# On induced L-infinity action of diffeomorphisms on Cochains

Andrey Losev  Shanghai Institute for Mathematics and Interdisciplinary
Sciences  Dmitrii Sheptunov  Moscow Institute of Physics and Technology  Xin
Geng  University of Science and Technology of China

###### Abstract

One of the approaches to quantum gravity is to formulate it in terms of De
Rham algebra, choose a triangulation of space-time, and replace differential
forms by cochains (that form a finite dimensional vector space). The key issue
of general relativity is the action of diffeomorphisms of space-time on
fields. In this paper, we induce the action of diffeomorphisms on cochains by
homotopy transfer (or, equivalently, BV integral) that leads to a $L_{\infty}$
action. We explicitly compute this action for the space-time being an
interval, a circle, and a square.

##  1 Introduction

The quantum theory of gravity is one of the main goals of theoretical physics.
The main problem is in ultraviolet divergences that are present due to the
infinite dimension of the space of differential forms on the smooth manifold.
Therefore, to quantize gravity in the spirit of Feynman integral, we should
modify the space of differential forms. In ([1]) we proposed the replacement
of the differential forms by A-infinity algebra, that was called Feynman
geometry.

The particular class of Feynman geometries may be obtained in the following
construction. Consider triangulation of a manifold. Decompose the space of
differential forms as complexes as a direct sum of an infinite-dimensional
complex of differential forms that have zero integrals against all chains of
the triangulation, and a finite dimensional complex of cochains dual to such
chains of the triangulation. The latter would be a replacement of the algebra
of differential forms. In works of Mnev ([2, 3]) and Sullivan ([4]) it was
shown how wedge product on differential forms leads to higher products of
cochains in the procedure known as homotopic transfer, which in mathematical
physics is just a BV integral against a Lagrangian submanifold determined by
the homotopy.

Since we are planning to study gravity the important issue is the general
covariance. In particular, we should know how diffeomorphisms are acting on
cochains. The hint is that we know that diffeomorphisms are acting on
differential forms by Lie derivative. Thus, in this paper, we will apply the
BV integral (homotopic transfer) to get the action of vector fields on
manifold on cochains. It is not surprising that we will get not the
representation of the Lie algebra but L-infinity representation. This
construction in the context of transferring the action of supersymmetry to
cohomology was earlier done in ([5]).

We believe that explicit formulas for such representation would be an
important stepping stone in understanding of quantum gravity in this approach.

The paper is organized as follows. In section 2 we review relation between
$L_{\infty}$ representations and solutions to classical master equations (CME)
of the special form. We also recall in this section that BV integral leads to
induction of $L_{\infty}$ representation (also known as homotopic transfer).
These results are not new, but we present them for completeness.

In section 3 we proceed to explicit calculation of effective BV-action. To
begin with, we consider an interval as a manifold. Further, in sections 4 and
5, explicit formulas for the action in the case of a circle and a square are
obtained.

##  2 BV formalism, $L_{\infty}$ module structure, BV integral and homotopic
transfer

###  2.1 How to understand $L_{\infty}$ module

The concept of $L_{\infty}$ module ([6]) could be easily understood by analogy
with the concept of a module over a Lie algebra. The latter may be explained
as follows. Take a Lie algebra $L$, a vector space $M$ (considered as an
abelian Lie algebra) and study $L\oplus M$. Now deform the Lie algebra
structure on $L\oplus M$ considered as a vector space keeping the structure
constants of $L$ . This would lead to structure constants

$T:\;\;L\otimes M\rightarrow M$

and condition that $T$ gives a structure of the Lie algebra module is
equivalent to Jacobi identity for the deformed Lie algebra.

If we apply the same procedure to $L_{\infty}$ algebra, namely, extend it by a
complex $M$ and deform, we will get a set of maps

$T_{p}:\;\;\bigwedge\,^{p}L\rightarrow End(M)$

subject to quadratic relations that follow from the $L_{\infty}$ relations.

###  2.2 BV formalism

A convenient way to write down these relations and to operate with them is
provided by BV formalism (see [7]).

In this formalism we study a supermanifold equipped with the odd simplectic
structure and a function $S_{BV}$ on it, that is called BV action. $S_{BV}$
should satisfy quantum master equation, but in this work we will restrict
ourselves by its classical limit, that we will write explicitly below.

For our purpose it is enough to consider as a supermanifold $T^{*}[1]W$, where
$W$ is a supervector space and simplectic structure is a canonical one. In BV
language linear coordinates on $W$ are called fields ( and we will denote them
as $\eta^{a}$) and dual linear coordinates on the fiber are called antifields
and we will denote them as $\eta_{a}^{*}$.

The classical master equation (CME) takes the form

$\sum_{a=1}^{dimV}(-1)^{|\eta^{a}|}\Bigg{(}\frac{\partial
S_{BV}}{\partial\eta^{a}}\frac{\partial
S_{BV}}{\partial\eta_{a}^{*}}+\frac{\partial
S_{BV}}{\partial\eta_{a}^{*}}\frac{\partial
S_{BV}}{\partial\eta^{a}}\Bigg{)}=0$ (1)

where $|\eta^{a}|$ is parity of $\eta^{a}$.

Using BV formalism is useful because one can reformulate the concept of
$L_{\infty}$ algebra and the concept of $L_{\infty}$ module as a particular
solutions to CME. Moreover, the BV formalism contains the procedure of BV
integral allowing to get new solutions of CME from the old ones.

Namely, if $W=W_{1}\oplus W_{2}$, and $\mathcal{L}$ is the Lagrangian
submanifold in $T^{*}[1]W_{1}$, then the integral

$\int_{\mathcal{L}}\exp(S_{BV}/\hbar)=\exp(S_{BV}^{ind}(\eta_{2},\eta_{2}^{*},\hbar)/\hbar)$
(2)

produces the induced BV action $S_{BV}^{ind}(\hbar)$ that satisfies quantum
master equation. For our purpose it is enough to consider the value of
$S^{ind}$ at $\hbar=0$ that gives a solution to CME on the space
$T^{*}[1]W_{2}$. Note, that

$S_{BV}^{ind}(\eta_{2},\eta_{2}^{*},0)=extr_{\mathcal{L}}S_{BV}(\eta,\eta^{*})$
(3)

###  2.3 $L_{\infty}$ algebra and $L_{\infty}$ module from the BV point of
view

Consider the Lie algebra $L$ with a basis { $t_{a}$ } and structure constants
$f_{ab}^{e}$:

$[t_{a},t_{b}]=f_{ab}^{d}t_{d}$ (4)

Consider $L[1]$, it means that $L[1]$ is an odd space with odd coordinates
$c^{a}$.

Correspondingly, dual coordinates on the fiber $T^{*}[1](L[1])$ are even and
according to our conventions we denote them as $c_{a}^{*}$

It is easy to show that CME for the action

$S_{L}=f_{ab}^{d}c^{a}c^{b}c_{d}^{*}$ (5)

is equivalent to Jacobi identity for the structure constants $f_{ab}^{d}$.

Consider an (even) vector space $M$ with the basis $e_{i}$ and linear
coordinates $\varphi^{i}$. The fiber $T^{*}[1]M$ has odd dual coordinates
$\varphi_{i}^{*}$. It is equally easy to check that $M$ is a module over Lie
algebra iff the map

$L\rightarrow End(M),t_{a}\mapsto T_{aj}^{i}$ (6)

is such that

$S_{M}=c^{a}T_{aj}^{i}\varphi_{i}^{*}\varphi^{j}+\frac{1}{2}f_{ab}^{d}c^{a}c^{b}c_{d}^{*}$
(7)

solves CME.

Note, that the latter case may be considered as a particular case of the
former one for the superalgebra obtained by extension of $L$ by the odd module
$M$.

It is instructive to notice that the $L_{\infty}$ algebra structure, i.e. the
collection of maps

$\bigwedge L\rightarrow L,\;\;\;t_{a_{1}}\wedge\ldots\wedge t_{a_{p}}\mapsto
f_{a_{1}\ldots a_{p}}^{b}t_{b}$ (8)

satisfying quadratic relations can be packed into a solution to CME for

$S_{L_{\infty}}=\sum\limits_{p=1}^{n}f_{a_{1}\ldots a_{p}}^{b}c^{a_{1}}\ldots
c^{a_{p}}c_{b}^{*}$ (9)

Now we are ready to explain the concept of $L_{\infty}$ module in BV
formalism.

Let us replace a supervector space $M$ by a complex with the differential $Q$
and introduce the matrix $Q_{i}^{j}$ as follows:

$Q(e_{i})=Q_{i}^{j}e_{j},\;\;\;Q^{2}=0$ (10)

Define a map

$\bigwedge L\rightarrow End(M),\;\;t_{a_{1}}\wedge\ldots\wedge
t_{a_{p}}\mapsto(T_{a_{1}\ldots a_{p}})_{i}^{j}$ (11)

Then one can show that this map gives a stucture of $L_{\infty}$ module iff
the follwing action $S_{M_{\infty}}$ solves CME:

$S_{M_{\infty}}=\varphi^{*}_{i}Q^{i}_{j}\varphi^{j}+\sum_{p=1}^{n}\frac{1}{p!}c^{a_{1}}\cdots
c^{a_{p}}(T_{a_{1}\cdots
a_{p}})^{i}_{j}\varphi^{*}_{i}\varphi^{j}+\frac{1}{2}f_{ab}^{d}c^{a}c^{b}c^{*}_{d}$
(12)

Like in the case of ordinary module over a Lie algebra, the structure of
$L_{\infty}$ module may be considered as a special case of $L_{\infty}$
algebra on the space $L\oplus M$.

###  2.4 Induction of the $L_{\infty}$ module by contraction of the acyclic
subcomplex in the BV formalism

Kadeishvili-like theorems ([8]) state that by contraction of the acyclic
subcomplex we can pass from the Lie algebra to $L_{\infty}$ algebra. Here we
will describe how in works in the particular case of $L$ modules and
$L_{\infty}$ modules.

Let us apply the BV induction procedure for the Lie algebra acting on a
complex $M$ with the action commuting with the differential $Q$ in the
complex:

$[T_{a},Q]=0$ (13)

Let $M$ be a direct sum of two subcomplexes:

$M=M_{Z}\oplus M_{A},$ (14)

such that $M_{A}$ is acyclic, and $h$ is the contracting homotopy:

$hQ+Qh=Proj_{M_{A}},\;\;h^{2}=0,\;\;hM_{Z}=0$ (15)

Take

$W_{1}=M_{A},\;\;W_{2}=L[1]\oplus M_{Z}$ (16)

and consider linear Lagrangian submanifold given by equations

$hw=0;\;\;h^{*}w^{*}=0$ (17)

where $h^{*}$ is a conjugated homotopy operator on the dual space.

Since $S_{M}$ is quadratic polynomial in $w_{A}\in M_{A}$ and $w_{A}^{*}\in
M_{A}^{*}[1]$, it is easy to compute its critical value on the Lagrangian
submanifold.

It is given by

$S^{ind}=<w_{Z}^{*},Qw_{Z}>+\frac{1}{2}f_{ab}^{d}c^{a}c^{b}c_{d}^{*}+\sum_{k=1}^{\infty}c^{a_{1}}\ldots
c^{a_{k}}<w_{Z}^{*},T_{a_{1}}h\ldots hT_{a_{k}}w_{Z}>$ (18)

where $w_{Z}\in M_{Z}$ and $w_{Z}^{*}\in M_{Z}^{*}[1]$. If we introduce an
universal action operator

$T_{c}=c^{a}T_{a}$

the formula above takes the simple form

$S^{ind}=<w_{Z}^{*},Qw_{Z}>+\frac{1}{2}f_{ab}^{d}c^{a}c^{b}c_{d}^{*}+\sum_{k=1}^{\infty}<w_{Z}^{*},\underbrace{T_{c}h\ldots
hT_{c}}_{\text{k times}}w_{Z}>$ (19)

Thus we have an induced structure of $L_{\infty}$ module also known as
homotopy transfer, see for example ([9]).

###  2.5 Application to the action of diffeomorphisms on cochains

In our approach to discretized gravity, we take as a module $M$ the space
$\Omega_{X}$ of differential forms on a manifold $X$, as a differential - De
Rham operator $d$ on these forms, as a Lie algebra - the algebra of
diffeomorphisms on $X$.

Now we proceed to discretization. We consider a triangulation of $X$, and take
as $M_{A}$ the subspace $\Omega_{A}$ of differential forms $\omega$ having
zero integrals against all the chains $z$ of the triangulation:

$\int_{z}\omega=0$ (20)

As a second summand we take differential forms $\Omega_{Z}$ dual to chains of
the triangulation. Such choice is not unique. Whitney considered polynomial
forms. However, in this paper we allow more general choices,restricted by
condition that

$d\Omega_{Z}\in\Omega_{Z}$

Applying the construction described above, we get $L_{\infty}$ module
structure on $\Omega_{Z}$. In the following sections we will study explicit
examples.

##  3 Interval

In this section we consider the manifold $X=[0,1]$ with the coordinate $t$,
$0\leq t\leq 1$. The corresponding Lie algebra $L=Vect([0,1])$ is spanned by

$v_{k}=t^{k}\frac{d}{dt},k\in\mathbb{Z},k\geq 0$ (21)

The structure constants can be written in this basis as

$\displaystyle[v_{k},v_{l}]=(l-k)t^{k+l-1}\frac{d}{dt}=(l-k)v_{k+l-1}$ (22)
$\displaystyle f_{kl}^{m}=(l-k)\delta_{k+l-1}^{m}$ (23)

We will also use the notation like in (2.4)

$\displaystyle
v_{c}=\sum\limits_{k}c^{k}v_{k}=\sum\limits_{k}c^{k}t^{k}\frac{d}{dt}=\nu_{c}(t)\frac{d}{dt}$
(24)

The action of $v_{c}\in L[1]$ on $\Omega_{X}$ is given by the Lie derivative
which we denote $\mathcal{L}_{v_{c}}$. So the action (12) takes the form

$S=\langle\varphi^{*},d\varphi\rangle+(-1)^{|\varphi^{*}|}\langle\varphi^{*},\mathcal{L}_{v_{c}}\varphi\rangle+\frac{1}{2}f_{ab}^{d}c^{a}c^{b}c_{d}^{*}$
(25)

We construct the triangulation of $[0,1]$ as follows. Choose $n+1$ points
$t_{0}<t_{1}<...<t_{n}$ such that $t_{0}=0,t_{n}=1$. Using this we set the
space of chains

$Z=Span\langle\alpha^{0}=[t_{0}],...,\alpha^{n}=[t_{n}],\beta^{0}=[t_{0},t_{1}],...,\beta^{n-1}=[t_{n-1},t_{n}]\rangle$
(26)

Further consider $n$ functions $\alpha_{1},...,\alpha_{n}$ with the condition
$\alpha_{j}(t_{i})=\delta_{j}^{i}$ and define
$\alpha_{0}=1-\sum\limits_{k=1}^{n}\alpha_{k}$. 1-forms are introduced as

$\beta_{j}=d\Bigg{(}\sum\limits_{k=j+1}^{n}\alpha_{k}\Bigg{)},0\leq j\leq n-1$
(27)

Note that:

$\int\limits_{t_{i}}^{t_{i+1}}\beta_{j}=\int\limits_{t_{i}}^{t_{i+1}}d\Bigg{(}\sum\limits_{k=j+1}^{n}\alpha_{k}\Bigg{)}=\sum\limits_{k=j+1}^{n}\alpha_{k}(t_{i+1})-\alpha_{k}(t_{i})=\sum\limits_{k=j+1}^{n}\delta_{k}^{i+1}-\delta_{k}^{i}=\delta_{j}^{i}$
(28)

Hence we get
$\langle\alpha^{i},\alpha_{j}\rangle=\langle\beta^{i},\beta_{j}\rangle=\delta_{j}^{i}$,
$\langle\alpha^{i},\beta_{j}\rangle=\langle\beta^{i},\alpha_{j}\rangle=0$,
where $\langle,\rangle$ is the pairing between chains and differential forms
given by integration, and the space of forms dual to chains can be written as:

$\Omega_{Z}=Span(\alpha_{0},...,\alpha_{n},\beta_{0},...,\beta_{n-1})$ (29)

Therefore, fields and antifields are

$\displaystyle\varphi_{Z}=\sum\limits_{i=0}^{n}\alpha_{i}\varphi^{i}+\sum\limits_{j=0}^{n-1}\beta_{j}\psi^{j}$
(30)
$\displaystyle\varphi_{Z}^{*}=\sum\limits_{i=0}^{n}\varphi_{i}^{*}\alpha^{i}+\sum\limits_{j=0}^{n-1}\psi_{j}^{*}\beta^{j}$
(31)

To construct the $L_{\infty}$ module structure on $\Omega_{Z}$, we need to
provide a homotopy $h$ satisfying (15). We write this condition in the form
$dh+hd=id-P_{Z}$, where the projector $P_{Z}:\Omega_{X}\to\Omega_{Z}$ is
defined as

$\displaystyle P_{Z}(g(t))=\sum\limits_{j}g(t_{j})\alpha_{j}$ (32)

for a 0-form $g(t)$ and

$\displaystyle
P_{Z}(f(t)dt)=\sum\limits_{j}\Bigg{(}\int\limits_{t_{j}}^{t_{j+1}}f(\tau)d\tau\Bigg{)}\beta_{j}$
(33)

for a 1-form $f(t)dt$.

Let us set now

$\displaystyle
h(f(t)dt)=\int\limits_{0}^{t}f(\tau)d\tau-\sum\limits_{j}\Bigg{(}\int\limits_{t_{j}}^{t_{j+1}}f(\tau)d\tau\Bigg{)}\int\limits_{0}^{t}\beta_{j}$
(34)

and check the relation $dh+hd=id-P_{Z}$ for 0-forms:

$\displaystyle(dh+hd)g(t)$
$\displaystyle=h(g^{\prime}(t)dt)=\int\limits_{0}^{t}g^{\prime}(\tau)d\tau-\sum\limits_{j}\Bigg{(}\int\limits_{t_{j}}^{t_{j+1}}g^{\prime}(\tau)d\tau\Bigg{)}\int\limits_{0}^{t}\beta_{j}=$
(35)
$\displaystyle=g(t)-g(0)-\sum\limits_{j}\Big{(}g(t_{j+1})-g(t_{j})\Big{)}\int\limits_{0}^{t}d\Bigg{(}\sum\limits_{k=j+1}^{n}\alpha_{k}\Bigg{)}=$
(36)
$\displaystyle=g(t)-g(0)-\sum\limits_{j}\Big{(}g(t_{j+1})-g(t_{j})\Big{)}\sum\limits_{k=j+1}^{n}\alpha_{k}(t)=$
(37)
$\displaystyle=g(t)-g(0)-\sum\limits_{k=1}^{n}\alpha_{k}(t)\sum\limits_{j=0}^{k-1}\Big{(}g(t_{j+1})-g(t_{j})\Big{)}=$
(38)
$\displaystyle=g(t)-g(0)-\sum\limits_{k=1}^{n}\alpha_{k}(t)\Big{(}g(t_{k})-g(t_{0})\Big{)}=$
(39)
$\displaystyle=g(t)-g(t_{0})(1-\sum\limits_{k=1}^{n}\alpha_{k})-\sum\limits_{k=1}^{n}g(t_{k})\alpha_{k}=(id-
P_{Z})g(t)$ (40)

and for 1-forms:

$\displaystyle(dh+hd)f(t)dt$
$\displaystyle=d\Bigg{[}\int\limits_{0}^{t}f(\tau)d\tau-\sum\limits_{j}\Bigg{(}\int\limits_{t_{j}}^{t_{j+1}}f(\tau)d\tau\Bigg{)}\int\limits_{0}^{t}\beta_{j}\Bigg{]}=$
(41)
$\displaystyle=f(t)dt-\sum\limits_{j}\Bigg{(}\int\limits_{t_{j}}^{t_{j+1}}f(\tau)d\tau\Bigg{)}\beta_{j}=(id-
P_{Z})f(t)dt$ (42)

Finally, the induced action (18) becomes

$S^{ind}=\langle\varphi^{*}_{Z},d\varphi_{Z}\rangle+(-1)^{|\varphi_{Z}^{*}|}\langle\varphi^{*}_{Z},\mathcal{L}_{v_{c}}\varphi_{Z}\rangle+\langle\varphi^{*}_{Z},\mathcal{L}_{v_{c}}h\mathcal{L}_{v_{c}}\varphi_{Z}\rangle+\frac{1}{2}f_{ab}^{d}c^{a}c^{b}c_{d}^{*}$
(43)

All the terms containing more than one operator $h$ vanish since $h$ lowers
form’s degree by 1.

To compute the induced action we introduce notations

$\displaystyle\beta_{j}^{i}=\sum\limits_{k=j+1}^{n}\alpha^{\prime}_{k}(t_{i})$
(44) $\displaystyle\alpha^{j}\nu_{c}=\nu_{c}(t_{j})=\nu_{c}^{j}$ (45)
$\displaystyle\alpha^{j}\nu^{\prime}_{c}=\nu^{\prime}_{c}(t_{j})=\nu^{\prime
j}_{c}$ (46)

What remains is to calculate the distinct components of the expression for the
induced action

$\displaystyle\langle\beta^{j},d\alpha_{i}\rangle=\int\limits_{t_{j}}^{t_{j+1}}d\alpha_{i}=\alpha_{i}(t_{j+1})-\alpha_{i}(t_{j})=\delta_{i}^{j+1}-\delta_{i}^{j}$
(47)
$\displaystyle\langle\alpha^{i},\mathcal{L}_{v_{c}}\alpha_{j}\rangle=\alpha^{i}\big{(}\nu_{c}(t)\frac{d\alpha_{j}}{dt}\big{)}=\nu_{c}(t_{i})\alpha^{\prime}_{j}(t_{i})=\nu_{c}^{i}(\beta_{j-1}^{i}-\beta_{j}^{i})$
(48)
$\displaystyle\begin{split}\langle\beta^{i},\mathcal{L}_{v_{c}}\beta_{j}\rangle&=\beta^{i}d\Bigg{(}\nu_{c}(t)\sum\limits_{k=j+1}^{n}\alpha^{\prime}_{k}\Bigg{)}=\int\limits_{t_{i}}^{t_{i+1}}d\Bigg{(}\nu_{c}(t)\sum\limits_{k=j+1}^{n}\alpha^{\prime}_{k}\Bigg{)}\\\
&=\sum\limits_{k=j+1}^{n}\nu_{c}(t_{i+1})\alpha^{\prime}_{k}(t_{i+1})-\nu_{c}(t_{i})\alpha^{\prime}_{k}(t_{i})=\nu_{c}^{i+1}\beta_{j}^{i+1}-\nu_{c}^{i}\beta_{j}^{i}\\\
\end{split}$ (49)
$\displaystyle\begin{split}\langle\alpha^{i},\mathcal{L}_{v_{c}}h\mathcal{L}_{v_{c}}\beta_{j}\rangle&=\alpha^{i}\mathcal{L}_{v_{c}}hd\Bigg{(}\nu_{c}(t)\sum\limits_{k=j+1}^{n}\alpha^{\prime}_{k}\Bigg{)}=\alpha^{i}\mathcal{L}_{v_{c}}(id-
P_{Z})\Bigg{(}\nu_{c}(t)\sum\limits_{k=j+1}^{n}\alpha^{\prime}_{k}\Bigg{)}=\\\
&=\alpha^{i}\mathcal{L}_{v_{c}}\Bigg{(}\nu_{c}(t)\sum\limits_{k=j+1}^{n}\alpha^{\prime}_{k}-\sum\limits_{m}\nu_{c}^{m}\beta_{j}^{m}\alpha_{m}\Bigg{)}=\\\
&=\nu_{c}^{i}\nu^{\prime
i}_{c}\beta_{j}^{i}-\nu_{c}^{i}\sum\limits_{m}\nu_{c}^{m}\beta_{j}^{m}(\beta_{m-1}^{i}-\beta_{m}^{i})\\\
\end{split}$ (50)

Thus, the resulting $L_{\infty}$ module structure is

$\displaystyle\begin{split}S^{ind}&=\psi_{i}^{*}(\delta_{j}^{i+1}-\delta_{j}^{i})\varphi^{j}-\varphi_{i}^{*}\nu_{c}^{i}(\beta_{j-1}^{i}-\beta_{j}^{i})\varphi^{j}+\psi_{i}^{*}(\nu_{c}^{i+1}\beta_{j}^{i+1}-\nu_{c}^{i}\beta_{j}^{i})\psi^{j}+\\\
&+\varphi_{i}^{*}\Big{(}\nu_{c}^{i}\nu^{\prime
i}_{c}\beta_{j}^{i}-\nu_{c}^{i}\sum\limits_{m}\nu_{c}^{m}\beta_{j}^{m}(\beta_{m-1}^{i}-\beta_{m}^{i})\Big{)}\psi^{j}+\frac{1}{2}f_{ab}^{d}c^{a}c^{b}c_{d}^{*}\end{split}$
(51)

##  4 Circle

In this section we consider the manifold $X=S^{1}$ with the coordinate $0\leq
t<2\pi$ and the Lie algebra $L=Vect(S^{1})$. Let

$v_{k}=e^{ikt}\frac{d}{dt},k\in\mathbb{Z}$ (52)

be a basis in $L=Vect(S^{1})$. Then, the corresponding structure constants are

$\displaystyle[v_{k},v_{l}]=i(l-k)e^{i(m+n)t}\frac{d}{dt}=i(l-k)v_{k+l}$ (53)
$\displaystyle f_{kl}^{m}=i(l-k)\delta_{k+l}^{m}$ (54)

Let us use the following notation for the subsequent computations

$\displaystyle
v_{c}=\sum\limits_{k}c^{k}v_{k}=\sum\limits_{k}c^{k}e^{ikt}\frac{d}{dt}=\nu_{c}(t)\frac{d}{dt}$
(55)

The action (12) takes the form analogous to the interval case

$S=\langle\varphi^{*},d\varphi\rangle+(-1)^{|\varphi^{*}|}\langle\varphi^{*},\mathcal{L}_{v_{c}}\varphi\rangle+\frac{1}{2}f_{ab}^{d}c^{a}c^{b}c_{d}^{*}$
(56)

The triangulation of $S^{1}$ is constructed as follows. Choose $n$ points
$t_{0}=0,...,t_{n-1}$ and 1-forms $\beta_{0},...,\beta_{n-1}$ such that
$\int\limits_{t_{i}}^{t_{i+1}}\beta_{j}=\delta_{j}^{i}$. Denote
$\beta_{j}^{i}:=\vartheta_{j}(t_{i})$, where $\beta_{j}=\vartheta_{j}(t)dt$.
Define the 0-forms:

$\displaystyle\alpha_{0}(t)=1+\int\limits_{0}^{t}(\beta_{n-1}-\beta_{0})$ (57)
$\displaystyle\alpha_{j}(t)=\int\limits_{0}^{t}(\beta_{j-1}-\beta_{j}),j\neq
0$ (58)

One can check that $\alpha_{j}(t_{i})=\delta_{j}^{i}$.

Then, the spaces of chains and cochains are

$\displaystyle
Z=Span\langle\alpha^{0}=[t_{0}],...,\alpha^{n-1}=[t_{n-1}],\beta^{0}=[t_{0},t_{1}],...,\beta^{n-1}=[t_{n-1},t_{0}]\rangle$
(59)
$\displaystyle\Omega^{\bullet}_{Z}=Span\langle\alpha_{0},...,\alpha_{n-1},\beta_{0},...,\beta_{n-1}\rangle$
(60)

The corresponding fields and antifields are

$\displaystyle\varphi_{Z}^{*}=\sum\limits_{i=0}^{n-1}\varphi_{i}^{*}\alpha^{i}+\sum\limits_{j=0}^{n-1}\psi_{j}^{*}\beta^{j}$
(61)
$\displaystyle\varphi_{Z}=\sum\limits_{i=0}^{n-1}\alpha_{i}\varphi^{i}+\sum\limits_{j=0}^{n-1}\beta_{j}\psi^{j}$
(62)

As mentioned earlier in section 3, it is necessary to provide $h$ satisfying
the condition $dh+hd=id-P_{Z}$, where the projector
$P_{Z}:\Omega_{X}\to\Omega_{Z}$ is defined in this case as

$\displaystyle P_{Z}(g(t))=\sum\limits_{j}g(t_{j})\alpha_{j}$ (63)

for a 0-form $g(t)$ and

$\displaystyle
P_{Z}(f(t)dt)=\sum\limits_{j}\Bigg{(}\int\limits_{t_{j}}^{t_{j+1}}f(\tau)d\tau\Bigg{)}\beta_{j}$
(64)

for a 1-form $f(t)dt$.

Let us set the homotopy $h$

$\displaystyle
h(f(t)dt)=\int\limits_{0}^{t}f(\tau)d\tau-\sum\limits_{j}\Bigg{(}\int\limits_{t_{j}}^{t_{j+1}}f(\tau)d\tau\Bigg{)}\int\limits_{0}^{t}\beta_{j}$
(65)

and check the the relation $dh+hd=id-P_{Z}$ for 1-forms:

$\displaystyle\begin{split}(dh+hd)(f(t)dt)&=d\Bigg{[}\int\limits_{0}^{t}f(\tau)d\tau-\sum\limits_{j}\Bigg{(}\int\limits_{t_{j}}^{t_{j+1}}f(\tau)d\tau\Bigg{)}\int\limits_{0}^{t}\beta_{j}\Bigg{]}=\\\
&=f(t)dt-\sum\limits_{j}\Bigg{(}\int\limits_{t_{j}}^{t_{j+1}}f(\tau)d\tau\Bigg{)}\beta_{j}=(id-
P_{Z})f(t)dt\\\ \end{split}$ (66)

and for 0-forms:

$\displaystyle\begin{split}(dh+hd)(g(t))&=\int\limits_{0}^{t}g^{\prime}(\tau)d\tau-\sum\limits_{j}\Bigg{(}\int\limits_{t_{j}}^{t_{j+1}}g^{\prime}(\tau)d\tau\Bigg{)}\int\limits_{0}^{t}\beta_{j}=\\\
&=g(t)-g(0)-\sum\limits_{j}(g(t_{j+1})-g(t_{j}))\int\limits_{0}^{t}\beta_{j}=\\\
&=g(t)-g(0)-g(0)\Bigg{(}\int\limits_{0}^{t}(\beta_{n-1}-\beta_{0})\Bigg{)}-\\\
&-\sum\limits_{j\neq
0}g(t_{j})\Bigg{(}\int\limits_{0}^{t}(\beta_{j-1}-\beta_{j})\Bigg{)}=\\\
&=g(t)-\sum\limits_{j}g(t_{j})\alpha_{j}=(id-P_{Z})g(t)\end{split}$ (67)

The induced action (18) is analogous to the case of the interval

$S^{ind}=\langle\varphi^{*}_{Z},d\varphi_{Z}\rangle+(-1)^{|\varphi_{Z}^{*}|}\langle\varphi^{*}_{Z},\mathcal{L}_{v_{c}}\varphi_{Z}\rangle+\langle\varphi^{*}_{Z},\mathcal{L}_{v_{c}}h\mathcal{L}_{v_{c}}\varphi_{Z}\rangle+\frac{1}{2}f_{ab}^{d}c^{a}c^{b}c_{d}^{*}$
(68)

We introduce notations

$\displaystyle\nu_{c}(t_{j})=\nu_{c}^{j}$ (69)
$\displaystyle\nu^{\prime}_{c}(t_{j})=\nu^{\prime j}_{c}$ (70)

Finally, let us compute the action

$\displaystyle d\alpha_{j}=\beta_{j-1}-\beta_{j}$ (71)
$\displaystyle\langle\beta^{i},d\alpha_{j}\rangle=\delta_{j-1}^{i}-\delta_{j}^{i}$
(72)
$\displaystyle\langle\alpha^{i},\mathcal{L}_{v_{c}}\alpha_{j}\rangle=\nu_{c}^{i}(\beta_{j-1}^{i}-\beta_{j}^{i})$
(73)
$\displaystyle\langle\beta^{i},\mathcal{L}_{v_{c}}\beta_{j}\rangle=\int\limits_{t_{i}}^{t_{i+1}}d\Big{(}\nu_{c}(t)\vartheta_{j}(t)\Big{)}=\nu_{c}^{i+1}\beta_{j}^{i+1}-\nu_{c}^{i}\beta_{j}^{i}$
(74)
$\displaystyle\begin{split}h\mathcal{L}_{v_{c}}\beta_{j}&=\int\limits_{0}^{t}d\Big{(}\nu_{c}\vartheta_{j}\Big{)}-\sum\limits_{m}\Bigg{(}\int\limits_{t_{m}}^{t_{m+1}}d\Big{(}\nu_{c}\vartheta_{j}\Big{)}\Bigg{)}\int\limits_{0}^{t}\beta_{m}=\\\
&=\nu_{c}(t)\vartheta_{j}(t)-\nu_{c}^{0}\beta_{j}^{0}-\sum\limits_{m}\Big{(}\nu_{c}^{m+1}\beta_{j}^{m+1}-\nu_{c}^{m}\beta_{j}^{m}\Big{)}\int\limits_{0}^{t}\beta_{m}=\\\
&=\nu_{c}(t)\vartheta_{j}(t)-\sum\limits_{m}\nu_{c}^{m}\beta_{j}^{m}\alpha_{m}(t)\\\
\end{split}$ (75)
$\displaystyle\begin{split}\mathcal{L}_{v_{c}}h\mathcal{L}_{v_{c}}\beta_{j}&=\nu_{c}(t)\frac{\partial}{\partial
t}\Big{(}\nu_{c}(t)\vartheta_{j}(t)-\sum\limits_{m}\nu_{c}^{m}\beta_{j}^{m}\alpha_{m}(t)\Big{)}=\\\
&=\nu_{c}(t)\nu^{\prime}_{c}(t)\vartheta_{j}(t)+\nu_{c}(t)\nu_{c}(t)\vartheta^{\prime}_{j}(t)-\\\
&-\nu_{c}(t)\sum\limits_{m}\nu_{c}^{m}\beta_{j}^{m}(\vartheta_{m-1}(t)-\vartheta_{m}(t))\\\
\end{split}$ (76)
$\displaystyle\langle\alpha^{i},\mathcal{L}_{v_{c}}h\mathcal{L}_{v_{c}}\beta_{j}\rangle=\nu_{c}^{i}\nu^{\prime
i}_{c}\beta_{j}^{i}-\nu_{c}^{i}\sum\limits_{m}\nu_{c}^{m}\beta_{j}^{m}(\beta_{m-1}^{i}-\beta_{m}^{i})$
(77)

Thus, we get the $L_{\infty}$ module structure

$\displaystyle\begin{split}S^{ind}&=\psi_{i}^{*}(\delta_{j-1}^{i}-\delta_{j}^{i})\varphi^{j}-\varphi_{i}^{*}\nu_{c}^{i}(\beta_{j-1}^{i}-\beta_{j}^{i})\varphi^{j}+\psi_{i}^{*}(\nu_{c}^{i+1}\beta_{j}^{i+1}-\nu_{c}^{i}\beta_{j}^{i})\psi^{j}+\\\
&+\varphi_{i}^{*}\Big{(}\nu_{c}^{i}\nu^{\prime
i}_{c}\beta_{j}^{i}-\nu_{c}^{i}\sum\limits_{m}\nu_{c}^{m}\beta_{j}^{m}(\beta_{m-1}^{i}-\beta_{m}^{i})\Big{)}\psi^{j}+\frac{1}{2}f_{ab}^{d}c^{a}c^{b}c_{d}^{*}\end{split}$
(78)

##  5 Square

In this section we consider the manifold $X=[0,1]\times[0,1]$ with the
coordinates $x,y$. The vector field $v_{c}$ takes the form

$\displaystyle v_{c}(x,y)=\nu_{c}^{x}(x,y)\frac{\partial}{\partial
x}+\nu_{c}^{y}(x,y)\frac{\partial}{\partial y},$ (79)
$\displaystyle\nu_{c}^{x}=\sum\limits_{n,m}c^{x,nm}x^{n}y^{m},$ (80)
$\displaystyle\nu_{c}^{y}=\sum\limits_{n,m}c^{y,nm}x^{n}y^{m}$ (81)

The triangulation is given by

$\displaystyle\gamma^{*}=[0,1]\times[0,1]$ (82)
$\displaystyle\beta^{0}=[(0,0),(1,0)],\beta^{1}=[(1,0),(1,1)],\beta^{2}=[(1,1),(0,1)],\beta^{3}=[(0,1),(0,0)]$
(83)
$\displaystyle\alpha^{0}=(0,0),\alpha^{1}=(1,0),\alpha^{2}=(1,1),\alpha^{3}=(0,1)$
(84) $\displaystyle
Z=Span(\alpha^{0},...,\alpha^{3},\beta^{0},...,\beta^{3},\gamma^{*})$ (85)

Let us define the constructions that willl be used in further calculations:
2-form $\gamma=dxdy$; 1-forms:

$\displaystyle\beta_{0}=\Bigg{(}\int\limits_{y}^{1}d\tilde{y}\Bigg{)}dx=(1-y)dx$
(86)
$\displaystyle\beta_{1}=\Bigg{(}\int\limits_{0}^{x}d\tilde{x}\Bigg{)}dy=xdy$
(87)
$\displaystyle\beta_{2}=-\Bigg{(}\int\limits_{0}^{y}d\tilde{y}\Bigg{)}dx=-ydx$
(88)
$\displaystyle\beta_{3}=-\Bigg{(}\int\limits_{x}^{1}d\tilde{x}\Bigg{)}dy=-(1-x)dy;$
(89)

and functions:

$\displaystyle\alpha_{0}=1+\int(\beta_{3}-\beta_{0})=(1-x)(1-y)$ (90)
$\displaystyle\alpha_{1}=\int(\beta_{0}-\beta_{1})=x(1-y)$ (91)
$\displaystyle\alpha_{2}=\int(\beta_{1}-\beta_{2})=xy$ (92)
$\displaystyle\alpha_{3}=\int(\beta_{2}-\beta_{3})=y(1-x).$ (93)

Since $d\beta_{j}=\gamma$, forms $\beta_{i}-\beta_{j}$ are closed, and the
functions $\alpha_{i}$ are defined correctly. Note that
$\langle\gamma^{*},\gamma\rangle=1,\langle\beta^{i},\beta_{j}\rangle=\delta_{j}^{i},\langle\alpha^{i},\alpha_{j}\rangle=\delta_{j}^{i}$.
Then the space dual to the space of chains is

$\displaystyle\Omega_{Z}=Span(\alpha_{0},...,\alpha^{3},\beta_{0},...,\beta^{3},\gamma)$
(94)

The fields under consideration are

$\displaystyle\varphi_{Z}=\sum\limits_{i=0}^{3}\alpha_{i}\varphi^{i}+\sum\limits_{j=0}^{3}\beta_{j}\psi^{j}+\gamma\omega$
(95)
$\displaystyle\varphi_{Z}^{*}=\sum\limits_{i=0}^{3}\varphi_{i}^{*}\alpha^{i}+\sum\limits_{j=0}^{3}\psi_{j}^{*}\beta^{j}+\omega^{*}\gamma^{*}$
(96)

We introduce the projector $P_{Z}$ ($g$, $f$ and $e$ are 0, 1 and 2 -forms
respectively) analogously to the previous sections

$\displaystyle P_{Z}(g)=\sum\limits_{j}\langle\alpha^{j},g\rangle\alpha_{j}$
(97) $\displaystyle
P_{Z}(f)=\sum\limits_{j}\langle\beta^{j},f\rangle\beta_{j}$ (98)
$\displaystyle P_{Z}(e)=\langle\gamma^{*},e\rangle\gamma$ (99)

To define the homotopy consider the operator $I:\Omega_{X}\to\Omega_{X}$; for
2-forms:

$\displaystyle
I(e(x,y)dxdy)=\frac{1}{2}\Bigg{(}\int\limits_{0}^{x}e(\tilde{x},y)d\tilde{x}\Bigg{)}dy-\frac{1}{2}\Bigg{(}\int\limits_{0}^{y}e(x,\tilde{y})d\tilde{y}\Bigg{)}dx$
(100)

for 1-forms:

$\displaystyle\begin{split}I(f_{x}(x,y)dx+f_{y}(x,y)dy)=&\frac{1}{2}\Bigg{(}\int\limits_{0}^{x}f_{x}(\tilde{x},0)d\tilde{x}+\int\limits_{0}^{y}f_{y}(x,\tilde{y})d\tilde{y}+\\\
&+\int\limits_{0}^{y}f_{y}(0,\tilde{y})d\tilde{y}+\int\limits_{0}^{x}f_{x}(\tilde{x},y)d\tilde{x}\Bigg{)}\end{split}$
(101)

The last expression is simply the sum of the integrals of an 1-form $f$ along
the edges of the rectangle with vertices $(0,0),(0,x),(0,y),(x,y)$. For
0-forms $g(t)$ $I(g)=0$. Using $I$, we can now define the homotopy (e and f
are 2- and 1- forms respectively):

$\displaystyle h(e)=I(e)-\sum\limits_{i}\langle\beta^{i},I(e)\rangle\beta_{i}$
(102) $\displaystyle
h(f)=I\Big{(}f-\sum\limits_{i}\langle\beta^{i},f\rangle\beta_{i}\Big{)}$ (103)

Let us check the condition (15) in the form $dh+hd=id-P_{Z}$:

$\displaystyle\begin{split}(dh+hd)g(x,y)&=h(dg)=I\Big{(}dg-\sum\limits_{i}\langle\beta^{i},dg\rangle\beta_{i}\Big{)}=\\\
&=g(x,y)-g(0,0)-I\Big{(}\sum\limits_{i}\big{(}\langle\alpha^{i+1},g\rangle-\langle\alpha^{i},g\rangle\big{)}\beta_{i}\Big{)}=\\\
&=g(x,y)-g(0,0)-\sum\limits_{i}\langle\alpha^{i},g\rangle
I(\beta_{i-1}-\beta_{i})=\\\
&=g(x,y)-\sum\limits_{i}\langle\alpha^{i},g\rangle\alpha_{i}=(id-P_{Z})g\\\
\end{split}$ (104)
$\displaystyle\begin{split}(dh+hd)e&=d(h(e))=d\Big{(}I(e)-\sum\limits_{i}\langle\beta^{i},I(e)\rangle\beta_{i}\Big{)}=\\\
&=e-\Big{(}\sum\limits_{i}\langle\beta^{i},I(e)\rangle\Big{)}\gamma=e-\langle\gamma^{*},e\rangle\gamma=(id-
P_{Z})e\\\ \end{split}$ (105) $\displaystyle
hdf=hd(f_{x}dx+f_{y}dy)=h(df)=I(df)-\sum\limits_{i}\langle\beta^{i},I(df)\rangle\beta_{i}$
(106)
$\displaystyle\begin{split}2I(df)&=\Bigg{(}f_{x}(x,y)-f_{x}(x,0)-\int\limits_{0}^{y}\frac{\partial
f_{y}}{\partial x}d\tilde{y}\Bigg{)}dx+\\\
&+\Bigg{(}f_{y}(x,y)-f_{y}(0,y)-\int\limits_{0}^{x}\frac{\partial
f_{x}}{\partial y}d\tilde{x}\Bigg{)}dy\\\ \end{split}$ (107)
$\displaystyle\sum\limits_{i}\langle\beta^{i},I(df)\rangle\beta_{i}=\Big{(}\sum\limits_{i}\langle\beta^{i},f\rangle\Big{)}\frac{\beta_{1}+\beta_{2}}{2}=\langle\gamma^{*},df\rangle
I(\gamma)$ (108)
$\displaystyle\begin{split}2dI(f)&=f_{x}(x,y)dx+f_{y}(x,y)dy+f_{x}(x,0)dx+f_{y}(0,y)dy+\\\
&+\Bigg{(}\int\limits_{0}^{y}\frac{\partial f_{y}}{\partial
x}d\tilde{y}\Bigg{)}dx+\Bigg{(}\int\limits_{0}^{x}\frac{\partial
f_{x}}{\partial y}d\tilde{x}\Bigg{)}dy\\\ \end{split}$ (109) $\displaystyle
I(df)+dI(f)=f$ (110)
$\displaystyle\begin{split}(hd+dh)f&=I(df)+dI(f)-\sum\limits_{i}\langle\beta^{i},I(df)\rangle\beta_{i}-\sum\limits_{i}\langle\beta^{i},f\rangle
dI(\beta_{i})=\\\
&=f-\sum\limits_{i}\langle\beta^{i},f\rangle\beta_{i}-\langle\gamma^{*},df\rangle
I(\gamma)+\sum\limits_{i}\langle\beta^{i},f\rangle I(d\beta_{i})=\\\
&=(1-P_{Z})f+\sum\limits_{i}\langle\beta^{i},f\rangle
I(\gamma)-\langle\gamma^{*},df\rangle I(\gamma)=(1-P_{Z})f\\\ \end{split}$
(111)

Let us compute the components of action

$\displaystyle d\alpha_{j}=\beta_{j-1}-\beta_{j}$ (112)
$\displaystyle\langle\beta^{i},d\alpha_{j}\rangle=\delta_{j-1}^{i}-\delta_{j}^{i}$
(113) $\displaystyle d\beta_{j}=\gamma$ (114)
$\displaystyle\langle\gamma^{*},d\beta_{j}\rangle=1$ (115)
$\displaystyle\begin{split}\langle\alpha^{i},\mathcal{L}_{v_{c}}\alpha_{j}\rangle&=\langle\alpha^{i},\nu_{c}^{x}\frac{\partial}{\partial
x}\int(\beta_{j-1}-\beta_{j})+\nu_{c}^{y}\frac{\partial}{\partial
y}\int(\beta_{j-1}-\beta_{j})\rangle\\\
&=\nu_{i}(\beta_{j-1}^{i}-\beta_{j}^{i})\\\ \end{split}$ (116)
$\displaystyle\begin{split}\langle\beta^{i},\mathcal{L}_{v_{c}}\beta_{j}\rangle&=\langle\beta^{i},d(v_{c}\beta_{j})+i_{v_{c}}d\beta_{j}\rangle=\nu_{i+1}\beta_{j}^{i+1}-\nu_{i}\beta_{j}^{i}+\\\
&+\int\limits_{\beta^{i}}\nu_{c}^{x}dy-\nu_{c}^{y}dx\\\ \end{split}$ (117)
$\displaystyle\langle\gamma^{*},\mathcal{L}_{v_{c}}\gamma\rangle=\langle\gamma^{*},d(\nu_{c}^{x}dy-\nu_{c}^{y}dx)\rangle=\int\limits_{[0,1]^{2}}\mathbf{div}(v_{c})dxdy$
(118)
$\displaystyle\begin{split}\langle\alpha^{i},\mathcal{L}_{v_{c}}h\mathcal{L}_{v_{c}}\beta_{j}\rangle&=\langle\alpha^{i},\mathcal{L}_{v_{c}}h\big{(}d(v_{c}\beta_{j})+i_{v_{c}}d\beta_{j}\big{)}\rangle=\langle\alpha^{i},\mathcal{L}_{v_{c}}(hd+dh)(v_{c}\beta_{j})\rangle+\\\
&+\langle\alpha^{i},\mathcal{L}_{v_{c}}h\big{(}\nu_{c}^{x}dy-\nu_{c}^{y}dx\big{)}\rangle=\langle\alpha^{i},\mathcal{L}_{v_{c}}(v_{c}\beta_{j}-\sum\limits_{m}\nu_{m}\beta_{j}^{m}\alpha_{m})\rangle+\\\
&+\langle\alpha^{i},\mathcal{L}_{v_{c}}h\big{(}\nu_{c}^{x}dy-\nu_{c}^{y}dx\big{)}\rangle\\\
&=\nu_{x,i}\frac{\partial(\nu_{i}\beta_{j})}{\partial
x}+\nu_{y,i}\frac{\partial(\nu_{i}\beta_{j})}{\partial
y}-\sum\limits_{m}\nu_{i}(\beta_{m-1}^{i}-\beta_{m}^{i})\nu_{m}\beta_{j}^{m}+\\\
&+\frac{1}{2}\langle\alpha^{i},\nu_{c}^{x}(x,y)\nu_{c}^{x}(x,0)-\nu_{c}^{x}\int\limits_{0}^{y}\frac{\partial\nu_{c}^{y}}{\partial
x}(x,\tilde{y})d{y}\rangle+\\\
&+\frac{1}{2}\langle\alpha^{i},-\nu_{c}^{y}(x,y)\nu_{c}^{y}(0,y)+\nu_{c}^{y}\int\limits_{0}^{y}\frac{\partial\nu_{c}^{x}}{\partial
x}(\tilde{x},y)d{x}\rangle-\\\
&-\langle\alpha^{i},\sum\limits_{k}\Big{(}\nu_{c}^{x}\frac{\partial
I(\beta_{k})}{\partial x}+\nu_{c}^{y}\frac{\partial I(\beta_{k})}{\partial
y}\Big{)}\int\limits_{\beta^{k}}\nu_{c}^{x}dy-\nu_{c}^{y}dx\rangle\\\
\end{split}$ (119)

Denote

$\displaystyle\begin{split}\xi:=\mathcal{L}_{v_{c}}h\mathcal{L}_{v_{c}}\gamma&=\mathcal{L}_{v_{c}}hd(\nu_{c}^{x}dy-\nu_{c}^{y}dx)=\mathcal{L}_{v_{c}}h\big{(}\mathbf{div}(v_{c})dxdy\big{)}\\\
&=\frac{1}{2}\Bigg{(}d\Big{(}\nu_{c}^{y}\big{(}\nu_{c}^{x}(x,y)-\nu_{c}^{x}(0,y)+\int\limits_{0}^{x}\frac{\partial\nu_{c}^{y}}{\partial
y}d\tilde{x}\big{)}-\\\
&-\nu_{c}^{x}\big{(}\nu_{c}^{y}(x,y)-\nu_{c}^{y}(x,0)+\int\limits_{0}^{y}\frac{\partial\nu_{c}^{x}}{\partial
x}d{y}\big{)}\Big{)}+\mathbf{div}(v_{c})\Big{(}\nu_{c}^{x}dy-\nu_{c}^{y}dx\Big{)}-\\\
&-\Big{(}\int\limits_{[0,1]^{2}}\mathbf{div}(v_{c})dxdy\Big{)}\Big{(}d\big{(}v_{c}(\beta_{1}+\beta_{2})\big{)}+2(\nu_{c}^{x}dy-\nu_{c}^{y}dx)\Big{)}\Bigg{)}\\\
\end{split}$ (120)

The most complicated term (cubic in $c^{a}$) has the form

$\displaystyle\begin{split}\langle\alpha^{i},\mathcal{L}_{v_{c}}h\mathcal{L}_{v_{c}}h\mathcal{L}_{v_{c}}\gamma\rangle&=\langle\alpha^{i},\mathcal{L}_{v_{c}}h\xi\rangle=\langle\alpha^{i},\mathcal{L}_{v_{c}}\Big{(}I(\xi)-\sum\limits_{i}I(\beta_{i})\int\limits_{\beta^{i}}\xi\Big{)}\rangle\\\
\end{split}$ (121)

Thus, from (112)-(121) we get the $L_{\infty}$ module structure

$\displaystyle\begin{split}S^{ind}&=\psi_{i}^{*}\langle\beta^{i},d\alpha_{j}\rangle\varphi^{j}+\omega^{*}\langle\gamma^{*},d\beta_{j}\rangle\psi^{j}-\varphi_{i}^{*}\langle\alpha^{i},\mathcal{L}_{v_{c}}\alpha_{j}\rangle\varphi^{j}+\psi_{i}^{*}\langle\beta^{i},\mathcal{L}_{v_{c}}\beta_{j}\rangle\psi^{j}-\\\
&-\omega^{*}\langle\gamma^{*},\mathcal{L}_{v_{c}}\gamma\rangle\omega+\varphi_{i}^{*}\langle\alpha^{i},\mathcal{L}_{v_{c}}h\mathcal{L}_{v_{c}}\beta_{j}\rangle\psi^{j}+\psi_{i}^{*}\langle\beta^{i},\mathcal{L}_{v_{c}}h\mathcal{L}_{v_{c}}\gamma\rangle\omega+\\\
&+\varphi_{i}^{*}\langle\alpha^{i},\mathcal{L}_{v_{c}}h\mathcal{L}_{v_{c}}h\mathcal{L}_{v_{c}}\gamma\rangle\omega+\frac{1}{2}f_{ab}^{d}c^{a}c^{b}c_{d}^{*}\end{split}$
(122)

## References

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# Crossing Numbers of Billiard Curves in the Multidimensional Box via
Translation Surfaces

Felix Christian Clemen111Department of Mathematics, Karlsruhe Institute of
Technology, 76131 Karlsruhe, Germany, E-mail<EMAIL_ADDRESS>Peter
Kaiser222Department of Mathematics, Karlsruhe Institute of Technology, 76131
Karlsruhe, Germany, E-mail<EMAIL_ADDRESS>
###### Abstract

The billiard table is modeled as an $n$-dimensional box
$[0,a_{1}]\times[0,a_{2}]\times\ldots\times[0,a_{n}]\subset\mathbb{R}^{n}$,
with each side having real-valued lengths $a_{i}$ that are pairwise
commensurable. A ball is launched from the origin in direction
$d=(1,1,\ldots,1)$. The ball is reflected if it hits the boundary of the
billiard table. It comes to a halt when reaching a corner. We show that the
number of intersections of the billiard curve at any given point on the table
is either $0$ or a power of $2$. To prove this, we use algebraic and number
theoretic tools to establish a bijection between the number of intersections
of the billiard curve and the number of satisfying assignments of a specific
constraint satisfaction problem.

##  1 Introduction

The trajectory of a ball on a billiard table is studied in various areas of
mathematics, for example in ergodic theory, Teichmüller theory [5, 9], from a
geometry perspective [8, 10], or through a combinatorial lens in the setting
of arithmetic billiards [2, 4, 6].

We model the _billiard curve_ as follows. Let
$\mathcal{T}\subset\mathbb{R}^{n}$ be a domain and $p\in\mathcal{T}$ a point
inside this domain. Additionally, let $d\in\mathbb{R}^{n}$ be a direction. The
billiard curve is the trajectory starting at $p$ moving in direction $d$,
where the direction is reflected if the trajectory reaches the boundary, in
the sense of an elastic collision. If it reaches a corner, the motion ceases.
A natural question is to determine if for a given domain there exists a
periodic trajectory. For a specific domain, it might be challenging to decide
whether a trajectory is periodic, stops, or has infinite length [1]. We are
considering a setting where all trajectories have finite length.

A set of positive reals $a_{1},a_{2},\ldots,a_{n}\in\mathbb{R}$ is called
_pairwise commensurable_ if all ratios $a_{i}/a_{j}$ are rational. In this
paper, the domain $\mathcal{T}$ is always the box
$\mathcal{T}=[0,a_{1}]\times[0,a_{2}]\times\ldots\times[0,a_{n}]$ with
pairwise commensurable side lengths $a_{i}$, the starting point is the origin,
i.e. $p=(0,\ldots,0)$, and the direction $d=(1,\ldots,1)$. A _corner_ is a
point $w\in\mathcal{T}$ with $w_{i}\in\\{0,a_{i}\\}$ for every $i\in[n]$. In
this setting, the trajectory always stops at a corner different from the
origin.

We ask how often the trajectory crosses a given point within the domain. Given
a point $v\in\mathcal{T}$ let the _crossing number of_ $v$, denoted by
$m_{\mathcal{T}}(v)$, be the number of times the trajectory crosses $v$. In
the 2-dimensional scenario, i.e. $\mathcal{T}=[0,a_{1}]\times[0,a_{2}]$, for
every $v\in D$, trivially $m_{\mathcal{T}}(v)\in\\{0,1,2\\}$ simply because
the trajectory is the union of lines with only two directions.

Our results concern the multi-dimensional case. For any positive real numbers
$a_{1},\ldots,a_{n}$ and
$\mathcal{T}=[0,a_{1}]\times[0,a_{2}]\times\ldots\times[0,a_{n}]$, we have
$m_{\mathcal{T}}(v)\leq 2^{n-1}$ for a given point $v\in\mathcal{T}$ because
$v$ can only be crossed in one of the two direction $d$ or $-d$ for a given
$d\in\\{-1,1\\}^{n}$. We give a more precise description of crossing numbers.

######  Theorem 1.1.

Let $T=[0,a_{1}]\times[0,a_{2}]\times\ldots\times[0,a_{n}]$ be the box with
pairwise commensurable side lengths $a_{i}>0$. Then for every
$v\in\mathcal{T}$, we have $m_{\mathcal{T}}(v)\in\\{0\\}\cup\\{2^{k}:k\geq
0\\}$.

For our proof, we initially establish a discrete structure, so that we can
analyze the problem by using algebraic and number theoretic methods. In the
second step, we formalize the setting in the language of translation surfaces.
Following this, we establish a bijection to the number of solutions of a
specific boolean constraint satisfaction problem. Finally, we count those
solutions.

In the case of pairwise coprime side lengths, we determine the crossing number
of a given $v\in\mathcal{T}$ precisely.

######  Theorem 1.2.

Let $a_{1},\ldots,a_{n}$ be pairwise coprime positive integers, and let
$v\in\mathcal{T}\cap\mathbb{N}^{n}$ not be a corner. Then

$\displaystyle m_{\mathcal{T}}(v)=2^{n-1-|\\{i:\ v_{i}=0\text{ or
}v_{i}=a_{i}\\}|}.$

In the context of arithmetic billiards, the number of self-intersecting points
in the two-dimensional case has been explored by Perucca, Reguengo De Sousa
and Tronto [6], as well as by Da Veiga Bruno [2]. Another counting question
asks for the number of bouncing points, i.e. the number of points in which the
trajectory meets the boundary of the domain, this has been studied by
Steinhaus [7], Gardner [3] and also in the multi-dimensional case in [2].

The paper is organized as follows. In the first step, in Section 2, we
discretize the problem. As a warm-up, we will present an elementary proof of
Theorem 1.2 in Section 3. In Section 4 we introduce translation surfaces.
Finally, in Section 5, we present the proof of our main result, Theorem 1.1.

##  2 Discretization of the billiard table

For our further reasoning, we rely on a discretization of the problem. Our
first observation is that scaling the billiard table $\mathcal{T}$ does not
change the form of the billiard curve. Since the side lengths $a_{i}$ are
pairwise commensurable, after scaling, we can assume that $a_{i}\in\mathbb{N}$
for all $i\in[n]$. From now on all $a_{i}$ will be assumed to be integers.

Observe that the set of _intersection points_ , i.e. the points satisfying
$m(v)\geq 2$, is a subset of $\mathbb{N}^{n}\cap\mathcal{T}$. This means, it
suffices to prove 1.1 and 1.2 for $v\in\mathbb{N}^{n}\cap\mathcal{T}$. This is
crucial for our argumentation since we will heavily rely on number theoretic
machinery. The following lemma gives a necessary condition for a point
$v\in\mathbb{N}^{n}\cap\mathcal{T}$ to be on the billiard curve.

######  Lemma 2.1.

Let $v\in\mathcal{T}\cap\mathbb{N}^{n}$ with $m_{\mathcal{T}}(v)>0$. Then
$v_{i}\equiv 0\mod 2$ for all $i\in[n]$ or $v_{i}\equiv 1\mod 2$ for all
$i\in[n]$.

###### Proof.

The starting point $(0,\ldots,0)$ satisfies $v_{i}\equiv 0\mod 2$ trivially.
The current direction $d$ of the trajectory is always of the form
$d\in\\{-1,1\\}^{n}$. For each $v\in\mathcal{T}\cap\mathbb{N}^{n}$ with
$v+d\in\mathcal{T}$ and $v_{i}\equiv v_{j}\mod 2$ for all $i,j\in[n]$, we have
$v+d\in\mathcal{T}\cap\mathbb{N}^{n}$ and $v_{i}+d_{i}\equiv v_{j}+d_{j}\mod
2$ since the parity in all entries is changed. The claim follows inductively.
∎

The following lemma determines the length of the trajectory, i.e. the sum of
the lengths of the linear segments of the trajectory.

######  Lemma 2.2 (see e.g. Carvalho da Veiga [2]).

Let $a_{i}\in\mathbb{N}$ for all $i\in[n]$. Then the trajectory has a length
of $\sqrt{n}\ell$, where $\ell=\operatorname{lcm}(a_{1},\ldots,a_{n})$.

###### Proof.

Whenever the initial segment of the trajectory has length a multiple of
$a_{i}\sqrt{n}$, the current point has $a_{i}$ or $0$ as the $i$-th
coordinate. Since every corner $p$ satisfies $p_{i}\equiv 0\mod a_{i}$ for
every $i\in[n]$, the length of the trajectory is $\ell\sqrt{n}$. ∎

An advantage of the discretization of the billiard table is that we can use
the Chinese Remainder Theorem.

######  Theorem 2.3 (Chinese Remainder Theorem).

Let $a_{1},\ldots,a_{n}$ be integers greater than $1$ and
$\ell=\operatorname{lcm}(a_{1},a_{2},\ldots,a_{n})$. Further, let
$v_{1},\ldots,v_{n}$ be integers such that $0\leq v_{i}<a_{i}$ for every
$i\in[n]$. Iff for all $i\neq j$ we have $v_{i}\equiv v_{j}\mod
gcd(a_{i},a_{j})$, then there is a unique integer $x$, such that $0\leq
x<\ell$ and $x\equiv v_{i}\mod a_{i}$.

##  3 The coprime case: Proof of Theorem 1.2

Let $a_{1},a_{2},\ldots,a_{n}$ be pairwise coprime integers. Given
$v\in\mathcal{T}\cap\mathbb{N}^{n}$, define

$I_{1}=\\{i\in[n]:v_{i}=0\\},\quad
I_{2}=\\{i\in[n]:v_{i}=a_{i}\\}\quad\text{and}\quad I(v)=I_{1}\cup I_{2}.$

First, we present an upper bound on $m_{\mathcal{T}}(v)$ depending on
$|I(v)|$.

######  Lemma 3.1.

Let $v\in\mathcal{T}\cap\mathbb{N}^{n}$ be a non-corner. Then
$m_{\mathcal{T}}(v)\leq 2^{n-1-|I(v)|}$.

###### Proof.

Let $W_{v}\subseteq\\{-1,1\\}^{n}$ be the set of directions which from $v$
point towards the board, i.e. the set of vectors $w\in\\{-1,1\\}^{n}$ such
that $v+\varepsilon w\in\mathcal{T}$ for some $\varepsilon>0$. Note that every
$w\in W_{v}$ satisfies $w_{i}=1$ if $i\in I_{1}$ and $w_{i}=-1$ if $i\in
I_{2}$. Thus, $|W_{v}|=2^{n-|I(v)|}$. Since the trajectory enters and leaves
the point $v$, we conclude $m_{\mathcal{T}}(v)\leq|W_{v}|/2\leq
2^{n-|I(v)|-1}$ for every $v\in\mathcal{T}\cap\mathbb{N}^{n}$. ∎

Recall that, by definition, $m_{\mathcal{T}}(v)=1$ for $v$ being the start
corner or the end corner. For all other corners $v\in\mathcal{T}$, we have
$m_{\mathcal{T}}(v)=0$. Set $b_{k}$ to be the number of _bouncing points_ on a
$k$-dimensional boarder, i.e. the number of
$v\in\mathcal{T}\cap\mathbb{N}^{n}$ such that $|I(v)|=k$ and
$m_{\mathcal{T}}(v)>0$.

######  Lemma 3.2.

Let $0\leq k\leq n$. Then

$\displaystyle b_{k}=2^{1-n+k}\sum_{\begin{subarray}{c}J\subseteq[n]:\\\
|J|=k\end{subarray}}\ \prod_{\begin{subarray}{c}j\in[n]\setminus
J\end{subarray}}(a_{j}-1).$

###### Proof.

We distinguish two cases.

Case 1: $2\nmid a_{i}$ for every $i\in[n]$.

Let $v\in\mathcal{T}\cap\mathbb{N}^{n}$ such that $|I(v)|=k$ and
$m_{\mathcal{T}}(v)>0$. By Lemma 2.1 we know that either $v_{i}\equiv 0\mod 2$
for all $i\in[n]$ or $v_{i}\equiv 1\mod 2$ for all $i\in[n]$. Because
$|I(v)|=k$ there are exactly $k$ coordinates of $v$ which are either $0$ or
$a_{i}$. Further, since $a_{i}\equiv 1\mod 2$ for all $i\in[n]$, these $k$
coordinates are simultaneously either $0$ or $a_{i}$. The $n-k$ coordinates,
which are not on the border, must have the same parity as the coordinates on
the border. Therefore, they can obtain precisely $\frac{a_{i}-1}{2}$ possible
values. This yields, for $0\leq k\leq n$,

$\displaystyle b_{k}=\sum_{\begin{subarray}{c}J\subseteq[n]:\\\
|J|=k\end{subarray}}\ 2\ \prod_{\begin{subarray}{c}j\in[n]\setminus
J\end{subarray}}\frac{a_{j}-1}{2}=2^{1-n+k}\sum_{\begin{subarray}{c}J\subseteq[n]:\\\
|J|=k\end{subarray}}\ \prod_{\begin{subarray}{c}j\in[n]\setminus
J\end{subarray}}(a_{j}-1).$

Case 2: $2\mid a_{i}$ for some $i\in[n]$.

Note that there is at most one even $a_{i}$ since the $a_{i}$ are assumed to
be coprime. Without loss of generality, we can assume $2\mid a_{1}$. We count
the number of $v\in\mathcal{T}\cap\mathbb{N}^{n}$ such that $|I(v)|=k$ and
$m_{\mathcal{T}}(v)>0$ by looking at the cases $v_{1}\in\\{0,a_{1}\\}$ and
$v_{1}\notin\\{0,a_{1}\\}$ separately.

If the first coordinate is on the boarder, i.e. $v_{1}\in\\{0,a_{1}\\}$, then
by Lemma 2.1 $v_{i}\equiv 0\mod 2$ for all $i\in[n]$. Thus, $v_{j}=0$ for
$j\in I(v)$. There are two choices for $v_{1}$. The coordinates not on the
border must be even, and therefore can obtain $\frac{a_{j}-1}{2}$ possible
values. We conclude that the number of $v\in\mathcal{T}\cap\mathbb{N}^{n}$
such that $|I(v)|=k$, $m_{\mathcal{T}}(v)>0$ and $v_{1}\in\\{0,a_{1}\\}$ is

$\displaystyle\sum_{\begin{subarray}{c}J\subseteq[n]:\\\ |J|=k,\ 1\in
J\end{subarray}}\left(\ 2\ \prod_{j\in[n]\setminus
J}\frac{a_{j}-1}{2}\right).$ (1)

If the first coordinate is not on the border, i.e. $v_{1}\notin\\{0,a_{1}\\}$,
there are $a_{1}-1$ choices for $v_{1}$. By Lemma 2.1, this choice determines
the coordinates $v_{j}$ for $j\in I(v)$ uniquely, and also the parity of
$v_{j}$ for $j\notin I(v)$. There are $\frac{a_{j}-1}{2}$ possible choices for
non-boundary coordinates. Thus, the number of
$v\in\mathcal{T}\cap\mathbb{N}^{n}$ such that $|I(v)|=k$,
$m_{\mathcal{T}}(v)>0$ and $v_{1}\notin\\{0,a_{1}\\}$ is

$\displaystyle\sum_{\begin{subarray}{c}J\subseteq[n]:\\\ |J|=k,\ 1\not\in
J\end{subarray}}\left(\ (a_{1}-1)\ \prod_{\begin{subarray}{c}j\in[n]\setminus
J\\\ j\neq 1\end{subarray}}\frac{a_{j}-1}{2}\right).$ (2)

By adding up (1) and (2) we obtain, for $0\leq k\leq n$,

$\displaystyle b_{k}$
$\displaystyle=\sum_{\begin{subarray}{c}J\subseteq[n]:\\\ |J|=k,\ 1\in
J\end{subarray}}\left(\ 2\ \prod_{j\in[n]\setminus
J}\frac{a_{j}-1}{2}\right)+\sum_{\begin{subarray}{c}J\subseteq[n]:\\\ |J|=k,\
1\not\in J\end{subarray}}\left(\ (a_{1}-1)\
\prod_{\begin{subarray}{c}j\in[n]\setminus J\\\ j\neq
1\end{subarray}}\frac{a_{j}-1}{2}\right)$
$\displaystyle=2^{1-n+k}\sum_{\begin{subarray}{c}J\subseteq[n]:\\\
|J|=k\end{subarray}}\ \prod_{\begin{subarray}{c}j\in[n]\setminus
J\end{subarray}}(a_{j}-1),$

completing the proof of this lemma. ∎

###### Proof of Theorem 1.2.

The length of the trajectory is $\sqrt{n}\ell$ by Lemma 2.2, with
$\ell=\operatorname{lcm}(a_{1},\ldots,a_{n})$. Therefore, the trajectory
passes through $\ell$ points from $\mathcal{T}\cap\mathbb{N}^{n}$, when we
count with multiplicities but without the starting point. The length $\ell$
can be expressed as the sum over the crossed points in
$\mathcal{T}\cap\mathbb{N}^{n}$ times their multiplicity, i.e.

$\displaystyle\ell$
$\displaystyle=\sum_{v\in\mathcal{T}\cap\mathbb{N}^{n}}m_{\mathcal{T}}(v).$
(3)

Further, we partition the points by the dimension of the border they are lying
on,

$\displaystyle\ell$
$\displaystyle=\sum_{v\in\mathcal{T}\cap\mathbb{N}^{n}}m_{\mathcal{T}}(v)=\sum_{k=0}^{n}\
\sum_{\begin{subarray}{c}v\in\mathcal{T}\cap\mathbb{N}^{n}:\\\
|I(v)|=k\end{subarray}}m_{\mathcal{T}}(v).$ (4)

By combining Lemmas 3.1, 3.2 with (4), we obtain

$\displaystyle\ell\leq\sum_{k=0}^{n}\ b_{k}2^{n-1-k}=\sum_{k=0}^{n}\
\sum_{\begin{subarray}{c}J\subseteq[n]:\\\ |J|=k\end{subarray}}\
\prod_{\begin{subarray}{c}j\in[n]\setminus
J\end{subarray}}(a_{j}-1)=\sum_{\begin{subarray}{c}J\subseteq[n]\end{subarray}}\
\prod_{\begin{subarray}{c}j\in
J\end{subarray}}(a_{j}-1)=\prod_{i\in[n]}a_{i}=\ell.$ (5)

The second to last equality in (5) holds by a standard inclusion-exclusion
argument. We conclude that the inequality in (5) must hold with equality, and
thus $m_{\mathcal{T}}(v)=2^{n-1-|I(v)|}$ for every
$v\in\mathcal{T}\cap\mathbb{N}^{n}$, completing the proof of Theorem 1.2. ∎

##  4 Translation surfaces

A problem with analyzing the trajectory is that it changes its direction
whenever it hits the boundary. We will overcome this issue by artificially
extending the board in such a way that the trajectory on the new board is
simply a single straight line. This will be done in two steps. In the first
step, we extend the domain to $2\mathcal{T}$. On this new domain the
trajectory will not change its directions, but will not be connected in the
Euclidean sense as a trade-off. In the second step, we embed $2\mathcal{T}$
into $\mathbb{R}^{n}$. On this domain the trajectory will be a straight line.

We now consider $2\mathcal{T}=[0,2a_{1}]\times\ldots\times[0,2a_{n}]$ instead
of $\mathcal{T}$. Let $e_{i}\in\mathbb{R}^{n}$ be the standard basis vector
with entry $1$ in the $i$-th component. The domain $\mathcal{T}$ is bounded by
the hyperplanes

$P_{i}:=\mathrm{span}\\{e_{1},\ldots,e_{i-1},e_{i+1},\ldots,e_{n}\\}\quad\text{and}\quad
P_{i}^{\prime}:=a_{i}\cdot
e_{i}+\mathrm{span}\\{e_{1},\ldots,e_{i-1},e_{i+1},\ldots,e_{n}\\}.$

Let $\tau_{i}$ be the reflection on the hyperplanes $P_{i}^{\prime}$. We
identify all points that are mapped to each other by the concatenation of some
combination of the $\tau_{i}$, i.e.

$\displaystyle v\sim w\ \Leftrightarrow\,\exists I\subseteq[n],\
I=\\{i_{1},\ldots,i_{k}\\}:\tau_{i_{1}}\circ\ldots\circ\tau_{i_{k}}(v)=w.$ (6)

This yields that a point $v=(v_{1},\ldots,v_{n})\in\mathcal{T}$ is identified
with all points from the set

$\displaystyle Q(v)=\left\\{\begin{pmatrix}y_{1}\\\ \vdots\\\
y_{n}\end{pmatrix}\in
2\mathcal{T}\,\middle|\,y_{i}\in\\{v_{i},2a_{i}-v_{i}\\}\right\\},$

see Figure 1 for an illustration. Points in the interior of $\mathcal{T}$ have
$2^{n}$ copies in $2\mathcal{T}$. Points on a $k$-dimensional border only have
$2^{k}$ copies.

Figure 1: The points with the same color are identified via the reflections.
The old trajectory on $\mathcal{T}$ is displayed with a solid line. The dashed
line is the new trajectory on $2\mathcal{T}$. Corresponding line segments in
both trajectories are colored with the same color. Note that the green and
purple line segments are coinciding.

In $2\mathcal{T}$ we glue opposite sides together and obtain an
$n$-dimensional torus. Instead of reflecting the direction of the trajectory
on the border we continue the trajectory on the point opposite to the bouncing
point, see also Figure 1. The advantage of this setting is that the direction
of the trajectory is not changed at the border. This implies that each point
can be crossed in exactly one way, therefore we now can count the number of
identified points that are crossed to get $m_{\mathcal{T}}(v)$. The drawback
of this new setup is that the trajectory now is not connected (when displayed
in $\mathbb{R}^{n}$), see Figure 1.

In the next step of our analysis we will even forget the gluing of the
opposing sides and consider the whole space $\mathbb{R}^{n}$. We identify
points in $\mathbb{R}^{n}$ with points in $2\mathcal{T}$ via the function
$\psi:\mathbb{R}^{n}\to 2\mathcal{T}$, given by

$\psi(\begin{pmatrix}v_{1}\\\ \vdots\\\
v_{n}\end{pmatrix})=\begin{pmatrix}v_{1}\mod 2a_{1}\\\ \vdots\\\ v_{n}\mod
2a_{n}\end{pmatrix}.$

If $v$ is of the form $(x,\ldots,x)$ we also use the following notation

$\psi(x)=\begin{pmatrix}x\mod 2a_{1}\\\ \vdots\\\ x\mod 2a_{n}\end{pmatrix}.$

By concatenation of the two identifications, via (6) and $\psi$, we get
another identification of points, such that
$v=(v_{1},\ldots,v_{n})\in\mathcal{T}$ is identified with all points from the
set

$\displaystyle R(v)=\left\\{\begin{pmatrix}y_{1}\\\ \vdots\\\
y_{n}\end{pmatrix}\in\mathbb{R}^{n}\,\middle|\,y_{i}\equiv v_{i}\bmod 2a_{i}\
\ \text{or}\ \ y_{i}\equiv-v_{i}\bmod 2a_{i}\right\\},$

Now the entire trajectory in $\mathbb{R}^{n}$ is a straight line in direction
$d=(1,\ldots,1)$ which starts at the origin and, by Lemma 2.2, ends at
$\sqrt{n}\cdot(\ell,\ldots,\ell)$.

##  5 Determining crossing numbers via CSPs: Proof of Theorem 1.1

We use the language of _constraint satisfaction problems_ (CSP). Let
$\mathcal{P}$ be the family of all subsets of $\\{0,1\\}\times\\{0,1\\}$. The
elements of $\mathcal{P}$ are referred to as _constraints_. A (boolean) CSP on
a finite set $V\subseteq\mathbb{N}$ is a pair $(V,\phi)$, where
$\phi:\binom{V}{2}\rightarrow\mathcal{P}$ is a function assigning a constraint
for each pair of vertices. An _assignment_ on $V$ is a function
$g:V\rightarrow\\{0,1\\}$ which assigns for every $v\in V$ an integer from
$\\{0,1\\}$. An assignment $g:V\rightarrow\\{0,1\\}$ is _satisfying_ for
$(V,\phi)$ if $(g(a),g(b))\notin\phi(\\{a,b\\})$ for any pair $a,b\in V$ such
that $a<b$. For a CSP $G=(V,\phi)$, denote by $A(G)$ the set of satisfying
assignments for $G$. Define

$\displaystyle\mathcal{C}:=\bigg{\\{}\emptyset,\Big{\\{}(0,0),(1,1)\Big{\\}},\Big{\\{}(0,1),(1,0)\Big{\\}},\Big{\\{}(0,0),(0,1),(1,0),(1,1)\Big{\\}}\bigg{\\}}\subseteq\mathcal{P},$

and, given $v\in\mathbb{N}^{n}\cap\mathcal{T}$, a function
$\phi_{v}:\binom{[n]}{2}\rightarrow\mathcal{C}$ by

$\displaystyle\phi_{v}(ij):=\begin{cases}\emptyset&,\text{if}\ v_{i}\equiv-
v_{j}\ \text{and}\ v_{i}\equiv v_{j}\mod\gcd(2a_{i},2a_{j}),\\\
\\{(0,1),(1,0)\\}&,\text{if}\ v_{i}\equiv v_{j}\ \text{and}\ v_{i}\not\equiv-
v_{j}\mod\gcd(2a_{i},2a_{j}),\\\ \\{(0,0),(1,1)\\}&,\text{if}\ v_{i}\equiv-
v_{j}\ \text{and}\ v_{i}\not\equiv v_{j}\mod\gcd(2a_{i},2a_{j}),\\\
\\{(0,0),(0,1),(1,0),(1,1)\\}&,\text{if}\ v_{i}\not\equiv-v_{j}\ \text{and}\
v_{i}\not\equiv v_{j}\mod\gcd(2a_{i},2a_{j}).\end{cases}$

Let $v,x,a\in\mathbb{Z}^{n}$. We write $v\equiv x\mod a$ iff $v_{i}\equiv
x_{i}\mod a_{i}$ for all $i\in[n]$. Given $v\in\mathbb{N}^{n}\cap\mathcal{T}$,
define $J(v):=\\{i\in[n]:\ v_{i}\equiv-v_{i}\mod 2a_{i}\\}$. The following
lemma sets the crossing number in relation to the number of satisfying
assignments of a CSP.

######  Lemma 5.1.

Let $v\in\mathbb{N}^{n}\cap\mathcal{T}$ be a non-corner and
$G=([n],\phi_{v})$. Then,

$\displaystyle m_{\mathcal{T}}(v)=\frac{1}{2^{|J(v)|+1}}|A(G)|.$

###### Proof.

Let $a:=(a_{1},\ldots,a_{n})$ and
$\ell=\operatorname{lcm}(a_{1},a_{2},\ldots,a_{n})$. The trajectory in
$\mathcal{T}$ passes through $v$ exactly if the trajectory in $\mathbb{R}^{n}$
passes through a point which is in $R(v)$, i.e. if the following system of
equations has one solution $x\in\\{0,1,\ldots,\ell\\}$ for all $i\in[n]$:

$x\equiv v_{i}\bmod 2a_{i}\quad\text{or}\quad x\equiv-v_{i}\bmod 2a_{i}.$ (7)

Note that if $v$ satisfies $v_{i}\equiv-v_{i}\mod 2a_{i}$ for some $i$, then
there is only one equation in the $i$-th condition of (7). Observe that iff
$x\in\\{1,\ldots,\ell-1\\}$ satisfies (7), then also
$2\ell-x\in\\{\ell+1,\ldots,2\ell-1\\}$ satisfies (7). Also $x=0$ and $x=\ell$
do not satisfy (7), because $\psi(0)=0$ and $\psi(\ell)\equiv 0\mod a$.
Therefore,

$\displaystyle m_{\mathcal{T}}(v)$
$\displaystyle=\lvert\big{\\{}x\in\\{0,1,\ldots,\ell-1\\}:\ x\equiv v_{i}\bmod
2a_{i}\ \text{or}\ x\equiv-v_{i}\bmod 2a_{i}\quad\text{for all
}i\in[n]\big{\\}}\rvert$
$\displaystyle=\frac{1}{2}\lvert\big{\\{}x\in\\{0,1,\ldots,2\ell-1\\}:\
x\equiv v_{i}\bmod 2a_{i}\ \text{or}\ x\equiv-v_{i}\bmod 2a_{i}\quad\text{for
all }i\in[n]\big{\\}}\rvert.$ (8)

By the Chinese Remainder Theorem, 2.3, iff $g:[n]\rightarrow\\{0,1\\}$ is a
satisfying assignment for $G$, then there exists a unique solution
$x\in\\{0,1,\ldots,2\ell-1\\}$ to the system of equations

$\displaystyle x\equiv(-1)^{g(i)}v_{i}\bmod 2a_{i}\quad\text{for }i\in[n].$
(9)

Two satisfying assignments $g,g^{\prime}$ for $G$ correspond to the same set
of equations in (9) iff $g(i)=g^{\prime}(i)$ for all $i\in[n]\setminus J(v)$.
We conclude that

$\displaystyle m_{\mathcal{T}}(v)$
$\displaystyle=\frac{1}{2}\lvert\big{\\{}x\in\\{0,1,\ldots,2\ell-1\\}:\
x\equiv v_{i}\bmod 2a_{i}\ \text{or}\ x\equiv-v_{i}\bmod 2a_{i}\quad\text{for
all }i\in[n]\big{\\}}\rvert$
$\displaystyle=\frac{1}{2}\cdot\frac{1}{2^{|J(v)|}}|A(G)|=\frac{1}{2^{|J(v)|+1}}|A(G)|,$

completing the proof of this lemma. ∎

######  Lemma 5.2.

Let $n\in\mathbb{N}$, $\phi:\binom{[n]}{2}\rightarrow\mathcal{C}$ and
$G=([n],\phi)$ be a CSP. Then

$\displaystyle|A(G)|\in\\{0\\}\cup\\{2^{k}\mid k\geq 1\\}.$

###### Proof.

Construct an auxiliary graph $H$ from the CSP $G=([n],\phi)$ as follows. The
vertex set of $H$ is $V(H)=[n]$ and the edge set $E(H)=\\{e\in\binom{[n]}{2}:\
\phi(e)\neq\emptyset\\}$. Let $C_{1},\ldots,C_{k}\subseteq[m]$ be the
connected components of $H$. For $i\in[k]$, let

$G[C_{i}]:=\left(C_{i},{\left.\kern-1.2pt\phi\vphantom{\big{|}}\right|_{\binom{C_{i}}{2}}}\right)$

be the _induced CSP_ of $G$ on $C_{i}$. Then $g\in A(G)$ iff
${\left.\kern-1.2ptg\vphantom{\big{|}}\right|_{C_{i}}}\in A(G[C_{i}])$ for all
$i\in[k]$, simply because for every pair $ab$ with $a<b,a\in C_{i},b\in C_{j}$
for $i\neq j$ we have $\phi(ab)=\emptyset$. We conclude

$\displaystyle|A(G)|=\prod_{i=1}^{k}|A(G[C_{i}])|.$

Next, we show that $|A(G[C_{i}])|\in\\{0,2\\}$ for $i\in[k]$, completing the
proof of this lemma. Let $i\in[k]$. Then, since $H[C_{i}]$ is connected, there
exists a spanning tree $T\subseteq H[C_{i}]$. Let
$C_{i}=\\{w_{1},w_{2},\ldots,w_{|C_{i}|}\\}$ such that $T[W_{j}]$, where
$W_{j}:=\\{w_{1},\ldots,w_{j}\\}$, is a tree. We prove inductively, for
$j\in[|C_{i}|]$ that there exists at most one satisfying assignment $g$ of
$G[W_{j}]$ such that $g(w_{1})=0$. For $j=1$, this is trivial. Now, let
$j\in[|C_{i}|]$ such that there exists at most one satisfying assignment $g$
of $G[W_{j}]$ such that $g(w_{1})=0$. If there does not exist a satisfying
assignment of $G[W_{j}]$ such that $g(w_{1})=0$, then in particular there does
not exist a satisfying assignment of $G[W_{j+1}]$ such that $g(w_{1})=0$. Let
$e=xw_{j+1}\in E(T)$ be the unique edge such that $x\in W_{j}$. Now, assume
there exists a satisfying assignment $g$ of $G[W_{j}]$ such that $g(w_{1})=0$.
If $g^{\prime}$ is a satisfying assignment of $G[W_{j+1}]$, then
$g^{\prime}(w_{a})=g(w_{a})$ for all $a\in[j]$ and

$\displaystyle g^{\prime}(w_{j+1})=\begin{cases}0,&\text{if }g(x)=1\text{ and
}\phi(e)=\Big{\\{}(0,0),(1,1)\Big{\\}},\\\ 1,&\text{if }g(x)=1\text{ and
}\phi(e)=\Big{\\{}(0,1),(1,0)\Big{\\}},\\\ 1,&\text{if }g(x)=0\text{ and
}\phi(e)=\Big{\\{}(0,0),(1,1)\Big{\\}},\\\ 0,&\text{if }g(x)=0\text{ and
}\phi(e)=\Big{\\{}(0,1),(1,0)\Big{\\}}.\\\ \end{cases}$

Thus, there is at most one satisfying assignment $g^{\prime}$ of $G[W_{j+1}]$
such that $g^{\prime}(w_{1})=0$. Inductively, we conclude that there is at
most one satisfying assignment of $G[C_{i}]$ such that $g(w_{1})=0$.
Similarly, there is at most one satisfying assignment of $G[C_{i}]$ such that
$g(w_{1})=1$. Thus, $|A(G[C_{i}])|\leq 2$. Further, if $g$ is a satisfying
assignment of $G[C_{i}]$, then $g^{\prime}:C_{i}\rightarrow\\{0,1\\}$ defined
by $g^{\prime}(j)=1-g(j)$ for $j\in C_{i}$ is also a satisfying assignment of
$G[C_{i}]$. We conclude $|A(G[C_{i}])|\in\\{0,2\\}$. ∎

###### Proof of 1.1.

If $v$ is a corner, then $m_{\mathcal{T}}(v)=0$ or $m_{\mathcal{T}}(v)=1$
simply because the trajectory starts at the origin and its run is over when it
hits a corner the first time. If $v$ is a non-corner, then Theorem 1.1 follows
by combining Lemma 5.1 with Lemma 5.2. ∎

##  6 Concluding Remarks

The reader may ask how the crossing number varies as the setup of the board is
changed. We note that for different starting positions on the board, different
rational starting slopes $d$ and for parallelepipeds instead of boxes the
result is not changed, i.e. the crossing number of a given point is still $0$
or a power of $2$. This observation follows by considering a linear
transformation that transforms the new setup into the old one without changing
the crossing number.

## Acknowledgments

We thank the members of Kaffeerunde IAG at KIT as well as Cameron Gates Rudd
for stimulating discussions on the topic.

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