An averaging formula for Nielsen numbers on infra-solvmanifolds

Abstract

Until now only for special classes of infra-solvmanifolds, namely, infra-nilmanifolds and infra-solvmanifolds of type (R), there was a formula available for computing the Nielsen number of a self-map on those manifolds. In this paper, we provide a general averaging formula which works for all self-maps on all possible infra-solvmanifolds and which reduces to the old formulas in the case of infra-nilmanifolds or infra-solvmanifolds of type (R). Moreover, when viewing an infra-solvmanifold as a polynomial manifold, we recall that any map is homotopic to a polynomial map and we show how our formula can be translated in terms of the Jacobian of that polynomial map.

Appendix

We prove:

Lemma 5.7. Let $X\in \mathbb {Z}^{n\times n}$ and $\Phi \in \mathbb {Q}^{m\times m}$ be matrices and let $A:\mathbb {Z}^m\rightarrow {\text {SL}}_n(\mathbb {Z})$ be an endomorphism, such that A(v) is net for all $v\in \mathbb {Z}^m$. Suppose that $\Phi$ does not have 1 as an eigenvalue, and that there exists $k\in \mathbb {N}$, such that $\Phi (k\mathbb {Z}^m)\subseteq \mathbb {Z}^m$ and $XA(kv)=A(\Phi (kv))X$ for all $v\in \mathbb {Z}^m$. Then, $\textrm{det}(I-A(v)X)=\textrm{det}(I-X)$ for all $v\in \mathbb {Z}^m$.

Proof

We follow the matrix analysis carried out by Keppelmann and McCord [22, Section 4]. This analysis consist of three steps.

Step 1. Reduction to the unipotent and semisimple case.

For $v\in \mathbb {Z}^m$, write $A(v)=U(v)T(v)$ with U(v) unipotent, T(v) semisimple and $[U(v),T(v)]=1$. This defines morphisms U, $T:\mathbb {Z}^m\rightarrow \textrm{GL}_n(\mathbb {Q})$ and $[U(v),T(w)]=1$ for $v\ne w\in \mathbb {Z}^m$, too. We show that it is sufficient to prove the lemma for $A=U$ (the unipotent case) and $A=T$ (the semisimple case).

Following [22], we say that X almost $\Phi$- commutes with A if there exists $k\in \mathbb {N}$ such that $XA(kv)=A(\Phi (kv))X$ for all $v\in \mathbb {Z}^m$.

Observation: X almost $\Phi$- commutes with both U and T.

Set $B(v):=\left( {\begin{matrix} A(kv)&{}0\\ 0&{}A (\Phi (kv)) \end{matrix}}\right)$ for all v in $\mathbb {Z}^m$. Then

$$\begin{aligned} B_u(v):=\left( {\begin{matrix} U(kv)&{}0\\ 0&{}U (\Phi (kv)) \end{matrix}}\right) \quad \text {and} \quad B_s(v):=\left( {\begin{matrix} T(kv)&{}0\\ 0&{}T (\Phi (kv)) \end{matrix}}\right) \end{aligned}$$

are the unipotent and semisimple part of B(v), respectively. Consider the subgroup $B:=\{B(v)\mid v\in \mathbb {Z}^m\}$ of ${\text {GL}}_{2n}(\mathbb {C})$. The relation $XA(kv)=A(\Phi (kv))X$ is polynomial in the coefficients of B(v). As $B_u(v)$ and $B_s(v)$ are contained in the Zariski closure of B, they too must satisfy this relation. So X almost $\Phi$- commutes with U and with T. $\square$

Suppose now that

$$\begin{aligned} \begin{aligned}&(*)\ \textrm{det}(I-U(v)M)=\textrm{det}(I-M)\text { for every } M \text { almost }\Phi - \text {commuting with } U;\\&(\star )\ \textrm{det}(I-T(v)M)=\textrm{det}(I-M) \text { for every }M \text { almost }\Phi - \text {commuting with }T. \end{aligned} \end{aligned}$$

Say k satisfies $XU(kv)=U(\Phi (kv))X$ for all $v\in \mathbb {Z}^n$. Then, for all $v, w\in \mathbb {Z}^n$

$$\begin{aligned} T(v)XU(kw)&=T(v)U(\Phi (kw))X\\&=U(\Phi (kw))T(v)X, \end{aligned}$$

so in particular, T(v)X almost $\Phi$- commutes with U. Hence

$$\begin{aligned} \textrm{det}(I-A(v)X)&=\textrm{det}(I-U(v)T(v)X)\\&\overset{(*)}{=}\textrm{det}(I-T(v)X)\\&\overset{(**)}{=}\textrm{det}(I-X). \end{aligned}$$

We prove ($*$) in step 2 and ($**$) in step 3.

Step 2. The unipotent case.

Suppose that $M\in M_n(\mathbb {C})$ almost $\Phi$- commutes with U, say $d\in \mathbb {N}$ satisfies $MU(dv)=U(\Phi (dv))M$ for all $v\in \mathbb {Z}^m$. From the original version of this lemma, [22, Theorem 4.2], we already know that

$$\begin{aligned}&\textrm{det}(I-U(d\,v)M)=\textrm{det}(I-M) \end{aligned}$$

for all $v\in \mathbb {Z}^m$. Fix $x\in \mathbb {Z}^m.$ Then, for all $z\in \mathbb {Z},$ also

$$\begin{aligned} \textrm{det}(I-U(dz\, x)M)=\textrm{det}(I-M). \end{aligned}$$

However, as U(x) is unipotent, the entries of $U(t\, x)=U(x)^{t}$, $t\in \mathbb {Z}$, are polynomials in t (depending on the entries of U(x), of course). Hence

$$\begin{aligned} \textrm{det}(I-U(t\, x)M)-\textrm{det}(I-M)\in \mathbb {Q}[t] \end{aligned}$$

is a polynomial vanishing on $d\mathbb {Z}$, so it must be zero. We conclude that $\textrm{det}(I-U(x)M)=\textrm{det}(I-M)$, as required.

Step 3. The semisimple case.

Suppose that $M\in M_n(\mathbb {C})$ almost $\Phi$- commutes with T; we are to show that $\textrm{det}(I-T(v)M)=\textrm{det}(I-M)$ for all $v\in \mathbb {Z}^m$.

Choose a basis of common eigenvectors $\{f_j\}_{j=1,\ldots , n}$ of the T(v)’s. Let $[x_{ij}]\in M_n(\mathbb {C})$ represent M with respect to the basis $\{f_j\}$, and let $\lambda _j(v)$ be the eigenvalue of T(v) associated to $f_j$. Then, in this notation

$$\begin{aligned} \textrm{det}(I-T(v)M)=\sum _{\sigma \in \mathcal {S}_n}\left( {\text {sgn}}(\sigma )\prod _{i=1}^n \, \delta _{i\sigma (i)}-\lambda _i(v)\, x_{i\sigma (i)}\right) \end{aligned}$$

denoting $\delta _{ij}$ the Kronecker delta. It is, therefore, sufficient to prove the following:

Reduction 1. $\forall \sigma \in \mathcal {S}_n: \prod _{i=1}^n \, \delta _{i\sigma (i)}-\lambda _i(v) \, x_{i\sigma (i)}=\prod _{i=1}^n \, \delta _{i\sigma (i)}-x_{i\sigma (i)}$

Take $\sigma \in \mathcal {S}_n.$ Write $\sigma =\sigma _1\circ \cdots \circ \sigma _l$ as a product of disjoint cycles $\sigma _i$. For each element in ${\text {fix}}(\sigma )$, we add a ‘cycle’ of length 1. Formally, denoting ${\text {fix}}(\sigma )=\{e_1,\ldots , e_d\}$ with $d=\#{\text {fix}}(\sigma )$, we set $\sigma _{l+j}:=(e_j)$ for $j\in \{1,\ldots , d\}$.

We can now partition the set $\{1,\ldots , n\}$ according to these cycles: define

$$\begin{aligned} V(\sigma _j):= {\left\{ \begin{array}{ll} \{1,\ldots ,n\}\setminus {\text {fix}}(\sigma _j) &{}\text { if }j\le l,\\ \{e_j\} &{}\text { if }j>l. \end{array}\right. } \end{aligned}$$

Setting $r=d+l$, we see that $\{1,\ldots , n\}=\cup _{j=1}^r V(\sigma _j)$, so we can further reduce to

Reduction 2. $\forall j\in \{1,\ldots r\}: \displaystyle \prod _{i\in V(\sigma _j)} \delta _{i\sigma _j(i)}-\lambda _i(v) \, x_{i\sigma _j(i)}=\displaystyle \prod _{i\in V(\sigma _j)} \delta _{i\sigma _j(i)}-x_{i\sigma _j(i)}$

Take $j\in \{1,\ldots , r\}$. Write

$$\begin{aligned} \sigma _j=\left( h\ \sigma _j(h)\ \sigma _j^2(h)\ \ldots \ \sigma _j^{s-1}(h)\right) . \end{aligned}$$

To shorten notation, we write $\sigma ^i:=\sigma _j^i(h)$. So $\sigma ^{s+1}=\sigma ^1$. We have to show that

We examine this statement more closely by distinguishing the cases $s=1$ and $s>1$.

• If $s=1$, statement ($\circ$) reads $1-\lambda _h(v)\,x_{hh}=1-x_{hh}$, or equivalently, $x_{hh}=0$ or $\lambda _h(v)=1$.

• If $s>1$, statement ($\circ$) reads $\prod _{i=1}^s\lambda _{\sigma ^i}(v)\,x_{\sigma ^i\sigma ^{i+1}}=\prod _{i=1}^s x_{\sigma ^i\sigma ^{i+1}}$, or equivalently, $\prod _{i=1}^s x_{\sigma ^i\sigma ^{i+1}}=0$ or $\prod _{i=1}^s\lambda _{\sigma ^i}(v)=1$.

So in both cases, statement ($\circ$) is equivalent to $\prod _{i=1}^s x_{\sigma ^i\sigma ^{i+1}}=0$ or $\prod _{i=1}^s\lambda _{\sigma ^i}(v)=1$. We will show the following:

Reduction 3. If $\prod _{i=1}^s x_{\sigma ^i\sigma ^{i+1}}\ne 0$, then $\prod _{i=1}^s\lambda _{\sigma ^i}(v)=1$.

Therefore, assume that $\prod _{i=1}^s x_{\sigma ^i\sigma ^{i+1}}\ne 0$. As M almost $\Phi$- commutes with T, there exists $k\in \mathbb {N}$ such that $MT(kw)=T(\Phi (kw))M$ for every $w\in \mathbb {Z}^m$. Take $i\in \{1,\ldots , s\}$. Then

$$\begin{aligned} MT(kw)f_{\sigma ^{i+1}}&=M\lambda _{\sigma ^{i+1}}(kw)f_{\sigma ^{i+1}}\\&=\sum _{j=1}^n \lambda _{\sigma ^{i+1}}(kw)x_{j\sigma ^{i+1}}f_j\\ \text {and} T(\Phi (kw))Mf_{\sigma ^{i+1}}&=T(\Phi (kw))\left( \sum _{j=1}^nx_{j\sigma ^{i+1}}f_j\right) \\&=\sum _{j=1}^n x_{j\sigma ^{i+1}} T(\Phi (kw))(f_j)\\&=\sum _{j=1}^n x_{j\sigma ^{i+1}} \lambda _j(\Phi (kw))f_j. \end{aligned}$$

Equating $MT(kw)=T(\Phi (kw))M$, we see that $x_{j\sigma ^{i+1}}\ne 0$ implies $\lambda _{\sigma ^{i+1}}(kw)=\lambda _j(\Phi (kw))$ for all $j\in \{1,\ldots , n\}$. As $x_{\sigma ^i\sigma ^{i+1}}\ne 0$ by assumption

for all $i\in \{1,\ldots , s\}$ and $w\in \mathbb {Z}^m$.

Take a basis $\{e_1,\ldots , e_m\}$ of $\mathbb {Z}^m$ such that $\{e_1,\ldots , e_q\}$ (viewed as subset of $\mathbb {R}^m)$ spans $\ker ((\Phi ^T)^s-I)$ for some $q\in \{0,\ldots , m\}$. Write $\Phi =[\phi _{ij}]$ with respect to the basis $\{e_j\}$, and write $\lambda _{i,j}:=\lambda _{\sigma ^i}(e_j)$. (So again $\lambda _{s+1,j}=\lambda _{1,j}$.)

We have to show that $\prod _{i=1}^s\lambda _{\sigma ^i}(v)=1$ for every $v\in \mathbb {Z}^m$. If $v=\sum _{j=1}^m \alpha _j e_j$, then $T(v)=\prod _{j=1}^m T(e_j)^{\alpha _j}$, hence $\lambda _{\sigma ^i}(v)=\prod _{j=1}^m \lambda _{i,j}^{\alpha _j}$ for every $i\in \{1,\ldots , s\}$. Therefore, it is sufficient to prove that

Reduction 4. $\prod _{i=1}^s \lambda _{i,j}=1$ for all $j\in \{1,\ldots ,m\}$.

We know exploit condition ($\bullet$). Thereto, take $j\in \{1,\ldots , m\}$ and choose $r_{i,j}$, $\theta _{i,j}\in \mathbb {R}$, $i\in \{1,\ldots , s\}$, satisfying $\lambda _{i,j}=e^{r_{i,j}+2\pi i \,\theta _{i,j}}$. In this notation

$$\begin{aligned} \lambda _{\sigma ^{i+1}}(ke_j)&={\lambda _{i+1, j}}^k\\&=e^{kr_{i+1,j}}e^{2\pi i\, k\theta _{i+1,j}} \end{aligned}$$

when we agree that $r_{s+1,j}:=r_{1,j}$ and $\theta _{s+1,j}:=\theta _{1,j}$. Furthermore, since $\Phi (ke_j)=\sum _{\alpha =1}^m k\phi _{\alpha j} e_\alpha ,$

$$\begin{aligned} \lambda _{\sigma ^i}(\Phi (ke_j))&=\prod _{\alpha =1}^m\lambda _{i,\alpha }^{k\phi _{\alpha j}}\\&=\prod _{\alpha =1}^m e^{k\phi _{\alpha j}r_{i,\alpha }+2\pi i \, k\phi _{\alpha j}\theta _{i, \alpha }}\\&=e^{k\sum _{\alpha =1}^m \phi _{\alpha j }r_{i,\alpha }} e^{2\pi i\, k \sum _{\alpha =1}^m\phi _{\alpha j}\theta _{i,\alpha }}. \end{aligned}$$

Imposing condition ($\bullet$) implies that for all $i\in \{1,\ldots , s\}$

$$\begin{aligned} kr_{i+1, j}&=k\sum _{\alpha =1}^m \phi _{\alpha j }r_{i,\alpha }{} & {} \text {and}&k\theta _{i+1, j}&\equiv k\sum _{\alpha =1}^m \phi _{\alpha j }\theta _{i,\alpha } \mod \mathbb {Z}. \end{aligned}$$

Define $R_i, \Theta _i\in \mathbb {R}^m$ as the vectors with j-th component equal to $r_{i,j}$ and $\theta _{i,j},$ respectively. Then

$$\begin{aligned} R_{i+1}&=\Phi ^T R_i{} & {} \text {and}&\Theta _{i+1}&\equiv \Phi ^T\Theta _i \mod \mathbb {Q}^m. \end{aligned}$$

Note that $R_{s+1}=R_1$ and $\Theta _{s+1}=\Theta _1$. Therefore, the above implies that

• $\Phi ^T\left( \sum _{i=1}^s R_i\right) =\sum _{i=1}^s R_i$;

• ${\Phi ^T}^s(R_i)=R_i$;

• ${\Phi ^T}^s(\Theta _i)\equiv \Theta _i \mod \mathbb {Q}^m$.

It follows from the first item that $\sum _{i=1}^s R_i=0$ for $\Phi ^{T}$ (and $\Phi$) does not have eigenvalue 1.

The second item implies that $R_i\in \ker ({\Phi ^T}^s-I)$, hence $r_{i,j}=0$ if $j>q$. As $\lambda _{i,j}$ cannot be a nontrivial root of unity, $\theta _{i,j}$ must either be 0 or irrational if $j>q$. In fact, $\theta _{i,j}=0$ as $({\Phi ^T}^s-I)(\Theta _i)$ must lie inside $\mathbb {Q}^m$ and

$$\begin{aligned} {\Phi ^T}^s-I= \begin{pmatrix} 0 &{}\quad *\\ 0 &{}\quad \Phi ' \end{pmatrix} \end{aligned}$$

for some $\Phi '\in \textrm{GL}_{m-q}(\mathbb {Q})$ (with respect to the basis $\{e_j\}$). So ${\Phi ^T}^s(\Theta _i)=\Theta _i$ as well.

However, then $\Phi ^T$ fixes $\sum _{i=0}^{s-1}{\Phi ^T}^i(\Theta _1)$, implying $\sum _{i=0}^{s-1}{\Phi ^T}^i(\Theta _1)=0$ for $\Phi$ does not have 1 as an eigenvalue. As $\Theta _{i+1}\equiv {\Phi ^T}^i(\Theta _1) \mod \mathbb {Q}^m$, we conclude that $\sum _{i=1}^s \Theta _i\in \mathbb {Q}^m$.

Translating $\sum _{i=1}^s R_i=0$ and $\sum _{i=1}^s \Theta _i\in \mathbb {Q}^m$ back to the $r_{i,j}$/$\theta _{i,j}$-notation gives

$$\begin{aligned} \sum _{i=1}^s r_{i,j}=0 \quad \text {and}\quad \sum _{i=1}^s \theta _{i,j}\in \mathbb {Q}\end{aligned}$$

for all $j\in \{1,\ldots , m\}$. Hence

$$\begin{aligned} \prod _{i=1}^s \lambda _{i,j}&=\prod _{i=1}^s e^{r_{i,j}+2\pi i\, \theta _{i,j}}\\&=e^{\sum _{i=1}^s r_{i,j} + 2\pi i\sum _{i=1}^s \theta _{i,j}}\\&\in e^{2\pi i \,\mathbb {Q}} \end{aligned}$$

is a root of unity. As $T(e_j)$ is net, this implies that $\prod _{i=1}^s \lambda _{i,j}=1$, concluding the proof. $\square$