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# Bipartite and Series-Parallel Graphs
Without Planar Lombardi Drawings

David Eppstein Department of Computer Science, University of California,
Irvine. Research supported in part by NSF grants CCF-1618301 and CCF-1616248.

###### Abstract

We find a family of planar bipartite graphs all of whose Lombardi drawings
(drawings with circular arcs for edges, meeting at equal angles at the
vertices) are nonplanar. We also find families of embedded series-parallel
graphs and apex-trees (graphs formed by adding one vertex to a tree) for which
there is no planar Lombardi drawing consistent with the given embedding.

##  1 Introduction

_Lombardi drawing_ is a style of graph drawing using curved edges. In this
style, each edge must be drawn as a circular arc, and consecutive edges around
each vertex must meet at equal angles. Many classes of graphs are known to
have such drawings, including regular bipartite graphs and all $2$-degenerate
graphs (graphs that can be reduced to the empty graph by repeatedly removing
vertices of degree at most two) [5]. This drawing style can significantly
reduce the area usage of tree drawings [6], and display many of the symmetries
of more general graphs [5].

When a given graph is planar, we would like to find a planar Lombardi drawing
of it. When this is possible, the resulting drawings have simple edge shapes,
no crossings, and optimal angular resolution, all of which are properties that
lead to more readable drawings. It is known that all Halin graphs have planar
Lombardi drawings [5], that all 3-regular planar graphs [7] and all 4-regular
polyhedral graphs [8] have planar Lombardi drawings, and that all outerpaths
have planar Lombardi drawings [4]. For some other classes of planar graphs,
even when a Lombardi drawing exists, it might not be planar. Classes of planar
graphs that are known to not always be drawable planarly in Lombardi style
include the nested triangle graphs [5], 4-regular planar graphs [7], planar
3-trees [4], and the graphs of knot and link diagrams [8].

However, for several other important classes of planar graphs, the existence
of a planar Lombardi drawing has remained open. These include the outerplanar
graphs, the series-parallel graphs, and the planar bipartite graphs.
Outerplanar and series-parallel graphs are 2-degenerate, and always have
Lombardi drawings. Planar bipartite graphs are 3-degenerate and such graphs
usually have Lombardi drawings.111The only obstacle to Lombardi drawing for
3-degenerate graphs is the forced placement of two vertices on the same point,
but the only examples for which this is known to happen are neither planar nor
bipartite [5]. However the known Lombardi drawings for these graphs are not
necessarily planar. In this paper we settle this open problem for two of these
classes of graphs, the planar bipartite graphs and the (embedded) series-
parallel graphs. We construct a family of planar bipartite graphs whose
Lombardi drawings are all nonplanar. We also construct a family of series-
parallel graphs with a given embedding such that no planar Lombardi drawing
respects that embedding. Our construction for series-parallel graphs can be
extended to maximal series-parallel graphs, to bipartite series-parallel
graphs and to apex-trees, the graphs formed by adding a single vertex to a
tree.

##  2 The graphs

Figure 1: The bipartite graph $B(5)$ formed by our construction. Although
drawn planarly with curved edges, this is not a Lombardi drawing: the edges
are arcs of ellipses rather than of circles, and pairs of consecutive edges at
the same vertex do not all have the same angles.

We begin by describing the family of planar bipartite graphs $B(k)$ that we
will prove (for sufficiently large $k$) do not have a planar Lombardi drawing.
Each vertex in $B(k)$ has degree either $2$ or $2k$. To construct $B(k)$,
begin with a complete bipartite graph $K_{2,2k}$ and its unique planar
embedding; in Figure 1, the two-vertex side of the bipartition of this graph
is shown by the yellow vertices and the $2k$-vertex side is shown by the blue
vertices. Each yellow vertex has exactly $2k$ blue neighbors.

Next, partition the blue vertices into $k$ pairs of vertices, each sharing a
face. For each pair of blue vertices in this partition, add another complete
bipartite graph $K_{2,2k-2}$ connecting these two blue vertices to $2k-2$
additional vertices (shown as red in the figure). After this addition, each
blue vertex has exactly $2k$ neighbors, two of them yellow and the rest red.
Each red vertex has exactly two neighbors. There are two yellow vertices, $2k$
blue vertices, and $k(2k-2)$ red vertices, for a total of $2k^{2}+2$ vertices
in the overall graph.

Clearly, the graphs $B(k)$ are all planar, because they are formed by
attaching together planar subgraphs (complete bipartite graphs where one side
has two vertices) on pairs of vertices that are cofacial in both subgraphs.
They are bipartite, with the yellow and red vertices on one side of their
bipartition and the blue vertices on the other side. Although they are not
$3$-vertex-connected, all of their planar embeddings are isomorphic.

Figure 2: The embedded series-parallel graph $S(3)$ formed by our
construction.

Analogously, we define a family of embedded series-parallel graphs $S(k)$.
Again, each such graph will have two yellow vertices and $2k$ blue vertices,
connected in the pattern of a complete bipartite graph $K_{2,2k}$. For each
yellow–blue edge $e$ of this graph, we add a path of $2k-1$ red vertices. We
connect every vertex in this path to the blue endpoint of $e$, and we connect
one endpoint of the path to the yellow endpoint of $e$. We fix an embedding of
$S(k)$ in which every yellow–blue quadrilateral contains either zero or four
red paths (Figure 2). The resulting graph has two yellow vertices, $2k$ blue
vertices, and $4k(2k-1)$ red vertices, for a total of $8k^{2}-2k+2$ vertices.
The yellow and blue vertices have degree $4k$, while the red vertices have
degrees two or three.

We claim that, for sufficiently large values of $k$, the graphs $G(k)$ and
$S(k)$ do not have planar Lombardi drawings. Therefore, neither every planar
bipartite graph nor every series-parallel graph has a planar Lombardi drawing.
In the remainder of this paper we prove this claim.

##  3 Equiangular arc-quadrilaterals

The key feature of both of our graph constructions $B(k)$ and $S(k)$ is the
existence of many yellow–blue quadrilateral faces in which all vertices have
equal and high degree (this degree is $d=2k$ in $B(k)$ and $d=4k$ in $S(k)$).
If such a graph is to have a Lombardi drawing, each of these faces must
necessarily be drawn as a quadrilateral with circular-arc sides and with the
same interior angle $2\pi/d$ at all four of its vertices. Equiangular arc-
quadrilaterals have been investigated before from the point of view of
conformal mapping [2]; in this section we investigate some of their additional
properties.

Our main tool is the following lemma:

######  Lemma .

Let $abcd$ be a non-self-crossing quadrilateral in the plane with circular-arc
sides and equal interior angles. Then the four points $abcd$ lie on a circle
and the quadrilateral $abcd$ either lies entirely inside or entirely outside
the circle.

###### Proof.

We abbreviate the conclusion of the lemma by saying that $abcd$ is cyclic. The
properties of being an equiangular non-self-crossing circular-arc
quadrilateral and of being cyclic are both invariant under Möbius
transformations, which preserve both circularity and the crossing angles of
curves. Therefore, if we can find a Möbius transformation of a given
equiangular circular-arc quadrilateral such that the transformed quadrilateral
is cyclic, the original quadrilateral will also be cyclic, as the lemma states
it to be.

Start by finding a Möbius transformations which makes two opposite arcs $ab$
and $cd$ come from circles with the same radius as each other. Because of the
equality of crossing angles, and by symmetry, both of the other two circular
arcs $bc$ and $ad$ must come from circles whose centers lie on the
perpendicular bisector of $ab$ and $cd$. There remains a one–dimensional
family of Möbius transformations that preserve the position of the circles
containing the transformed copies of arcs $ab$ and $cd$ but that move the
other two circles along the bisector of these two fixed circles. We can use
this remaining degree of freedom to move the other two circles so that their
centers are equidistant from the midpoint of the centers of the two fixed
circles.

Figure 3: Illustration for section 3: Four circles with centers on a rhombus,
and with opposite pairs of circles having equal radii, define two rectangles
of pairwise intersection points.

After this transformation, it follows from the equality of crossing angles
that the circles containing the transformed copies of arcs $bc$ and $ad$ have
the same radii as each other, and the four circles have been transformed into
a position centered at the vertices of a rhombus with opposite pairs having
the same radius as each other. By symmetry, the transformed copies of vertices
$abcd$ must lie on one of the two rectangles defined by the crossing points of
these four transformed circles. (Figure 3).

If the interior angle of $abcd$ is less than $\pi$, then the transformed copy
of $abcd$ must lie within the circle that circumscribes the inner rectangle,
forming the boundary of a hole in the union of the four transformed disks. If
the interior angle is greater than $\pi$, it must lie outside the outer
circle, forming the outer boundary of the union of the four disks. In either
case, the transformed copy of $abcd$ is cyclic, so $abcd$ itself must be
cyclic. ∎

A special case of this lemma for right-angled arc-quadrilaterals was used
previously by the author to prove that some 4-regular planar graphs have no
planar Lombardi drawing [7]. Another special case, for arc-quadrilaterals in
which all interior angles are zero, has been used previously in mesh
generation [1].

######  Definition .

We define the _tilt_ of an equiangular circular-arc quadrilateral to be the
maximum interior angle of any of the four circular-arc bigons between the
quadrilateral and its enclosing circle.

Each of the four bigons has equal angles at its two vertices. At each of the
four vertices, the two bigon angles and the interior angle of the
quadrilateral add to $\pi$. It follows that opposite bigons have the same
angles as each other, and each vertex of the quadrilateral is incident to a
bigon with vertices of the tilt angle.

##  4 Bipolar coordinates

Figure 4: Curves of constant and evenly-spaced coordinate values for bipolar
coordinates, forming two orthogonal pencils of circles.

To describe a second parameter of equiangular circular-arc quadrilaterals, it
is convenient to introduce the _bipolar coordinate system_ , defined from a
pair of points $s$ and $t$, the _foci_ of the coordinate system. These
coordinates are conventionally denoted $\sigma$ and $\tau$. The
$\sigma$-coordinate $\sigma_{p}$ of a point $p$ is the (oriented) angle $spt$,
whose level sets are the blue circular arcs through the two foci in Figure 4.
The $\tau$-coordinate $\tau_{p}$ of $p$ is the logarithm of the ratio of the
two distances from $p$ to the two foci, whose level sets are the red circles
separating the two foci in Figure 4. This coordinate system has the convenient
property that any (orientation-preserving) Möbius transformation that
preserves the location of the two foci acts by translation on the
coordinates.222This property can be seen as a reflection of the fact that the
bipolar coordinate system comes from a conformal mapping of a rectangular
grid; see, e.g., [3].

######  Lemma .

Any Möbius transformation that preserves the location of the two foci acts on
the bipolar coordinates of any point by adding a fixed value to its
$\sigma$-coordinate (modulo $2\pi$) and adding another fixed value to its
$\tau$-coordinate, with the added values depending on the transformation but
not on the point.

###### Proof.

All Möbius transformations preserves circles, incidences between points and
curves, and angles between pairs of incident curves. Therefore, any focus-
preserving Möbius transformation takes circles through the two foci (the level
sets for $\sigma$-coordinates) to other circles through the two foci, and it
takes the perpendicular family of circles (the level sets for
$\tau$-coordinates) to other circles in the same family. Therefore it acts
separately on the $\sigma$\- and $\tau$-coordinates. The additivity of its
action on the $\sigma$-coordinates follows from the preservation of angles
between pairs of circles through the two foci.

To show that the transformation acts additively on $\tau$-coordinate (the
logarithm of the ratio of distances of a point from the two foci), we can
assume without loss of generality (by scaling, translating, and rotating the
plane if necessary) that the two foci are at the two points $q=\pm 1$ of the
complex plane. Consider the general form $q\mapsto(aq+b)/(cq+d)$ of a Möbius
transformation as a fractional linear transformation of the complex plane. For
a transformation to fix $q=1$ we need $a+b=c+d$ and for it to fix $q=-1$ we
need $b-a=-(d-c)$. Solving these two equations in four unknowns gives $a=d$
and $b=c$. Therefore, the transformations fixing the foci take the special
form $q\mapsto(aq+b)/(bq+a)$.

For a transformation of this form, and for any point $x$ on the interval
$[-1,1]$ of real numbers with distance ratio $(1+x)/(1-x)$, the image of $x$
has distance ratio

$\frac{1+(ax+b)(bx+a)}{1-(ax+b)(bx+a)}=\frac{a+b}{a-b}\cdot\frac{1+x}{1-x},$

multiplying the original distance ratio of $x$ by a value that depends only on
the transformation. Because the transformation acts separately on $\sigma$\-
and $\tau$-coordinates, we obtain the same multiplicative action on distance
ratios for any other point on the complex plane with the same
$\tau$-coordinate as $x$. This multiplicative action on distance ratios
translates into an additive action on their logarithms. ∎

Figure 5: Illustration for section 4: If quadrilateral $sptq$ has high tilt,
and $r$ lies between the quadrilateral and its enclosing circle, to the left
of the red arc, then $r$ must have a higher $\tau$-coordinate value than $p$.

Another advantage of bipolar coordinates is that they provide a way of
comparing angles at the two foci, that will be convenient for relating the
tilts of different quadrilaterals to each other:

######  Observation .

Let $sptq$ be an equiangular arc-quadrilateral with interior angle $\theta$
and tilt $\varphi$. Then, in the bipolar coordinate system with foci $s$ and
$t$, the angle (difference between the $\sigma$-coordinates) of arc $tp$ in
the limit as it approaches $t$ and of arc $sq$ as it approaches $s$ is exactly
$2\theta+2\varphi-\pi$.

We can also use bipolar coordinates to show that heavily tilted quadrilaterals
lead to an increase in $\tau$-coordinate:

######  Lemma .

Let $sptq$ be an equilateral arc-quadrilateral with tilt at least $3\pi/4$,
such that the large angle between vertex $s$ and the circle $C$ containing the
quadrilateral is on the clockwise side of $s$ (the side closest to $p$). Let
$r$ be a point in the bigon between arc $tq$ and $C$, such that circular arc
$srt$ makes an angle of at most $\pi/2$ with circular arc $spt$. Then, in the
bipolar coordinate system for foci $s$ and $t$, $\tau_{r}>\tau_{p}$.

###### Proof.

Because of the Möbius invariance of coordinate differences in the bipolar
coordinate system, we can without loss of generality perform a Möbius
transformation so that $s$, $p$, and $t$ are the bottom, left, and topmost
points of $C$, as shown in Figure 5. After this transformation, points above
the horizontal line through $p$ will have higher $\tau$-coordinate than $o$,
and points below the horizontal line through $o$ will have lower
$\tau$-coordinate.

As the figure shows, an arc with tilt exactly $3\pi/4$ through $p$ and $s$
passes through the center of circle $C$, causing the region in which $r$ may
lie to be bounded by a vertical line segment (red) from the circle’s center to
$t$. All points within this region have higher $\tau$-coordinate than $p$. For
tilt values greater than $3\pi/4$, the arc from $p$ to $s$ with that tilt
extends even farther beyond the center of $C$, so (although arc $tq$ may also
extend farther to the left) the region in which $r$ may lie remains bounded
within the upper left quarter of $C$, within which all $\tau$-coordinates are
greater than that of $p$. ∎

##  5 Nonplanarity

We are now ready for our main theorems.

######  Theorem 1.

For $k>8$, the bipartite graph $B(k)$ does not have a planar Lombardi drawing.

###### Proof.

Let $s$ and $t$ be the two yellow vertices of the graph $B(k)$. We will
consider a bipolar coordinate system with foci $s$ and $t$. Note that graph
$B(k)$ contains $k$ quadrilateral faces $sp_{i}tq_{i}$, where $p_{i}$ and
$q_{i}$ are blue vertices. Because all four vertices of these quadrilaterals
have degree $2k$, these quadrilaterals must be drawn (if a planar Lombardi
drawing is to exist) as equiangular arc-quadrilaterals with interior angle
$\pi/k$.

Figure 6: Two arc-quadrilaterals with sharp angles between them at their
shared vertices reach into each other’s pockets to touch their circumscribing
circles. The two smaller pockets on the outer arcs of the circles have
significantly different $\tau$-coordinates from each other in bipolar
coordinates with the shared vertices as foci.

Consider the two consecutive quadrilaterals $sp_{i}tq_{i}$ and
$sp_{i+1}tq_{i+1}$ whose enclosing circles $C_{i}$ and $C_{i+1}$ meet each
other at the sharpest angle of any two consecutive enclosing circles. The sum
of the angles between the $k$ consecutive circles is $2\pi$ so this minimum
angle is at most $2\pi/k$. In order for point $q_{i}$ to lie on circle
$C_{i}$, some arc of circle $C_{i}$ must lie on the same side as $q_{i}$ of
quadrilateral $sp_{i+1}tq_{i+1}$. This arc must stay outside of quadrilateral
$sp_{i+1}tq_{i+1}$ from its crossing point with the quadrilateral until
terminating at either $s$ or $t$; by symmetry, we can assume without loss of
generality that it terminates at the lower vertex $s$, as shown in Figure 6.
Then, near $s$, quadrilateral $sp_{i+1}tq_{i+1}$ lies between circles $C_{i}$
and $C_{i+1}$, so it must have tilt at least $\pi(1-\tfrac{2}{k})$. By section
4 and the equal spacing of angles around $s$ and $t$, all quadrilaterals
$sp_{i}tq_{i}$ must have the same tilt.

Because $k>8$, this tilt is $\geq 3\pi/4$, so each quadrilateral
$sp_{i}tp_{i}$ meets the precondition of having high tilt of section 4. For
any quadrilateral $sp_{i}tq_{i}$, the point $p_{i+1}$ of the next
quadrilateral is connected to $s$ by an arc of quadrilateral
$sp_{i+1}tq_{i+1}$ that lies entirely within $C_{i+1}$ and makes an angle of
$\pi/k$ to arc $sq_{i}$, so the arc of $C_{i+1}$ containing $p_{i+1}$ makes an
angle of at most $3\pi/k$ to the arc of $C_{i}$ containing $p_{i}$. Thus,
point $p_{i+1}$ meets the other precondition of section 4 for the position of
the point $r$ with respect to the quadrilateral. By this lemma, each point
$p_{i+1}$ has a greater $\tau$-coordinate than $p_{i}$. But it is impossible
for this monotonic increase in $\tau$-coordinates to continue all the way
around the circle of quadrilaterals surrounding the two foci and back to the
starting point. This impossibility shows that the drawing cannot exist. ∎

######  Theorem 2.

For $k>8$ the series-parallel graph $S(k)$, embedded as shown in Figure 2,
does not have a planar Lombardi drawing.

###### Proof.

As with $B(k)$, this graph contains $k$ quadrilateral faces, sharing the same
two opposite yellow vertices, in which all vertices have equal degree ($4k$ in
$S(k)$ instead of $2k$ in $B(k)$). The proof of Theorem 1 used only this
property of $B(k)$, and not the precise value of the interior angle of these
quadrilaterals, so it applies equally well to $S(k)$. ∎

We remark that the construction of $S(k)$ can be adjusted in several different
ways to obtain more constrained families of embedded series-parallel and
related graphs that, again, have no planar Lombardi drawing:

  * •

If we add an edge between the two yellow vertices, and adjust the lengths of
the red chains to keep the yellow and blue degrees equal, we obtain a family
of embedded maximal series-parallel graphs (that is, embedded 2-trees) with no
planar Lombardi drawing.

  * •

If we subdivide the yellow–red and red–red edges of $S(k)$, we obtain a family
of embedded bipartite series-parallel graphs with no planar Lombardi drawing.

  * •

If we replace the red chains of $S(k)$ by an appropriate number of degree-one
red vertices, connected to the blue vertices, we obtain a family of embedded
apex-trees (graphs formed by adding a single vertex to a tree) with no planar
Lombardi drawing. The apex vertex (the vertex whose removal produces a tree)
can be chosen to be either of the two yellow vertices.

We omit the details.

##  6 Conclusions

We have shown that bipartite planar graphs, and series-parallel graphs with a
fixed planar embedding, do not always have planar Lombardi drawings, even
though their low degeneracy implies that they always have (nonplanar) Lombardi
drawings. In the question of which important subfamilies of planar graphs have
planar Lombardi drawings, several important cases remain unsolved. These
include the outerplanar graphs, both with and without assuming an outerplanar
embedding, the cactus graphs, and the series-parallel graphs without a fixed
choice of embedding. We leave these as open for future research.

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#  Geometry of submanifolds of all classes of third-order ODEs as a Riemannian
manifold

Z. Bakhshandeh-Chamazkoti†111Corresponding Author, A. Behzadi§, R.
Bakhshandeh-Chamazkoti‡,
and M. Rafie-Rad§§
†,§,§§Department of Mathematics, Faculty of Mathematical Sciences, University
of Mazandaran, Babolsar, Iran.
‡Department of Mathematics, Faculty of Basic Sciences, Babol Noshirvani
University of Technology, Babol, Iran.
Emails<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract

In this paper, we prove that any surface corresponding to linear second-order
ODEs as a submanifold is minimal in all classes of third-order ODEs
$y^{\prime\prime\prime}=f(x,y,p,q)$ as a Riemannian manifold where
$y^{\prime}=p$ and $y^{\prime\prime}=q$, if and only if $q_{yy}=0$. Moreover,
we will see the linear second-order ODE with general form
$y^{\prime\prime}=\pm y+\beta(x)$ is the only case that is defined a minimal
surface and is also totally geodesic.

Keywords: Levi-Civita connection, minimal surface, moving frame, Riemannian
manifold, Riemann curvature tensor, totally geodesic.

##  1 Introduction

Riemannian geometry is characterized, and research is oriented towards and
shaped by concepts for examples geodesics connections, curvature [12].
Originally, the geometry of submanifolds was only a part of Riemannian
geometry but today it is one of several independent aspects of multi-
dimensional generalizations of the classical theory of surfaces. While
Riemannian geometry is the development of Gauss’ idea on intrinsic geometry,
the geometry of submanifolds starts from the idea of the extrinsic geometry of
a surface. This theory is devoted to the study of the position and properties
of a submanifold in ambient space, both in their local and global aspects [2,
3, 19].

On the other hand, the general equivalence problem is about studding when two
geometrical objects are mapped on each other by a certain class of
diffeomorphisms. $\rm\acute{E}$lie Cartan developed the general equivalence
problem and provided a systematic procedure for determining the necessary and
sufficient conditions. There are many papers as applications of Cartan method
for third-order ODEs [11, 13, 18] and fourth-order differential operators [4].
The theory of moving frames is most closely associated with the name of
Cartan, that is a powerful and algorithmic tool for studying the geometric
properties of submanifolds and their invariants under the action of a
transformation group [16]. Now if we apply the moving frame method for
Riemannian manifold and geometry of submanifolds, we can obtain very
interesting results. Recently, a lot of research has been carried out about
the minimal surfaces in a three-dimensional Riemannian manifold [9, 10, 17].

The graph of the function $u(x,y)$ is minimal if and only if $u$ satisfies:

$\displaystyle(1+u_{x}^{2})\;u_{yy}-2u_{x}u_{y}u_{xy}+(1+u_{y}^{2})\;u_{xx}=0.$
(1)

The PDE (1) which is non-linear but depends linearly on the second derivatives
is called quasi-linear. We will not discuss the general theory of existence of
solutions to this equation, but rather describe a couple of special solutions.
Of course, a function $u(x,y)$ with vanishing second derivatives is solution
of Equation (1). Although these are not very exciting solutions, the resulting
surfaces $\cal S$ being planes, they are important because of the Calabi
theorem [3]. According to this theorem, if $u$ be a solution of Equation (1),
on $U={\mathbb{R}}^{2}$ such that $\|\nabla u\|_{0}<1$, (hence $\cal S$ is
called spacelike) then $u$ is an affine function that means $u$ is a function
with vanishing second derivatives. There is an analogous theorem (which
actually predates Calabi’s) in the case of the Riemannian metric
$\langle.,.\rangle_{0}=dx_{1}^{2}+dx_{2}^{2}+dx_{3}^{2}$ and is known as the
Bernstein theorem.

It is a classical result that any simply connected minimal surface in
Euclidean space ${\mathbb{R}}^{3}$ admits a one-parameter family of minimal
isometric deformations, called the _associate family_. Conversely, two minimal
isometric immersions of the same Riemannian surface into ${\mathbb{R}}^{3}$
are associate. These are easy consequences of the Gauss and Codazzi equations
in ${\mathbb{R}}^{3}$. More generally, analogous results hold for constant
mean curvature (CMC) surfaces in $3$-dimensional space forms.

Benoit Daniel in [9] investigated extensions of these results and related
questions for minimal surfaces in the product manifolds
${\mathbb{S}}^{2}\times{\mathbb{R}}$ and ${\mathbb{H}}^{2}\times{\mathbb{R}}$,
where ${\mathbb{S}}^{2}$ is the $2$-sphere of curvature $1$ and
${\mathbb{H}}^{2}$ is the hyperbolic plane of curvature $-1$.

The systematic study of minimal surfaces in
${\mathbb{S}}^{2}\times{\mathbb{R}}$ and ${\mathbb{H}}^{2}\times{\mathbb{R}}$
was initiated by H. Rosenberg and W. Meeks [14, 17] and has been very active
since then. The existence of an associate family for simply connected minimal
surfaces in ${\mathbb{S}}^{2}\times{\mathbb{R}}$ and
${\mathbb{H}}^{2}\times{\mathbb{R}}$ was proved in [8].

It has been shown in Bayrakdar et al. [6], the Gaussian curvature of a surface
corresponding to a first-order ODE is given by certain Burgers’ equations and
it is possible to obtain two-dimensional spaces of constant curvature from
some integrable PDEs.

The main idea for writing present paper is raised from the papers [5], written
by T. Bayrakdar and A. A. Ergin. They proved that a surface corresponding to a
first-order ODE is minimal in three-dimensional Riemannian manifold which is
determined by the first prolongation of $y^{\prime}=p(x,y)$, if and only if
$p_{yy}=0$. Hence any linear first-order ODE characterizes a minimal surface
which is not necessarily totally geodesic. Z. O. Bayrakdar and T. Bayrakdar
used the same idea and got good results [7].

In present paper, we prove that any surface corresponding to linear second-
order ODEs $y^{\prime\prime}=\alpha(x)y+\beta(x)$ where $\alpha$ and $\beta$
are two smooth functions in term of $x$, as a submanifold is minimal in the
class of third-order ODEs corresponding to the third-order equation
$y^{\prime\prime\prime}=\alpha(x)y^{\prime}+\alpha^{\prime}(x)y+\beta^{\prime}(x)$.
Furthermore, we will show the linear second-order ODE $y^{\prime\prime}=\pm
y+\beta(x)$ is the only case that is defined a minimal surface and is also
totally geodesic.

##  2 Geometry of submanifolds via moving frame method

In this section we consider the method of moving frames to investigation of
geometry of submanifolds. For more details we refer to [19].

Assume $N$ be a $n$-dimensional Riemannian manifold equipped with metric $g$.
Let $M$ is a $m$-dimensional submanifold of $N$, locally imbedded in $N$. The
submanifold of the orthogonal frame bundle over $N$, denoted by $F(N,M)$,
which includes of adapted frames $\\{{\bf e}_{A}\\},~{}A=1,\ldots,n$ of which
$\\{{\bf e}_{i}\\},~{}i=1,\ldots,m$ are tangent to $M$ and $\\{{\bf
e}_{\alpha}\\},~{}\alpha=m+1,\ldots,n$ are normal to $M$. We have the matrix
equation $\widetilde{\bf E}={\bf E}K$ between two adapted frames
$\widetilde{\bf E}$ and ${\bf E}$ where $K$ is the matrix

$K=\left[\begin{array}[]{cccc}A&{\rm O}\\\\[5.69054pt] {\rm
O}&B\end{array}\right],$

where $A\in{\rm O}(m)$ and $B\in{\rm O}(n-m)$.

The matrix of 1-forms $(\omega_{B}^{A})$ is the Levi-Civita connection for $N$
where we have

$\displaystyle d\omega^{A}+\omega_{B}^{A}\wedge\omega^{B}$ $\displaystyle=$
$\displaystyle 0,$ (2) $\displaystyle\omega_{B}^{A}+\omega_{A}^{B}$
$\displaystyle=$ $\displaystyle 0.$ (3)

We can find a connection over $M$ by a restriction $\omega_{B}^{A}$ to
$\omega_{j}^{i}$ over $M$. Namely, when we restrict the equation

$\displaystyle d\omega^{A}=-\sum_{B}\omega_{B}^{A}\wedge\omega^{B},$ (4)

to $M$ gives

$\displaystyle
d\omega^{i}=-\sum_{B}\omega_{B}^{i}\wedge\omega^{B}=-\sum_{j}\omega_{j}^{i}\wedge\omega^{j},$
(5)

because $\omega^{\alpha}=0$ for tangent vectors on $M$. Furthermore
$\omega_{j}^{i}=-\omega_{i}^{j}$, so the matrix $(\omega_{j}^{i})$ is the
uniquely defined Levi-Civita connection corresponding to the metric on $M$
induced from that on $N$. Takeing the exterior derivative of
$\omega^{\alpha}=0$, leads to

$\displaystyle
0=d\omega^{\alpha}=-\sum_{A}\omega_{A}^{\alpha}\wedge\omega^{A}=-\sum_{i}\omega_{i}^{\alpha}\wedge\omega^{i}.$
(6)

From Cartan’s lemma it follows that

$\displaystyle\omega^{\alpha}_{i}=\sum_{j}h_{ij}^{\alpha}\;\omega^{i},$ (7)

for smooth functions $h_{ij}^{\alpha}=h_{ji}^{\alpha}$.

##  3 Submanifolds in Riemannian manifold

Let $(S,g)$ be a Riemannian manifold and $U\subset S$ be an open subset. For
each arbitrary $p\in U$, we can define an orthonormal frame ${\bf E}=({\bf
e}_{1},{\bf e}_{2},{\bf e}_{3},{\bf e}_{4})$ where ${\bf e}_{i}\in T_{p}U$ and
its corresponding dual, the coframe
$\omega=\\{\omega^{1},\omega^{2},\omega^{3},\omega^{4}\\}$ where
$\omega^{i}\in T_{p}^{*}U$ and we have $\omega^{i}({\bf e}_{j})=\delta_{ij}$.
Therefore the metric tensor is defined by

$\displaystyle g=\sum_{i,j=1}^{4}\delta_{ij}~{}\omega^{i}\otimes\omega^{j}.$
(8)

By applying the exterior derivatives on these 1-forms we obtain the first
structural equations

$\displaystyle
d\omega^{i}=-\sum_{j=1}^{4}~{}\theta_{j}^{i}\wedge\omega^{j},\;\;\;\;\;\;\;1\leq
i\leq 4,$ (9)

where the skew-symmetric matrix of 1-forms $\theta=(\theta_{j}^{i})$ is called
the $\mathfrak{o}(3,{\mathbb{R}})$-valued torsion-free connection. Now
computing the exterior derivative of $\theta_{j}^{i}$ gets the second
structural equations as following

$\displaystyle\Omega^{i}_{j}=d\theta_{j}^{i}+\sum_{k=1}^{4}~{}\theta_{k}^{i}\wedge\theta_{j}^{k},$
(10)

where the skew-symmetric matrix $\Omega=(\Omega_{j}^{i})$ is called the
Riemannian curvature tensor. We can rewrite the $\Omega^{i}_{j}$ with respect
to coframe $\omega$ as follows

$\displaystyle\Omega^{i}_{j}=\sum_{k<l}R_{jkl}^{i}~{}\omega^{k}\wedge\omega^{l}.$
(11)

Let $(\widetilde{S},\widetilde{g})$ be an isometrically immersed submanifold
of the surface of $(S,g)$ by inclusion map
$\sigma:\widetilde{S}\longrightarrow S$ satisfies $\sigma^{*}g=\widetilde{g}$
and $\widetilde{\omega}^{4}=0$. Suppose that $\widetilde{\bf
E}=(\widetilde{\bf e}_{1},\widetilde{\bf e}_{2},\widetilde{\bf e}_{2},{\bf
n})$ is an adapted frame on $S$ with corresponding coframe
$\widetilde{\omega}=(\widetilde{\omega}^{1},\widetilde{\omega}^{2},\widetilde{\omega}^{3},\widetilde{\omega}^{4})$,
such that $(\widetilde{\bf e}_{1},\widetilde{\bf e}_{2},\widetilde{\bf
e}_{3})$ is tangent to $\widetilde{S}$ and ${\bf n}$ is normal to
$\widetilde{S}$ and ${\bf E}$ and $\widetilde{\bf E}$ have the same
orientation. As a result, the surface $\widetilde{S}$ is equipped with metric
tensor

$\displaystyle\widetilde{g}=\sum_{i=1}^{4}\widetilde{\omega}^{i}\otimes\widetilde{\omega}^{i}.$
(12)

The orthogonal frames $\widetilde{\bf E}$ and ${\bf E}$ are related by the
equation $\widetilde{\bf E}={\bf E}A$, where $A\in SO(4,{\mathbb{R}})$, thus
we can write the associated connection 1-forms as follows

$\displaystyle\widetilde{\theta}=A^{-1}dA+A^{-1}\theta A.$ (13)

Using (5), we can deduce the first structure equations for coframe
$\widetilde{\omega}$ by following formula

$\displaystyle
d\widetilde{\omega}^{i}=-\sum_{j=1}^{4}~{}\widetilde{\theta}_{j}^{i}\wedge\widetilde{\omega}^{j}.\;\;\;\;\;\;\;1\leq
i\leq 4$ (14)

Take a look at the formula (6), we derive the second structure equations

$\displaystyle d\widetilde{\omega}^{i}$ $\displaystyle=$
$\displaystyle-\sum_{j}~{}\widetilde{\theta}_{j}^{i}\wedge\widetilde{\omega}^{j},\;\;\;\;\;\;\;i=1,2,3$
(15) $\displaystyle 0$ $\displaystyle=$
$\displaystyle\widetilde{\theta}_{1}^{4}\wedge\widetilde{\omega}^{1}+\widetilde{\theta}_{2}^{4}\wedge\widetilde{\omega}^{2}+\widetilde{\theta}_{3}^{4}\wedge\widetilde{\omega}^{3},$
(16)

on the $\widetilde{S}$. It is necessary to mention the (16) concludes by
$\widetilde{\omega}^{4}=0$. An immediate consequences of the equation (16) are

$\displaystyle\begin{split}\widetilde{\theta}_{1}^{4}&=h_{11}\widetilde{\omega}^{1}+h_{12}\widetilde{\omega}^{2}+h_{13}\widetilde{\omega}^{3},\\\
\widetilde{\theta}_{2}^{4}&=h_{21}\widetilde{\omega}^{1}+h_{22}\widetilde{\omega}^{2}+h_{23}\widetilde{\omega}^{3},\\\
\widetilde{\theta}_{3}^{4}&=h_{31}\widetilde{\omega}^{1}+h_{32}\widetilde{\omega}^{2}+h_{33}\widetilde{\omega}^{3},\end{split}$
(17)

where $h_{ij}=h_{ji}$. These functions are the coefficients of the second
fundamental form

$\displaystyle{\rm
II}=\widetilde{\theta}_{1}^{4}\otimes\widetilde{\omega}^{1}+\widetilde{\theta}_{2}^{4}\otimes\widetilde{\omega}^{2}+\widetilde{\theta}_{3}^{4}\otimes\widetilde{\omega}^{3},$
(18)

Now one can evaluate the $\widetilde{\Omega}^{i}_{j}$ by following formula

$\displaystyle\widetilde{\Omega}^{i}_{j}=\sum_{k<l}\widetilde{R}_{jkl}^{i}~{}\widetilde{\omega}^{k}\wedge\widetilde{\omega}^{l},$
(19)

for metric $\widetilde{g}$ on $\widetilde{S}$. According to Gauss formula we
have

$\displaystyle R_{ijrs}-\widetilde{R}_{ijrs}=h_{is}h_{jr}-h_{ir}h_{js}.$ (20)

Also in an orthonormal frame we have

$\displaystyle\sum_{k}\delta_{ik}R_{jrs}^{k}=R_{ijrs}.$ (21)

We shall define the Weingarten operator ${\cal A}:TM\longrightarrow TM$, for
each $X,Y\in T_{p}M$, by

$\displaystyle\langle{\cal A}X,Y\rangle=\langle{\rm II}(X,Y),{\bf
n}\rangle_{N},$ (22)

componentwise this means that

$\displaystyle{\cal
A}=\sum_{i,j}h_{ij}\widetilde{\omega}^{i}\otimes\widetilde{\bf e}_{j}.$ (23)

In fact, often we shall not distinguish between ${\cal A}$ and the second
fundamental tensor in the direction of ${\bf n}$, that is, the map
$\langle{\rm II}(,),~{}{\bf n}\rangle_{N}:TM\times
TM\longrightarrow{\mathbb{R}}$. The $k$-th mean curvatures of the hypersurface
in the direction of ${\bf n}$ are given by

$\displaystyle H_{k}=\binom{m}{k}^{-1}S_{k},$ (24)

where $S_{0}=1$ and, for $1\leq k\leq m$, $S_{k}$ is the $k$-th elementary
symmetric function of the eigenvalues of ${\cal A}=(h_{ij})$. In particular
$H_{1}=H$ is the mean curvature that

$\displaystyle H=\frac{1}{3}{\rm tr}~{}{\cal A},$ (25)

and $H_{m}$ is the Gauss-Kronecker curvature that equals to

$\displaystyle H_{m}={\rm det}~{}{\cal A},$ (26)

and $H_{2}$ is strictly related to the scalar curvature of $M$, [1].

We remind the reader that the surface $\widetilde{S}$ is said to be totally
geodesic if the second fundamental form identically vanishes on
$\widetilde{S}$ and is said to be minimal if $H=0$.

##  4 The class of third-order ODEs as a Riemannian geometry

Geometrically, one can consider the third-order ODE with the following form

$\displaystyle\dfrac{d^{3}y}{dx^{3}}=f(x,y,y^{\prime},y^{\prime\prime}),$ (27)

as a submanifold ${\cal S}$ in the third-order jet bundle ${\rm J}^{3}$, which
has local coordinates

$\Upsilon=\\{(x,y,p,q,r)\in{\rm
J}^{3}:p=y^{\prime},q=y^{\prime\prime},r=y^{\prime\prime\prime}\\},$

and it is denoted by the zero set of the function $F(x,y,p,q,r)=r-f(x,y,p,q)$,
that means ${\cal S}=F^{-1}(0)$ is presented on ${\rm J}^{2}$ as the graph of
the function $r=f(x,y,p,q)$.

From the geometric theory of such equations, it follows that there exists a
collection of independent $1$-forms as a coframe

$\displaystyle\omega^{1}$ $\displaystyle=$ $\displaystyle dx,$
$\displaystyle\omega^{2}$ $\displaystyle=$ $\displaystyle dy-p\;dx,$
$\displaystyle\omega^{3}$ $\displaystyle=$ $\displaystyle dp-q\;dx,$ (28)
$\displaystyle\omega^{4}$ $\displaystyle=$ $\displaystyle dq-f\;dx,$

on ${\mathbb{R}}^{5}$ with coordinates $(x,y,p,q,r)$. The third prolongation
of a solution curve of differential equation (27) is a curve on ${\cal S}$
represented by 3-jet of a smooth section $\sigma(x,y,p)=(x,y,p,q,r)$ of the
trivial bundle $\pi:{\mathbb{R}}\times{\mathbb{R}}\longrightarrow{\mathbb{R}}$
on which the contact forms $\omega^{2}$, $\omega^{3}$ and $\omega^{4}$ vanish.
Since the local coframe
$\omega=\\{\omega^{1},\omega^{2},\omega^{3},\omega^{4}\\}$ is dual to the
frame of the vector fields $\Omega=\\{{\bf e}_{1},{\bf e}_{2},{\bf e}_{3},{\bf
e}_{4}\\}$ that means $\omega^{i}({\bf e}_{j})=\delta_{ij}$ therefore we have

$\displaystyle{\bf
e}_{1}=\partial_{x}+p\partial_{y}+q\partial_{p}+f\partial_{q},\;\;\;\;\;\;\;\;{\bf
e}_{2}=\partial_{y},\;\;\;\;\;\;\;\;\;\;{\bf
e}_{3}=\partial_{p},\;\;\;\;\;\;\;\;{\bf e}_{4}=\partial_{q}.$ (29)

The Riemannian metric

$\displaystyle g=\sum_{i,j=1}^{4}\delta_{ij}~{}\omega^{i}\otimes\omega^{j},$
(30)

on ${\cal S}$ is given in coordinates $(x,y,p,q)$ as

$\displaystyle
ds^{2}=(1+p^{2}+q^{2}+f^{2})dx^{2}-2pdxdy-2qdxdp-2fdxdq+dy^{2}+dp^{2}+dq^{2}.$
(31)

Differentiating the coframe (28), we have

$\displaystyle\begin{split}d\omega^{1}&=0,\\\
d\omega^{2}&=\omega^{1}\wedge\omega^{3},\\\
d\omega^{3}&=\omega^{1}\wedge\omega^{4},\\\
d\omega^{4}&=f_{y}~{}\omega^{1}\wedge\omega^{2}+f_{p}~{}\omega^{1}\wedge\omega^{3}+f_{q}~{}\omega^{1}\wedge\omega^{4}.\end{split}$
(32)

Now using the (9) formula, we can write

$\left[\begin{array}[]{cccc}d\omega^{1}\\\\[5.69054pt]
d\omega^{2}\\\\[5.69054pt] d\omega^{3}\\\\[5.69054pt]
d\omega^{4}\\\\[5.69054pt]
\end{array}\right]=-\left[\begin{array}[]{cccc}0&\theta^{1}_{2}&\theta^{1}_{3}&\theta^{1}_{4}\\\\[5.69054pt]
\theta_{1}^{2}&0&\theta^{2}_{3}&\theta^{2}_{4}\\\\[5.69054pt]
\theta_{1}^{3}&\theta_{2}^{3}&0&\theta^{3}_{4}\\\\[5.69054pt]
\theta_{1}^{4}&\theta_{2}^{4}&\theta_{3}^{4}&0\\\\[5.69054pt]
\end{array}\right]~{}\wedge~{}\left[\begin{array}[]{cccc}\omega^{1}\\\\[5.69054pt]
\omega^{2}\\\\[5.69054pt] \omega^{3}\\\\[5.69054pt] \omega^{4}\\\\[5.69054pt]
\end{array}\right],$

where $\Theta=(\theta_{j}^{i})$ is a ${\mathfrak{o}}(4,{\mathbb{R}})$-valued
torsion-free connection is a antisymmetric matrix, where

$\displaystyle\begin{split}\theta_{2}^{1}&=-\theta_{1}^{2}=-\frac{1}{2}\
\omega^{3}-\frac{1}{2}f_{y}\ \omega^{4},\\\
\theta_{3}^{1}&=-\theta_{1}^{3}=-\frac{1}{2}\ \omega^{2}-\frac{1}{2}(1+f_{p})\
\omega^{4},\\\ \theta_{4}^{1}&=-\theta_{1}^{4}=-\frac{1}{2}f_{y}\
\omega^{2}-\frac{1}{2}(1+f_{p})\ \omega^{3}-f_{q}\ \omega^{4},\\\
\theta_{3}^{2}&=-\theta_{2}^{3}=-\frac{1}{2}\ \omega^{1},\\\
\theta_{4}^{2}&=-\theta_{2}^{4}=\frac{1}{2}f_{y}\ \omega^{1},\\\
\theta_{4}^{3}&=-\theta_{3}^{4}=-\frac{1}{2}(1-f_{p})\ \omega^{1}.\end{split}$
(33)

Now according to (11), the components of the curvature $2$-form are

$\displaystyle\begin{split}\Omega_{2}^{1}&=-\frac{1}{2}\Big{[}\frac{1}{2}(-1+3f^{2}_{y})\
\omega^{1}\wedge\omega^{2}+\frac{1}{2}f_{y}(1+3f_{p})\
\omega^{1}\wedge\omega^{3}+f_{yy}\ \omega^{2}\wedge\omega^{4}\\\
&~{}~{}~{}~{}~{}~{}+f_{yp}\
\omega^{3}\wedge\omega^{4}+\Big{(}\frac{1}{2}+2f_{y}f_{q}-\frac{f_{p}}{2}+pf_{yy}+f_{xy}+ff_{yq}+qf_{yp}\Big{)}\
\omega^{1}\wedge\omega^{4}\Big{]},\\\
\Omega_{3}^{1}&=\frac{1}{2}\Big{[}-\frac{1}{2}f_{y}(1+3f_{p})\
\omega^{1}\wedge\omega^{2}-(1+f_{p}+\frac{3}{2}f_{p}^{2})\
\omega^{1}\wedge\omega^{3}\\\
&~{}~{}~{}~{}~{}~{}-(2f_{p}f_{q}+pf_{yp}+ff_{pq}+\frac{1}{2}f_{y}+f_{xp})\
\omega^{1}\wedge\omega^{4}-f_{yp}\ \omega^{2}\wedge\omega^{4}-f_{pp}\
\omega^{3}\wedge\omega^{4}\Big{]},\\\
\Omega_{4}^{1}&=-\frac{1}{2}\Big{(}2f_{y}f_{q}+\frac{1}{2}(1-f_{p})+pf_{yy}+f_{xy}+ff_{yq}+qf_{yp}\Big{)}\
\omega^{1}\wedge\omega^{2}-\frac{1}{2}\Big{(}2f_{p}f_{q}+pf_{yp}+ff_{pq}\\\
&~{}~{}~{}~{}~{}~{}+qf_{pp}+\frac{1}{2}f_{y}+f_{xp}\Big{)}\
\omega^{1}\wedge\omega^{3}-\frac{1}{2}\Big{(}\frac{3}{2}-\frac{1}{2}f_{p}^{2}-\frac{1}{2}f_{y}^{2}+2f_{q}^{2}+f_{p}+2qf_{pq}+2ff_{qq}+2pf_{yq}\\\
&~{}~{}~{}~{}~{}~{}+2f_{xq}\Big{)}\
\omega^{1}\wedge\omega^{4}-\frac{1}{2}f_{yq}\
\omega^{2}\wedge\omega^{4}-\frac{1}{2}f_{pq}\ \omega^{3}\wedge\omega^{4},\\\
\Omega_{3}^{2}&=\frac{1}{2}\Big{[}\frac{1}{2}\
\omega^{2}\wedge\omega^{3}+\frac{1}{2}f_{y}\
\omega^{2}\wedge\omega^{4}-\frac{1}{2}(1+f_{p})\
\omega^{3}\wedge\omega^{4}\Big{]},\\\\[5.69054pt]
\Omega_{4}^{2}&=\frac{1}{2}\Big{[}-f_{yy}\;\omega^{1}\wedge\omega^{2}-f_{yp}\
\omega^{1}\wedge\omega^{3}-f_{yq}\
\omega^{1}\wedge\omega^{4}+\frac{1}{2}f_{y}\;\omega^{2}\wedge\omega^{3}+\frac{1}{2}f_{y}^{2}\;\omega^{2}\wedge\omega^{4}\\\
&~{}~{}~{}~{}~{}~{}-\Big{(}f_{q}-\frac{1}{2}f_{y}-\frac{1}{2}f_{y}f_{q}\Big{)}\;\omega^{3}\wedge\omega^{4}\Big{]},\\\
\Omega_{4}^{3}&=\frac{1}{2}\Big{[}-f_{yp}\ \omega^{1}\wedge\omega^{2}-f_{pp}\
\omega^{1}\wedge\omega^{3}-f_{pq}\
\omega^{1}\wedge\omega^{4}-\frac{1}{2}(1+f_{p})\
\omega^{2}\wedge\omega^{3}+\frac{1}{2}\Big{(}f_{y}(1+f_{p})\\\
&~{}~{}~{}~{}~{}~{}-2f_{q}\Big{)}\
\omega^{2}\wedge\omega^{4}+\frac{1}{2}(1+f_{p})^{2}\
\omega^{3}\wedge\omega^{4}\Big{]}.\end{split}$

Thus independent components of the Riemann curvature tensor are given below

$\displaystyle R^{1}_{212}$ $\displaystyle=$
$\displaystyle\frac{1}{4}\Big{(}1-3f^{2}_{y}\Big{)},\;\;\;\;R^{1}_{213}=-\frac{1}{4}f_{y}(1+3f_{p}),$
$\displaystyle R^{1}_{214}$ $\displaystyle=$
$\displaystyle-\frac{1}{2}\Big{(}\frac{1}{2}+2f_{y}f_{q}-\frac{f_{p}}{2}+pf_{yy}+f_{xy}+ff_{yq}+qf_{yp}\Big{)},$
$\displaystyle R^{1}_{224}$ $\displaystyle=$
$\displaystyle-\frac{1}{2}f_{yy},\;\;\;\;\;\;\;\;R^{1}_{234}=-\frac{1}{2}f_{yp},\;\;\;$
$\displaystyle R^{1}_{312}$ $\displaystyle=$
$\displaystyle-\frac{1}{4}f_{y}(1+3f_{p}),\;\;\;\;R^{1}_{313}=-\frac{1}{2}(1+f_{p}+\frac{3}{2}f_{p}^{2}),$
$\displaystyle R^{1}_{314}$ $\displaystyle=$
$\displaystyle-\frac{1}{2}(2f_{p}f_{q}+pf_{yp}+ff_{pq}+\frac{1}{2}f_{y}+f_{xp}),\;\;\;R^{1}_{324}=-\frac{1}{2}f_{yp},\;\;\;\;R^{1}_{334}=-\frac{1}{2}f_{pp},$
$\displaystyle R^{1}_{412}$ $\displaystyle=$
$\displaystyle-\frac{1}{2}\Big{(}2f_{y}f_{q}+\frac{1}{2}(1-f_{p})+pf_{yy}+f_{xy}+ff_{yq}+qf_{yp}\Big{)},$
$\displaystyle R^{1}_{413}$ $\displaystyle=$
$\displaystyle\frac{1}{2}\Big{(}2f_{p}f_{q}+pf_{yp}+ff_{pq}+qf_{pp}+\frac{1}{2}f_{y}+f_{xp}\Big{)},$
$\displaystyle R^{1}_{414}$ $\displaystyle=$
$\displaystyle-\frac{1}{2}\Big{(}\frac{3}{2}-\frac{1}{2}f_{p}^{2}-\frac{1}{2}f_{y}^{2}+2f_{q}^{2}+f_{p}+2qf_{pq}+2ff_{qq}+2pf_{yq}+2f_{xq}\Big{)},$
$\displaystyle R^{1}_{424}$ $\displaystyle=$
$\displaystyle-\frac{1}{2}f_{yq},\;\;\;\;\;\;\;R^{1}_{434}=-\frac{1}{2}f_{pq},$
$\displaystyle R^{2}_{323}$ $\displaystyle=$
$\displaystyle\frac{1}{4},\;\;\;\;\;\;\;R^{2}_{324}=\frac{1}{4}f_{y},\;\;\;\;\;\;\;R^{2}_{334}=-\frac{1}{4}(1+f_{p}),$
$\displaystyle R^{2}_{412}$ $\displaystyle=$
$\displaystyle-\frac{1}{2}f_{yy},\;\;\;\;\;R^{2}_{413}=-\frac{1}{2}f_{yp},\;\;\;\;\;R^{2}_{414}=-\frac{1}{2}f_{yq},\;\;\;\;R^{2}_{423}=\frac{1}{4}f_{y},\;\;\;\;\;R^{2}_{424}=\frac{1}{4}f_{y}^{2},$
$\displaystyle R^{2}_{434}$ $\displaystyle=$
$\displaystyle-\frac{1}{4}\Big{(}f_{q}-\frac{1}{2}f_{y}-\frac{1}{2}f_{y}f_{q}\Big{)},\;\;\;R^{3}_{412}=-\frac{1}{2}f_{yp},\;\;\;R^{3}_{413}=-\frac{1}{2}f_{pp},\;\;\;R^{3}_{414}=-\frac{1}{2}f_{pq},$
$\displaystyle\;\;\;\;\;R^{3}_{423}$ $\displaystyle=$
$\displaystyle-\frac{1}{4}(1+f_{p}),\;\;\;\;R^{3}_{424}=\frac{1}{4}\Big{(}f_{y}(1+f_{p})-2f_{q}\Big{)},\;\;\;\;\;R^{3}_{434}=\frac{1}{4}(1+f_{p})^{2}.$

##  5 Geometry of submanifolds

Let $\widetilde{\cal S}$ is a submanifold in $\cal S$ determined by the smooth
section

$\displaystyle\sigma:(x,y,p)\longrightarrow(x,y,p,q(x,y,p)),$ (34)

corresponding to the equation $\displaystyle{y^{\prime\prime}=q(x,y,p)}$ such
that $dp-qdx$ vanishes on the first prolongation of an integral curve of (27).
In fact, 3-graph of an integral curve of (27) is specified by

$y^{\prime\prime\prime}=q_{x}+q_{y}y^{\prime}+q_{p}p^{\prime}=q_{x}+pq_{y}+qq_{p},$

lies on $\widetilde{\cal S}$. We can compute the pullbacks of $\omega^{i}$ by
$\sigma$ as follow

$\displaystyle\sigma^{*}\omega^{1}$ $\displaystyle=$
$\displaystyle\omega^{1},$ $\displaystyle\sigma^{*}\omega^{2}$
$\displaystyle=$ $\displaystyle\omega^{2},$
$\displaystyle\sigma^{*}\omega^{3}$ $\displaystyle=$
$\displaystyle\omega^{3},$ $\displaystyle\sigma^{*}\omega^{4}$
$\displaystyle=$ $\displaystyle q_{y}\;\omega^{2}+q_{p}\;\omega^{3}.$

With a simple calculation we find

$\displaystyle\sigma^{*}(d\omega^{4})$ $\displaystyle=$
$\displaystyle(q_{xy}+pq_{yy}+qq_{yp}+q_{y}q_{p}+q_{y}\sigma^{*}(f_{q}))\;\omega^{1}\wedge\omega^{2}+(q_{xp}+q_{y}+pq_{yp}+qq_{pp}$
$\displaystyle~{}~{}~{}+q_{p}^{2}+q_{p}\sigma^{*}(f_{q}))\;\omega^{1}\wedge\omega^{3},$
$\displaystyle d(\sigma^{*}\omega^{4})$ $\displaystyle=$
$\displaystyle(q_{xy}+pq_{yy}+qq_{yp})\;\omega^{1}\wedge\omega^{2}+(q_{y}+q_{xp}+pq_{yp}+qq_{pp})\;\omega^{1}\wedge\omega^{3}+q_{p}\;\omega^{1}\wedge\omega^{4}.$

The equality $\sigma^{*}(d\omega^{4})=d(\sigma^{*}\omega^{4})$, leads to
$q_{p}=0$ and $q_{y}\sigma^{*}(f_{q})=0$. Now we can consider two different
cases:

###  5.1 Case of $q_{y}\neq 0$

Since $\sigma^{*}(f_{q})=0$, that is, $f_{q}=0$ on the submanifold
$\widetilde{\cal S}$, the induced metric on $\widetilde{\cal S}$ is readily
found as

$\displaystyle\sigma^{*}ds^{2}=\omega^{1}\otimes\omega^{1}+(1+q_{y}^{2})\;\omega^{2}\otimes\omega^{2}+\;\omega^{3}\otimes\omega^{3}.$
(35)

sur In local coordinates $(x,y,p)$, we can rewrite the metric tensor by
$\widetilde{g}$ with following form

$\displaystyle\widetilde{g}=[1+p^{2}(1+q_{y})^{2}+q^{2}]\;dx^{2}+(1+q_{y}^{2})\;dy^{2}+\;dp^{2}-2p(1+q_{y}^{2})\;dxdy-2q\;dxdp.$

For each point in a coordinate neighborhood $(\widetilde{U};x,y,p)$ of
$T\widetilde{\cal S}$, the tangent space $T\widetilde{\cal S}$ is spanned by
the vector fields $\widehat{{\bf
e}}_{1}=\sigma_{*}\partial_{x},\;\widehat{{\bf e}}_{2}=\sigma_{*}\partial_{y}$
and $\widehat{{\bf e}}_{3}=\sigma_{*}\partial_{p}$,

$\displaystyle\widehat{{\bf
e}}_{1}=\partial_{x}+q_{x}\partial_{q},\;\;\;\;\;\;\;\;\widehat{{\bf
e}}_{2}=\partial_{y}+q_{y}\partial_{q},\;\;\;\;\;\;\;\;\widehat{{\bf
e}}_{3}=\partial_{p}.$ (36)

Now we can rewrite the vector fields $\widehat{{\bf e}}_{1},\widehat{{\bf
e}}_{2}$ and $\widehat{{\bf e}}_{3}$ in terms of ${\bf e}_{1},{\bf e}_{2},{\bf
e}_{3},{\bf e}_{4}$ with following form

$\displaystyle\widehat{{\bf e}}_{1}={\bf e}_{1}-p{\bf e}_{2}-q{\bf
e}_{3}+(q_{x}-f){\bf e}_{4},\;\;\;\widehat{{\bf e}}_{2}={\bf e}_{2}+q_{y}{\bf
e}_{4},\;\;\;\widehat{{\bf e}}_{3}={\bf e}_{3}.$ (37)

To find unit normal vector field to $\widetilde{\cal S}$, we can consider the
cross product of $X,Y,Z\in T{\cal S}$ at a given point of a four-dimensional
Riemannian manifold. The cross product on ${\cal S}$ is defined in terms of
the Riemannian metric (31) and the volume form by

$\displaystyle g(X\times Y\times Z,W):={\rm
vol}_{g}(X,Y,Z,W),\;\;\;\forall~{}W\in T{\cal S},$ (38)

where the positive definite matrix $(g_{ij})$ is defined by

$(g_{ij})=\left[\begin{array}[]{cccc}1+p^{2}+q^{2}+f^{2}&-p&-q&-f\\\
-p&1&0&0\\\ -q&0&1&0\\\ -f&0&0&1\\\ \end{array}\right].$

Let $\displaystyle X=\sum_{i=1}^{4}x_{i}\;{\bf
e}_{i},\;\;Y=\sum_{i=1}^{4}y_{i}\;{\bf e}_{i}$ and $\displaystyle
Z=\sum_{i=1}^{4}z_{i}\;{\bf e}_{i}$ in $T{\cal S}$, using [15] we can define
the cross product ,

$\displaystyle\begin{split}X\times Y\times Z&=\left|\begin{array}[]{cccc}{\bf
e}_{1}&{\bf e}_{2}&{\bf e}_{3}&{\bf e}_{4}\\\ x_{1}&x_{2}&x_{3}&x_{4}\\\
y_{1}&y_{2}&y_{3}&y_{4}\\\ z_{1}&z_{2}&z_{3}&z_{4}\\\ \end{array}\right|\\\
&=\left|\begin{array}[]{cccc}x_{2}&x_{3}&x_{4}\\\ y_{2}&y_{3}&y_{4}\\\
z_{2}&z_{3}&z_{4}\\\ \end{array}\right|\;{\bf
e}_{1}+\left|\begin{array}[]{cccc}x_{1}&x_{3}&x_{4}\\\ y_{1}&y_{3}&y_{4}\\\
z_{1}&z_{3}&z_{4}\\\ \end{array}\right|\;{\bf
e}_{2}+\left|\begin{array}[]{cccc}x_{1}&x_{2}&x_{4}\\\ y_{1}&y_{2}&y_{4}\\\
z_{1}&z_{2}&z_{4}\\\ \end{array}\right|\;{\bf
e}_{3}+\left|\begin{array}[]{cccc}x_{1}&x_{2}&x_{3}\\\ y_{1}&y_{2}&y_{3}\\\
z_{1}&z_{2}&z_{3}\\\ \end{array}\right|\;{\bf e}_{4}.\end{split}$ (39)

According to (39), we can compute

$\widehat{{\bf e}}_{1}\times\widehat{{\bf e}}_{2}\times\widehat{{\bf
e}}_{3}=q_{y}{\bf e}_{2}-{\bf e}_{4},$

and therefore we obtain the unit normal vector field on $\widetilde{\cal S}$
as follow

$\displaystyle{\bf n}=\frac{1}{\sqrt{1+q_{y}^{2}}}\left(-q_{y}{\bf e}_{2}+{\bf
e}_{4}\right).$ (40)

Thus the adapted orthogonal frame $\widetilde{{\bf E}}=(\widetilde{{\bf
e}}_{1},\widetilde{{\bf e}}_{2},\widetilde{{\bf e}}_{3},{\bf n})$ on
$\widetilde{\cal S}$ is

$\displaystyle\widetilde{{\bf e}}_{1}={\bf e}_{1},\;\;\;\;\;\widetilde{{\bf
e}}_{2}=\frac{1}{\sqrt{1+q_{y}^{2}}}\left({\bf e}_{2}+q_{y}{\bf
e}_{4}\right),\;\;\;\;\;\;\widetilde{{\bf e}}_{3}={\bf
e}_{3},\;\;\;\;\;\;\;{\bf n}=\frac{1}{\sqrt{1+q_{y}^{2}}}\left(-q_{y}{\bf
e}_{2}+{\bf e}_{4}\right),$ (41)

that frame
$\widetilde{{\bf\Omega}}=(\widetilde{\omega}^{1},\widetilde{\omega}^{2},\widetilde{\omega}^{3},\widetilde{\omega}^{4})$
on $\widetilde{\cal S}$ is

$\displaystyle\widetilde{\omega}^{1}={\omega}^{1},\;\;\;\;\widetilde{\omega}^{2}=\frac{1}{\sqrt{1+q_{y}^{2}}}\left({\omega}^{2}+q_{y}{\omega}^{4}\right),\;\;\;\;\;\widetilde{\omega}^{3}={\omega}^{3},\;\;\;\;\widetilde{\omega}^{4}=\frac{1}{\sqrt{1+q_{y}^{2}}}\left(-q_{y}{\omega}^{2}+{\omega}^{4}\right).$
(42)

Now since we have $\sigma^{*}\omega^{4}=q_{y}\;\omega^{2}$ on the submanifold
$\widetilde{\cal S}$, we will have $\widetilde{\omega}^{4}=0$ on it. Therefore
the Riemannian metric (35) can be rewritten based on the members of coframe
$\widetilde{{\bf\Omega}}$ as follows

$\displaystyle\widetilde{g}=\widetilde{\omega}^{1}\otimes\widetilde{\omega}^{1}+\widetilde{\omega}^{2}\otimes\widetilde{\omega}^{2}+\widetilde{\omega}^{3}\otimes\widetilde{\omega}^{3},$
(43)

on the submanifold $\widetilde{\cal S}$, here $\widetilde{\omega}^{2}$ is
equal to $\sqrt{1+q_{y}^{2}}\;\omega^{2}$.

We have the relation $\widetilde{{\bf E}}={\bf E}A$ between the frames ${\bf
E}=({\bf e}_{1},{\bf e}_{2},{\bf e}_{3},{\bf e}_{4})$ and $\widetilde{{\bf
E}}=(\widetilde{{\bf e}}_{1},\widetilde{{\bf e}}_{2},\widetilde{{\bf
e}}_{3},{\bf n})$ where $A\in{\rm SO}(4,{\mathbb{R}})$, with following
representation

$\displaystyle A=\left[\begin{array}[]{cccc}1&0&0&0\\\
0&\cos(\varepsilon)&0&-\sin(\varepsilon)\\\ 0&0&1&0\\\
0&\sin(\varepsilon)&0&\cos(\varepsilon)\\\ \end{array}\right],$ (48)

is defined by putting $q_{y}=\tan(\varepsilon)$ for the sufficiently small
values of $\varepsilon$. Using the formula (13), the connection matrix
$\widetilde{\theta}$ is equal to

$\widetilde{\theta}=\left[\begin{smallmatrix}0&\cos(\varepsilon)\;\theta^{1}_{2}+\sin(\varepsilon)\;\theta^{1}_{4}&\theta^{1}_{3}&-\sin(\varepsilon)\;\theta^{1}_{2}+\cos(\varepsilon)\;\theta^{1}_{4}\\\\[8.53581pt]
-\cos(\varepsilon)\;\theta^{1}_{2}-\sin(\varepsilon)\;\theta^{1}_{4}&0&\cos(\varepsilon)\;\theta^{2}_{3}-\sin(\varepsilon)\;\theta^{3}_{4}&-d\varepsilon+\theta^{2}_{4}\\\\[8.53581pt]
-\theta^{1}_{3}&-\cos(\varepsilon)\;\theta^{2}_{3}+\sin(\varepsilon)\;\theta^{3}_{4}&0&\sin(\varepsilon)\;\theta^{2}_{3}+\cos(\varepsilon)\;\theta^{3}_{4}\\\\[8.53581pt]
\sin(\varepsilon)\;\theta^{1}_{2}-\cos(\varepsilon)\;\theta^{1}_{4}&d\varepsilon-\theta^{2}_{4}&-\sin(\varepsilon)\;\theta^{2}_{3}-\cos(\varepsilon)\;\theta^{3}_{4}&0\end{smallmatrix}\right],$

where the 1-forms $\theta_{i}^{j}$s are defined in (33). Since
$q_{y}=\tan(\varepsilon)$, thus

$\displaystyle
d\varepsilon=\dfrac{(q_{xy}+pq_{yy})\widetilde{\omega}^{1}+q_{yy}(1+q_{y}^{2})^{-\frac{1}{2}}\;\widetilde{\omega}^{2}}{1+q_{y}^{2}},$
(49)

and by putting $\sin(\varepsilon)=q_{y}(1+q_{y}^{2})^{-\frac{1}{2}}$ and
$\cos(\varepsilon)=(1+q_{y}^{2})^{-\frac{1}{2}}$, we can rewrite the arrays of
the matrice $\widetilde{\theta}$ as follow

$\displaystyle\begin{split}\widetilde{\theta}_{2}^{1}&=-\frac{1}{2}\Big{[}f_{y}\sin(\varepsilon)\cos(\varepsilon)\;\widetilde{\omega}^{2}+(\cos(\varepsilon)+(1+f_{p})\sin(\varepsilon))\;\widetilde{\omega}^{3}\Big{]},\\\
\widetilde{\theta}_{3}^{1}&=-\frac{1}{2}\Big{[}(\cos(\varepsilon)+(1+f_{p})\sin(\varepsilon))\;\widetilde{\omega}^{2}\Big{]},\\\
\widetilde{\theta}_{4}^{1}&=-\frac{1}{2}\Big{[}f_{y}\cos(2\varepsilon)\;\widetilde{\omega}^{2}+(\sin(\varepsilon)-(1+f_{p})\cos(\varepsilon))\;\widetilde{\omega}^{3}\Big{]},\\\
\widetilde{\theta}_{3}^{2}&=-\frac{1}{2}\Big{[}\cos(\varepsilon)-(1-f_{p})\sin(\varepsilon)\Big{]}\widetilde{\omega}^{1},\\\
\widetilde{\theta_{4}^{2}}&=-d\varepsilon+\frac{1}{2}f_{y}\;\widetilde{\omega}^{1},\\\
\widetilde{\theta_{4}^{3}}&=-\frac{1}{2}\Big{[}\sin(\varepsilon)+(1-f_{p})\cos(\varepsilon)\Big{]}\widetilde{\omega}^{1}.\end{split}$
(50)

Looking at (17), the coefficients of the second fundamental form are
determined by

$\displaystyle h_{11}=0,$ $\displaystyle
h_{12}=h_{21}=\frac{1-q_{y}^{2}}{1+q_{y}^{2}}\;(q_{x}+pq_{y})_{y},$
$\displaystyle h_{13}=h_{31}=-\frac{1}{2\sqrt{1+q_{y}^{2}}},$ $\displaystyle
h_{22}=-\frac{q_{yy}}{(1+q_{y}^{2})^{\frac{3}{2}}},$ $\displaystyle
h_{23}=h_{32}=0,$ $\displaystyle h_{33}=0,$

here ${\cal A}=(h_{ij})$. Now Using the formula (18), the second fundamental
form is

$\displaystyle\begin{split}{\rm
II}&=-\left(\frac{1-q_{y}^{2}}{1+q_{y}^{2}}\right)(q_{x}+pq_{y})_{y}\;\widetilde{\omega}^{1}\otimes\widetilde{\omega}^{2}-\left(\frac{q_{y}-(q_{x}+pq_{y})_{p}}{\sqrt{1+q_{y}^{2}}}\right)\;\widetilde{\omega}^{1}\otimes\widetilde{\omega}^{3}-\dfrac{q_{yy}}{(1+q_{y}^{2})^{3/2}}\;\widetilde{\omega}^{2}\otimes\widetilde{\omega}^{2}\end{split}$
(51)

Then using (26) and (25), we have

$\displaystyle{\cal K}_{e}$ $\displaystyle=$
$\displaystyle\frac{q_{yy}}{4(1+q_{y}^{2})^{\frac{5}{2}}},$ (52)
$\displaystyle H$ $\displaystyle=$
$\displaystyle-\frac{q_{yy}}{3(1+q_{y}^{2})^{\frac{3}{2}}}.$ (53)

Using above findings we can summarize following theorems:

###### thm 1

A submanifold $\widetilde{\cal S}\subset\cal S$ which is determined by the
section (34) is minimal if and only if

$\displaystyle q_{yy}=0.$ (54)

In this case, the Gauss-Kronecker curvature of a minimal surface equals to
zero.

###### thm 2

Let $\widetilde{\cal S}$ be a submanifold corresponding to linear second-order
differential equation

$\displaystyle y^{\prime\prime}=\alpha(x)y+\beta(x)$ (55)

where $\alpha$ and $\beta$ are two smooth functions in term of $x$, then
$\widetilde{\cal S}$ determines a minimal surface in four-dimensional manifold
corresponding to the following third-order equation

$\displaystyle
y^{\prime\prime\prime}=\alpha(x)y^{\prime}+\alpha^{\prime}(x)y+\beta^{\prime}(x),$
(56)

###### thm 3

A submanifold $\widetilde{\cal S}\subset\cal S$, which is defined by the
section

$(x,y,p)\mapsto(x,y,p,\alpha(x)y+\beta(x)),$

is totally geodesic if and only if $\alpha(x)=\pm 1$.

###  5.2 Case of $q_{y}=0$

In this case, we want to consider the submanifold determined with
$\sigma^{*}\omega^{4}=0$, on the submanifold $\widetilde{\cal S}$. Since
$\sigma^{*}\omega^{4}=q_{y}\;\omega^{2}$, this case concludes to $q_{y}=0$.
Thus the matrix (48) reduces to the identity and then we have
$\widetilde{\theta}=\theta$. According to above discussion $\omega^{4}$ equals
to zero on the submanifold $\widetilde{\cal S}$.

$\displaystyle\begin{split}\theta_{2}^{1}&=-\theta_{1}^{2}=-\frac{1}{2}\
\omega^{3},\\\ \theta_{3}^{1}&=-\theta_{1}^{3}=-\frac{1}{2}\ \omega^{2},\\\
\theta_{4}^{1}&=-\theta_{1}^{4}=-\frac{1}{2}f_{y}\
\omega^{2}-\frac{1}{2}(1+f_{p})\ \omega^{3},\\\
\theta_{3}^{2}&=-\theta_{2}^{3}=-\frac{1}{2}\ \omega^{1},\\\
\theta_{4}^{2}&=-\theta_{2}^{4}=\frac{1}{2}f_{y}\ \omega^{1},\\\
\theta_{4}^{3}&=-\theta_{3}^{4}=-\frac{1}{2}(1-f_{p})\ \omega^{1},\end{split}$
(57)

Therefore

$\displaystyle\begin{split}\widetilde{\theta}_{2}^{1}&=\frac{1}{2}\Big{[}f_{y}\;\widetilde{\omega}^{2}+(1+f_{p})\;\widetilde{\omega}^{3}\Big{]},\\\
\widetilde{\theta}_{3}^{1}&=\frac{1}{2}(1+f_{p})\;\widetilde{\omega}^{2},\\\
\widetilde{\theta}_{4}^{1}&=\frac{1}{2}\Big{[}f_{y}\;\widetilde{\omega}^{2}+\widetilde{\omega}^{3}\Big{]},\\\
\widetilde{\theta}_{3}^{2}&=-\frac{1}{2}(1-f_{p})\widetilde{\omega}^{1},\\\
\widetilde{\theta_{4}^{2}}&=-\frac{1}{2}f_{y}\;\widetilde{\omega}^{1},\\\
\widetilde{\theta_{4}^{3}}&=-\frac{1}{2}\widetilde{\omega}^{1}.\end{split}$
(58)

That means $h_{11}=h_{22}=h_{33}=h_{23}=0$ and
$h_{12}=\frac{1}{2}f_{y},\;h_{13}=\frac{1}{2}$, and therefore $H={\cal
K}_{e}=0$ and the second fundamental form is

$\displaystyle{\rm
II}=\frac{1}{2}(q_{x}+pq_{y})_{y}\;\widetilde{\omega}^{1}\otimes\widetilde{\omega}^{2}+\frac{1}{2}\;\widetilde{\omega}^{1}\otimes\widetilde{\omega}^{3},$
(59)

###### thm 4

A submanifold $\widetilde{\cal S}$ of $\cal S$ which is determined by the
section (34) is a minimal but not totally geodesic. Its Gauss-Kronecker
curvature equals to zero.

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