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Seeking a more symmetric realization of a configuration of 10 planes, 25 lines and 15 points in projective space
Clash Royale CLAN TAG#URR8PPP
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I've got ten (projective) planes in projective 3-space:
beginalign
&x=0\
&z=0\
&t=0\
&x+y=0\
&x-y=0\
&z+t=0\
&x-y-z=0\
&x+y+z=0\
&x-y+t=0\
&x+y-t=0
endalign
If I did not make a mistake somewhere, their intersections produce $25$ lines and $15$ points, each line containing $3$ points, each plane containing $6$ lines and $7$ points. Ten of the points belong to $4$ lines and $4$ planes each and five of them belong to $7$ lines and $6$ planes each. $15$ lines belong to $2$ of the planes and $10$ of them belong to $3$ of the planes.
The above indicates that this configuration is highly symmetric, is it known? How to compute its automorphism group? Where to look? My goal is to find another more symmetric realization.
co.combinatorics projective-geometry hyperplane-arrangements
asked 4 hours ago
მáƒÂმუკრჯიბლáƒÂძე
7,511242109
add a comment |Â
up vote
7
down vote
favorite
I've got ten (projective) planes in projective 3-space:
beginalign
&x=0\
&z=0\
&t=0\
&x+y=0\
&x-y=0\
&z+t=0\
&x-y-z=0\
&x+y+z=0\
&x-y+t=0\
&x+y-t=0
endalign
If I did not make a mistake somewhere, their intersections produce $25$ lines and $15$ points, each line containing $3$ points, each plane containing $6$ lines and $7$ points. Ten of the points belong to $4$ lines and $4$ planes each and five of them belong to $7$ lines and $6$ planes each. $15$ lines belong to $2$ of the planes and $10$ of them belong to $3$ of the planes.
The above indicates that this configuration is highly symmetric, is it known? How to compute its automorphism group? Where to look? My goal is to find another more symmetric realization.
co.combinatorics projective-geometry hyperplane-arrangements
asked 4 hours ago
მáƒÂმუკრჯიბლáƒÂძე
7,511242109
add a comment |Â
up vote
7
down vote
favorite
up vote
7
down vote
favorite
I've got ten (projective) planes in projective 3-space:
beginalign
&x=0\
&z=0\
&t=0\
&x+y=0\
&x-y=0\
&z+t=0\
&x-y-z=0\
&x+y+z=0\
&x-y+t=0\
&x+y-t=0
endalign
If I did not make a mistake somewhere, their intersections produce $25$ lines and $15$ points, each line containing $3$ points, each plane containing $6$ lines and $7$ points. Ten of the points belong to $4$ lines and $4$ planes each and five of them belong to $7$ lines and $6$ planes each. $15$ lines belong to $2$ of the planes and $10$ of them belong to $3$ of the planes.
The above indicates that this configuration is highly symmetric, is it known? How to compute its automorphism group? Where to look? My goal is to find another more symmetric realization.
co.combinatorics projective-geometry hyperplane-arrangements
asked 4 hours ago
მáƒÂმუკრჯიბლáƒÂძე
7,511242109
I've got ten (projective) planes in projective 3-space:
beginalign
&x=0\
&z=0\
&t=0\
&x+y=0\
&x-y=0\
&z+t=0\
&x-y-z=0\
&x+y+z=0\
&x-y+t=0\
&x+y-t=0
endalign
If I did not make a mistake somewhere, their intersections produce $25$ lines and $15$ points, each line containing $3$ points, each plane containing $6$ lines and $7$ points. Ten of the points belong to $4$ lines and $4$ planes each and five of them belong to $7$ lines and $6$ planes each. $15$ lines belong to $2$ of the planes and $10$ of them belong to $3$ of the planes.
The above indicates that this configuration is highly symmetric, is it known? How to compute its automorphism group? Where to look? My goal is to find another more symmetric realization.
co.combinatorics projective-geometry hyperplane-arrangements
co.combinatorics projective-geometry hyperplane-arrangements
asked 4 hours ago
მáƒÂმუკრჯიბლáƒÂძე
7,511242109
asked 4 hours ago
მáƒÂმუკრჯიბლáƒÂძე
7,511242109
asked 4 hours ago
მáƒÂმუკრჯიბლáƒÂძე
7,511242109
asked 4 hours ago
მáƒÂმუკრჯიბლáƒÂძე
7,511242109
asked 4 hours ago
მáƒÂმუკრჯიბლáƒÂძე
7,511242109
7,511242109
add a comment |Â
add a comment |Â
1 Answer
1
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oldest
votes
up vote
4
down vote
This configuration has automorphisms by the symmetric group $S_5$,
and can be identified with the planes $a_i = a_j$ ($0 leq i < j leq 4$)
in the projective 3-space $a_0+a_1+a_2+a_3+a_4 = 0$, by setting
$$
(x,z,t,y) = (a_0 - a_1, 2(a_2 - a_3), 2(a_4 - a_2), a_0 + a_1 - 2a_2).
$$
the ten planes then have $(i,j)$ in the order
$$
(0,1),,(2,3),,(2,4),,(0,2),,(1,2),,(3,4),,(1,3),,(0,3),,(1,4),,(0,4).
$$
The 25 lines include the ten lines $a_i=a_j=a_k$ where three of the planes
$a_i=a_j$ meet; these ten lines together with the ten planes $a_i = a_j$ encode
Desargues
configurations: intersecting with a generic plane $Pi$ in the 3-space
yields the ten points and ten lines of a Desargues configuration,
and any Desargues configuration can be obtained this way for some
choice of $Pi$.
answered 40 mins ago
Noam D. Elkies
55.2k10195280
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
This configuration has automorphisms by the symmetric group $S_5$,
and can be identified with the planes $a_i = a_j$ ($0 leq i < j leq 4$)
in the projective 3-space $a_0+a_1+a_2+a_3+a_4 = 0$, by setting
$$
(x,z,t,y) = (a_0 - a_1, 2(a_2 - a_3), 2(a_4 - a_2), a_0 + a_1 - 2a_2).
$$
the ten planes then have $(i,j)$ in the order
$$
(0,1),,(2,3),,(2,4),,(0,2),,(1,2),,(3,4),,(1,3),,(0,3),,(1,4),,(0,4).
$$
The 25 lines include the ten lines $a_i=a_j=a_k$ where three of the planes
$a_i=a_j$ meet; these ten lines together with the ten planes $a_i = a_j$ encode
Desargues
configurations: intersecting with a generic plane $Pi$ in the 3-space
yields the ten points and ten lines of a Desargues configuration,
and any Desargues configuration can be obtained this way for some
choice of $Pi$.
answered 40 mins ago
Noam D. Elkies
55.2k10195280
add a comment |Â
up vote
4
down vote
This configuration has automorphisms by the symmetric group $S_5$,
and can be identified with the planes $a_i = a_j$ ($0 leq i < j leq 4$)
in the projective 3-space $a_0+a_1+a_2+a_3+a_4 = 0$, by setting
$$
(x,z,t,y) = (a_0 - a_1, 2(a_2 - a_3), 2(a_4 - a_2), a_0 + a_1 - 2a_2).
$$
the ten planes then have $(i,j)$ in the order
$$
(0,1),,(2,3),,(2,4),,(0,2),,(1,2),,(3,4),,(1,3),,(0,3),,(1,4),,(0,4).
$$
The 25 lines include the ten lines $a_i=a_j=a_k$ where three of the planes
$a_i=a_j$ meet; these ten lines together with the ten planes $a_i = a_j$ encode
Desargues
configurations: intersecting with a generic plane $Pi$ in the 3-space
yields the ten points and ten lines of a Desargues configuration,
and any Desargues configuration can be obtained this way for some
choice of $Pi$.
answered 40 mins ago
Noam D. Elkies
55.2k10195280
add a comment |Â
up vote
4
down vote
up vote
4
down vote
This configuration has automorphisms by the symmetric group $S_5$,
and can be identified with the planes $a_i = a_j$ ($0 leq i < j leq 4$)
in the projective 3-space $a_0+a_1+a_2+a_3+a_4 = 0$, by setting
$$
(x,z,t,y) = (a_0 - a_1, 2(a_2 - a_3), 2(a_4 - a_2), a_0 + a_1 - 2a_2).
$$
the ten planes then have $(i,j)$ in the order
$$
(0,1),,(2,3),,(2,4),,(0,2),,(1,2),,(3,4),,(1,3),,(0,3),,(1,4),,(0,4).
$$
The 25 lines include the ten lines $a_i=a_j=a_k$ where three of the planes
$a_i=a_j$ meet; these ten lines together with the ten planes $a_i = a_j$ encode
Desargues
configurations: intersecting with a generic plane $Pi$ in the 3-space
yields the ten points and ten lines of a Desargues configuration,
and any Desargues configuration can be obtained this way for some
choice of $Pi$.
answered 40 mins ago
Noam D. Elkies
55.2k10195280
This configuration has automorphisms by the symmetric group $S_5$,
and can be identified with the planes $a_i = a_j$ ($0 leq i < j leq 4$)
in the projective 3-space $a_0+a_1+a_2+a_3+a_4 = 0$, by setting
$$
(x,z,t,y) = (a_0 - a_1, 2(a_2 - a_3), 2(a_4 - a_2), a_0 + a_1 - 2a_2).
$$
the ten planes then have $(i,j)$ in the order
$$
(0,1),,(2,3),,(2,4),,(0,2),,(1,2),,(3,4),,(1,3),,(0,3),,(1,4),,(0,4).
$$
The 25 lines include the ten lines $a_i=a_j=a_k$ where three of the planes
$a_i=a_j$ meet; these ten lines together with the ten planes $a_i = a_j$ encode
Desargues
configurations: intersecting with a generic plane $Pi$ in the 3-space
yields the ten points and ten lines of a Desargues configuration,
and any Desargues configuration can be obtained this way for some
choice of $Pi$.
answered 40 mins ago
Noam D. Elkies
55.2k10195280
answered 40 mins ago
Noam D. Elkies
55.2k10195280
answered 40 mins ago
Noam D. Elkies
55.2k10195280
answered 40 mins ago
Noam D. Elkies
55.2k10195280
55.2k10195280
add a comment |Â
add a comment |Â
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