Gauss map of surfaces in 3-dimentional Heisenberg group
Author
Christiam Figueroa
Title
Gauss map of surfaces in 3-dimentional Heisenberg group
Description
We establishes a relationship between the tension field of the Gauss map and mean curvature of a surface in the Heisenberg group.
Category
Essays, Posts & Presentations
Keywords
Gauss map, Tension field, Heisenberg group
URL
http://www.notebookarchive.org/2020-11-cfrnklh/
DOI
https://notebookarchive.org/2020-11-cfrnklh
Date Added
2020-11-27
Date Last Modified
2020-11-27
File Size
266.07 kilobytes
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Tension field of the Gauss map in the Heisenberg group
Tension field of the Gauss map in the Heisenberg group
Christiam Figueroa
In this paper we study the Gauss map of surfaces in 3-dimensional Heisenberg group using the Gans model of the hyperbolic plane. We establishes a relationship between the tension field of the Gauss map and mean curvature of a surface in .
ℋ
3
Introduction
Introduction
It is well-known the classical Weierstrass representation formula describes minimal surfaces in in terms of their Gauss map. More generally, Kenmotsu shows a representation formula for arbitrary surfaces with non-vanishing mean curvature, which describes these surfaces in terms of their Gauss map and mean curvature functions. Similar result have been obtained for minimal surface in the Heisenberg group, see .
3
R
[1]
[2]
Motivated by these results, we wish to investigate a relationship between the tension field of the Gauss map and mean curvature of a surface in .
We have organized the paper as follows. In section 2 we present the Gans model of the hyperbolic plane where a geodesic is a line or a branch of a hyperbola.
Section 3 contains the basic Riemannian geometry of equipped with a left-invariant metric.
In section 4 we review some of the standard facts on the nonparametric surface in. We compute the coefficients of the first and second fundamental form of these surfaces.
Section 5 provides a detailed exposition of the Gauss map of nonparametric surfaces. And section 6 we describe the relationship between the tension field of the Gauss map and mean curvature of a surface in.
Finally, we include an appendix with a Mathematica program that allows us to compute the tension field of the Gauss map.
ℋ
3
We have organized the paper as follows. In section 2 we present the Gans model of the hyperbolic plane where a geodesic is a line or a branch of a hyperbola.
Section 3 contains the basic Riemannian geometry of
ℋ
3
In section 4 we review some of the standard facts on the nonparametric surface in
ℋ
3
Section 5 provides a detailed exposition of the Gauss map of nonparametric surfaces. And section 6 we describe the relationship between the tension field of the Gauss map and mean curvature of a surface in
ℋ
3
Finally, we include an appendix with a Mathematica program that allows us to compute the tension field of the Gauss map.
The Gans model
The Gans model
This is a model of the hyperbolic geometry, developed by David Gans, (see ), unlike the other models, utilizes the entire plane. We shall present the basic concepts of this geometry, that is, isometries group, connection and geodesics. Consider the Poincare Disk
[3]
={(x,y):+<1}
2
x
2
y
endowed with the metric
g(x,y)=
4d+d
2
x
2
y
2
1--
2
x
2
y
We first define a diffeomorphism between the Poincare disk and the plane : z = 1 . Using the stereographic projection from the south pole (0, 0, -1) of the unite sphere, we have the following diffeomorphism ϕ between the upper hemisphere onto the disk ,
S
+
ϕ(x,y,z)=,
x
z+1
y
z+1
In the same way, considering the stereographic projection from the origin (0, 0, 0) of the unite sphere, we define a diffeomorphism ψ of , onto the plane : z = 1,
S
+
ψ(x,y,z)=,,1
x
z
y
z
(
1
)
Then, F(x,y)=ψ∘ is a diffeomorphism from the disk onto the plane , where
-1
ϕ
F(x,y)=,,1
2x
1--
2
x
2
y
2y
1--
2
x
2
y
(
2
)and the inverse is given by
-1
F
u
1+
1++
2
u
2
v
v
1+
1++
2
u
2
v
Then the metric induced on by F is
h(u,v)
(1+)d-2uvdudv+(1+)d
2
v
2
u
2
u
2
v
1++
2
u
2
v
The riemannian space (,h) is the Gans model of the hyperbolic geometry.
The Geometry of the Heisenberg group
The Geometry of the Heisenberg group
The 3-dimensional Heisenberg group is a two-step nilpotent Lie group. It has the following standard representation in ()
ℋ
3
GL
3
1 | r | t |
0 | 1 | s |
0 | 0 | 1 |
with r,s,t ∈ .
In order to describe a left-invariant metric on , we note that the Lie algebra of is given by the matrices
ℋ
3
3
ℋ
3
A=
0 | x | z |
0 | 0 | y |
0 | 0 | 0 |
with x,y,z real. The exponential map exp:⟶ is a global diffeomorphism, and is given by
3
ℋ
3
exp(A)=I+A+=
2
A
2
1 | x | z+ xy 2 |
0 | 1 | y |
0 | 0 | 1 |
Using the exponential map as a global parametrization, with the identification of the Lie algebra with given by
Using the exponential map as a global parametrization, with the identification of the Lie algebra
3
3
(x,y,z)⟶
0 | x | z |
0 | 0 | y |
0 | 0 | 0 |
the group structure of is given by
ℋ
3
(a,b,c)(x,y,z)=a+x,b+y,c+z+.
ay-bx
2
(
3
)From now on, modulo the identification given by exp, we consider as with the product given in (,,} of as the orthonormal frame at the identity, we have the following left-invariant metric d in ,
ℋ
3
3
3
). Now, using {e
1
e
2
e
3
3
2
s
ℋ
3
2
ds
2
dx
2
dy
2
y
2
x
2
(
4
)And the basis of the orthonormal left-invariant vector fields is given by
=-; =+; =
E
1
∂
∂x
y
2
∂
∂z
E
2
∂
∂y
x
2
∂
∂z
E
3
∂
∂z
Graphs in ℋ3
Graphs in
ℋ
3
Let S be a graph of a smooth function f:Ω⟶ where Ω is an open set of . We consider the following parametrization of S
3
2
X(x,y)=(x,y,f(x,y)),(x,y)∈Ω
A basis of the tangent space S associated to this parametrization is given by
T
p
X x | = | (1,0, f x | = | E 1 f x y 2 E 3 |
X y | = | 0,1, f y | = | E 2 f y x 2 E 3 |
and its unit normal vector is given by
and its unit normal vector is given by
η(x,y)=-+--+
f
x
y
2
w
E
1
f
y
x
2
w
E
2
1
w
E
3
where w=. Then the coefficients of the first fundamental form of S are given by
1+++-
2
f
x
y
2
2
f
y
x
2
E | = | < X x X x | = | 1+ 2 f x y 2 |
F | = | < X y X x | = | f x y 2 f y x 2 |
G | = | < X y X y | = | 1+ 2 f y x 2 |
(
5
)
If ∇ is the Riemannian connection of, by Weingarten formula for hypersurfaces, we have that
If ∇ is the Riemannian connection of
ℋ
3
A
η
∇
v
T
p
and the coefficients of the second fundamental form are given by
and the coefficients of the second fundamental form are given by
L | = | -< ∇ X x X x | = | f xx y 2 f x x 2 f y w |
M | = | -< ∇ X x X y | = | - 1 2 2 y 2 f x f xy 1 2 2 - x 2 f y w |
N | = | -< ∇ X y X y | = | - y 2 f x x 2 f y f yy w |
(
6
)The Gauss map
The Gauss map
Recall that the Gauss map is a function from an oriented surface, S⊂, to the unit sphere in the Euclidean space. It associates to every point on the surface its oriented unit normal vector. Considering the Euclidean space as a commutative Lie group, the Gauss map is just the translation of the unit normal vector at any point of the surface to the origin, the identity element of . Reasoning in this way we define a Gauss map in the following form:
3
E
3
Definition
Let S⊂ G be an orientable hypersurface of a n-dimensional Lie group G provided with a left invariant metric. The map
γ:S⟶={v∈:v=1}
n-1
S
g
where γ(p)=d ◦ η(p) , the Lie algebra of G and η the unitary normal vector field of S is called the Gauss map of S
()
-1
L
p
g
We observe that
dγ(S)⊆=γd(S)
T
p
T
γ(p)
n-1
S
⊥
(p)
-1
L
p
T
p
therefore
d◦dγ[S]⊆S
L
p
T
p
T
p
Now we obtain a local expression of the Gauss map γ. In fact, we consider the following sequence of maps
Now we obtain a local expression of the Gauss map γ. In fact, we consider the following sequence of maps
ϕ:ΩX(Ω)⊂
X
→
ℋ
3
γ
→
2
S
ψ
→
where, X is a parametrization of S and ψ is given by (
6
)When S is the graph of a smooth function f(x,y) with (x,y) in a domain Ω ⊂. Then
2
ϕ(x,y)=-+,-
f
x
y
2
f
y
x
2
and the Jacobian matrix of ϕ is
dϕ(x,y)=
- f xx | - f xy 1 2 |
- f xy 1 2 | - f yy |
(
7
)
Notice that
det(dϕ(x,y))=-+
f
xx
f
yy
2
f
xy
1
4
(
8
)
and we will call this expression, the determinant of the Gauss map at the point (x,y). If Ω =, the greatest lower bound of the absolute value of det(dϕ(x,y)) is zero. This was proved by A. Borisenko and E. Petrov .
and we will call this expression, the determinant of the Gauss map at the point (x,y). If Ω =
2
[4]
Tension field and mean curvature of surface in ℋ3
Tension field and mean curvature of surface in
ℋ
3
In this section we establishes a relationship between the tension field of the Gauss map and mean curvature of a surface in . We recall firstly the mean curvature formula of any surface of in terms of the coefficients of their first and second fundamental forms in some parametrization.
ℋ
3
ℋ
3
H=
EN+GL-2FM
2(EG-)
2
F
When the surface is graph of a smooth function f, using (
8
) and (8
) into the above equation1+-2pq+1+
2
q
f
xx
f
xy
2
p
f
yy
3/2
1++
2
p
2
q
(
9
)where p = and q = - . Unlike the minimal surface case, for graphs of non - zero constant mean curvature, we have a Bernstein type theorem, see for more details.
f
x
+
y
2
f
y
x
2
[5]
Theorem 1
There are no complete graphs of constant mean curvature H≠0.
When H=0 we obtain the equation of the minimal graphs in
ℋ
3
1+-2pq+1+=0
2
q
f
xx
f
xy
2
p
f
yy
(
10
)
Thisequationappearsforthefirsttimein.Beforepresentingsomeconsequencesoftheaboveequation,weshallshowsomeexamplesofcompleteminimalgraphs,thatisΩ=,andusingformulas()and()tofindtheimageandtherankoftheirGaussmap,
[6]
2
Example 1
As in Euclidean space , the plane f(x,y)= ax + by + c is a minimal graph of . Then ϕ() = and the rank of the Gauss map is 2.
3
E
ℋ
3
2
2
Example 2
Another minimal graph may be obtained by searching for solutions of Scherk type, i.e for solutions of the form f(x,y) = . From this method we find, among others, the following example, see .
g(x)+h(y)+
xy
2
[7]
f(x,y)+ky
xy
2
1+
+Logy+2
y
1+
2
y
where k ∈ . .Notice that this minimal surface is ruled by affine lines, i.e.translations of 1 - parameter subgroups.Then
ϕ()=
2
|
That is, the image of ϕ are geodesics of the hyperbolic plane and its rank is 1 . As we have seen, we have several solutions of (, the following theorem relates the mean curvature and the tension field of the Gauss map of the surface.
10
) defined on the entire xy- plane. Unlike the case of Euclidean spaces, where the only complete minimal graphs are linear (Bernstein' s theorem). On the other hand, is known that the Gauss map of a minimal surface in the Euclidean space is antiholomorphic.For any graph in the Heisenberg group ℋ
3
Theorem 2
Let S⊂ be a graph of a smooth function f:Ω⟶ where Ω is an open set of . Suppose that H and ϕ are the mean curvature and the Gauss map of S, respectively.Then the tension field of the Gauss map τ(ϕ) =(τ(),τ()) satisfy
ℋ
3
2
1
ϕ
2
ϕ
τ 1 ϕ wH x | = | H 2 f y x 2 w w x f x y 2 2 Hw |
τ 2 ϕ wH y | = | H- 2 f x y 2 w w y f y x 2 2 Hw |
where w=.
1+++-
2
f
x
y
2
2
f
y
x
2
Proof
We begin by recalling that the Gauss map of S is given by ϕ (x, y) = -( + , - ) and the metric tensors of S is g = () where g is as (}), the tension field of ϕ is given by τ () = △ + where α,β = 1, 2 , (x, y) = -( + ), (x,y)= - ), △ is the Riemannian Laplace on S, are the Christoffel symbols of the metric h of , evaluated in the Gauss map and and = . First we compute the first component of the tension field and add twice the x - derivative of wH (see and Appendix) . Substituting the mean curvature (
f
x
y
2
f
y
x
2
g
ij
10
). By definition (see for instance [8]
α
ϕ
α
ϕ
α
Γ
βγ
β
ϕ
j
γ
ϕ
i
ij
g
1
ϕ
f
x
y
2
2
ϕ
f-
(
y
x
2
α
Γ
βγ
β
ϕ
1
=
∂
β
ϕ
∂x
β
ϕ
2
∂
β
ϕ
∂y
10
) we obtain τ+2(2+y)+2(2+y)(x-2)+4++(4-6)(x-2)
1
ϕ
(wH)
x
H
4w
f
x
2
(x-2)
f
y
f
xx
f
x
f
y
f
xy
2
(2+y)
f
x
f
yy
f
xy
f
y
Replacing again the mean curvature in the above equation, we get the desired formula. We now apply this argument again, to obtain the second formula, which completes the proof.
Let us mention an important consequence of the above theorem.
Corollary
Let S ⊂ a minimal surface.Then the Gauss map of S is harmonic.
ℋ
3
Appendix: Tension field of the Gauss map
Appendix: Tension field of the Gauss map
This is a Mathematica program to compute the Tension Field of the Gauss map between the graph of a function f:Ω⟶ and the Gans model of the hyperbolic plane. Let S be the graphic of a function f. We need the partial derivatives of f up to order three:
In[]:=
Clear[u,v,x,y]
In[]:=
f[u_,v_];[u_,v_];[u_,v_];[u_,v_];[u_,v_];[u_,v_];[u_,v_];[u_,v_];[u_,v_];[u_,v_];[u_,v_];
f
x
f
y
f
xx
f
xy
f
yx
f
yy
f
xyy
f
yyy
f
xxx
f
xxy
In[]:=
Derivative[1,0][f][u,v]:=[u,v]
f
x
In[]:=
Derivative[0,1][f][u,v]:=[u,v]
f
y
In[]:=
Derivative[1,0][][u,v]:=[u,v]
f
x
f
xx
In[]:=
Derivative[0,1][][u,v]:=[u,v]
f
x
f
xy
In[]:=
Derivative[1,0][][u,v]:=[u,v]
f
y
f
xy
In[]:=
Derivative[0,1][][u,v]:=[u,v]
f
y
f
yy
In[]:=
Derivative[0,1][][u,v]:=
f
xy
f
xyy
In[]:=
Derivative[1,0][][u,v]:=
f
yy
f
xyy
In[]:=
Derivative[0,1][][u,v]:=
f
yy
f
yyy
In[]:=
Derivative[1,0][][u,v]:=
f
xx
f
xxx
In[]:=
Derivative[0,1][][u,v]:=
f
xx
f
xxy
In[]:=
Derivative[1,0][][u,v]:=
f
xy
f
xxy
In[]:=
X[u_,v_]:={u,v,f[u,v]}
Let {,} be the basis of the tangent space S associated to the parametrization X. The components of and with respect the left-invariant vector fields {} of the Heisenberg are:
X
x
X
y
T
p
X
x
X
y
E
i
In[]:=
X
x
f
x
v
2
Out[]=
1,0,+[u,v]
v
2
f
x
In[]:=
X
y
f
y
u
2
Out[]=
0,1,-+[u,v]
u
2
f
y
The normal field to the surface S is given by
In[]:=
N1=Cross[,]
X
x
X
y
Out[]=
--[u,v],-[u,v],1
v
2
f
x
u
2
f
y
The Gauss map ϕ:S→ is given by:
In[]:=
ϕ1[u_,v_]:=-+[u,v]
v
2
f
x
In[]:=
ϕ2[u_,v_]:=-[u,v]-
f
y
u
2
In[]:=
ϕ={ϕ1[u,v],ϕ2[u,v]}
Out[]=
--[u,v],-[u,v]
v
2
f
x
u
2
f
y
The coefficients of the first fundamental form of the graph of f are:
In[]:=
g11=Dot[,]
X
x
X
x
Out[]=
1++[u,v]
2
v
2
f
x
In[]:=
g12=Dot[,]
X
x
X
y
Out[]=
v
2
f
x
u
2
f
y
In[]:=
g21=Dot[,]
X
y
X
x
Out[]=
v
2
f
x
u
2
f
y
In[]:=
g22=Dot[,]
X
y
X
y
Out[]=
1+
2
-+[u,v]
u
2
f
y
Laplacian of the Gauss map of the graph of the function f(x,y)
Laplacian of the Gauss map of the graph of the function f(x,y)
This is a Mathematica program to compute the Laplacian of the Gauss map. The Laplacian of a differential function h:S→ is calculated from the formula
Δh=h
1
Det(g)
2
∑
i,j=2
∂
x
i
Det(g)
ij
g
∂
j
x
where(S,g)isariemanniansurfaceandisthematrixinverseofcalledtheinversemetric.
ij
g
g
ij
In[]:=
Clear[coord,metric,inversemetric]
In[]:=
coord={u,v};
We input the metric of the surface S as a matrix.
In[]:=
metric={{g11,g21},{g12,g22}};
In matrix form:
In[]:=
metric//MatrixForm;
The inverse metric is obtained through matrix inversion.
In[]:=
inversemetric=Inverse[metric];
The inverse metric can also be displayed in matrix form:
In[]:=
inversemetric//MatrixForm;
The Laplacian of the first component of the map Gauss,
1
ϕ
In[]:=
LA1=Simplify*SumD
1
Det[metric]
Det[metric]
*inversemetric[[i,j]]*D[ϕ[[1]],coord[[i]]],coord[[j]],{i,1,2},{j,1,2};The Laplacian of the second component of the map Gauss,
2
ϕ
In[]:=
LA2=Simplify*SumD
1
Det[metric]
Det[metric]
*inversemetric[[i,j]]*D[ϕ[[2]],coord[[i]]],coord[[j]],{i,1,2},{j,1,2};Christoffel Symbols of the Gans model
Christoffel Symbols of the Gans model
This is a Mathematica program to compute the Christoffel symbols . The Christoffel symbols are calculated from the formula
λ
Γ
μν
1
2
λσ
g
∂
μ
g
σν
∂
ν
g
σμ
∂
σ
g
μν
whereisthematrixinverseofcalledtheinversemetric.Thisprogramwasadaptedfrom.ThisprogramwasadaptedfromthenotebookCurvatureandtheEinsteinequationwrittenbyLeonardParker
λσ
g
g
λσ
In[]:=
Clear[coord,metricgg,inversemetricgg,affine,u,v]
In[]:=
coord={u,v};
We input the metric of the Gans model of the Hyperbolic space as a matrix.
In[]:=
metricgg=,,,;
1+
2
v
1++
2
u
2
v
-uv
1++
2
u
2
v
-uv
1++
2
u
2
v
1+
2
u
1++
2
u
2
v
In matrix form:
In[]:=
metricgg//MatrixForm;
The inverse metric is obtained through matrix inversion.
In[]:=
inversemetricgg=Simplify[Inverse[metricgg]];
The inverse metric can also be displayed in matrix form:
In[]:=
inversemetricgg//MatrixForm;
The calculation of the components of the affine connection is done by transcribing the definition given earlier into the notation of Mathematica and using the Mathematica functions D for taking partial derivatives, Sum for summing over repeated indices, Table for forming a list of components, and Simplify for simplifying the result.
In[]:=
affine:=affine=Simplify[Table[(1/2)*Sum[(inversemetricgg[[i,s]])*(D[metricgg[[s,j]],coord[[k]]]+D[metricgg[[s,k]],coord[[j]]]-D[metricgg[[j,k]],coord[[s]]]),{s,1,2}],{i,1,2},{j,1,2},{k,1,2}]]
The components of the affine connections of the Gans model are displayed below. Because the affine connection is symmetric under interchange of the last two indices, only the independent components are displayed.
In[]:=
listaffine:=Table[If[UnsameQ[affine[[i,j,k]],0],{ToString[Γ[i,j,k]],affine[[i,j,k]]}],{i,1,2},{j,1,2},{k,1,j}]
In[]:=
TableForm[Partition[DeleteCases[Flatten[listaffine],Null],2],TableSpacing{2,2}]
Out[]//TableForm=
Γ[1, 1, 1] | - u(1+ 2 v 1+ 2 u 2 v |
Γ[1, 2, 1] | 2 u 1+ 2 u 2 v |
Γ[1, 2, 2] | - u+ 3 u 1+ 2 u 2 v |
Γ[2, 1, 1] | - v+ 3 v 1+ 2 u 2 v |
Γ[2, 2, 1] | u 2 v 1+ 2 u 2 v |
Γ[2, 2, 2] | - (1+ 2 u 1+ 2 u 2 v |
In matrix form:
In[]:=
l1={{affine[[1,1,1]],affine[[1,1,2]]},{affine[[1,2,1]],affine[[1,2,2]]}};
In[]:=
l2={{affine[[2,1,1]],affine[[2,1,2]]},{affine[[2,2,1]],affine[[2,2,2]]}};
Tension field of the Gauss map ϕ
Tension field of the Gauss map ϕ
This is a Mathematica program to compute the Tension field of the Gauss map. The tension field is calculated from the formula
τ()=Δ+
α
ϕ
α
ϕ
2
∑
i,j=1
2
∑
β,γ=1
α
Γ
βγ
∂
i
β
ϕ
∂
j
γ
ϕ
ij
g
WhereΔistheS-laplacianofthecomponent,theChristoffelsymbolsoftheHyperbolicspaceevaluateatGaussmap,isthepartialderivativeofwhithrespecttouandthepartialderivativeofwithrespecttov.
α
ϕ
α
ϕ
α
Γ
βγ
∂
1
β
ϕ
β
ϕ
∂
2
β
ϕ
β
ϕ
First, we evaluate the Christoffel symbols of the Gans model at the Gauss map ϕ:
In[]:=
AFF1=l1/.{uϕ[[1]],vϕ[[2]]};
In[]:=
AFF2=l2/.{uϕ[[1]],vϕ[[2]]};
Let D1=
2
∑
i,j=1
2
∑
β,γ=1
1
Γ
βγ
∂
i
β
ϕ
∂
j
γ
ϕ
ij
g
In[]:=
D1=Sum[inversemetric[[i,j]]*AFF1[[k,l]]*D[ϕ[[k]],coord[[i]]]*D[ϕ[[l]],coord[[j]]],{i,1,2},{j,1,2},{k,1,2},{l,1,2}];
In[]:=
DD1=Simplify[D1];
Let D2=
2
∑
i,j=1
2
∑
β,γ=1
2
Γ
βγ
∂
i
β
ϕ
∂
j
γ
ϕ
ij
g
In[]:=
D2=Sum[inversemetric[[i,j]]*AFF2[[k,l]]*D[ϕ[[k]],coord[[i]]]*D[ϕ[[l]],coord[[j]]],{i,1,2},{j,1,2},{k,1,2},{l,1,2}];
In[]:=
DD2=Simplify[D2];
The first component of the tension field of the map Gauss, τ( ) is given by
1
ϕ
In[]:=
Ten1=Simplify[LA1+DD1];
In the same manner we can see the second componnent of the tension field of the Gauss map, τ() is given by
2
ϕ
In[]:=
Ten2=Simplify[LA2+DD2];
The mean curvature equation
The mean curvature equation
Let f:Ω⟶ be a differential function in the Heisenberg group. The mean curvature equation of f is given by
ℋ
3
1+[u,v]-[u,v]-2[u,v]+[u,v]-[u,v]+1+[u,v]+[u,v]1+[u,v]++[u,v]-=2Hw
2
f
y
u
2
f
xx
f
x
v
2
f
y
u
2
f
xy
2
f
x
v
2
f
yy
2
f
x
v
2
2
f
y
u
2
The derivative of 2(wH) with respect to x
In[]:=
MU=D1+[u,v]-[u,v]-2[u,v]+[u,v]-[u,v]+1+[u,v]+[u,v]1+[u,v]++[u,v]-,{u,1};
2
f
y
u
2
f
xx
f
x
v
2
f
y
u
2
f
xy
2
f
x
v
2
f
yy
2
f
x
v
2
2
f
y
u
2
The simplify espression of τ( )+2 is:
1
ϕ
(wH)
x
In[]:=
FullSimplify[Ten1+MU];
In[]:=
ReplaceAll[%,{[u,v],[u,v],[u,v],[u,v],[u,v]}]
f
x
f
x
f
y
f
y
f
xx
f
xx
f
xy
f
xy
f
yy
f
yy
Out[]=
(-6+4+(v+2)(2(v+2)+(u-2)))(u-2)+(v+2)(4++4(v+))4++4(-u+)+4+(v+2)(2(u-2)+(v+2))2
f
xy
f
x
f
x
f
xy
f
xx
f
y
f
y
f
x
2
v
f
x
f
x
f
yy
f
xx
2
u
f
y
f
y
f
yy
f
x
f
xy
f
y
f
x
f
yy
2
4+++4(v+)+4(-u+)
2
u
2
v
f
x
f
x
f
y
f
y
In[]:=
Numerator[%]
Out[]=
(-6+4+(v+2)(2(v+2)+(u-2)))(u-2)+(v+2)(4++4(v+))4++4(-u+)+4+(v+2)(2(u-2)+(v+2))
f
xy
f
x
f
x
f
xy
f
xx
f
y
f
y
f
x
2
v
f
x
f
x
f
yy
f
xx
2
u
f
y
f
y
f
yy
f
x
f
xy
f
y
f
x
f
yy
We now apply this argument again, with u replaced with v, to obtain the derivative of 2(wH) with respect to y
In[]:=
MV=D1+[u,v]-[u,v]-2[u,v]+[u,v]-[u,v]+1+[u,v]+[u,v]1+[u,v]++[u,v]-,{v,1};
2
f
y
u
2
f
xx
f
x
v
2
f
y
u
2
f
xy
2
f
x
v
2
f
yy
2
f
x
v
2
2
f
y
u
2
The simplify expression of
τ+2is:
2
ϕ
(wH)
y
In[]:=
FullSimplify[Ten2+MV];
In[]:=
ReplaceAll[%,{[u,v],[u,v],[u,v],[u,v],[u,v]}]
f
x
f
x
f
y
f
y
f
xx
f
xx
f
xy
f
xy
f
yy
f
yy
Out[]=
-4++4(-u+)+4+(v+2)(2(u-2)+(v+2))2v(3+2)+4(u-2)+43+2++4(-u+)+v(u-2)+(u-2)4++4(-u+)+v(2(u-2)+v)2
f
xx
2
u
f
y
f
y
f
yy
f
x
f
xy
f
y
f
x
f
yy
f
xy
2
f
x
f
y
f
yy
f
x
f
xy
2
u
f
y
f
y
f
y
f
yy
f
y
f
xx
2
u
f
y
f
y
f
xy
f
y
f
yy
2
4+++4(v+)+4(-u+)
2
u
2
v
f
x
f
x
f
y
f
y
In[]:=
Numerator[%]
Out[]=
-4++4(-u+)+4+(v+2)(2(u-2)+(v+2))2v(3+2)+4(u-2)+43+2++4(-u+)+v(u-2)+(u-2)4++4(-u+)+v(2(u-2)+v)
f
xx
2
u
f
y
f
y
f
yy
f
x
f
xy
f
y
f
x
f
yy
f
xy
2
f
x
f
y
f
yy
f
x
f
xy
2
u
f
y
f
y
f
y
f
yy
f
y
f
xx
2
u
f
y
f
y
f
xy
f
y
f
yy
[9]References
[9]
1. Kenmotsu, Katsuei, Weierstrass formula for surfaces of prescribed mean curvature. Mathematische Annalen, 1979.
2. Figueroa, CB, On the Gauss map of a minimal surface in the Heisenberg group. Mat. Contemp, 2007.
3. Gans, David, A new model of the hyperbolic plane. The American Mathematical Monthly, 1966.
4. Borisenko, Alexander Andreevich and Petrov, Eugene V, Surfaces in the three-dimensional Heisenberg group on which the Gauss map has bounded Jacobian. Mathematical Notes, 2011.
5. Figueroa, Christiam and others, The Gauss map of Minimal graphs in the Heisenberg group. Journal of Geometry and Symmetry in Physics, 2012.
6. Bekkar, M, Exemples de surfaces minimales dans l’espace de Heisenberg. Rend. Sem. Fac. Sci. Univ. Cagliari, 1991.
7. Sari, M Bekkar-T, . Rend. Sem. Mat. Univ. Pol. Torino, 1992.
Surfaces minimales reglées dans l’espace de Heisenberg H 3
8. Eells, James and Lemaire, Luc, A report on harmonic maps. Bulletin of the London mathematical society, 1978.
9. , Holomorphic quadratic differentials and the Bernstein problem in Heisenberg space. Transactions of the American Mathematical Society, 2009.
Fernández, Isabel and Mira, Pablo
Cite this as: Christiam Figueroa, "Gauss map of surfaces in 3-dimentional Heisenberg group" from the Notebook Archive (2020), https://notebookarchive.org/2020-11-cfrnklh
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