSR-Sb. 1 (1967),
> 429-444).
> [432] E. B. Vinberg, Discrete linear groups generated by reflections,
> Izv. Akad. Nauk. SSSR Ser. Mat. 35 (1971) 1072-1112 (= Math. USSR-Izv. 5
> (1971), 1083-1119).
> [433] E. B. Vinberg, Hyperbolic reflection groups. Uspekhi Mat. Nauk 40
> (1985), 29-66 (= Russian Math. Surveys 40 (1985), 31-75).
>
> ______________
> Now I can only say "to the best of our knowledge, I haven't seen any
> discussion about it".

I noticed one further possible reference in Coxeter's "reguar honeycombs in
hyperbolic space" paper-- on the first page, he refers to:
"... (Coxeter 1933), not insisting on finite fundamental regions,
was somewhat lacking in rigour"
where (Coxeter 1933) is "The densities of the regular polytopes, Part 3".
I believe that paper is in the collection "Kaleidoscopes: Selected
writings of H.S.M. Coxeter" (my copy of which is buried in a box in
storage somewhere :-( ). I suspect that one *will* mention the {7,3,3};
I'd be interested to know what he says about it, now that we're thinking
along those lines.


> Some more thoughts by Roice:
> __________
>
> We know that for {n,3,3), as n -> 6 from higher values of n, the {n,3}
> tiling approaches a horosphere, reaching it at n = 6.

Right... or more precisely,
the circumsphere, edge-tangency-sphere, face-tangency-sphere, and in-sphere
all approach horospheres
(different horospheres, but sharing the same center-at-infinity)...

> For {7,3,n), as n -> infinity, does the {7,3} tiling approach a
> horosphere as well?

I'm not completely confident that this will stay meaningful
as we lose the locations of the vertices (for n >= 7).
The circum-sphere certainly becomes ill-defined...
however one or more of the other tangency spheres
might stay well-defined.
One concievable outcome might be
that the in-sphere and mid-spheres approach different limits--
maybe the in-sphere approaches a horosphere but the face-tangency
mid-sphere doesn't, and maybe the edge-tangency mid-sphere
is ill-defined just like the circumsphere is.

In thinking about this,
I have to first think about the significant
events that happen for smaller n...
{7,3,2} two cells, the wall between them tiled with {7,3}, cell centers are imaginary/complex
{7,3,3} finite vertex figure and local structure, although cell centers are imaginary/complex
{7,3,4} same
{7,3,5} same
{7,3,6} infinite vertex figure, vertices are at infinity (and cell centers still imaginary/complex)
{7,3,7} self-dual; both vertices and cell centers are now imaginary/complex
And now, like Nan, I've lost my intuition...
{7,3,7} is the one for me to ponder at this point.

> The curvature definitely flattens out as n
> increases.

right (i.e. the curvature increases, i.e. becomes less negative)

> If cells are a horosphere in the limit, a {7,3,infinity}
> tiling would have finite cells.

hmm? why?

> It would have an infinite edge-figure,
> in addition to an infinite vertex-figure, but as Coxeter did an
> enumeration allowing the latter, why not allow the former?

well, for things like {3,infinity} and {3,3,6}
with infinite vertex-figure, you can still draw them and measure things
about them even though the vertex figures are infinite-- the vertices
are isolated, at least. it seems to me that if the edge figure is
infinite, then it can no longer have isolated vertices (if the vertices
are even accessible at all),
so it's hard to draw a definite picture of anything any more,
or make any measurements... so we have less and less we can say about
the thing, I guess.

> I'd like to
> understand where in Coxeter's analysis a {7,3,infinity} tiling does not
> fit in. One guess is that even if the {7,3} approaches a horosphere,
> it's volume also goes to 0, so is trivial. The heptagons get smaller
> for larger n,

Are you sure?
The edge length is finite for {7,3,2...5},
and infinite for {7,3,6}... that makes me think the heptagons are probably *growing*,
not shrinking, at least for n in that range...
and after that, the edge length is the acosh of an imaginary number,
so it's hard to say whether it's growing or shrinking or what.
To verify, the formula for the half-edge-length is:

acosh(cos(pi/p)*sin(pi/r)/sqrt(1-cos(pi/q)^2-cos(pi/r)^2))

{7,3,2} -> acosh(1.0403492368298681) = 0.28312815336765745
{7,3,3} -> acosh(1.1034570002469741) = 0.45104488629937328
{7,3,4} -> acosh(1.2741623922635352) = 0.72453736133879376
{7,3,5} -> acosh(1.7137446255953275) = 1.1331675164780453
{7,3,6} -> acosh(+infinity) = +infinity
{7,3,7} -> acosh(+-1.5731951893240572 i) = (1.2346906773191777 +- 1.5707963267948966 i)
{7,3,8} -> acosh(+-1.0714385881055031 i) = (0.9309971259601171 +- 1.5707963267948966 i)
{7,3,9} -> acosh(+-0.8448884457716658 i) = (0.7673378247178905 +- 1.5707963267948966 i)

All that said, I still don't have a clear picture of what happens
when n goes to infinity. We certainly lose the vertices
at n=7, so the circum-sphere isn't well-defined...
and I think we must lose the edges eventually as well? in which case the
edge-tangency mid-sphere isn't well-defined either...
but I'm guessing we *don't* lose the face centers...
so the face-tangency sphere and in-sphere may still be well-defined,
and may approach a limit,
in which case if your question has meaning,
one of those limits would be its meaning (I think). And I don't know the answer.

> so I suppose they must approach 0 size as well.
> It would also be interesting to consider how curvature changes for
> {n,3,3} as n-> infinity, especially since we already know what the
> {infinity,3} tiling looks like.
>
> _______________
> Currently I can't imagine what {7,3,n} like when n>=6. So I really cannot
> comment on {7,3,infinity}. But {infinity, 3, 3} seems to be a good thing
> to study.
> My formula for the edge length of {n,3,3} is as follows. Following
> Coxeter's notation, if 2*phi is the length of an edge of {n,3,3} (n>=6),
> then
> cosh(2*phi) = 3*cos^2(pi/n) - 1
> Sanity check: when n = 6, this formula gives cosh(2*phi)=5/4, which is
> consistent with the number in Coxeter's table: cosh^2(phi)=9/8.
> By sending n to infinity, the edge length of {infinity, 3, 3} is
> arccosh(2). I should be able to plot it soon.
> By the way, in the applet there's a "Clifford Torus". It looks much more
> beautiful than the polytopes, because the colors of the edges work pretty
> well here. Imagine you can fly around a donut, or go into the donut. The
> amazing thing is if the space is 3-sphere, the view inside the donut is
> exactly as same as the outside.
> Nan
>

Don

--
Don Hatch
hatch@plunk.org
http://www.plunk.org/~hatch/

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Hi Don,

Thanks for your enlightening email, and for correcting some speculations I
made without thinking deeply enough.

- I was totally wrong about horosphere cells being finite.
- When considering {7,3,n} as n increases, I see it was incorrect to
conclude the heptagon size decreases (some flawed internal reasoning).
Interesting that the magnitude of the complex edge length starts
decreasing when n>=7, although I guess those complex outputs are pretty
meaningless (then again, maybe not!). Since I got the trend backwards for
low n, I had no idea about the vertices going to infinity at n = 6. I
should have noticed that {3,3,6}, {4,3,6}, {5,3,6}, and {6,3,6} all do the
same thing. It's noteworthy that the "vertices -> infinity" pattern holds
for any {p,3,6}, which makes sense since the vertex figure is an infinite
tiling.

Offline, Nan and I also discussed the 2D analogue of the {3,3,7}, something
akin to the {3,inf} tiling where the triangle vertices are no longer
accessible. Let me know if you're interested in that discussion, as it
would be cool to hear your thoughts.

Anyway, thanks again. Very cool stuff,
Roice


On Fri, Jun 29, 2012 at 7:32 PM, Don Hatch wrote:

> Hi Nan,
>
> I just wanted to address a couple of points that caught my eye
> in your message (and in the part of Roice's that you quoted)...
>
> On Sun, Jun 24, 2012 at 08:14:31AM -0000, schuma wrote:
> >
> >
> > Hi everyone,
> > I'm continuing talking about my honeycomb/polytope viewer applet. I
> added
> > a new honeycomb, and I think it deserves a new topic. This is {7,3,3}.
> > Each cell is a hyperbolic tiling {7,3}. Please check it here:
> > http://people.bu.edu/nanma/InsideH3/H3.html
> > I first heard of this thing together with {3,3,7} in emails with Roice
> > Nelson. He had been exchanging emails with Andrey Astrelin about
> them. We
> > have NOT seen any publication talking about these honeycombs. Even
> when
> > Coxeter enumerate the hyperbolic honeycombs, he stopped at honeycombs
> like
> > {6,3,3}, where each cell is at most an Euclidean tessellation like
> {6,3}.
> > He said, "we shall restrict consideration to cases where the
> fundamental
> > region of the symmetry group has a finite content" (content =
> volume?),
>
> Right. The fundamental region is the characteristic simplex,
> so (since even ideal simplices have finite volume)
> this is the same as saying that all the vertices
> of the characteristic simplex (i.e. the honeycomb vertex, edge center,
> face center, cell center) are "accessible" (either finite, or infinite
> i.e. at some definite location on the boundary of the poincare ball).
> So you're examining some cases where that condition is partially
> relaxed, i.e. the fundamental region contains more of the horizon than just
> isolated points there... and the characteristic tetrahedron
> is actually missing one or more of its vertices.
>
> > and hence didn't consider {7,3,3}, where each cell is a hyperbolic
> > tessellation {7,3}.
> > We think {3,3,7} and {7,3,3} and other similar objects are
> constructable.
>
> {7,3,3} yes, in the sense that the vertices/edges/faces are finite,
> and there's clear local structure around them, and, as you observe,
> the edge length formula works out fine
> (but not the cell in-radius nor circum-radius formula)
> and you can render it (as you have-- nice!)...
>
> {3,3,7} less so... its vertices are not simply at infinity (as in {3,3,6}),
> they are "beyond infinity"...
> If you try to draw this one, none of the edges will meet at all (not even
> at
> infinity)... they all diverge! You'll see each edge
> emerging from somewhere on the horizon (although there's no vertex
> there) and leaving somewhere else on the horizon...
> so nothing meets up, which kind of makes the picture less satisfying.
> If you run the formula for edge length or cell circumradius, you'll get,
> not infinity,
> but an imaginary or complex number (although the cell in-radius is finite,
> of
> course, being equal to the half-edge-length of the dual {7,3,3}).
>
> It may be Coxeter refrained omitted these figures
> because the "beyond infinity" parts are awkward to talk about,
> and if you insist on running the formulas and completing the tables,
> a lot of it will consist of imaginary and complex numbers
> that aren't all that meaningful physically, and might scare some readers
> away
> (even though, as you've noted, some of the entries
> are perfectly fine finite numbers or plain old infinity).
>
>
>
> > I derived the edge length of {n,3,3} for general n, and then computed
> the
> > coordinates of several vertices of {7,3,3}, then I plotted them.
> There's
> > really nothing so weird about this honeycomb. It looks just like, or,
> as
> > weird as, {6,3,3}. The volume of the fundamental region of {7,3,3}
> may be
> > infinite, but as long as we talk about the edge length, face area,
> > everything is finite and looks normal.
> > I could go and make {8,3,3}, {9,3,3} etc. I also believe {7,3,4} and
> > {7,3,5} are also pretty well behaved, and looks just like {6,3,4} and
> > {6,3,5} respectively. Or even {7,4,3}. As long as the vertex figure is
> > finite (not like {3,4,4}), the image shouldn't be crazy. Since we are
> > facing an infinite number of honeycombs here, I feel I should stop at
> some
> > point. After all we don't understand {7,3,3} well, which is the
> smallest
> > representative of them. I'd like to spend more energy making sense of
> > {7,3,3} rather than go further.
> > It's not clear for me whether we can identify some heptagons in {7,3}
> to
> > make it Klein Quartic, in {7,3,3}. For example, in the hypercube
> {4,3,3},
> > we can replace each cubic cell by hemi-cube by identification. The
> result
> > is that all the vertices end up identified as only one vertex. I don't
> > know what'll happen if I replace {7,3} by Klein Quartic ({7,3}_8). It
> will
> > be awesome if we can fit three KQ around each edge to make a polytope
> > based on {7,3,3}. If "three" doesn't work, maybe the one based on
> {7,3,4}
> > or {7,3,5} works. I actually also don't know what'll happen if I
> replace
> > the dodecahedral cells of 120-cell by hemi-dodecahedra. Does anyone
> know?
> > I still suspect people have discussed it somewhere in literature. But
> I
> > haven't found anything really related. Roice found the following
> statement
> > and references. I don't haven't check them yet.
> > __________
> >
> > I checked 'Abstract Regular Polytopes', and was not able to find
> > anything on the {7,3,3}. H3 honeycombs make several appearances at
> > various places in the book, but the language seems to be similar to
> > Coxeter, and their charts also limited to the same ones. On page
> 77,
> > they distinguish between "compact" and "non-compact" hyperbolic
> types,
> > and say:
> > Coxeter groups of hyperbolic type exist only in ranks 3 to 10, and
> there
> > are only finitely many such groups in ranks 4 to 10. Groups of
> compact
> > hyperbolic type exist only in ranks 3, 4, and 5.
> > But as best I can tell, "non-compact" still only refers to the same
> > infinite honeycombs Coxeter enumerated. They reference the
> following
> > book:
> > J. E. Humphreys, Reflection Groups and Coxeter Groups, Cambridge
> > University Press (Cambridge, 1990).
> > When researching just now on wikipedia, the page on uniform
> hyperbolic
> > honeycombs has a short section on noncompact hyperbolic honeycombs,
> and
> > also references the same book by Humphreys. So maybe this book
> could be
> > a good reference to dig up, even though I suspect it will still not
> > mention the {7,3,3}.
> > Also: Abstract Regular Polytopes, p78:
> > For the general theory of hyperbolic reflexion groups, the reader is
> > referred to Vinberg [431-433]. We remark that there are examples of
> > discrete groups generated by hyperplane reflexions in a hyperbolic
> space
> > which are Coxeter groups, but do not have a simplex as a fundamental
> > region.
> > These honeycombs fall into that category.
> > Here are those references:
> > [431] E. B. Vinberg, Discrete groups in Lobachevskii spaces
> generated by
> > reflections, Mat. Sb. 72 (1967), 471-488 (= Math. USSR-Sb. 1 (1967),
> > 429-444).
> > [432] E. B. Vinberg, Discrete linear groups generated by
> reflections,
> > Izv. Akad. Nauk. SSSR Ser. Mat. 35 (1971) 1072-1112 (= Math.
> USSR-Izv. 5
> > (1971), 1083-1119).
> > [433] E. B. Vinberg, Hyperbolic reflection groups. Uspekhi Mat.
> Nauk 40
> > (1985), 29-66 (= Russian Math. Surveys 40 (1985), 31-75).
> >
> > ______________
> > Now I can only say "to the best of our knowledge, I haven't seen any
> > discussion about it".
>
> I noticed one further possible reference in Coxeter's "reguar honeycombs in
> hyperbolic space" paper-- on the first page, he refers to:
> "... (Coxeter 1933), not insisting on finite fundamental regions,
> was somewhat lacking in rigour"
> where (Coxeter 1933) is "The densities of the regular polytopes, Part 3".
> I believe that paper is in the collection "Kaleidoscopes: Selected
> writings of H.S.M. Coxeter" (my copy of which is buried in a box in
> storage somewhere :-( ). I suspect that one *will* mention the {7,3,3};
> I'd be interested to know what he says about it, now that we're thinking
> along those lines.
>
>
> > Some more thoughts by Roice:
> > __________
> >
> > We know that for {n,3,3), as n -> 6 from higher values of n, the
> {n,3}
> > tiling approaches a horosphere, reaching it at n = 6.
>
> Right... or more precisely,
> the circumsphere, edge-tangency-sphere, face-tangency-sphere, and in-sphere
> all approach horospheres
> (different horospheres, but sharing the same center-at-infinity)...
>
> > For {7,3,n), as n -> infinity, does the {7,3} tiling approach a
> > horosphere as well?
>
> I'm not completely confident that this will stay meaningful
> as we lose the locations of the vertices (for n >= 7).
> The circum-sphere certainly becomes ill-defined...
> however one or more of the other tangency spheres
> might stay well-defined.
> One concievable outcome might be
> that the in-sphere and mid-spheres approach different limits--
> maybe the in-sphere approaches a horosphere but the face-tangency
> mid-sphere doesn't, and maybe the edge-tangency mid-sphere
> is ill-defined just like the circumsphere is.
>
> In thinking about this,
> I have to first think about the significant
> events that happen for smaller n...
> {7,3,2} two cells, the wall between them tiled with {7,3}, cell
> centers are imaginary/complex
> {7,3,3} finite vertex figure and local structure, although cell
> centers are imaginary/complex
> {7,3,4} same
> {7,3,5} same
> {7,3,6} infinite vertex figure, vertices are at infinity (and
> cell centers still imaginary/complex)
> {7,3,7} self-dual; both vertices and cell centers are now
> imaginary/complex
> And now, like Nan, I've lost my intuition...
> {7,3,7} is the one for me to ponder at this point.
>
> > The curvature definitely flattens out as n
> > increases.
>
> right (i.e. the curvature increases, i.e. becomes less negative)
>
> > If cells are a horosphere in the limit, a {7,3,infinity}
> > tiling would have finite cells.
>
> hmm? why?
>
> > It would have an infinite edge-figure,
> > in addition to an infinite vertex-figure, but as Coxeter did an
> > enumeration allowing the latter, why not allow the former?
>
> well, for things like {3,infinity} and {3,3,6}
> with infinite vertex-figure, you can still draw them and measure things
> about them even though the vertex figures are infinite-- the vertices
> are isolated, at least. it seems to me that if the edge figure is
> infinite, then it can no longer have isolated vertices (if the vertices
> are even accessible at all),
> so it's hard to draw a definite picture of anything any more,
> or make any measurements... so we have less and less we can say about
> the thing, I guess.
>
> > I'd like to
> > understand where in Coxeter's analysis a {7,3,infinity} tiling
> does not
> > fit in. One guess is that even if the {7,3} approaches a
> horosphere,
> > it's volume also goes to 0, so is trivial. The heptagons get
> smaller
> > for larger n,
>
> Are you sure?
> The edge length is finite for {7,3,2...5},
> and infinite for {7,3,6}... that makes me think the heptagons are probably
> *growing*,
> not shrinking, at least for n in that range...
> and after that, the edge length is the acosh of an imaginary number,
> so it's hard to say whether it's growing or shrinking or what.
> To verify, the formula for the half-edge-length is:
>
> acosh(cos(pi/p)*sin(pi/r)/sqrt(1-cos(pi/q)^2-cos(pi/r)^2))
>
> {7,3,2} -> acosh(1.0403492368298681) = 0.28312815336765745
> {7,3,3} -> acosh(1.1034570002469741) = 0.45104488629937328
> {7,3,4} -> acosh(1.2741623922635352) = 0.72453736133879376
> {7,3,5} -> acosh(1.7137446255953275) = 1.1331675164780453
> {7,3,6} -> acosh(+infinity) = +infinity
> {7,3,7} -> acosh(+-1.5731951893240572 i) = (1.2346906773191777 +-
> 1.5707963267948966 i)
> {7,3,8} -> acosh(+-1.0714385881055031 i) = (0.9309971259601171 +-
> 1.5707963267948966 i)
> {7,3,9} -> acosh(+-0.8448884457716658 i) = (0.7673378247178905 +-
> 1.5707963267948966 i)
>
> All that said, I still don't have a clear picture of what happens
> when n goes to infinity. We certainly lose the vertices
> at n=7, so the circum-sphere isn't well-defined...
> and I think we must lose the edges eventually as well? in which case the
> edge-tangency mid-sphere isn't well-defined either...
> but I'm guessing we *don't* lose the face centers...
> so the face-tangency sphere and in-sphere may still be well-defined,
> and may approach a limit,
> in which case if your question has meaning,
> one of those limits would be its meaning (I think). And I don't know the
> answer.
>
> > so I suppose they must approach 0 size as well.
> > It would also be interesting to consider how curvature changes for
> > {n,3,3} as n-> infinity, especially since we already know what the
> > {infinity,3} tiling looks like.
> >
> > _______________
> > Currently I can't imagine what {7,3,n} like when n>=6. So I really
> cannot
> > comment on {7,3,infinity}. But {infinity, 3, 3} seems to be a good
> thing
> > to study.
> > My formula for the edge length of {n,3,3} is as follows. Following
> > Coxeter's notation, if 2*phi is the length of an edge of {n,3,3}
> (n>=6),
> > then
> > cosh(2*phi) = 3*cos^2(pi/n) - 1
> > Sanity check: when n = 6, this formula gives cosh(2*phi)=5/4, which is
> > consistent with the number in Coxeter's table: cosh^2(phi)=9/8.
> > By sending n to infinity, the edge length of {infinity, 3, 3} is
> > arccosh(2). I should be able to plot it soon.
> > By the way, in the applet there's a "Clifford Torus". It looks much
> more
> > beautiful than the polytopes, because the colors of the edges work
> pretty
> > well here. Imagine you can fly around a donut, or go into the donut.
> The
> > amazing thing is if the space is 3-sphere, the view inside the donut
> is
> > exactly as same as the outside.
> > Nan
> >
>
> Don
>
> --
> Don Hatch
> hatch@plunk.org
> http://www.plunk.org/~hatch/
>
>
> ------------------------------------
>
> Yahoo! Groups Links
>
>
>
>

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Content-Type: text/html; charset=ISO-8859-1
Content-Transfer-Encoding: quoted-printable

Hi Don,

Hi Don,

Nice to see you here. Here are my thoughts about {3,3,7} and the things sim=
ilar to it.=20

Just like when we constructed {7,3,3} we were not able locate the cell cent=
ers, when we consider {3,3,7} we have to sacrifice the vertices. Let's star=
t by considering something simpler in lower dimensions.

For example, in 2D, we could consider a hyperbolic "triangle" for which the=
sides don't meet even at the circle of infinity. The sides are ultraparall=
el. Since there's no "angle", the name "triangle" is not appropriate any mo=
re. I'll call it a "trilateral", because it does have three sides (the comm=
on triangle is also a trilateral in my notation). Here's a tessellation of =
H2 using trilaterals, in which different colors indicate different trilater=
als.

http://games.groups.yahoo.com/group/4D_Cubing/files/Nan%20Ma/figure3.gif

I constructed it as follows:

In R2, a hexagon can be regarded as a truncated triangle, that is, when you=
extend the first, third, and fifth side of a hexagon, you get a triangle. =
In H3, when you extend the sides of a hexagon, you don't always get a trian=
gle in the common sense: sometimes the extensions don't meet. But I claim y=
ou always get a trilateral. So I started with a regular {6,4} tiling, and a=
pplied the extensions to get the tiling of trilaterals.

And I believe we can do similar things in H3: extend a properly scaled trun=
cated tetrahedron to construct a "tetrahedron" with no vertices. Fortunatel=
y the name "tetrahedron" remains valid because hedron means face rather tha=
n vertices. But I have never done an illustration of it yet. Then, maybe we=
can go ahead and put seven of them around an edge and make a {3,3,7}.=20

I agree that these objects are not conventional at all. We lost something l=
ike the vertices. But just like the above image, they do have nice patterns=
and are something worth considering.=20

So, what do you think about them? Do they sound more legit now?

Nan

--- In 4D_Cubing@yahoogroups.com, Don Hatch wrote:
> {3,3,7} less so... its vertices are not simply at infinity (as in {3,3,6}=
),
> they are "beyond infinity"...
> If you try to draw this one, none of the edges will meet at all (not even=
at
> infinity)... they all diverge! You'll see each edge
> emerging from somewhere on the horizon (although there's no vertex
> there) and leaving somewhere else on the horizon...
> so nothing meets up, which kind of makes the picture less satisfying.
> If you run the formula for edge length or cell circumradius, you'll get, =
not infinity,
> but an imaginary or complex number (although the cell in-radius is finite=
, of
> course, being equal to the half-edge-length of the dual {7,3,3}).

Hi Nan,

Heh, I try not to make judgements...
I think {3,3,7} is as legit as {7,3,3}
(in fact I'd go so far as to say they are the same object,
with different names given to the components).
But pragmatically, {7,3,3} seems easier to get a grip on,
in a viewer program such as yours which focuses naturally
on the vertices and edges.

Perhaps the best way to get a feeling for {3,3,7}
would be to view it together with the {7,3,3}?
Maybe one color for the {7,3,3} edges,
another color for the {3,3,7} edges,
and a third color for the edges formed where
the faces of one intersect the faces of the other.
And then perhaps, optionally,
the full outlines of the characteristic tetrahedra?
There are 6 types of edges in all (6 edges of a characteristic tet);
I wonder if there's a natural coloring scheme
using the 6 primary and secondary colors.

Don


On Sat, Jun 30, 2012 at 11:29:32PM -0000, schuma wrote:
>
>
> Hi Don,
>
> Nice to see you here. Here are my thoughts about {3,3,7} and the things
> similar to it.
>
> Just like when we constructed {7,3,3} we were not able locate the cell
> centers, when we consider {3,3,7} we have to sacrifice the vertices. Let's
> start by considering something simpler in lower dimensions.
>
> For example, in 2D, we could consider a hyperbolic "triangle" for which
> the sides don't meet even at the circle of infinity. The sides are
> ultraparallel. Since there's no "angle", the name "triangle" is not
> appropriate any more. I'll call it a "trilateral", because it does have
> three sides (the common triangle is also a trilateral in my notation).
> Here's a tessellation of H2 using trilaterals, in which different colors
> indicate different trilaterals.
>
> http://games.groups.yahoo.com/group/4D_Cubing/files/Nan%20Ma/figure3.gif
>
> I constructed it as follows:
>
> In R2, a hexagon can be regarded as a truncated triangle, that is, when
> you extend the first, third, and fifth side of a hexagon, you get a
> triangle. In H3, when you extend the sides of a hexagon, you don't always
> get a triangle in the common sense: sometimes the extensions don't meet.
> But I claim you always get a trilateral. So I started with a regular {6,4}
> tiling, and applied the extensions to get the tiling of trilaterals.
>
> And I believe we can do similar things in H3: extend a properly scaled
> truncated tetrahedron to construct a "tetrahedron" with no vertices.
> Fortunately the name "tetrahedron" remains valid because hedron means face
> rather than vertices. But I have never done an illustration of it yet.
> Then, maybe we can go ahead and put seven of them around an edge and make
> a {3,3,7}.
>
> I agree that these objects are not conventional at all. We lost something
> like the vertices. But just like the above image, they do have nice
> patterns and are something worth considering.
>
> So, what do you think about them? Do they sound more legit now?
>
> Nan
>
> --- In 4D_Cubing@yahoogroups.com, Don Hatch wrote:
> > {3,3,7} less so... its vertices are not simply at infinity (as in
> {3,3,6}),
> > they are "beyond infinity"...
> > If you try to draw this one, none of the edges will meet at all (not
> even at
> > infinity)... they all diverge! You'll see each edge
> > emerging from somewhere on the horizon (although there's no vertex
> > there) and leaving somewhere else on the horizon...
> > so nothing meets up, which kind of makes the picture less satisfying.
> > If you run the formula for edge length or cell circumradius, you'll get,
> not infinity,
> > but an imaginary or complex number (although the cell in-radius is
> finite, of
> > course, being equal to the half-edge-length of the dual {7,3,3}).
>
>

--
Don Hatch
hatch@plunk.org
http://www.plunk.org/~hatch/

Hi all,
About {3,3,7} I had some idea (but it was long ago...). We know that its =
cell is a tetragedron that expands infinitely beyond "vertices". For each 3=
its faces we have a plane perpendicular to them (that cuts them is the nar=
rowest place). It we cut {3,3,7} cell by these planes, we get truncated tet=
rahedron with behedral angles =3D 180 deg. What happens if we reflect it ab=
out triangle faces, and continue this process to infinity? It will be some =
"fractal-like" network inscribed in the cell of {3,3,7} - regular polyhedro=
n with infinite numer of infinite faces but with no vertices. I'm sure that=
it has enough regular patterns of face coloring, and it may be a good base=
for 3D puzzle.

Andrey







--- In 4D_Cubing@yahoogroups.com, Don Hatch wrote:
>
> Hi Nan,
>=20
> Heh, I try not to make judgements...
> I think {3,3,7} is as legit as {7,3,3}
> (in fact I'd go so far as to say they are the same object,
> with different names given to the components).
> But pragmatically, {7,3,3} seems easier to get a grip on,
> in a viewer program such as yours which focuses naturally
> on the vertices and edges.
>=20
> Perhaps the best way to get a feeling for {3,3,7}
> would be to view it together with the {7,3,3}?
> Maybe one color for the {7,3,3} edges,
> another color for the {3,3,7} edges,
> and a third color for the edges formed where
> the faces of one intersect the faces of the other.
> And then perhaps, optionally,
> the full outlines of the characteristic tetrahedra?
> There are 6 types of edges in all (6 edges of a characteristic tet);
> I wonder if there's a natural coloring scheme
> using the 6 primary and secondary colors.
>=20
> Don
>=20
>=20
> On Sat, Jun 30, 2012 at 11:29:32PM -0000, schuma wrote:
> >=20=20=20=20=20
> >=20
> > Hi Don,
> >=20
> > Nice to see you here. Here are my thoughts about {3,3,7} and the thi=
ngs
> > similar to it.
> >=20
> > Just like when we constructed {7,3,3} we were not able locate the ce=
ll
> > centers, when we consider {3,3,7} we have to sacrifice the vertices.=
Let's
> > start by considering something simpler in lower dimensions.
> >=20
> > For example, in 2D, we could consider a hyperbolic "triangle" for wh=
ich
> > the sides don't meet even at the circle of infinity. The sides are
> > ultraparallel. Since there's no "angle", the name "triangle" is not
> > appropriate any more. I'll call it a "trilateral", because it does h=
ave
> > three sides (the common triangle is also a trilateral in my notation=
).
> > Here's a tessellation of H2 using trilaterals, in which different co=
lors
> > indicate different trilaterals.
> >=20
> > http://games.groups.yahoo.com/group/4D_Cubing/files/Nan%20Ma/figure3=
.gif
> >=20
> > I constructed it as follows:
> >=20
> > In R2, a hexagon can be regarded as a truncated triangle, that is, w=
hen
> > you extend the first, third, and fifth side of a hexagon, you get a
> > triangle. In H3, when you extend the sides of a hexagon, you don't a=
lways
> > get a triangle in the common sense: sometimes the extensions don't m=
eet.
> > But I claim you always get a trilateral. So I started with a regular=
{6,4}
> > tiling, and applied the extensions to get the tiling of trilaterals.
> >=20
> > And I believe we can do similar things in H3: extend a properly scal=
ed
> > truncated tetrahedron to construct a "tetrahedron" with no vertices.
> > Fortunately the name "tetrahedron" remains valid because hedron mean=
s face
> > rather than vertices. But I have never done an illustration of it ye=
t.
> > Then, maybe we can go ahead and put seven of them around an edge and=
make
> > a {3,3,7}.
> >=20
> > I agree that these objects are not conventional at all. We lost some=
thing
> > like the vertices. But just like the above image, they do have nice
> > patterns and are something worth considering.
> >=20
> > So, what do you think about them? Do they sound more legit now?
> >=20
> > Nan
> >=20
> > --- In 4D_Cubing@yahoogroups.com, Don Hatch wrote:
> > > {3,3,7} less so... its vertices are not simply at infinity (as in
> > {3,3,6}),
> > > they are "beyond infinity"...
> > > If you try to draw this one, none of the edges will meet at all (n=
ot
> > even at
> > > infinity)... they all diverge! You'll see each edge
> > > emerging from somewhere on the horizon (although there's no vertex
> > > there) and leaving somewhere else on the horizon...
> > > so nothing meets up, which kind of makes the picture less satisfyi=
ng.
> > > If you run the formula for edge length or cell circumradius, you'l=
l get,
> > not infinity,
> > > but an imaginary or complex number (although the cell in-radius is
> > finite, of
> > > course, being equal to the half-edge-length of the dual {7,3,3}).
> >=20
> >=20=20=20=20
>=20
> --=20
> Don Hatch
> hatch@...
> http://www.plunk.org/~hatch/
>

Hi Roice,

Yeah, I was wondering if there's a meaningful interpretation
of the complex edge lengths too.
I was thinking maybe it's more helpful to just look
at cosh(half edge length) instead of the edge length itself...
that would be a pure imaginary number instead of a complex number,
which is a *little* easier to think about...
and then I was thinking maybe sinh(half edge length) might be
more meaningful than cosh(half edge length)...
I think that's something akin to a half-chord length,
analagous to sin(half edge length) of a spherical tiling
(though I still have a lot of trouble visualizing what that means
in the hyperbolic case).

Sure, I'm interested in what you guys came up with
along the lines of a {3,ultrainfinity}...
I guess it would look like the picture Nan included in his previous e-mail
(obtained by erasing some edges of the {6,4})
however you're free to choose any triangle in-radius
in the range (in-radius of {3,infinity}, infinity], right?
Is there a nicer parametrization of that one degree of freedom?
Or is there some special value which could be regarded as the canonical one?

Don

On Sat, Jun 30, 2012 at 12:39:16PM -0500, Roice Nelson wrote:
>
>
> Hi Don,
> Thanks for your enlightening email, and for correcting some speculations I
> made without thinking deeply enough.
> * I was totally wrong about horosphere cells being finite.
> * When considering {7,3,n} as n increases, I see it was incorrect to
> conclude the heptagon size decreases (some flawed internal reasoning).
> Interesting that the magnitude of the complex edge length starts
> decreasing when n>=7, although I guess those complex outputs are
> pretty meaningless (then again, maybe not!). Since I got the trend
> backwards for low n, I had no idea about the vertices going to
> infinity at n = 6. I should have noticed that {3,3,6}, {4,3,6},
> {5,3,6}, and {6,3,6} all do the same thing. It's noteworthy that the
> "vertices -> infinity" pattern holds for any {p,3,6}, which makes
> sense since the vertex figure is an infinite tiling.
> Offline, Nan and I also discussed the 2D analogue of the {3,3,7},
> something akin to the {3,inf} tiling where the triangle vertices are no
> longer accessible. Let me know if you're interested in that discussion,
> as it would be cool to hear your thoughts.
> Anyway, thanks again. Very cool stuff,
> Roice
> On Fri, Jun 29, 2012 at 7:32 PM, Don Hatch wrote:
>
> Hi Nan,
>
> I just wanted to address a couple of points that caught my eye
> in your message (and in the part of Roice's that you quoted)...
> On Sun, Jun 24, 2012 at 08:14:31AM -0000, schuma wrote:
> >
> >
> > Hi everyone,
> > I'm continuing talking about my honeycomb/polytope viewer applet. I
> added
> > a new honeycomb, and I think it deserves a new topic. This is
> {7,3,3}.
> > Each cell is a hyperbolic tiling {7,3}. Please check it here:
> > http://people.bu.edu/nanma/InsideH3/H3.html
> > I first heard of this thing together with {3,3,7} in emails with
> Roice
> > Nelson. He had been exchanging emails with Andrey Astrelin about
> them. We
> > have NOT seen any publication talking about these honeycombs. Even
> when
> > Coxeter enumerate the hyperbolic honeycombs, he stopped at
> honeycombs like
> > {6,3,3}, where each cell is at most an Euclidean tessellation like
> {6,3}.
> > He said, "we shall restrict consideration to cases where the
> fundamental
> > region of the symmetry group has a finite content" (content =
> volume?),
>
> Right. The fundamental region is the characteristic simplex,
> so (since even ideal simplices have finite volume)
> this is the same as saying that all the vertices
> of the characteristic simplex (i.e. the honeycomb vertex, edge center,
> face center, cell center) are "accessible" (either finite, or infinite
> i.e. at some definite location on the boundary of the poincare ball).
> So you're examining some cases where that condition is partially
> relaxed, i.e. the fundamental region contains more of the horizon than
> just
> isolated points there... and the characteristic tetrahedron
> is actually missing one or more of its vertices.
> > and hence didn't consider {7,3,3}, where each cell is a hyperbolic
> > tessellation {7,3}.
> > We think {3,3,7} and {7,3,3} and other similar objects are
> constructable.
>
> {7,3,3} yes, in the sense that the vertices/edges/faces are finite,
> and there's clear local structure around them, and, as you observe,
> the edge length formula works out fine
> (but not the cell in-radius nor circum-radius formula)
> and you can render it (as you have-- nice!)...
>
> {3,3,7} less so... its vertices are not simply at infinity (as in
> {3,3,6}),
> they are "beyond infinity"...
> If you try to draw this one, none of the edges will meet at all (not
> even at
> infinity)... they all diverge! You'll see each edge
> emerging from somewhere on the horizon (although there's no vertex
> there) and leaving somewhere else on the horizon...
> so nothing meets up, which kind of makes the picture less satisfying.
> If you run the formula for edge length or cell circumradius, you'll get,
> not infinity,
> but an imaginary or complex number (although the cell in-radius is
> finite, of
> course, being equal to the half-edge-length of the dual {7,3,3}).
>
> It may be Coxeter refrained omitted these figures
> because the "beyond infinity" parts are awkward to talk about,
> and if you insist on running the formulas and completing the tables,
> a lot of it will consist of imaginary and complex numbers
> that aren't all that meaningful physically, and might scare some readers
> away
> (even though, as you've noted, some of the entries
> are perfectly fine finite numbers or plain old infinity).
>
> > I derived the edge length of {n,3,3} for general n, and then
> computed the
> > coordinates of several vertices of {7,3,3}, then I plotted them.
> There's
> > really nothing so weird about this honeycomb. It looks just like,
> or, as
> > weird as, {6,3,3}. The volume of the fundamental region of {7,3,3}
> may be
> > infinite, but as long as we talk about the edge length, face area,
> > everything is finite and looks normal.
> > I could go and make {8,3,3}, {9,3,3} etc. I also believe {7,3,4}
> and
> > {7,3,5} are also pretty well behaved, and looks just like {6,3,4}
> and
> > {6,3,5} respectively. Or even {7,4,3}. As long as the vertex figure
> is
> > finite (not like {3,4,4}), the image shouldn't be crazy. Since we
> are
> > facing an infinite number of honeycombs here, I feel I should stop
> at some
> > point. After all we don't understand {7,3,3} well, which is the
> smallest
> > representative of them. I'd like to spend more energy making sense
> of
> > {7,3,3} rather than go further.
> > It's not clear for me whether we can identify some heptagons in
> {7,3} to
> > make it Klein Quartic, in {7,3,3}. For example, in the hypercube
> {4,3,3},
> > we can replace each cubic cell by hemi-cube by identification. The
> result
> > is that all the vertices end up identified as only one vertex. I
> don't
> > know what'll happen if I replace {7,3} by Klein Quartic ({7,3}_8).
> It will
> > be awesome if we can fit three KQ around each edge to make a
> polytope
> > based on {7,3,3}. If "three" doesn't work, maybe the one based on
> {7,3,4}
> > or {7,3,5} works. I actually also don't know what'll happen if I
> replace
> > the dodecahedral cells of 120-cell by hemi-dodecahedra. Does anyone
> know?
> > I still suspect people have discussed it somewhere in literature.
> But I
> > haven't found anything really related. Roice found the following
> statement
> > and references. I don't haven't check them yet.
> > __________
> >
> > I checked 'Abstract Regular Polytopes', and was not able to find
> > anything on the {7,3,3}. H3 honeycombs make several appearances
> at
> > various places in the book, but the language seems to be similar
> to
> > Coxeter, and their charts also limited to the same ones. On page
> 77,
> > they distinguish between "compact" and "non-compact" hyperbolic
> types,
> > and say:
> > Coxeter groups of hyperbolic type exist only in ranks 3 to 10,
> and there
> > are only finitely many such groups in ranks 4 to 10. Groups of
> compact
> > hyperbolic type exist only in ranks 3, 4, and 5.
> > But as best I can tell, "non-compact" still only refers to the
> same
> > infinite honeycombs Coxeter enumerated. They reference the
> following
> > book:
> > J. E. Humphreys, Reflection Groups and Coxeter Groups, Cambridge
> > University Press (Cambridge, 1990).
> > When researching just now on wikipedia, the page on uniform
> hyperbolic
> > honeycombs has a short section on noncompact hyperbolic
> honeycombs, and
> > also references the same book by Humphreys. So maybe this book
> could be
> > a good reference to dig up, even though I suspect it will still
> not
> > mention the {7,3,3}.
> > Also: Abstract Regular Polytopes, p78:
> > For the general theory of hyperbolic reflexion groups, the reader
> is
> > referred to Vinberg [431-433]. We remark that there are examples
> of
> > discrete groups generated by hyperplane reflexions in a
> hyperbolic space
> > which are Coxeter groups, but do not have a simplex as a
> fundamental
> > region.
> > These honeycombs fall into that category.
> > Here are those references:
> > [431] E. B. Vinberg, Discrete groups in Lobachevskii spaces
> generated by
> > reflections, Mat. Sb. 72 (1967), 471-488 (= Math. USSR-Sb. 1
> (1967),
> > 429-444).
> > [432] E. B. Vinberg, Discrete linear groups generated by
> reflections,
> > Izv. Akad. Nauk. SSSR Ser. Mat. 35 (1971) 1072-1112 (= Math.
> USSR-Izv. 5
> > (1971), 1083-1119).
> > [433] E. B. Vinberg, Hyperbolic reflection groups. Uspekhi Mat.
> Nauk 40
> > (1985), 29-66 (= Russian Math. Surveys 40 (1985), 31-75).
> >
> > ______________
> > Now I can only say "to the best of our knowledge, I haven't seen
> any
> > discussion about it".
>
> I noticed one further possible reference in Coxeter's "reguar honeycombs
> in
> hyperbolic space" paper-- on the first page, he refers to:
> "... (Coxeter 1933), not insisting on finite fundamental regions,
> was somewhat lacking in rigour"
> where (Coxeter 1933) is "The densities of the regular polytopes, Part
> 3".
> I believe that paper is in the collection "Kaleidoscopes: Selected
> writings of H.S.M. Coxeter" (my copy of which is buried in a box in
> storage somewhere :-( ). I suspect that one *will* mention the {7,3,3};
> I'd be interested to know what he says about it, now that we're thinking
> along those lines.
>
> > Some more thoughts by Roice:
> > __________
> >
> > We know that for {n,3,3), as n -> 6 from higher values of n, the
> {n,3}
> > tiling approaches a horosphere, reaching it at n = 6.
>
> Right... or more precisely,
> the circumsphere, edge-tangency-sphere, face-tangency-sphere, and
> in-sphere
> all approach horospheres
> (different horospheres, but sharing the same center-at-infinity)...
> > For {7,3,n), as n -> infinity, does the {7,3} tiling approach a
> > horosphere as well?
>
> I'm not completely confident that this will stay meaningful
> as we lose the locations of the vertices (for n >= 7).
> The circum-sphere certainly becomes ill-defined...
> however one or more of the other tangency spheres
> might stay well-defined.
> One concievable outcome might be
> that the in-sphere and mid-spheres approach different limits--
> maybe the in-sphere approaches a horosphere but the face-tangency
> mid-sphere doesn't, and maybe the edge-tangency mid-sphere
> is ill-defined just like the circumsphere is.
>
> In thinking about this,
> I have to first think about the significant
> events that happen for smaller n...
> {7,3,2} two cells, the wall between them tiled with {7,3},
> cell centers are imaginary/complex
> {7,3,3} finite vertex figure and local structure, although
> cell centers are imaginary/complex
> {7,3,4} same
> {7,3,5} same
> {7,3,6} infinite vertex figure, vertices are at infinity (and
> cell centers still imaginary/complex)
> {7,3,7} self-dual; both vertices and cell centers are now
> imaginary/complex
> And now, like Nan, I've lost my intuition...
> {7,3,7} is the one for me to ponder at this point.
> > The curvature definitely flattens out as n
> > increases.
>
> right (i.e. the curvature increases, i.e. becomes less negative)
> > If cells are a horosphere in the limit, a {7,3,infinity}
> > tiling would have finite cells.
>
> hmm? why?
> > It would have an infinite edge-figure,
> > in addition to an infinite vertex-figure, but as Coxeter did an
> > enumeration allowing the latter, why not allow the former?
>
> well, for things like {3,infinity} and {3,3,6}
> with infinite vertex-figure, you can still draw them and measure things
> about them even though the vertex figures are infinite-- the vertices
> are isolated, at least. it seems to me that if the edge figure is
> infinite, then it can no longer have isolated vertices (if the vertices
> are even accessible at all),
> so it's hard to draw a definite picture of anything any more,
> or make any measurements... so we have less and less we can say about
> the thing, I guess.
> > I'd like to
> > understand where in Coxeter's analysis a {7,3,infinity} tiling
> does not
> > fit in. One guess is that even if the {7,3} approaches a
> horosphere,
> > it's volume also goes to 0, so is trivial. The heptagons get
> smaller
> > for larger n,
>
> Are you sure?
> The edge length is finite for {7,3,2...5},
> and infinite for {7,3,6}... that makes me think the heptagons are
> probably *growing*,
> not shrinking, at least for n in that range...
> and after that, the edge length is the acosh of an imaginary number,
> so it's hard to say whether it's growing or shrinking or what.
> To verify, the formula for the half-edge-length is:
>
> acosh(cos(pi/p)*sin(pi/r)/sqrt(1-cos(pi/q)^2-cos(pi/r)^2))
>
> {7,3,2} -> acosh(1.0403492368298681) = 0.28312815336765745
> {7,3,3} -> acosh(1.1034570002469741) = 0.45104488629937328
> {7,3,4} -> acosh(1.2741623922635352) = 0.72453736133879376
> {7,3,5} -> acosh(1.7137446255953275) = 1.1331675164780453
> {7,3,6} -> acosh(+infinity) = +infinity
> {7,3,7} -> acosh(+-1.5731951893240572 i) = (1.2346906773191777 +-
> 1.5707963267948966 i)
> {7,3,8} -> acosh(+-1.0714385881055031 i) = (0.9309971259601171 +-
> 1.5707963267948966 i)
> {7,3,9} -> acosh(+-0.8448884457716658 i) = (0.7673378247178905 +-
> 1.5707963267948966 i)
>
> All that said, I still don't have a clear picture of what happens
> when n goes to infinity. We certainly lose the vertices
> at n=7, so the circum-sphere isn't well-defined...
> and I think we must lose the edges eventually as well? in which case the
> edge-tangency mid-sphere isn't well-defined either...
> but I'm guessing we *don't* lose the face centers...
> so the face-tangency sphere and in-sphere may still be well-defined,
> and may approach a limit,
> in which case if your question has meaning,
> one of those limits would be its meaning (I think). And I don't know
> the answer.
> > so I suppose they must approach 0 size as well.
> > It would also be interesting to consider how curvature changes
> for
> > {n,3,3} as n-> infinity, especially since we already know what
> the
> > {infinity,3} tiling looks like.
> >
> > _______________
> > Currently I can't imagine what {7,3,n} like when n>=6. So I really
> cannot
> > comment on {7,3,infinity}. But {infinity, 3, 3} seems to be a good
> thing
> > to study.
> > My formula for the edge length of {n,3,3} is as follows. Following
> > Coxeter's notation, if 2*phi is the length of an edge of {n,3,3}
> (n>=6),
> > then
> > cosh(2*phi) = 3*cos^2(pi/n) - 1
> > Sanity check: when n = 6, this formula gives cosh(2*phi)=5/4, which
> is
> > consistent with the number in Coxeter's table: cosh^2(phi)=9/8.
> > By sending n to infinity, the edge length of {infinity, 3, 3} is
> > arccosh(2). I should be able to plot it soon.
> > By the way, in the applet there's a "Clifford Torus". It looks much
> more
> > beautiful than the polytopes, because the colors of the edges work
> pretty
> > well here. Imagine you can fly around a donut, or go into the
> donut. The
> > amazing thing is if the space is 3-sphere, the view inside the
> donut is
> > exactly as same as the outside.
> > Nan
> >
>
> Don
>
> --
> Don Hatch
> hatch@plunk.org
> http://www.plunk.org/~hatch/
>
> ------------------------------------
>
> Yahoo! Groups Links
>
>
>
>

--
Don Hatch
hatch@plunk.org
http://www.plunk.org/~hatch/

Hi Andrey,

I'm not sure if I'm understanding correctly...
is "behedral angle" the same as "dihedral angle"?
If so, isn't the dihedral angle going to be 2*pi/7,
since, by definition, 7 tetrahedra surround each edge?

Don


On Tue, Jul 03, 2012 at 05:14:47AM -0000, Andrey wrote:
>
>
> Hi all,
> About {3,3,7} I had some idea (but it was long ago...). We know that its
> cell is a tetragedron that expands infinitely beyond "vertices". For each
> 3 its faces we have a plane perpendicular to them (that cuts them is the
> narrowest place). It we cut {3,3,7} cell by these planes, we get truncated
> tetrahedron with behedral angles = 180 deg. What happens if we reflect it
> about triangle faces, and continue this process to infinity? It will be
> some "fractal-like" network inscribed in the cell of {3,3,7} - regular
> polyhedron with infinite numer of infinite faces but with no vertices. I'm
> sure that it has enough regular patterns of face coloring, and it may be a
> good base for 3D puzzle.
>
> Andrey
>
> --- In 4D_Cubing@yahoogroups.com, Don Hatch wrote:
> >
> > Hi Nan,
> >
> > Heh, I try not to make judgements...
> > I think {3,3,7} is as legit as {7,3,3}
> > (in fact I'd go so far as to say they are the same object,
> > with different names given to the components).
> > But pragmatically, {7,3,3} seems easier to get a grip on,
> > in a viewer program such as yours which focuses naturally
> > on the vertices and edges.
> >
> > Perhaps the best way to get a feeling for {3,3,7}
> > would be to view it together with the {7,3,3}?
> > Maybe one color for the {7,3,3} edges,
> > another color for the {3,3,7} edges,
> > and a third color for the edges formed where
> > the faces of one intersect the faces of the other.
> > And then perhaps, optionally,
> > the full outlines of the characteristic tetrahedra?
> > There are 6 types of edges in all (6 edges of a characteristic tet);
> > I wonder if there's a natural coloring scheme
> > using the 6 primary and secondary colors.
> >
> > Don
> >
> >
> > On Sat, Jun 30, 2012 at 11:29:32PM -0000, schuma wrote:
> > >
> > >
> > > Hi Don,
> > >
> > > Nice to see you here. Here are my thoughts about {3,3,7} and the
> things
> > > similar to it.
> > >
> > > Just like when we constructed {7,3,3} we were not able locate the cell
> > > centers, when we consider {3,3,7} we have to sacrifice the vertices.
> Let's
> > > start by considering something simpler in lower dimensions.
> > >
> > > For example, in 2D, we could consider a hyperbolic "triangle" for
> which
> > > the sides don't meet even at the circle of infinity. The sides are
> > > ultraparallel. Since there's no "angle", the name "triangle" is not
> > > appropriate any more. I'll call it a "trilateral", because it does
> have
> > > three sides (the common triangle is also a trilateral in my notation).
> > > Here's a tessellation of H2 using trilaterals, in which different
> colors
> > > indicate different trilaterals.
> > >
> > >
> http://games.groups.yahoo.com/group/4D_Cubing/files/Nan%20Ma/figure3.gif
> > >
> > > I constructed it as follows:
> > >
> > > In R2, a hexagon can be regarded as a truncated triangle, that is,
> when
> > > you extend the first, third, and fifth side of a hexagon, you get a
> > > triangle. In H3, when you extend the sides of a hexagon, you don't
> always
> > > get a triangle in the common sense: sometimes the extensions don't
> meet.
> > > But I claim you always get a trilateral. So I started with a regular
> {6,4}
> > > tiling, and applied the extensions to get the tiling of trilaterals.
> > >
> > > And I believe we can do similar things in H3: extend a properly scaled
> > > truncated tetrahedron to construct a "tetrahedron" with no vertices.
> > > Fortunately the name "tetrahedron" remains valid because hedron means
> face
> > > rather than vertices. But I have never done an illustration of it yet.
> > > Then, maybe we can go ahead and put seven of them around an edge and
> make
> > > a {3,3,7}.
> > >
> > > I agree that these objects are not conventional at all. We lost
> something
> > > like the vertices. But just like the above image, they do have nice
> > > patterns and are something worth considering.
> > >
> > > So, what do you think about them? Do they sound more legit now?
> > >
> > > Nan
> > >
> > > --- In 4D_Cubing@yahoogroups.com, Don Hatch wrote:
> > > > {3,3,7} less so... its vertices are not simply at infinity (as in
> > > {3,3,6}),
> > > > they are "beyond infinity"...
> > > > If you try to draw this one, none of the edges will meet at all (not
> > > even at
> > > > infinity)... they all diverge! You'll see each edge
> > > > emerging from somewhere on the horizon (although there's no vertex
> > > > there) and leaving somewhere else on the horizon...
> > > > so nothing meets up, which kind of makes the picture less
> satisfying.
> > > > If you run the formula for edge length or cell circumradius, you'll
> get,
> > > not infinity,
> > > > but an imaginary or complex number (although the cell in-radius is
> > > finite, of
> > > > course, being equal to the half-edge-length of the dual {7,3,3}).
> > >
> > >
> >
> > --
> > Don Hatch
> > hatch@...
> > http://www.plunk.org/~hatch/
> >
>
>

--
Don Hatch
hatch@plunk.org
http://www.plunk.org/~hatch/

Yes, dihedral... I mean angle between hexagonal and triangular faces of tru=
ncated tetrahedron. By the selection of truncating planes it will be pi/2. =
And angles between hexagonal faces are 2*pi/7.

Andrey


--- In 4D_Cubing@yahoogroups.com, Don Hatch wrote:
>
> Hi Andrey,
>=20
> I'm not sure if I'm understanding correctly...
> is "behedral angle" the same as "dihedral angle"?
> If so, isn't the dihedral angle going to be 2*pi/7,
> since, by definition, 7 tetrahedra surround each edge?
>=20
> Don
>=20
>=20
> On Tue, Jul 03, 2012 at 05:14:47AM -0000, Andrey wrote:
> >=20=20=20=20=20
> >=20
> > Hi all,
> > About {3,3,7} I had some idea (but it was long ago...). We know that=
its
> > cell is a tetragedron that expands infinitely beyond "vertices". For=
each
> > 3 its faces we have a plane perpendicular to them (that cuts them is=
the
> > narrowest place). It we cut {3,3,7} cell by these planes, we get tru=
ncated
> > tetrahedron with behedral angles =3D 180 deg. What happens if we ref=
lect it
> > about triangle faces, and continue this process to infinity? It will=
be
> > some "fractal-like" network inscribed in the cell of {3,3,7} - regul=
ar
> > polyhedron with infinite numer of infinite faces but with no vertice=
s. I'm
> > sure that it has enough regular patterns of face coloring, and it ma=
y be a
> > good base for 3D puzzle.
> >=20
> > Andrey
> >=20
> > --- In 4D_Cubing@yahoogroups.com, Don Hatch wrote:
> > >
> > > Hi Nan,
> > >
> > > Heh, I try not to make judgements...
> > > I think {3,3,7} is as legit as {7,3,3}
> > > (in fact I'd go so far as to say they are the same object,
> > > with different names given to the components).
> > > But pragmatically, {7,3,3} seems easier to get a grip on,
> > > in a viewer program such as yours which focuses naturally
> > > on the vertices and edges.
> > >
> > > Perhaps the best way to get a feeling for {3,3,7}
> > > would be to view it together with the {7,3,3}?
> > > Maybe one color for the {7,3,3} edges,
> > > another color for the {3,3,7} edges,
> > > and a third color for the edges formed where
> > > the faces of one intersect the faces of the other.
> > > And then perhaps, optionally,
> > > the full outlines of the characteristic tetrahedra?
> > > There are 6 types of edges in all (6 edges of a characteristic tet=
);
> > > I wonder if there's a natural coloring scheme
> > > using the 6 primary and secondary colors.
> > >
> > > Don
> > >
> > >
> > > On Sat, Jun 30, 2012 at 11:29:32PM -0000, schuma wrote:
> > > >
> > > >
> > > > Hi Don,
> > > >
> > > > Nice to see you here. Here are my thoughts about {3,3,7} and the
> > things
> > > > similar to it.
> > > >
> > > > Just like when we constructed {7,3,3} we were not able locate th=
e cell
> > > > centers, when we consider {3,3,7} we have to sacrifice the verti=
ces.
> > Let's
> > > > start by considering something simpler in lower dimensions.
> > > >
> > > > For example, in 2D, we could consider a hyperbolic "triangle" fo=
r
> > which
> > > > the sides don't meet even at the circle of infinity. The sides a=
re
> > > > ultraparallel. Since there's no "angle", the name "triangle" is =
not
> > > > appropriate any more. I'll call it a "trilateral", because it do=
es
> > have
> > > > three sides (the common triangle is also a trilateral in my nota=
tion).
> > > > Here's a tessellation of H2 using trilaterals, in which differen=
t
> > colors
> > > > indicate different trilaterals.
> > > >
> > > >
> > http://games.groups.yahoo.com/group/4D_Cubing/files/Nan%20Ma/figure3=
.gif
> > > >
> > > > I constructed it as follows:
> > > >
> > > > In R2, a hexagon can be regarded as a truncated triangle, that i=
s,
> > when
> > > > you extend the first, third, and fifth side of a hexagon, you ge=
t a
> > > > triangle. In H3, when you extend the sides of a hexagon, you don=
't
> > always
> > > > get a triangle in the common sense: sometimes the extensions don=
't
> > meet.
> > > > But I claim you always get a trilateral. So I started with a reg=
ular
> > {6,4}
> > > > tiling, and applied the extensions to get the tiling of trilater=
als.
> > > >
> > > > And I believe we can do similar things in H3: extend a properly =
scaled
> > > > truncated tetrahedron to construct a "tetrahedron" with no verti=
ces.
> > > > Fortunately the name "tetrahedron" remains valid because hedron =
means
> > face
> > > > rather than vertices. But I have never done an illustration of i=
t yet.
> > > > Then, maybe we can go ahead and put seven of them around an edge=
and
> > make
> > > > a {3,3,7}.
> > > >
> > > > I agree that these objects are not conventional at all. We lost
> > something
> > > > like the vertices. But just like the above image, they do have n=
ice
> > > > patterns and are something worth considering.
> > > >
> > > > So, what do you think about them? Do they sound more legit now?
> > > >
> > > > Nan
> > > >
> > > > --- In 4D_Cubing@yahoogroups.com, Don Hatch wrote:
> > > > > {3,3,7} less so... its vertices are not simply at infinity (as=
in
> > > > {3,3,6}),
> > > > > they are "beyond infinity"...
> > > > > If you try to draw this one, none of the edges will meet at al=
l (not
> > > > even at
> > > > > infinity)... they all diverge! You'll see each edge
> > > > > emerging from somewhere on the horizon (although there's no ve=
rtex
> > > > > there) and leaving somewhere else on the horizon...
> > > > > so nothing meets up, which kind of makes the picture less
> > satisfying.
> > > > > If you run the formula for edge length or cell circumradius, y=
ou'll
> > get,
> > > > not infinity,
> > > > > but an imaginary or complex number (although the cell in-radiu=
s is
> > > > finite, of
> > > > > course, being equal to the half-edge-length of the dual {7,3,3=
}).
> > > >
> > > >
> > >
> > > --
> > > Don Hatch
> > > hatch@
> > > http://www.plunk.org/~hatch/
> > >
> >=20
> >=20=20=20=20
>=20
> --=20
> Don Hatch
> hatch@...
> http://www.plunk.org/~hatch/
>

Natural parameter of ultrainfinite trangles is a distance between its sides=
(but radius of the inscribed circle also works). In H2 pictures with diffe=
rent parameters look equivalent, but faces of cells of different {3,3,n} ho=
neycombs will have different parameters.

Andrey


--- In 4D_Cubing@yahoogroups.com, Don Hatch wrote:
>
> Hi Roice,
>=20
> Yeah, I was wondering if there's a meaningful interpretation
> of the complex edge lengths too.
> I was thinking maybe it's more helpful to just look
> at cosh(half edge length) instead of the edge length itself...
> that would be a pure imaginary number instead of a complex number,
> which is a *little* easier to think about...
> and then I was thinking maybe sinh(half edge length) might be
> more meaningful than cosh(half edge length)...
> I think that's something akin to a half-chord length,
> analagous to sin(half edge length) of a spherical tiling
> (though I still have a lot of trouble visualizing what that means
> in the hyperbolic case).
>=20
> Sure, I'm interested in what you guys came up with
> along the lines of a {3,ultrainfinity}...
> I guess it would look like the picture Nan included in his previous e-mai=
l
> (obtained by erasing some edges of the {6,4})
> however you're free to choose any triangle in-radius
> in the range (in-radius of {3,infinity}, infinity], right?
> Is there a nicer parametrization of that one degree of freedom?
> Or is there some special value which could be regarded as the canonical o=
ne?
>=20
> Don
>=20
> On Sat, Jun 30, 2012 at 12:39:16PM -0500, Roice Nelson wrote:
> >=20=20=20=20=20
> >=20
> > Hi Don,
> > Thanks for your enlightening email, and for correcting some speculat=
ions I
> > made without thinking deeply enough.=20=20
> > * I was totally wrong about horosphere cells being finite.=20=20
> > * When considering {7,3,n} as n increases, I see it was incorrect =
to
> > conclude the heptagon size decreases (some flawed internal reaso=
ning).
> > Interesting that the magnitude of the complex edge length start=
s
> > decreasing when n>=3D7, although I guess those complex outputs a=
re
> > pretty meaningless (then again, maybe not!). Since I got the tr=
end
> > backwards for low n, I had no idea about the vertices going to
> > infinity at n =3D 6. I should have noticed that {3,3,6}, {4,3,6=
},
> > {5,3,6}, and {6,3,6} all do the same thing. It's noteworthy tha=
t the
> > "vertices -> infinity" pattern holds for any {p,3,6}, which make=
s
> > sense since the vertex figure is an infinite tiling.
> > Offline, Nan and I also discussed the 2D analogue of the {3,3,7},
> > something akin to the {3,inf} tiling where the triangle vertices are=
no
> > longer accessible. Let me know if you're interested in that discuss=
ion,
> > as it would be cool to hear your thoughts.
> > Anyway, thanks again. Very cool stuff,
> > Roice
> > On Fri, Jun 29, 2012 at 7:32 PM, Don Hatch wrote:
> >=20
> > Hi Nan,
> >=20
> > I just wanted to address a couple of points that caught my eye
> > in your message (and in the part of Roice's that you quoted)...
> > On Sun, Jun 24, 2012 at 08:14:31AM -0000, schuma wrote:
> > >
> > >
> > > Hi everyone,
> > > I'm continuing talking about my honeycomb/polytope viewer app=
let. I
> > added
> > > a new honeycomb, and I think it deserves a new topic. This is
> > {7,3,3}.
> > > Each cell is a hyperbolic tiling {7,3}. Please check it here:
> > > http://people.bu.edu/nanma/InsideH3/H3.html
> > > I first heard of this thing together with {3,3,7} in emails w=
ith
> > Roice
> > > Nelson. He had been exchanging emails with Andrey Astrelin ab=
out
> > them. We
> > > have NOT seen any publication talking about these honeycombs.=
Even
> > when
> > > Coxeter enumerate the hyperbolic honeycombs, he stopped at
> > honeycombs like
> > > {6,3,3}, where each cell is at most an Euclidean tessellation=
like
> > {6,3}.
> > > He said, "we shall restrict consideration to cases where the
> > fundamental
> > > region of the symmetry group has a finite content" (content =
=3D
> > volume?),
> >=20
> > Right. The fundamental region is the characteristic simplex,
> > so (since even ideal simplices have finite volume)
> > this is the same as saying that all the vertices
> > of the characteristic simplex (i.e. the honeycomb vertex, edge cen=
ter,
> > face center, cell center) are "accessible" (either finite, or infi=
nite
> > i.e. at some definite location on the boundary of the poincare bal=
l).
> > So you're examining some cases where that condition is partially
> > relaxed, i.e. the fundamental region contains more of the horizon =
than
> > just
> > isolated points there... and the characteristic tetrahedron
> > is actually missing one or more of its vertices.
> > > and hence didn't consider {7,3,3}, where each cell is a hyper=
bolic
> > > tessellation {7,3}.
> > > We think {3,3,7} and {7,3,3} and other similar objects are
> > constructable.
> >=20
> > {7,3,3} yes, in the sense that the vertices/edges/faces are finite=
,
> > and there's clear local structure around them, and, as you observe=
,
> > the edge length formula works out fine
> > (but not the cell in-radius nor circum-radius formula)
> > and you can render it (as you have-- nice!)...
> >=20
> > {3,3,7} less so... its vertices are not simply at infinity (as in
> > {3,3,6}),
> > they are "beyond infinity"...
> > If you try to draw this one, none of the edges will meet at all (n=
ot
> > even at
> > infinity)... they all diverge! You'll see each edge
> > emerging from somewhere on the horizon (although there's no vertex
> > there) and leaving somewhere else on the horizon...
> > so nothing meets up, which kind of makes the picture less satisfyi=
ng.
> > If you run the formula for edge length or cell circumradius, you'l=
l get,
> > not infinity,
> > but an imaginary or complex number (although the cell in-radius is
> > finite, of
> > course, being equal to the half-edge-length of the dual {7,3,3}).
> >=20
> > It may be Coxeter refrained omitted these figures
> > because the "beyond infinity" parts are awkward to talk about,
> > and if you insist on running the formulas and completing the table=
s,
> > a lot of it will consist of imaginary and complex numbers
> > that aren't all that meaningful physically, and might scare some r=
eaders
> > away
> > (even though, as you've noted, some of the entries
> > are perfectly fine finite numbers or plain old infinity).
> >=20
> > > I derived the edge length of {n,3,3} for general n, and then
> > computed the
> > > coordinates of several vertices of {7,3,3}, then I plotted th=
em.
> > There's
> > > really nothing so weird about this honeycomb. It looks just l=
ike,
> > or, as
> > > weird as, {6,3,3}. The volume of the fundamental region of {7=
,3,3}
> > may be
> > > infinite, but as long as we talk about the edge length, face =
area,
> > > everything is finite and looks normal.
> > > I could go and make {8,3,3}, {9,3,3} etc. I also believe {7,3=
,4}
> > and
> > > {7,3,5} are also pretty well behaved, and looks just like {6,=
3,4}
> > and
> > > {6,3,5} respectively. Or even {7,4,3}. As long as the vertex =
figure
> > is
> > > finite (not like {3,4,4}), the image shouldn't be crazy. Sinc=
e we
> > are
> > > facing an infinite number of honeycombs here, I feel I should=
stop
> > at some
> > > point. After all we don't understand {7,3,3} well, which is t=
he
> > smallest
> > > representative of them. I'd like to spend more energy making =
sense
> > of
> > > {7,3,3} rather than go further.
> > > It's not clear for me whether we can identify some heptagons =
in
> > {7,3} to
> > > make it Klein Quartic, in {7,3,3}. For example, in the hyperc=
ube
> > {4,3,3},
> > > we can replace each cubic cell by hemi-cube by identification=
. The
> > result
> > > is that all the vertices end up identified as only one vertex=
. I
> > don't
> > > know what'll happen if I replace {7,3} by Klein Quartic ({7,3=
}_8).
> > It will
> > > be awesome if we can fit three KQ around each edge to make a
> > polytope
> > > based on {7,3,3}. If "three" doesn't work, maybe the one base=
d on
> > {7,3,4}
> > > or {7,3,5} works. I actually also don't know what'll happen i=
f I
> > replace
> > > the dodecahedral cells of 120-cell by hemi-dodecahedra. Does =
anyone
> > know?
> > > I still suspect people have discussed it somewhere in literat=
ure.
> > But I
> > > haven't found anything really related. Roice found the follow=
ing
> > statement
> > > and references. I don't haven't check them yet.
> > > __________
> > >
> > > I checked 'Abstract Regular Polytopes', and was not able to=
find
> > > anything on the {7,3,3}. H3 honeycombs make several appear=
ances
> > at
> > > various places in the book, but the language seems to be si=
milar
> > to
> > > Coxeter, and their charts also limited to the same ones. O=
n page
> > 77,
> > > they distinguish between "compact" and "non-compact" hyperb=
olic
> > types,
> > > and say:
> > > Coxeter groups of hyperbolic type exist only in ranks 3 to =
10,
> > and there
> > > are only finitely many such groups in ranks 4 to 10. Group=
s of
> > compact
> > > hyperbolic type exist only in ranks 3, 4, and 5.
> > > But as best I can tell, "non-compact" still only refers to =
the
> > same
> > > infinite honeycombs Coxeter enumerated. They reference the
> > following
> > > book:
> > > J. E. Humphreys, Reflection Groups and Coxeter Groups, Camb=
ridge
> > > University Press (Cambridge, 1990).
> > > When researching just now on wikipedia, the page on uniform
> > hyperbolic
> > > honeycombs has a short section on noncompact hyperbolic
> > honeycombs, and
> > > also references the same book by Humphreys. So maybe this =
book
> > could be
> > > a good reference to dig up, even though I suspect it will s=
till
> > not
> > > mention the {7,3,3}.
> > > Also: Abstract Regular Polytopes, p78:
> > > For the general theory of hyperbolic reflexion groups, the =
reader
> > is
> > > referred to Vinberg [431-433]. We remark that there are ex=
amples
> > of
> > > discrete groups generated by hyperplane reflexions in a
> > hyperbolic space
> > > which are Coxeter groups, but do not have a simplex as a
> > fundamental
> > > region.
> > > These honeycombs fall into that category.
> > > Here are those references:
> > > [431] E. B. Vinberg, Discrete groups in Lobachevskii spaces
> > generated by
> > > reflections, Mat. Sb. 72 (1967), 471-488 (=3D Math. USSR-Sb=
. 1
> > (1967),
> > > 429-444).
> > > [432] E. B. Vinberg, Discrete linear groups generated by
> > reflections,
> > > Izv. Akad. Nauk. SSSR Ser. Mat. 35 (1971) 1072-1112 (=3D Ma=
th.
> > USSR-Izv. 5
> > > (1971), 1083-1119).
> > > [433] E. B. Vinberg, Hyperbolic reflection groups. Uspekhi =
Mat.
> > Nauk 40
> > > (1985), 29-66 (=3D Russian Math. Surveys 40 (1985), 31-75).
> > >
> > > ______________
> > > Now I can only say "to the best of our knowledge, I haven't s=
een
> > any
> > > discussion about it".
> >=20
> > I noticed one further possible reference in Coxeter's "reguar hone=
ycombs
> > in
> > hyperbolic space" paper-- on the first page, he refers to:
> > "... (Coxeter 1933), not insisting on finite fundamental region=
s,
> > was somewhat lacking in rigour"
> > where (Coxeter 1933) is "The densities of the regular polytopes, P=
art
> > 3".
> > I believe that paper is in the collection "Kaleidoscopes: Selected
> > writings of H.S.M. Coxeter" (my copy of which is buried in a box i=
n
> > storage somewhere :-( ). I suspect that one *will* mention the {7=
,3,3};
> > I'd be interested to know what he says about it, now that we're th=
inking
> > along those lines.
> >=20
> > > Some more thoughts by Roice:
> > > __________
> > >
> > > We know that for {n,3,3), as n -> 6 from higher values of n=
, the
> > {n,3}
> > > tiling approaches a horosphere, reaching it at n =3D 6.
> >=20
> > Right... or more precisely,
> > the circumsphere, edge-tangency-sphere, face-tangency-sphere, and
> > in-sphere
> > all approach horospheres
> > (different horospheres, but sharing the same center-at-infinity)..=
.
> > > For {7,3,n), as n -> infinity, does the {7,3} tiling approa=
ch a
> > > horosphere as well?
> >=20
> > I'm not completely confident that this will stay meaningful
> > as we lose the locations of the vertices (for n >=3D 7).
> > The circum-sphere certainly becomes ill-defined...
> > however one or more of the other tangency spheres
> > might stay well-defined.
> > One concievable outcome might be
> > that the in-sphere and mid-spheres approach different limits--
> > maybe the in-sphere approaches a horosphere but the face-tangency
> > mid-sphere doesn't, and maybe the edge-tangency mid-sphere
> > is ill-defined just like the circumsphere is.
> >=20
> > In thinking about this,
> > I have to first think about the significant
> > events that happen for smaller n...
> > {7,3,2} two cells, the wall between them tiled with {7,3=
},
> > cell centers are imaginary/complex
> > {7,3,3} finite vertex figure and local structure, althou=
gh
> > cell centers are imaginary/complex
> > {7,3,4} same
> > {7,3,5} same
> > {7,3,6} infinite vertex figure, vertices are at infinity=
(and
> > cell centers still imaginary/complex)
> > {7,3,7} self-dual; both vertices and cell centers are no=
w
> > imaginary/complex
> > And now, like Nan, I've lost my intuition...
> > {7,3,7} is the one for me to ponder at this point.
> > > The curvature definitely flattens out as n
> > > increases.
> >=20
> > right (i.e. the curvature increases, i.e. becomes less negative)
> > > If cells are a horosphere in the limit, a {7,3,infinity}
> > > tiling would have finite cells.
> >=20
> > hmm? why?
> > > It would have an infinite edge-figure,
> > > in addition to an infinite vertex-figure, but as Coxeter di=
d an
> > > enumeration allowing the latter, why not allow the former?
> >=20
> > well, for things like {3,infinity} and {3,3,6}
> > with infinite vertex-figure, you can still draw them and measure t=
hings
> > about them even though the vertex figures are infinite-- the verti=
ces
> > are isolated, at least. it seems to me that if the edge figure is
> > infinite, then it can no longer have isolated vertices (if the ver=
tices
> > are even accessible at all),
> > so it's hard to draw a definite picture of anything any more,
> > or make any measurements... so we have less and less we can say ab=
out
> > the thing, I guess.
> > > I'd like to
> > > understand where in Coxeter's analysis a {7,3,infinity} ti=
ling
> > does not
> > > fit in. One guess is that even if the {7,3} approaches a
> > horosphere,
> > > it's volume also goes to 0, so is trivial. The heptagons g=
et
> > smaller
> > > for larger n,
> >=20
> > Are you sure?
> > The edge length is finite for {7,3,2...5},
> > and infinite for {7,3,6}... that makes me think the heptagons are
> > probably *growing*,
> > not shrinking, at least for n in that range...
> > and after that, the edge length is the acosh of an imaginary numbe=
r,
> > so it's hard to say whether it's growing or shrinking or what.
> > To verify, the formula for the half-edge-length is:
> >=20
> > acosh(cos(pi/p)*sin(pi/r)/sqrt(1-cos(pi/q)^2-cos(pi/r)^2))
> >=20
> > {7,3,2} -> acosh(1.0403492368298681) =3D 0.28312815336765745
> > {7,3,3} -> acosh(1.1034570002469741) =3D 0.45104488629937328
> > {7,3,4} -> acosh(1.2741623922635352) =3D 0.72453736133879376
> > {7,3,5} -> acosh(1.7137446255953275) =3D 1.1331675164780453
> > {7,3,6} -> acosh(+infinity) =3D +infinity
> > {7,3,7} -> acosh(+-1.5731951893240572 i) =3D (1.234690677319177=
7 +-
> > 1.5707963267948966 i)
> > {7,3,8} -> acosh(+-1.0714385881055031 i) =3D (0.930997125960117=
1 +-
> > 1.5707963267948966 i)
> > {7,3,9} -> acosh(+-0.8448884457716658 i) =3D (0.767337824717890=
5 +-
> > 1.5707963267948966 i)
> >=20
> > All that said, I still don't have a clear picture of what happens
> > when n goes to infinity. We certainly lose the vertices
> > at n=3D7, so the circum-sphere isn't well-defined...
> > and I think we must lose the edges eventually as well? in which ca=
se the
> > edge-tangency mid-sphere isn't well-defined either...
> > but I'm guessing we *don't* lose the face centers...
> > so the face-tangency sphere and in-sphere may still be well-define=
d,
> > and may approach a limit,
> > in which case if your question has meaning,
> > one of those limits would be its meaning (I think). And I don't k=
now
> > the answer.
> > > so I suppose they must approach 0 size as well.
> > > It would also be interesting to consider how curvature chan=
ges
> > for
> > > {n,3,3} as n-> infinity, especially since we already know w=
hat
> > the
> > > {infinity,3} tiling looks like.
> > >
> > > _______________
> > > Currently I can't imagine what {7,3,n} like when n>=3D6. So I=
really
> > cannot
> > > comment on {7,3,infinity}. But {infinity, 3, 3} seems to be a=
good
> > thing
> > > to study.
> > > My formula for the edge length of {n,3,3} is as follows. Foll=
owing
> > > Coxeter's notation, if 2*phi is the length of an edge of {n,3=
,3}
> > (n>=3D6),
> > > then
> > > cosh(2*phi) =3D 3*cos^2(pi/n) - 1
> > > Sanity check: when n =3D 6, this formula gives cosh(2*phi)=3D=
5/4, which
> > is
> > > consistent with the number in Coxeter's table: cosh^2(phi)=3D=
9/8.
> > > By sending n to infinity, the edge length of {infinity, 3, 3}=
is
> > > arccosh(2). I should be able to plot it soon.
> > > By the way, in the applet there's a "Clifford Torus". It look=
s much
> > more
> > > beautiful than the polytopes, because the colors of the edges=
work
> > pretty
> > > well here. Imagine you can fly around a donut, or go into the
> > donut. The
> > > amazing thing is if the space is 3-sphere, the view inside th=
e
> > donut is
> > > exactly as same as the outside.
> > > Nan
> > >
> >=20
> > Don
> >=20
> > --
> > Don Hatch
> > hatch@...
> > http://www.plunk.org/~hatch/
> >=20
> > ------------------------------------
> >=20
> > Yahoo! Groups Links
> >=20
> >=20
> >=20
> >=20=20=20=20
>=20
> --=20
> Don Hatch
> hatch@...
> http://www.plunk.org/~hatch/
>

Okay I think maybe I follow you now...
But each face formed by truncation...
it's a triangle, not a hexagon, right?
In fact it's a spherical triangle, on the sphere at infinity, right?
All of these spherical triangles, of different apparent sizes,
would tile the sphere, 7 at each vertex...
but with some "cluster points" which are the limit points
of infinitely many of these triangles of decreasing size.
I'd like to see a picture of this-- it shouldn't be too hard to generate
(together with the spherical circles
that are the intersection of the dual {7,3,3} with the sphere
at infinity, in a different color... I think each cluster point
would be the center of one of these circles).

Don

On Tue, Jul 03, 2012 at 02:43:52PM -0000, Andrey wrote:
>
>
> Yes, dihedral... I mean angle between hexagonal and triangular faces of
> truncated tetrahedron. By the selection of truncating planes it will be
> pi/2. And angles between hexagonal faces are 2*pi/7.
>
> Andrey
>
> --- In 4D_Cubing@yahoogroups.com, Don Hatch wrote:
> >
> > Hi Andrey,
> >
> > I'm not sure if I'm understanding correctly...
> > is "behedral angle" the same as "dihedral angle"?
> > If so, isn't the dihedral angle going to be 2*pi/7,
> > since, by definition, 7 tetrahedra surround each edge?
> >
> > Don
> >
> >
> > On Tue, Jul 03, 2012 at 05:14:47AM -0000, Andrey wrote:
> > >
> > >
> > > Hi all,
> > > About {3,3,7} I had some idea (but it was long ago...). We know that
> its
> > > cell is a tetragedron that expands infinitely beyond "vertices". For
> each
> > > 3 its faces we have a plane perpendicular to them (that cuts them is
> the
> > > narrowest place). It we cut {3,3,7} cell by these planes, we get
> truncated
> > > tetrahedron with behedral angles = 180 deg. What happens if we reflect
> it
> > > about triangle faces, and continue this process to infinity? It will
> be
> > > some "fractal-like" network inscribed in the cell of {3,3,7} - regular
> > > polyhedron with infinite numer of infinite faces but with no vertices.
> I'm
> > > sure that it has enough regular patterns of face coloring, and it may
> be a
> > > good base for 3D puzzle.
> > >
> > > Andrey
> > >
> > > --- In 4D_Cubing@yahoogroups.com, Don Hatch wrote:
> > > >
> > > > Hi Nan,
> > > >
> > > > Heh, I try not to make judgements...
> > > > I think {3,3,7} is as legit as {7,3,3}
> > > > (in fact I'd go so far as to say they are the same object,
> > > > with different names given to the components).
> > > > But pragmatically, {7,3,3} seems easier to get a grip on,
> > > > in a viewer program such as yours which focuses naturally
> > > > on the vertices and edges.
> > > >
> > > > Perhaps the best way to get a feeling for {3,3,7}
> > > > would be to view it together with the {7,3,3}?
> > > > Maybe one color for the {7,3,3} edges,
> > > > another color for the {3,3,7} edges,
> > > > and a third color for the edges formed where
> > > > the faces of one intersect the faces of the other.
> > > > And then perhaps, optionally,
> > > > the full outlines of the characteristic tetrahedra?
> > > > There are 6 types of edges in all (6 edges of a characteristic tet);
> > > > I wonder if there's a natural coloring scheme
> > > > using the 6 primary and secondary colors.
> > > >
> > > > Don
> > > >
> > > >
> > > > On Sat, Jun 30, 2012 at 11:29:32PM -0000, schuma wrote:
> > > > >
> > > > >
> > > > > Hi Don,
> > > > >
> > > > > Nice to see you here. Here are my thoughts about {3,3,7} and the
> > > things
> > > > > similar to it.
> > > > >
> > > > > Just like when we constructed {7,3,3} we were not able locate the
> cell
> > > > > centers, when we consider {3,3,7} we have to sacrifice the
> vertices.
> > > Let's
> > > > > start by considering something simpler in lower dimensions.
> > > > >
> > > > > For example, in 2D, we could consider a hyperbolic "triangle" for
> > > which
> > > > > the sides don't meet even at the circle of infinity. The sides are
> > > > > ultraparallel. Since there's no "angle", the name "triangle" is
> not
> > > > > appropriate any more. I'll call it a "trilateral", because it does
> > > have
> > > > > three sides (the common triangle is also a trilateral in my
> notation).
> > > > > Here's a tessellation of H2 using trilaterals, in which different
> > > colors
> > > > > indicate different trilaterals.
> > > > >
> > > > >
> > >
> http://games.groups.yahoo.com/group/4D_Cubing/files/Nan%20Ma/figure3.gif
> > > > >
> > > > > I constructed it as follows:
> > > > >
> > > > > In R2, a hexagon can be regarded as a truncated triangle, that is,
> > > when
> > > > > you extend the first, third, and fifth side of a hexagon, you get
> a
> > > > > triangle. In H3, when you extend the sides of a hexagon, you don't
> > > always
> > > > > get a triangle in the common sense: sometimes the extensions don't
> > > meet.
> > > > > But I claim you always get a trilateral. So I started with a
> regular
> > > {6,4}
> > > > > tiling, and applied the extensions to get the tiling of
> trilaterals.
> > > > >
> > > > > And I believe we can do similar things in H3: extend a properly
> scaled
> > > > > truncated tetrahedron to construct a "tetrahedron" with no
> vertices.
> > > > > Fortunately the name "tetrahedron" remains valid because hedron
> means
> > > face
> > > > > rather than vertices. But I have never done an illustration of it
> yet.
> > > > > Then, maybe we can go ahead and put seven of them around an edge
> and
> > > make
> > > > > a {3,3,7}.
> > > > >
> > > > > I agree that these objects are not conventional at all. We lost
> > > something
> > > > > like the vertices. But just like the above image, they do have
> nice
> > > > > patterns and are something worth considering.
> > > > >
> > > > > So, what do you think about them? Do they sound more legit now?
> > > > >
> > > > > Nan
> > > > >
> > > > > --- In 4D_Cubing@yahoogroups.com, Don Hatch wrote:
> > > > > > {3,3,7} less so... its vertices are not simply at infinity (as
> in
> > > > > {3,3,6}),
> > > > > > they are "beyond infinity"...
> > > > > > If you try to draw this one, none of the edges will meet at all
> (not
> > > > > even at
> > > > > > infinity)... they all diverge! You'll see each edge
> > > > > > emerging from somewhere on the horizon (although there's no
> vertex
> > > > > > there) and leaving somewhere else on the horizon...
> > > > > > so nothing meets up, which kind of makes the picture less
> > > satisfying.
> > > > > > If you run the formula for edge length or cell circumradius,
> you'll
> > > get,
> > > > > not infinity,
> > > > > > but an imaginary or complex number (although the cell in-radius
> is
> > > > > finite, of
> > > > > > course, being equal to the half-edge-length of the dual
> {7,3,3}).
> > > > >
> > > > >
> > > >
> > > > --
> > > > Don Hatch
> > > > hatch@
> > > > http://www.plunk.org/~hatch/
> > > >
> > >
> > >
> >
> > --
> > Don Hatch
> > hatch@...
> > http://www.plunk.org/~hatch/
> >
>
>

--
Don Hatch
hatch@plunk.org
http://www.plunk.org/~hatch/

Oh wait!
I realized I got last part wrong, just after I hit the "send" button :-)

The picture would start with an Apollonian gasket (see wikipedia article)
of circles on the sphere;
this is the intersection of the {7,3,3} with the sphere at infinity.
Then each circle in the gasket is filled with a {3,7},
the final result being the intersection of the {3,3,7} with the sphere
at infinity.
So, it isn't true that there are isolated cluster points
in the *center* of each circle; the clustering is
towards the *boundary* of each circle. I think I have
a clear picture in my head of what this looks like now.

A cell of the {3,3,7} would touch the sphere
in 4 spherical triangles (its "feet"),
each foot in a different one of four mutually kissing circles
of the gasket, I think.

Don


On Tue, Jul 03, 2012 at 03:29:38PM -0400, Don Hatch wrote:
>
>
> Okay I think maybe I follow you now...
> But each face formed by truncation...
> it's a triangle, not a hexagon, right?
> In fact it's a spherical triangle, on the sphere at infinity, right?
> All of these spherical triangles, of different apparent sizes,
> would tile the sphere, 7 at each vertex...
> but with some "cluster points" which are the limit points
> of infinitely many of these triangles of decreasing size.
> I'd like to see a picture of this-- it shouldn't be too hard to generate
> (together with the spherical circles
> that are the intersection of the dual {7,3,3} with the sphere
> at infinity, in a different color... I think each cluster point
> would be the center of one of these circles).
>
> Don
>
> On Tue, Jul 03, 2012 at 02:43:52PM -0000, Andrey wrote:
> >
> >
> > Yes, dihedral... I mean angle between hexagonal and triangular faces of
> > truncated tetrahedron. By the selection of truncating planes it will be
> > pi/2. And angles between hexagonal faces are 2*pi/7.
> >
> > Andrey
> >
> > --- In 4D_Cubing@yahoogroups.com, Don Hatch wrote:
> > >
> > > Hi Andrey,
> > >
> > > I'm not sure if I'm understanding correctly...
> > > is "behedral angle" the same as "dihedral angle"?
> > > If so, isn't the dihedral angle going to be 2*pi/7,
> > > since, by definition, 7 tetrahedra surround each edge?
> > >
> > > Don
> > >
> > >
> > > On Tue, Jul 03, 2012 at 05:14:47AM -0000, Andrey wrote:
> > > >
> > > >
> > > > Hi all,
> > > > About {3,3,7} I had some idea (but it was long ago...). We know that
> > its
> > > > cell is a tetragedron that expands infinitely beyond "vertices". For
> > each
> > > > 3 its faces we have a plane perpendicular to them (that cuts them is
> > the
> > > > narrowest place). It we cut {3,3,7} cell by these planes, we get
> > truncated
> > > > tetrahedron with behedral angles = 180 deg. What happens if we
> reflect
> > it
> > > > about triangle faces, and continue this process to infinity? It will
> > be
> > > > some "fractal-like" network inscribed in the cell of {3,3,7} -
> regular
> > > > polyhedron with infinite numer of infinite faces but with no
> vertices.
> > I'm
> > > > sure that it has enough regular patterns of face coloring, and it
> may
> > be a
> > > > good base for 3D puzzle.
> > > >
> > > > Andrey
> > > >
> > > > --- In 4D_Cubing@yahoogroups.com, Don Hatch wrote:
> > > > >
> > > > > Hi Nan,
> > > > >
> > > > > Heh, I try not to make judgements...
> > > > > I think {3,3,7} is as legit as {7,3,3}
> > > > > (in fact I'd go so far as to say they are the same object,
> > > > > with different names given to the components).
> > > > > But pragmatically, {7,3,3} seems easier to get a grip on,
> > > > > in a viewer program such as yours which focuses naturally
> > > > > on the vertices and edges.
> > > > >
> > > > > Perhaps the best way to get a feeling for {3,3,7}
> > > > > would be to view it together with the {7,3,3}?
> > > > > Maybe one color for the {7,3,3} edges,
> > > > > another color for the {3,3,7} edges,
> > > > > and a third color for the edges formed where
> > > > > the faces of one intersect the faces of the other.
> > > > > And then perhaps, optionally,
> > > > > the full outlines of the characteristic tetrahedra?
> > > > > There are 6 types of edges in all (6 edges of a characteristic
> tet);
> > > > > I wonder if there's a natural coloring scheme
> > > > > using the 6 primary and secondary colors.
> > > > >
> > > > > Don
> > > > >
> > > > >
> > > > > On Sat, Jun 30, 2012 at 11:29:32PM -0000, schuma wrote:
> > > > > >
> > > > > >
> > > > > > Hi Don,
> > > > > >
> > > > > > Nice to see you here. Here are my thoughts about {3,3,7} and the
> > > > things
> > > > > > similar to it.
> > > > > >
> > > > > > Just like when we constructed {7,3,3} we were not able locate
> the
> > cell
> > > > > > centers, when we consider {3,3,7} we have to sacrifice the
> > vertices.
> > > > Let's
> > > > > > start by considering something simpler in lower dimensions.
> > > > > >
> > > > > > For example, in 2D, we could consider a hyperbolic "triangle"
> for
> > > > which
> > > > > > the sides don't meet even at the circle of infinity. The sides
> are
> > > > > > ultraparallel. Since there's no "angle", the name "triangle" is
> > not
> > > > > > appropriate any more. I'll call it a "trilateral", because it
> does
> > > > have
> > > > > > three sides (the common triangle is also a trilateral in my
> > notation).
> > > > > > Here's a tessellation of H2 using trilaterals, in which
> different
> > > > colors
> > > > > > indicate different trilaterals.
> > > > > >
> > > > > >
> > > >
> > http://games.groups.yahoo.com/group/4D_Cubing/files/Nan%20Ma/figure3.gif
> > > > > >
> > > > > > I constructed it as follows:
> > > > > >
> > > > > > In R2, a hexagon can be regarded as a truncated triangle, that
> is,
> > > > when
> > > > > > you extend the first, third, and fifth side of a hexagon, you
> get
> > a
> > > > > > triangle. In H3, when you extend the sides of a hexagon, you
> don't
> > > > always
> > > > > > get a triangle in the common sense: sometimes the extensions
> don't
> > > > meet.
> > > > > > But I claim you always get a trilateral. So I started with a
> > regular
> > > > {6,4}
> > > > > > tiling, and applied the extensions to get the tiling of
> > trilaterals.
> > > > > >
> > > > > > And I believe we can do similar things in H3: extend a properly
> > scaled
> > > > > > truncated tetrahedron to construct a "tetrahedron" with no
> > vertices.
> > > > > > Fortunately the name "tetrahedron" remains valid because hedron
> > means
> > > > face
> > > > > > rather than vertices. But I have never done an illustration of
> it
> > yet.
> > > > > > Then, maybe we can go ahead and put seven of them around an edge
> > and
> > > > make
> > > > > > a {3,3,7}.
> > > > > >
> > > > > > I agree that these objects are not conventional at all. We lost
> > > > something
> > > > > > like the vertices. But just like the above image, they do have
> > nice
> > > > > > patterns and are something worth considering.
> > > > > >
> > > > > > So, what do you think about them? Do they sound more legit now?
> > > > > >
> > > > > > Nan
> > > > > >
> > > > > > --- In 4D_Cubing@yahoogroups.com, Don Hatch wrote:
> > > > > > > {3,3,7} less so... its vertices are not simply at infinity (as
> > in
> > > > > > {3,3,6}),
> > > > > > > they are "beyond infinity"...
> > > > > > > If you try to draw this one, none of the edges will meet at
> all
> > (not
> > > > > > even at
> > > > > > > infinity)... they all diverge! You'll see each edge
> > > > > > > emerging from somewhere on the horizon (although there's no
> > vertex
> > > > > > > there) and leaving somewhere else on the horizon...
> > > > > > > so nothing meets up, which kind of makes the picture less
> > > > satisfying.
> > > > > > > If you run the formula for edge length or cell circumradius,
> > you'll
> > > > get,
> > > > > > not infinity,
> > > > > > > but an imaginary or complex number (although the cell
> in-radius
> > is
> > > > > > finite, of
> > > > > > > course, being equal to the half-edge-length of the dual
> > {7,3,3}).
> > > > > >
> > > > > >
> > > > >
> > > > > --
> > > > > Don Hatch
> > > > > hatch@
> > > > > http://www.plunk.org/~hatch/
> > > > >
> > > >
> > > >
> > >
> > > --
> > > Don Hatch
> > > hatch@...
> > > http://www.plunk.org/~hatch/
> > >
> >
> >
>
> --
> Don Hatch
> hatch@plunk.org
> http://www.plunk.org/~hatch/
>
>

--
Don Hatch
hatch@plunk.org
http://www.plunk.org/~hatch/

On Tue, Jul 03, 2012 at 03:29:38PM -0400, I wrote:
> But each face formed by truncation...
> it's a triangle, not a hexagon, right?

Sorry, mental lapse on my part!
The hexagons you're talking about
are what's left of the *original* faces when you truncate
a tetrahedron.
Don't know where my mind went.

Don

Yes, it'correct. And cutting triangles are planar (i.e. H2), not sperical. =
Their position is selected so that triangles have minimal possible size (in=
the narrowest point of the "vertex"). I'll try to draw one face of the obj=
ect, but it will be not easy.
Looks like this combination of truncated tetrahedra will be convex in H3 =
(and have a structure of regular infinite 4-graph without loops).

Andrey


--- In 4D_Cubing@yahoogroups.com, Don Hatch wrote:
>
> On Tue, Jul 03, 2012 at 03:29:38PM -0400, I wrote:
> > But each face formed by truncation...
> > it's a triangle, not a hexagon, right?
>=20
> Sorry, mental lapse on my part!
> The hexagons you're talking about
> are what's left of the *original* faces when you truncate
> a tetrahedron.
> Don't know where my mind went.
>=20
> Don
>

Hi Andrey,

Maybe I finally get what you mean now...
there is a unique plane that contains
the three points at infinity where three edges of the tet come closest to meeting;
I didn't see that before.
But are you sure the tet faces meet that plane
at a right angle as you claimed?
I believe the tet faces meet the sphere-at-infinity at right angles;
I don't think both can be true.

And I don't understand what you mean by "a structure of regular infinite
4-graph without loops" at all.
By "4-graph", do you mean every node has degree 4,
and by "without loops", do you mean a tree?
But I don't see any such tree in what we're talking about, so I'm lost :-(

Don

On Wed, Jul 04, 2012 at 04:03:38AM -0000, Andrey wrote:
>
>
> And cutting triangles are planar (i.e. H2), not sperical.
> Their position is selected so that triangles have minimal possible size
> (in the narrowest point of the "vertex"). I'll try to draw one face of the
> object, but it will be not easy.
> Looks like this combination of truncated tetrahedra will be convex in H3
> (and have a structure of regular infinite 4-graph without loops).
>
> Andrey
>
> --- In 4D_Cubing@yahoogroups.com, Don Hatch wrote:
> >
> > On Tue, Jul 03, 2012 at 03:29:38PM -0400, I wrote:
> > > But each face formed by truncation...
> > > it's a triangle, not a hexagon, right?
> >
> > Sorry, mental lapse on my part!
> > The hexagons you're talking about
> > are what's left of the *original* faces when you truncate
> > a tetrahedron.
> > Don't know where my mind went.
> >
> > Don
> >
>
>

--
Don Hatch
hatch@plunk.org
http://www.plunk.org/~hatch/

Hi, Don
Let's start with triangle with vertices "beyond infinity". Two its edges =
are "ultraparallel" and they have common perpendicular. If you cut triange =
by three such perpendiculars, you'll get irregular hexagon with 6 right ang=
les. Now you can reflect it about its short sides and continue this process=
to infinity. In result you'll get very strange object - it is convex, it's=
infinite, all its boundaries are straight lines and it's regular - you for=
every two edges there is a movement of H2 that moves one edge to another a=
nd moves the whole object to itself.=20
Something like this:
http://groups.yahoo.com/group/4D_Cubing/photos/album/772706687/pic/85638639=
4/view
Red lines are sides of hexagons.
You see that each hexagon is connected to three another and they make an =
acyclic graph (tree) together.

The same is for tetrahedra in H3. You have a convex objects, bounded by p=
lanes (and faces of this object are exactly like H2 pattern), each truncate=
d tetrahedron is connected to 4 others and together they make regular polyh=
edron with tree structure (all nodes of graph are equivalent).

Points on edges of tetrahedron with the minimal distance are not infinite=
- they are inside H3, so whole object is "real".

Andrey

--- In 4D_Cubing@yahoogroups.com, Don Hatch wrote:
>
> Hi Andrey,
>=20
> Maybe I finally get what you mean now...
> there is a unique plane that contains
> the three points at infinity where three edges of the tet come closest to=
meeting;
> I didn't see that before.
> But are you sure the tet faces meet that plane
> at a right angle as you claimed?
> I believe the tet faces meet the sphere-at-infinity at right angles;
> I don't think both can be true.
>=20
> And I don't understand what you mean by "a structure of regular infinite
> 4-graph without loops" at all.
> By "4-graph", do you mean every node has degree 4,
> and by "without loops", do you mean a tree?
> But I don't see any such tree in what we're talking about, so I'm lost :-=
(
>=20
> Don
>=20
> On Wed, Jul 04, 2012 at 04:03:38AM -0000, Andrey wrote:
> >=20=20=20=20=20
> >=20
> > And cutting triangles are planar (i.e. H2), not sperical.
> > Their position is selected so that triangles have minimal possible s=
ize
> > (in the narrowest point of the "vertex"). I'll try to draw one face =
of the
> > object, but it will be not easy.
> > Looks like this combination of truncated tetrahedra will be convex i=
n H3
> > (and have a structure of regular infinite 4-graph without loops).
> >=20
> > Andrey
> >=20
> > --- In 4D_Cubing@yahoogroups.com, Don Hatch wrote:
> > >
> > > On Tue, Jul 03, 2012 at 03:29:38PM -0400, I wrote:
> > > > But each face formed by truncation...
> > > > it's a triangle, not a hexagon, right?
> > >
> > > Sorry, mental lapse on my part!
> > > The hexagons you're talking about
> > > are what's left of the *original* faces when you truncate
> > > a tetrahedron.
> > > Don't know where my mind went.
> > >
> > > Don
> > >
> >=20
> >=20=20=20=20
>=20
> --=20
> Don Hatch
> hatch@...
> http://www.plunk.org/~hatch/
>

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> Sure, I'm interested in what you guys came up with
> along the lines of a {3,ultrainfinity}...
> I guess it would look like the picture Nan included in his previous e-mail
> (obtained by erasing some edges of the {6,4})
> however you're free to choose any triangle in-radius
> in the range (in-radius of {3,infinity}, infinity], right?
>

yep, our discussion finished on that picture, so Nan already shared most of
what we talked about. I like your thought to use the inradius as the
parameter for {3,ultrainf}, and that range sounds right to me.


> Is there a nicer parametrization of that one degree of freedom?
> Or is there some special value which could be regarded as the canonical
> one?
>
>
Nan and I had discussed the parametrization Andrey mentions, the (closest)
perpendicular distance between pairs of the the 3 ultraparallel lines.
Since "trilaterals" have no vertices, these distances can somewhat play
the role of angle - if they are all the same you have a regular trilateral.
The trilateral derived from the {6,4} tiling that Nan shared is even more
regular in a sense. Even though trilaterals have infinite edge length, we
can consider the edge lengths between the perpendicular lines above. Only
for the trilateral based on the {6,4} are those lengths equal to the
"angles". So perhaps it is the best canonical example for {3,ultrainf}.

seeya,
Roice

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