Configuration space of equilateral and
equiangular n-gons (n 6)
Jun O’Hara
[email protected]
Department of Mathematics, Tokyo Metropolitan University
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Introduction
An equilateral and equiangular polygon is a mathematical
model of a cycloalkane.
Jun O’Hara, Aug. 2008 – p.2/21
Introduction
An equilateral and equiangular polygon is a mathematical
model of a cycloalkane.
Examples of cyclohexane:
Jun O’Hara, Aug. 2008 – p.2/21
Introduction
An equilateral and equiangular polygon is a mathematical
model of a cycloalkane.
Examples of cyclohexane: a boat
Jun O’Hara, Aug. 2008 – p.2/21
Introduction
An equilateral and equiangular polygon is a mathematical
model of a cycloalkane.
Examples of cyclohexane: a boat and a chair.
Jun O’Hara, Aug. 2008 – p.2/21
Introduction
An equilateral and equiangular polygon is a mathematical
model of a cycloalkane.
Examples of cyclohexane: a boat and a chair.
Problem
What are all the possible shapes?
Jun O’Hara, Aug. 2008 – p.2/21
Introduction
An equilateral and equiangular polygon is a mathematical
model of a cycloalkane.
Examples of cyclohexane: a boat and a chair.
What are all the possible shapes?
The space of all such polygons is called the configuration
(or conformation) space.
Problem
Jun O’Hara, Aug. 2008 – p.2/21
Notations
n : the number of vertices.
: the “bond angle”, i.e. the angle between adjacent edges
(0 < ).
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Notations
n : the number of vertices.
: the “bond angle”, i.e. the angle between adjacent edges
(0 < 8).
9
jP
Pi+1j = 1; =
<
i
P
=(
P0 ; ; Pn 1 ) f
Mn():= : (Pi 2 R 3 ) \Pi 1PiPi+1 = ;
(8i (mod. n))
Jun O’Hara, Aug. 2008 – p.3/21
Notations
n : the number of vertices.
: the “bond angle”, i.e. the angle between adjacent edges
(0 < 8).
9
jP
Pi+1j = 1; =
<
i
P
=(
P0 ; ; Pn 1 ) f
Mn():= : (Pi 2 R 3 ) \Pi 1PiPi+1 = ;
(8i (mod. n))
G : the group of motions of R 3 .
Mn() := Mn()=G : the configuration space of unit
f
equilateral and -equiangular n-gons.
Jun O’Hara, Aug. 2008 – p.3/21
Remarks
8
(
P ; ;P
Mn()= : (Pi 2 )
<
0
j
Pi
)
n 1 R3
Pi+1j = 1;
\Pi 1PiPi+1 = ;
(8i (mod. n))
9
,
=
G
Jun O’Hara, Aug. 2008 – p.4/21
Remarks
8
(
P ; ;P
Mn()= : (Pi 2 )
<
0
j
Pi
)
n 1 R3
Pi+1j = 1;
\Pi 1PiPi+1 = ;
(8i (mod. n))
9
,
=
G
We allow intersections of edges and overlapping of vertices.
Jun O’Hara, Aug. 2008 – p.4/21
Remarks
8
(
P ; ;P
Mn()= : (Pi 2 )
<
0
j
Pi
)
n 1 R3
Pi+1j = 1;
\Pi 1PiPi+1 = ;
(8i (mod. n))
9
,
=
G
We allow intersections of edges and overlapping of vertices.
Remark
We distinguish
and
in
M , although their shapes are the same.
6
Jun O’Hara, Aug. 2008 – p.4/21
Config. sp. of equilateral planar polygon
9 Great number of studies of the configuration spaces of
“linkages” from various fields.
The edge lengths are fixed, but not the angles.
Planar (2D) or spatial (3D).
Jun O’Hara, Aug. 2008 – p.5/21
Config. sp. of equilateral planar polygon
9 Great number of studies of the configuration spaces of
“linkages” from various fields.
The edge lengths are fixed, but not the angles.
Planar (2D) or spatial (3D).
Some examples of configuration spaces n of planar
equilateral polygons.
P
Jun O’Hara, Aug. 2008 – p.5/21
Config. sp. of equilateral planar polygon
9 Great number of studies of the configuration spaces of
“linkages” from various fields.
The edge lengths are fixed, but not the angles.
Planar (2D) or spatial (3D).
Some examples of configuration spaces n of planar
equilateral polygons.
equilateral planar triangles
points .
f
g = f2
P
g
Jun O’Hara, Aug. 2008 – p.5/21
Config. sp. of equilateral planar polygon
9 Great number of studies of the configuration spaces of
“linkages” from various fields.
The edge lengths are fixed, but not the angles.
Planar (2D) or spatial (3D).
Some examples of configuration spaces n of planar
equilateral polygons.
equilateral planar triangles
points .
P
f
g = f2
fequilateral planar quadrilateralsg =
g
.
Jun O’Hara, Aug. 2008 – p.5/21
Config. sp. of equilateral planar polygon
9 Great number of studies of the configuration spaces of
“linkages” from various fields.
The edge lengths are fixed, but not the angles.
Planar (2D) or spatial (3D).
Some examples of configuration spaces n of planar
equilateral polygons.
equilateral planar triangles
points .
P
f
g = f2
g
fequilateral planar quadrilateralsg = .
fequilateral planar pentagonsg = (Havel 1991).
4
Jun O’Hara, Aug. 2008 – p.5/21
Config. sp. of equilateral planar polygon
9 Great number of studies of the configuration spaces of
“linkages” from various fields.
The edge lengths are fixed, but not the angles.
Planar (2D) or spatial (3D).
Some examples of configuration spaces n of planar
equilateral polygons.
equilateral planar triangles
points .
P
f
g = f2
g
fequilateral planar quadrilateralsg = .
fequilateral planar pentagonsg = (Havel 1991).
Pn is of dimension n 3, and smooth if n is odd.
4
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f Equilateral quadrilaterals g
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f Equilateral quadrilaterals g
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f Equilateral quadrilaterals g
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f Equilateral quadrilaterals g
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f Equilateral quadrilaterals g
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f Equilateral quadrilaterals g
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f Equilateral quadrilaterals g
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Config. space of pentagons
fequilateral planar pentagonsg = 4
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Config. space of pentagons
fequilateral planar pentagonsg = Configuration spaces of planar pentagons of edge lengths
a ; ; a is diffeomorphic to one of S ; T ; ; , and
4
1
4
5
2
2
2
3
when it is a smooth surface,
Jun O’Hara, Aug. 2008 – p.7/21
Config. space of pentagons
fequilateral planar pentagonsg = Configuration spaces of planar pentagons of edge lengths
a ; ; a is diffeomorphic to one of S ; T ; ; , and
when it is a smooth surface,
i.e. a closed orientable surface of genus g (g = 0; ; 4).
The genus is determined by a ; ; a .
4
1
2
5
2
2
3
4
1
5
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Config. space of pentagons
fequilateral planar pentagonsg = Configuration spaces of planar pentagons of edge lengths
a ; ; a is diffeomorphic to one of S ; T ; ; , and
when it is a smooth surface,
i.e. a closed orientable surface of genus g (g = 0; ; 4).
The genus is determined by a ; ; a .
4
1
2
5
2
2
3
4
1
5
Mn() = fequilateral equiangular n-gonsg= in the
Now back to
next slide.
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Range of the “bond angle” Upper bound for . Cauchy’s Arm Lemma implies
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Range of the “bond angle” Upper bound for . Cauchy’s Arm Lemma implies
2
If n then the only possible configuration is
a planar regular n-gon.
2
If >
.
n n then
= (1
)
(1
)
M ( )=;
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Range of the “bond angle” Upper bound for . Cauchy’s Arm Lemma implies
2
If n then the only possible configuration is
a planar regular n-gon.
2
If >
.
n n then
= (1
)
(1
)
M ( )=;
Lower bound for :
If the only possible configuration is
can happen only if n is even.
=0
, which
Jun O’Hara, Aug. 2008 – p.8/21
Range of the “bond angle” Upper bound for . Cauchy’s Arm Lemma implies
2
If n then the only possible configuration is
a planar regular n-gon.
2
If >
.
n n then
= (1
)
(1
)
M ( )=;
Lower bound for :
If the only possible configuration is
can happen only if n is even.
=0
If n is odd then star-shape.
, which
n . When = n we have a planar
Jun O’Hara, Aug. 2008 – p.8/21
Expected dimension of
Mn()
We can fix the first two edges P0 P1 and P1 P2 , namely, we
can fix 3 points, 2 edges, and 1 angle.
Jun O’Hara, Aug. 2008 – p.9/21
Expected dimension of
Mn()
We can fix the first two edges P0 P1 and P1 P2 , namely, we
can fix 3 points, 2 edges, and 1 angle.
We have n
more points to be fixed.
(
3)
Jun O’Hara, Aug. 2008 – p.9/21
Expected dimension of
Mn()
We can fix the first two edges P0 P1 and P1 P2 , namely, we
can fix 3 points, 2 edges, and 1 angle.
We have n
more points to be fixed.
(
3)
(
The constraints are given by the conditions of n
edges and n
angles.
(
1)
2)
Jun O’Hara, Aug. 2008 – p.9/21
Expected dimension of
Mn()
We can fix the first two edges P0 P1 and P1 P2 , namely, we
can fix 3 points, 2 edges, and 1 angle.
We have n
more points to be fixed.
(
3)
(
The constraints are given by the conditions of n
edges and n
angles.
Therefore, generically speaking,
(
1)
2)
dim Mn()“=”3(n 3) (n 2) (n 1) = n 6:
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Gordon Crippen’s results
Gordon Crippen studied the space of shapes of cycloalkanes
(not exactly same as
(J. Computational
n ) for n
Chemistry 1992).
M
7
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Gordon Crippen’s results
Gordon Crippen studied the space of shapes of cycloalkanes
(not exactly same as
(J. Computational
n ) for n
Chemistry 1992).
Cyclobutane (n
,
4 )
M
7
=4 0 Cyclopentane (n = 5, )
different from my result?
5
3
5
Jun O’Hara, Aug. 2008 – p.10/21
Gordon Crippen’s results
Gordon Crippen studied the space of shapes of cycloalkanes
(not exactly same as
(J. Computational
n ) for n
Chemistry 1992).
Cyclobutane (n
,
4 )
M
7
=4 0 Cyclopentane (n = 5, )
different from my result?
3
5
5
By numerical experiment, searching out all the possibile
values with 0.05 step size,
1
1
Cyclohexane (n
,
)
3
= 6 = os ( )
Cycloheptane (n = 7, = os ( ))
fshapesg = S
[ S
1
1
"
boat/twist-boat
1
"
1
3
chair/twist-chair
11:57?
Jun O’Hara, Aug. 2008 – p.10/21
The case when n = 3; 4
=3
=
When n
If then
1 point .
3 3
The conformation is a regular triangle.
If then
.
3 3
6=
M(
M(
)
=f
)
=;
g
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The case when n = 3; 4
=3
=
When n
If then
1 point .
3 3
The conformation is a regular triangle.
If then
.
3 3
6=
=4
=0
M(
M(
)
=f
)
=;
g
When n
If ; 4 then
1 point .
4 The conformation is a 4-times-covered edge or a square.
M ( ) = f
g
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The case when n = 3; 4
=3
=
When n
If then
1 point .
3 3
The conformation is a regular triangle.
If then
.
3 3
6=
=4
=0
M(
M(
)
=f
)
=;
g
When n
If ; 4 then
1 point .
4 The conformation is a 4-times-covered edge or a square.
0
M ( ) = f
g
M ( ) = f
g
If < < 4 then
2 points .
4 The conformation is not planar. The shapes are the same.
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The case when n = 5
When n
= 5 (different from Crippen’s result)
Jun O’Hara, Aug. 2008 – p.12/21
The case when n = 5
=5
=
When n
(different from Crippen’s result)
3
If ; 5 then
1 point .
5 5
The conformations are planar;
M ( ) = f
g
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The case when n = 5
=5
=
When n
(different from Crippen’s result)
3
If ; 5 then
1 point .
5 5
The conformations are planar;
)
a star shape (
5
M ( ) = f
g
=
Jun O’Hara, Aug. 2008 – p.12/21
The case when n = 5
=5
=
When n
(different from Crippen’s result)
3
If ; 5 then
1 point .
5 5
The conformations are planar;
) or a regular pentagon (
a star shape (
5
M ( ) = f
=
g
=
3
5
).
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The case when n = 5
=5
=
When n
(different from Crippen’s result)
3
If ; 5 then
1 point .
5 5
The conformations are planar;
) or a regular pentagon (
a star shape (
5
M ( ) = f
=
If 6= ;
5
3
5
then
g
=
3
5
).
M () = ;.
5
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n = 6, large angle case
If = 0;
then M ( ) = f1 pointg.
2
3
6
The conformations are planar;
a 6-times-covered edge or a regular hexagon.
Jun O’Hara, Aug. 2008 – p.13/21
n = 6, large angle case
If = 0;
then M ( ) = f1 pointg.
2
3
6
The conformations are planar;
a 6-times-covered edge or a regular hexagon.
<<
3
M6() =
If
2
3
, e.g. the cyclohexane case =os ( ),
1
1
3
Jun O’Hara, Aug. 2008 – p.13/21
n = 6, large angle case
If = 0;
then M ( ) = f1 pointg.
2
3
6
The conformations are planar;
a 6-times-covered edge or a regular hexagon.
2
, e.g. the cyclohexane case =os
<
<
3
3
M6() = S1
1
( ),
1
3
2
If
boat
Jun O’Hara, Aug. 2008 – p.13/21
n = 6, large angle case
If = 0;
then M ( ) = f1 pointg.
2
3
6
The conformations are planar;
a 6-times-covered edge or a regular hexagon.
2
, e.g. the cyclohexane case =os
<
<
3
3
M6() = S1 [ f2 pointsg
2
If
boat
1
( ),
1
3
"
chair, its mirror
Jun O’Hara, Aug. 2008 – p.13/21
n = 6, large angle case
If = 0;
then M ( ) = f1 pointg.
2
3
6
The conformations are planar;
a 6-times-covered edge or a regular hexagon.
2
, e.g. the cyclohexane case =os
<
<
3
3
M6() = S1 [ f2 pointsg
2
If
boat
1
( ),
1
3
"
chair, its mirror
A boat and its mirror image can be joined by a path in
6 .
M( )
Jun O’Hara, Aug. 2008 – p.13/21
n = 6, small angle case
If 0 < < then
chair
M () = S [ S [ mirror chair [f2 pointsg
3
boat
1
2
2
6
1
mirror boat
Jun O’Hara, Aug. 2008 – p.14/21
n = 6, small angle case
If 0 < < then
chair
M () = S [ S [ mirror chair [f2 pointsg
3
boat
2
1
2
6
1
mirror boat
The last 2 points are
in a “prism”,
and its mirror image. Edges intersect each other.
Jun O’Hara, Aug. 2008 – p.14/21
n = 6, small angle case
If 0 < < then
chair
M () = S [ S [ mirror chair [f2 pointsg
3
boat
2
1
2
6
1
mirror boat
The last 2 points are
in a “prism”,
and its mirror image. Edges intersect each other.
A boat and its mirror image cannot be joined by a path in
M ()
6
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Boat component
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Boat component
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Boat component
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Boat component
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Boat component
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Boat component
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Boat component
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Boat component
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Boat component
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Boat component
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Boat component
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Boat component
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Exceptional case when n = 6
= then
M ( ) =
If 3
6 3
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Exceptional case when n = 6
= then
1-skelton of tetrahedron
M ( ) = with each edge doubled
If 3
6 3
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Exceptional case when n = 6
= then
chair
1-skelton of tetrahedron
M ( ) = with each edge doubled [ mirror chair
If 3
6 3
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M6( 3 ) n fhairg
4
2
0
5
4
3
4
0
2
0
2
4
2
0
1
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M6( 3 ) n fhairg
4
1
3
4
3
0
5
2
1
1
3
0
5
0
2
2
1
4
5
4
3
2
0
5
Jun O’Hara, Aug. 2008 – p.17/21
M6( 3 ) n fhairg
4
5
0
1
2
3
4
5
5
1
4
3
3
0
2
4
2
0
1
5
3
0
2
1
Jun O’Hara, Aug. 2008 – p.17/21
M6( 3 ) n fhairg
4
2
0
4
4
0
2
0
2
4
0
2
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Conclusion
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Conclusion
The configuration spaces of linkages and polygons are rich!
Jun O’Hara, Aug. 2008 – p.18/21
Proof
how to express polygons
P = (P ; ; P ) 2 M ()
) jPi Pi j = 2 sin .
0
5
+2
6
2
Jun O’Hara, Aug. 2008 – p.19/21
Proof
how to express polygons
P = (P ; ; P ) 2 M ()
) jPi Pi j = 2 sin .
0
5
+2
6
2
) (P ; P ; P ) forms a regular triangle.
Suppose 4P P P lies on the xy -plane.
0
2
4
0
2
4
Jun O’Hara, Aug. 2008 – p.19/21
Proof
how to express polygons
P = (P ; ; P ) 2 M ()
) jPi Pi j = 2 sin .
0
5
+2
6
2
) (P ; P ; P ) forms a regular triangle.
Suppose 4P P P lies on the xy -plane.
Consider “double cones” around the three edges of
4P P P .
0
2
4
0
0
2
2
4
4
Jun O’Hara, Aug. 2008 – p.19/21
Proof
how to express polygons
P = (P ; ; P ) 2 M ()
) jPi Pi j = 2 sin .
0
5
+2
6
2
) (P ; P ; P ) forms a regular triangle.
Suppose 4P P P lies on the xy -plane.
Consider “double cones” around the three edges of
4P P P .
Then P lies on this “double-coned” regular triangle.
0
2
4
0
0
2
2
4
4
Jun O’Hara, Aug. 2008 – p.19/21
Double cone expression
P1 ; P3 , and P5 rotate in the middle circles of the double
cones.
Let '1 ; '2 , and '3 be the angles of P1 ; P3 , and P5 from
the xy -plane.
Jun O’Hara, Aug. 2008 – p.20/21
Double cone expression
P1 ; P3 , and P5 rotate in the middle circles of the double
cones.
Let '1 ; '2 , and '3 be the angles of P1 ; P3 , and P5 from
the xy -plane.
=
=
For a given '1 , \P0
\P2 '2 and '3 can take 0, 1, or 2 values f '1 .
)
( )
Jun O’Hara, Aug. 2008 – p.20/21
Double cone expression
P1 ; P3 , and P5 rotate in the middle circles of the double
cones.
Let '1 ; '2 , and '3 be the angles of P1 ; P3 , and P5 from
the xy -plane.
=
=
For a given '1 , \P0
\P2 '2 and '3 can take 0, 1, or 2 values f '1 .
What is exceptional in the case when n
is:
If '2
f+ '1 and '3 f '1 then \P4 .
)
= ( )
( )
=6
= ( )
=
Jun O’Hara, Aug. 2008 – p.20/21
Double cone expression
P1 ; P3 , and P5 rotate in the middle circles of the double
cones.
Let '1 ; '2 , and '3 be the angles of P1 ; P3 , and P5 from
the xy -plane.
=
=
For a given '1 , \P0
\P2 '2 and '3 can take 0, 1, or 2 values f '1 .
What is exceptional in the case when n
is:
If '2
f+ '1 and '3 f '1 then \P4 .
)
( )
=6
= ( )
= ( )
=
; dim M () = 1 > 0 = 6 6, which is generic.
6
Jun O’Hara, Aug. 2008 – p.20/21
Methods to express polygons
!
Fix the first two edges.
We have n
more points (or vectors Pi 1 Pi ),
and n
conditions
for edge lengths and n
for angles).
(n
2
2
(
3
3)
1
Jun O’Hara, Aug. 2008 – p.21/21
Methods to express polygons
!
Fix the first two edges.
We have n
more points (or vectors Pi 1 Pi ),
and n
conditions
for edge lengths and n
for angles).
(n
1
3(n 3)
2n
F
for
some
F
R
R
n
(
2 3
2
M ( )=
3)
(0)
:
1
!
3
.
Jun O’Hara, Aug. 2008 – p.21/21
Methods to express polygons
!
Fix the first two edges.
We have n
more points (or vectors Pi 1 Pi ),
and n
conditions
for edge lengths and n
for angles).
(n
1
3(n 3)
2n 3
F
for
some
F
R
R
.
n
When n
m, “double-coned” equilateral m-gon
P0 ; P2 ;
; Pn 2
m angles.
(
3)
2 3
2
M ( ) = (0)
=2
(
)+
:
1
!
Jun O’Hara, Aug. 2008 – p.21/21
Methods to express polygons
!
Fix the first two edges.
We have n
more points (or vectors Pi 1 Pi ),
and n
conditions
for edge lengths and n
for angles).
(n
1
3(n 3)
2n 3
F
for
some
F
R
R
.
n
When n
m, “double-coned” equilateral m-gon
P0 ; P2 ;
; Pn 2
m angles.
Metric matrix (Crippen-Havel, Distance Geometry)
n n matrix M
vi vj , where vi Pi 1 Pi.
(
3)
2 3
2
M ( ) = (0)
=2
(
)+
=( )
:
1
!
=
!
Jun O’Hara, Aug. 2008 – p.21/21
Methods to express polygons
!
Fix the first two edges.
We have n
more points (or vectors Pi 1 Pi ),
and n
conditions
for edge lengths and n
for angles).
(n
1
3(n 3)
2n 3
F
for
some
F
R
R
.
n
When n
m, “double-coned” equilateral m-gon
P0 ; P2 ;
; Pn 2
m angles.
Metric matrix (Crippen-Havel, Distance Geometry)
n n matrix M
vi vj , where vi Pi 1 Pi.
Dihedral angles
(
3)
2 3
2
M ( ) = (0)
=2
(
)+
=( )
:
1
!
=
!
Jun O’Hara, Aug. 2008 – p.21/21