# Plant leaves, jupes à godets, and conformal embeddings

```On the plant leaf’s boundary,
“jupe &agrave; godets” and
isometric embeddings
Sergei Nechaev
LPTMS (Orsay); FIAN (Moscow)
lettuce (Lactuca sativa)
What are the geometrical reasons for the boundary of
plant leaves to fluctuate in the direction transverse to
the leaf’s surface?
How to describe the corresponding stable profile?
Some more examples…
Hierarchical construction of an
unbounded hyperbolic surface
jupe &agrave; godets –
“surface &agrave; godets” (SG)
We are interested in the embedding of unbounded
(open) surfaces only!
Hyperbolic surfaces with a boundary can be
isometrically embedded in a Euclidean space. The
pseudosphere (surface of revolution of the
tractrix) is an example of such an embedding:
 x  cos u sin u

 y  sin u sin v
 z  cos v  log tan v / 2

Bounded vs open hyperbolic surfaces (M. Esher)
Precise formulation of the problem
1. The SG, being the hyperbolic structure, admits
Cayley trees as possible discretizations. Cover the
SG by a lattice – the 4-branching Cayley tree. The
Cayley trees cover the SG isometrically, i.e.
without gaps and selfintersections, preserving
angles and distances.
2. Our aim consists in the embedding a 4-branching
Cayley tree (isometrically covering the SG) into a
3D Euclidean metric space with a signature
{+1,+1,+1}.
Example
Tesselate isometrically the open disc |ζ | &lt; 1 with all
images of the some triangle. This way one gets a
graph isometrically embedded into the unit disc
endowed with the Poincar&eacute; metric
*
d

d

2
ds 
* 2
(1   )
This structure is invariant
under conformal transforms
of the Poincar&eacute; disc onto
itself
  0

  1
*
0
The stereographic projection of an open disc gives the
2D hyperboloid with the metric tensor
gik 
1
0
0 sinh 
2
,
ds2  d  2  sinh2  d 2
1 r
where   ln
is the hyperbolic distance.
1 r
The tessellation of the Poincar&eacute; disc by circular
triangles (rectangles) is uniform in a surface of a twodimensional hyperboloid. It can be embedded into a
3D space with a Minkovski metric, i.e. with a metric
tensor of signature {+1, +1,−1}.
The main statement
Tesselate isometrically the disc |ζ | &lt; 1 with all images
of zero-angled rectangle AζBζA’ζCζ. Connecting the
centres of the neighbouring rectangles, one gets a 4branching Cayley tree isometrically covering the unit
Poincar&eacute; disc.
The requested isometric embedding of a 4-branching
Cayley tree into a 3D Euclidean space is realized via
conformal transform z=z(ζ ) which maps a flat square
AzBzA’zCz in the complex plane z to a circular zeroangled rectangle AζBζA’ζCζ in the Poincar&eacute; disc |ζ|&lt;1.
The main statement (continuation)
The relief of the corresponding surface is encoded in
2
the so-called coefficient of deformation, J ( )  dz / d 
coinciding with the Jacobian of the conformal
transform z(ζ ):
 Re z
 Re 
2
J ( )  z '( ) 
 Im z
 Re 
 Re z
 Im 
 Im z
 Im 
Finding explicit form of z(ζ ) is our main goal.
Remark:
In which sense our surface is optimal?
All the images of the zero-angled rectangle AζBζA’ζCζ
in the complex plane ζ have the same area. The areas
of elementary cells AzBzA’zCz tessellating the plane z
are
S
Az Bz A'z Cz
dzdz  
*
A B A' C
J ( z )d d
*
Thus, all zero-angled rectangles have the same area S
in ζ.
Consequence. Under requested constraints the
surface J(z)=|z’(ζ )|2 is minimal.
Explicit construction by conformal maps
Explicit expression in terms of elliptic functions

dz( )
4
'
J ( ) 
 2
1  0, e
4

d
  i

2
1  , ei   2e
1'  0, ei  
1i  1
i
2  i
2


 2  0, e


1i  1
i
2  i

i  
4
n i n ( n 1) 
(

1)
sin(2n  1)
 e
n 0
d1  , ei 
2  0, ei   2e
d

i  
4
 2e

i  
4
 0
n i n ( n 1) 
(

1)
 e
n 0
n
i n ( n 1) 
(

1)
(2
n

1)
e

n 0



2
3D plots of SG
Sample plots of J ( ,  ) for 0 &lt;&lt;max and 0&lt;ϕ&lt;/2
1 r
Im 
  ln
;   arctan
1 r
Re 
max = 1.2
max = 1.5
max = 1.8
The boundary of the SG is a fractal
0.2
0.4
0.6
0.8
max = 2.4
1.5
1.8
2.1
1.0
1.2
1.4
0&lt;ϕ&lt;/2
SG is a “continuous Cayley tree”
Plot the “normalized” surface
g ( )  J ( ) 
1
(1   )
* 2
  r exp(i )
1 r
Im 
  ln
;  arctan
1 r
Re 
This surface can be considered as a “continuous
analog” of the Cayley tree.
```