[math-fun] Model of the Fano plane on the sphere?
I.e. can you assign each point in the Fano plane to antipodal points on the sphere and each line in the Fano plane to a great circle such that the incidence relation is preserved? -- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
Very interesting question. (Are you requiring that all intersections between great circles be nodes? Or are extraneous intersections allowed?) Let tau = exp(2pi*i/6). There's a highly symmetrical version of the Fano plane on the torus T = C/L, where C = complexes and L = Z[2+tau]. Let S = {0, tau^k | 0 <= k < 6} and draw a circle of radius sqrt(1/3) about each z in the image of S in T = C/L. These 7 circles each intersect 3 others, for a total of 7 intersection points in T. The incidence relations among these 7 circles and 7 intersection points is isomorphic to the Fano plane. (Which leaves Mike's question unanswered, alas.) --Dan On Oct 1, 2014, at 2:47 PM, Mike Stay <metaweta@gmail.com> wrote:
. . . can you assign each point in the Fano plane to antipodal points on the sphere and each line in the Fano plane to a great circle such that the incidence relation is preserved?
Arggh. I think that was rather mistaken of me. --Dan On Oct 1, 2014, at 7:04 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Very interesting question. (Are you requiring that all intersections between great circles be nodes? Or are extraneous intersections allowed?)
Let tau = exp(2pi*i/6). There's a highly symmetrical version of the Fano plane on the torus T = C/L, where C = complexes and L = Z[2+tau].
Let S = {0, tau^k | 0 <= k < 6} and draw a circle of radius sqrt(1/3) about each z in the image of S in T = C/L.
These 7 circles each intersect 3 others, for a total of 7 intersection points in T.
The incidence relations among these 7 circles and 7 intersection points is isomorphic to the Fano plane.
(Which leaves Mike's question unanswered, alas.)
--Dan
On Oct 1, 2014, at 2:47 PM, Mike Stay <metaweta@gmail.com> wrote:
. . . can you assign each point in the Fano plane to antipodal points on the sphere and each line in the Fano plane to a great circle such that the incidence relation is preserved?
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Arggh^2. I think it's OK after all. --Dan On Oct 1, 2014, at 7:36 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Arggh. I think that was rather mistaken of me.
--Dan
On Oct 1, 2014, at 7:04 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Very interesting question. (Are you requiring that all intersections between great circles be nodes? Or are extraneous intersections allowed?)
Let tau = exp(2pi*i/6). There's a highly symmetrical version of the Fano plane on the torus T = C/L, where C = complexes and L = Z[2+tau].
Let S = {0, tau^k | 0 <= k < 6} and draw a circle of radius sqrt(1/3) about each z in the image of S in T = C/L.
These 7 circles each intersect 3 others, for a total of 7 intersection points in T.
The incidence relations among these 7 circles and 7 intersection points is isomorphic to the Fano plane.
(Which leaves Mike's question unanswered, alas.)
--Dan
On Oct 1, 2014, at 2:47 PM, Mike Stay <metaweta@gmail.com> wrote:
. . . can you assign each point in the Fano plane to antipodal points on the sphere and each line in the Fano plane to a great circle such that the incidence relation is preserved?
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The 7 circles on T = C/L do work on the torus as a model of the Fano plane. But where I wrote "These 7 circles each intersect 3 others, for a total of 7 intersection points in T" the part about "3 others" wasn't right. The 7 circles intersect in a total of 7 intersection points, with 3 of these points on each circle, and each point lying on 3 circles. (And just as with the Fano plane, each circle intersects 6 others.) --Dan On Oct 1, 2014, at 8:16 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Very interesting question. (Are you requiring that all intersections between great circles be nodes? Or are extraneous intersections allowed?)
Let tau = exp(2pi*i/6). There's a highly symmetrical version of the Fano plane on the torus T = C/L, where C = complexes and L = Z[2+tau].
Let S = {0, tau^k | 0 <= k < 6} and draw a circle of radius sqrt(1/3) about each z in the image of S in T = C/L.
These 7 circles each intersect 3 others, for a total of 7 intersection points in T.
The incidence relations among these 7 circles and 7 intersection points is isomorphic to the Fano plane.
(Which leaves Mike's question unanswered, alas.)
--Dan
On Oct 1, 2014, at 2:47 PM, Mike Stay <metaweta@gmail.com> wrote:
. . . can you assign each point in the Fano plane to antipodal points on the sphere and each line in the Fano plane to a great circle such that the incidence relation is preserved?
On Wed, Oct 1, 2014 at 7:04 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Very interesting question. (Are you requiring that all intersections between great circles be nodes? Or are extraneous intersections allowed?)
I'd prefer to have just the proper intersections, if possible.
Let tau = exp(2pi*i/6). There's a highly symmetrical version of the Fano plane on the torus T = C/L, where C = complexes and L = Z[2+tau].
Let S = {0, tau^k | 0 <= k < 6} and draw a circle of radius sqrt(1/3) about each z in the image of S in T = C/L.
These 7 circles each intersect 3 others, for a total of 7 intersection points in T.
The incidence relations among these 7 circles and 7 intersection points is isomorphic to the Fano plane.
Interesting!
(Which leaves Mike's question unanswered, alas.)
--Dan
On Oct 1, 2014, at 2:47 PM, Mike Stay <metaweta@gmail.com> wrote:
. . . can you assign each point in the Fano plane to antipodal points on the sphere and each line in the Fano plane to a great circle such that the incidence relation is preserved?
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-- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
Can anyone supply a picture of Dan's construction, transplanted from T to a polygonal domain in the plane with identifications along the boundary? Jim On Wednesday, October 1, 2014, Dan Asimov <dasimov@earthlink.net> wrote:
Very interesting question. (Are you requiring that all intersections between great circles be nodes? Or are extraneous intersections allowed?)
Let tau = exp(2pi*i/6). There's a highly symmetrical version of the Fano plane on the torus T = C/L, where C = complexes and L = Z[2+tau].
Let S = {0, tau^k | 0 <= k < 6} and draw a circle of radius sqrt(1/3) about each z in the image of S in T = C/L.
These 7 circles each intersect 3 others, for a total of 7 intersection points in T.
The incidence relations among these 7 circles and 7 intersection points is isomorphic to the Fano plane.
(Which leaves Mike's question unanswered, alas.)
--Dan
On Oct 1, 2014, at 2:47 PM, Mike Stay <metaweta@gmail.com <javascript:;>> wrote:
. . . can you assign each point in the Fano plane to antipodal points on the sphere and each line in the Fano plane to a great circle such that the incidence relation is preserved?
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Can anyone supply a picture of Dan's construction, transplanted from T to a polygonal domain in the plane with identifications along the boundary?
Here's a picture generated from a 100-character Mathematica program: https://twitter.com/wolframtap/status/517631202837405696 Identify discs of equal colour. Large discs represent lines; small discs represent points. Adjacency represents incidence. Sincerely, Adam P. Goucher
Thanks, Adam! Jim On Thursday, October 2, 2014, Adam P. Goucher <apgoucher@gmx.com> wrote:
Can anyone supply a picture of Dan's construction, transplanted from T to a polygonal domain in the plane with identifications along the boundary?
Here's a picture generated from a 100-character Mathematica program:
https://twitter.com/wolframtap/status/517631202837405696
Identify discs of equal colour. Large discs represent lines; small discs represent points. Adjacency represents incidence.
Sincerely,
Adam P. Goucher
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Adam's picture is very nice, but I think having the "points" and "lines" colored with the same or similar colors is a bit misleading. Maybe 14 distinct colors, 7 for the points and 7 for the lines might be better. The universal cover of the picture I originally intended would be what you get if you start with a triangular lattice of circle centers on the plane, and then increase the circles' common radius until they first have triple intersections. It would resemble a picture of chain mail. Then the circles are the lines, and their intersections are the points. (Ignore the centers.) Now 7-color the "lines", and separately 7-color the "points", in a repeating way. --Dan On Oct 2, 2014, at 4:19 AM, Adam P. Goucher <apgoucher@gmx.com> wrote:
Can anyone supply a picture of Dan's construction, transplanted from T to a polygonal domain in the plane with identifications along the boundary?
Here's a picture generated from a 100-character Mathematica program:
https://twitter.com/wolframtap/status/517631202837405696
Identify discs of equal colour. Large discs represent lines; small discs represent points. Adjacency represents incidence.
Actually, the chain mail thing should overlook every other (triple) intersection point, contrary to what I wrote below. Here's a bare-bones, line-point symmetric way to see it, perhaps: Consider a triangular lattice L in the plane, and 3-color it, say by numbering the nodes 0, 1, 2 so that every smallest triangle has all three numbers. Briefly imagine Voronoifying this lattice with hexagonal cells numbered 0, 1, 2. Now forget about the points labeled 2. Do not think of the number 2 !!! The points labeled k, for k = 0 or for k = 1, each form a triangular lattice themselves. Call these L_0 and L_1. Now 7-color, separately, each of L_0 and L_1 (using 14 colors in all). Redraw the previous hexagonal Voronoi cells of the points of L_0 and of L_1. (Just in case we need them later, pick a third heptad of colors to 7-color the hexagons of L_2.) Finally, mod out the plane by the appropriate sublattice of the original lattice L, so that (the images of) exactly 7 hexagons of L_0 and 7 hexagons of L_1 appear in the quotient torus T. The (images of the) 7 hexagons of L_0 may be thought of as the 7 points of the Fano plane, and the (images of the) 7 hexagons of L_1 as the 7 lines. This higlghts the duality between points and lines. Now we can bring back the (images of the) 7 hexagons of L_2 to get a *third* group of 7 in the torus T, with 3-way symmetry among the 3 groups. --Dan On Oct 2, 2014, at 1:46 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Adam's picture is very nice, but I think having the "points" and "lines" colored with the same or similar colors is a bit misleading.
Maybe 14 distinct colors, 7 for the points and 7 for the lines might be better.
The universal cover of the picture I originally intended would be what you get if you start with a triangular lattice of circle centers on the plane, and then increase the circles' common radius until they first have triple intersections. It would resemble a picture of chain mail. Then the circles are the lines, and their intersections are the points. (Ignore the centers.) Now 7-color the "lines", and separately 7-color the "points", in a repeating way.
--Dan
On Oct 2, 2014, at 4:19 AM, Adam P. Goucher <apgoucher@gmx.com> wrote:
Can anyone supply a picture of Dan's construction, transplanted from T to a polygonal domain in the plane with identifications along the boundary?
Here's a picture generated from a 100-character Mathematica program:
https://twitter.com/wolframtap/status/517631202837405696
Identify discs of equal colour. Large discs represent lines; small discs represent points. Adjacency represents incidence.
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Firstly, what's the common name for A_2? I usually say `hexagonal lattice' (using the Voronoi partition), whereas Dan appears to use `triangular lattice' (taking the dual partition). My copy of SPLAG is 110 miles away, and that tends to be the authoritative resource on this topic. Sloane? Dan Asimov wrote:
Actually, the chain mail thing should overlook every other (triple) intersection point, contrary to what I wrote below.
Here's a bare-bones, line-point symmetric way to see it, perhaps:
Consider a triangular lattice L in the plane, and 3-color it, say by numbering the nodes 0, 1, 2 so that every smallest triangle has all three numbers. Briefly imagine Voronoifying this lattice with hexagonal cells numbered 0, 1, 2.
Now forget about the points labeled 2. Do not think of the number 2 !!!
The points labeled k, for k = 0 or for k = 1, each form a triangular lattice themselves. Call these L_0 and L_1.
Now 7-color, separately, each of L_0 and L_1 (using 14 colors in all).
Redraw the previous hexagonal Voronoi cells of the points of L_0 and of L_1. (Just in case we need them later, pick a third heptad of colors to 7-color the hexagons of L_2.)
Your description so far is precisely what my program does, apart from the trivial difference between hexagons and discs. There is a palette of 21 colours, as implied by the /21 in the argument to the Hue command. The third group is invisible since those discs have radius 0 in the program. (I cheated in several places to make the code as compact as possible -- 100 characters is not many at all.)
Finally, mod out the plane by the appropriate sublattice of the original lattice L, so that (the images of) exactly 7 hexagons of L_0 and 7 hexagons of L_1 appear in the quotient torus T.
The (images of the) 7 hexagons of L_0 may be thought of as the 7 points of the Fano plane, and the (images of the) 7 hexagons of L_1 as the 7 lines. This higlghts the duality between points and lines.
Now we can bring back the (images of the) 7 hexagons of L_2 to get a *third* group of 7 in the torus T, with 3-way symmetry among the 3 groups.
Yes, indeed: S_3 symmetry, in fact. Bizarrely, I think that this reduces the overall amount of symmetry of the graph from PGL(2,7) (order 336) to order-252.
--Dan
On Oct 2, 2014, at 1:46 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Adam's picture is very nice, but I think having the "points" and "lines" colored with the same or similar colors is a bit misleading.
Maybe 14 distinct colors, 7 for the points and 7 for the lines might be better.
The universal cover of the picture I originally intended would be what you get if you start with a triangular lattice of circle centers on the plane, and then increase the circles' common radius until they first have triple intersections. It would resemble a picture of chain mail. Then the circles are the lines, and their intersections are the points. (Ignore the centers.) Now 7-color the "lines", and separately 7-color the "points", in a repeating way.
--Dan
On Oct 2, 2014, at 4:19 AM, Adam P. Goucher <apgoucher@gmx.com> wrote:
Can anyone supply a picture of Dan's construction, transplanted from T to a polygonal domain in the plane with identifications along the boundary?
Here's a picture generated from a 100-character Mathematica program:
https://twitter.com/wolframtap/status/517631202837405696
Identify discs of equal colour. Large discs represent lines; small discs represent points. Adjacency represents incidence.
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Adam P. Goucher wrote:
Firstly, what's the common name for A_2? I usually say `hexagonal lattice' (using the Voronoi partition), whereas Dan appears to use `triangular lattice' (taking the dual partition). My copy of SPLAG is 110 miles away, and that tends to be the authoritative resource on this topic.
Sloane?
There might be a transponder difference; I think triangular is a bit more popular among math'ns here, but hexagonal is widely understood to mean the same thing.
Dan Asimov wrote:
Actually, the chain mail thing should overlook every other (triple) intersection point, contrary to what I wrote below.
Here's a bare-bones, line-point symmetric way to see it, perhaps:
Consider a triangular lattice L in the plane, and 3-color it, say by numbering the nodes 0, 1, 2 so that every smallest triangle has all three numbers. Briefly imagine Voronoifying this lattice with hexagonal cells numbered 0, 1, 2.
Now forget about the points labeled 2. Do not think of the number 2 !!!
The points labeled k, for k = 0 or for k = 1, each form a triangular lattice themselves. Call these L_0 and L_1.
Now 7-color, separately, each of L_0 and L_1 (using 14 colors in all).
Redraw the previous hexagonal Voronoi cells of the points of L_0 and of L_1. (Just in case we need them later, pick a third heptad of colors to 7-color the hexagons of L_2.)
Your description so far is precisely what my program does, apart from the trivial difference between hexagons and discs. There is a palette of 21 colours, as implied by the /21 in the argument to the Hue command. The third group is invisible since those discs have radius 0 in the program.
(I cheated in several places to make the code as compact as possible -- 100 characters is not many at all.)
The only real difference is that all hexagons are the same size. I'd look for 14 colors that seem visually well-separated. Maybe take the vertices and face centers of the RGB cube.
Finally, mod out the plane by the appropriate sublattice of the original lattice L, so that (the images of) exactly 7 hexagons of L_0 and 7 hexagons of L_1 appear in the quotient torus T.
The (images of the) 7 hexagons of L_0 may be thought of as the 7 points of the Fano plane, and the (images of the) 7 hexagons of L_1 as the 7 lines. This higlghts the duality between points and lines.
Now we can bring back the (images of the) 7 hexagons of L_2 to get a *third* group of 7 in the torus T, with 3-way symmetry among the 3 groups.
Yes, indeed: S_3 symmetry, in fact. Bizarrely, I think that this reduces the overall amount of symmetry of the graph from PGL(2,7) (order 336) to order-252.
I may be confused about this, but I don't think you can turn the thing over (the 7-color tiling of the hexagonal torus by 7 hexagons is chiral). If so I see a total symmetry group of size 21*6 = 126. (Unless you allow permuting the within-group colors, in which case multiply by (7!^3)/42 = 3048192000.) --Dan
On Oct 2, 2014, at 1:46 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Adam's picture is very nice, but I think having the "points" and "lines" colored with the same or similar colors is a bit misleading.
Maybe 14 distinct colors, 7 for the points and 7 for the lines might be better.
The universal cover of the picture I originally intended would be what you get if you start with a triangular lattice of circle centers on the plane, and then increase the circles' common radius until they first have triple intersections. It would resemble a picture of chain mail. Then the circles are the lines, and their intersections are the points. (Ignore the centers.) Now 7-color the "lines", and separately 7-color the "points", in a repeating way.
--Dan
On Oct 2, 2014, at 4:19 AM, Adam P. Goucher <apgoucher@gmx.com> wrote:
Can anyone supply a picture of Dan's construction, transplanted from T to a polygonal domain in the plane with identifications along the boundary?
Here's a picture generated from a 100-character Mathematica program:
https://twitter.com/wolframtap/status/517631202837405696
Identify discs of equal colour. Large discs represent lines; small discs represent points. Adjacency represents incidence.
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It's the A_2 root lattice, and it is variously called the triangular or the hexagonal lattice. It depends on where you focus your eyes. If you look at the lattice points, and join each point to its six nbrs, you see a network of triangles. If you are tiler and you have a lot of hexagonal tiles, you see a network of hexagons (and the lattice points at the centers of the tiles are invisible) Take your pick! Neil On Thu, Oct 2, 2014 at 8:57 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Adam P. Goucher wrote:
Firstly, what's the common name for A_2? I usually say `hexagonal lattice' (using the Voronoi partition), whereas Dan appears to use `triangular lattice' (taking the dual partition). My copy of SPLAG is 110 miles away, and that tends to be the authoritative resource on this topic.
Sloane?
There might be a transponder difference; I think triangular is a bit more popular among math'ns here, but hexagonal is widely understood to mean the same thing.
Dan Asimov wrote:
Actually, the chain mail thing should overlook every other (triple) intersection point, contrary to what I wrote below.
Here's a bare-bones, line-point symmetric way to see it, perhaps:
Consider a triangular lattice L in the plane, and 3-color it, say by numbering the nodes 0, 1, 2 so that every smallest triangle has all three numbers. Briefly imagine Voronoifying this lattice with hexagonal cells numbered 0, 1, 2.
Now forget about the points labeled 2. Do not think of the number 2 !!!
The points labeled k, for k = 0 or for k = 1, each form a triangular lattice themselves. Call these L_0 and L_1.
Now 7-color, separately, each of L_0 and L_1 (using 14 colors in all).
Redraw the previous hexagonal Voronoi cells of the points of L_0 and of L_1. (Just in case we need them later, pick a third heptad of colors to 7-color the hexagons of L_2.)
Your description so far is precisely what my program does, apart from the trivial difference between hexagons and discs. There is a palette of 21 colours, as implied by the /21 in the argument to the Hue command. The third group is invisible since those discs have radius 0 in the program.
(I cheated in several places to make the code as compact as possible -- 100 characters is not many at all.)
The only real difference is that all hexagons are the same size.
I'd look for 14 colors that seem visually well-separated. Maybe take the vertices and face centers of the RGB cube.
Finally, mod out the plane by the appropriate sublattice of the original lattice L, so that (the images of) exactly 7 hexagons of L_0 and 7 hexagons of L_1 appear in the quotient torus T.
The (images of the) 7 hexagons of L_0 may be thought of as the 7 points of the Fano plane, and the (images of the) 7 hexagons of L_1 as the 7 lines. This higlghts the duality between points and lines.
Now we can bring back the (images of the) 7 hexagons of L_2 to get a *third* group of 7 in the torus T, with 3-way symmetry among the 3 groups.
Yes, indeed: S_3 symmetry, in fact. Bizarrely, I think that this reduces the overall amount of symmetry of the graph from PGL(2,7) (order 336) to order-252.
I may be confused about this, but I don't think you can turn the thing over (the 7-color tiling of the hexagonal torus by 7 hexagons is chiral). If so I see a total symmetry group of size 21*6 = 126. (Unless you allow permuting the within-group colors, in which case multiply by (7!^3)/42 = 3048192000.)
--Dan
On Oct 2, 2014, at 1:46 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Adam's picture is very nice, but I think having the "points" and "lines" colored with the same or similar colors is a bit misleading.
Maybe 14 distinct colors, 7 for the points and 7 for the lines might be better.
The universal cover of the picture I originally intended would be what you get if you start with a triangular lattice of circle centers on the plane, and then increase the circles' common radius until they first have triple intersections. It would resemble a picture of chain mail. Then the circles are the lines, and their intersections are the points. (Ignore the centers.) Now 7-color the "lines", and separately 7-color the "points", in a repeating way.
--Dan
On Oct 2, 2014, at 4:19 AM, Adam P. Goucher <apgoucher@gmx.com> wrote:
Can anyone supply a picture of Dan's construction, transplanted from T to a polygonal domain in the plane with identifications along the boundary?
Here's a picture generated from a 100-character Mathematica program:
https://twitter.com/wolframtap/status/517631202837405696
Identify discs of equal colour. Large discs represent lines; small discs represent points. Adjacency represents incidence.
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-- Dear Friends, I have now retired from AT&T. New coordinates: Neil J. A. Sloane, President, OEIS Foundation 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com
(I cheated in several places to make the code as compact as possible -- 100 characters is not many at all.)
The only real difference is that all hexagons are the same size.
Indeed. I wanted to specifically distinguish between lines and points, so that the picture definitely represents `the Fano plane' rather than `the Heawood graph': http://en.wikipedia.org/wiki/Heawood_graph
I'd look for 14 colors that seem visually well-separated. Maybe take the vertices and face centers of the RGB cube.
That should work, but it would vastly increase the length of code to beyond the @wolframtap limit of 128 characters.
Yes, indeed: S_3 symmetry, in fact. Bizarrely, I think that this reduces the overall amount of symmetry of the graph from PGL(2,7) (order 336) to order-252.
I may be confused about this, but I don't think you can turn the thing over (the 7-color tiling of the hexagonal torus by 7 hexagons is chiral). If so I see a total symmetry group of size 21*6 = 126.
Ahh, yes, you're completely correct. Sincerely, Adam P. Goucher
participants (5)
-
Adam P. Goucher -
Dan Asimov -
James Propp -
Mike Stay -
Neil Sloane