# domination by geometric 4-manifolds

Jonathan A. Hillman  School of Mathematics and Statistics
University of Sydney, NSW 2006
Australia<EMAIL_ADDRESS>
###### Abstract.

We consider aspects of the question “when is an orientable closed 4-manifold
$Y$ dominated by another such manifold $X$?”, focusing on the cases when $X$
is geometric or fibres non-trivially over an orientable surface.

###### Key words and phrases:

4-manifold, aspherical, bundle, dominate, geometry

S.-C. Wang has considered in detail properties of maps of non-zero degree
between 3-manifolds. In particular, he grouped aspherical 3-manifolds into 8
families, according to the nature of their JSJ decompositions, and determined
which pairs allowed maps of non-zero degree between representative 3-manifolds
[13]. Purely algebraic arguments for $PD_{n}$-groups with JSJ decompositions
and all $n$ were given in [6]. Sharper results for maps between aspherical
geometric 4-manifolds were given in [10]. We shall complement this work in
dimension 4 by considering cases where the domain is geometric but not
aspherical, or fibres non-trivially over a surface. In the latter case the
strongest results are when the range is aspherical.

We begin by reviewing the results of [6] and [10] for aspherical geometric
4-manifolds. (Some of these earlier results are recovered below in passing.)
In §2 we make some basic observations and give four simple lemmas. The
remaining sections are organized in terms of the geometry of the dominating
space $X$. The geometries $\mathbb{S}^{4}$, $\mathbb{CP}^{2}$,
$\mathbb{S}^{2}\times\mathbb{S}^{2}$, $\mathbb{S}^{3}\times\mathbb{E}^{1}$,
$\mathbb{S}^{2}\times\mathbb{E}^{2}$ and $\mathbb{S}^{2}\times\mathbb{H}^{2}$
are considered in §3, and the geometries of solvable Lie type in §4. All total
spaces of bundles with base and fibre $S^{2}$ or the torus $T$ have such
geometries, excepting only $S^{2}\tilde{\times}{S^{2}}$. In §5 we consider
domination by total spaces of $T$-bundles with base a hyperbolic surface.
Among these are manifolds with geometry $\mathbb{H}^{2}\times\mathbb{E}^{2}$
or $\widetilde{\mathbb{SL}}\times\mathbb{E}^{1}$, but there are also non-
geometric $T$-bundle spaces. The next section considers aspherical domains
which are (virtually) both $S^{1}$-bundle spaces and mapping tori. (This
includes $\mathbb{H}^{3}\times\mathbb{E}^{1}$-manifolds.) In §7 we consider
bundles with fibre a hyperbolic closed surface, and we show that if a bundle
space $Y$ is dominated by a product $B\times{F}$ then it is also a product.
Most bundle spaces are not geometric. In the final section we consider
dominations of aspherical geometric 4-manifolds by bundles with hyperbolic
fibre. These include reducible $\mathbb{H}^{2}\times\mathbb{H}^{2}$-manifolds,
which are finitely covered by products, and
$\mathbb{H}^{3}\times\mathbb{E}^{1}$-manifolds. We have little to say about
irreducible $\mathbb{H}^{2}\times\mathbb{H}^{2}$-manifolds, or about the
geometries $\mathbb{H}^{4}$ and $\mathbb{H}^{2}(\mathbb{C})$.

Our main interest is in the existence of maps of non-zero degree. For this
purpose we may replace the domain $X$ and and range $Y$ by more convenient
finite covering spaces, and assume that the map induces an epimorphism on the
fundamental group. In several places we ask whether a specific 4-manifold $Y$
has such a domination by one of the representative geometric 4-manifolds under
consideration. Note that although geometric 4-manifolds have natural smooth
structures, the use of topological surgery implicit in some of our arguments
only justifies identifications of the possible range spaces up to
homeomorphism.

This note was prompted by a query from R. İ. Baykur, arising from [1]. In that
paper the authors consider the more specific question of which closed
4-manifolds have branched coverings by the total spaces of surface bundles.
Their main results are that every 1-connected closed 4-manifold has a branched
covering of degree $\leqslant 16$ by a product $B\times{T}$, with $B$ a closed
surface and $T$ the torus, and every product $B\times{S^{1}}$ with $B$ a
closed orientable 3-manifold has a 2-fold branched cover by a symplectic
4-manifold which fibres over $T$. The maps considered below often have degree
$1$, and then are either homotopy equivalences or not homotopic to branched
covers.

I would like to thank R. İ. Baykur for his question and for his comments on an
early draft, and S. Vidussi for pointing out a blunder in §7 of the original
arXival submission.

##  1\. dominations between aspherical geometric 4-manifolds

In the aspherical case the underlying question is essentially one of group
theory. In [6] it is shown that $PD_{n}$-groups with max-$c$ may be
partitioned into families, analogous to those of Wang, and that the pattern of
possible maps of non-zero degree is very similar. A group has max-$c$ if all
chains of centralizers in the group are finite. A $PD_{n}$-pair of groups
$(G,\partial{G})$ is atoroidal if every polycyclic subgroup of Hirsch length
$n-1$ is conjugate into a boundary component, and is of Seifert type if it has
a normal polycyclic subgroup of Hirsch length $n-2$. Kropholler showed that
all $PD_{n}$-groups with max-$c$ have JSJ decompositions along virtually
polycyclic subgroups of Hirsch length $n-1$ into pieces which are either
atoroidal or of Seifert type [9].

When $n=4$ the qualification “of Seifert type” reduces to “having a normal
$\mathbb{Z}^{2}$ subgroup”, and these families of groups are either

  1. (1)

atoroidal;

  2. (2)

have a non-trivial JSJ decomposition with at least one atoroidal piece;

  3. (3)

have a non-trivial JSJ decomposition with all pieces of Seifert type;

  4. (4)

(a) virtually polycyclic, but not virtually of Seifert type; or (b) virtually
polycyclic and of Seifert type but not virtually nilpotent:

  5. (5)

virtually a product $G\times\mathbb{Z}^{2}$ with $G$ a $PD_{2}$-group and
$\chi(G)<0$;

  6. (6)

Seifert type but not virtually a product nor virtually polycyclic;

  7. (7)

virtually nilpotent but not virtually abelian; or

  8. (8)

virtually abelian.

We have preserved Wang’s enumeration, but in higher dimensions it is useful to
subdivide type (4), the analogue of the class of $\mathbb{S}ol^{3}$-manifolds.

The fundamental groups of aspherical $n$-manifolds are $PD_{n}$-groups, and
the groups of geometric $4$-manifolds satisfy max-$c$. Not all aspherical
4-manifolds with groups of types (1) or (6) are geometric, and there are no
geometric 4-manifolds with groups of type (2) or (3). The correspondence with
geometries is (1) $\mathbb{H}^{4}$, $\mathbb{H}^{2}(\mathbb{C})$,
$\mathbb{H}^{2}\times\mathbb{H}^{2}$ and $\mathbb{H}^{3}\times\mathbb{E}^{1}$;
(4.a) $\mathbb{S}ol_{m,n}^{4}$ (with $m\not=n$) and $\mathbb{S}ol_{1}^{4}$;
(4.b) $\mathbb{S}ol^{3}\times{E}^{1}$; (5)
$\mathbb{H}^{2}\times\mathbb{E}^{2}$; (6)
$\widetilde{\mathbb{SL}}\times\mathbb{E}^{1}$; (7)
$\mathbb{N}il^{3}\times\mathbb{E}^{1}$ and $\mathbb{N}il^{4}$; and (8)
$\mathbb{E}^{4}$.

For geometric 4-manifolds in the families (4–8) the conclusion of [6] is that
all maps between groups of different types have degree 0, except for maps from
groups of type (5) to groups of type (8) and from (6) to
$\mathbb{N}il^{3}\times\mathbb{E}^{1}$-groups in type (7). (The assertion
there that there are maps of nonzero degree from groups of type (5) to groups
of type (4c) is wrong.) On the other hand every
$\mathbb{N}il^{3}\times\mathbb{E}^{1}$-group and every $\mathbb{E}^{4}$-group
is so dominated. Theorem 1.1 of [10] is slightly sharper, in that it shows
that there are no such maps between groups of distinct geometries within types
(4.a) and in (7).

##  2\. some general observations

If $X$ and $Y$ are closed orientable $n$-manifolds then $X$ dominates $Y$ if
there is a map $f:X\to{Y}$ with nonzero degree. If so, then

  1. (1)

$d$-fold (branched) finite covers have degree $d$;

  2. (2)

the image of $\pi_{1}(f)$ has finite index in $\pi_{1}(Y)$, so $f$ factors
through a map $\widehat{f}:X\to\widehat{Y}$, where $\widehat{Y}$ covers $Y$,
and $\pi_{1}(\widehat{f})$ is an epimorphism;

  3. (3)

if $R$ is a ring in which $d=deg(f)$ is invertible then $H^{*}(Y;R)$ is a
subring of $H^{*}(X;R)$, and is a direct summand as an $R$-module:

  4. (4)

hence if $n=4$ then $\chi(Y)\leqslant 2+\beta_{2}(X;\mathbb{Q})$.

If $f$ has degree $d$ (for some choice of orientations) then we shall say that
$X$ $d$-dominates $Y$. If $X$ 1-dominates $Y$ then $\pi_{1}(f)$ is an
epimorphism, and $H^{*}(Y;\mathbb{Z})$ is a direct summand of
$H^{*}(X;\mathbb{Z})$. We say that $X$ essentially dominates $Y$ if $f$ has
non-zero degree and $\pi_{1}(f)$ is an epimorphism. (Such maps need not have
degree 1, as is already clear when $X=S^{2}$. Self maps of $S^{2}$ of degree
$>1$ induce isomorphisms on $\pi_{1}$, but do not induce splittings of
$H^{2}(S^{2};\mathbb{Z})$.) Clearly $X$ dominates $Y$ if and only if $X$
essentially dominates some finite cover of $Y$.

We note also that if $X$ and $V$ are orientable closed 4-manifolds with
$\chi(X)=\chi(Y)$ then a map $f:X\to{Y}$ is a homotopy equivalence if and only
if it has degree 1 and $\pi_{1}(f)$ is an isomorphism [5, Theorem 3.2].

######  Lemma 1.

Let $f:X\to{Y}$ be a degree-$1$ map between orientable closed $4$-manifolds.
If $H_{1}(X;\mathbb{Z})$ is torsion-free then so is $H_{1}(Y;\mathbb{Z})$.

###### Proof.

This follows from the Universal Coefficient Theorem, since torsion in
$H_{1}(Y;\mathbb{Z})$ is detected by torsion in $H^{2}(Y;\mathbb{Z})$, and
$H^{2}(Y;\mathbb{Z})$ is a direct summand of $H^{2}(X;\mathbb{Z})$. ∎

######  Lemma 2.

Let $f:X\to{Y}$ be a map of non-zero degree between orientable closed
$4$-manifolds. Suppose that there is an integer $D\geqslant 0$ such that
$\beta_{2}(\hat{X};\mathbb{Q})\leqslant{D}$, for all finite covering spaces
$\hat{X}$ of $X$. Then

  1. (1)

if $\pi_{1}(Y)$ is infinite and has subgroups of arbitrarily large finite
index then $\chi(Y)\leqslant 0$;

  2. (2)

if $\pi_{1}(Y)$ is finite then $|\pi_{1}(Y)|\chi(Y)\leqslant{D+2}$, and so
$|\pi_{1}(Y)|\leqslant\frac{1}{2}(D+2)$.

###### Proof.

If $:\hat{Y}\to{Y}$ is a finite covering and $:\hat{X}\to{X}$ is the induced
covering then $f$ lifts to a dominating map $\hat{f}:\hat{X}\to\hat{Y}$. On
the one hand $\chi(\hat{Y})=[\pi_{1}(Y):\pi_{1}(\hat{Y})]\chi(Y)$; on the
other, $\chi(\hat{Y})\leqslant{D+2}$. The first assertion follows easily.

The second assertion has a similar proof. (Note that $\chi(Y)\geqslant 2$,
since $Y$ is orientable and $\pi_{1}(Y)$ is finite.) ∎

In conjunction with this lemma, note that if the $L^{2}$-Betti numbers
$\beta_{i}^{(2)}(Y)=0$ for $i\leqslant 1$ then
$\chi(Y)=\beta_{2}^{(2)}(Y)\geqslant 0$, by the $L^{2}$-Euler characteristic
formula. (This is the case if $\pi_{1}(Y)$ is infinite and amenable, or has a
finitely generated infinite normal subgroup of infinite index.)

######  Lemma 3.

Let $f:X\to{Y}$ be a map between orientable closed $4$-manifolds. Then

  1. (1)

if $f$ has odd degree and $w_{2}(X)=0$ then $w_{2}(Y)=0$;

  2. (2)

if $Y$ is $1$-connected and $w_{2}(Y)=0$ then $\chi(Y)$ is even.

###### Proof.

The map $f$ induces a monomorphism $H^{*}(f)$ from
$H^{*}(Y;\mathbb{Z}/2\mathbb{Z})$ to $H^{*}(X;\mathbb{Z}/2\mathbb{Z})$. Since
$\xi^{2}=0$ for all $\xi\in{H^{*}(X;\mathbb{Z}/2\mathbb{Z})}$ the same is true
for $H^{*}(Y;\mathbb{Z}/2\mathbb{Z})$, and so $w_{2}(Y)=0$.

If $M$ is an orientable 4-manifold then $w_{2}(M)^{2}=w_{4}(M)$, by the Wu
formulae, and so $[M]\cap{w_{2}(M)^{2}}\equiv\chi(M)$ mod (2). Hence if $Y$ is
1-connected then $\chi(Y)$ is even. ∎

On the other hand, there is a degree-1 map from $CP^{2}$ to $S^{4}$, and so
$w_{2}(Y)=0$ does not imply that $w_{2}(X)=0$.

If $X$ is a cell complex of dimension $\leqslant 4$ then
$[X,CP^{2}]=[X,K(\mathbb{Z},2)]$, by general position, since we may construct
$K(\mathbb{Z},2)\simeq{CP^{\infty}}$ by adding cells of dimension $\geqslant
6$ to $CP^{2}$. Hence if $u$ is a generator of $H^{2}(CP^{2};\mathbb{Z})$ then
$f\mapsto{f^{*}u}$ defines a bijection $[X,CP^{2}]\to{H^{2}(X;\mathbb{Z})}$.
If $X$ is a closed orientable 4-manifold the degree of $f$ is given by
$d=[X]\cap(f^{*}u)^{2}$.

An element $\xi\in{H^{2}(X;\mathbb{Z})}$ is in the image of
$[X,S^{2}]=[X,CP^{1}]$ if and only if $\xi^{2}=0$ [12, Theorem 8.11].

There is a similarly defined surjection from $[X,S^{3}]$ to
$H^{3}(X;\mathbb{Z})$.

######  Lemma 4.

Let $X=M\times{S^{1}}$, where $M$ is a $3$-manifold, and let $f:X\to{Y}$ be an
essentially dominating map. If the image of the $S^{1}$-factor in $\pi_{Y}$ is
infinite then $Y\simeq{P\times{S^{1}}}$, where $P$ is a $PD_{3}$-complex.

###### Proof.

The hypotheses imply that $\pi_{Y}$ is a product $\sigma\times\mathbb{Z}$, for
some group $\sigma$. We may apply the argument of Lemma 2 to the covering
spaces $X_{n}$ and $Y_{n}$ associated to the subgroups of the form
$\pi_{1}(M)\times{n\mathbb{Z}}$ and $\sigma\times{n\mathbb{Z}}$ to show that
$\chi(Y)\leqslant 0$. On the other hand,

$\beta_{2}(Y_{n})\geqslant\beta_{2}(\sigma\times\mathbb{Z})\geqslant\beta_{1}(\sigma)=\beta_{1}(Y_{n})-1,$

and so $n\chi(Y)=\chi(Y_{n})\geqslant-\beta_{1}(\sigma)-1$ for all $n$. Hence
$\chi(Y)=0$, and so $Y\simeq{P}\times{S^{1}}$, where $P$ is a $PD_{3}$-complex
[5, Theorem 4.5]. ∎

######  Lemma 5.

Let $M$ be the mapping torus of a self-homeomorphism $\varphi$ of an
$n$-manifold $N$. Then $M$ $1$-dominates ${S^{n}}\times{S^{1}}$.

###### Proof.

We may assume that $\varphi$ fixes a disc $D^{n}\subset{N}$. Collapsing the
image of $\overline{N\setminus{D^{n}}}$ to a point in each fibre induces a map
from $M$ to $S^{n}\times{S^{1}}$ which clearly has degree 1. ∎

The following simple lemma is based on [4, Lemma 2.3].

######  Lemma 6.

Let $P$ be a $PD_{3}$-complex with finite fundamental group $\rho$. If $M$ is
an orientable $3$-manifold such that $\pi_{1}(M)$ maps onto $\rho$ then there
is a map $f:M\to{P}$ with non-zero degree.

###### Proof.

Since $\rho$ is finite, $P$ is orientable and has universal cover
$\widetilde{P}\simeq{S^{3}}$. Hence $\pi_{2}(P)=0$, and so we may construct a
$K(\rho,1)$ space $K=P\cup{e^{\geqslant 4}}$ by adding cells of dimension
$\geqslant 4$ to $P$. Let $g:M\to{K}$ be a map such that $\pi_{1}(g)$ is an
epimorphism. We may assume that $g$ has image in $P$, by cellular
approximation. If $g:M\to{P}$ has non-zero degree then we set $f=g$.
Otherwise, let $p:M\to{M\vee{S^{3}}}$ be the pinch map, and let
$c:S^{3}\to{P}$ be the composition of a homotopy equivalence
$S^{3}\simeq\widetilde{P}$ with the universal covering projection. Then
$f=(g\vee{c})\circ{p}$ has the desired properties. ∎

When $\rho$ is cyclic and $H^{1}(M;\mathbb{Z})\not=0$ we may assume that $g$
factors through $S^{1}$, and so has degree 0. In this case $f$ has degree
$|\rho|$, as in [4].

######  Lemma 7.

Let $f:X\to{Y}$ be a map between orientable closed $4$-manifolds. If $Y$ is
aspherical and $\pi_{1}(f)$ factors through a group $G$ such that
$H_{4}(G;\mathbb{Q})=0$ then $f$ has degree $0$.

###### Proof.

If $Y$ is aspherical then $f$ is determined by $\pi_{1}(f)$, and so factors
through $K(G,1)$. Since $H_{4}(G;\mathbb{Q})=0$ the lemma follows. ∎

We shall henceforth assume that all manifolds considered are closed and
(excepting $RP^{2}$) orientable, and that $X$ and $Y$ are 4-manifolds,
$\pi_{1}(f)$ is an epimorphism and $f:X\to{Y}$ has degree $d\not=0$. All
homology and cohomology groups have coefficients $\mathbb{Q}$, unless
otherwise specified. If $F$ is a subgroup of $\pi_{1}(Y)$ then $Y_{F}$ is the
associated covering space. If $W$ and $Z$ are topological spaces then we write
$W\simeq{Z}$ if they are homotopy equivalent and $W\cong{Z}$ if they are
homeomorphic.

##  3\. compact or mixed compact-aspherical

Suppose that $X$ has one of the compact or mixed compact-solvable geometries
$\mathbb{S}^{4}$, $\mathbb{CP}^{2}$, $\mathbb{S}^{2}\times\mathbb{S}^{2}$,
$\mathbb{S}^{3}\times\mathbb{E}^{1}$ or $\mathbb{S}^{2}\times\mathbb{E}^{2}$.
Then $X$ is finitely covered by one of $S^{4}$, $CP^{2}$,
$S^{2}\times{S^{2}}$, $S^{3}\times{S^{1}}$ or $S^{2}\times{T}$, respectively.
With these geometries we shall also consider the bundle space
$S^{2}\tilde{\times}{S^{2}}$ and the mixed compact-aspherical geometry
$\mathbb{S}^{2}\times\mathbb{H}^{2}$. (See [5, Chapters 10–12].)

$\mathbb{S}^{4}$. We may assume that $X=S^{4}$. Then $\pi_{1}(Y)=1$ and
$\beta_{2}(Y)=0$, and so $Y\simeq{S^{4}}$. If $d=1$ then $f$ is homotopic to a
homeomorphism.

$\mathbb{CP}^{2}$. We may assume that $X=CP^{2}$. Then $\pi_{1}(Y)=1$ and
$\beta_{2}(Y)=1$ or 0. Hence $Y$ is homeomorphic to one of $CP^{2}$,
$Ch=*CP^{2}$ (the fake complex projective plane) or $S^{4}$.

$\mathbb{S}^{2}\times\mathbb{S}^{2}$. We may assume that
$X=S^{2}\times{S^{2}}$. Then $\pi_{1}(Y)=1$ and $\beta_{2}(Y)=2,1$ or 0. If
$\beta_{2}(Y)=2$ then $Y\cong{X}$, $S^{2}\tilde{\times}{S^{2}}=CP^{2}\sharp-
CP^{2}$ or $CP^{2}\sharp{CP^{2}}$. If also $d=1$ then $f$ is homotopic to a
homeomorphism. Let $q:S^{2}\tilde{\times}{S^{2}}\to{S^{2}}$ be the nontrivial
$S^{2}$-bundle over $S^{2}$. The pullback of $q$ over a degree-2 map from
$S^{2}$ to $S^{2}$ is a trivial bundle. Hence there is a degree-2 map from $X$
to $S^{2}\tilde{\times}{S^{2}}$. (There is no map of degree 1 [5, Theorem
3.2], and the degree must be even, by Lemma 3.)

If $\beta_{2}(Y)=1$ then $Y\simeq{CP^{2}}$. There are maps
$f:S^{2}\times{S^{2}}\to{CP^{2}}$ of every even degree, but none of degree 1.

If $\beta_{2}(Y)=0$ then $Y$ is homeomorphic to $S^{4}$.

$S^{2}\tilde{\times}{S^{2}}$. There are degree-1 maps from
$S^{2}\tilde{\times}{S^{2}}$ to $CP^{2}$ and to $S^{4}$, and there is a
degree-2 map to $S^{2}\times{S^{2}}$. (There is no map of degree 1 [5, Theorem
3.2].)

Maps between $S^{2}\times{S^{2}}$ or $S^{2}\tilde{\times}{S}^{2}$ and
$CP^{2}\sharp{CP^{2}}$ have degree 0, since the former have signature 0 while
$CP^{2}\sharp{CP^{2}}$ has signature 2, and a map $f$ of non-zero degree would
induce a ring isomorphism $H^{*}(f;\mathbb{Q})$.

$\mathbb{S}^{3}\times\mathbb{E}^{1}$. We may assume that
$X=S^{3}\times{S^{1}}$. Then $\pi_{1}(Y)$ is cyclic, and $\beta_{2}(Y)=0$.

If $\pi_{1}(Y)\cong\mathbb{Z}$ then $Y\cong{S^{3}\times{S^{1}}}$ [5, Theorem
11.1]. If also $d=1$ then $f$ is homotopic to a homeomorphism.

If $\pi_{1}(Y)\cong\mathbb{Z}/n\mathbb{Z}$ then $n=1$, by Lemma 2. Hence $Y$
is homeomorphic to $S^{4}$.

$\mathbb{S}^{2}\times\mathbb{E}^{2}$. We may assume that $X=S^{2}\times{T}$.
If $\pi_{1}(Y)$ is infinite then $\chi(Y)=0$, by Lemma 2 and the subsequent
remark.

If $\beta_{1}(Y)=2$ then $\pi_{1}(Y)\cong\mathbb{Z}^{2}$, and so $\pi_{1}(f)$
is an isomorphism. Since $H^{2}(Y;\mathbb{Z})$ is a direct summand of
$H^{2}(X;\mathbb{Z})$, it follows that $H^{*}(f)$ is an isomorphism of
cohomology rings. In particular, $f$ has degree 1. Since $\chi(Y)=\chi(X)$,
these observations imply that $f$ is a homotopy equivalence [5, Theorem 3.2].
Hence $f$ is homotopic to a homeomorphism [5, Theorem 6.16].

If $\beta_{1}(Y)=1$ then $\chi(Y)=0$ and
$\pi_{1}(Y)\cong\mathbb{Z}\oplus\mathbb{Z}/n\mathbb{Z}$, for some $n\geqslant
1$. Hence $Y\simeq{L}\times{S^{1}}$, where $L$ is a lens space [5, Theorem
11.1], and so $S^{2}\times{T}=(S^{2}\times{S^{1}})\times{S^{1}}$ essentially
dominates $Y$, by Lemma 6.

If $\beta_{1}(Y)=0$ then $|\pi_{1}(Y)|\chi(Y)\leqslant 4$, by Lemma 2. There
are obvious degree-1 maps to $S^{2}\times{S^{2}}$ (since $T$ 1-dominates
$S^{2}$) and $S^{4}$. There are degree-2 maps to $S^{2}\tilde{\times}{S^{2}}$
and to $CP^{2}$, but none of degree 1.

If $\chi(Y)=2$ and $\pi_{1}(Y)=\mathbb{Z}/2\mathbb{Z}$ then $Y$ is homotopy
equivalent to one of the two orientable 4-manifolds which are total spaces of
$S^{2}$-bundles over $RP^{2}$ [5, Chapter 12]. A map $f:X\to{Y}$ of non-zero
degree lifts to a map $f^{+}:X^{+}\to\widetilde{Y}=S^{2}\times{S^{2}}$ of non-
zero degree, which induces an isomorphism
$H_{2}(f^{+};\mathbb{Q}):H_{2}(X^{+};\mathbb{Q})\to{H_{2}(\widetilde{Y};\mathbb{Q})}=\mathbb{Q}\otimes\pi_{2}(Y)$.
Since $\pi_{1}(X$ acts trivially on $H_{2}(X^{+};\mathbb{Q})$ but $\pi_{1}(Y)$
acts non-trivially on $\pi_{2}(Y)$ it follows that $\pi_{1}(f)$ cannot be an
epimorphism. Thus $S^{2}\times{T}$ does not essentially dominate either of
these bundle spaces.

$\mathbb{S}^{2}\times\mathbb{H}^{2}$. If $X$ is a
$\mathbb{S}^{2}\times\mathbb{H}^{2}$-manifold then it is finitely covered by
$S^{2}\times{B}$, where $B$ is a hyperbolic surface. Since there is a degree-1
map from $B$ to $T$, there is a degree-1 map from $S^{2}\times{B}$ to
$S^{2}\times{T}$, and hence there are such maps to $S^{3}\times{S^{1}}$,
$S^{2}\times{S^{2}}$ and $S^{4}$. Thus every
$\mathbb{S}^{2}\times\mathbb{H}^{2}$-manifold and every 4-manifold with one of
the five compact or mixed compact-solvable geometries above is dominated by
such a product. On the other hand, any map from $S^{2}\times\hat{B}$ to an
aspherical 4-manifold factors through the projection to $\hat{B}$, and so has
degree 0.

##  4\. solvable lie type

There are six solvable Lie geometries. One is an infinite family
$\mathbb{S}ol_{m,n}^{4}$ of closely related geometries, which includes the
product geometry $\mathbb{S}ol^{3}\times\mathbb{E}^{1}$ as the equal parameter
case $m=n$. This product geometry needs separate consideration here. (See [5,
Chapters 7–8] for details of these geometries and the associated fundamental
groups. It appears to be unknown when different pairs $(m,n)$ and
$(m^{\prime},n^{\prime})$ determine the same geometry
$\mathbb{S}ol_{m,n}^{4}$. See [5, page 137].)

Suppose that $X$ has a solvable Lie geometry. Then $\pi_{X}=\pi_{1}(X)$ is
polycyclic of Hirsch length 4 and $\chi(X)=0$, so $\beta_{2}(X)\leqslant 6$.
After passing to a finite covering space, if necessary, we may assume that $X$
is a solvmanifold (i.e., a coset space of a 1-connected solvable Lie group )
and either $X=T^{4}$ or $\pi_{X}$ is not virtually abelian. Thus $X$ is
parallelizable. Moreover, $X$ is a mapping torus $N\rtimes{S^{1}}$, where
$N=T^{3}$ or is a coset space of $Nil^{3}$. If $\pi_{X}$ is nilpotent and
$S=\\{g_{1},\dots,g_{\beta}\\}$ represents a basis for the maximal torsion
free quotient of $\pi_{X}/\pi_{X}^{\prime}$ then the subgroup generated by $S$
has finite index in $\pi_{X}$ and $S/S^{\prime}\cong\mathbb{Z}^{\beta}$. Hence
in this case we may also assume that $H_{1}(X;\mathbb{Z})$ is torsion-free.

######  Lemma 8.

Let $f:X\to{Y}$ be a map of non-zero degree between orientable closed
$4$-manifolds. If $X$ is a solvmanifold and $\pi_{1}(f)$ is an epimorphism
then

  1. (1)

$\pi_{1}(Y)$ is polycyclic of Hirsch length $h_{Y}\leqslant 4$, and
$w_{2}(Y)=0$;

  2. (2)

if $h_{Y}=0$ then $\pi_{1}(Y)$ is finite and $2\leqslant\chi(Y)\leqslant
2+\beta_{2}(X)$;

  3. (3)

if $h_{Y}=1$ and $F$ is the maximal finite normal subgroup of $\pi_{1}(Y)$
then $Y_{F}$ is a $PD_{3}$-complex, and $Y$ is finitely covered by
$S^{3}\times{S^{1}}$;

  4. (4)

if $h_{Y}=2$ then $Y$ is finitely covered by $S^{2}\times{T}$;

  5. (5)

if $h_{Y}>2$ then $f$ is homotopic to a homeomorphism.

###### Proof.

Since $\pi_{Y}=\pi_{1}(Y)$ is a quotient of $\pi_{X}$, it is polycyclic and
$h_{Y}\leqslant 4$. Since $X$ is parallelizable, $w_{2}(X)=0$ and so
$w_{2}(Y)=0$, by Lemma 3.

It follows from Lemma 2 and the subsequent remark that if $h_{Y}=0$ then
$2\leqslant\chi(Y)\leqslant 2+\beta_{2}(X)$, while if $h_{Y}>0$ then
$\chi(Y)=0$.

Part (3) is taken from [5, Theorem 11.1].

If $h_{Y}=2$ then $Y$ has a covering space of degree dividing 4 which is
homeomorphic to $S^{2}\times{T}$ [5, Theorem 10.1].

If $h_{Y}>2$ then $H^{i}(\pi_{Y};\mathbb{Z}[\pi_{Y}])=0$ for $i\leqslant 2$.
Since we also have $\chi(Y)=0$, it follows that $Y$ is aspherical and
$h_{Y}=4$. In this case $\pi_{1}(f)$ is an isomorphism, and so $f$ is
homotopic to a homeomorphism [5, Theorem 8.1]. ∎

In particular, there are no maps of non-zero degree between manifolds having
distinct solvable Lie geometries.

######  Corollary 9.

If $\pi_{Y}$ is nilpotent, $h_{Y}=2$, $H_{1}(Y;\mathbb{Z})$ is torsion-free
and $w_{2}(Y)=0$ then $Y\cong{S^{2}}\times{T}$.

###### Proof.

Since $Y$ has a finite cover homeomorphic to $S^{2}\times{T}$, the other
hypotheses imply that $\pi_{Y}\cong\mathbb{Z}^{2}$ [5, Theorem 10.14].
Therefore $Y\cong{S^{2}}\times{T}$, since $Y$ is orientable and $w_{2}(Y)=0$.
∎

We shall consider the possibilities for $X$ in decreasing order of
$\beta_{1}(X)$.

$\mathbb{E}^{4}$. Every flat 4-manifold is finitely covered by $T^{4}$, so we
may assume that $X=T^{4}$. Then $\pi_{Y}$ is abelian, $\beta_{1}(Y)\leqslant
4$ and $\beta_{2}(Y)\leqslant 6$.

If $f$ has degree 1 then $\pi_{Y}$ is torsion-free, by Lemma 1, and so
$\pi_{Y}\cong\mathbb{Z}^{r}$, where $r=4,2,1$ or 0. There are obvious degree-1
maps to $T^{4}$, $S^{2}\times{T}$, $S^{3}\times{S^{1}}$, $S^{2}\times{S^{2}}$
and $S^{4}$. The first three are the only possibilities for $Y$ with $\pi_{Y}$
infinite, since $w_{2}(Y)=0$, by Lemma 3. If $\pi_{Y}=1$ then $\chi(Y)=2$, 4,
6 or 8, by Lemmas 2 and 3.

Suppose now that $f$ is essentially dominating, but has degree $>1$. If
$h_{Y}=2$ then $\pi_{Y}\cong\mathbb{Z}^{2}$, by [5, Theorems 10.14 and 10.16],
since $Y$ is orientable and $\pi_{Y}$ is abelian. Thus $Y$ is the total space
of an $S^{2}$-bundle over $T$ [5, Theorem 10.10].

If $h_{Y}=1$ then $\pi_{Y}\cong\mathbb{Z}\oplus{A}$, where $A$ is a finite
abelian group. Hence $Y\simeq{L\times{S^{1}}}$, for some lens space $L$ with
$\pi_{1}(L)\cong\mathbb{Z}/n\mathbb{Z}$. Now $L\times{S^{1}}$ is essentially
$n$-dominated by $S^{2}\times{T}$ (as in §3 above), and so there is an
essentially dominating map $f:X\to{L\times{S^{1}}}$ of degree $n$.

If $h_{Y}=0$ then $|\pi_{1}(Y)|\chi(Y)\leqslant 8$, by Lemma 2. Suppose that
$f:X\to{Y}$ has nonzero degree, and let $X^{+}\to{X}$ be the covering induced
from the universal cover $\widetilde{Y}\to{Y}$. The lift
$f^{+}:X^{+}\to\widetilde{Y}$ induces an epimorphism
$H_{2}(f^{+};\mathbb{Q})$. Since $\beta_{2}(X^{+})=\beta_{2}(X)$, the covering
projection induces an isomorphism
$H_{2}(X^{+};\mathbb{Q})\cong{H_{2}(X;\mathbb{Q})}$, and so $\pi_{X}$ acts
trivially on $H_{2}(X^{+};\mathbb{Q})$. But if $\pi_{Y}\not=1$ then $\pi_{Y}$
acts non-trivially on $H_{2}(\widetilde{Y};\mathbb{Q})$, and so $\pi_{1}(f)$
cannot be an epimorphism. Thus we may assume that $\pi_{Y}=1$. Since there are
degree-1 maps from $X$ to $S^{2}\times{T}$, there are degree-2 maps from $X$
to $CP^{2}$ and to $S^{2}\tilde{\times}{S^{2}}$.

We have no examples with $\pi_{1}(Y)=1$ and $5\leqslant\chi(Y)\leqslant 8$.

$\mathbb{N}il^{3}\times\mathbb{E}^{1}$. Every closed
$\mathbb{N}il^{3}\times\mathbb{E}^{1}$-manifold is finitely covered by
$N\times{S^{1}}$, where $N$ is the total space of the $S^{1}$ bundle over the
torus $T$ with Euler class the generator of $H^{2}(T;\mathbb{Z})$, so we may
assume that $X=N\times{S^{1}}$ and $H_{1}(X;\mathbb{Z})\cong\mathbb{Z}^{3}$.
Moreover, if $\hat{X}$ is a finite covering space of $X$ then
$\beta_{2}(\hat{X};\mathbb{Q})=4$. In this case $\pi_{Y}$ is nilpotent,
$\beta_{1}(Y)\leqslant 3$ and $\beta_{2}(Y)\leqslant 4$.

If $f$ has degree 1 then $H_{1}(Y;\mathbb{Z})$ is torsion-free, by Lemma 1.
Hence if $h_{Y}=2$ then $Y\cong{S^{2}}\times{T}$, by Corollary 9. If $h_{Y}=1$
then $\pi_{Y}\cong\mathbb{Z}$ and so $Y\cong{S^{3}}\times{S^{1}}$ [5, Theorem
11.1]. If $h_{Y}=0$ then $\pi_{Y}=1$ and $\chi(Y)=2$, 4 or 6, and
$w_{2}(Y)=0$, by Lemmas 2 and 3. Since $N$ 1-dominates $S^{2}\times{S^{1}}$,
by Lemma 5, there is a degree-1 map from $X$ to $S^{2}\times{T}$. Hence there
also degree-1 maps to $S^{3}\times{S^{1}}$, $S^{2}\times{S^{2}}$ and $S^{4}$.

Suppose now that $f$ is essentially dominating, but has degree $>1$. If
$h_{Y}=2$ then $Y$ is the total space of an $S^{2}$-bundle over $T$ (as for
$\mathbb{E}^{4}$).

If $h_{Y}=1$ then the maximal finite normal subgroup $F$ of $\pi_{Y}$ is
nilpotent, and $Y$ is homotopy equivalent to the mapping torus of a self-
homotopy equivalence of $Y_{F}$. In particular, if the image of the $S^{1}$
factor of $X$ has infinite order in $\pi_{Y}$ then $F$ is a quotient of
$\pi_{1}(N)$ and $Y\simeq{Y_{F}}\times{S^{1}}$. Since the commutator subgroup
$\pi_{1}(N)^{\prime}$ is central, $F^{\prime}$ is central, and so $F$ is
cyclic or is the product of $Q(8)$ with a cyclic group of odd order. (Thus $F$
is a 3-manifold group.) There are essentially dominating maps $f:X\to{Y}$, by
Lemma 6.

On the other hand, if the image of the $S^{1}$ factor is finite then the image
of $\pi_{1}(N)$ in $\pi_{1}(Y)$ is infinite. Hence $F$ is abelian, and so
$Y_{F}$ is homotopy equivalent to a lens space. We do not know whether there
are examples of this type (other than when $Y$ is a product).

If $h_{Y}=0$ then $|\pi_{1}(Y)|\chi(Y)\leqslant 6$, by Lemma 2. As in the case
$\mathbb{E}^{4}$, we may assume that $\pi_{Y}=1$, and there are again degree-2
maps from $X$ to $CP^{2}$ and to $S^{2}\tilde{\times}{S^{2}}$. We have no
examples with $\chi(Y)=5$ or 6,

$\mathbb{N}il^{4}$. There is again a canonical choice for $X$. There is an
unique torsion-free nilpotent group of Hirsch length 4 which can be generated
by 2 elements [7, Corollary 11]. This group has the presentation

$\langle{t,u}\mid[t,[t,[t,u]]]=[u,[t,u]]=1\rangle.$

We may assume that $X$ is the corresponding $\mathbb{N}il^{4}$-manifold, which
is the mapping torus of a self-homeomorphism of $T^{3}$, and so
$H_{1}(X;\mathbb{Z})\cong\mathbb{Z}^{2}$. Moreover, if $\hat{X}$ is a finite
covering space of $X$ then $\beta_{2}(\hat{X};\mathbb{Q})=2$. In this case
$\pi_{Y}$ is nilpotent, $\beta_{1}(Y)\leqslant 2$ and $\beta_{2}(Y)\leqslant
2$.

If $f$ has degree 1 and $h_{Y}>0$ then $Y\cong{S^{2}}\times{T}$ or
$S^{3}\times{S^{1}}$ (as in the $\mathbb{N}il^{3}\times\mathbb{E}^{1}$ case).
There are degree-1 maps to $S^{2}\times{T}$ and to $S^{3}\times{S^{1}}$, and
hence also to $S^{2}\times{S^{2}}$ and $S^{4}$. If $h_{Y}=0$ then $\pi_{Y}=1$
and $\chi(Y)=2$ or 4, and $w_{2}(Y)=0$, by Lemmas 2 and 3. Hence
$Y\cong{S^{4}}$ or $S^{2}\times{S^{2}}$.

Suppose now that $f$ is essentially dominating, but has degree $>1$. If
$h_{Y}=2$ then $Y$ is the total space of an $S^{2}$-bundle over $T$ (as for
$\mathbb{E}^{4}$).

If $h_{Y}=1$ then the maximal finite normal subgroup $F$ of $\pi_{Y}$ is
nilpotent and $Y$ is homotopy equivalent to the mapping torus of a self-
homotopy equivalence of $Y_{F}$. Since the kernels of maps from $\pi_{1}(X)$
to $\mathbb{Z}$ have nilpotency class $\leqslant 2$, either
$F\cong{Q(8)}\times{C}$ with $C$ cyclic of odd order or $F$ is cyclic. We do
not know whether there are examples of this type.

If $h_{Y}=0$ then $|\pi_{1}(Y)|\chi(Y)\leqslant 4$, by Lemma 2. As in the case
$\mathbb{E}^{4}$, we may assume that $\pi_{Y}=1$, and there are again degree-2
maps from $X$ to $CP^{2}$ and to $S^{2}\tilde{\times}{S^{2}}$.

In the remaining cases there are no natural choices for $X$, and Corollary 9
does not apply.

$\mathbb{S}ol^{3}\times\mathbb{E}^{1}$. Every
$\mathbb{S}ol^{3}\times\mathbb{E}^{1}$-manifold is finitely covered by a
product $X=P\times{S^{1}}$, where $P$ is orientable and is the total space of
a $T$-bundle over $S^{1}$, and $\beta_{1}(X)=\beta_{2}(X)=2$. (The latter
conditions hold for all finite covering spaces $\hat{X}$ of $X$.) The manifold
$X$ is the mapping torus of a self-homeomorphism of $T^{3}$.

Since $P$ 1-dominates $S^{2}\times{S^{1}}$, by Lemma 5, there is a degree-1
map from $X$ to $S^{2}\times{T}$, and hence also to $S^{3}\times{S^{1}}$,
$S^{2}\times{S^{2}}$ and $S^{4}$.

Suppose now that $f$ is essentially dominating, but has degree $>1$. If
$h_{Y}=2$ then $\pi_{Y}$ maps onto $\mathbb{Z}^{2}$, since the non-trivial
normal subgroups of infinite index in $\pi_{1}(P)$ are commensurate with
$\pi_{1}(P)^{\prime}$. We again find that $Y$ is the total space of an
$S^{2}$-bundle over $T$ (as for $\mathbb{E}^{4}$).

If $h_{Y}=1$ and $F$ is the maximal finite normal subgroup of $\pi_{Y}$ then
$F^{\prime}$ is abelian (so $F$ is cyclic, metacyclic or generalized
quaternionic), and $Y$ is homotopy equivalent to the mapping torus of a self-
homotopy equivalence of $Y_{F}$. In particular, if the image of the $S^{1}$
factor of $X$ has infinite order in $\pi_{Y}$ then $F$ is a quotient of
$\pi_{1}(N)$ and $Y\simeq{Y_{F}}\times{S^{1}}$. There are essentially
dominating maps $f:X\to{Y}$, by Lemma 6.

On the other hand, if the image of the $S^{1}$ factor is finite then the image
of $\pi_{1}(N)$ in $\pi_{1}(Y)$ is infinite. Hence $F$ is abelian, and so
$Y_{F}$ is homotopy equivalent to a lens space. We do not know whether there
are examples of this type (other than when $Y$ is a product).

If $h_{Y}=0$ then $|\pi_{1}(Y)|\chi(Y)\leqslant 4$, by Lemma 2. As in the case
$\mathbb{E}^{4}$, we may assume that $\pi_{Y}=1$, and there are again degree-2
maps from $X$ to $CP^{2}$ and to $S^{2}\tilde{\times}{S^{2}}$.

$\mathbb{S}ol^{4}_{m,n}$ with $m\not=n$, $\mathbb{S}ol^{4}_{0}$ and
$\mathbb{S}ol^{4}_{1}$. In these cases we may assume that $X$ is the mapping
torus of a self-homeomorphism of $T^{3}$ or of a closed
$\mathbb{N}il^{3}$-manifold, and that $\beta_{1}(X)=1$ and $\beta_{2}(X)=0$.
(The latter conditions hold for all finite covering spaces $\hat{X}$ of $X$.)
There are degree-1 maps from $X$ to $S^{3}\times{S^{1}}$ (by Lemma 5) and to
$S^{4}$.

If $f$ is an essentially dominating map but is not homotopic to a
homeomorphism then $h_{Y}<2$, since no quotient of $\pi_{X}$ has Hirsch length
2. If $h_{Y}=1$ and $F$ is the maximal finite normal subgroup of $\pi_{Y}$
then either $F^{\prime}$ is abelian (if the geometry is
$\mathbb{S}ol^{4}_{m,n}$ or $\mathbb{S}ol^{4}_{0}$), or $F$ is cyclic or
$Q(8)\times{C}$ with $C$ cyclic of odd order (if the geometry is
$\mathbb{S}ol^{4}_{1}$), and $Y$ is homotopy equivalent to the mapping torus
of a self-homotopy equivalence of $Y_{F}$. We do not know whether there are
examples of this type (other than $Y=S^{3}\times{S^{1}}$).

If $h_{Y}=0$ then $\chi(Y)=2$ and $\pi_{Y}=1$, by Lemma 2, and so
$Y\cong{S^{4}}$.

##  5\. $\mathbb{H}^{2}\times\mathbb{E}^{2}$, $\widetilde{\mathbb{SL}}$ and
$T$-bundles

In this section we assume that the domain $X$ is the total space of a bundle
$p:X\to{B}$, with base a hyperbolic surface $B$ and fibre $T$. (Bundles with
hyperbolic base and fibre $S^{2}$ are
$\mathbb{S}^{2}\times\mathbb{H}^{2}$-manifolds, and were considered in §3
above. The case with hyperbolic base and fibre is considered in §7 below.) For
brevity, we may call $X$ a $T$-bundle space.

Clearly $\chi(X)=\chi(B)\chi(F)=0$. The bundle $p$ is determined by its
monodromy $\theta:\pi_{1}(B)\to{GL(2,\mathbb{Z})}$ and a class
$[p]\in{H^{2}(B;\mathcal{T})}$, where $\mathcal{T}$ is the $\pi_{1}(B)$-module
determined by $\theta$. Such a bundle has a section if and only if $[p]=0$.
The total space $X$ is geometric if and only if the monodromy has finite
image. If this is so and the bundle has a section then $X$ is finitely covered
by $\hat{B}\times{T}$, where $\hat{B}$ is a hyperbolic surface,. In this case
the geometry is $\mathbb{H}^{2}\times\mathbb{E}^{2}$. The other possible
geometry is $\widetilde{\mathbb{SL}}\times\mathbb{E}^{1}$, and then $X$ is
finitely covered by $M\times{S^{1}}$, where $M$ is a
$\widetilde{\mathbb{SL}}$-manifold [5, Corollary 7.3.1 and Theorem 9.3].

If the monodromy is infinite then either it preserves a flag
${\mathbb{Z}<\mathbb{Z}^{2}}$, or its image contains a matrix whose
eigenvalues are not roots of unity. In the first case, after passing to a
double covering space if necessary, the total space $X$ is the total space of
a principal $S^{1}$-bundle over a 3-manifold $M$, which is in turn the total
space of a principal $S^{1}$-bundle over a surface. If $p$ has a section then
$M$ is a $\mathbb{H}^{2}\times\mathbb{E}^{2}$-manifold; otherwise $M$ is a
$\widetilde{\mathbb{SL}}$-manifold. If the monodromy does not preserve a flag
then the extension class group $H^{2}(B;\mathcal{T})$ is finite.

Let $A$ be the image of $\pi_{1}(T)$ in $\pi_{Y}$. Then either
$\pi_{1}(f|_{T})$ is injective, or $A$ has rank 1 or it is finite. If
$A\cong\mathbb{Z}^{2}$, $\pi_{Y}/A$ is infinite and $\chi(Y)=0$ then $Y$ is
aspherical [5, Theorem 9.2]. Conversely, if $Y$ is aspherical then
$A\cong\mathbb{Z}^{2}$, for otherwise $f$ would have degree 0, by Lemma 7. It
then follows that $\pi_{Y}/A$ is virtually a $PD_{2}$-group [5, Theorem 3.10],
and so $\chi(Y)=\chi(A)\chi^{virt}(\pi_{Y}/A)=0$. If $A$ has rank 1 and finite
index in $\pi_{Y}$ then $Y$ is finitely covered by $S^{3}\times{S^{1}}$, by
Lemma 4.

If $X$ is not geometric then $\pi_{X}$ has no $\mathbb{Z}$ normal subgroup,
and so either $A\cong\mathbb{Z}^{2}$ or $A$ is finite.

In the next theorem we shall assume that $A\cong\mathbb{Z}^{2}$ and that $Y$
is dominated by a $T$-bundle space. Having an infinite abelian normal subgroup
implies that $\chi(Y)\geqslant 0$ [5, Theorem 3.4], but something more seems
needed to ensure that $\chi(Y)=0$.

######  Theorem 10.

Let $X$ be the total space of a $T$-bundle $p:X\to{B}$ over a hyperbolic
surface $B$, and let $f:X\to{Y}$ be a map with non-zero degree and such that
$\pi_{1}(f|_{T})$ is injective. Then

  1. (1)

if $X\cong{M}\times{S^{1}}$ then $Y\simeq{P}\times{S^{1}}$, where $P$ is a
Seifert fibred $3$-manifold;

  2. (2)

if $X\cong{B\times{T}}$ then $Y\simeq{C}\times{T}$, where $C$ is a surface;

  3. (3)

if $p$ has a section but $X$ is not geometric then $Y$ is aspherical, and has
a finite covering which is homotopy equivalent to the total space of a
$T$-bundle with a section.

In each case, if $q:Y\to\overline{B}$ is a $T$-bundle then the monodromy for
$p$ maps isomorphically onto the monodromy for $q$, and $f$ is homotopic to
the map induced by a map of bases.

###### Proof.

If $X$ is geometric then we may assume that
$\pi_{X}\cong\rho\times\mathbb{Z}$, where
$\rho\cong\pi_{1}(B)\times\mathbb{Z}$ or $\pi_{1}(M)$, while if $p$ has a
section then $\pi_{X}$ is a semidirect product $\mathbb{Z}^{2}\rtimes\beta$.
We may assume also that $f_{*}=\pi_{1}(f)$ is an epimorphism. Let
$A\cong\mathbb{Z}^{2}$ be the image of $\pi_{1}(T)$ in $\pi_{Y}$. Then
$\pi_{Y}\cong{f_{*}\rho}\times\mathbb{Z}$ is the product of two finitely
presentable proper normal subgroups (if $X$ is geometric) and $\pi_{Y}$ is a
semidirect product $A\rtimes(\pi_{Y}/A)$ (if $p$ has a section). In each case
we may apply the argument of Lemmas 2 and 4 to the covering spaces associated
to the subgroups of the form $\rho\times{n\mathbb{Z}}$ and
$f_{*}\rho\times{n\mathbb{Z}}$, and $n\mathbb{Z}^{2}\rtimes\beta$ and
$nA\rtimes(\pi_{Y}/A)$, respectively, to conclude that $\chi(Y)=0$.

It follows that if $\pi_{Y}/A$ is infinite then $Y$ is aspherical [5, Theorem
9.2]. If $\pi_{Y}/A$ is finite and $\chi(Y)=0$ then $Y$ is finitely covered by
$S^{2}\times{T}$ [5, Theorem 10.13]. We may then assume that $A=\pi_{Y}$, and
so the inclusion of $\pi_{1}(T)$ into $\pi_{X}$ splits. Hence
$\pi_{X}\cong\mathbb{Z}^{2}\times\pi_{1}(B)$. Thus if $p$ has a section but
$X$ is not geometric $Y$ must be aspherical.

The further detail in cases (1) and (2) is clear. In case (3), let
$G=\pi_{Y}/A$. Since $A\cong\mathbb{Z}^{2}$ and $Y$ is aspherical, $G$ is
virtually a $PD_{2}$-group [5, Theorem 3.10]. Hence $Y$ is virtually homotopy
equivalent to the total space of a $T$-bundle $q:Y\to{C}$. Since $f_{*}$ maps
$\pi_{1}(T)$ isomorphically onto $A$ any section for $p$ induces a section for
$A$.

If $f_{*}(g)=1$ then $gt=tg$, for all $t\in\pi_{1}(T)$, since
$\pi_{1}(f|_{T})$ is injective. Hence $\mathrm{Ker}(f_{*})$ is contained in
the centralizer of $\pi_{1}(T)$, and $f_{*}$ induces an isomorphism of
monodromy groups. ∎

In a similar vein, if an aspherical 4-manifold $Y$ is dominated by the total
space of a $T$-bundle such that the monodromy preserves a flag then it is
virtually homotopy equivalent to such a $T$-bundle space.

Every solvmanifold of type $\mathbb{E}^{4}$,
$\mathbb{N}il^{3}\times\mathbb{E}^{1}$, $\mathbb{N}il^{4}$ or
$\mathbb{S}ol^{3}\times\mathbb{E}^{1}$ is the total space of a $T$-bundle over
$T$, and so is 1-dominated by $T$-bundles over hyperbolic bases. Of these,
only $T^{4}$ is dominated by $\mathbb{H}^{2}\times\mathbb{E}^{2}$-manifolds.
(However, $\mathbb{N}il^{3}\times\mathbb{E}^{1}$-manifolds are dominated by
$\widetilde{\mathbb{SL}}\times\mathbb{E}^{1}$-manifolds.)

On the other hand, no $T$-bundle space can dominate a solvmanifold of type
$\mathbb{S}ol^{4}_{m,n}$ with $m\not=n$, $\mathbb{S}ol^{4}_{0}$ or
$\mathbb{S}ol^{4}_{1}$.

##  6\. ${\mathbb{H}^{3}\times\mathbb{E}^{1}}$ and mapping tori

The total space $X$ of an $S^{1}$-bundle over a 3-manifold $M$ has a mixed
compact-aspherical or solvable Lie geometry if $M$ is an $\mathbb{S}^{3}$-,
$\mathbb{S}^{2}\times\mathbb{E}^{1}$-, $\mathbb{E}^{3}$-, $\mathbb{N}il^{3}$\-
or $\mathbb{S}ol^{3}$-manifold. If $M$ is a
$\mathbb{H}^{2}\times\mathbb{E}^{1}$\- or $\widetilde{\mathbb{SL}}$-manifold
then $X$ is a $T$-bundle space, but is not necessarily geometric. Domination
by such 4-manifolds was considered in §3 , §4 and §5 above.

Here we shall consider the remaining possibility: dominations by aspherical
4-manifolds which are total spaces of $S^{1}$-bundles over
$\mathbb{H}^{3}$-manifolds. Such manifolds are finitely covered by mapping
tori of self-homeomorphisms of $\mathbb{H}^{2}\times\mathbb{E}^{1}$\- or
$\widetilde{\mathbb{SL}}$-manifolds, by the Virtual Fibration Theorem of Agol.
This feature is used in Theorem 13 below.

######  Theorem 11.

An aspherical $4$-manifold $Y$ is dominated by a
${\mathbb{H}^{3}\times\mathbb{E}^{1}}$-manifold if and only if $\pi_{Y}$ is
virtually a product $G\times\mathbb{Z}$, where $G$ is a $PD_{3}$-group.

###### Proof.

Every $\mathbb{H}^{3}\times\mathbb{E}^{1}$-manifold is finitely covered by a
product $B\times{S^{1}}$, where $B$ is a closed $\mathbb{H}^{3}$-manifold. If
$X=B\times{S^{1}}$ is such a product and $f:X\to{Y}$ is an essentially
dominating map then the image of $\pi_{1}(S^{1})$ in $\pi_{Y}$ must be an
infinite cyclic direct factor, by Lemma 7. Hence
$\pi_{Y}\cong{G}\times\mathbb{Z}$, and the first factor must be a
$PD_{3}$-group, since $Y$ is aspherical.

Conversely, if $Y$ is aspherical and $\pi_{Y}\cong{G}\times\mathbb{Z}$, then
$G$ is a finitely presentable $PD_{3}$-group, and so $P=K(G,1)$ is a
$PD_{3}$-complex. An application of the Atiyah-Hirzebruch spectral sequence
for oriented bordism $\Omega_{*}(P)$ (or direct desingularization of a
geometric 3-cycle representing a fundamental class) shows that $P$ is
1-dominated by a closed orientable 3-manifold. This is in turn 1-dominated by
a closed $\mathbb{H}^{3}$-manifold [13]. Hence $Y\simeq{P}\times{S^{1}}$ is
1-dominated by a $\mathbb{H}^{3}\times\mathbb{E}^{1}$-manifold. ∎

Hence solvmanifolds of type $\mathbb{E}^{4}$,
$\mathbb{N}il^{3}\times\mathbb{E}^{1}$ and
$\mathbb{S}ol^{3}\times\mathbb{E}^{1}$ are 1-dominated by
$\mathbb{H}^{3}\times\mathbb{E}^{1}$-manifolds. Similarly,
$\mathbb{H}^{2}\times\mathbb{E}^{2}$\- and
$\widetilde{\mathbb{SL}}\times\mathbb{E}^{1}$-manifolds are so dominated.
However no $\mathbb{N}il^{4}$-manifold or solvmanifold of type
$\mathbb{S}ol^{4}_{m,n}$ with $m\not=n$, $\mathbb{S}ol^{4}_{0}$ or
$\mathbb{S}ol^{4}_{1}$ can be so dominated.

The following lemma and theorem extend Theorem 11 to a wider class of
manifolds which have two such fibrations.

######  Lemma 12.

If an aspherical $4$-manifold $Y$ is dominated by the mapping torus of a self-
homeomorphism of an aspherical $3$-manifold then $\pi_{Y}$ is virtually a
semidirect product $K\rtimes\mathbb{Z}$, with $K$ finitely generated.

###### Proof.

Let $X=M(\varphi)$ be the mapping torus of a self-homeomorphism $\varphi$ of
an aspherical 3-manifold $M$, and let $f:X\to{Y}$ be a map of non-zero degree.
After we may assume that $f_{*}=\pi_{1}(f)$ is an epimorphism. Let $K$ be the
image of $\pi_{1}(M)$ in $\pi_{Y}$. Then $K$ is finitely generated and
$\pi_{Y}/K$ is cyclic.

Suppose that $\pi/K$ is finite. Then after passing to a finite cover, if
necessary, we may assume that $K=\pi_{Y}$. But $f_{*}$ induces a homomorphism
of spectral sequences from the LHS spectral sequence for $\pi_{X}$ as an
extension of $\mathbb{Z}$ by $\pi_{1}(M)$ and for $\pi_{Y}$ as an extension of
1 by $\pi_{Y}$. Since $H_{4}(X)\cong{H_{1}(\mathbb{Z};H_{3}(\pi_{1}(M)))}$,
while $H_{1}(1;H_{3}(\pi_{Y}))=0$, it follows that $f$ has degree 0, contrary
to hypothesis. Therefore $\pi_{Y}/K\cong\mathbb{Z}$, and so $\pi_{Y}$ is a
semidirect product ${K}\rtimes\mathbb{Z}$. ∎

######  Theorem 13.

An aspherical $4$-manifold $Y$ is essentially dominated by the mapping torus
of a self-homeomorphism of an aspherical $3$-manifold $M$ such that
$\pi_{1}(M)$ has non-trivial centre if and only if $Y$ is homotopy equivalent
to such a mapping torus.

###### Proof.

Let $M$ be a 3-manifold such that $\pi_{1}(M)$ has non-trivial centre $C$, and
let $X=M(\varphi)$ be the mapping torus of a self-homeomorphism $\varphi$ of
$M$. If $f:X\to{Y}$ is a map with non-zero degree and such that
$f_{*}=\pi_{1}(f)$ is an epimorphism then $A=f_{*}C$ and $K=f_{*}\pi_{1}(M)$
are normal subgroups of $\pi_{Y}$ such that $A\leqslant{K}$. Moreover, $K$ is
finitely generated and $\pi_{Y}/K\cong\mathbb{Z}$, by Lemma 12.

If $C\cong\mathbb{Z}$ then $A\cong\mathbb{Z}$ also, for otherwise $f$ would
factor through $K(\pi_{X}/C,1)$, and so have degree 0, by Lemma 7. Hence
$\beta_{i}^{(2)}(\pi_{Y})=0$ for all $i$, and so $\chi(Y)=0$, by a result of
Gromov. (See [5, Corollary 2.3.1].) Therefore $K$ is a $PD_{3}$-group [5,
Theorem 4.5]. Since $K$ has non-trivial centre it is the fundamental group of
an aspherical Seifert fibred 3-manifold, and the claim follows.

If $C$ has rank $>1$ then $M=T^{3}$, so $X$ is a solvmanifold and the claim
follows from the discussion in the first three paragraphs of §4.

The condition is clearly sufficient, since we may take $f=id_{Y}$. ∎

##  7\. $\mathbb{H}^{2}\times\mathbb{H}^{2}$ and surface bundles

A Surface bundle  is a bundle projection $p:X\to{B}$ with fibre $F$, where $B$
and $F$ are closed surfaces, $\chi(B)\leqslant 0$ and $\chi(F)<0$. A Surface
bundle space is the total space of a Surface bundle, and a Surface bundle
group is the fundamental group of a Surface bundle space. Such Surface bundles
may be partitioned into three types. Type I consists of such bundles for which
the monodromy has infinite image, but is not injective, type II are those
which are virtually products and type III have injective monodromy [8]. The
fibration is essentially unique when the bundle is of type I; product bundles
have only the two obvious bundle projections, and a surface bundle space $X$
with a fibration of type III may have many inequivalent fibrations. (See the
“Johnson trichotomy” in [5, Chapter 5.2].)

If a Surface bundle space $X$ is geometric but is not a
$\mathbb{H}^{4}$-manifold then either $\chi(X)=0$ (so $B=T$) and the geometry
is $\mathbb{H}^{2}\times\mathbb{E}^{2}$ or
$\mathbb{H}^{3}\times\mathbb{E}^{1}$, or $X$ is a reducible
$\mathbb{H}^{2}\times\mathbb{H}^{2}$-manifold [5, Theorems 9.10, 13.5 and
13.6]. Every reducible $\mathbb{H}^{2}\times\mathbb{H}^{2}$-manifold is
finitely covered by a product $B\times{F}$, while every
$\mathbb{H}^{3}\times\mathbb{E}^{1}$-manifold is finitely covered by a Surface
bundle space of type I, with base $T$. No Surface bundle space has the
geometry $\mathbb{H}^{2}(\mathbb{C})$ [5, Corollary 13.7.2], and it is
believed that $\mathbb{H}^{4}$ is also impossible.

Pullback along a degree-1 map of bases induces a degree-1 map of Surface
bundle spaces. If the original bundle is of type I or III and the base change
map is not a homotopy equivalence then the induced bundle is of type I. If the
monodromy fixes a separating curve in the fibre $F$ then we may construct a
degree-1 map by crushing one of the two complementary regions of this curve in
$F$ to a point. In this way we may show that every Surface bundle space of
type I or II is 1-dominated by Surface bundle spaces of types I and III. On
the other hand, we shall see that Surface bundle spaces of type II can only
dominate bundle spaces of the same type.

######  Lemma 14.

Let $f:\pi\to{G}$ be an epimorphism of $PD_{4}$-groups, where $\pi$ has a
normal subgroup $\kappa$ such that $\kappa$ and $\pi/\kappa$ are
$PD_{2}$-groups. If $f(\kappa)$ or $G/f(\kappa)$ is virtually free then $f$
has degree $0$.

###### Proof.

If $f(\phi)=1$ then $f$ factors though the $PD_{2}$-group $\rho=\pi/\kappa$,
and so $f$ has degree 0, by Lemma 7. In general, $f$ induces a map between the
LHS spectral sequences for $\mathbb{Q}$-homology of $\pi$ and $G$ as
extensions or $\rho$ by $\kappa$ and of $G/f(\kappa)$ by $f(\kappa)$,
respectively. If $f(\kappa)$ or $G/f(\kappa)$ is virtually free then the
homomorphisms between $E^{2}_{p,q}$ terms with $p+q=4$ either have domain 0 or
codomain 0. Hence they are all trivial, and so $H_{4}(f)=0$. ∎

We have a nearly complete understanding of the possibilities when $X$ is of
type II and $Y$ is aspherical.

######  Theorem 15.

Let $X=B\times{F}$, where $B$ and $F$ are hyperbolic surfaces, and let
$f:X\to{Y}$ be an essentially dominating map with aspherical codomain $Y$. If
$\chi(Y)\not=0$ or if $Y$ is the total space of a bundle with base or fibre
$T$ then $Y$ is homotopy equivalent to a product of surfaces.

###### Proof.

Let $G$ and $N$ be the images of $\pi_{1}(B)$ and $\pi_{1}(F)$, respectively.
Then $G$ and $N$ are finitely generated normal subgroups of $\pi_{Y}$ such
that $gn=ng$ for all $g\in{G}$ and $n\in{N}$, and $\pi_{Y}=GN$. Hence $gh=hg$
for all $g\in{G\cap{N}}$ and all $h\in\pi_{Y}$, and so $G\cap{N}$ is a
subgroup of the centre of $\pi_{Y}$. Let $\overline{\pi}=\pi_{Y}/(G\cap{N})$.
Then $\overline{\pi}\cong\pi/G\times\pi/N$.

If $G\cap{N}\not=1$ then $\beta_{i}^{(2)}(Y)=0$ for all $i$, and so
$\chi(Y)=0$. Since we are assuming in this case that $Y$ is a bundle space,
$G\cap{N}\cong\mathbb{Z}^{r}$, for some $r>0$. Then $\overline{\pi}$ is $FP$,
and hence so are $\pi/G$ and $\pi/N$. Consideration of the LHS spectral
sequence for $\pi_{Y}$ as an extension of $\overline{\pi}$ by $\mathbb{Z}^{r}$
then shows that
$H^{4-r}(\overline{\pi};\mathbb{Z}[\overline{\pi}])\cong\mathbb{Z}$. (Compare
[5, Theorem 3.10].) It then follows from the Künneth Theorem for
$\overline{\pi}\cong\pi/G\times\pi/N$ that either $\pi/G$ or $\pi/N$ two ends
or one is finite. This contradicts Lemma 14.

Therefore $G\cap{N}=1$, so $\pi_{Y}\cong{G}\times{N}$, and so $G$ and $N$ are
finitely presentable. Hence they are in fact $FP_{3}$ (each being a quotient
of an $FP$ group by a finitely presentable group). Neither $G$ nor $N$ is
trivial or $\mathbb{Z}$, by Lemma 14, and so each is a $PD_{2}$-group [5,
theorem 3.10]. ∎

We can push the argument for this theorem a little further if $\chi(Y)=0$ and
$G\cap{N}\not=1$. We note first that $c.d.G\leqslant 3$ and $c.d.N\leqslant
3$, since $G$ and $N$ each have infinite index in $\pi_{Y}$. Hence $G\cap{N}$
has rank $r\leqslant 3$. The argument of the theorem shows that $G\cap{N}$ is
not finitely generated Hence it has rank $r<c.d.G\cap{N}<4$ and so $r=1$ or 2.

Suppose that $r=2$. Then $c.d.G\cap{N}=3$, since $G\cap{N}$ is not finitely
generated. But $c.d.G=3$ also, and so $G$ is abelian [2, Theorem 8.8 ]. Since
$G$ is finitely generated, $G\cap{N}$ must be finitely generated also. This is
a contradiction, and so $r=1$. It remains an open question whether the centre
of a $PD_{n}$-group must be finitely generated, if $n>3$.

General results on essential dominations of aspherical 4-manifolds by Surface
bundle spaces of types I or III appear to be hard to find. When $Y$ is also a
bundle space we should ask whether such a domination need be compatible with a
fibration, so that the normal subgroup $N=f_{*}\pi_{1}(F)$ of $\pi_{Y}$ and
the quotient $\pi_{Y}/N$ are each $PD_{2}$-groups.

######  Lemma 16.

Let $E$ be a Surface bundle group. If $N$ is a finitely generated infinite
normal subgroup of infinite index in $E$ then $N$ is either a $PD_{2}$-group
or is an extension of a $PD_{2}$-group by a free group of infinite rank.
Moreover either $N$ or $E/N$ has one end.

###### Proof.

Let $p:E\to{G}$ be the projection of $E$ onto $G=\pi_{1}(B)$, with kernel
$K=\pi_{1}(F)$. If $p(N)\not=1$ then $[G:p(N)]<\infty$, and
$\mathrm{Ker}(p|_{N})$ has infinite index in $K$. Hence either $p|_{N}$ is a
monomorphism or $\mathrm{ker}(p|_{N})$ is free of infinite rank. (This uses
the fact that $F$ is hyperbolic!)

The second assertion follows from consideration of the LHS spectral sequence
for $E$ as an extension of $E/N$ by $N$. Since $E/N$ and $N$ are finitely
generated and infinite, $H^{p}(E/N;H^{q}(N;\mathbb{Z}[E]))=0$ if $q=0$ or if
$p=0$ and $q=1$, while

$H^{1}(E/N;H^{1}(N;\mathbb{Z}[E]))\cong{H^{1}(E/N;\mathbb{Z}[E/N])}\otimes{H^{1}(N;\mathbb{Z}[N])}.$

Since $H^{i}(E;\mathbb{Z}[E])=0$ for $i\leqslant 3$, a spectral sequence
corner argument shows that $H^{1}(E/N;H^{1}(N;\mathbb{Z}[E]))=0$. Since the
terms in the tensor product are free abelian groups, at least one must be 0. ∎

If $\beta_{1}(E)>\beta_{1}(B)$ then there are finitely generated normal
subgroups $N$ such that $E/N\cong\mathbb{Z}$ and $\mathrm{ker}(p|_{N})$ is
nontrivial, and if, moreover, $\chi(E)\not=0$ then such a subgroup $N$ cannot
be $FP_{2}$ [3].

##  8\. domination of aspherical geometric manifolds by surface bundle spaces

Suppose that $X$ is a Surface bundle space which dominates a
$\mathbb{H}^{2}\times\mathbb{E}^{2}$-manifold $Y$. After passing to finite
covers we may assume that $Y$ is a product $C\times{T}$, where $C$ is a
hyperbolic surface. If $N$ is a non-trivial finitely generated normal subgroup
of infinite index in $\pi_{Y}$ then either $N$ has finite index in
$\pi_{1}(C)$ or $\pi_{1}(T)$, or $N\cong\mathbb{Z}<\pi_{1}(T)$ or
$N\cap\pi_{1}(C)$ has finite index in $\pi_{1}(C)$ and $\pi_{Y}/N$ is
virtually $\mathbb{Z}$. The latter two possibilities are ruled out by Lemma
14, and so $f$ must be compatible with one of the two projections from $Y$ to
$C$ or $T$.

If $Y$ is a $\widetilde{\mathbb{SL}}\times\mathbb{E}^{1}$-manifold then after
passing to finite covers we may assume that $Y$ is the total space of a
$T$-bundle over a hyperbolic surface. A similar argument applies, except that
in this case the bundle fibration is unique and the only normal
$PD_{2}$-groups in $\pi_{Y}$ are subgroups of finite index in $\pi_{1}(T)$.
Hence $f$ must be compatible with the fibration of $Y$. In particular, no
$\widetilde{\mathbb{SL}}\times\mathbb{E}^{1}$-manifold is dominated by a
product of surfaces.

Suppose now that $Y$ is a solvmanifold. The image of $\pi_{1}(F)$ in $\pi_{Y}$
is a finitely generated normal subgroup, and is torsion-free polycyclic. It
must have Hirsch length 2, by Lemma 14, and so $\pi_{1}(f)$ factors though the
group of a $T$-bundle over $B$. Every solvmanifold with geometry
$\mathbb{E}^{4}$, $\mathbb{N}il^{3}\times\mathbb{E}^{1}$, $\mathbb{N}il^{4}$
or $\mathbb{S}ol^{3}\times\mathbb{E}^{1}$ is the total space of a $T$-bundle
over $T$. Hence such manifolds are 1-dominated by Surface bundle spaces.

There are epimorphisms from Surface bundle groups to semidirect products
$\mathbb{Z}^{3}\rtimes\mathbb{Z}$, with $\pi_{1}(F)$ mapping onto
$\mathbb{Z}^{3}$. However all such maps have degree 0, and no solvmanifold of
type $\mathbb{S}ol^{4}_{m,n}$ or $\mathbb{S}ol^{4}_{0}$ is dominated by a
Surface bundle space. Similarly, no $\mathbb{S}ol^{4}_{1}$-manifold is
dominated by a Surface bundle space.

If $Y$ is aspherical then the image of $\pi_{1}(F)$ in $\pi_{Y}$ is a finitely
generated normal subgroup, and $G=\pi_{Y}/K$ is finitely presentable. Lemma 14
implies that $K$ is not a free group and that $G$ is not virtually free, so
$[\pi_{Y}:K]=\infty$. Hence $c.d.K=2$ or 3.

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