# Torsion of digraphs and path complexes

Alexander Grigor’yan  Alexander Grigor’yan, Department of Mathematics,
University of Bielefeld, 33501 Bielefeld, Germany<EMAIL_ADDRESS>bielefeld.de ,  Yong Lin  Yong Lin, Yau Mathematical Sciences Center, Tsinghua
University, Beijing, 100084, China<EMAIL_ADDRESS>and  Shing-Tung
Yau  Shing-Tung Yau, Department of Mathematics, Harvard University, Cambridge,
Massachusetts, USA<EMAIL_ADDRESS>
(Date: December 2020)

###### Abstract.

We define the notions of Reidemeister torsion and analytic torsion for
directed graphs by means of the path homology theory introduced by the authors
in [3, 4, 5, 7]. We prove the identity of the two notions of torsions as well
as obtain formulas for torsions of Cartesian products and joins of digraphs.

AG is supported by the SFB1283 of the German Research Council. YL is supported
by the National Science Foundation of China (Grant No. 12071245 and
11761131002)

###### Contents

  1. 1 Introduction
  2. 2 Path complexes and path homology
    1. 2.1 Path complex
    2. 2.2 Digraphs
    3. 2.3 Path homology
  3. 3 Finite chain complexes
    1. 3.1 Hodge Laplacian
    2. 3.2 R-torsion
    3. 3.3 Examples
    4. 3.4 Analytic torsion
  4. 4 Cartesian product of path complexes
    1. 4.1 Product of paths
    2. 4.2 Product of path complexes and digraphs
    3. 4.3 Operators $\partial^{\ast}$ and $\Delta$ on products
    4. 4.4 Torsion of products
  5. 5 Join of path complexes
    1. 5.1 Augmented chain complex
    2. 5.2 Join of path complexes and digraphs
    3. 5.3 Operators $\partial^{\ast}$ and $\Delta$ on joins
    4. 5.4 Torsion of joins

##  1\. Introduction

Let $M$ be a compact oriented Riemannian manifold. Assume that $M$ is
triangulated by a simplicial complex $K$. Let $\rho$ be a acyclic
representation of $\pi_{1}(K)$ by orthogonal matrices, i.e., the twisted
cohomology group $H^{p}(K;\rho)$ is trivial for all $p$. The Reidemeister
torsion $\tau_{\rho}(M)$ is defined from the cochain complex of $K$ by taking
a alternating product of determinants [10, 11]. It is a manifold invariant and
is used to distinguish homotopy equivalent spaces [2].

To describe the Reidemeister torsion in analytic terms, Ray and Singer [9]
defined an analytic torsion $T_{\rho}(M)$ for any compact oriented manifold
$M$ and orthogonal representation $\rho$ of the fundamental group
$\pi_{1}(M)$. Their definition used the spectrum of the Hodge Laplacian on
twisted forms. When $\rho$ is acyclic and orthogonal, Cheeger [1] and Müller
[12] proved that $\tau_{\rho}(M)=T_{\rho}(M)$. When $\rho$ is orthogonal but
not acyclic, one still can define the analytic torsion $T_{\rho}(M)$ [8].

In this paper we introduce the notions of Reidemeister and analytic torsions
on finite digraphs by means of the path homology theory of Grigoryan, Lin,
Muranov and Yau [3], [4], [5], [7]. Namely, we use the homology basis to
construct a preferred basis of the path complex on a digraph $G$, which leads
to the definition of the Reidemeister torsion $\tau(G)$. Next, we define the
Hodge Laplace operator $\Delta_{p}$ acting on $p$-paths and use the positive
eigenvalues of $\Delta_{p}$ in order to define the analytic torsion $T(G)$.
Although the homology groups can be nontrivial in our case, we still can prove
that $\tau(G)=T(G)$ (Theorem 3.14) by using an extension of the argument of
[9, Proposition 1.7].

Given two finite digraphs $X$ and $Y$, we obtain formulas for the torsions of
their Cartesian product $X\Box Y$ and join $X\ast Y$ (Theorems 4.8 and 5.7).
Our proofs rely essentially on the Künneth formulas for chain complexes of
$X\Box Y$ and $X\ast Y$ proved in [6] and [7]. The approach to the proof is
borrowed from [9, Thm. 2.5] but our setting is more complicated in the
following sense. The notion of torsion depends on the choice of an inner
product in the chain spaces, and the cases of the Cartesian product and join
require usage of different inner products. Besides, the case of join requires
usage of an augmented chain complex. For that reason, the final formulas for
$\tau\left(X\Box Y\right)$ and $\tau\left(X\ast Y\right)$ stated in
Corollaries 4.15 and 5.8 are more complicated than one could expect.

In Section 2 we revise the path homology theory. In Section 3 we introduce the
notions of the Hodge Laplacian on an arbitrary finite-dimensional chain
complex, prove the Hodge decomposition, define the notions of R-torsion and
analytic torsion, and prove their identity (Theorem 3.14).

In Section 4 we revise the notion of the Cartesian product of digraphs, the
Künneth formula for the Cartesian product, and use it to prove the formula for
the torsion of $X\Box Y$ (Theorem 4.8 and Corollaries 4.12, 4.15). In Section
5 we fulfil a similar program for the join of digraphs (Theorem 5.7 and
Corollaries 5.8, 5.9).

We give numerous examples of application of our results by computing torsions
of various digraphs including simplices, cubes, spheres, cycles, prism, etc.

##  2\. Path complexes and path homology

Let us briefly revise the definition of _path complex_ and _path homology_
introduced by Grigoryan, Lin, Muranov and Yau in [7] (see also [3]).

###  2.1. Path complex

Let $V$ be a finite set. For any $p\geq 0$, an _elementary_ $p$-_path_ is any
(ordered) sequence $i_{0},\ldots,i_{p}$ of $p+1$ vertices of $V$ that will be
denoted simply by $i_{0}\ldots i_{p}$ or by $e_{i_{0}\ldots i_{p}}$. The
number $p$ is called the length of the path $i_{0}...i_{p}.$

Formal $\mathbb{R}$-linear combinations of $e_{i_{0}...i_{p}}$ are called
$p$-_paths_. Denote by $\Lambda_{p}=\Lambda_{p}\left(V\right)$ the linear
space of all $p$-paths; that is, the elements of $\Lambda_{p}$ are

$v=\sum_{i_{0}i_{1}\ldots i_{p}}v^{{}_{i_{0}i_{1}\ldots
i_{p}}}e_{i_{0}i_{1}...i_{p}},\ \ \text{where}\ \ v^{{}_{i_{0}i_{1}\ldots
i_{p}}}\in\mathbb{R}.$

######  Definition 2.1.

For any $p\geq 0,$ define the _boundary operator_
$\partial:\Lambda_{p+1}\rightarrow\Lambda_{p}$ by

$\left(\partial v\right)^{i_{0}\ldots i_{p}}=\sum\limits_{k\in
V}\sum\limits_{q=0}^{p+1}\left(-1\right)^{q}v^{i_{0}\ldots i_{q-1}ki_{q}\ldots
i_{p}},$ (2.1)

where the index $k$ is inserted so that it is preceded by $q$ indices
$i_{0}...i_{q-1}.$ Set also $\Lambda_{-1}=\left\\{0\right\\}$ and define the
operator $\partial:\Lambda_{0}\rightarrow\Lambda_{-1}$ by setting $\partial
v=0$ for all $v\in\Lambda_{0}$.

It follows from (2.1) that

$\partial e_{j_{0}\ldots
j_{p+1}}=\sum\limits_{q=0}^{p+1}\left(-1\right)^{q}e_{j_{0}\ldots\widehat{j_{q}}\ldots
j_{p+1}},$ (2.2)

where $\widehat{\cdot}$ means omission of the index.

It is easy to show that $\partial^{2}v=0$ for any $v\in\Lambda_{p}$ ([7, Lemma
2.1]). Hence, the family of linear spaces $\left\\{\Lambda_{p}\right\\}$ with
the boundary operator $\partial$ determine a chain complex that will be
denoted by $\Lambda\left(V\right)$.

######  Definition 2.2.

An elementary $p$-path $e_{i_{0}\ldots i_{p}}$ on a set $V$ is called
_regular_ if $i_{k}\neq i_{k+1}$ for all $k=0,\ldots,p-1$, and _irregular_
otherwise.

Let $I_{p}$ be the subspace of $\Lambda_{p}$ that is spanned by all irregular
$e_{i_{0}\ldots i_{p}}.$ It is easy to verify that $\partial I_{p}\subset
I_{p-1}$ (cf. [7]). Hence, the boundary operator $\partial$ is well-defined on
the quotient space $\mathcal{R}_{p}:=\Lambda_{p}/I_{p}$:

$\partial:\mathcal{R}_{p}\rightarrow\mathcal{R}_{p-1}$

for all $p\geq 0.$ Clearly, $\mathcal{R}_{p}$ is linearly isomorphic to the
space of all regular $p$-paths:

$\mathcal{R}_{p}\cong\mathrm{span}\left\\{e_{i_{0}\ldots i_{p}}:i_{0}\ldots
i_{p}\text{ is regular}\right\\}.$ (2.3)

For simplicity of notation, we will identify $\mathcal{R}_{p}$ with the space
of all regular $p$-paths. With this identification, the formula (2.1) for the
operator $\partial:\mathcal{R}_{p+1}\rightarrow\mathcal{R}_{p}$ is true only
for regular paths $i_{0}...i_{p}$ whereas $\left(\partial
v\right)^{i_{0}\ldots i_{p}}=0$ if $i_{0}...i_{p}$ is irregular. The identity
(2.2) remains true if we replace by $0$ each irregular path on the right hand
side.

Denote by $\mathcal{R}\left(V\right)$ the chain complex
$\left\\{\mathcal{R}_{p}\right\\}$ with the boundary operator $\partial.$

######  Definition 2.3.

A _path complex_ over a set $V$ is a non-empty collection $P$ of regular
elementary paths on $V$ with the following property:

$\text{if\ }\ e_{i_{0}...i_{n}}\in P\ \text{then\ }e_{i_{0}...i_{n-1}}\in P\
\text{and\ }e_{i_{1}...i_{n}}\in P.$ (2.4)

When a path complex $P$ is fixed, all the paths from $P$ are called _allowed_
, whereas the elementary paths that are not in $P$ are called _non-allowed_.
Condition (2.4) means that if we remove the first or the last element of an
allowed $n$-path then the resulting $\left(n-1\right)$-path is also allowed.

The set of all $n$-paths from $P$ is denoted by $P_{n}$. The set $P_{-1}$
consists of a single empty path $e$. The elements of $P_{0}$ (that is, allowed
$0$-paths) are called the _vertices_ of $P$. Clearly, $P_{0}$ is a subset of
$V$. By the property (2.4), if $i_{0}...i_{n}\in P$ then all $i_{k}$ are
vertices of $P$. Hence, we can (and will) remove from the set $V$ all non-
vertices so that $V=P_{0}.$

There are two natural families/examples of path complexes. Any abstract finite
simplicial complex $S$ is a collection of subsets of a finite vertex set $V$
that satisfies the following property:

$\text{if\ }\sigma\in S\ \text{then any subset of }\sigma\ \text{is also in
}S.$

Let us enumerate the elements of $V$ by distinct reals and identify any subset
$s$ of $V$ with the elementary path that consists of the elements of $s$ put
in the (strictly) increasing order. Denote by $P\left(S\right)$ this
collections of elementary paths on $V$ that uniquely determines $S$. The
defining property of a simplex can be restated the following:

$\text{if \ }v\in P\left(S\right)\text{ then any subsequence of \ }v\text{ is
also in \ }P\left(S\right)\text{.}$ (2.5)

Consequently, the family $P\left(S\right)$ satisfies the property (2.4) so
that $P\left(S\right)$ is a path complex. The allowed $n$-paths in
$P\left(S\right)$ are exactly the $n$-simplexes.

###  2.2. Digraphs

Another natural family of path complexes comes from digraphs.

######  Definition 2.4.

A _digraph_ $G=(V,E)$ is a couple, where $V$ is a set, whose elements are
called the _vertices_ , and $E$ is a subset of $\\{V\times
V\setminus\mathrm{diag}\\}$ that consists of ordered pairs of vertices called
(directed) _edges_ or _arrows_. The fact that a pair $\left(x,y\right)$ is an
arrow will be denoted by $x\rightarrow y.$

An elementary $n$-path $i_{0}...i_{n}$ on the vertex set $V$ of a digraph is
called allowed if $i_{k-1}\rightarrow i_{k}$ for any $k=1,...,n$. Denote by
$P_{n}=P_{n}\left(G\right)$ the set of all allowed $n$-paths. In particular,
we have $P_{0}=V$ and $P_{1}=E$. Clearly, the collection
$P=\mathop{\textstyle\bigcup}_{n}P_{n}$ of all allowed paths satisfies the
condition (2.4) so that $P$ is a path complex. This path complex is naturally
associated with the digraph $G$ and will be denoted by $P\left(G\right)$.

###  2.3. Path homology

Let us return to an arbitrary path complex $P$ over $V$. Denote by
$\mathcal{A}_{p}\left(P\right)$ the subspace of
$\mathcal{R}_{p}\left(V\right)$ spanned by the allowed elementary $p$-paths,
that is,

$\mathcal{A}_{p}=\mathrm{span}{}\left\\{e_{i_{0}\ldots i_{p}}:i_{0}\ldots
i_{p}\in E_{p}\right\\}.$ (2.6)

The elements of $\mathcal{A}_{p}$ are called _allowed_ $p$-paths.

Note that the spaces $\mathcal{A}_{p}$ of allowed paths are in general _not_
invariant for $\partial$. Consider the following subspace of $\mathcal{A}_{p}$

$\Omega_{p}\equiv\Omega_{p}\left(P\right):=\left\\{v\in\mathcal{A}_{p}:\partial
v\in\mathcal{A}_{p-1}\right\\}.$ (2.7)

The spaces $\Omega_{p}$ are $\partial$-invariant. Indeed, $v\in\Omega_{p}$
implies $\partial v\in\mathcal{A}_{p-1}$ and $\partial\left(\partial
v\right)=0\in\mathcal{A}_{p-2}$, whence $\partial v\in\Omega_{p-1}$. The
elements of $\Omega_{p}$ are called $\partial$-_invariant_ $p$-paths.

Hence, we obtain a chain complex $\Omega=\Omega\left(P\right):$

$\begin{array}[]{cccccccccccc}0&\leftarrow&\Omega_{0}&\overset{\partial}{\leftarrow}&\Omega_{1}&\overset{\partial}{\leftarrow}&\cdots&\overset{\partial}{\leftarrow}&\Omega_{p-1}&\overset{\partial}{\leftarrow}&\Omega_{p}&\overset{\partial}{\leftarrow}\cdots\end{array}$
(2.8)

By construction we have $\Omega_{0}=\mathcal{A}_{0}$ and
$\Omega_{1}=\mathcal{A}_{1}$, while in general
$\Omega_{p}\subset\mathcal{A}_{p}$.

Set

$Z_{p}=\ker\partial|_{\Omega_{p}}\ \text{and}\ \ B_{p}=\partial\Omega_{p+1}.$

######  Definition 2.5.

Define for all $p\geq 0$ the _path homology groups_ $H_{p}\left(P\right)$ of
the path complex $P$ by

$H_{p}\left(P\right):=H_{p}\left(\Omega\left(P\right)\right)=Z_{p}/B_{p}.$
(2.9)

Let us note that the spaces $H_{p}\left(P\right)$ (as well as the spaces
$\Omega_{p}\left(P\right)$) can be computed directly by definition using
simple tools of linear algebra, in particular, those implemented in modern
computational software. On the other hand, some theoretical tools for
computation of homology groups, like homotopy theory and Künneth formulas,
were developed in [4], [6], [7].

In particular, for any digraph $G$ define its path homology groups by

$H_{p}\left(G\right)=H_{p}\left(P\left(G\right)\right).$

In what follows we are going to deal with only finite chain complexes:

$\begin{array}[]{cccccccccccccccc}0&\leftarrow&\Omega_{0}&\overset{\partial}{\leftarrow}&\Omega_{1}&\overset{\partial}{\leftarrow}&\cdots&\overset{\partial}{\leftarrow}&\Omega_{p-1}&\overset{\partial}{\leftarrow}&\Omega_{p}&\overset{\partial}{\leftarrow}\cdots&\overset{\partial}{\leftarrow}&\Omega_{N}&\leftarrow&0\end{array}$
(2.10)

where $N\in\mathbb{N}.$ Clearly, any chain complex (2.8) can be truncated to
the form (2.10).

For path complexes and digraphs this means that we restrict the length of
allowed paths to $N$. There is a large family of digraphs where the chain
complex $\Omega$ is finite naturally because $\Omega_{N}=\left\\{0\right\\}$
for some $N$ (and, hence, $\Omega_{n}=\left\\{0\right\\}$ for all $n\geq N$).
All examples of digraphs that are considered in this paper have naturally
finite chain complex $\Omega.$

If this is not the case then we can choose $N$ arbitrarily and truncate the
chain complex $\Omega$ to (2.10). The number $N$ will be referred to as the
dimension of the chain complex (2.10) or that of the underlying path complex.

Some examples of chain complexes $\Omega$ and homology groups of digraphs will
be given in Section 3.3.

##  3\. Finite chain complexes

Let us fix a finite chain complex $\Omega$ (2.10) of finite dimensional linear
spaces $\Omega_{p}.$ We are interested in chain complexes that are coming from
path complexes as described above, but in this section we revise rather well
known facts about general chain complexes $\Omega.$

Let us choose arbitrarily an inner product $\langle,\rangle$ in each linear
space $\Omega_{p}.$ In the case when $\Omega$ comes from a path complex, an
inner product in $\Omega_{p}$ can be taken from the ambient space
$\mathcal{R}_{p}.$ In this paper we use two different inner products in
$\mathcal{R}_{p}$. Let $u,v\in\mathcal{R}_{p}$ and

$u=\sum_{\mathbf{i}}u^{\mathbf{i}}e_{\mathbf{i}}\ \ \ \text{and\ \
}v=\sum_{\mathbf{i}}v^{\mathbf{i}}e_{\mathbf{i}}$

where $\mathbf{i}=i_{0}...i_{p}$. The first (_standard_) inner product is

$\langle u,v\rangle=\sum_{\mathbf{i}}u^{\mathbf{i}}v^{\mathbf{i}},$ (3.1)

and the second (_normalized_) inner product is

$\langle
u,v\rangle=\frac{1}{p!}\sum_{\mathbf{i}}u^{\mathbf{i}}v^{\mathbf{i}}.$ (3.2)

These inner products will be used in examples and in Section 4, but in general
we do not impose any restriction on the choice of inner products in the spaces
$\Omega_{p}.$

###  3.1. Hodge Laplacian

Denote by $\partial_{p}$ the operator
$\partial:\Omega_{p}\rightarrow\Omega_{p-1}.$ Assuming that the inner product
structure in $\Omega$ is chosen, consider the operator
$\partial_{p}^{\ast}:\Omega_{p-1}\rightarrow\Omega_{p}$ that is the adjoint
operator of $\partial_{p}$ with respect to the inner products in $\Omega_{p}$
and $\Omega_{p-1}$.

######  Definition 3.1.

Define the _Hodge-Laplace operator_
$\Delta_{p}:\Omega_{p}\rightarrow\Omega_{p}$ by

$\Delta_{p}u=\partial_{p}^{\ast}\partial_{p}u+\partial_{p+1}\partial_{p+1}^{\ast}u.$
(3.3)

We will use a shorter notation

$\Delta_{p}u=\partial^{\ast}\partial u+\partial\partial^{\ast}u$

since it is clear from this expression in which spaces $\Omega_{p}$ act the
operators $\partial$ and $\partial^{\ast}.$

An element $u\in\Omega_{p}$ is called _harmonic_ if $\Delta_{p}u=0.$

######  Lemma 3.2.

An element $u\in\Omega_{p}$ is harmonic if and only if $\partial u=0$ and
$\partial^{\ast}u=0.$

###### Proof.

If $\partial u=0$ and $\partial^{\ast}u=0$ then by (3.3) we have
$\Delta_{p}u=0.$ Conversely, if $\Delta_{p}u=0$ then we obtain

$0=\left\langle\Delta_{p}u,u\right\rangle=\left\langle\partial^{\ast}\partial
u,u\right\rangle+\left\langle\partial\partial^{\ast}u,u\right\rangle=\left\langle\partial
u,\partial
u\right\rangle+\left\langle\partial^{\ast}u,\partial^{\ast}u\right\rangle,$

whence $\left\|\partial u\right\|=\left\|\partial^{\ast}u\right\|=0.$

Denote by $\mathcal{H}_{p}$ the set of all harmonic elements in $\Omega_{p}$
so that $\mathcal{H}_{p}$ is a subspace of $\Omega_{p}.$

######  Lemma 3.3.

_(Hodge decomposition)_ The space $\Omega_{p}$ is an orthogonal sum of three
subspaces as follows:

$\Omega_{p}=\partial\Omega_{p+1}\mathop{\displaystyle\bigoplus}\partial^{\ast}\Omega_{p-1}\mathop{\displaystyle\bigoplus}\mathcal{H}_{p}.$
(3.4)

###### Proof.

If $u\in\partial\Omega_{p+1}$ and $v\in\partial^{\ast}\Omega_{p-1}$ then
$u=\partial u^{\prime}$ and $v=\partial^{\ast}v^{\prime}$, and we have

$\langle u,v\rangle=\langle\partial
u^{\prime},\partial^{\ast}v^{\prime}\rangle=\left\langle\partial^{2}u^{\prime},v^{\prime}\right\rangle=0$

so that the subspaces $\partial\Omega_{p+1}$ and $\partial^{\ast}\Omega_{p-1}$
are orthogonal. Denote by $K$ the orthogonal complement of
$\partial\Omega_{p+1}\mathop{\displaystyle\bigoplus}\partial^{\ast}\Omega_{p-1}$
in $\Omega_{p}.$ Then we have

$u\in K\Leftrightarrow\left\langle u,v\right\rangle=0\ \ \text{for all
}v\in\partial\Omega_{p+1}\ \text{and }v\in\partial^{\ast}\Omega_{p-1}$

that is,

$\displaystyle u\overset{}{\in}K$ $\displaystyle\Leftrightarrow$
$\displaystyle\left\langle u,\partial v^{\prime}\right\rangle=0\ \ \forall
v^{\prime}\in\Omega_{p+1}\ \ \text{and\ \ }\left\langle
u,\partial^{\ast}v\right\rangle\ \ \forall v\in\Omega_{p-1}$
$\displaystyle\Leftrightarrow$
$\displaystyle\left\langle\partial^{\ast}u,v^{\prime}\right\rangle=0\ \
\forall v^{\prime}\in\Omega_{p+1}\ \ \text{and\ \ }\left\langle\partial
u,v\right\rangle\ \ \forall v\in\Omega_{p-1}$ $\displaystyle\Leftrightarrow$
$\displaystyle\partial^{\ast}u=0\ \ \text{and\ \ }\partial u=0$
$\displaystyle\Leftrightarrow$ $\displaystyle u\in\mathcal{H}_{p}.$

Hence, $K=\mathcal{H}_{p}$ which finishes the proof.

######  Corollary 3.4.

There is a natural isomorphism

$H_{p}\cong\mathcal{H}_{p}.$ (3.5)

###### Proof.

Observe first that $Z_{p}:=\ker\partial_{p}$ is the orthogonal complement of
$\partial^{\ast}\Omega_{p-1}$ in $\Omega_{p}$ because, for any
$u\in\Omega_{p}$,

$u\in Z_{p}\Leftrightarrow\partial u=0\Leftrightarrow\left\langle
u,\partial^{\ast}v\right\rangle=0\ \forall v\in\Omega_{p-1}\Leftrightarrow
u\bot\partial^{\ast}\Omega_{p-1}.$

It follows from (3.4) that

$Z_{p}=\partial\Omega_{p+1}\mathop{\displaystyle\bigoplus}\mathcal{H}_{p}=B_{p}\mathop{\displaystyle\bigoplus}\mathcal{H}_{p}$
(3.6)

whence

$\mathcal{H}_{p}\cong Z_{p}/B_{p}=H_{p}.$

######  Remark 3.5.

It follows from this argument that $\mathcal{H}_{p}$ is an orthogonal
complement of $B_{p}$ in $Z_{p}$ and that a harmonic form
$u\in\mathcal{H}_{p}$ that corresponds to a homology class $\omega\in H_{p}$,
minimizes the norm $\left\|\cdot\right\|$ among all elements of $\omega.$

###  3.2. R-torsion

Let $\Omega$ be a finite chain complex of finite dimensional linear spaces
over $\mathbb{R}:$

$0\overset{\partial}{\leftarrow}\Omega_{0}\overset{\partial}{\leftarrow}\Omega_{1}\overset{\partial}{\leftarrow}...\overset{\partial}{\leftarrow}\Omega_{p-1}\overset{\partial}{\leftarrow}\Omega_{p}\overset{\partial}{\leftarrow}...\overset{\partial}{\leftarrow}\Omega_{N}\overset{\partial}{\leftarrow}0.$

Denote $B_{p}=\partial\Omega_{p+1}$, $Z_{p}=\ker\partial|_{\Omega_{p}}$ and
$H_{p}=Z_{p}/B_{p}.$

In any $\Omega_{p}$ choose a basis $\omega_{p}$ and a basis $h_{p}$ in
$H_{p}$. For each element of $h_{p}$ choose its representative in $Z_{p}$ and
denote the resulting independent set by $\widetilde{h}_{p}.$

Let $b_{p}$ be _any_ basis in $B_{p}$. For each element $w\in b_{p-1}$ choose
one element $v\in\partial^{-1}w\subset\Omega_{p}$ so that $\partial v=w$. Let
$\widetilde{b}_{p}$ be the collection of chosen elements $v$ so that

$b_{p-1}=\partial\widetilde{b}_{p}.$ (3.7)

Note that always $\widetilde{b}_{0}=\emptyset.$ Since $b_{p-1}$ is linearly
independent, the set $\widetilde{b}_{p}$ is also linearly independent.
Clearly, the union $(b_{p},\widetilde{h}_{p})$ is a basis in $Z_{p}$. Since
the subspaces $Z_{p}$ and $\mathop{\mathrm{s}pan}(\widetilde{b}_{p})$ of
$\Omega_{p}$ have a trivial intersection $\left\\{0\right\\}$, by the rank-
nullity theorem we conclude that the direct sum of these subspaces is
$\Omega_{p}$. Hence, the union $(b_{p},\widetilde{h}_{p},\widetilde{b}_{p})$
of the thee sequences is a basis in $\Omega_{p}$.

If $U$ and $W$ are two bases in an $n$-dimensional linear space, then denote
by $\left(U/W\right)$ the transformation matrix from $W$ to $U$ and set

$[U/W]=\left|\det\left(U/W\right)\right|.$

In the case $n=0$ set $[U/W]=1.$

Denote $\omega$ the collection $\left\\{\omega_{p}\right\\}$ of the bases in
$\Omega_{p}$ and similarly let $h=\left\\{h_{p}\right\\}$ be the collection of
the bases in $H_{p}.$

######  Definition 3.6.

The _R-torsion_ $\tau(\Omega,\omega,h)$ of the chain complex $\Omega$ with the
preferred bases $\omega$ and $h$ is a positive real number defined by

$\log\tau(\Omega,\omega,h)=\sum_{p=0}^{N}(-1)^{p}\log[b_{p},\widetilde{h}_{p},\widetilde{b}_{p}\,/\,\omega_{p}].$
(3.8)

We justify this definition in the following statement.

######  Lemma 3.7.

  * $\left(a\right)$

The value of $\tau\left(\Omega,\omega,h\right)$ does not depend on the choice
of the bases $b_{p}$, the representatives in $\widetilde{b}_{p}$ and the
representatives in $\widetilde{h}_{p}$ (which justifies the notation
$\tau(\Omega,\omega,h)$).

  * $\left(b\right)$

If $\omega^{\prime}$ and $h^{\prime}$ are other collections of bases in
$\Omega$ and $H$ respectively, then

$\log\tau\left(\Omega,\omega^{\prime},h^{\prime}\right)=\log\tau\left(\Omega,\omega,h\right)+\sum_{p=0}^{N}\left(-1\right)^{p}\left(\log\left[\omega_{p}/\omega_{p}^{\prime}\right]+\log\left[h_{p}^{\prime}/h_{p}\right]\right).$
(3.9)

The relation $\left[U/W\right]=1$ for bases $U$ and $W$ is an equivalence
relation, and each equivalence class determines a _volume form_ in the
underlying linear space. We see from (3.9) that
$\tau\left(\Omega,\omega,h\right)$ depends only on the volume forms determined
by $\omega$ and $h$ in the spaces $\Omega_{p}$ and $H_{p}$, respectively,

###### Proof of Lemma 3.7.

$\left(a\right)$ Let $b_{p}^{\prime}$ be another basis in $B_{p}$ with the
corresponding set $\widetilde{b}_{p}^{\prime}$, and
$\widetilde{h}_{p}^{\prime}$ be another set of representatives of $h_{p}$. Let
us first verify that

$[b_{p}^{\prime},\widetilde{h}_{p}^{\prime},\widetilde{b}_{p}^{\prime}\,/\,b_{p},\widetilde{h}_{p},\widetilde{b}_{p}]=[b_{p}^{\prime}/b_{p}][{b}_{p-1}^{\prime}/{b}_{p-1}].$
(3.10)

Let $\widetilde{h}_{p}=\left\\{u_{1},u_{2},...\right\\}$ and
$\widetilde{h}_{p}^{\prime}=\left\\{u_{1}^{\prime},u_{2}^{\prime},...\right\\}.$
Since $u_{i}^{\prime}$ and $u_{i}$ represent the same homology class, we have

$u_{i}^{\prime}=u_{i}+b_{i}\ \ \text{for some }b_{i}\in B_{p}.$ (3.11)

Let $\widetilde{b}_{p}=\left\\{v_{1},v_{2},...\right\\}$ and
$\widetilde{b}_{p}^{\prime}=\left\\{v_{1}^{\prime},v_{2}^{\prime},...\right\\}$
so that

$b_{p-1}=\partial\widetilde{b}_{p}=\left\\{\partial v_{1},\partial
v_{2},...\right\\}\ \ \text{and\ \ \
}b_{p-1}^{\prime}=\partial\widetilde{b}_{p}^{\prime}=\left\\{\partial
v_{1}^{\prime},\partial v_{2}^{\prime},...\right\\}.$

Since $\,b_{p-1}$ and $b_{p-1}^{\prime}$ are bases in the same subspace
$B_{p-1}$, the transformation matrix
$\left(c_{ij}\right)=\left({b}_{p-1}^{\prime}/{b}_{p-1}\right)$ is well
defined so that

$\partial v_{i}^{\prime}=\sum_{j}c_{ij}\partial v_{j}.$

It follows that

$v_{i}^{\prime}=z_{i}+\sum_{j}c_{ij}v_{j}\ \ \text{for some }z_{i}\in Z_{p}.$
(3.12)

Since $Z_{p}=\mathop{\mathrm{s}pan}\left(b_{p},h_{p}\right)$, we obtain from
(3.11) and (3.12) that

$(b_{p}^{\prime},\widetilde{h}_{p}^{\prime},\widetilde{b}_{p}^{\prime}\,/\,b_{p},\widetilde{h}_{p},\widetilde{b}_{p})=\left(\begin{array}[]{ccc}(b_{p}^{\prime}/b_{p})&\vdots&\vdots\\\
0&\mathop{\mathrm{i}d}&\vdots\\\
0&0&(b_{p-1}^{\prime}/b_{p-1})\end{array}\right)$

where the dots $\vdots$ denote the terms coming from $b_{i}$ and $z_{i}$.
Since this matrix is upper block-diagonal, we obtain (3.10).

Consequently, we have

$\displaystyle[b_{p}^{\prime},\widetilde{h}_{p}^{\prime},\widetilde{b}_{p}^{\prime}\,/\,\omega_{p}]$
$\displaystyle=$
$\displaystyle[b_{p}^{\prime},\widetilde{h}_{p}^{\prime},\widetilde{b}_{p}^{\prime}\,/\,b_{p},\widetilde{h}_{p},\widetilde{b}_{p}][b_{p},\widetilde{h}_{p},\widetilde{b}_{p}\,/\,\omega_{p}]$
$\displaystyle=$
$\displaystyle[b_{p}^{\prime}/b_{p}][{b}_{p-1}^{\prime}/{b}_{p-1}][b_{p},\widetilde{h}_{p},\widetilde{b}_{p}\,/\,\omega_{p}].$

Computing the sum in (3.8) we obtain

$\displaystyle\sum_{p=0}^{N}(-1)^{p}\log[b_{p}^{\prime},\widetilde{h}_{p}^{\prime},\widetilde{b}_{p}^{\prime}\,/\,\omega_{p}]$
$\displaystyle=$
$\displaystyle\sum_{p=0}^{N}\left(-1\right)^{p}\log[b_{p},\widetilde{h}_{p},\widetilde{b}_{p}/\omega_{p}]$
(3.13)
$\displaystyle+\sum_{p=0}^{N}\left(-1\right)^{p}\log[b_{p}^{\prime}/b_{p}]+\sum_{p=0}^{N}\left(-1\right)^{p}\log[{b}_{p-1}^{\prime}/{b}_{p-1}].$

It remains to observe that the expression in (3.13) vanishes because it is
equal to

$\sum_{p=0}^{N}\left(-1\right)^{p}\log[b_{p}^{\prime}/b_{p}]+\sum_{q=-1}^{N-1}\left(-1\right)^{q+1}\log[{b}_{q}^{\prime}/{b}_{q}]=\left(-1\right)^{N}\log[b_{N}^{\prime}/b_{N}]=0.$

$\left(b\right)$ Let $h_{p}=\left\\{\eta_{1},\eta_{2},...\right\\}$ and
$h_{p}^{\prime}=\left\\{\eta_{1}^{\prime},\eta_{2}^{\prime},...\right\\}$ so
that

$\eta_{i}^{\prime}=\sum_{j}c_{ij}\eta_{j}$

where $\left(c_{ij}\right)=\left(h_{p}^{\prime}/h_{p}\right).$ For the
representatives $u_{i}\in\widetilde{h}_{p}$ of $\eta_{i}$ and
$u_{i}^{\prime}\in\widetilde{h}_{p}^{\prime}$ of $\eta_{i}^{\prime}$ in
$Z_{p}$ we have then

$u_{i}^{\prime}=\sum_{j}c_{ij}u_{ij}+b_{i}\ \ \text{for some }b_{i}\in B_{p}.$

It follows that

$(b_{p},\widetilde{h}_{p}^{\prime},\widetilde{b}_{p}\,/\,b_{p},\widetilde{h}_{p},\widetilde{b}_{p}\,)=\left(\begin{array}[]{ccc}\mathop{\mathrm{i}d}&\vdots&0\\\
0&(h_{p}^{\prime}/h_{p})&0\\\ 0&0&\mathop{\mathrm{i}d}\end{array}\right)$

where the dots $\vdots$ denote the terms coming from $b_{i}$. Hence, we obtain

$\displaystyle[b_{p},\widetilde{h}_{p}^{\prime},\widetilde{b}_{p}\,/\,\omega_{p}^{\prime}]$
$\displaystyle=$
$\displaystyle[b_{p},\widetilde{h}_{p}^{\prime},\widetilde{b}_{p}\,/\,b_{p},\widetilde{h}_{p},\widetilde{b}_{p}][b_{p},\widetilde{h}_{p},\widetilde{b}_{p}\,/\,\omega_{p}]\left[\omega_{p}/\omega_{p}^{\prime}\right]$
$\displaystyle=$
$\displaystyle[h_{p}^{\prime}/h_{p}][b_{p},\widetilde{h}_{p},\widetilde{b}_{p}\,/\,\omega_{p}]\left[\omega_{p}/\omega_{p}^{\prime}\right],$

whence (3.9) follows.

Let us fix an inner product in each space $\Omega_{p}$ and denote by $\iota$
the _inner product structure_ in $\Omega,$ that is, the collection of all
inner products for $p=0,...,N.$ Then we have the induced inner product in the
subspaces $B_{p}$, $Z_{p}$ and $\mathcal{H}_{p}.$ Using the isomorphism
$H_{p}\cong\mathcal{H}_{p}$ we transfer the inner product to $H_{p}.$ Hence,
in this case we have a canonical choice of volume forms $\omega$ in
$\Omega_{\ast}$ and $h$ in $H_{\ast}$ as we prefer orthonormal bases
$\omega_{p}$ in $\Omega_{p}$ and $h_{p}$ in $H_{p}$. In fact, we can identify
$h_{p}$ with an orthonormal basis in $\mathcal{H}_{p}$ and set
$\widetilde{h}_{p}=h_{p}.$ With this choice of $\omega$ and $h$, we define the
R-torsion of $\left(\Omega,\iota\right)$ by

$\tau\left(\Omega,\iota\right)=\tau\left(\Omega,\omega,h\right).$

By (3.9) the right hand side does not depend on the choice of orthonormal
bases $\omega$ and $h$.

######  Corollary 3.8.

Let $\iota$ and $\iota^{\prime}$ be two inner product structures in $\Omega.$
Assume that there are positive reals $c_{p}$, $p=0,...,N$, such that, for all
$u,v\in\Omega_{p},$

$\iota^{\prime}\left(u,v\right)=c_{p}\iota\left(u,v\right).$

Then

$\tau\left(\Omega,\iota^{\prime}\right)=\tau\left(\Omega,\iota\right)\prod_{p=0}^{N}c_{p}^{\frac{1}{2}\left(-1\right)^{p}\left(\dim\Omega_{p}-\dim
H_{p}\right)}.$ (3.14)

In particular, if all $c_{p}=c$ are equal, then we obtain

$\tau\left(\Omega,\iota^{\prime}\right)=\tau\left(\Omega,\iota\right)$

because

$\sum_{p=0}^{N}\left(-1\right)^{p}\dim\Omega_{p}=\sum_{p=0}^{N}\left(-1\right)^{p}\dim
H_{p}=\chi\left(\Omega\right),$

where $\chi\left(\Omega\right)$ is the Euler characteristic of $\Omega.$

###### Proof.

Since the notion of orthogonality is the same for $\iota$ and
$\iota^{\prime}$, the space $\mathcal{H}_{p}$ is also the same. If
$\omega_{p}$ and $h_{p}$ are $\iota$-orthonormal bases in $\Omega_{p}$ and
$\mathcal{H}_{p}$, respectively, then
$\omega_{p}^{\prime}=\frac{1}{\sqrt{c_{p}}}\omega_{p}$ and
$h_{p}^{\prime}=\frac{1}{\sqrt{c_{p}}}h_{p}$ are $\iota^{\prime}$-orthonormal
bases. Since

$\left[\omega_{p}/\omega_{p}^{\prime}\right]=c_{p}^{\dim\Omega_{p}/2}\ \ \
\text{and\ \ \ }\left[h_{p}^{\prime}/h_{p}\right]=c_{p}^{-\dim H_{p}/2},$

we obtain from (3.9)

$\log\tau\left(\Omega,\iota^{\prime}\right)=\log\tau\left(\Omega,\iota\right)+\frac{1}{2}\sum_{p=0}^{N}\left(-1\right)^{p}\left(\dim\Omega_{p}-\dim
H_{p}\right)\log c_{p},$

whence (3.14) follows.

Let $\Omega$ be a chain complex that comes from a path complex $P.$ Then the
inner product in $\Omega_{p}$ can be taken from the ambient space
$\mathcal{R}_{p}.$ The so obtained R-torsion
$\tau\left(\Omega\left(P\right),\iota\right)$ will also be denoted by
$\tau\left(P,\iota\right).$

Let $\iota$ be the standard inner product (3.1) in $\mathcal{R}_{p}$ and
$\iota^{\prime}$ be the normalized inner product (3.2) in $\mathcal{R}_{p}.$
We set

$\tau\left(P\right)=\tau\left(P,\iota\right)\ \ \text{and\ \
}\tau^{\prime}\left(P\right)=\tau\left(P,\iota^{\prime}\right).$

In this case $c_{p}=\frac{1}{p!}$ and we obtain from (3.14)

$\tau^{\prime}\left(P\right)=\tau\left(P\right)\prod_{p=0}^{N}\left(p!\right)^{\frac{1}{2}\left(-1\right)^{p+1}\left(\dim\Omega_{p}-\dim
H_{p}\right)}.$ (3.15)

Denoting

$r_{p}=\dim\Omega_{p}-\dim H_{p}$

and observing that

$\sum_{p=0}^{N}\left(-1\right)^{p}r_{p}=0$

and

$\displaystyle\sum_{p=2}^{N}\left(-1\right)^{p+1}r_{p}\log\left(p!\right)$
$\displaystyle=$
$\displaystyle\sum_{p=2}^{N}\left(-1\right)^{p+1}r_{p}\sum_{k=2}^{p}\log k$
$\displaystyle=$ $\displaystyle\sum_{k=2}^{N}\log
k\sum_{p=k}^{N}\left(-1\right)^{p+1}r_{p}$ $\displaystyle=$
$\displaystyle\sum_{k=2}^{N}\log k\sum_{p=0}^{k-1}\left(-1\right)^{p}r_{p}$
$\displaystyle=$ $\displaystyle\sum_{k=2}^{N}\log
k^{r_{0}-r_{1}+...+\left(-1\right)^{k-1}r_{k-1}},$

we obtain from (3.15)

$\tau^{\prime}\left(P\right)=\tau\left(P\right)\left(2^{r_{0}-r_{1}}\cdot
3^{r_{0}-r_{1}+r_{2}}\cdot 4^{r_{0}-r_{1}+r_{2}-r_{3}}\cdot...\right)^{1/2}.$
(3.16)

###  3.3. Examples

Let us give some examples of computation of R-torsion by definition.

######  Example 3.9.

Consider a _line_ digraph $G=\left(V,E\right)$ that consists of $m$ vertices
$V=\left\\{0,1,...,m-1\right\\}$ and $m-1$ arrows having the form either
$i\rightarrow i+1$ or $i+1\rightarrow i$, for $i=0,...,m-2.$ An example of a
line digraph is shown in Fig. 1.

Figure 1. A line digraph with $m=5$

Denote

$\overline{e}_{i\left(i+1\right)}=\left\\{\begin{array}[]{ll}e_{i\left(i+1\right)}&\text{if
}i\rightarrow i+1\\\ e_{\left(i+1\right)i}&\text{if }i+1\rightarrow
i\end{array}\right.$ (3.17)

so that $\overline{e}_{i\left(i+1\right)}\in\Omega_{1}$, and set

$\sigma_{i}=\left\\{\begin{array}[]{ll}1,&\text{if }i\rightarrow i+1\\\
-1,&\text{if }i+1\rightarrow i\end{array}\right.$ (3.18)

so that

$\partial\overline{e}_{i\left(i+1\right)}=\sigma_{i}\left(e_{i+1}-e_{i}\right).$

Choose the following $\iota$-orthonormal bases in $\Omega_{0}$ and
$\Omega_{1}$:

$\omega_{0}=\left\\{e_{i}:i=0,...,m-1\right\\}$

and

$\omega_{1}=\left\\{\overline{e}_{i\left(i+1\right)}:i=0,...,m-2\right\\}.$

Clearly, $\Omega_{p}=\left\\{0\right\\}$ for $p\geq 2.$ In particular, we have
$\chi\left(G\right)=1.$ Since $\dim H_{0}=1$ (as for any connected graph) and
$\dim H_{p}=0$ for $p\geq 2$, it follows that $\dim H_{1}=0.$

Since $B_{1}=\partial\Omega_{2}=\left\\{0\right\\}$, it follows that also
$Z_{1}=\left\\{0\right\\}.$ We have $Z_{0}=\Omega_{0}$ and, hence, $\dim
B_{0}=m-1$. Choose in $B_{0}=\partial\Omega_{1}$ the basis

$b_{0}=\left\\{\sigma_{i}\left(e_{i+1}-e_{i}\right),\ \ i=0,...,m-2\right\\}$

and set, respectively,

$\ \widetilde{b}_{1}=\left\\{\overline{e}_{i\left(i+1\right)},\
i=0,...,m-2\right\\}.$

The orthogonal complement of $B_{0}$ in $Z_{0}$ is one-dimensional:

$\mathcal{H}_{0}=\mathop{\mathrm{s}pan}\left\\{e_{0}+...+e_{m-1}\right\\},$

so that

$h_{0}=\left\\{\tfrac{1}{\sqrt{m}}\left(e_{0}+...+e_{m-1}\right)\right\\}.$

We see that

$[b_{0},h_{0},\widetilde{b}_{0}\,/\,\omega_{0}]=\left|\det\right|\left(\begin{array}[]{ccccccc}-\sigma_{0}&&&&&0&\tfrac{1}{\sqrt{m}}\\\
\sigma_{0}&-\sigma_{1}&&&&&\tfrac{1}{\sqrt{m}}\\\
&\sigma_{1}&\ddots&&&&\vdots\\\ &&\ddots&\ddots&&&\vdots\\\
&&&\ddots&\ddots&&\vdots\\\ &&&&\ddots&-\sigma_{m-2}&\vdots\\\
0&&&&&\sigma_{m-2}&\tfrac{1}{\sqrt{m}}\end{array}\right)=\sqrt{m},$ (3.19)

because expanding the determinant in the last column, we obtain that is equal
to

$\left(-1\right)^{m+1}m\sigma_{0}...\sigma_{m-2}\frac{1}{\sqrt{m}}.$

Since $(b_{1},h_{1},\widetilde{b}_{1})\,=\omega_{1}$, it follows that

$\tau\left(G\right)=\mathop{\displaystyle\prod}\limits_{p=0}^{1}[b_{p},h_{p},\widetilde{b}_{p}\,/\,\omega_{p}]^{(-1)^{p}}=\sqrt{m}.$

For the normalized inner product $\iota^{\prime}$ we have the same value
$\tau^{\prime}\left(G\right)=\sqrt{m}$ since $\Omega_{p}=\left\\{0\right\\}$
for all $p\geq 2$.

######  Example 3.10.

Consider a digraph $G=\left(V,E\right)$ with the vertex set
$V=\left\\{0,1,2\right\\}$ and with the edge set $E=\left\\{01,12,02\right\\}$
(Fig. 2). This digraph is called a _triangle_.

Figure 2. A triangle digraph

We have

$\Omega_{0}=\mathop{\mathrm{s}pan}\left\\{e_{0},e_{1},e_{2}\right\\},\ \
\Omega_{1}=\mathop{\mathrm{s}pan}\left\\{e_{01},e_{12},e_{02}\right\\},\
\Omega_{2}=\left\\{e_{012}\right\\}$

and $\Omega_{p}=\left\\{0\right\\}$ otherwise. Hence,

$B_{0}=\partial\Omega_{1}=\mathop{\mathrm{s}pan}\left\\{e_{1}-e_{0},e_{2}-e_{1}\right\\},\
\ \
B_{1}=\partial\Omega_{2}=\mathop{\mathrm{s}pan}\left\\{e_{01}-e_{02}+e_{12}\right\\}$

and $B_{p}=\left\\{0\right\\}$ otherwise. Next, we have

$Z_{0}=\mathop{\mathrm{s}pan}\left\\{e_{0},e_{1},e_{2}\right\\},\ \ \
Z_{1}=\mathop{\mathrm{s}pan}\left\\{e_{01}-e_{02}+e_{12}\right\\}$

and $Z_{p}=\left\\{0\right\\}$ otherwise. It follows that $\dim H_{0}=1$ and
$\dim H_{p}=0$ otherwise.

We choose the following $\iota$-orthonormal bases in $\Omega_{p}$:

$\omega_{0}=\left\\{e_{0},e_{1},e_{2}\right\\},\ \ \
\omega_{1}=\left\\{e_{01},e_{12},e_{02}\right\\},\ \
\omega_{2}=\left\\{e_{012}\right\\}.$

Choose also

$\displaystyle b_{0}$ $\displaystyle=$
$\displaystyle\left\\{e_{1}-e_{0},e_{2}-e_{1}\right\\},\ \ \
\widetilde{b}_{1}=\left\\{e_{01},e_{12}\right\\}$ $\displaystyle b_{1}$
$\displaystyle=$ $\displaystyle\left\\{e_{01}-e_{02}+e_{12}\right\\},\ \
\widetilde{b}_{2}=\left\\{e_{012}\right\\}.$

The orthogonal complement of $B_{0}$ in $Z_{0}$ is

$\mathcal{H}_{0}=\mathop{\mathrm{s}pan}\left\\{e_{0}+e_{1}+e_{2}\right\\},$

so that

$h_{0}=\left\\{\tfrac{1}{\sqrt{3}}\left(e_{0}+e_{1}+e_{2}\right)\right\\}.$

We see that

$(b_{p},h_{p},\widetilde{b}_{p})=\left\\{\begin{array}[]{ll}\\{e_{1}-e_{0},e_{2}-e_{1},\frac{1}{\sqrt{3}}\left(e_{0}+e_{1}+e_{2}\right)\\},&p=0\\\
\left\\{e_{01}-e_{02}+e_{12},e_{01},e_{12}\right\\},&p=1\\\
\left\\{e_{012}\right\\},&p=2.\end{array}\right.$

It follows that

$[b_{0},h_{0},\widetilde{b}_{0}\,/\,\omega_{0}]=\left|\det\right|\left(\begin{array}[]{ccc}-1&0&\frac{1}{\sqrt{3}}\\\
1&-1&\frac{1}{\sqrt{3}}\\\ 0&1&\frac{1}{\sqrt{3}}\end{array}\right)=\sqrt{3},$
$[b_{1},h_{1},\widetilde{b}_{1}\,/\,\omega_{1}]=\left|\det\right|\left(\begin{array}[]{ccc}1&1&0\\\
1&0&1\\\ -1&0&0\end{array}\right)=1$

and

$[b_{2},h_{2},\widetilde{b}_{2}\,/\,\omega_{2}]=\left|\det\right|\left(1\right)=1.$

Hence, we obtain

$\tau\left(G\right)=\mathop{\displaystyle\prod}\limits_{p=0}^{2}[b_{p},h_{p},\widetilde{b}_{p}\,/\,\omega_{p}]^{(-1)^{p}}=\sqrt{3}.$

For the normalized inner product $\iota^{\prime}$ we obtain from (3.14)

$\tau^{\prime}\left(G\right)=\tau\left(G\right)\prod_{p=0}^{2}\left(p!\right)^{\frac{1}{2}\left(-1\right)^{p+1}\left(\dim\Omega_{p}-\dim
H_{p}\right)}=\sqrt{3}2^{-\frac{1}{2}}=\sqrt{3/2}.$

######  Example 3.11.

Consider a digraph $G=\left(V,E\right)$ with the set of vertices
$V=\left\\{0,1,2,3\right\\}$ and the set of edges
$E=\left\\{01,02,13,23\right\\}$ (Fig. 3). This digraph is called a _square_.

Figure 3. A square digraph

We have

$\Omega_{0}=\mathop{\mathrm{s}pan}\left\\{e_{0},e_{1},e_{2},e_{3}\right\\},\
\Omega_{1}=\mathop{\mathrm{s}pan}\left\\{e_{01},e_{02},e_{13},e_{23}\right\\},\
\ \Omega_{2}=\mathop{\mathrm{s}pan}\left\\{e_{013}-e_{023}\right\\}$

and $\Omega_{p}=\left\\{0\right\\}$ otherwise. Hence,

$\displaystyle B_{0}$ $\displaystyle=$
$\displaystyle\partial\Omega_{1}=\mathop{\mathrm{s}pan}\left\\{e_{1}-e_{0},e_{2}-e_{0},e_{3}-e_{1}\right\\}$
$\displaystyle B_{1}$ $\displaystyle=$
$\displaystyle\partial\Omega_{2}=\mathop{\mathrm{s}pan}\left\\{e_{01}+e_{13}-e_{02}-e_{23}\right\\}$

and $B_{p}=\left\\{0\right\\}$ otherwise. Next we have

$Z_{0}=\mathop{\mathrm{s}pan}\left\\{e_{0},e_{1},e_{2},e_{3}\right\\},\ \ \
Z_{1}=\mathop{\mathrm{s}pan}\left\\{e_{01}+e_{13}-e_{02}-e_{23}\right\\}$

and $Z_{p}=\left\\{0\right\\}$ otherwise. Consequently, $\,\dim H_{0}=1$ and
$\dim H_{0}=0$ for $p\geq 1.$

We choose the following $\iota$-orthonormal bases in $\Omega_{p}$:

$\omega_{0}=\left\\{e_{0},e_{1},e_{2},e_{3}\right\\},\ \ \
\omega_{1}=\left\\{e_{01},e_{02},e_{13},e_{23}\right\\},\ \
\omega_{2}=\\{\tfrac{1}{\sqrt{2}}\left(e_{013}-e_{023}\right)\\}.$

Choose also

$\displaystyle b_{0}$ $\displaystyle=$
$\displaystyle\left\\{e_{1}-e_{0},e_{2}-e_{0},e_{3}-e_{1}\right\\},\ \ \
\widetilde{b}_{1}=\left\\{e_{01},e_{02},e_{13}\right\\}$ $\displaystyle b_{1}$
$\displaystyle=$ $\displaystyle\left\\{e_{01}-e_{02}+e_{13}-e_{23}\right\\},\
\ \widetilde{b}_{2}=\left\\{e_{013}-e_{023}\right\\}.$

The orthogonal complement of $B_{0}$ in $Z_{0}$ is

$\mathcal{H}_{0}=\mathop{\mathrm{s}pan}\left\\{e_{0}+e_{1}+e_{2}+e_{3}\right\\}$

and we take

$h_{0}=\left\\{\tfrac{1}{2}\left(e_{0}+e_{1}+e_{2}+e_{3}\right)\right\\}.$

It follows that

$(b_{p},h_{p},\widetilde{b}_{p})=\left\\{\begin{array}[]{ll}\left\\{e_{1}-e_{0},e_{2}-e_{0},e_{3}-e_{1},\frac{1}{2}\left(e_{0}+e_{1}+e_{2}+e_{3}\right)\right\\},&p=0\\\
\left\\{e_{01}-e_{02}+e_{13}-e_{23},e_{01},e_{02},e_{13}\right\\},&p=1\\\
\left\\{e_{013}-e_{023}\right\\},&p=2.\end{array}\right.$

Hence,

$[b_{0},h_{0},\widetilde{b}_{0}\,/\,\omega_{0}]=\left|\det\right|\left(\begin{array}[]{cccc}-1&-1&0&\frac{1}{2}\\\
1&0&-1&\frac{1}{2}\\\ 0&1&0&\frac{1}{2}\\\
0&0&1&\frac{1}{2}\end{array}\right)=2,$
$[b_{1},h_{1},\widetilde{b}_{1}\,/\,\omega_{1}]=\left|\det\right|\left(\begin{array}[]{cccc}1&1&0&0\\\
-1&0&-1&0\\\ 1&1&0&1\\\ -1&0&0&0\end{array}\right)=1$
$[b_{2},h_{2},\widetilde{b}_{2}\,/\,\omega_{2}]=\left|\det\right|(\sqrt{2})=\sqrt{2},$

and we obtain

$\tau\left(G\right)=\mathop{\displaystyle\prod}\limits_{p=0}^{2}[b_{p},h_{p},\widetilde{b}_{p}\,/\,\omega_{p}]^{(-1)^{p}}=2\sqrt{2}.$

For the normalized inner product $\iota^{\prime}$ we obtain from (3.14)

$\tau^{\prime}\left(G\right)=\tau\left(G\right)\prod_{p=0}^{2}\left(p!\right)^{\frac{1}{2}\left(-1\right)^{p+1}\left(\dim\Omega_{p}-\dim
H_{p}\right)}=2\sqrt{2}2^{-\frac{1}{2}}=2.$

Note that the triangle and square digraphs have the same homology groups and
are even homotopy equivalent (see [4]) but their torsions are different.
Moreover, the torsion is not preserved by covering mappings between digraphs,
which are surjective mappings that preserve arrows. For example, consider a
mapping $\Phi:X\rightarrow Y$ of a square $X$ on Fig. 3 onto a line digraph
$Y=\\{0\rightarrow 1\rightarrow 2\\}$ such that $\Phi\left(0\right)=0$,
$\Phi\left(1\right)=\Phi\left(2\right)=1$ and $\Phi\left(3\right)=2,$ which is
obviously covering but $\tau\left(X\right)=2\sqrt{2}$ while
$\tau\left(Y\right)=\sqrt{3}.$

######  Example 3.12.

We say that a digraph $G=\left(V,E\right)$ is _cyclic_ if it is connected (as
an undirected graph), every vertex had the degree $2,$ and there are no double
arrows. For example, the triangle from Example 3.10 and the square from
Example 3.11 are cyclic.

Here we assume that $G$ is neither triangle nor square. Some examples of such
digraphs are shown on Fig. 4.

Figure 4. Three cyclic digraphs with $3$, $4$ and $6$ vertices

Note that a triangular digraph on Fig. 4 is not a triangle in the sense of
Example 3.10 because of different orientation of the arrows, and the
quadrilateral digraph here is not a square for the same reason.

For a cyclic digraph that is neither triangle nor square, it is known that
$\Omega_{p}\left(G\right)=\left\\{0\right\\}$ and
$H_{p}\left(G\right)=\left\\{0\right\\}$ for all $p\geq 2$, whereas

$\dim H_{0}\left(G\right)=\dim H_{1}\left(G\right)=1$

and, hence, $\chi\left(G\right)=0$ (see [3, Sect. 4.5]). Assume that $G$ has
$m$ vertices $0,1,...,m-1$ that we identify with residues
$\mathop{\mathrm{m}od}m.$ The numeration of vertices can be chosen so that all
arrows have the form either $i\rightarrow i+1$ or $i+1\rightarrow i$, for
$i=0,....,m-1.$

Let us use notations $\overline{e}_{i\left(i+1\right)}$ from (3.17) and
$\sigma_{i}$ from (3.18) so that
$\overline{e}_{i\left(i+1\right)}\in\Omega_{1}$ and

$\partial\overline{e}_{i\left(i+1\right)}=\sigma_{i}\left(e_{i+1}-e_{i}\right).$

Choose the following $\iota$-orthonormal bases in $\Omega_{0}$ and
$\Omega_{1}$:

$\omega_{0}=\left\\{e_{i}:i=0,...,m-1\right\\}$

and

$\omega_{1}=\left\\{\overline{e}_{i\left(i+1\right)}:i=0,...,m-1\right\\}.$

Observe that $Z_{0}=\Omega_{0}$ and

$Z_{1}=\ker\partial|_{\Omega_{1}}=\mathop{\mathrm{s}pan}\left\\{\sum_{i=0}^{m-1}\sigma_{i}\overline{e}_{i\left(i+1\right)}\right\\}$

because

$\partial\left(\sum_{i}\alpha_{i}\overline{e}_{i\left(i+1\right)}\right)=\sum_{i}\alpha_{i}\sigma_{i}\left(e_{i+1}-e_{i}\right)=\sum_{i}\left(\alpha_{i-1}\sigma_{i-1}-\alpha_{i}\sigma_{i}\right)e_{i},$

which vanishes if $\alpha_{i}$ is proportional to $1/\sigma_{i}=\sigma_{i}.$
Then $B_{0}=\partial\Omega_{1}$ has dimension $m-1$ and we choose

$b_{0}=\left\\{\sigma_{i}\left(e_{i+1}-e_{i}\right),\ \ i=0,...,m-2\right\\}$

and, respectively,

$\ \widetilde{b}_{1}=\left\\{\overline{e}_{i\left(i+1\right)},\
i=0,...,m-2\right\\}.$

The orthogonal complement of $B_{0}$ in $Z_{0}=\Omega_{0}$ is

$\mathcal{H}_{0}=\mathop{\mathrm{s}pan}\left\\{e_{0}+...+e_{m-1}\right\\},$

so that

$h_{0}=\\{\tfrac{1}{\sqrt{m}}\left(e_{0}+...+e_{m-1}\right)\\}.$

Hence, as in (3.19), we obtain

$[b_{0},h_{0},\widetilde{b}_{0}\,/\,\omega_{0}]=\sqrt{m}.$

Next, we have $B_{1}=\partial\Omega_{2}=\left\\{0\right\\}$ whence
$b_{1}=\emptyset$, $\mathcal{H}_{1}=Z_{1}$ and

$h_{1}=\\{\tfrac{1}{\sqrt{m}}\sum_{i=0}^{m-1}\sigma_{i}\overline{e}_{i\left(i+1\right)}\\}.$

We see that

$[b_{1},h_{1},\widetilde{b}_{1}\,/\,\omega_{1}]=\left|\det\right|\left(\begin{array}[]{cccccc}\tfrac{1}{\sqrt{m}}\sigma_{0}&1&&&&0\\\
\tfrac{1}{\sqrt{m}}\sigma_{1}&&1&&&\\\ \vdots&&&\ddots&&\\\
\vdots&&&&\ddots&\\\ \vdots&&&&&1\\\
\frac{1}{\sqrt{m}}\sigma_{m-1}&0&&&&0\end{array}\right)=\frac{1}{\sqrt{m}}.$

It follows that

$\tau\left(G\right)=\mathop{\displaystyle\prod}\limits_{p=0}^{1}[b_{p},h_{p},\widetilde{b}_{p}\,/\,\omega_{p}]^{(-1)^{p}}=m$

and also $\tau^{\prime}\left(G\right)=m.$

###  3.4. Analytic torsion

Let $\Omega$ be a chain complex as above and $\iota$ be an inner product
structure on $\Omega$. It is easy to check from (3.3) that $\Delta_{p}$ is a
self-adjoint non-negative definite operator on $\Omega_{p}$. Hence, its
eigenvalues are non-negative reals, denote them by
$\left\\{\lambda_{i}\right\\}_{i=1}^{\dim\Omega_{p}}.$ The zeta function
$\zeta_{p}\left(s\right)$ of $\Delta_{p}$ is defined by

$\zeta_{p}(s)=\sum_{\lambda_{i}>0}\frac{1}{\lambda_{i}^{s}}.$

######  Definition 3.13.

The _analytic torsion_ $T(\Omega,\iota)$ of the chain complex $\Omega$ with an
inner product structure $\iota$ is defined by

$\log
T(\Omega,\iota)=\frac{1}{2}\sum_{p=0}^{N}(-1)^{p}\,p\,\zeta_{p}^{\prime}(0).$
(3.20)

The next theorem is one of the main results of this paper.

######  Theorem 3.14.

We have

$\tau(\Omega,\iota)=T\left(\Omega,\iota\right).$

This theorem was proved in [9, Proposition 1.7] for a special case when the
homology groups $H_{p}$ are trivial. We use a modification of the argument of
[9] that works with arbitrary homology groups.

###### Proof.

Observe that

$\zeta_{p}^{\prime}(s)=-\sum_{\lambda_{i}>0}\left(\log\lambda_{i}\right)\lambda_{i}^{-s},$

whence

$\zeta_{p}^{\prime}\left(0\right)=-\sum_{\lambda_{i}>0}\left(\log\lambda_{i}\right)=-\log
D_{p},$ (3.21)

where

$D_{p}:=\mathop{\displaystyle\prod}\limits_{\lambda_{i}>0}\lambda_{i}$

is the determinant of $\Delta_{p}$ restricted on the direct sum of the
eigenspaces with positive eigenvalues. In the view of (3.20) and (3.21), it
suffices to prove that

$\log\tau(\Omega,\iota)=-\frac{1}{2}\sum_{p=0}^{N}(-1)^{p}\,p\,\log D_{p}.$
(3.22)

As before, we use notations
$B_{p}=\mathop{\mathrm{i}m}\partial_{p+1}=\partial\Omega_{p+1}$ and
$Z_{p}=\ker\partial_{p}$ so that $H_{p}=Z_{p}/B_{p}$. Since any element of
$u\in B_{p}$ has the form $u=\partial v$ for some $v\in\Omega_{p+1}$, we have

$\Delta_{p}u=\partial^{\ast}\partial\partial
v+\partial\partial^{\ast}u=\partial\left(\partial^{\ast}u\right)\in B_{p}.$
(3.23)

Hence, $B_{p}$ is an invariant subspace of $\Delta_{p}$. Therefore, there
exists an orthonormal basis $b_{p}=\\{b_{p}^{i}\\}$ of $B_{p}$ that consists
of the eigenvectors of $\Delta_{p}$:

$\Delta_{p}b_{p}^{i}=\beta_{p}^{i}b_{p}^{i},$

where $\beta_{p}^{i}$ are the corresponding eigenvalues. Since by (3.4)
$B_{p}$ is orthogonal to $\mathcal{H}_{p}$ and all the eigenvectors of
$\Delta_{p}$ with eigenvalue $0$ belong to $\mathcal{H}_{p}$, we have
$\beta_{p}^{i}>0.$

By (3.23) we have $\Delta_{p}b_{p}^{i}=\partial\partial^{\ast}b_{p}^{i}$,
whence

$\partial\partial^{\ast}b_{p}^{i}=\beta_{p}^{i}b_{p}^{i}.$ (3.24)

Set

$\widetilde{b}_{p}^{i}:=\frac{1}{\beta_{p-1}^{i}}\partial^{\ast}b_{p-1}^{i}\in\Omega_{p}.$

We have by (3.24)

$\partial\widetilde{b}_{p}^{i}=\frac{1}{\beta_{p-1}^{i}}\partial\partial^{\ast}b_{p-1}^{i}=\frac{1}{\beta_{p-1}^{i}}\cdot\beta_{p-1}^{i}b_{p-1}^{i}=b_{p-1}^{i}$

so that the sequences $\widetilde{b}_{p}=\\{\widetilde{b}_{p}^{i}\\}$ and
$b_{p-1}=\\{b_{p-1}^{i}\\}$ satisfy the identity (3.7) and, hence, can be used
in the definition of R-torsion. Since also

$\partial^{\ast}\widetilde{b}_{p}^{i}=\frac{1}{\beta_{p-1}^{i}}\partial^{\ast}\partial^{\ast}b_{p-1}^{i}=0,$

we obtain

$\Delta_{p}\widetilde{b}_{p}^{i}=\partial^{\ast}\partial\widetilde{b}_{p}^{i}+\partial\partial^{\ast}\widetilde{b}_{p}^{i}=\partial^{\ast}b_{p-1}^{i}+0=\beta_{p-1}^{i}\widetilde{b}_{p}^{i}.$

Hence, $\widetilde{b}_{p}^{i}$ are the eigenvectors of $\Delta_{p}$ with
eigenvalues $\beta_{p-1}^{i}.$ Moreover, the sequence
$\,\\{\widetilde{b}_{p}^{i}\\}$ is orthogonal because by (3.24) for $i\neq j$

$\displaystyle\langle\widetilde{b}_{p}^{i},\widetilde{b}_{p}^{j}\rangle$
$\displaystyle=$
$\displaystyle\frac{1}{\beta_{p-1}^{i}\beta_{p-1}^{j}}\langle\partial^{\ast}b_{p-1}^{i},\partial^{\ast}b_{p-1}^{j}\rangle$
$\displaystyle=$
$\displaystyle\frac{1}{\beta_{p-1}^{i}\beta_{p-1}^{j}}\langle\partial\partial^{\ast}b_{p-1}^{i},b_{p-1}^{j}\rangle$
$\displaystyle=$ $\displaystyle\frac{1}{\beta_{p-1}^{j}}\langle
b_{p-1}^{i},b_{p-1}^{j}\rangle$ $\displaystyle=$ $\displaystyle 0.$

In the case $i=j$ we obtain similarly

$\|\widetilde{b}_{p}^{i}\|^{2}=\frac{1}{\beta_{p-1}^{i}}\langle
b_{p-1}^{i},b_{p-1}^{i}\rangle=\frac{1}{\beta_{p-1}^{i}}.$

Note also that the vectors $b_{p}^{i}$ and $\widetilde{b}_{p}^{j}$ are
necessarily orthogonal since

$\langle
b_{p}^{i},\widetilde{b}_{p}^{j}\rangle=\frac{1}{\beta_{p-1}^{i}}\langle
b_{p}^{i},\partial^{\ast}b_{p-1}^{j}\rangle=\frac{1}{\beta_{p-1}^{i}}\langle\partial
b_{p}^{i},b_{p-1}^{j}\rangle=0.$

Let $h_{p}=\\{h_{p}^{i}\\}$ be an orthonormal basis of $\mathcal{H}_{p}.$ Then
the following sequence

$\omega_{p}=b_{p}\cup\\{\sqrt{\beta_{p-1}^{i}}\,\widetilde{b}_{p}^{i}\\}\cup
h_{p}$ (3.25)

consists of the eigenvectors of $\Delta_{p}$ and is orthonormal. By
construction, this sequence is a basis in $\Omega_{p}$ (see Section 3.2). It
follows that all the positive eigenvalues of $\Delta_{p}$ are

$\left\\{\beta_{p-1}^{i}\right\\}\cup\left\\{\beta_{p}^{i}\right\\},$

whence

$D_{p}=\prod_{i}\beta_{p-1}^{i}\prod_{i}\beta_{p}^{i}.$

Setting

$L_{k}:=\log\prod_{i}\beta_{k}^{i}$

we obtain

$\log D_{p}=L_{p-1}+L_{p}.$

Using that $L_{N}=0$, we obtain

$\displaystyle L_{p-1}$ $\displaystyle=$
$\displaystyle\left(L_{p-1}+L_{p}\right)-\left(L_{p}+L_{p+1}\right)+...$
(3.26) $\displaystyle=$
$\displaystyle\sum_{q=p}^{N}\left(-1\right)^{q-p}\left(L_{q-1}+L_{q}\right)=\sum_{q=p}^{N}(-1)^{q-p}\log
D_{q}{.}$

It follows from (3.25) that

$[b_{p},\widetilde{b}_{p},h_{p}\,/\,\omega_{p}]=\prod_{i}\left(\beta_{p-1}^{i}\right)^{-1/2}=\left(L_{p-1}\right)^{-1/2}.$

Using the definition of $\tau\left(\Omega,\iota\right)$ and (3.26), we obtain

$\displaystyle\log\tau(\Omega,\iota)$ $\displaystyle=$
$\displaystyle\sum_{p=0}^{N}(-1)^{p}\log[b_{p},\widetilde{b}_{p},h_{p}\,/\,\omega_{p}]$
$\displaystyle=$ $\displaystyle-\frac{1}{2}\sum_{p=1}^{N}(-1)^{p}L_{p-1}$
$\displaystyle=$
$\displaystyle-\frac{1}{2}\sum_{p=1}^{N}(-1)^{p}\sum_{q=p}^{N}(-1)^{q-p}\log
D_{q}$ $\displaystyle=$
$\displaystyle-\frac{1}{2}\sum_{q=1}^{N}\sum_{p=1}^{q}(-1)^{q}\log D_{q}$
$\displaystyle=$ $\displaystyle-\frac{1}{2}\sum_{q=1}^{N}(-1)^{q}q\log
D_{q}{,}$

which finishes the proof of (3.22).

##  4\. Cartesian product of path complexes

###  4.1. Product of paths

Given two finite sets $X,Y$, consider their Cartesian product $Z=X\times Y.$
Let $z=z_{0}z_{1}...z_{r}$ be a regular elementary $r$-path on $Z$, where
$z_{k}=\left(x_{k},y_{k}\right)$ with $x_{k}\in X$ and $y_{k}\in Y$.

######  Definition 4.1.

We say that the path $z$ is _step-like_ if, for any $k=1,...,r$, either
$x_{k-1}=x_{k}$ or $y_{k-1}=y_{k}$. In fact, exactly one of these conditions
holds as $z$ is regular.

Any step-like path $z$ on $Z$ determines by projection regular elementary
paths $x$ on $X$ and $y$ on $Y$. More precisely, $x$ is obtained from $z$ by
taking the sequence of all $X$-components of the vertices of $z$ and then by
collapsing in it any subsequence of repeated vertices to one vertex. The same
rule applies to $y$. By construction, the projections $x$ and $y$ are regular
elementary paths on $X$ and $Y$, respectively. If the projections of
$z=z_{0}...z_{r}$ are $x=x_{0}...x_{p}$ and $y=y_{0}...y_{q}$ then $p+q=r$
(cf. Fig. 5(left)).

Figure 5. Left: a step-like path $z$ and its projections $x$ and $y$. Right: a
staircase $S\left(z\right)$ and its elevation $L\left(z\right)$ (here
$L\left(z\right)=30$).

Every vertex $z_{k}=\left(x_{i},y_{j}\right)$ of a step-like path $z$ can be
represented as a point $\left(i,j\right)$ of $\mathbb{Z}^{2}$ so that the
whole path $z$ is represented by a _staircase_ $S\left(z\right)$ in
$\mathbb{Z}^{2}$ connecting the points $\left(0,0\right)$ and
$\left(p,q\right)$.

######  Definition 4.2.

Define the _elevation_ $L\left(z\right)$ of the path $z$ as the number of
cells in $\mathbb{Z}_{+}^{2}$ below the staircase $S\left(z\right)$ (the
shaded area on Fig. 5(right)).

For given elementary regular $p$-path $x$ on $X$ and $q$-path $y$ on $Y$,
denote by $\Pi_{x,y}$ the set of all step-like paths $z$ on $Z$ whose
projections on $X$ and $Y$ are $x$ and $y$, respectively.

######  Definition 4.3.

For regular elementary paths $e_{x}$ on $X$ and $e_{y}$ on $Y$ define their
cross product $e_{x}\times e_{y}$ as a path on $Z$ as follows:

$e_{x}\times
e_{y}=\sum_{z\in\Pi_{x,y}}\left(-1\right)^{L\left(z\right)}e_{z}.$ (4.1)

Then extend by linearly the definition of $u\times v$ to all regular paths $u$
on $X$ and $v$ on $Y.$

Clearly, if $u\in\mathcal{R}_{p}\left(X\right)$ and
$v\in\mathcal{R}_{q}\left(Y\right)$ then $u\times
v\in\mathcal{R}_{p+q}\left(Z\right).$ Moreover, the cross product satisfies
the product rule with respect to the boundary operator $\partial$:

$\partial\left(u\times v\right)=\left(\partial u\right)\times
v+\left(-1\right)^{p}u\times\left(\partial v\right)$ (4.2)

(see [4, Prop. 6.3]).

###  4.2. Product of path complexes and digraphs

######  Definition 4.4.

Given two finite sets $X$ and $Y$ with path complexes $P\left(X\right)$ and
$P\left(Y\right)$ over $X$ and $Y$, respectively, define a path complex
$P\left(Z\right)$ over the set $Z=X\times Y$ as follows: the elements of
$P\left(Z\right)$ are step-like paths on $Z$ whose projections on $X$ and $Y$
belong to $P\left(X\right)$ and $P\left(Y\right)$, respectively. The path
complex $P\left(Z\right)$ is called the _Cartesian product_ of the path
complexes $P\left(X\right)$ and $P\left(Y\right)$ and is denoted by
$P\left(X\right)\Box P\left(Y\right).$

In short: a path $z$ on $Z$ is allowed if it is step-like and if its
projections on $X$ and $Y$ are allowed. In particular, if $x$ and $y$ are
elementary allowed paths on $X$ and $Y$, respectively, then all the paths
$z\in\Pi_{x,y}$ are allowed on $Z$.

######  Definition 4.5.

Let $X$ and $Y$ be digraphs. The Cartesian product $Z=X\Box Y$ of the digraphs
$X$ and $Y$ is defined as a digraph with the vertices $\left(x,y\right)$ where
$x\in X$ and $y\in Y$, and arrows
$\left(x,y\right)\rightarrow\left(x^{\prime},y^{\prime}\right)$ where either
$x\rightarrow x^{\prime}$ and $y=y^{\prime}$ or $x=x^{\prime}$ and
$y\rightarrow y^{\prime}$.

For example, if $a\rightarrow a^{\prime}$ is an arrow in $X$ and $b\rightarrow
b^{\prime}$ is an arrow in $Y$ then they induce the following arrows in $Z$:

$\begin{array}[]{ccc}\overset{\left(a,b^{\prime}\right)}{\bullet}&\longrightarrow&\overset{\left(a^{\prime},b^{\prime}\right)}{\bullet}\\\
\uparrow&&\uparrow\\\
\overset{\left(a,b\right)}{\bullet}&\longrightarrow&\overset{\left(a^{\prime},b\right)}{\bullet}\end{array}\
$

Let $P\left(X\right)$ and $P\left(Y\right)$ be the path complexes in $X$ and
$Y$, respectively, coming from the digraph structures. It is easy to see that

$P\left(X\Box Y\right)=P\left(X\right)\Box P\left(Y\right),$

that is, the Cartesian product of the path complexes is compatible with the
Cartesian product of digraphs. The reader who is interested only in digraphs
can always think of $X$ and $Y$ as digraphs and of $Z$ as their Cartesian
product.

For a general path complex $P\left(V\right)$ over a set $V$ we use the short
notations

$\mathcal{A}_{p}\left(P\left(V\right)\right)\equiv\mathcal{A}_{p}\left(V\right)\
\ \text{and\ \
}\Omega_{p}\left(P\left(V\right)\right)\equiv\Omega_{p}\left(V\right).$

It follows from (4.1)

$u\in\mathcal{A}_{p}\left(X\right)\ \text{and\
}v\in\mathcal{A}_{q}\left(Y\right)\ \ \Rightarrow\ \ u\times
v\in\mathcal{A}_{p+q}\left(Z\right).$

Moreover, (4.2) implies that

$u\in\Omega_{p}\left(X\right)\ \text{and\ }v\in\Omega_{q}\left(Y\right)\ \
\Rightarrow\ \ u\times v\in\Omega_{p+q}\left(Z\right)$

(see [4, Prop. 6.5], [6, Prop. 4.6]). Furthermore, the following Künneth
formula is true: for any $r\geq 0$,

$\Omega_{r}\left(Z\right)=\bigoplus_{\left\\{p,q\geq
0:p+q=r\right\\}}\Omega_{p}\left(X\right)\otimes\Omega_{q}\left(Y\right),$
(4.3)

where $\otimes$ denotes the tensor product of linear spaces, and $u\otimes v$
for $u\in\Omega_{p}\left(X\right)$ and $v\in\Omega_{q}\left(Y\right)$ is
identified with the element $u\times v$ of $\Omega_{r}\left(Z\right)$ (see [4,
Thm. 6.6] and [7, Thm 6.6]).

###  4.3. Operators $\partial^{\ast}$ and $\Delta$ on products

For the standard inner product $\iota$ defined by (3.1) on each of the space
$\mathcal{R}\left(X\right)$, $\mathcal{R}\left(Y\right)$ and
$\mathcal{R}\left(Z\right)$ the following identity is known: if $\
u\in\mathcal{A}_{p}\left(X\right)$, $v\in\mathcal{A}_{q}\left(Y\right)$,
$\varphi\in\mathcal{A}_{p^{\prime}}\left(X\right)$ and
$\psi\in\mathcal{A}_{q^{\prime}}\left(Y\right)$, then

$\langle u\times v,\varphi\times\psi\rangle_{\iota}=\tbinom{p+q}{p}\langle
u,\varphi\rangle_{\iota}\langle v,\psi\rangle_{\iota}$

(see [6, Lemma 4.13]). This identity includes also the case when two paths in
the inner product have different length - in this case their inner product is
zero by definition. Hence, we have

$\frac{1}{\left(p+q\right)!}\langle u\times
v,\varphi\times\psi\rangle_{\iota}=\frac{1}{p!}\langle
u,\varphi\rangle_{\iota}\frac{1}{q!}\langle v,\psi\rangle_{\iota}.$

In the case $p^{\prime}=p$ and $q^{\prime}=q$ we pass to the normalized inner
product $\iota^{\prime}$ given by (3.2) and obtain

$\langle u\times v,\varphi\times\psi\rangle_{\iota^{\prime}}=\langle
u,\varphi\rangle_{i^{\prime}}\langle v,\psi\rangle_{\iota^{\prime}}~{}.$ (4.4)

This identity is true also if $p^{\prime}\neq p$ or $q^{\prime}\neq q$ as in
these cases the both sides vanish.

In the rest of this section we use the normalized inner product

$\left\langle,\right\rangle=\left\langle,\right\rangle_{\iota^{\prime}}$

unless otherwise specified. In particular, we define the adjoint operator
$\partial^{\ast}$ and the Hodge Laplacian with respect to the normalized inner
product and refer to them as _normalized_.

######  Lemma 4.6.

Let $u\in\Omega_{p}\left(X\right)$ and $v\in\Omega_{q}\left(Y\right).$ Then
for the normalized adjoint operator we have

$\partial^{\ast}\left(u\times v\right)=\partial^{\ast}u\times
v+\left(-1\right)^{p}u\times\left(\partial^{\ast}v\right).$ (4.5)

###### Proof.

By definition, we have, for any $w\in\Omega_{p+q+1}\left(Z\right)$

$\langle\partial^{\ast}\left(u\times v\right),w\rangle=\langle u\times
v,\partial w\rangle.$

Any $w\in\Omega_{\ast}\left(Z\right)$ admits a representation

$w=\sum_{k}\varphi_{k}\times\psi_{k}$

where the sum is finite and

$\varphi_{k}\in\Omega_{p_{k}}\left(X\right)\ \ \text{and
}\psi_{k}\in\Omega_{q_{k}}\left(Y\right)$

with $p_{k}+q_{k}=p+q+1$ (see [4, Thm. 6.12], [6, Theorem 5.1]).

Then we have

$\displaystyle\langle\partial^{\ast}\left(u\times v\right),w\rangle$
$\displaystyle=$ $\displaystyle\langle u\times
v,\sum\partial\left(\varphi_{k}\times\psi_{k}\right)\rangle$ $\displaystyle=$
$\displaystyle\langle u\times
v,\sum\left(\partial\varphi_{k}\times\psi_{k}+\left(-1\right)^{p_{k}}\varphi_{k}\times\partial\psi_{k}\right)\rangle$
$\displaystyle=$ $\displaystyle\sum\langle u\times
v,\partial\varphi_{k}\times\psi_{k}\rangle+\left(-1\right)^{p_{k}}\langle
u\times v,\varphi_{k}\times\partial\psi_{k}\rangle$ $\displaystyle=$
$\displaystyle\sum\langle u,\partial\varphi_{k}\rangle\langle
v,\psi_{k}\rangle+\left(-1\right)^{p_{k}}\langle u,\varphi_{k}\rangle\langle
v,\partial\psi_{k}\rangle$ $\displaystyle=$
$\displaystyle\sum\langle\partial^{\ast}u,\varphi_{k}\rangle\langle
v,\psi_{k}\rangle+\left(-1\right)^{p_{k}}\langle
u,\varphi_{k}\rangle\langle\partial^{\ast}v,\psi_{k}\rangle.$

Note that if $p_{k}\neq p$ then

$\langle u,\varphi_{k}\rangle=0.$

Hence, we can replace $p_{k}$ everywhere by $p$ and obtain

$\displaystyle\langle\partial^{\ast}\left(u\times v\right),w\rangle$
$\displaystyle=$
$\displaystyle\sum\langle\partial^{\ast}u,\varphi_{k}\rangle\langle
v,\psi_{k}\rangle+\left(-1\right)^{p}\langle
u,\varphi_{k}\rangle\langle\partial^{\ast}v,\psi_{k}\rangle$ $\displaystyle=$
$\displaystyle\sum\langle\partial^{\ast}u\times
v,\varphi_{k}\times\psi_{k}\rangle+\left(-1\right)^{p}\langle
u\times\partial^{\ast}v,\varphi_{k}\times\psi_{k}\rangle$ $\displaystyle=$
$\displaystyle\langle\partial^{\ast}u\times
v+\left(-1\right)^{p}u\times\partial^{\ast}v,\sum\varphi_{k}\times\psi_{k}\rangle$
$\displaystyle=$ $\displaystyle\langle\partial^{\ast}u\times
v+\left(-1\right)^{p}u\times\partial^{\ast}v,w\rangle,$

whence

$\partial^{\ast}\left(u\times v\right)=\partial^{\ast}u\times
v+\left(-1\right)^{p}u\times\partial^{\ast}v.$

######  Lemma 4.7.

For the normalized Hodge Laplacian we have

$\Delta\left(u\times v\right)=\left(\Delta u\right)\times v+u\times\Delta v.$
(4.6)

###### Proof.

Let $u\in\Omega_{p}\left(X\right)$ and $v\in\Omega_{q}\left(X\right)$. Then we
have

$\displaystyle\partial\partial^{\ast}\left(u\times v\right)$ $\displaystyle=$
$\displaystyle\partial\left(\partial^{\ast}u\times
v+\left(-1\right)^{p}u\times\partial^{\ast}v\right)$ $\displaystyle=$
$\displaystyle\partial\left(\partial^{\ast}u\times
v\right)+\left(-1\right)^{p}\partial\left(u\times\partial^{\ast}v\right)$
$\displaystyle=$ $\displaystyle\partial\partial^{\ast}u\times
v+\left(-1\right)^{p+1}\partial^{\ast}u\times\partial v$
$\displaystyle+\left(-1\right)^{p}\left(\partial
u\times\partial^{\ast}v+\left(-1\right)^{p}u\times\partial\partial^{\ast}v\right)$
$\displaystyle=$ $\displaystyle\partial\partial^{\ast}u\times
v+\left(-1\right)^{p+1}\partial^{\ast}u\times\partial
v+\left(-1\right)^{p}\partial
u\times\partial^{\ast}v+u\times\partial\partial^{\ast}v$

and

$\displaystyle\partial^{\ast}\partial\left(u\times v\right)$ $\displaystyle=$
$\displaystyle\partial^{\ast}\left(\partial u\times
v+\left(-1\right)^{p}u\times\partial v\right)$ $\displaystyle=$
$\displaystyle\partial^{\ast}\left(\partial u\times
v\right)+\left(-1\right)^{p}\partial^{\ast}\left(u\times\partial v\right)$
$\displaystyle=$ $\displaystyle\partial^{\ast}\partial u\times
v+\left(-1\right)^{p-1}\partial u\times\partial^{\ast}v$
$\displaystyle+\left(-1\right)^{p}\left(\partial^{\ast}u\times\partial
v+\left(-1\right)^{p}u\times\partial^{\ast}\partial v\right)$ $\displaystyle=$
$\displaystyle\partial^{\ast}\partial u\times v+\left(-1\right)^{p-1}\partial
u\times\partial^{\ast}v+\left(-1\right)^{p}\partial^{\ast}u\times\partial
v+u\times\partial^{\ast}\partial v.$

Adding up the two identities and noticing that the terms
$\partial^{\ast}u\times\partial v$ and $\partial u\times\partial^{\ast}v$
cancel out, we obtain

$\Delta\left(u\times v\right)=\left(\Delta u\right)\times
v+u\times\left(\Delta v\right).$

###  4.4. Torsion of products

Let $P\left(V\right)$ be a path complex on a set $V$ with the maximal length
$N$ . As before, let $\iota$ be the standard inner product structure on
$P\left(V\right)$ given by (3.1) and $\iota^{\prime}$ be the normalized inner
product structure on $P\left(V\right)$ given by (3.2). Consider the
corresponding standard and normalized torsions:

$T\left(V\right)=T\left(\Omega\left(V\right),\iota\right)\ \ \ \text{and\ \ \
}T^{\prime}\left(V\right)=T\left(\Omega\left(V\right),\iota^{\prime}\right).$

In the same way we will use notation $\tau\left(V\right)$ and
$\tau^{\prime}\left(V\right)$ for R-torsions with respect to $\iota$ and
$\iota^{\prime}$, respectively. Since $T\left(V\right)=\tau\left(V\right)$ and
$T^{\prime}\left(V\right)=\tau^{\prime}\left(V\right)$, the relation between
$T\left(V\right)$ and $T^{\prime}\left(V\right)$ is given by (3.15) and
(3.16).

Although the main object of interest for us is the standard torsion
$T\left(V\right)$, in this section we make an essential use of
$T^{\prime}\left(V\right)$ as it behaves better with respect to the Cartesian
product.

We need also the Euler characteristic of $\Omega\left(V\right)$:

$\chi\left(V\right)=\chi\left(\Omega\left(V\right)\right)=\sum_{p=0}^{N}\left(-1\right)^{p}\dim\Omega_{p}\left(V\right)=\sum_{p=0}^{N}\left(-1\right)^{p}\dim
H_{p}\left(V\right).$

The next theorem is our main result about torsion on the product of path
complexes.

######  Theorem 4.8.

If $P\left(Z\right)=P\left(X\right)\Box P\left(Z\right)$ then

$\log T^{\prime}(Z)=\chi(Y)\log T^{\prime}(X)+\chi(X)\log T^{\prime}(Y).$
(4.7)

Before the proof of Theorem 4.8, we need to do some preparations. In the next
lemmas we work with an arbitrary chain complex $\Omega$ with some inner
product structure $\iota$. Let $\lambda$ be an eigenvalue of the Hodge
Laplacian $\Delta_{p}$ on some chain complex $\Omega.$ Consider the eigenspace
of $\lambda$ and its subspaces:

$\displaystyle E_{p}(\lambda)$ $\displaystyle=$
$\displaystyle\\{\varphi\in\Omega_{p}:\Delta_{p}\varphi=\lambda\varphi\\},$
$\displaystyle E_{p}^{\prime}(\lambda)$ $\displaystyle=$
$\displaystyle\\{\varphi\in E_{p}(\lambda):\partial\varphi=0\\},$
$\displaystyle E_{p}^{\prime\prime}(\lambda)$ $\displaystyle=$
$\displaystyle\\{\varphi\in E_{p}(\lambda):\partial^{\ast}\varphi=0\\}.$

In the case $\lambda=0$ these three spaces are identical by Lemma 3.2. In the
case $\lambda\neq 0$ the situation is different.

######  Lemma 4.9.

Assume that $\lambda\neq 0$. Then we have

$\displaystyle E_{p}^{\prime}\left(\lambda\right)$ $\displaystyle=$
$\displaystyle\\{\varphi\in\Omega_{p}:\partial\partial^{\ast}\varphi=\lambda\varphi\\}$
(4.8) $\displaystyle E_{p}^{\prime\prime}(\lambda)$ $\displaystyle=$
$\displaystyle\\{\varphi\in\Omega_{p}:\partial^{\ast}\partial\varphi=\lambda\varphi\\}$
(4.9)

and

$E_{p}(\lambda)=E_{p}^{\prime}(\lambda)\bigoplus
E_{p}^{\prime\prime}(\lambda).$ (4.10)

###### Proof.

Let us first prove (4.8). If $\varphi\in E_{p}^{\prime}\left(\lambda\right)$
then

$\lambda\varphi=\Delta\varphi=\partial^{\ast}\partial\varphi+\partial\partial^{\ast}\varphi=\partial\partial^{\ast}\varphi.$

Conversely, if $\partial\partial^{\ast}\varphi=\lambda\varphi$ then

$\partial\varphi=\frac{1}{\lambda}\partial\left(\partial\partial^{\ast}\varphi\right)=0$

and, hence, $\Delta\varphi=\partial\partial^{\ast}\varphi=\lambda\varphi$ so
that $\varphi\in E_{\lambda}^{\prime}.$ In the same way one proves (4.9).

In order to verify (4.10), observe first that the space
$E_{p}^{\prime}\left(\lambda\right)$ and
$E_{p}^{\prime\prime}\left(\lambda\right)$ are orthogonal because for any
$\varphi\in E_{p}^{\prime}\left(\lambda\right)$ and $\psi\in
E_{p}^{\prime\prime}\left(\lambda\right)$ we have

$\left\langle\varphi,\psi\right\rangle=\frac{1}{\lambda^{2}}\left\langle\partial\partial^{\ast}\varphi,\partial^{\ast}\partial\psi\right\rangle=\frac{1}{\lambda^{2}}\left\langle\partial\partial\partial^{\ast}\varphi,\partial\psi\right\rangle=0.$

For any $\varphi\in E_{p}\left(\lambda\right)$ we have

$\left(\partial\partial^{\ast}\right)^{2}\varphi=\partial\partial^{\ast}\left(\partial\partial^{\ast}\varphi+\partial^{\ast}\partial\varphi\right)=\partial\partial^{\ast}\Delta\varphi=\lambda\partial\partial^{\ast}\varphi,$

which implies by (4.8) that $\partial\partial^{\ast}\varphi\in
E_{p}^{\prime}.$ Similarly, we have $\partial^{\ast}\partial\varphi\in
E_{p}^{\prime\prime}.$ Finally, for any $\varphi\in E_{p}\left(\lambda\right)$
we have

$\varphi=\frac{1}{\lambda}\Delta\varphi=\frac{1}{\lambda}\partial\partial^{\ast}\varphi+\frac{1}{\lambda}\partial^{\ast}\partial\varphi,$

whence (4.10) follows.

######  Lemma 4.10.

The operator $\lambda^{-1/2}\partial$ is an isometry of
$E_{p}^{\prime\prime}(\lambda)$ onto $E_{p-1}^{\prime}(\lambda)$ with the
inverse $\lambda^{-1/2}\partial^{\ast}$.

###### Proof.

Let $\varphi\in E_{p}^{\prime\prime}(\lambda)$ so that
$\partial^{\ast}\varphi=0$ and
$\partial^{\ast}\partial\varphi=\lambda\varphi$. For $\psi=\partial\varphi$ we
have

$\partial\partial^{\ast}\psi=\partial\partial^{\ast}\partial\varphi=\lambda\partial\varphi=\lambda\psi$

whence $\psi\in E_{p-1}^{\prime}\left(\lambda\right).$ Hence, $\partial$ maps
$E_{p}^{\prime\prime}\left(\lambda\right)$ into
$E_{p-1}^{\prime}\left(\lambda\right).$ Let us verify that
$\lambda^{-1/2}\partial$ is an isometry. For $\varphi\in
E_{p}^{\prime\prime}\left(\lambda\right)$ and
$\psi=\lambda^{-1/2}\partial\varphi$ we have

$\left\langle\psi,\psi\right\rangle=\frac{1}{\lambda}\left\langle\partial\varphi,\partial\varphi\right\rangle=\frac{1}{\lambda}\left\langle\partial^{\ast}\partial\varphi,\varphi\right\rangle=\left\langle\varphi,\varphi\right\rangle.$

It remains to show that the mapping $\lambda^{-1/2}\partial$ is onto and has
the inverse $\lambda^{-1/2}\partial^{\ast}.$ For any $\psi\in
E_{p-1}^{\prime}\left(\lambda\right)$ we have
$\partial^{\ast}\left(\partial^{\ast}\psi\right)=0$ and

$\Delta_{p}(\partial^{\ast}\psi)=\left(\partial^{\ast}\partial+\partial\partial^{\ast}\right)\partial^{\ast}\psi=\partial^{\ast}\partial\partial^{\ast}\psi=\partial^{\ast}(\lambda\psi)=\lambda\partial^{\ast}\psi,$

which implies $\partial^{\ast}\psi\in E_{p}^{\prime\prime}(\lambda).$ Since by
(4.8) $\partial\partial^{\ast}\psi=\lambda\psi,$ we obtain

$\lambda^{-1/2}\partial(\lambda^{-1/2}\partial^{\ast}\psi)=\frac{1}{\lambda}\partial\partial^{\ast}\psi=\psi.$

and we conclude that $\lambda^{-1/2}\partial$ and
$\lambda^{-1/2}\partial^{\ast}$ are mutually inverse.

Let $n_{p}(\lambda)$, $n_{p}^{\prime}(\lambda)$,
$n_{p}^{\prime\prime}(\lambda)$ be the dimensions of spaces $E_{p}(\lambda)$,
$E_{p}^{\prime}(\lambda)$, $E_{p}^{\prime\prime}(\lambda)$, respectively It
follows from Lemmas 4.9 and 4.10 that

$\displaystyle n_{p-1}^{\prime}(\lambda)$ $\displaystyle=$ $\displaystyle
n_{p}^{\prime\prime}(\lambda),$ $\displaystyle n_{p}(\lambda)$
$\displaystyle=$ $\displaystyle
n_{p}^{\prime}(\lambda)+n_{p}^{\prime\prime}(\lambda).$

As it follows from the definition of the Euler characteristic
$\chi\left(\Omega\right)$ and $H_{p}\cong\mathcal{H}_{p}$, we have

$\chi\left(\Omega\right)=\sum_{p=0}^{N}\left(-1\right)^{p}n_{p}\left(0\right).$

######  Lemma 4.11.

If $\lambda>0$ then

$\sum\limits_{p=0}^{N}(-1)^{p}n_{p}(\lambda)=0.$

###### Proof.

We have

$\displaystyle\sum_{p=0}^{N}(-1)^{p}n_{p}(\lambda)$ $\displaystyle=$
$\displaystyle\sum_{p=0}^{N}(-1)^{p}n_{p}^{\prime}(\lambda)+\sum_{p=0}^{N}(-1)^{p}n_{p}^{\prime\prime}(\lambda)$
$\displaystyle=$
$\displaystyle\sum_{p=0}^{N}(-1)^{p}n_{p}^{\prime}(\lambda)+\sum_{p=1}^{N}(-1)^{p}n_{p-1}^{\prime}(\lambda)+n_{0}^{\prime\prime}(\lambda)$
$\displaystyle=$
$\displaystyle(-1)^{N}n_{N}^{\prime}(\lambda)+n_{0}^{\prime\prime}(\lambda)$
$\displaystyle=$ $\displaystyle 0.$

Here $n_{N}^{\prime}\left(\lambda\right)=0$ because for every vector
$\varphi\in E_{N}^{\prime}\left(\lambda\right)$ we have
$\partial^{\ast}\varphi=0$ and, hence,

$\varphi=\frac{1}{\lambda}\partial\partial^{\ast}\varphi=0,$

and $n_{0}^{\prime\prime}\left(\lambda\right)=0$ because for any $\varphi\in
E_{0}^{\prime\prime}\left(\lambda\right)$ we have $\partial\varphi=0$ and,
hence,

$\varphi=\frac{1}{\lambda}\partial^{\ast}\partial\varphi=0.$

Now we can prove Theorem 4.8. The idea of proof is borrowed from [9, Thm.
2.5].

###### Proof of Theorem 4.8.

The zeta function $\zeta_{p,X}\left(s\right)$ of $\Delta_{p}$ on $X$ can be
represented in the form

$\zeta_{p,X}\left(s\right)=\sum_{\lambda>0}\lambda^{-s}n_{p}\left(\lambda,X\right),$

where the sum is taken over all distinct positive eigenvalues $\lambda$ of
$\Delta_{p}$ and $n_{p}\left(\lambda,X\right)$ is the multiplicity of
$\lambda$. Similar formulas hold for $\zeta_{q,Y}\left(s\right)$ and
$\zeta_{r,Z}\left(s\right).$

Let $u\in\Omega_{p}\left(X\right)$ be an eigenvector of $\Delta_{p}$ with
eigenvalue $\lambda$ and $v\in\Omega_{q}\left(Y\right)$ be an eigenvector of
$\Delta_{q}$ with eigenvalue $\mu$. It follows from (4.6) that $u\times
v\in\Omega_{p+q}\left(Z\right)$ is an eigenvector of $\Delta_{p+q}$ with
eigenvalue $\lambda+\mu$. If $\left\\{u_{i}\right\\}$ is an orthonormal basis
in $\Omega_{p}\left(X\right)$ consisting of the eigenvectors of $\Delta_{p}$
and $\left\\{v_{j}\right\\}$ is an orthonormal basis in
$\Omega_{q}\left(Y\right)$ consisting of the eigenvectors of $\Delta_{q}$ then
the sequence $\left\\{u_{i}\times v_{j}\right\\}$ is orthonormal by (4.4) and,
hence, forms a basis in
$\Omega_{p}\left(X\right)\otimes\Omega_{q}\left(Y\right).$

Let us fix $r\geq 0$ and recall that by the Künneth formula (4.3)
$\Omega_{r}\left(Z\right)$ is a direct sum of the spaces
$\Omega_{p}\left(X\right)\otimes\Omega_{q}\left(Y\right)$ over all pairs
$p,q\geq 0$ with $p+q=r.$ Hence, collecting all the bases in
$\Omega_{p}\left(X\right)\otimes\Omega_{q}\left(Y\right)$ of the form
$\left\\{u_{i}\times v_{j}\right\\}$ we obtain an orthonormal basis in
$\Omega_{r}\left(Z\right)$. This basis consists of the eigenvectors of
$\Delta_{r}$ in $\Omega_{r}\left(Z\right).$ Hence, all the eigenvalues of
$\Delta_{r}$ in $\Omega_{r}\left(Z\right)$ have the form $\lambda+\mu$ where
$\lambda$ is an eigenvalue of $\Delta_{p}$ in $\Omega_{p}\left(X\right)$,
$\mu$ is an eigenvalue of $\Delta_{q}$ in $\Omega_{q}\left(Y\right)$, and the
multiplicity of $\lambda+\mu$ is
$n_{p}\left(\lambda,X\right)n_{q}\left(\mu,Y\right).$

Hence, we obtain

$\zeta_{r,Z}(s)=\sum_{\lambda+\mu>0}\sum_{p+q=r}(\lambda+\mu)^{-s}n_{p}(\lambda,X)n_{q}(\mu,Y).$
(4.11)

It follows that

$\displaystyle\sum_{r\geq 0}(-1)^{r}r\zeta_{r,Z}(s)$ (4.12) $\displaystyle=$
$\displaystyle\sum_{\lambda+\mu>0}(\lambda+\mu)^{-s}\sum_{p\geq 0}\sum_{q\geq
0}(-1)^{p+q}(p+q)n_{p}(\lambda,X)n_{q}(\mu,Y)$ $\displaystyle=$
$\displaystyle\sum_{\lambda+\mu>0}(\lambda+\mu)^{-s}\left(\sum_{p\geq
0}(-1)^{p}p\,n_{p}(\lambda,X)\right)\left(\sum_{q\geq
0}(-1)^{q}n_{q}(\mu,Y)\right)$ $\displaystyle+$
$\displaystyle\sum_{\lambda+\mu>0}(\lambda+\mu)^{-s}\left(\sum_{p\geq
0}(-1)^{p}n_{p}(\lambda,X)\right)\left(\sum_{q\geq
0}(-1)^{q}q\,n_{q}(\mu,Y)\right).$ (4.13)

By Lemma 4.11, if $\lambda>0$ then

$\sum_{p\geq 0}(-1)^{p}n_{p}(\lambda,X)=0$

and if $\mu>0$ then

$\sum_{q\geq 0}(-1)^{q}n_{q}(\mu,Y)=0.$

Hence, in (4.12)-(4.13) all the terms with $\lambda>0$ and $\mu>0$ vanish, and
we obtain

$\displaystyle\sum_{r\geq 0}(-1)^{r}r\zeta_{r,Z}(s)$ $\displaystyle=$
$\displaystyle\sum_{\lambda>0,\mu=0}\lambda^{-s}\left(\sum_{p\geq
0}(-1)^{p}p\,n_{p}(\lambda,X)\right)\left(\sum_{q\geq
0}(-1)^{q}n_{q}(0,Y)\right)$ $\displaystyle+$
$\displaystyle\sum_{\mu>0,\lambda=0}\mu^{-s}\left(\sum_{p\geq
0}(-1)^{p}n_{p}(0,X)\right)\left(\sum_{q\geq 0}(-1)^{q}q\,n_{q}(\mu,Y)\right)$
$\displaystyle=$ $\displaystyle\chi(Y)\sum_{p\geq
0}(-1)^{p}p\zeta_{p,X}(s)+\chi(X)\sum_{q\geq 0}(-1)^{q}q\zeta_{q,Y}(s).$

Taking derivative of the both sides at $s=0$ and using the definition of
analytic torsion, we obtain

$\log T^{\prime}(Z)=\chi(Y)\log T^{\prime}(X)+\chi(X)\log T^{\prime}(Y).$
(4.14)

The Künneth formula (4.3) implies that

$\chi\left(Z\right)=\chi\left(X\right)\chi\left(Y\right),$

which will be used in the next statement.

For any $n\geq 2$ define on the set $X^{\Box n}=\underset{n\
\mathrm{times}}{\underbrace{X\times...\times X}}$ the following path complex

$P(X^{\Box n})=\underset{n\
\mathrm{times}}{\underbrace{P\left(X\right)\Box...\Box
P\left(X\right)}}=P\left(X\right)^{\Box n}.$

######  Corollary 4.12.

We have

$\log T^{\prime}(X^{\Box n})=n\chi\left(X\right)^{n-1}\log
T^{\prime}\left(X\right).$ (4.15)

###### Proof.

Denote $\log T^{\prime}\left(X^{\Box n}\right)=x_{n}$ and
$\chi\left(X\right)=a.$ Then $\chi\left(X^{n}\right)=a^{n}$, and we have by
(4.14)

$x_{n+1}=ax_{n}+a^{n}x_{1}.$

For $n=1$ (4.15) is trivial. Assuming the induction hypothesis
$x_{n}=na^{n-1}x_{1},$ we obtain

$x_{n+1}=na^{n}x_{1}+a^{n}x_{1}=\left(n+1\right)a^{n}x_{1},$

which finishes the proof by induction.

######  Example 4.13.

Let $G$ be a cyclic digraph with $m$ vertices from Example 3.12. For $n\geq 2$
the product $G^{\Box n}$ can be regarded as an analogue of a torus. Since
$\chi\left(G\right)=0$, we obtain from (4.15) that, for any $n\geq 2$,

$T^{\prime}(G^{\Box n})=1.$

Recall for comparison that $T^{\prime}\left(G\right)=T\left(G\right)=m.$

Before we can compute $T(G^{\Box n})$, let us verify that

$\dim\Omega_{p}(G^{\Box n})=\tbinom{n}{p}m^{n}.$ (4.16)

Indeed, for $n=1$ this is true because

$\dim\Omega_{0}\left(G\right)=\dim\Omega_{1}\left(G\right)=m.$

Assuming that (4.16) is true for some $n$, we obtain by the Künneth formula
(4.3)

$\displaystyle\dim\Omega_{r}(G^{\Box\left(n+1\right)})$ $\displaystyle=$
$\displaystyle\sum_{p+q=r}\dim\Omega_{p}(G^{\Box
n})\dim\Omega_{q}\left(G\right)$ $\displaystyle=$ $\displaystyle
m\dim\Omega_{r}(G^{\Box n})+m\dim\Omega_{r-1}(G^{\Box n})$ $\displaystyle=$
$\displaystyle m\tbinom{n}{r}m^{n}+m\tbinom{n}{r-1}m^{n}$ $\displaystyle=$
$\displaystyle m^{n+1}\tbinom{n+1}{r}.$

In the same way, using that

$\dim H_{0}\left(G\right)=\dim H_{1}\left(G\right)=1,$

we obtain that

$\dim H_{p}(G^{\Box n})=\tbinom{n}{p}.$

Hence, by (3.15) we obtain

$\displaystyle T(G^{\Box n})$ $\displaystyle=$ $\displaystyle
T^{\prime}(G^{\Box
n})\prod_{p=0}^{n}\left(p!\right)^{\frac{1}{2}\left(-1\right)^{p}\left(\dim\Omega_{p}-\dim
H_{p}\right)}$ $\displaystyle=$
$\displaystyle\prod_{p=2}^{n}\left(p!\right)^{\frac{1}{2}\left(-1\right)^{p}\binom{n}{p}\left(m^{n}-1\right)}.$

In particular, we have $T(G^{\Box 2})=2^{\frac{1}{2}\left(m^{2}-1\right)}.$

######  Example 4.14.

For the _interval_ $I=\left.{}^{0}\bullet\rightarrow\bullet^{1}\right.$ we
have by Example 3.9 $T^{\prime}\left(I\right)=\sqrt{2}$ and
$\chi\left(I\right)=1.$ Consider the $n$-dimensional digraph _cube_ $I^{\Box
n}.$ In the case $n=2$ it coincides with the square from Example 3.11, in the
case $n=3$ this digraph is shown on Fig. 6.

Figure 6. The cube $I^{\Box 3}$

By (4.15) we obtain

$\log T^{\prime}(I^{\Box n})=n\log\sqrt{2}$

whence

$T^{\prime}(I^{\Box n})=2^{n/2}.$

Let us compute the torsion $T(I^{\Box n})$ of the cube with respect to the
standard inner product $\iota$. For that let us first verify that

$\dim\Omega_{p}(I^{\Box n})=2^{n-p}\tbinom{n}{p},$ (4.17)

We have

$\dim\Omega_{0}\left(I\right)=2,\ \ \dim\Omega_{1}\left(I\right)=1\ \
\text{and\ \ }\dim\Omega_{p}\left(I\right)=0\ \ \text{for }p\geq 2$

so that (4.17) holds for $n=1.$ For the inductive step from $n$ to $n+1$,
observe that $I^{\Box\left(n+1\right)}=I^{\Box n}\Box I$. By the Künneth
formula (4.3) we have

$\displaystyle\dim\Omega_{r}(I^{\Box\left(n+1\right)})$ $\displaystyle=$
$\displaystyle\sum_{p+q=r}\dim\Omega_{p}(I^{\Box
n})\dim\Omega_{q}\left(I\right)$ $\displaystyle=$ $\displaystyle
2\dim\Omega_{r}(I^{\Box n})+\dim\Omega_{r-1}(I^{\Box n})$ $\displaystyle=$
$\displaystyle 2^{n+1-r}\tbinom{n}{r}+2^{n-r+1}\tbinom{n}{r-1}$
$\displaystyle=$ $\displaystyle 2^{n+1-r}\tbinom{n+1}{r},$

which finishes the proof of (4.17). In the same way one obtains that $\dim
H_{0}(I^{\Box n})=1$ and $\dim H_{p}(I^{\Box n})=0$ for all $p\geq 1.$ Hence,
by (3.15) we obtain

$\displaystyle T(I^{\Box n})$ $\displaystyle=$ $\displaystyle
T^{\prime}(I^{\Box
n})\prod_{p=0}^{n}\left(p!\right)^{\frac{1}{2}\left(-1\right)^{p}\left(\dim\Omega_{p}-\dim
H_{p}\right)}$ $\displaystyle=$ $\displaystyle
2^{n/2}\prod_{p=2}^{n}\left(p!\right)^{\frac{1}{2}\left(-1\right)^{p}2^{n-p}\binom{n}{p}}.$

For example, we have

$T(I^{\Box 2})=2\sqrt{2},\ \ \ \ T(I^{\Box 3})=\frac{16}{3}\sqrt{3},\ \ \ \
T(I^{\Box 4})=\frac{2048}{81}\sqrt{6},$

etc.

######  Corollary 4.15.

If $P\left(Z\right)=P\left(X\right)\Box P\left(Z\right)$ then

$\displaystyle\log T(Z)\underset{}{=}\chi(Y)\log T(X)+\chi(X)\log T(Y)$
$\displaystyle+\frac{1}{2}\sum_{p,q\geq
1}\left(-1\right)^{p+q}\log\tbinom{p+q}{p}\left(\dim\Omega_{p}(X)\dim\Omega_{q}(Y)-\dim
H_{p}(X)\dim H_{q}(Y)\right).$ (4.18)

###### Proof.

We use the R-torsions $\tau$ and $\tau^{\prime}$ defined with respect to the
inner products $\iota$ and $\iota^{\prime}$, respectively. By Theorems 3.14
and 4.8 we have

$\log\tau^{\prime}(Z)=\chi(Y)\log\tau^{\prime}(X)+\chi(X)\log\tau^{\prime}(Y).$
(4.19)

By (3.15) we have

$\log\tau^{\prime}(X)=\log\tau(X)-\frac{1}{2}\sum_{p}\left(-1\right)^{p}\dim\Omega_{p}\left(X\right)\log\left(p!\right)+\frac{1}{2}\sum_{p}\left(-1\right)^{p}\dim
H_{p}\left(X\right)\log\left(p!\right),$

where summation is taken over all $p\geq 0$. Similar identities hold for $Y$
and $Z$. Substituting into (4.19), we obtain

$\displaystyle\log\tau(Z)-\chi(Y)\log\tau(X)-\chi(X)\log\tau(Y)$ (4.20)
$\displaystyle=$
$\displaystyle\frac{1}{2}\sum_{r}\left(-1\right)^{r}\dim\Omega_{r}\left(Z\right)\log\left(r!\right)-\frac{1}{2}\sum_{r}\left(-1\right)^{r}\dim
H_{r}\left(Z\right)\log\left(r!\right)$
$\displaystyle-\frac{1}{2}\chi(Y)\sum_{p}\left(-1\right)^{p}\dim\Omega_{p}\left(X\right)\log\left(p!\right)+\frac{1}{2}\chi(Y)\sum_{p}\left(-1\right)^{p}\dim
H_{p}\left(X\right)\log\left(p!\right)$
$\displaystyle-\frac{1}{2}\chi(X)\sum_{q}\left(-1\right)^{q}\dim\Omega_{q}\left(Y\right)\log\left(q!\right)+\frac{1}{2}\chi(X)\sum_{q}\left(-1\right)^{q}\dim
H_{q}\left(Y\right)\log\left(q!\right)$

Denote for simplicity

$x_{p}=\dim\Omega_{p}\left(X\right),\ \ \ y_{q}=\dim\Omega_{q}\left(Y\right),\
\ \ z_{r}=\dim\Omega_{r}\left(Z\right).$

By the Künneth formula we have

$z_{r}=\sum_{p+q=r}x_{p}y_{q}.$

It follows that

$\displaystyle\sum_{r}\left(-1\right)^{r}z_{r}\log\left(r!\right)-\chi\left(Y\right)\sum_{p}\left(-1\right)^{p}x_{p}\log\left(p!\right)-\chi\left(X\right)\sum_{q}\left(-1\right)^{q}y_{p}\log\left(q!\right)$
$\displaystyle=$
$\displaystyle\sum_{r}\left(-1\right)^{r}\sum_{p+q=r}x_{p}y_{q}\log\left(r!\right)-\sum_{q}\left(-1\right)^{q}y_{q}\sum_{p}\left(-1\right)^{p}x_{p}\log\left(p!\right)-\sum_{p}\left(-1\right)^{p}x_{p}\sum_{q\geq
0}\left(-1\right)^{q}y_{p}\log\left(q!\right)$ $\displaystyle=$
$\displaystyle\sum_{p,q}\left(-1\right)^{p+q}x_{p}y_{q}\log\left(\left(p+q\right)!\right)-\sum_{p,q}\left(-1\right)^{p+q}x_{p}y_{q}\log\left(p!\right)-\sum_{p,q}\left(-1\right)^{p+q}x_{p}y_{q}\log\left(q!\right)$
$\displaystyle=$
$\displaystyle\sum_{p,q}\left(-1\right)^{p+q}x_{p}y_{q}\log\tbinom{p+q}{p}.$

Note that the summation here can be restricted to $p,q\geq 1$ since otherwise
$\log\binom{p+q}{p}=0.$ A similar formula takes place for $\dim H_{p}$ instead
of $\dim\Omega_{p}.$ Substituting into (4.20) we obtain (4.18).

######  Example 4.16.

Let us compute the torsions of the digraph $Z=I\Box Y$ where $I$ is the
interval from Example 4.14 and $Y$ is the triangle from Example 3.10 (see Fig.
7).

Figure 7. A prism digraph $I\Box Y$

By Examples 3.9 and 3.10, we have $\chi\left(I\right)=\chi\left(Y\right)=1$
and

$T^{\prime}\left(I\right)=\sqrt{2},\ \ \ \
T^{\prime}\left(Y\right)=\sqrt{3/2}.$

Hence, we obtain by (4.7)

$T^{\prime}\left(Z\right)=T^{\prime}\left(I\right)^{\chi\left(Y\right)}T^{\prime}\left(Y\right)^{\chi\left(I\right)}=\sqrt{2}\sqrt{3/2}=\sqrt{3}.$

Since $H_{p}\left(I\right)$ and $H_{p}\left(Y\right)$ are non-trivial only for
$p=0,$ we obtain by (4.18)

$\displaystyle\log T(Z)$ $\displaystyle=$ $\displaystyle\chi(Y)\log
T(I)+\chi(I)\log T(Y)$
$\displaystyle+\frac{1}{2}\sum_{p=1}^{1}\sum_{q=1}^{2}\left(-1\right)^{p+q}\log\tbinom{p+q}{p}\dim\Omega_{p}\left(I\right)\dim\Omega_{q}\left(Y\right)$
$\displaystyle=$
$\displaystyle\log\sqrt{2}+\log\sqrt{3}+\frac{1}{2}\log\binom{2}{1}\cdot
1\cdot 3-\frac{1}{2}\log\binom{3}{1}\cdot 1\cdot 1$ $\displaystyle=$
$\displaystyle\log\left(\sqrt{2}\sqrt{3}2^{3/2}3^{-1/2}\right)=\log 4,$

so that

$T\left(Z\right)=4.$

##  5\. Join of path complexes

###  5.1. Augmented chain complex

Let $P$ be a path complex over a set $V$ as in Section 2.1. In that section we
have constructed a chain complex
$\mathcal{R}=\left\\{\mathcal{R}_{p}\right\\}_{p\geq 0}$ with the boundary
operator $\partial$, and the space $\mathcal{R}_{-1}$ was defined as
$\left\\{0\right\\}.$ In this section we change the definition of
$\mathcal{R}_{-1}$ as follows. For any elementary $0$-paths $e_{i}$ redefine
$\partial$ by

$\partial e_{i}=e$

where $e$ is an empty path that by definition has the length $-1.$ Set
$\mathcal{R}_{-1}=\mathop{\mathrm{s}pan}_{\mathbb{R}}\left\\{e\right\\}\cong\mathbb{R}$
and consider the _augmented_ chain complex
$\widetilde{\mathcal{R}}=\left\\{\mathcal{R}_{p}\right\\}_{p\geq-1}$ where the
operator $\partial$ still satisfies $\partial^{2}=0.$ Consequently, we obtain
also the _augmented_ chain complex
$\widetilde{\Omega}=\left\\{\Omega_{p}\right\\}_{p\geq-1}$ of
$\partial$-invariant paths, where $\Omega_{p}$ with $\,p\geq 0$ is as before
and $\Omega_{-1}=\mathcal{R}_{-1},$ as well as the _reduced_ homology groups
$\\{\widetilde{H}_{p}\\}_{p\geq-1}$ where
$\widetilde{H}_{-1}=\left\\{0\right\\}$, $\widetilde{H}_{p}=H_{p}$ for $p\geq
1$ and $H_{0}\cong\widetilde{H}_{0}\oplus\mathbb{R}.$ The reduced Euler
characteristic is

$\widetilde{\chi}\left(P\right)=\sum_{p\geq-1}\left(-1\right)^{p}\dim\Omega_{p}=\sum_{p\geq-1}\left(-1\right)^{p}\dim\widetilde{H}_{p}=\chi\left(P\right)-1.$
(5.1)

Denote by $\left\langle,\right\rangle$ the standard inner product in
$\mathcal{R}_{p}$ defined by (3.1) in the case $p\geq 0$ and by $\left\langle
e,e\right\rangle=1$ for $p=-1.$ If $u\in\mathcal{R}_{p}$ and
$v\in\mathcal{R}_{q}$ with $p\neq q$ then set $\left\langle
u,v\right\rangle=0.$ As before, we denote by $\iota$ the standard inner
product structure in $\widetilde{\mathcal{R}}$.

###  5.2. Join of path complexes and digraphs

Let $V$ be a finite set.

######  Definition 5.1.

For any paths $u\in\mathcal{R}_{p}\left(V\right)$ and
$v\in\mathcal{R}_{q}\left(V\right)$ with $p,q\geq-1$ define their _join_
$u\cdot v\in\mathcal{R}_{p+q+1}\left(V\right)$ as follows: first set

$e_{i_{0}...i_{p}}\cdot e_{j_{0}...j_{q}}=e_{i_{0}...i_{p}j_{0}...j_{q}}$

and then extend this definition by linearity.

In particular, we have $e_{i_{0}...i_{p}}\cdot e=e_{i_{0}...i_{p}}.$ The
following product formula holds for the chain complex
$\widetilde{\mathcal{R}}\left(V\right)$:

$\partial\left(u\cdot v\right)=\left(\partial u\right)\cdot
v+\left(-1\right)^{p+1}u\cdot\left(\partial v\right).$ (5.2)

(see [7, Lemma 2.2] and [6, Lemma 2.4]).

Let $X,Y$ be two finite disjoint sets, set $Z=X\sqcup Y.$ Then all paths on
$X$ and $Y$ can be considered as paths on $Z$. It follows easily from
definition of $u\cdot v$ that

$u\in\mathcal{R}_{p}\left(X\right)\ \text{and\
}v\in\mathcal{R}_{q}\left(Y\right)\ \ \Rightarrow\ \ u\cdot
v\in\mathcal{R}_{p+q+1}\left(Z\right).$

Also, for the standard inner product $\langle,\rangle$ given by (3.1), we have

$\langle u\cdot v,\varphi\cdot\psi\rangle=\langle u,\varphi\rangle\langle
v,\psi\rangle,$ (5.3)

for all $u\in\mathcal{R}_{p}\left(X\right)$,
$v\in\mathcal{R}_{q}\left(Y\right)$,
$\varphi\in\mathcal{R}_{p^{\prime}}\left(X\right)$ and
$\psi\in\mathcal{R}_{q^{\prime}}\left(X\right)$ (see also [6, Lemma 3.10]).

Let us extend the property (2.4) of the definition of a path complex $P$ also
to $n$-paths with $n=0$, that is, we allow in (2.4) also $n=0.$ Then
necessarily the empty path $e$ belongs to $P.$

######  Definition 5.2.

Let $P\left(X\right)$ and $P\left(Y\right)$ be two path complexes over finite
disjoint sets $X$ and $Y$, respectively. Define the chain complex
$P\left(Z\right)$ over the set $Z=X\sqcup Y$ as follows: $P\left(Z\right)$
consists of all the paths of the form $\,u\cdot v$ where $u\in
P\left(X\right)$ and $v\in P\left(Y\right).$ The path complex
$P\left(Z\right)$ is called the _join_ of $P\left(X\right),P\left(Y\right)$
and is denoted by $P\left(Z\right)=P\left(X\right)\ast P\left(Z\right).$

Clearly, $P\left(X\right)$ and $P\left(Y\right)$ are subsets of
$P\left(Z\right)$.

######  Definition 5.3.

If $X$ and $Y$ are digraphs then define their _join_ as a digraph $Z=X\ast Y$
where the set of vertices is $X\sqcup Y$ and the set of arrows consists of all
arrows of $X$, all arrows of $Y$ as well as of all arrows of the form
$x\rightarrow y$ where $x\in X$ and $y\in Y.$

It is easy to see that

$P\left(X\ast Y\right)=P\left(X\right)\ast P\left(Y\right)$

so that the operation of join of digraphs is compatible with join of path
complexes (cf. [7]).

It is clear from the definition of $P\left(Z\right)$ that

$u\in\mathcal{A}_{p}\left(X\right)\ \text{and\
}v\in\mathcal{A}_{q}\left(Y\right)\ \ \Rightarrow\ \ u\cdot
v\in\mathcal{A}_{p+q+1}\left(Z\right).$

It follows from (5.2) that, for all $p,q\geq-1$,

$u\in\Omega_{p}\left(X\right)\ \ \text{and\ \ }v\in\Omega_{q}\left(Y\right)\ \
\Rightarrow\ \ u\cdot v\in\Omega_{p+q+1}\left(Z\right)$

(see [7, Prop 5.4]). Furthermore, the following version of the Künneth formula
is true for join: for any $r\geq-1$

$\Omega_{r}\left(Z\right)=\bigoplus_{\left\\{p,q\geq-1:p+q+1=r\right\\}}\Omega_{p}\left(X\right)\otimes\Omega_{q}\left(Y\right)$
(5.4)

where $u\otimes v$ for $u\in\Omega_{p}\left(X\right)$ and
$v\in\Omega_{q}\left(Y\right)$ is identified with the element $u\cdot v$ of
$\Omega_{r}\left(Z\right)$ (see [7, Thm 5.5]).

As a consequence of (5.4) we obtain that

$\widetilde{\chi}\left(Z\right)=-\widetilde{\chi}\left(X\right)\widetilde{\chi}\left(Y\right),$
(5.5)

where the minus comes from the additional $1$ in $r=p+q+1$.

###  5.3. Operators $\partial^{\ast}$ and $\Delta$ on joins

We always assume in what follows that all the spaces $\mathcal{R}_{p}$ under
consideration are endowed with the standard inner product (3.1), in
particular, (5.3) is satisfied.

######  Lemma 5.4.

Let $u\in\Omega_{p}\left(X\right)$ and $v\in\Omega_{q}\left(Y\right).$ Then

$\partial^{\ast}\left(u\cdot v\right)=\left(\partial^{\ast}u\right)\cdot
v+\left(-1\right)^{p+1}u\cdot\left(\partial^{\ast}v\right).$ (5.6)

###### Proof.

By definition, we have, for any $w\in\Omega_{p+q+2}\left(Z\right)$

$\langle\partial^{\ast}\left(u\cdot v\right),w\rangle=\langle u\cdot
v,\partial w\rangle$

Any $w\in\Omega_{\ast}\left(Z\right)$ admits a representation

$w=\sum_{k}\varphi_{k}\cdot\psi_{k}$

where the sum is finite and

$\varphi_{k}\in\Omega_{p_{k}}\left(X\right)\ \ \text{and
}\psi_{k}\in\Omega_{q_{k}}\left(Y\right)$

with $p_{k}+q_{k}+1=p+q+2$ (see [6, Thm 5.1] and [7, Thm 5.15]). Then we have
using (5.3)

$\displaystyle\langle\partial^{\ast}\left(u\cdot v\right),w\rangle$
$\displaystyle=$ $\displaystyle\langle u\cdot
v,\sum\partial\left(\varphi_{k}\cdot\psi_{k}\right)\rangle$ $\displaystyle=$
$\displaystyle\langle u\cdot
v,\sum\left(\partial\varphi_{k}\cdot\psi_{k}+\left(-1\right)^{p_{k}+1}\varphi_{k}\cdot\partial\psi_{k}\right)\rangle$
$\displaystyle=$ $\displaystyle\sum\langle u\cdot
v,\partial\varphi_{k}\cdot\psi_{k}\rangle+\left(-1\right)^{p_{k}+1}\langle
u\cdot v,\varphi_{k}\cdot\partial\psi_{k}\rangle$ $\displaystyle=$
$\displaystyle\sum\langle u,\partial\varphi_{k}\rangle\langle
v,\psi_{k}\rangle+\left(-1\right)^{p_{k}+1}\langle u,\varphi_{k}\rangle\langle
v,\partial\psi_{k}\rangle$ $\displaystyle=$
$\displaystyle\sum\langle\partial^{\ast}u,\varphi_{k}\rangle\langle
v,\psi_{k}\rangle+\left(-1\right)^{p_{k}+1}\langle
u,\varphi_{k}\rangle\langle\partial^{\ast}v,\psi_{k}\rangle.$

Note that if $p_{k}\neq p$ then $\langle u,\varphi_{k}\rangle=0.$ Hence, we
can replace $p_{k}$ everywhere by $p$ and obtain

$\displaystyle\langle\partial^{\ast}\left(u\cdot v\right),w\rangle$
$\displaystyle=$
$\displaystyle\sum\langle\partial^{\ast}u,\varphi_{k}\rangle\langle
v,\psi_{k}\rangle+\left(-1\right)^{p+1}\langle
u,\varphi_{k}\rangle\langle\partial^{\ast}v,\psi_{k}\rangle$ $\displaystyle=$
$\displaystyle\sum\langle\partial^{\ast}u\cdot
v,\varphi_{k}\cdot\psi_{k}\rangle+\langle\left(-1\right)^{p+1}u\cdot\partial^{\ast}v,\varphi_{k}\cdot\psi_{k}\rangle$
$\displaystyle=$ $\displaystyle\langle\partial^{\ast}u\cdot
v+\left(-1\right)^{p+1}u\cdot\partial^{\ast}v,\sum\varphi_{k}\cdot\psi_{k}\rangle$
$\displaystyle=$ $\displaystyle\langle\partial^{\ast}u\cdot
v+\left(-1\right)^{p+1}u\cdot\partial^{\ast}v,w\rangle,$

whence (5.6) follows.

For the _Hodge Laplacian_

$\Delta u=\partial\partial^{\ast}u+\partial^{\ast}\partial u$

we have then the following identity.

######  Lemma 5.5.

For all $u\in\Omega_{p}\left(X\right)$ and $v\in\Omega_{q}\left(X\right)$ we
have

$\Delta\left(u\cdot v\right)=\left(\Delta u\right)\cdot v+u\cdot\Delta v.$
(5.7)

###### Proof.

Indeed, by (5.6) we have

$\displaystyle\partial\partial^{\ast}\left(u\cdot v\right)$ $\displaystyle=$
$\displaystyle\partial\left(\partial^{\ast}u\cdot
v+\left(-1\right)^{p+1}u\cdot\partial^{\ast}v\right)$ $\displaystyle=$
$\displaystyle\partial\left(\partial^{\ast}u\cdot
v\right)+\left(-1\right)^{p+1}\partial\left(u\cdot\partial^{\ast}v\right)$
$\displaystyle=$ $\displaystyle\partial\partial^{\ast}u\cdot
v+\left(-1\right)^{p+2}\partial^{\ast}u\cdot\partial v$
$\displaystyle+\left(-1\right)^{p+1}\left(\partial
u\cdot\partial^{\ast}v+\left(-1\right)^{p+1}u\cdot\partial\partial^{\ast}v\right)$
$\displaystyle=$ $\displaystyle\partial\partial^{\ast}u\cdot
v+\left(-1\right)^{p}\partial^{\ast}u\cdot\partial
v+\left(-1\right)^{p+1}\partial
u\cdot\partial^{\ast}v+u\cdot\partial\partial^{\ast}v$

and by (5.2)

$\displaystyle\partial^{\ast}\partial\left(u\cdot v\right)$ $\displaystyle=$
$\displaystyle\partial^{\ast}\left(\partial u\cdot
v+\left(-1\right)^{p+1}u\cdot\partial v\right)$ $\displaystyle=$
$\displaystyle\partial^{\ast}\left(\partial u\cdot
v\right)+\left(-1\right)^{p+1}\partial^{\ast}\left(u\cdot\partial v\right)$
$\displaystyle=$ $\displaystyle\partial^{\ast}\partial u\cdot
v+\left(-1\right)^{p}\partial u\cdot\partial^{\ast}v$
$\displaystyle+\left(-1\right)^{p+1}\left(\partial^{\ast}u\cdot\partial
v+\left(-1\right)^{p+1}u\cdot\partial^{\ast}\partial v\right)$
$\displaystyle=$ $\displaystyle\partial^{\ast}\partial u\cdot
v+\left(-1\right)^{p}\partial
u\cdot\partial^{\ast}v+\left(-1\right)^{p+1}\partial^{\ast}u\cdot\partial
v+u\cdot\partial^{\ast}\partial v.$

Adding up the two identities, we see that the terms
$\partial^{\ast}u\cdot\partial v$ and $\partial u\cdot\partial^{\ast}v$ cancel
out, and we obtain (5.7).

###  5.4. Torsion of joins

Let $P$ be a path complex over a set $V$ with the standard inner product
structure $\iota$ given by (3.1). By means of the augmented chain complex
$\widetilde{\Omega}\left(P\right)$, let us define the _reduced_ analytic
torsion $\widetilde{T}\left(P\right)$ by

$\log\widetilde{T}\left(P\right)=\log
T(\widetilde{\Omega}\left(P\right),\iota)=\frac{1}{2}\sum_{p=-1}^{N}(-1)^{p}\,p\,\zeta_{p}^{\prime}(0).$

In the previous sections we used the standard analytic torsion
$T\left(P\right)$ given by

$\log T\left(P\right)=\log
T(\Omega\left(P\right),\iota)=\frac{1}{2}\sum_{p=0}^{N}(-1)^{p}\,p\,\zeta_{p}^{\prime}(0)$

The relation between $T\left(P\right)$ and $\widetilde{T}\left(P\right)$ is
given by the following formula.

######  Lemma 5.6.

We have

$T\left(P\right)=\sqrt{\left|V\right|}\widetilde{T}\left(P\right).$ (5.8)

###### Proof.

The zeta function $\zeta_{p}\left(s\right)$ is determined by the operator
$\Delta_{p}$ that is the same for the chain complexes $\Omega$ and
$\widetilde{\Omega}$ for all $p\geq 1.$ For $p=0$ the operators $\Delta_{p}$
are different for these two complexes, but the value $p=0$ does not give any
contribution to the analytic torsions. Hence, the difference is determined by
$p=-1,$ that is,

$\log\widetilde{T}\left(P\right)=\log
T\left(P\right)+\frac{1}{2}\,\zeta_{-1}^{\prime}(0).$ (5.9)

For $e\in\Omega_{-1}$ we have $\partial e=0$ and

$\partial^{\ast}e=\sum_{i\in V}e_{i},$

because for any $i$

$\left\langle\partial^{\ast}e,e_{i}\right\rangle=\left\langle e,\partial
e_{i}\right\rangle=\left\langle e,e\right\rangle=1.$

Hence,

$\Delta e=\partial\partial^{\ast}e+\partial^{\ast}\partial
e=\partial\sum_{i\in V}e_{i}=\left|V\right|e.$

Therefore, $\zeta_{-1}\left(s\right)=\left|V\right|^{-s}$ and
$\zeta_{-1}^{\prime}(0)=-\log\left|V\right|$. Substituting into (5.9) we
obtain

$\log\widetilde{T}\left(P\right)=\log
T\left(P\right)-\frac{1}{2}\log\left|V\right|,$ (5.10)

which is equivalent to (5.8).

The next theorem is our main result about torsion on joins.

######  Theorem 5.7.

For the join path complex $P\left(Z\right)=P\left(X\right)\ast
P\left(Y\right)$ we have

$\log\widetilde{T}(Z)=-\widetilde{\chi}(Y)\log\widetilde{T}(X)-\widetilde{\chi}(X)\log\widetilde{T}(Y)$
(5.11)

where $\widetilde{\chi}$ is the reduced Euler characteristic.

###### Proof of Theorem 5.7.

The proof is similar to that of Theorem 4.8. Let $E_{p}\left(\lambda\right)$
be the eigenspace of $\Delta_{p}$ with the eigenvalue $\lambda$, and set

$n_{p}\left(\lambda\right)=\dim E_{p}\left(\lambda\right).$

Lemmas 4.9 and 4.10 go unchanged also for the augmented chain complex
$\widetilde{\Omega}=\\{\Omega_{p}\\}_{p\geq-1}.$ The same argument as in the
proof of Lemma 4.11 gives for any $\lambda>0$ that

$\sum\limits_{p=-1}^{N}(-1)^{p}n_{p}(\lambda)=0,$ (5.12)

because $n_{-1}^{\prime\prime}\left(\lambda\right)=0$: indeed, for any
$\varphi\in E_{-1}^{\prime\prime}\left(\lambda\right)$ we have
$\partial\varphi=0$ and, hence,
$\varphi=\frac{1}{\lambda}\partial^{\ast}\partial\varphi=0.$

Arguing as in the proof of Theorem 4.8 and using the Künneth formula (5.4) we
obtain in place of (4.11) the following identity:

$\zeta_{r,Z}(s)=\sum_{\lambda+\mu>0}\sum_{p+q+1=r}(\lambda+\mu)^{-s}n_{p}(\lambda,X)n_{q}(\mu,Y)$

for any $r\geq-1.$ It follows that

$\displaystyle\sum_{r\geq-1}(-1)^{r}r\zeta_{r,Z}(s)$ (5.15) $\displaystyle=$
$\displaystyle\sum_{\lambda+\mu>0}(\lambda+\mu)^{-s}\sum_{p\geq-1}\sum_{q\geq-1}(-1)^{p+q+1}(p+q+1)n_{p}(\lambda,X)n_{q}(\mu,Y)$
$\displaystyle=$
$\displaystyle-\sum_{\lambda+\mu>0}(\lambda+\mu)^{-s}\left(\sum_{p\geq-1}(-1)^{p}p\,n_{p}(\lambda,X)\right)\left(\sum_{q\geq-1}(-1)^{q}n_{q}(\mu,Y)\right)$
$\displaystyle-\sum_{\lambda+\mu>0}(\lambda+\mu)^{-s}\left(\sum_{p\geq-1}(-1)^{p}n_{p}(\lambda,X)\right)\left(\sum_{q\geq-1}(-1)^{q}q\,n_{q}(\mu,Y)\right)$
$\displaystyle-\sum_{\lambda+\mu>0}(\lambda+\mu)^{-s}\sum_{p\geq-1}\sum_{q\geq-1}(-1)^{p+q}n_{p}(\lambda,X)n_{q}(\mu,Y).$

By (5.12), if $\lambda>0$ then

$\sum_{p\geq-1}(-1)^{p}n_{p}(\lambda,X)=0$

and if $\mu>0$ then

$\sum_{q\geq-1}(-1)^{q}n_{q}(\mu,Y)=0.$

Hence, the double sum in (5.15) is equal to zero, while in (5.15)-(5.15) all
the terms with $\lambda>0$ and $\mu>0$ vanish. We obtain

$\displaystyle\sum_{r\geq-1}(-1)^{r}r\zeta_{r,Z}(s)$ $\displaystyle=$
$\displaystyle-\sum_{\lambda>0,\mu=0}\lambda^{-s}\left(\sum_{p\geq-1}(-1)^{p}p\,n_{p}(\lambda,X)\right)\left(\sum_{q\geq-1}(-1)^{q}n_{q}(0,Y)\right)$
$\displaystyle-\sum_{\mu>0,\lambda=0}\mu^{-s}\left(\sum_{p\geq-1}(-1)^{p}n_{p}(0,X)\right)\left(\sum_{q\geq-1}(-1)^{q}q\,n_{q}(\mu,Y)\right)$
$\displaystyle=$
$\displaystyle-\widetilde{\chi}(Y)\sum_{p\geq-1}(-1)^{p}p\zeta_{p,X}(s)-\widetilde{\chi}(X)\sum_{q\geq-1}(-1)^{q}q\zeta_{q,Y}(s).$

Taking derivative of the both sides at $s=0$ and using the definition of
analytic torsion, we obtain (5.11).

For the standard analytic torsion $T$ we obtain the following.

######  Corollary 5.8.

We have

$\displaystyle\log T(Z)$ $\displaystyle=$
$\displaystyle-\widetilde{\chi}(Y)\log T(X)-\widetilde{\chi}(X)\log T(Y)$
(5.16)
$\displaystyle+\frac{1}{2}\log\left|Z\right|+\frac{\widetilde{\chi}\left(Y\right)}{2}\log\left|X\right|+\frac{\widetilde{\chi}\left(X\right)}{2}\log\left|Y\right|.$

###### Proof.

Using (5.8) and (5.1), we obtain

$\displaystyle\log T\left(Z\right)$ $\displaystyle=$
$\displaystyle\frac{1}{2}\log\left|Z\right|+\log\widetilde{T}\left(Z\right)$
$\displaystyle=$
$\displaystyle\frac{1}{2}\log\left|Z\right|-\left(\widetilde{\chi}(Y)\log\widetilde{T}(X)+\widetilde{\chi}(X)\log\widetilde{T}(Y)\right)$
$\displaystyle=$
$\displaystyle\frac{1}{2}\log\left|Z\right|-\widetilde{\chi}(Y)\left(\log
T(X)-\frac{1}{2}\log\left|X\right|\right)-\widetilde{\chi}(X)\left(\log
T(Y)-\frac{1}{2}\log\left|Y\right|\right)$ $\displaystyle=$
$\displaystyle\frac{1}{2}\log\left|Z\right|+\widetilde{\chi}(Y)\frac{1}{2}\log\left|X\right|+\widetilde{\chi}(X)\frac{1}{2}\log\left|Y\right|$
$\displaystyle-\widetilde{\chi}(Y)\log T(X)-\widetilde{\chi}(X)\log T(Y).$

For any $n\geq 2$ define on the set $X^{\ast n}=\underset{n\
\mathrm{times}}{\underbrace{X\sqcup...\sqcup X}}$ the following path complex

$P\left(X^{\ast n}\right)=\underset{n\
\mathrm{times}}{\underbrace{P\left(X\right)\ast...\ast
P\left(X\right)}}=P(X)^{\ast n}.$

######  Corollary 5.9.

We have

$\log\widetilde{T}(X^{\ast
n})=n\left(-\widetilde{\chi}\left(X\right)\right)^{n-1}\log\widetilde{T}\left(X\right)$
(5.17)

and

$\log T(X^{\ast n})=n\left(1-\chi\left(X\right)\right)^{n-1}\log
T\left(X\right)-\frac{1}{2}n\left(1-\chi\left(X\right)\right)^{n-1}\log\left|X\right|+\frac{1}{2}\log\left(n\left|X\right|\right).$
(5.18)

###### Proof.

Denote $\log\widetilde{T}(X^{\ast n})=x_{n}$ and
$-\widetilde{\chi}\left(X\right)=a.$ Then $-\widetilde{\chi}(X^{\ast
n})=a^{n}$, and we have by (5.11)

$x_{n+1}=ax_{n}+a^{n}x_{1}.$

By induction we obtain $x_{n}=na^{n-1}x_{1}$, which proves (5.17).

Using (5.8) (or (5.10)) and (5.1), we obtain from (5.17)

$\displaystyle\log T(X^{\ast n})$ $\displaystyle=$
$\displaystyle\log\widetilde{T}(X^{\ast n})+\frac{1}{2}\log\left|X^{\ast
n}\right|$ $\displaystyle=$ $\displaystyle
n\left(1-\chi\left(X\right)\right)^{n-1}(\log
T\left(X\right)-\frac{1}{2}\log\left|X\right|)+\frac{1}{2}\log\left(n\left|X\right|\right)$
$\displaystyle=$ $\displaystyle n\left(1-\chi\left(X\right)\right)^{n-1}\log
T\left(X\right)-\frac{1}{2}n\left(1-\chi\left(X\right)\right)^{n-1}\log\left|X\right|+\frac{1}{2}\log\left(n\left|X\right|\right),$

which proves (5.18).

For example, if $\chi\left(X\right)=0$ then (5.18) yields

$\log T(X^{\ast n})=n\log
T\left(X\right)-\frac{n-1}{2}\log\left|X\right|+\frac{1}{2}\log n,$ (5.19)

if $\chi\left(X\right)=1$ then (5.18) yields

$T(X^{\ast n})=\sqrt{n\left|X\right|},$ (5.20)

and if $\chi\left(X\right)=2$ then

$\log T(X^{\ast n})=n\left(-1\right)^{n-1}\log
T\left(X\right)+\frac{1+n\left(-1\right)^{n}}{2}\log\left|X\right|+\frac{1}{2}\log
n.$ (5.21)

######  Example 5.10.

Let $O=\left\\{\bullet\right\\}$ be a trivial digraph of one vertex. It is
easy to see that $\chi\left(O\right)=1.$ The join $O\ast O=O^{\ast 2}$ is the
interval $I$ from Example 3.9, the join $O\ast I=O^{\ast 3}$ is the triangle
from Example 3.10. More generally, $O^{\ast n}$ can be regarded as an
$\left(n-1\right)$-dimensional digraph _simplex_ (see Fig. 8).

Figure 8. Simplex $O^{\ast 4}=I^{\ast 2}$

From (5.20) we obtain that

$T(O^{\ast n})=\sqrt{n}.$

For example, $T(O^{\ast 3})=\sqrt{3}$ as we have seen in Example 3.10.

######  Example 5.11.

Consider the digraph $D=\left\\{\bullet,\bullet\right\\}$ consisting of two
disjoint vertices. The join $D^{\ast 2}$ is a quadrilateral, and $D^{\ast 3}$
is an octahedron (see Fig. 9). The digraph $D^{\ast n}$ can be regarded as a
digraph analogue of an $\left(n-1\right)$-dimensional sphere.

Figure 9. The octahedron $D^{\ast 3}$ is shown in two ways. The green subgraph
is $D^{\ast 2}.$

We have $\chi\left(D\right)=2$ and $\tau\left(D\right)=T\left(D\right)=1.$ By
(5.21) we obtain that

$\log T((D^{\ast n})=\frac{1+n\left(-1\right)^{n}}{2}\log 2+\frac{1}{2}\log
n,$

that is

$T(D^{\ast n})=\sqrt{n}2^{\frac{1+n\left(-1\right)^{n}}{2}}.$

For example,

$T(D^{\ast 2})=\sqrt{2}2^{\frac{3}{2}}=4$

and

$T(D^{\ast 3})=\sqrt{3}2^{-1}=\frac{\sqrt{3}}{2}.$

######  Example 5.12.

Let $G$ be a cyclic digraph from Example 3.12 with $m$ vertices. Since
$\chi\left(G\right)=0$ and $T\left(G\right)=m$, we obtain by (5.19)

$\log T(G^{\ast n})=n\log m-\frac{n-1}{2}\log m+\frac{1}{2}\log n$

and

$T(G^{\ast n})=\sqrt{n}m^{\frac{n+1}{2}}.$

Acknowledgments. We would like to thank Wang Chong for the computation of
torsions of digraphs in some Examples.

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