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\newcommand{\dps} {\displaystyle}
\newcommand{\pn} {\par\noindent}
\newcommand{\vs} {\vskip}
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\title {The coincidence Reidemeister classes\\ of maps on Nilmanifolds}


\author{Daciberg L. Gon\c calves\\
Departamento de Matem\'atica - IME-USP\\
Caixa Postal 66281 - CEP: 05315-970\\
S\~ao Paulo - SP, Brasil\\
e-mail: dlgoncal@ime.usp.br}


\date{}

\maketitle
 
      
  


\baselineskip=8.5mm

\pn {\bf Introduction:} Given a pair of maps $f, g: N_1\to N_2$ where $N_1 , N_2$ are compact nilmanifolds of the same dimension, in
[Mc], C. K. McCord has very recently shown that $N(f,g)= |L(f,g)|$ where $N(f,g), L(f,g)$ mean the coincidence Nielsen number and Lefschetz coincidence number respectively. Furthermore, he has also shown that the essential coincidence Nielsen classes have the same coincidence index which is either +1 or -1. In the fixed point situation, or even more general in the coincidence case where $N_1 = N_2$, several authors, have exploit the relation among $N(f,g), L(f,g),\ coin (f_\# , g_\# )$ and ${\cal R} (f,g)$ where $f_\# , g_\#$ are the induced homomorphism on the fundamental group by $f, g$ respectively and ${\cal R} (f,g)$ are the Reidemeister classes. See for example [BBPT], [FHW] and [G]. For the general situation $f,g : N_1 \to N_2$, the main part which is missed so far, is the relation between $coin\ ( f_\# , g_\# )$ and the $\#{\cal R} (f,g)$. The purpose of this work is to study such relation including the case where the two compact nilmanifolds $N_1$ and $N_2$ does not have the same dimension. Then we prove Theorem 2.5 which says: {\it Given $f,g: N_1 \to N_2$ then the two conditions below are equivalent:

a) The Hirsch lenght of $coin(f_\#,g_\#)$ is $dim N_1-dimN_2$;

b) $\#{\cal R} [f, g] < \infty$.}

Based on this we can prove Theorem  2.6
which says: {\it Given $f,g : N_1 \to N_2$ where $N_1$, $N_2$ are compact Nilmanifolds of the same dimension, then the three conditions below
are equivalent:

a) $N(f,g) \ne 0$;

b) coin$(f_{\#}, g_{\#}) = 1$;

c) $\#{\cal R} [f, g] < \infty$;

If one of the three conditions holds, then
$N(f,g) = \#{\cal R} [f,g] = |L(f,g)|$}


This paper is organized in two sections. In Section 1 we present some general facts about maps on Lie groups and give the background to transform our question in a question about maps on Lie groups. In Section 2 we consider for given $N_1 , N_2$ compact nilmanifolds, the Lie Groups which are the  universal covers respectively. Then we solve the related problem for these Lie Groups. 

Certainly the result here will have implication in the coincidence theory of solvmanifolds. This will be analized somewhere.

The Theorem 2.5 for the case where both nilmanifolds have the same dimension, has been obtained independently by P. Wong using different method, see [W], at the same time.

We would like to thank Rui de Almeida for many helpful conversations about Lie Groups.

\vs 1cm

\large

\par\noindent {\bf Section 1 - Preliminaries}

\normalsize
\baselineskip=8.5mm

\vs 0.5cm

Let $G_1 , G_2$ be two simply connected Lie groups, and $\Gamma_1 \hookrightarrow G_1$, \  $\Gamma_2 \hookrightarrow G_2$ two uniform (discrete and co-compact) subgroups. Suppose $\varphi_1 , \varphi_2 : G_1 \rightarrow G_2$ are two group homomorphisms such that $\varphi_i (\Gamma_1) \subset \Gamma_2 \  \  \  i= 1,2$, and call $f,g: G_1/_{\Gamma_1} \to G_2/_{\Gamma_2}$ the induced maps from $\varphi_1 , \varphi_2$ respectively on the homogeneous spaces.

The following elementary fact is true:

\vs 0.2cm

\pn {\bf Proposition 1.1} - We have that $\pi_1 (G_1/_{\Gamma_1}) = \Gamma_1$, \  $\pi_1 (G_2/_{\Gamma_2}) = \Gamma_2$ \ and \ $f_\# , g_\# : \pi_1 (G_1/_{\Gamma_1}) \to \pi_1 (G_2/_{\Gamma_2})$ are the homomorphism $\varphi_1|_{\Gamma_1} ,\  \varphi_2|_{\Gamma_2}$ respectively. $\Box$

As before let $G$ be a simply connected Lie Group.

\vs 0.2cm

\pn {\bf Definition 1.2} - We say that $G$ has the $P_1$ property, if the map $\psi : H_1 \times H_2 \to G$ given by $\psi (h_1,h_2) = h_2h_1^{-1}$ is surjective for any pair of closed Lie subgroups $H_1$ and $H_2$ which are transversal at the identity.

For the next definition we will consider $H_1, H_2$ any two simply connected closed Lie subgroups of $G$ and also $\Gamma_1\subset H_1,\ \Gamma_2\subset H_2,\ \Gamma\subset G$ uniform subgroups such that $\Gamma_1, \Gamma_2\subset \Gamma$.

\vs 0.2cm

\pn {\bf Definition 1.3} - The Reidemeister class of $(\Gamma_1,\Gamma_2)$ on $\Gamma$, denoted by   ${\cal R} (\Gamma_1,\Gamma_2; \Gamma)$, are the classes of elements of $\Gamma$ given by the relation $\alpha \sim h_2\alpha h_1^{-1} \  \ h_1\in \Gamma_1,\ h_2\in \Gamma_2$.

\pn {\bf Remark} - We can define in a similar way ${\cal R} (\Gamma_1,\Gamma_2; \Gamma)$ without the assumption $\Gamma_i \subset \Gamma \ \ i=1,2$. For this we have the relation
$\alpha \sim h_2\alpha h_1^{-1}$ whenever both elements belong to $\Gamma$.
\vs 0.2cm

\pn {\bf Definition 1.4} - A pair $(G,\Gamma)$, where $G$ is a simply connected Lie group and $\Gamma\subset G$ is a uniform subgroup, has the $P_2$ property, if for any two pairs $(H_1,\Gamma_1),\ (H_2,\Gamma_2)$ as above, $\#{\cal R} (\Gamma_1,\Gamma_2; \Gamma)<\infty$ implies that $\psi : H_1\times H_2 \to G$ is surjective. Finally we say that $G$ has the $P_2$ property if $(G,\Gamma)$ has the $P_2$ property  for all uniform subgroups $\Gamma$.

The map $\psi$ plays an important role in our approach and it has the following nice property, which is going to be used later.

\vs 0.2cm

\pn {\bf Proposition 1.5} - The map $\psi$ has constant rank.

\vs 0.2cm

\pn {\bf Proof:} Let us show that the rank of $d\psi_{(e_1,e_2)}$ is the same as the rank of $d\psi_{(h_1,h_2)}$ for any point $(h_1,h_2)\in H_1\times H_2$. Consider the commutative diagram:

\renewcommand{\arraystretch}{1.4}
\begin{center}
 $\begin{array}{cccc}
   H_1\times H_2 & 
   {\stackrel{\psi}{\longrightarrow}} &
   \hskip 0.5cm G & \\
   %
   L_{(h_1,h_2)} \bigdownarrow \hskip 1.3cm & & 
   \hskip 0.5cm  \bigdownarrow & L_{h_2} . (  ) .R_{h_1^{-1}} \\
   %
   H_1\times H_2 & 
   {\stackrel{\psi}{\longrightarrow}} &
   \hskip 0.5cm G &  \\
  \end{array}$
\end{center}
\renewcommand{\arraystretch}{1}



So at the tangent space level we have the commutative diagram:


\renewcommand{\arraystretch}{1.4}
\begin{center}
 $\begin{array}{cccc}
   T_{(e_1,e_2)}(H_1\times H_2) & 
   {\stackrel{d\psi}{\longrightarrow}} &
   T_eG & \\
   %
   dL_{(h_1,h_2)} \hskip 0.3cm \bigdownarrow \hskip 0.5cm & & 
   \bigdownarrow & d(L_{h_2} . (  ) . R_{h_1^{-1}}) \\
   %
   T_{(h_1,h_2)}(H_1\times H_2) & 
   {\stackrel{d\psi}{\longrightarrow}} &
   T_{h_2 h_1^{-1}}G &  \\
  \end{array}$
\end{center}
\renewcommand{\arraystretch}{1}

Since the maps $L_{(h_1,h_2)}$ and $L_{h_2} . (  ) .R_{h_1^{-1}}$ are diffeomorphisms, we have that $dL_{(h_1,h_2)}$ and  
$d(L_{h_2} . (  ) .R_{h_1^{-1}})$ are isomorphisms and the result follows. $\Box$

\vs 0.2cm

\pn {\bf Remark:} By complete analogous argument, if $\varphi_1,\varphi_2 : G_1\to G_2$ are two group homomorphisms, then the map $\varphi=  \varphi_2 \cdot \varphi^{-1}_1 : G_1\to G_2$ has also constant rank.

Now we are ready to show the two main results of this section. For the purpose of proposition 1.6, we will assume that $G_1$ is diffeomorfic to the Euclidean Space $R^n$. The is certainly the case for $G$ a  simply connected solvable Lie Group. Also this is the case for all simply connected Lie groups which does not have a compact connected subgroup besides the subgroup consisting of the identity element.

Finally, let us consider the groups $\Gamma$ which are uniform subgroups of a Lie Group $G$ as in the last paragraph. Define $l(\Gamma)$ to be the dimension of the Lie group $G$. This is certainly a well defined number. Further, if $\Gamma$ is a finitely generated torsion free nilpotent  group, by [Ma] we have that $l(\Gamma)$ coincide with the Hirsh lenght of $\Gamma$. See [R] for the definition of the Hirsh lenght. Let $dim(G_1)=m$ and $dim(G_2)=n$.

\vs 0.2cm

\pn {\bf Proposition 1.6} - If $G_1\times G_2$ satisfies the property $P_1$, then $l(coin\ (f_\# ,g_\#))=m-n$ implies $\varphi : G_1\to G_2$ is surjective.

\vs 0.2cm

\pn {\bf Proof:} Let $G=G_1\times G_2$ and $H_1, H_2$ the graphs of $\varphi_1 , \varphi_2$ respectively. Certainly $H_1,H_2$ are closed subgroups. Since $dim\ H_1+dim\ H_2 = 2dim\ G_1 =2m$, in order to show that $H_1,H_2$  are transversal at the identity, it suffices to show that  $T_{e}H_1 \cap T_{e}H_2=T_e(H_1 \cap H_2)$ is a subspace of dimension $m-n$. 

Certainly $H_1\cap H_2 = \{(x, \varphi_1(x))| x \in coin (\varphi_1,\varphi_2)\}$. 

But $coin\ (\varphi_1 ,\varphi_2)$ is certainly a closed subgroup which has $coin\ (f_\# ,g_\#)$ as a uniform subgroup (see the proof of
Lemma 2.2 in [Mc]). Therefore $coin\ (\varphi_1 ,\varphi_2)$ has dimension $m-n$, since $G_1$ has no nontrivial compact subgroup. 
So $H_1, H_2$ are transversal and $\psi$ is surjective. So $\varphi$ is also surjective.                $\Box$


\vs 0.2cm

\pn {\bf Proposition 1.7} - If $G_1\times G_2$ satisfies the property $P_2$, then $\# {\cal R} (f_\# ,g_\#) < \infty$ implies that $l(coin\ (\varphi_1 , \varphi_2))=m-n$.

\vs 0.2cm

\pn {\bf Proof:} As before, let $H_1, H_2$ be the closed Lie subgroups of $G$ which are the graphs of $\varphi_1 , \varphi_2$ respectively. It is straightforward to see that the inclusion $G_2 \hookrightarrow G_1\times G_2 \ \ i(g)=(e_1,g)$ induces a map $\overline i : {\cal R} (f_\# , g_\# ) \to {\cal R} (\Gamma'_1, \Gamma'_2;\ \Gamma_1\times \Gamma_2)$ which is a bijection where $\Gamma'_1 = (\Gamma_1, f_\#(\Gamma_1)) \ and\ \Gamma'_2=(\Gamma_1, g_\#(\Gamma_1))$. So $\# {\cal R} (\Gamma'_1, \Gamma'_2;\ \Gamma_1\times \Gamma_2) < \infty$. Since $G_1\times G_2$ satisfies the $P_2$ condition  follows by definition that $\psi : H_1\times H_2 \to G$ is surjective. By Proposition 1.5, $\psi$ has constant rank. Since it is surjective implies by Sard's Theorem that $\psi$ is a submersion and therefore $dim(Kern(d\psi_{(e,e)}))$ is $m-n$. Let $C_0 = coin\ (\varphi_1,\varphi_2)$ and $C'_0$ the connected component which contains the identity. If $dim(C'_0)>m-n$ let ${v_1,v_2,...,v_k}$ be a set of linearly independent vector field at $T_eG_1$ and $\gamma_i i=1,...,k$ $k$ curves in $C'_0$ having $v_i$ as tangent vector respectively. So the curve $(\gamma_i (t), f(\gamma_i(t));\ (\gamma_i (t), g(\gamma_i(t)))$ has non null  tangent vector and $d\psi(\dot\gamma_i (0))=0$. So we must have $dim\ C'_0=m-n$ and the result follows.     $\Box$



\vs 1cm

\large

\par\noindent {\bf Section 2 - The nilmanifold case}

\normalsize
\baselineskip=8.5mm

\vs 0.5cm

In this section we specialize for the case where $G_1, G_2$ are simply connected nilpotent Lie groups, and proof our main result for compact nilmanifolds.

\vs 0.2cm

\pn {\bf Proposition 2.1} - If $G$ is a simply connected nilpotent Lie group then $G$ satisfies the property $P_1$.

\vs 0.2cm

\pn {\bf Proof:} The proof is by induction on the dimension of $G$. If $dim\ G=2$ then $G = R^2$ with the standart Lie Group structure and the result is clear. So, suppose that the result is true for simply connected nilpotent Lie Group of dimension less or equal to $n$. Let $dim\ G=n+1$ and $H_1, H_2 \subset G$ be two closed subgroups which are transversal. Consider the sequence.

\begin{center}
$1 \longrightarrow C(G)\longrightarrow G \stackrel{p}{\longrightarrow}  {G \over C(G)} \longrightarrow  1$
\end{center}

\pn where $C(G)$ is the center of $G$. It is known that $dim\ (C(G))>0$ therefore $dim {G \over C(G)} \leq n$. Since the two closed Lie subgroups $H_1, H_2\subset G$  are transversal, then the  subgroups $H'_1 = p(H_1),\ H'_2 = p(H_2) \subset {G \over C(G)}$ are also closed subgroups which are transversal. This follows from the diagram:


\renewcommand{\arraystretch}{1.4}
\begin{center}
 $\begin{array}{cccc}
    T(H_1\times H_2) & 
   {\stackrel{d\psi}{\longrightarrow}} &
   TG & \\
   %
   \hskip 0.5cm dp\hskip 0.2cm \bigdownarrow & & 
   \bigdownarrow & \\
   %
   T(H'_1\times H'_2) & 
   {\stackrel{d\psi'}{\longrightarrow}} &
   T{G \over C(G)} &  \\
  \end{array}$
\end{center}
\renewcommand{\arraystretch}{1}

\pn since the right vertical map and the top horizontal map are surjective, where $\psi'$ is the induced map from $\psi$.

By induction hypothesis we have that $im (\psi') = {G \over C(G)}$. So it suffices to show the $C(G) \subset im(\psi )$. Since $d\psi_e : T(H_1\times H_2) \to T_eG$ is surjective, by the local submersion Theorem (See [GP] Section 4) it follows  that $im \psi$ contains an open neighborhood $U$ of the identity. So $U_C= U\cap C(G)$ is an open neighborhood of $e$ in $C(G)$. If we show that $im\ \psi \cap C(G)$ is closed under the group operation, follows by [C] Chapter II \S IV Theorem 1,  that $im\ \psi \cap C(G) = C(G)$. So let us show that $im\ \psi \cap C(G)$ is closed under the group operation. For, let $c_1, c_2 \in im\ \psi \cap C(G)$. We have $c_1=g_2g_1^{-1} \ \ c_2=h_2h_1^{-1} \ \ and \ \ c_1.c_2 = (g_2g_1^{-1}) (h_2h_1^{-1}) = g_2(g_1^{-1}(h_2h_1^{-1})) = g_2(h_2h_1^{-1})g_1^{-1} = (g_2h_2) . (h_1^{-1}g_1^{-1}) = (g_2h_2) (g_1h_1)^{-1}$, where the third equality follows from the fact that $(h_2h_1^{-1})$ belongs to the center. So the result follows. $\Box$


\vs 0.2cm

\pn {\bf Proposition 2.2} - If $G$ is a simply connected Nilpotent Lie Group, then $G$ satisfies the property $P_2$.

\vs 0.2cm

\pn {\bf Proof:} Let $\Gamma \subset G$ be any uniform subgroup. We will show that $(G,\Gamma)$ satifies the property $P_2$. The proof is by induction on the dimension of $G$. If $dim\ G$ is 2 the result is easy. Let us assume that the result is true if $dim\ G\leq n$. Let $dim\ G=n+1$. In order to apply the induction hypothese, we will define a Lie subgroup $H$ of $C(G)$ of dimensional one.  As before let us consider a short exact sequence:


  In order to define $H$, let $\Gamma_0$ be the center of $\Gamma$. Take $H_1, H_2\subset G$ and $\Gamma_1, \Gamma_2\subset \Gamma \subset G$ where $\Gamma_1, \Gamma_2, \Gamma$ are uniform subgroups of $H_1, H_2, G$ respectively, and $\#{\cal R} (\Gamma_1,\Gamma_2; \Gamma)<\infty$.  Since $\Gamma_0$ is abelian, we have that $\Gamma_0\cap [e]$, where $[e]\in{\cal R}(\Gamma_1,\Gamma_2; \Gamma)$ is the class which contains the identity, is a subgroup and the set of classes on $\Gamma_0$ given by $\Gamma_0\cap [g] \ \ for \ \ [g]\in{\cal R}(\Gamma_1,\Gamma_2; \Gamma)$, is finite because ${\cal R}(\Gamma_1, \Gamma_2; \Gamma)$ is finite. That $\Gamma_0\cap [e]$ is a subgroup can be proved as follows: given $c= h_2 h_1^{-1}$ \ \ then \ \ $c^{-1} = h_1h_2^{-1} = h_1h_2^{-1}   h_1^{-1}h_2 h_2^{-1}h_1 = h_1^{-1} . h_1h_2^{-1}. h_2h_2^{-1}h_1 = h_2^{-1}h_1$. If $c_1=h_2h_1^{-1} \ c_2=g_2g_1^{-1}$ then $c_1c_2 = h_2h_1^{-1} g_2g_1^{-1} = h_2g_2 g_1^{-1}h_1^{-1} = h_2g_2 (h_1g_1)^{-1}$. So this is a subgroup. Further if $c_0\in \Gamma_0$ then $c_0(\Gamma_0\cap [e]) = \Gamma_0 \cap [c_0]$. For $c_0h_2h_1^{-1} = h_2c_0h_1^{-1} \in \Gamma_0\cap [c_0]$. On the other hand if $g\in \Gamma_0\cap [c_0]$, then $g = h_2c_0h_1^{-1}$ where $g\in \Gamma_0$. So $g=c_0 h_2h_1^{-1}$ which implies that $h_2h_1^{-1}\in \Gamma_0$. Therefore $h_2h_1^{-1}\in \Gamma_0 \cap [e]$ and we have $\Gamma_0\cap [c_0] \subset c_0 (\Gamma_0\cap [e])$. So we conclude that $\Gamma_0/_{\sim}$ where $\sim$ is the relation induced by the one which gives ${\cal R}(\Gamma_1, \Gamma_2; \Gamma)$, is the coset classes of $\Gamma_0 \over {\Gamma_0\cap [e]}$. Since $ {\Gamma_0 \over \Gamma_0 \cap [e]}$ is finite, it means that we have an element $g \in [e]$ where $g \ne e$. So $g=g_2g_1^{-1}$ for some $g_i \in \Gamma_i$ $i=1,2$. If $\gamma_x$ denote the one-parameter subgroup through $x$, we define $H=\gamma_g$.

 As before let us consider a short exact sequence:

\begin{center}
$1 \longrightarrow H\longrightarrow G \stackrel{p}{\longrightarrow} {G \over H} \longrightarrow  1$.
\end{center}
Certainly, $p(H_1), p(H_2)\subset {G \over H}$ are closed subgroups and $p(\Gamma_1), p(\Gamma_2) \subset p(\Gamma)$ where $p(\Gamma_1), p(\Gamma_2), p(\Gamma)$ are uniform subgroups of $p(H_1), p(H_2), G$ respectively. Certainly the projection induces a map $\overline p : {\cal R} (\Gamma_1, \Gamma_2 ; \Gamma) \to {\cal R}(p(\Gamma_1), p(\Gamma_2) ; p(\Gamma))$ which is  surjective. By induction hypothesis $\psi : p(H_1) \times p(H_2) \to {G \over H}$ is surjective. So it suffices to show that $im\ \psi \supset H$. 

In order to show that $im\ \psi \supset H$, it suffices  
to show that $ \gamma_g = \gamma_{g_2} . \gamma_{g_1}^{-1}$. First we are show that for every rational $p \over q$ less than one, the two curves coincide. First we consider the parameter $t= {1 \over q}$. Let $\Gamma'$ be  the subgroup generated by ${\Gamma, w, w_1 \ \ and \ \ w_2}$ where $w=\gamma_g({1 \over q}), w_1=\gamma_{g_1}({1 \over q}) \ \ and \ \ w_2=\gamma_{g_2}({1 \over q})$. The center of $\Gamma'$, denoted by $\Gamma_0'$, certainly contains $\Gamma_0, w$.  We would like to show that $w=w_2w_1^{-1}$. We know that $w^q=w_2^qw_1^{-q}$. By a result of Mal'cev (see [R] chapter 5 statement 5.2.19) the quotient of a Nilpotent group by the center is torsion free. So the quotient of $\Gamma'$ by the center $\Gamma_0'$ is torsion free and we have
$[w_1]^q=[w_2]^q$ since $w^q \in \Gamma_0'$. Also in a torsion free nilpotent group the qth-root is unique. Therefore $[w_1]=[w_2]$ and $w_2w_1^{-1} \in \Gamma_0'$. Since $w^q=w_2^qw_1^{-q}=(w_1w_2^{-1})^q$, where the last equality follows because $w_2w_1^{-1} \in \Gamma_0'$, we have $w=w_2w_1^{-1}$. It follows that $\gamma_g( {1 \over q})=\gamma_{g_2}( {1 \over q})\gamma_{g_1}^{-1}( {1  \over q})$. By similar argument, and simpler, we show that in fact the two curves coincide for all $t={p \over q}$. By continuity follows that $\gamma_x=\gamma_{g_2}\gamma_{g_1}^{-1}$ for all $t$. Therefore we conclude that $H \subset im(\psi)$. So the result follows.   $\Box$






 
Now we can prove:


\pn {\bf Proposition 2.3} - If $l(coin (f_\# , g_\# )) = m-n$ then $\#{\cal R}(f,g)<\infty$.

\vs 0.2cm

\pn {\bf Proof:} By Lemma 2.7 of [Mc], the maps $f,g$ can be covered by homomorphism $\varphi_1 , \varphi_2 : G_1 \to G_2$.

By Proposition 1.6 $\varphi$ is surjective. So given any element $y \in \Gamma$ there exists $g \in G_1$ such that $\varphi_1(g)\varphi_2(g^{-1})=y^{-1}$ or $\varphi_2(g^{-1})y=\varphi_1(g^{-1})$. Therefore $g^{-1}\in coin\ (\varphi_2y, \varphi_1)$ and  consequently $coin\ (\varphi_2y, \varphi_1)$ is non empty. Therefore for each Reidemeister class, there is a Nielsen class (non empty one) which correspond this Reidemeister class. Since the number of  Nielsen classes
is finite, we must have only a finite number of Reidemeister classes and the result follows.      $\Box$

\pn {\bf Proposition 2.4} - If $\#{\cal R}(f,g) < \infty$ then $l(coin\ (f_\# , g_\#)) = m-n$.

\vs 0.2cm

\pn {\bf Proof:} As in Proposition 2.3 let $\varphi_1, \varphi_2 : G_1\to G_2$ be homomorphisms which cover $f$ and $g$ respectively.

Consider the uniform subgroups $(\Gamma_1,\varphi_1(\Gamma_1)), \ (\Gamma_1,\varphi_2(\Gamma_1)), \Gamma_1\times \Gamma_2$ of $H_1, H_2, G$ respectively. We have that $R(f_\#, g_\#) < \infty$ implies $\#{\cal R} (\Gamma'_1,\Gamma'_2, \Gamma_1\times \Gamma_2)<\infty$. Since $G_1\times G_2$ satisfies the property $P_2$ by Proposition 1.7, we have that $\psi : H_1 \times H_2 \to G$ is surjective. This implies that $\psi$ is a submersion. By Proposition 1.7 we have that $dim(coin\ (\varphi_1, \varphi_2))=m-n$. But by Lemma 2.2 of [Mc] it is also connected. So we must have  $l(coin\ (f_\#, g_\#)) = m-n$ and the result follows. $\Box$

Now we come to the main result.

\pn {\bf Theorem 2.5}  Given $f,g: N_1 \to N_2$ where $N_1$, $N_2$ are compact  Nilmanifolds, then the two conditions below are equivalent:

a) The Hirsch lenght of $coin(f_\#,g_\#)$ is $dim N_1-dimN_2$;

b) $\#{\cal R} [f, g] < \infty.$ 

\pn {\bf Proof:} The equivalence follows imediatly from proposations 2.3 and 2.4. $\Box$

\pn {\bf Theorem 2.6}  Given $f,g : N_1 \to N_2$ where $N_1$, $N_2$ are compact Nilmanifolds of the same dimension, then the three conditions below
are equivalent:

a) $N(f,g) \ne 0$;

b) coin$(f_{\#}, g_{\#}) = 1$;

c) $\#{\cal R} [f, g] < \infty$;

If one of the three conditions holds, then
$N(f,g) = \#{\cal R} [f,g] = |L (f,g)|$


\pn {\bf Proof:} That a) is equivalent to b) has been proved in [Mc].
That b) is equivalent to c) follows Theorem 2.5.  That if one of the two conditions, a) and c), holds implies $N(f,g) = |L(f,g)|$, also follows from [Mc]. Finally if c) holds we know that the map $\psi$ is surjective. Therefore every Reidemeister class comes from a non empty Nielsen class. By Lemma 2.6 of [Mc], we have that in fact this Reidemeister class represents an  essential Nielsen class. So the remain equality follows.  $\Box$

\pn {\bf Comment} In Theorem 2.5 one should expect that the condition a) and b) are also equivalent to say that the pair $(f,g)$ can't be deformed to coincidence free. The usual type  argument to show this, can not be applied, because at the present there are difficults to define a suitable Nielsen coincidence number or a local index. 
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