Abstract
We study the variational structure of the discrete KadomtsevPetviashvili (dKP) equation by means of its pluriLagrangian formulation. We consider the dKP equation and its variational formulation on the cubic lattice \(\mathbb Z^{N}\) as well as on the root lattice \(Q(A_{N})\). We prove that, on a lattice of dimension at least four, the corresponding EulerLagrange equations are equivalent to the dKP equation.
Keywords
 Root Lattice
 Elementary Cell
 Cyclic Permutation
 Exterior Derivative
 White Triangle
These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.
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1 Introduction
We developed the theory of pluriLagrangian problems (integrable systems of variational origin) in recent papers [2–6, 15, 16], influenced by the fundamental insight of [11–13, 17]. In the present paper, we consider the pluriLagrangian formulation of the discrete bilinear KadomtsevPetviashvili (dKP) equation on threedimensional lattices and its consistent extension to higher dimensional lattices. This equation belongs to integrable octahedrontype equations which were classified in [1]. A Lagrangian formulation of this equation was given in [13]. There, the authors consider a discrete 3form on the lattice \(\mathbb Z^{3}\) together with the corresponding EulerLagrange equations which are shown to be satisfied on solutions of the dKP equation. They also show that this 3form is closed on solutions of the dKP equation, namely, the socalled 4D closure relation is satisfied. The main goal of the present paper is to provide a more precise understanding of the findings in that paper. More concretely:

In the framework of the pluriLagrangian formulation, we construct the elementary building blocks of EulerLagrange equations, which, in the present situation, are the socalled 4D corner equations.

In the twodimensional case, as noticed in [4], the corresponding 3D corner equations build a consistent system. Its solutions are more general then the solutions of the underlying hyperbolic system of quadequations. On the contrary, in the present threedimensional situation, the system of 4D corner equations is not consistent in the usual sense (i.e., it does not allow to determine general solutions with the maximal number of initial data). However, this system turns out to be equivalent, in a sense which we are going to explain later, to the corresponding hyperbolic system, namely the dKP equation.

We provide a rigorous consideration of the branches of the logarithm functions involved in the EulerLagrange equations. This leads to the following more precise result: the system of 4D corner equations is equivalent, and thus provides a variational formulation, to two different hyperbolic equations, namely the dKP equation itself and its version obtained under inversion \(x\mapsto x^{1}\) of all fields which will be denoted by \(\mathrm {dKP}^{}\).
One can consider the dKP equation on the cubic lattice \(\mathbb Z^{3}\) and its higher dimensional analogues \(\mathbb Z^{N}\), but, as discussed in [1, 8, 9] another natural setting the dKP equation (and related octahedrontype equations) is the threedimensional root lattice
Also in this setting, the dKP equation can be extended in a consistent way to the higher dimensional lattices \(Q(A_{N})\) with \(N>3\).
Both lattices have their advantages and disadvantages. The cubic lattice \(\mathbb Z^{N}\), on the one hand, is more manageable and easier to visualize. Its cell structure is very simple: for every dimension N, all Ndimensional elementary cells are Ndimensional cubes. On the other hand, it is less natural to consider dKP on the lattice \(\mathbb Z^{3}\), because this equation depends on the variables assigned to six out of eight vertices of a (threedimensional) cube.
The root lattice \(Q(A_{N})\), in contrast, has a more complicated cell structure, because the number of different Ndimensional elementary cells increases with the dimension N. For instance, for \(N=3\) there are two types of elementary cells octahedra and tetrahedra. Moreover, especially in higher dimensions, a visualization of the elementary cells is difficult, if not impossible. However, this lattice is more natural for the consideration of dKP from the combinatorial point of view, because this equation depends on variables which can be assigned to the six vertices of an octahedron, one of the elementary cells of the lattice. Furthermore, the fourdimensional elementary cells are combinatorially smaller (they contain only 10 vertices, as compared with 16 vertices of a fourdimensional cube) and possess higher symmetry than the cubic ones. Since they support the equations which serve as variational analogue of the dKP equation, this leads to a simpler situation.
We will see that a fourdimensional cube is combinatorially equivalent to the sum of four elementary cells of the root lattice \(Q(A_{4})\). Therefore, several results in the cubic case can be seen as direct consequences of results of the more fundamental \(Q(A_{N})\)case.
Let us start with some concrete definitions valid for an arbitrary Ndimensional lattice \(\mathscr {X}\).
Definition 1.1
(Discrete 3form) A discrete 3form on \(\mathscr {X}\) is a realvalued function \(\mathscr {L}\) of oriented 3cells \(\sigma \) depending on some field \(x:\mathscr {X}\rightarrow \mathbb R\), such that \(\mathscr {L}\) changes the sign by changing the orientation of \(\sigma \).
For instance, in \(Q(A_{N})\), the 3cells are tetrahedra and octahedra, and, in \(\mathbb Z^{N}\), the 3cells are 3D cubes.
Definition 1.2
(3dimensional pluriLagrangian problem) Let \(\mathscr {L}\) be a discrete 3form on \(\mathscr {X}\) depending on \(x:\mathscr {X}\rightarrow \mathbb R\).

To an arbitrary 3manifold \(\varSigma \subset \mathscr {X}\), i.e., a union of oriented 3cells which forms an oriented threedimensional topological manifold, there corresponds the action functional, which assigns to \(x_{V(\varSigma )}\), i.e., to the fields in the set of the vertices \(V(\varSigma )\) of \(\varSigma \), the number
$$\begin{aligned} S_{\varSigma }:=\sum _{\sigma \in \varSigma }\mathscr {L}(\sigma ). \end{aligned}$$ 
We say that the field \(x:V(\varSigma )\rightarrow \mathbb R\) is a critical point of \(S_{\varSigma }\), if at any interior point \(n\in V(\varSigma )\), we have
$$\begin{aligned} \frac{\partial S_{\varSigma }}{\partial x(n)}=0. \end{aligned}$$(1)Equation (1) are called discrete EulerLagrange equations for the action \(S_{\varSigma }\).

We say that the field \(x:\mathscr {X}\rightarrow \mathbb R\) solves the pluriLagrangian problem for the Lagrangian 3form \(\mathscr {L}\) if, for any 3manifold \(\varSigma \subset \mathscr {X}\), the restriction \(x_{V(\varSigma )}\) is a critical point of the corresponding action \(S_{\varSigma }\).
In the present paper, we focus on the variational formulation of the dKP equation on \(Q(A_{N})\) and \(\mathbb Z^{N}\). Let us formulate the main results of the paper.
On the lattice \(Q(A_{N})\), we consider discrete 3forms vanishing on all tetrahedra. One can show (see Corollary 2.5) that, for an arbitrary interior vertex of any 3manifold in \(Q(A_{N})\), the EulerLagrange equations follow from certain elementary building blocks. These socalled 4D corner equations are the EulerLagrange equations for elementary 4cells of \(Q(A_{N})\) different from 4simplices, socalled 4ambosimplices. Such a 4ambosimplex has ten vertices. Therefore, the crucial issue is the study of the system consisting of the corresponding ten corner equations. In our case, each corner equation depends on all ten fields at the vertices of the 4ambosimplex. Therefore, one could call this system consistent if any two equations are functionally dependent. It turns out that this is not the case. We will prove the following statement:
Theorem 1.3
Every solution of the system of ten corner equations for a 4ambosimplex in \(Q(A_{N})\) satisfies either the system of five dKP equations or the system of five \( dKP ^{}\) equations on the five octahedral facets of the 4ambosimplex.
Thus, one can prescribe arbitrary initial values at seven vertices of a 4ambosimplex. We will also prove the following theorem:
Theorem 1.4
The discrete 3form \(\mathscr {L}\) is closed on any solution of the system of corner equations.
In [4, 15], it was shown that in dimensions 1 and 2 the analogues of the property formulated in Theorem 1.4 are related to more traditional integrability attributes.
For the case of the cubic lattice \(\mathbb Z^{N}\), the situation is similar: one can show (see Corollary 4.2) that, for an arbitrary interior vertex of any 3manifold in \(\mathbb Z^{3}\), the EulerLagrange equations follow from certain elementary building blocks. These socalled 4D corner equations are the EulerLagrange equations for elementary 4D cubes in \(\mathbb Z^{N}\). A 4D cube has sixteen vertices, but in our case the action on a 4D cube turns out to be independent of the fields on two of the vertices. Therefore, the crucial issue is the study of the system consisting of the corresponding fourteen corner equations. Six of the fourteen corner equations depend each on thirteen of the fourteen fields. There do not exist pairs of such equations which are independent of one and the same field. All other equations depend each on ten of the fourteen fields. Therefore, one could call this system consistent if it would have the minimal possible rank 2 (assign twelve fields arbitrarily and use two of the six corner equations—depending on thirteen fields—to determine the remaining two fields, then all twelve remaining equations should be satisfied automatically). It turns out that the system of the fourteen corner equations is not consistent in this sense. We will prove the following analogue of Theorem 1.3:
Theorem 1.5
Every solution of the system of fourteen corner equations for a 4D cube in \(\mathbb Z^{N}\) satisfies either the system of eight dKP equations or the system of eight \( dKP ^{}\) equations on the eight cubic facets of the 4D cube.
Thus, one can prescribe arbitrary initial values at nine vertices of a 4D cube. Correspondingly, we will also prove the following statement:
Theorem 1.6
The discrete 3form \(\mathscr {L}\) is closed on any solution of the system of corner equations.
The paper is organized as follows: we start with the root lattice \(Q(A_{N})\), thus considering the combinatorial issues and some general properties of pluriLagrangian systems. Then we introduce the dKP equation and its pluriLagrangian structure. In the second part of the paper the present similar considerations for the cubic lattice \(\mathbb Z^{N}\).
2 The Root Lattice \(Q(A_{N})\)
We consider the root lattice
where \(N\ge 3\). The threedimensional sublattices \(Q(A_{3})\) are given by
We consider fields \(x:Q(A_{N})\rightarrow \mathbb R\), and use the shorthand notations
where \(e_{i}\) is the unit vector in the ith coordinate direction. Furthermore, the shift functions \(T_{i}\) and \(T_{\bar{\imath }}\) are defined by
for a multiindex \(\alpha \). For simplicity, we sometimes abuse notations by identifying lattice points n with the corresponding fields x(n).
We now give a very brief introduction to the Delaunay cell structure of the ndimensional root lattice \(Q(A_{N})\) [7, 14]. Here, we restrict ourselves to a very elementary description which is appropriate to our purposes and follow the considerations in [1]. For each N there are N sorts of Ncells of \(Q(A_{N})\) denoted by P(k, N) with \(k=1,\ldots ,N\):

Two sorts of 2cells:
\(\begin{array}{ll} P(1,2):&{} \text {black triangles}\lfloor ijk\rfloor :=\{x_{i},x_{j},x_{k}\};\\ P(2,2):&{} \text {white triangles} \lceil ijk\rceil :=\{x_{ij},x_{ik},x_{jk}\}; \end{array}\)

Three sorts of 3cells:
\(\begin{array}{ll} P(1,3):&{} \text {black tetrahedra} \lfloor ijk\ell \rfloor :=\{x_{i},x_{j},x_{k},x_{\ell }\};\\ P(2,3):&{} \text {octahedra} [ijk\ell ]:=\{x_{ij},x_{ik},x_{i\ell },x_{jk},x_{j\ell },x_{k\ell }\};\\ P(3,3):&{} \text {white tetrahedra} \lceil ijk\ell \rceil :=\{x_{ijk},x_{ij\ell },x_{ik\ell },x_{jk\ell }\}; \end{array}\)

Four sorts of 4cells:
\({\begin{array}{ll} P(1,4): \text {black 4simplices} \lfloor \!\!\lfloor ijk\ell m\rfloor \!\!\rfloor :=\{x_{i},x_{j},x_{k},x_{\ell },x_{m}\};\\ P(2,4): \text {black 4ambosimplices} \lfloor ijk\ell m\rfloor :=\{x_{\alpha \beta }:\alpha ,\beta \in \{i,j,k,\ell ,m\}, \alpha \ne \beta \};\\ P(3,4): \text {white 4ambosimplices} \lceil ijk\ell m\rceil :=\{x_{\alpha \beta \gamma }:\alpha ,\beta ,\gamma \in \{i,j,k,\ell ,m\},\\ \alpha \ne \beta \ne \gamma \ne \alpha \};\\ P(4,4): \text {white 4simplices} \lceil \!\!\lceil ijk\ell m\rceil \!\!\rceil :=\{x_{ijk\ell },x_{ijkm},x_{ij\ell m},x_{ik\ell m},x_{jk\ell m}\}. \end{array}}\)
The facets of 3cells and 4cells can be found in Appendix 1.
In the present paper we will consider objects on oriented manifolds. We say that a black triangle \(\lfloor ijk\rfloor \) and white triangle \(\lceil ijk\rceil \) are positively oriented if \(i<j<k\) (see Fig. 1). Any permutation of two indices changes the orientation to the opposite one.
When we use the bracket notation, we always write the letters in brackets in increasing order, so, e.g., in writing \(\lfloor ijk\rfloor \) we assume that \(i<j<k\) and avoid the notation \(\lfloor jik\rfloor \) or \(\lfloor ikj\rfloor \) for the negatively oriented triangle \(\lfloor ijk\rfloor \).
There is a simple recipe to derive the orientation of facets of an Ncell: On every index in the brackets we put alternately a “\(+\)” or a “−” starting with a “\(+\)” on the last index. Then we get each of its facets by deleting one index and putting the corresponding sign in front of the bracket. For instance, the black 4ambosimplex
has the five octahedral facets \([ijk\ell ]\), \([ijkm]\), \([ij\ell m]\), \([ik\ell m]\), and \([jk\ell m]\).
The following two definitions are valid for arbitrary Ndimensional lattices \(\mathscr {X}\).
Definition 2.1
(Adjacent N cell) Given an Ncell \(\sigma \), another Ncell \(\bar{\sigma }\) is called adjacent to \(\sigma \) if \(\sigma \) and \(\bar{\sigma }\) share a common \((N1)\)cell. The orientation of this \((N1)\)cell in \(\sigma \) must be opposite to its orientation in \(\bar{\sigma }\).
The latter property guarantees that the orientations of the adjacent Ncells agree.
Definition 2.2
(Flower) A 3manifold in \(\mathscr {X}\) with exactly one interior vertex x is called a flower with center x. The flower at an interior vertex x of a given 3manifold is the flower with center x which lies completely in the 3manifold.
As a consequence, in \(Q(A_{N})\), in each flower every tetrahedron has exactly three adjacent 3cells and every octahedron has exactly four adjacent 3cells.
Examples for open 3manifolds in \(Q(A_{N})\) are the threedimensional sublattices \(Q(A_{3})\). Here, the flower at an interior vertex consists of eight tetrahedra (four black and four white ones) and six octahedra.
Examples of closed 3manifolds in \(Q(A_{N})\) are the set of facets of a 4ambosimplex (consisting of five tetrahedra) and the set of facets of a 4ambosimplex (consisting of five tetrahedra and five octahedra).
The elementary building blocks of 3manifolds are socalled 4D corners:
Definition 2.3
(4D corner) A 4D corner with center x is a 3manifold consisting of all facets of a 4cell adjacent to x.
In \(Q(A_{N})\), there are two different types of 4D corners: a corner on a 4simplex (consisting of a four tetrahedra) and a corner on a 4ambosimplex (consisting of two tetrahedra and three octahedra), see Appendix 2 for details.
The following combinatorial statement will be proven in Appendix 3:
Theorem 2.4
The flower at any interior vertex of any 3manifold in \(Q(A_{N})\) can be represented as a sum of 4D corners in \(Q(A_{N+2})\).
Let \(\mathscr {L}\) be a discrete 3form on \(Q(A_{N})\). The exterior derivative \(d\mathscr {L}\) is a discrete 4form whose value at any 4cell in \(Q(A_{N})\) is the action functional of \(\mathscr {L}\) on the 3manifold consisting of the facets of the 4cell. For our purposes, we consider discrete 3forms \(\mathscr {L}\) vanishing on all tetrahedra. In particular, we have
since a 4simplices only contain tetrahedra. The exterior derivative on a black 4ambosimplex \(\lfloor ijk\ell m\rfloor \) is given by
The exterior derivative on a white 4ambosimplex \(\lceil ijk\ell m\rceil \) is given by
Accordingly, the EulerLagrange equations on black 4ambosimplices \(\lfloor ijk\ell m\rfloor \) are
and the EulerLagrange equations on white 4ambosimplices \(\lceil ijk\ell m\rceil \) are
The last two systems are called corner equations.
The following statement is an immediate consequence of Theorem 2.4:
Theorem 2.5
For discrete every 3form on \(Q(A_{N})\) and every 3manifold in \(Q(A_{N})\) all corresponding EulerLagrange equations can be written as a sum of corner equations.
3 The dKP Equation on \(Q(A_{N})\)
We will now introduce the dKP equation on the root lattice \(Q(A_{3})\). Every oriented octahedron \([ijk\ell ]\) (\(i<j<k<\ell \)) in \(Q(A_{3})\) supports the equation
We can extend this system in a consistent way (see [1]) to the fourdimensional root lattice \(Q(A_{4})\) and higherdimensional analogues, such that the five octahedral facets \([ijk\ell ]\), \([jk\ell m]\), \([ik\ell m]\), \([ijm\ell ]\), and \([ijkm]\) of the black 4ambosimplex \(\lfloor ijk\ell m\rfloor \) support the equations
and the five octahedral facets \(T_{m}[ijk\ell ]\), \(T_{i}[jk\ell m]\), \(T_{j}[ik\ell m]\), \(T_{k}[ij\ell m]\), and \(T_{\ell }[ijkm]\) of the white 4ambosimplex \(\lceil ijk\ell m\rceil \) support the equations
In both systems one can derive one equation from another by cyclic permutations of indices \((ijk\ell m)\).
We propose the following discrete 3form \(\mathscr {L}\) defined on oriented octahedra \([ijk\ell ]\):
where
The discrete 3form (9) has its motivation in [13]. Indeed, in [13], the authors consider a similar discrete 3form on the cubic lattice \(\mathbb Z^{N}\). One can also consider our 3form on the cubic lattice \(\mathbb Z^{N}\). Then one would assign to each 3D cube the 3form at its inscribed octahedron. This 3form differs from their one by an additive constant and a slightly different definition of the function \(\lambda (z)\): they use the function
instead of \(\lambda (z)\). Our choice of \(\lambda (z)\) allows us for a more precise consideration of the branches of the occurring logarithm.
Observe that the expression (9) only changes its sign under the cyclic permutation of indices \((ijk\ell m)\). This follows from \(\varLambda (z)=\varLambda (z^{1})\). As a consequence, the exterior derivatives \(\underline{S}^{ijk\ell m}\) and \(\bar{S}^{ijk\ell m}\) defined in (2) and (3), respectively, are invariant under the cyclic permutation of indices \((ijk\ell m)\). Therefore, one can obtain all corner equations in (4) and (5) by (iterated) cyclic permutation \((ijk\ell m)\) from
Let us study separately the corner equations on black and white 4ambosimplices. The corner equations which live on the black 4ambosimplex \(\lfloor ijk\ell m\rfloor \) are given by
and
Explicitly, they read
where
and
For every corner equation (12) there are two classes of solutions, because any solution can either solve \(E_{ij}=1\) or \(E_{ij}=1\). Hereafter, we only consider solutions, where all fields \(x_{ij}\) are nonzero (we call such solutions nonsingular).
Theorem 3.1
Every solution of the system (4) solves either the system
or the system
Furthermore, the system (13) is equivalent to the system (7) (that is dKP on the corresponding black 4ambosimplex). The system (14) is equivalent to the system
which is the system (7) after the transformation \(x\mapsto x^{1}\) of fields (that is \( dKP ^{}\) on the corresponding black 4ambosimplex).
Proof
Consider a solution x of (4) that solves \(E_{ij}=1\) and \(E_{jk}=1\). We set
and
and use these equations to substitute \(x_{ij}\), \(x_{ik}\) and \(x_{jk}\) in \(E_{ij}=1\) and \(E_{jk}=1\). Writing down the result in polynomial form, we get
and
where \(e_{ij}\) and \(e_{jk}\) are certain polynomials. Since for every solutions of (4) all fields are nonzero this leads us to \(e_{ij}=0\) and \(e_{jk}=0\). Computing the difference of the latter two equations we get
and, with the use of (16) and (18),
which depends on seven independent fields, i.e., no subset of six fields belong to one octahedron. Then comparing coefficients leads to \(a_{ij}=a_{jk}=0\). Substituting
into \(E_{ij}=1\) and solving the resulting equation with respect to \(x_{ik}\), we get
Substituting \(x_{ij}\), \(x_{ik}\) and \(x_{jk}\) in \(E_{ik}\) by using the last three equations, we get \(E_{ik}=1\).
Analogously, one can prove that, for a solution x of (4) which solves \(E_{ij}=1\) and \(E_{ik}=1\), we have \(E_{jk}=1\), and for a solution x of (4) which solves \(E_{ik}=1\) and \(E_{i\ell }=1\), we have \(E_{k\ell }=1\). Therefore, for every solution x of (4) and for every white triangle \(\{x_{\alpha },x_{\beta },x_{\gamma }\}\) on the black 4ambosimplex \(\lfloor ijk\ell m\rfloor \) we proved the following: if \(E_{\alpha }=1\) and \(E_{\beta }=1\) then \(E_{\gamma }=1\), too.
On the other hand, one can easily see that x solves \(E_{ij}=1\) or \(E_{jk}=1\) if and only if \(x^{1}\) solves \(E_{ij}=1\) or \(E_{ik}=1\), respectively. Therefore, we also know that, if \(E_{\alpha }=1\) and \(E_{\beta }=1\) then \(E_{\gamma }=1\), too.
Summarizing, we proved that every solution x of (4) solves either (13) and then also (7) or (14) and then also (15).
Consider a nonsingular solution x of the system (7). Then
and
This proves the equivalence of (13) and (7) and also the equivalence of (14) and (15) since x solves \(E_{ij}=1\) or (7) if and only if \(x^{1}\) solves \(E_{ij}=1\) or (15), respectively.\(\square \)
We will present the closure relation which can be seen as a criterion of integrability:
Theorem 3.2
(Closure relation) There holds:
on all solutions of (13) and (14), respectively. Therefore, one can redefine the 3form \(\mathscr {L}\) as
in order to get \(\underline{S}^{ijk\ell m}=0\) on all solutions of (13) and (14), respectively.
Proof
The set of solutions \(\mathscr {S}^{+}\) of (13), as well as the set of solutions \(\mathscr {S}^{}\) (14), is a connected sevendimensional algebraic manifold which can be parametrized by the set of variables \(\{x_{ij},x_{ik},x_{i\ell },x_{im},x_{jk},x_{j\ell },x_{jm}\}\). We want to show that the directional derivatives of \(\underline{S}^{ijk\ell m}\) along tangent vectors of \(\mathscr {S}^{\pm }\) vanish. It is easy to see that the stronger property \(\text {grad}\underline{S}^{ijk\ell m}=0\) on \(\mathscr {S}^{\pm }\), where we \(\underline{S}^{ijk\ell m}\) is considered as a function of ten variables \(x_{ij}\), is a consequence of (13), respectively (14). Therefore, the function \(\underline{S}^{ijk\ell m}\) is constant on \(\mathscr {S}^{\pm }\).
To determine the value of \(\underline{S}^{ijk\ell m}\) on solutions of (13), we consider the constant solution of (7)
where
(Indeed, for this point every equation from (7) looks like \(a^{2}1a=0\).) Therefore, this point satisfies (13), because (7) and (13) are equivalent.
Consider the dilogarithm as defined in (11) and suppose that \(z>1\). According to [10], we derive:
and
where \(\lambda (z)\) is the same function as in (9). Therefore, we have
By using the following special values [10]
a straightforward computation gives
and
This is, because the expression for \(\mathscr {L}([ijk\ell ])\) (see (9)) changes the sign under the cyclic permutation of indices \((ijk\ell )\) and the solution is invariant under cyclic permutation of indices \((ijk\ell m)\).
Let us now consider the second branch of solutions: one can easily see that
with
is a solution of (14) and (15), because (19) is a solution of (13) and (7). Therefore, on the solution (20) as well as on all other solutions of (14), we have
where we used \(\varLambda (z)=\lambda (z)\lambda (z^{1})\), and, therefore, \(\varLambda (z^{1})=\varLambda (z)\).\(\square \)
Analogously, we get similar results for the white 4ambosimplex \(\lceil ijk\ell m\rceil \). Here, the corner equations are:
and
Explicitly, they read
where
and
The analogue of Theorem 3.1 reads:
Theorem 3.3
Every solution of the system (5) solves either the system
or the system
Furthermore the system (22) is equivalent to the system (8) (that is dKP on the corresponding white 4ambosimplex). The system (23) is equivalent to the system
which is the system (8) after the transformation \(x\mapsto x^{1}\) of fields (that is \( dKP ^{}\) on the corresponding white 4ambosimplex).
The analogue of Theorem 3.2 reads:
Theorem 3.4
(Closure relation) There holds:
on all solutions of (22) and (23), respectively. Therefore, one can redefine the 3form \(\mathscr {L}\) as
in order to get \(\bar{S}^{ijk\ell m}=0\) on all solutions of (22) and (23), respectively.
4 The Cubic Lattice \(\mathbb Z^{N}\)
We will now consider the relation between the elementary cells of the root lattice \(Q(A_{N})\) and the cubic lattice \(\mathbb Z^{N}\). The points of \(Q(A_{N})\) and of \(\mathbb Z^{N}\) are in a onetoone correspondence via
In the present paper, we will always apply \(P_{i}\) with \(i<j,k,\ell ,\ldots \)
We denote by
the oriented 3D cubes of \(\mathbb Z^{N}\). We say that the 3D cube \(\{jk\ell \}\) is positively oriented if \(j<k<\ell \). Any permutation of two indices changes the orientation to the opposite one. Also in this case, we always write the letters in the brackets in increasing order, so, e.g., in writing \(\{jk\ell \}\) we assume that \(j<k<\ell \) and avoid the notation \(\{kj\ell \}\) or \(\{j\ell k\}\) for the negatively oriented 3D cube \(\{jk\ell \}\).
The object in \(Q(A_{N})\) which corresponds to the 3D cube \(\{jk\ell \}\) is the sum of three adjacent 3cells, namely

the black tetrahedron \(T_{i}\lfloor ijk\ell \rfloor \) (see Fig. 2a),

the octahedron \([ijk\ell ]\) (see Fig. 2b),

and the white tetrahedron \(T_{\bar{\imath }}\lceil ijk\ell \rceil \) (see Fig. 2c).
It contains sixteen triangles and to every quadrilateral face of \(\{jkl\}\) there corresponds a pair of these triangles containing one black and one white triangle. Here, the map \(P_{i}\) reads as follows:
As a fourdimensional elementary cell of \(\mathbb Z^{N}\), we consider an oriented 4D cube
The 4D cube \(\{jk\ell m\}\) corresponds to the sum of four 4cells in \(Q(A_{N})\):

the black 4simplex \(T_{i}\lfloor \!\!\lfloor ijk\ell m\rfloor \!\!\rfloor \),

the black 4ambosimplex \(\lfloor ijk\ell m\rfloor \),

the white 4ambosimplex \(T_{\bar{\imath }}\lceil ijk\ell m\rceil \), and

the white 4simplex \(T_{\bar{\imath }}T_{\bar{\imath }}\lceil \!\!\lceil ijk\ell m\rceil \!\!\rceil \)
(see Fig. 3). It contains sixteen tetrahedra (eight black and eight white ones) and eight octahedra. Here, the map \(P_{i}\) reads as follows:
Also in the cubic case there is an easy recipe to obtain the orientation of the facets of an (oriented) 4D cube: on every index between the brackets we put alternately a “\(+\)” and a “−” starting with a “\(+\)” on the last index. Then we get each facet by deleting one index and putting the corresponding sign in front of the bracket. For instance., the 4D cube
has the eight 3D facets: \(\{jk\ell \}\), \(\{jkm\}\), \(\{j\ell m\}\), \(\{k\ell m\}\) and the opposite ones \(T_{m}\{jk\ell \}\), \(T_{\ell }\{jkm\}\), \(T_{k}\{j\ell m\}\), and \(T_{j}\{k\ell m\}\).
As a consequence of Definition 2.2, in each flower in \(\mathbb Z^{N}\), every 3D cube has exactly four adjacent 3D cubes.
We will now prove the analogue of Theorem 2.5. This proof is easier than the one for \(Q(A_{N})\), because of the simpler combinatorial structure.
Theorem 4.1
The flower at any interior vertex of any 3manifold in \(\mathbb Z^{N}\) can be represented as a sum of 4D corners in \(\mathbb Z^{N+1}\).
Proof
Set \(M:=N+1\) and consider the flower of an interior vertex x of an arbitrary 3manifold in \(\mathbb Z^{N}\). Over each 3D corner \(\{jk\ell \}\) (petal) of the flower, we can build a 4D corner adjacent to x on the 4D cube \(\{jk\ell M\}\). Then the vertical 3D cubes coming from two successive petals of the flower carry opposite orientations, so that all vertical squares cancel away from the sum of the 4D corners.\(\square \)
Let \(\mathfrak {L}\) be a discrete 3form on \(\mathbb Z^{N}\). The exterior derivative \(d\mathfrak {L}\) is a discrete 4form whose value at any 4D cube in \(\mathbb Z^{N}\) is the action functional of \(\mathfrak {L}\) on the 3manifold consisting of the facets of the 4D cube:
Accordingly, the EulerLagrange equations on the 4D cube \(\{jk\ell m\}\) are given by
They are called corner equations.
The following statement is an immediate consequence of Theorem 4.1:
Theorem 4.2
For every discrete 3form on \(\mathbb Z^{N}\) and every 3manifold in \(\mathbb Z^{N}\) all corresponding EulerLagrange equations can be written as a sum of corner equations.
5 The dKP Equation on \(\mathbb Z^{N}\)
On the 3D cube \(\{jk\ell \}\) in \(\mathbb Z^{3}\) (\(j<k<\ell \)) we put the equation
We can extend this system in a consistent way (see [1]) to the fourdimensional cubic lattice \(\mathbb Z^{4}\) and its higherdimensional analogues, such that the eight facets \(\{jk\ell \}\), \(\{jkm\}\), \(\{j\ell m\}\), \(\{k\ell m\}\), \(T_{m}\{jk\ell \}\), \(T_{\ell }\{jkm\}\), \(T_{k}\{j\ell m\}\), \(T_{j} \{k\ell m\}\) of a 4D cube \(\{jk\ell m\}\) carry the equations
Note that, in the four equations in the left column, the fields with one index always appear with increasing order of indices. The equations in the right column are shifted copies of the ones in the left column. One can derive the system (27) from the system of dKP equations (7) on the black 4ambosimplex \(\lfloor ijk\ell m\rfloor \) and the system of dKP equations (8) on the white 4ambosimplex \(T_{\bar{\imath }}\lceil ijkm\ell \rceil \), by removing the equations on the octahedra \([jk\ell m]\) and \([jkm\ell ]\), respectively, from both systems and applying the transformation \(P_{i}\) to the fields in the remaining eight equations.
We propose the discrete 3form \(\mathfrak {L}\) defined as
where \(\mathscr {L}\) is the discrete 3form on the root lattice \(Q(A_{N})\) (see (9)). Therefore, \(\mathfrak {L}\) evaluated at the 3D cube \(\{jk\ell \}\) reads as
For this discrete 3form, there are no corner equations on the 4D cube \(\{jk\ell m\}\) centered at x and \(x_{jk\ell m}\) since \(S^{jk\ell m}\) does not depend on these two variables. The remaining corner equations from (25) are given by
where
Hereafter, we only consider solutions, where all fields are nonzero (we call these solutions nonsingular). As in the case of the root lattice \(Q(A_{N})\) every corner equation has two classes of solutions.
Theorem 5.1
Every solution of the system (25) solves either the system
or the system
Furthermore the system (29) is equivalent to the system (27) (this is dKP on the corresponding 4D cube). The system (30) is equivalent to the system
which is the system (27) after the transformation \(x\mapsto x^{1}\) of fields (this is \( dKP ^{}\) on the corresponding 4D cube).
Proof
Let x be a solution of the system (25) such that \(\mathscr {E}_{j}=1\) and \(\mathscr {E}_{k}=1\). Then we know from the proof of Theorem 3.1 that
and that the latter system is equivalent to
On the other hand, if we consider a solution x of (25) such that \(\mathscr {E}_{j}=1\) and \(\mathscr {E}_{k}=1\), we know from the proof of Theorem 3.1 that
and that the latter system is equivalent to
Now, let x be a solution of the system (25) such that \(\mathscr {E}_{jk\ell }=1\) and \(\mathscr {E}_{jkm}=1\). Then we know from the proof of Theorem 3.3 that
and that the latter system is equivalent to
On the other hand, if we consider a solution x of (25) such that \(\mathscr {E}_{j}=1\) and \(\mathscr {E}_{k}=1\), we know from the proof of Theorem 3.3 that
and that the latter system is equivalent to
Since a solution x of (25) cannot solve
and
at the same time, this proves the theorem.\(\square \)
Theorem 5.2
(Closure relation) There holds \(S^{jk\ell m}=0\) on all solutions of (25).
Proof
Let x be a solution of (29) or (30). Then
due to Theorems 3.2 and 3.4 since every solution of (29) solves (13) and (22) after the transformation \(P_{i}\) of variables and every solution of (30) solves (14) and (23) after the transformation \(P_{i}\) of variables.\(\square \)
6 Conclusion
The fact that the threedimensional (hyperbolic) dKP equation is, in a sense, equivalent to the EulerLagrange equations of the corresponding action is rather surprising since for the twodimensional (hyperbolic) quadequations an analogous statement is not true (see [4, 6] for more details). On the other hand, in the continuous situation there is an example of a 2form whose EulerLagrange equations are equivalent to the set of equations consisting of the (hyperbolic) sineGordon equation and the (evolutionary) modified Kortewegde Vries equation (see [16] for more details). So, the general picture remains unclear.
In particular, the variational formulation for the other equations of octahedron type in the classification of [1] is still an open problem.
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Acknowledgments
This research was supported by the DFG Collaborative Research Center TRR 109 “Discretization in Geometry and Dynamics”.
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Appendices
Appendix 1: Facets of NCells of the Root Lattice \(Q(A_{N})\)
Facets of 3cells:
\({\begin{array}{ll} \text {Black tetrahedra} \lfloor ijk\ell \rfloor {:}&{} \text {four black triangles} \lfloor ijk\rfloor , \lfloor ij\ell \rfloor , \lfloor ik\ell \rfloor , \text {and} \lfloor jk\ell \rfloor ;\\ \text {Octahedra} [ijk\ell ]{:}&{} \text {four black triangles} T_{\ell }\lfloor ijk\rfloor , T_{k}\lfloor ij\ell \rfloor , T_{j}\lfloor ik\ell \rfloor ,\\ &{} \text {and} T_{i}\lfloor jk\ell \rfloor ,\\ &{} \text {four white triangles} \lceil ijk\rceil , \lceil ij\ell \rceil , \lceil ik\ell \rceil , \text {and} \lceil jk\ell \rceil ;\\ \text {White tetrahedra} \lceil ijk\ell \rceil {:}&{} \text {four white triangles} T_{\ell }\lceil ijk\rceil , T_{k}\lceil ij\ell \rceil , T_{j}\lceil ik\ell \rceil ,\\ &{} \text {and} T_{i}\lceil jk\ell \rceil ; \end{array}}\)
Facets of 4cells:
\({\begin{array}{ll} \text {Black 4simplices} \lfloor \!\!\lfloor ijk\ell m\rfloor \!\!\rfloor {:}&{} five black tetrahedra \lfloor ijk\ell \rfloor , \lfloor ijkm\rfloor , \lfloor ij\ell m\rfloor ,\\ &{} \lfloor ik\ell m\rfloor , \text {and} \lfloor jk\ell m\rfloor ;\\ \text {Black 4ambosimplices} \lfloor ijk\ell m\rfloor {:}&{} \text {five black tetrahedra} T_{m}\lfloor ijk\ell \rfloor , T_{\ell }\lfloor ijkm\rfloor ,\\ &{} T_{k}\lfloor ij\ell m\rfloor , T_{j}\lfloor ik\ell m\rfloor , \text {and} T_{i}\lfloor jk\ell m\rfloor ,\\ &{} \text {and five octahedra} [ijk\ell ], [ijkm], [ij\ell m],\\ &{} [ik\ell m], \text {and} [jk\ell m]; \\ \text {White 4ambosimplices} \lceil ijk\ell m\rceil {:}&{} \text {five octahedra} T_{m}[ijk\ell ], T_{\ell }[ijkm], T_{k}[ij\ell m],\\ &{} T_{j}[ik\ell m], \text {and} T_{i}[jk\ell m],\\ &{} \text {and five white tetrahedra} \lceil ijk\ell \rceil , \lceil ijkm\rceil ,\\ &{} \lceil ij\ell m\rceil , \lceil ik\ell m\rceil , \text {and} \lceil jk\ell m\rceil ; \\ \text {White 4simplices} \lceil \!\!\lceil ijk\ell m\rceil \!\!\rceil {:}&{} \text {five white tetrahedra} T_{m}\lceil ijk\ell \rceil , T_{\ell }\lceil ijkm\rceil ,\\ &{} T_{k}\lceil ij\ell m\rceil , T_{j}\lceil ik\ell m\rceil , \text {and} T_{i}\lceil jk\ell m\rceil . \end{array}}\)
Appendix 2: 4D Corners on 4Cells of the Root Lattice \(Q(A_{N})\)
Black 4simplex \(\lfloor \!\!\lfloor ijk\ell m\rfloor \!\!\rfloor \):
The 4D corner with center vertex \(x_{i}\) contains

the four black tetrahedra \(\lfloor ijk\ell \rfloor \), \(\lfloor ijkm\rfloor \), \(\lfloor ij\ell m\rfloor \), and \(\lfloor ik\ell m\rfloor \);
Black 4ambosimplex \(\lfloor ijk\ell m\rfloor \):
The 4D corner with center vertex \(x_{ij}\) contains

the two black tetrahedra \(T_{j}\lfloor ik\ell m\rfloor \), and \(T_{i}\lfloor jk\ell m\rfloor \),

and the three octahedra \([ijk\ell ]\), \([ijkm]\), and \([ij\ell m]\);
White 4ambosimplex \(\lceil ijk\ell m\rceil \):
The 4D corner with center vertex \(x_{ijk}\) contains

the three octahedra \(T_{k}[ij\ell m]\), \(T_{j}[ik\ell m]\), and \(T_{i}[jk\ell m]\),

and the two white tetrahedra \(\lceil ijk\ell \rceil \), and \(\lceil ijkm\rceil \);
White 4simplex \(\lceil \!\!\lceil ijk\ell m\rceil \!\!\rceil \):
The 4D corner with center vertex \(x_{ijk\ell }\) contains

the four white tetrahedra \(T_{\ell }\lceil ijkm\rceil \), \(T_{k}\lceil ij\ell m\rceil \), \(T_{j}\lceil ik\ell m\rceil \), and \(T_{i}\lceil jk\ell m\rceil \).
Appendix 3: Proof of Theorem 2.4
Set \(M:=N+1\) and \(L:=N+2\). Then, for the construction of the sum \(\varSigma \) of 4D corners representing the flower \(\sigma \) centered in X, we use the following algorithm:

(i)
For every black tetrahedron \(\pm \lfloor ijk\ell \rfloor \in \sigma \) at the interior vertex X we add the 4D corner with center vertex X on the black 4simplex \(\pm \lfloor \!\!\lfloor ijk\ell M\rfloor \!\!\rfloor \) to \(\varSigma \).

(ii)
For every octahedron \(\pm [ijk\ell ]\in \sigma \) we add the 4D corner with center vertex X on the black 4ambosimplex \(\pm \lfloor ijk\ell M\rfloor \) to \(\varSigma \).

(iii)
For every white tetrahedron \(\pm \lceil ijk\ell \rceil \in \sigma \) we add the 4D corner with center vertex X on the white 4ambosimplex \(\pm \lceil ijk\ell M\rceil \) to \(\varSigma \).

(iv)
For every white tetrahedron \(\pm \lceil ijkM\rceil \in \varSigma \setminus \sigma \) which appeared in \(\varSigma \) during the previous step we add the 4D corner with center vertex X on the white 4simplex to \(\varSigma \).
Therefore, we have to prove that \(\varSigma =\sigma \).
Assume that \(X=x_{i}\). Then for each black tetrahedron \(\pm \lfloor ijk\ell \rfloor \in \sigma \) we added the three black tetrahedra , \(\pm \lfloor ij\ell M\rfloor \), and to \(\varSigma \) which do not belong to \(\sigma \). Moreover, \(\pm \lfloor ijk\ell \rfloor \) has three black triangular facets adjacent to \(x_{i}\), namely \(\pm \lfloor ijk\rfloor \), which is the common triangle with (up to orientation), , which is the common triangle with \(\pm \lfloor ij\ell M\rfloor \), and \(\pm \lfloor ik\ell \rfloor \), which is the common triangle with . Therefore, each of these black tetrahedra has to cancel away with the corresponding black tetrahedra from the 4D corner which is coming from the 3cell adjacent to \(\pm \lfloor ijk\ell \rfloor \) via the corresponding black triangle.
Assume that \(X=x_{ij}\). Then for each octahedron \(\pm [ijk\ell ]\in \sigma \) we added the two black tetrahedra and \(\pm T_{i}\lfloor jk\ell M\rfloor \) as well as the two octahedra and \(\pm [ij\ell M]\) to \(\varSigma \) which do not belong to \(\sigma \). Moreover, \(\pm [ijk\ell ]\) has two black tetrahedral facets adjacent to \(x_{ij}\), namely \(\pm \lfloor ijk\rfloor \), which is the common triangle with , and , which is the common triangle with \(\pm T_{i}\lfloor jk\ell M\), as well as two white tetrahedral facets adjacent to \(x_{ij}\), namely \(\pm T_{j}\lceil ik\ell \rceil \), which is the common triangle with and \(\pm [ij\ell M]\), and , which is the common triangle with \(\pm [ij\ell M]\). Therefore, each of the black tetrahedra and \(\pm T_{i}\lfloor jk\ell M\) has to cancel away with the corresponding black tetrahedron from the 4D corner which is coming from the 3cell adjacent to\(\pm [ijk\ell ]\) via the corresponding black triangle, and each of the octahedra and \(\pm [ij\ell M]\) has to cancel away with the corresponding octahedron coming the 4D corner which is coming from the 3cell adjacent to \(\pm [ijk\ell ]\) via the corresponding white triangle.
Assume that \(X=x_{ijk}\). Then for each white tetrahedron \(\pm \lceil ijk\ell \rceil \in \sigma \) we added the three octahedra \(\pm T_{k}[ij\ell M]\), , and \(\pm T_{i}[jk\ell M]\) as well as the white tetrahedron to \(\varSigma \) which do not belong to \(\sigma \). Moreover, \(\pm \lceil ijk\ell \rceil \) has three white triangular facets adjacent to \(x_{ijk}\), namely , which is the common triangle with \(\pm T_{k}[ij\ell M]\), \(\pm T_{j}\lceil ik\ell \rceil \), which is the common triangle with , and , which is the common triangle with \(\pm T_{i}[jk\ell M]\). Therefore, each of these octahedra has to cancel away with the corresponding octahedron from the 4D corner which is coming from the 3cell adjacent to \(\pm \lceil ijk\ell \rceil \) via the corresponding white triangle.
Consider two 3cells \(\Omega ,\bar{\Omega }\in \sigma \) adjacent via the black triangle \(\lfloor ijk\rfloor \), say \(\lfloor ijk\rfloor \) belongs to \(\Omega \) and \(\lfloor ijk\rfloor \) belongs to \(\bar{\Omega }\). Then the 4D corner corresponding to \(\Omega \) contributes the black tetrahedron \(\lfloor ijkM\rfloor \) to \(\varSigma \), whereas the 4D corner corresponding to \(\bar{\Omega }\) contributes the black tetrahedron \(\lfloor ijkM\rfloor \) to \(\varSigma \). Therefore, the latter two black tetrahedra cancel out.
Consider two 3cells \(\Omega ,\bar{\Omega }\in \sigma \) adjacent via the white triangle \(\lceil ijk\rceil \), say \(\lceil ijk\rceil \) belongs to \(\Omega \) and \(\lceil ijk\rceil \) belongs to \(\bar{\Omega }\). Then the 4D corner corresponding to \(\Omega \) contributes the octahedron \([ijkM]\) to \(\varSigma \), whereas the 4D corner corresponding to \(\bar{\Omega }\) contributes the octahedron [ijkM] to \(\varSigma \). Therefore, the latter two octahedra cancel out.
Up to now we proved that all black tetrahedra and all octahedra in \(\varSigma \setminus \sigma \) cancel out. We will now consider with the white tetrahedra in \(\varSigma \setminus \sigma \).
Lemma 6.1
The white tetrahedra \(\lfloor ijk M\rfloor \) arising in the third step of the algorithm build flowers which only contain white tetrahedra.
Proof
We have two prove that each of these white tetrahedra has exactly three adjacent white tetrahedra in the flowers, one via each white triangle adjacent to X. They are not adjacent to the 3cells in \(\sigma \), but each of them has three common neighbors adjacent to X with the corresponding white tetrahedron in \(\sigma \). These common neighbors are octahedra which cancel out in the previous steps of the algorithm.
Consider now two white tetrahedra \(T,\bar{T}\in \varSigma \setminus \sigma \) after the third step of the algorithm, where the corresponding white tetrahedra in \(\sigma \) are adjacent, i.e., there is a pair of octahedra with the same set of points and different orientation, one adjacent to T and the other adjacent to \(\bar{T}\). Therefore, T and \(\bar{T}\) share a common white triangle (up to orientation), i.e., they are adjacent.
Consider the 4D corner which we add to \(\varSigma \) for an octahedron. Its two octahedra which do not belong to \(\sigma \) share a common white triangle (up to orientation) which does no lie in \(\sigma \). Furthermore, consider a sequence of adjacent octahedra in \(\sigma \), where the common triangles are all white triangles. Then the octahedra in the corresponding 4D corners which are not in \(\sigma \) all share a common white triangle (up to orientation).
Consider now two white tetrahedra \(T,\bar{T}\in \varSigma \setminus \sigma \) after the third step of the algorithm, where the corresponding tetrahedra in \(\sigma \) are connected by a sequence of octahedra adjacent via white triangles. Then, there is a pair of octahedra, one of them adjacent to T and the other one adjacent to \(\bar{T}\), which share a common white triangle (up to orientation). This triangle does not belong to any 3cell in \(\sigma \) and, therefore, is a common triangle of T and \(\bar{T}\), i.e., T and \(\bar{T}\) are adjacent.\(\square \)
Now we continue with the proof of Theorem 2.4. We already proved that a flower containing only black tetrahedra can be written as a sum of 3D corners on black 4simplices (see proof of step 1). Analogously, one can write every flower containing only white tetrahedra as a sum of 3D corners on white 4simplices. So we write for each of the flowers of white tetrahedra in \(\varSigma \setminus \sigma \) after the third step of the algorithm the flower of opposite orientation as a sum of 3D corners on white 4simplices and add this sums to \(\varSigma \). Then \(\varSigma =\sigma \).
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Boll, R., Petrera, M., Suris, Y.B. (2016). On the Variational Interpretation of the Discrete KP Equation. In: Bobenko, A. (eds) Advances in Discrete Differential Geometry. Springer, Berlin, Heidelberg. https://doi.org/10.1007/9783662504475_12
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