(Strict) hyperbolisation

 Today I would like to discuss hyperbolisation procedures, which are methods of producing non-positively curved spaces from a given polyhedral complex. Specifically, for each dimension $n$ fix a non-positively curved space $X_n$ and, given a polyhedral complex $K$, the broad idea is to replace $n$-faces of the complex $K$ with $X_n$ in a way which preserves the combinatorial incidence structure of $K$ (to be made precise) and results in a metric space of non-positive curvature in the sense of Gromov. The motivation for this is that in many areas of geometric group theory and geometric topology negative curvature results in much better behaviour. 

There are different notions of hyperbolisation procedure which have different assumptions on the input $K$ and produce different results. We will discuss one, known as strict hyperbolisation, due to Charney and Davis, which is the first one to produce metric spaces of curvature $\leq -1$ and hence with Gromov hyperbolic fundamental group, and at the end give some indication as to what other procedures are available. Charney-Davis also show that in general the previously known hyperbolisation procedures give rise to spaces with fundamental groups that contain $\mathbb{Z}^2$, so can't be strictly negatively curved. One might object that in the case where a non-positively curved metric which isn't strictly negatively curved is obtained the name hyperbolisation is misleading. Since the resulting complex is always aspherical some authors call these asphericalisation procedures. Note also that hyperbolisation doesn't mean the same thing as it does in the theory of locally symmetric spaces, so where I remember to do so I will try to distinguish between hyperbolic in the sense of locally symmetric spaces and strictly negatively curved resulting in Gromov hyperbolic fundamental group.

The input complexes for this procedure will be polyhedral complexes. Briefly, a convex polytope or cell is the non-empty compact intersection of a finite number of half spaces in a sphere, Euclidean space, or hyperbolic space of appropriate dimension and curvature. A polyhedral complex is built out of gluing such objects along faces, or more accurately I should call them facets in higher dimensions, along isometries. These are generalisations of the notions of polyhedra in 3 dimensions, and are scaled with the appropriate metric to get the right curvature. The reader's intuitions in low dimensions will probably carry over for the basic notions.

Note that cells have faces which are partially ordered by inclusion so one can forget all the geometry and just considers the combinatorics of this poset. One can perform the same construction on a more general polyhedral complex to get a poset which encodes the combinatorics of how faces intersect, and by making the new complex have incidence structure inducing the same poset it is in this sense that the new complex is built out of the old one.

In general, one should hope for hyperbolisation procedures which are well-behaved, just as e.g. one hopes that homology theories reflect in some way the topology of the input topological space. I will use the languages of functors below, because this is the language that the paper is written in, but really one should think of this in the same way as any other map or function which takes an input and gives an output in a sensible way. Here are some properties listed in Charney-Davis which I hope the reader agrees are reasonable things to ask for in a hyperbolisation procedure $\mathcal{H}$:

1. Functoriality: $\mathcal{H}$ is a functor from cell complexes to nonpositively curved polyhedra in the following sense: if $i : J \to K$ is an embedding onto a subcomplex, then there is a mapping $\mathcal{H}(i): \mathcal{H}(J) \to \mathcal{H}(K)$ which is an isometric embedding onto a totally geodesic subspace.
2.  Preservation of local structure: If $\sigma^n$ is an $n$-cell in $K$, then $\mathcal{H}((\sigma^n)$ is an $n$-manifold with boundary and the link of $\mathcal{H}(\sigma)$ in $\mathcal{H}(K)$ is PL-homeomorphic to the link of $\sigma^n$ in $K$.
3. $\mathcal{H}$ sends a point to a point.
4. For a cell $\sigma$ the manifold with boundary $\mathcal{H}(\sigma)$ is orientable.

If 1 and 2 hold then there is a continuous map $\varphi: \mathcal{H}(K) \to K$, well-defined up to homotopy, such that $\varphi(\mathcal{H}(\sigma) \subset \sigma$ for all $\sigma \in K$. It then turns out that 3 and 4 are equivalent to requiring that $\varphi$ induces a surjection on homology.

Conditions 2 and 4 imply that if $K$ is a manifold (resp. orientable manifold) then so is $\mathcal{H}(K)$. Another condition we might require is the following.

Covered by bundle map: If the underlying polyhedron of $K$ is a manifold, then$\varphi: \mathcal{H}(K) \to K$ is covered by a map between the stable tangent bundles.

A weaker version is the following: If K is a manifold, then the map $\varphi$ pulls back the rational Pontryagin classes of $K$ to those of $\mathcal{H}(K)$.

We now require the notion of cutting open a manifold along a system of submanifolds. Given a manifold $M^n$ and a codimension 1 submanifold $Y$ of $M^n$ the reader's first guess for what it means to cut $M$ open along $Y$ is likely to be correct, e.g. cutting $\mathbb{R^3}$ along $z=0$ gives rise to two disjoint halves corresponding to $z>0$ and $z<0$. We denote the result by $M \odot Y$. We consider more generally a collection $\mathcal{Y}=\{Y_1, \dots, Y_k\}$ of codimension one submanifolds of $M^n$ which intersect transversely, which we call \emph{a system of codimension one submanifolds}. The system is \emph{two-sided} if each $Y_i$ is two-sided in $M$. If a finite group $G$ acts on $M^n$ with the union $\cup Y_i$ fixed setwise by the action then the system is said to be \emph{G-stable}.

Up to technical details that the reader is welcome to ignore, one can repeatedly cut along the manifolds in a system to obtain a (smooth) manifold with corners. We denote the result $M \odot \mathcal{Y}$. This comes with a projection map $\pi: M \odot \mathcal{Y} \to M$. For example, the $n-$torus is given by gluing opposite facets of an $n$-cube, and cutting along these facets returns the original cube. It is left to the reader to see that if a finite group $G$ acts smoothly on $M$ stabilising $\mathcal{Y}$ then there is an induced $G$-action on $M \odot \mathcal{Y}$.

I think the following is intuitive so I state without proof

Lemma 1: The poset of non-empty faces of nonempty faces of $M \odot \mathcal{Y}$ can be identified with a subset of the poset of faces of the $n$-cube.

We will now observe that cutting open manifolds preserves a lot of nice properties.

Lemma 2:  Suppose that \(Y_{1}\cap\cdots\cap Y_{n}\) is a single point \(y\). For each \(i\), let \(I(i)=\{1,\ldots,n\}-\{i\}\) and let \(S_{i}\) denote the component of the one-dimensional intersection \(Y_{I(i)}\) which contains \(y\).

1.  \(\bigcup S_{i}\) is a bouquet of circles and \(\pi^{-1}(\bigcup S_{i})\) is isomorphic to the \(1\)-skeleton of \(\square^{n}\).
2. For any face \(F=\partial_{U,a}(M\odot\mathcal{Y})\), all \(0\)-dimensional faces of \(F\) lie in a single component \(\overline{F}\). In particular, \(\pi^{-1}(y)\) is contained in a single component of \(M\odot\mathcal{Y}\).

Proof: Since \(M\) is closed, \(S_{i}\) is a closed \(1\)-manifold; hence, a circle. Since \(S_{i}\cap S_{j}=\{y\}\) if \(i\neq j\), it is clear that \(\bigcup S_{i}\) is a bouquet of circles. Call the points of \(\pi^{-1}(y)\) \textit{vertices}. They are indexed by functions \(\varepsilon\colon\{1,\ldots,n\}\to\{\pm1\}\). Let \(v_{\varepsilon}\) denote the vertex corresponding to \(\varepsilon\). Let \(\varepsilon_{i}\) be the function defined by \(\varepsilon_{i}(j)=(-1)^{\delta_{ij}}\). A component of \(\pi^{-1}(S_{i})\) is an edge connecting two vertices and for any \(\varepsilon\) there is such an edge connecting \(v_{\varepsilon}\) and \(v_{\varepsilon\varepsilon_{i}}\). This proves (1). Statement (2) follows easily. $\blacksquare$

Lemma 3: Suppose that \(Y_{1}\cap\cdots\cap Y_{n}\) is a single point \(y\). Suppose further that \(M\) is connected and that there are nontrivial commuting involutions \(r_{1},\ldots,r_{n}\) on \(M\) (generating an action of \(G=(\mathbb{Z}/2)^{n}\)) such that \(Y_{i}\) is contained in the fixed point set \(R_{i}\) of \(r_{i}\) and such that \(G\) stabilizes \(\mathcal{Y}\). Then \(M\odot\mathcal{Y}\) is connected.

Proof: Let \(U=M-\bigcup Y_{i}\) denote the interior of \(M\odot\mathcal{Y}\). We must show that \(U\) is connected. Put \(W=M-\bigcup R_{i}\). Since \(W\) is open and dense in \(U\), \(\pi_{0}(W)\to\pi_{0}(U)\) is onto. Let \(p\colon M\to M/G\) be the projection. Since \(M\) is connected, so is \(M/G\). Since \(p(R_{i})\) cannot disconnect \(M/G\) locally, \(p(W)\) is connected. Choose a vector in \(T_{y}M\) not tangent to any \(R_{i}\) and use it to push \(y\) into \(W\). Call the resulting point \(w\). Since \(G\) acts freely on \(W\), \(p|W\colon W\to p(W)\) is a \(2^{n}\)-sheeted cover and each component of \(W\) contains at least one point in \(Gw\) (the \(G\)-orbit of \(w\)). Hence, each component of \(U\) contains at least one point in \(Gw\). By Lemma 2(2), \(\pi^{-1}(y)\) is contained in a single component of \(M\odot\mathcal{Y}\); hence, \(Gw\) is contained in a single component of \(U\). Therefore, \(U\) (and consequently \(M\odot\mathcal{Y}\)) is connected. $\blacksquare$


Lemma 4: Suppose that $\mathcal{Y} = \{Y_1, \dots, Y_n\}$ is a two-sided system of codimension one submanifolds of a smooth manifold $M^n$. Then there is a smooth map $\varphi: M^n \to T^n$ such that $\mathcal{Y}$ is the transverse inverse image of the standard system of subtori in $T^n$.
Suppose further that $M^n$ is closed and oriented and that $Y_1\cap \dots \cap Y_n$ is a single point $y$. For each subset $J \subset \{1, \dots, n\}$, let $\tilde{Y_j}$ denote the component of $Y_j$ which contains $y$. Then for each $J$ the map $\varphi|_{\tilde{Y_j}} \to T_J$ is degree one (for an appropriate choice of sides).

This is just some fiddling so we leave it to the reader to look up the proper proof if interested. To give a flavour of what sort of fiddling, to prove a one chooses tubular neighbourhoods of each $Y_i$ and applies the Pontryagin-Thom construction. Do this for all $i$ and put them together.  This map is compatible with the cutting-open operation; hence, we get a map from \(M\odot\mathcal{Y}\) to \(\square^{n}\). This gives us the following lemma, the proof of which is left to the reader.
Lemma 5: 

1. There is a smooth face-preserving map \(f\colon M\odot\mathcal{Y}\to\square^{n}\) (in fact, unique up to homotopy through such maps).
2. Suppose that \(M\) is closed and oriented and that \(Y_{1}\cap\cdots\cap Y_{n}\) is a single point \(y\). Let \(F\) be a \(k\)-dimensional face of \(M\odot\mathcal{Y}\) and \(\overline{F}\) the distinguished component of \(F\) containing \(y\) (as in Lemma \ref{lem:5.7}(2)); then \(f|\overline{F}\colon\overline{F}\to\square^{k}\) is degree one (and the other components are mapped by degree zero maps).

If \(M\) has a Riemannian metric, then there is an induced Riemannian metric on \(M\odot\mathcal{Y}\). In the next lemma we record two elementary observations.

Lemma 6: Suppose \(M\) is a Riemannian manifold and that \(\mathcal{Y}\) is a system of codimension one submanifolds.

1.  If each \(Y_{i}\) is totally geodesic in \(M\), then each face of \(M\odot\mathcal{Y}\) is totally geodesic.
2. If \(G\) is a finite group of isometries of \(M\) stabilizing \(\mathcal{Y}\), then \(G\) acts isometrically on \(M\odot\mathcal{Y}\).

Constructing the hyperbolic cells for replacement

Recall that \(B_{n}\) denotes the symmetry group of the \(n\)-cube. It has a \textit{standard} orthogonal action on \(\mathbb{R}^{n}\) generated by all permutations of coordinates and sign changes. The sign changes are generated by involutions \(r_{i}\), \(i=1,\ldots,n\), where \(r_{i}\) denotes the linear reflection across the hyperplane \(x_{i}=0\).


Our goal in this section is to prove the following theorem. 

Theorem 7: For each \(n>0\), there is a closed, connected hyperbolic \(n\)-manifold \(M^{n}\), a system \(\mathcal{G}=\{Y_{1},\ldots,Y_{n}\}\) of closed, connected submanifolds of codimension one in \(M^{n}\), and an isometric action of \(B_{n}\) on \(M^{n}\), stabilizing \(\mathcal{G}\), such that the following properties hold:

1.  \(Y_{i}\) is a component of the fixed point set of \(r_{i}\) on \(M^{n}\).
2.  Each \(Y_{i}\) is totally geodesic in \(M^{n}\).
3. The \(Y_{i}\)'s intersect orthogonally.
4.  \(Y_{1}\cap\cdots\cap Y_{n}\) is a single point \(y\).
5.  \(B_{n}\) fixes \(y\) and the representation of \(B_{n}\) on \(T_{y}M^{n}\) is equivalent to the standard representation.
6.  \(M^{n}\), as well as each \(Y_{i}\), is orientable.

If we put  \[X^{n}=M^{n}\odot\mathcal{G}\] then the previous lemmas give us the following corollary.


Corollary 8: For each \(n>0\), there is a compact, connected, orientable hyperbolic \(n\)-manifold with corners \(X^{n}\) together with an action of \(B_{n}\) on \(X^{n}\) by isometries so that the following properties hold.

1.  The poset of faces of \(X^{n}\) is \(B_{n}\)-equivariantly isomorphic to the poset of faces of \(\square^{n}\).
2.  Each face of \(X^{n}\) is totally geodesic.
3. The faces of \(X^{n}\) intersect orthogonally.
4.  Each \(0\)-dimensional face is a single point (i.e. \(X^{n}\) has precisely \(2^{n}\) vertices).
5.  The map \(f\colon X^{n}\to\square^{n}\) of Lemma \ref{lem:5.9} is degree one as is its restriction to each face of \(X^{n}\).

Remark: In this corollary the meaning of the word "face" is as in the previous section: a \(k\)-dimensional face of \(X^{n}\) is a union of \(k\)-dimensional strata. In particular, \textit{a face need not be connected}. Statement (4) asserts that each \(0\)-dimensional face is connected; however, in general we do not know if it is possible to find such \(X^{n}\) with all faces connected.


All currently known constructions of hyperbolic manifolds in higher dimensions involve arithmetic, so of course this one does too. Let $K= \mathbb{Q}(\sqrt{d})$ be a totally real quadratic extension of the rationals and let $\mathcal{O_K}$ denote its ring of integers. Denote by $\alpha \mapsto \overline{\alpha}$ the automorphism induced by conjugation. Fix an embedding of $K$ into $\mathbb{R}$ and choose $\epsilon >0 \in \mathcal{O}_K$ as well as a basis $e_0, \dots, e_n$. Define a symmetric bilinear form on $\mathcal{O}_K^{n+1}$ by $\varphi(e_i, e_j)= \delta_{ij}$ if $(i,j) \neq (0,0)$ and $-\epsilon$ otherwise.

By tensoring with $\mathbb{R}$ this induces a symmetric bilinear form $\varphi_{\mathbb{R}}$ on $\mathbb{R}^{n+1}$. One can embed the isometry group $O(\varphi)$ into the group of isometries of $\mathbb{H}^n$ as a discrete and cocompact subgroup by considering $\mathbb{H}^n$ as the space of solutions to a certain quadratic form.

$B_n$ is identified as a subgroup of $O(\varphi)$ generated by all permutations of basis vector and for $i>0$ reflections $r_i$ where $r_i$ flips the $i^{th}$ basis vector and fixes its orthogonal complement. This implies that $B_n$ fixes $e_0$.

For $i>0$ let $P_i$ denote the intersection of the orthogonal complement of $e_i$ in $\mathbb{R}^{n+1}$ with $\mathbb{H^n}$. the intersection of all $P_i$ is a single point $p$. $B_n$ fixes $p$ and acts on the tangent space $T_p\mathbb{H}^n$ via the standard representation.

From now let $\Gamma$ be a torsion-free normal subgroup of $O(\varphi)$ and let the corresponding quotient manifold be $M^n$ with universal covering map $\pi$, and set $Y_i=\pi(P_i), y=\pi(p)$. Since $B_n$ normalises $\Gamma$ it acts by isometries on $M^n$

Lemma 9:Let \(\Gamma\) be any torsion-free subgroup of \(O(\varphi)\). Let \(M^{n}=\mathbb{H}^{n}/\Gamma\) and \(\mathcal{Y}=\{Y_{1},\ldots,Y_{n}\}\), where the \(Y_{i}\) are as above. Then \(\mathcal{Y}\) is a system of closed, connected submanifolds of codimension one in \(M^{n}\) such that the following properties hold:

1. \(Y_{i}\) is a component of the fixed point set of \(r_{i}\) on \(M^{n}\).
2.  Each \(Y_{i}\) is totally geodesic in \(M^{n}\).
3. The \(Y_{i}\)'s intersect orthogonally.
4.  \(Y_{1}\cap\cdots\cap Y_{n}\) is a finite set which contains the point \(y\).
5. \(B_{n}\) fixes \(y\) and the representation of \(B_{n}\) on \(T_{y}M^{n}\) is equivalent to the standard representation.

Proof: Each \(Y_{i}\), being the image of a connected space, is connected, and \(Y_{i}\) is clearly contained in the fixed set of \(r_{i}\). Since the fixed set of any smooth involution is a submanifold, it follows that each \(Y_{i}\) is an embedded submanifold and a component of the fixed set of \(r_{i}\). Thus, \(\mathcal{Y}\) is a system of closed, connected, codimension one submanifolds. Since \(r_{i}\) is an isometry, (2) holds; since the \(r_{i}\) commute, (3) holds. Statements (4)' and (5) are obvious.

Hence, it remains to verify properties (4) and (6) of Theorem 7. With regard to (6), the following result is immediate from the fact that \(SO_{0}(n,1)\) is connected. $\blacksquare$

Lemma 10: Suppose that \(\Gamma\) is as in Lemma 9 and that \(\Gamma\subset SO_{0}(n,1)\). Then \(M^{n}\), as well as each \(Y_{i}\), is orientable.
\end{lemma}

We now focus on property (4) of Theorem 7. Suppose that \(z\in Y_{1}\cap\cdots\cap Y_{n}\) and that \(\tilde{z}\) is a lift of \(z\) to \(\mathbb{H}^{n}\). We seek a condition on \(\Gamma\) which will insure that \(z=y\), i.e. that \(\tilde{z}=y\gamma\) for some \(\gamma\in\Gamma\). Let \(\gamma_{i}\) in \(\Gamma\) be such that \(\gamma_{i}P_{i}\) is the component of \(\pi^{-1}(Y_{i})\) which contains \(\tilde{z}\). Since the vector \(e_{i}\) is a unit normal to \(P_{i}\), \(\gamma_{i}e_{i}\) is a unit normal to \(\gamma_{i}P_{i}\). Since the \(\gamma_{i}P_{i}\) intersect orthogonally at \(\tilde{z}\), we have that
\[
\varphi(\gamma_{i}e_{i},\gamma_{j}e_{j})=\delta_{ij},\quad 1\leq i,j\leq n.
\]
Hence, \((\gamma_{1}e_{1},\ldots,\gamma_{n}e_{n})\) is an orthogonal basis for a sublattice \(L\) of \(A^{n+1}\). Moreover, \(L\) is equivalent to \(\langle 1 \rangle\perp\cdots\perp\langle 1 \rangle\) where \(\langle a \rangle\) denotes the one-dimensional lattice generated by a basis vector of norm \(a\) and where \(\perp\) denotes orthogonal direct sum. It follows that \(A^{n+1}\) is the orthogonal direct sum of \(L\) and its orthogonal complement \(L^{\perp}\) and that \(L^{\perp}\cong\langle -e \rangle\). Hence, there is a vector in \(A^{n+1}\) orthogonal to \(L\) and of norm \(-e\). The only possibilities are \(\pm e\), where \(e=\sqrt{\varepsilon}\tilde{z}\). Let \([e,\gamma_{1}e_{1},\ldots,\gamma_{n}e_{n}]\) denote the \((n+1)\times(n+1)\) matrix with column vectors \(e,\gamma_{1}e_{1},\ldots,\gamma_{n}e_{n}\). Since \(\{e,\gamma_{1}e_{1},\ldots,\gamma_{n}e_{n}\}\) is an orthogonal basis for \(A^{n+1}\) with respect to \(\varphi\), the matrix \([e,\gamma_{1}e_{1},\ldots,\gamma_{n}e_{n}]\) lies in \(O(\varphi)\). Since it maps \(e_{0}\) to \(e\), it takes \(p=(1/\sqrt{\varepsilon})e_{0}\) to \(\tilde{z}=(1/\sqrt{\varepsilon})e\).
 

Lemma 11: Let \(\Gamma\) be a normal, torsion-free subgroup of \(O(\varphi)\). With notation as above, the following statements are equivalent:

1. \(Y_{1}\cap\cdots\cap Y_{n}=\{y\}\).
2. Given any \(n\)-tuple \((\gamma_{1},\ldots,\gamma_{n})\) in \(\Gamma\times\cdots\times\Gamma\) such that \(\gamma_{1}P_{1}\cap\cdots\cap\gamma_{n}P_{n}\) is nonempty, there is an element \(\gamma\) in \(\Gamma\) such that \(\gamma_{1}P_{1}\cap\cdots\cap\gamma_{n}P_{n}=\gamma(P_{1}\cap\cdots\cap P_{n})\).
3.  Given any \(n\)-tuple \((\gamma_{1},\ldots,\gamma_{n})\) in \(\Gamma\times\cdots\times\Gamma\) such that \(\varphi(\gamma_{i}e_{i},\gamma_{j}e_{j})=\delta_{ij}\), \(1\leq i,j\leq n\), one of the two matrices \(\{\pm e,\gamma_{1}e_{1},\ldots,\gamma_{n}e_{n}\}\) lies in \(B_{n}\Gamma\).

The lemma is obvious. Perhaps the only point which needs comment is the appearance
of the group $B_n$ in statement (iii). The reason is that in passing from (ii) to (iii) there is no
reason to choose $e_i$ as the unit normal to $P_i$ rather than $-e_i$. Similarly there is no preferred
ordering for $\{\gamma_1e_1, \ldots, \gamma_ne_n\}$. These sign changes and permutations of coordinates are accounted for by $B_n$.

Let $\mathfrak{p}$ be a prime ideal in $A$. Then $A/\mathfrak{p}$ is a finite field and the form $\varphi$ induces a nonsingular bilinear form $\varphi_{\mathfrak{p}}$ on $(A/\mathfrak{p})^{n+1}$. Its isometry group $O(\varphi_{\mathfrak{p}})$ is finite. The kernel of the natural projection $\pi : O(\varphi) \to O(\varphi_{\mathfrak{p}})$ is denoted by $\Gamma(\mathfrak{p})$ and called a congruence subgroup of $O(\varphi)$. It is, of course, normal and of finite index in $O(\varphi)$. Moreover, if $|\mathfrak{p}|$ is sufficiently large, then $\Gamma(\mathfrak{p})$ is torsion-free. More generally, if $\mathfrak{a}$ is any nonzero ideal in $A$, then
\[\Gamma(\mathfrak{a}) = \{\gamma \in O(\varphi) \mid \gamma \equiv 1 \ (\mathrm{mod}\ \mathfrak{a})\}\]
is also called a congruence subgroup: 
Lemma 12: Let $\Gamma$ be any torsion-free congruence subgroup of $O(\varphi)$. Then $Y_1 \cap \cdots \cap Y_n$ is a single point.

Proof: Suppose $\Gamma = \Gamma(\mathfrak{a})$. Using Lemma 6.5(iii) we consider an $n$-tuple $(\gamma_1, \ldots, \gamma_n)$
where $\gamma_i \in \Gamma$ and $\varphi(\gamma_ie_i, \gamma_je_j) = \delta_{ij}$. Since $\gamma_i \in \Gamma$, $\gamma_ie_i \equiv e_i \ (\mathrm{mod}\ \mathfrak{a})$. Hence, the vector $e$, which
generates the orthogonal complement to the lattice spanned by $\gamma_1e_1, \ldots, \gamma_n e_n$, satisfies
either $e \equiv +e_0 \ (\mathrm{mod}\ \mathfrak{a})$. Therefore,
the matrix $[\gamma_1e_1, \ldots, \gamma_n e_n]$ or $[-e_0, \gamma_1e_1, \ldots, \gamma_n e_n]$ lies in $\Gamma$. $\blacksquare$

We now quote without proof that such a subgroup does exist, since the construction involves non-trivial number theory which is orthogonal to the current discussion.

Putting the cells together

Denote by $\mathcal{PE}^n$ the category of piecewise Euclidean polyhedra of
dimension $\leq n$. Let $\mathcal{H}^n$ denote the category of piecewise hyperbolic polyhedra of dimension $\leq n$. Morphisms in both categories are isometries onto totally geodesic subcomplexes.
Let $X^n$ be the hyperbolic $n$-manifold with corners constructed in Corollary 8.

Proposition 13: Suppose that $K$ is a cubical Euclidean cell complex of dimension $\leq n$. Then there is a piecewise hyperbolic polyhedron $K_X$ and a map $q : K_X \to K$ such
that the following properties hold:

1.  For each $k$-cell $\sigma^k$ in $K$, $q^{-1}(\sigma^k)$ is isometric to a $k$-dimensional face of $X^n$. Furthermore, if $J$ is any subcomplex of $K$, then $q^{-1}(J)$ is isometric to $J_X$.
2. The directions normal to $q^{-1}(\sigma^k)$ naturally form a piecewise spherical polyhedron, denoted $\mathrm{Link}(q^{-1}(\sigma^k), K_X)$, and this polyhedron is isometric to $\mathrm{Link}(\sigma^k, K)$.
3. The construction of $K_X$ from $K$ defines a functor from $\mathcal{E}^n$ to $\mathcal{H}^n$. In particular, it takes totally geodesic subcomplexes to totally geodesic subcomplexes.
4. The map $q$ induces a surjection on homology.
5. If $K$ has curvature $\geq 0$, then $K_X$ has curvature $\leq -1$.

The polyhedron $K_X$ can be constructed as a quotient space of a disjoint union
$\widetilde{K}$ of copies of standard Euclidean cubes. Such a standard cube may be viewed as a face of
a fixed $n$-cube $\square^n$. The equivalence relation on $\widetilde{K}$ is defined by identifying various faces via isometries viewed as elements of $B_n$. By Corollary 8 (1), the poset of faces of $X^n$ is $B_n$-equivariantly isomorphic to the poset of faces of $\square^n$. To construct $K_X$, one replaces each $k$-cube in $\widetilde{K}$ by the corresponding $k$-face of $X^n$. Call the resulting disjoint union $\widetilde{K}_X$. To each isometry in the equivalence relation on $\widetilde{K}$ the associated element of $B_n$ gives an isometry between the corresponding faces of $X^n$. By definition $K_X$ is then the quotient space of $\widetilde{K}_X$ by the resulting equivalence relation. The map $q : K_X \to K$ is then defined on each face of $K_X$ via the map $f$ of Corollary 8 (5).


Definition: Suppose that \(K\) is a cubical cell complex of dimension \(\leq n\). A \emph{projection to} \(\square^n\) is a cellular map \(p : K \to \square^n\) such that the restriction of \(p\) to any cell is a combinatorial isomorphism.

If \(K\) admits a projection to \(\square^n\), then the piecewise hyperbolic polyhedron \(K_X\) of the previous proposition can be obtained as a fiber product:

That is to say, \(K_X\) can be identified with the subspace of \(K \times X^n\) consisting of all \((k,x)\) such that \(p(k) = f(x)\).

Proposition 14: Suppose that \(K\) is a cubical cell complex homeomorphic to a smooth or PL manifold (i.e.\ \(K\) is a smooth or PL ``cubization'' of a manifold).
Suppose further that \(K\) admits a projection \(p : K \to \square^n\). Then: 

1. \(K_X\) is embedded in \(K \times X^n\) with trivial normal bundle.
2. The rational Pontryagin classes of \(K_X\) are the pullbacks (via \(q\)) of those of \(K\).

Proof: In the smooth case the argument goes as follows: Viewing \(\square^n\) as \([0,1]^n \subset \mathbb{R}^n\), the space \(K_X\) is the inverse image of \(0 \in \mathbb{R}^n\) under the map \(\psi : K \times X^n \to \mathbb{R}^n\) defined by \((k,x) \mapsto p(k) - f(x)\).
Assuming, as we may, that \(f\) is transverse to each face of \(\square^n\), we have that \(0\) is a regular value of \(\psi\); statement (1) follows.
Statement (2) follows from the fact that the rational Pontryagin classes of any hyperbolic manifold (e.g.\ \(M^n\) or \(X^n\)) are trivial together with the Whitney product formula.
The other cases are analogous. $\blacksquare$

Theorem 15: Any triangulable manifold is cobordant to a triangulable manifold of strictly negative curvature.

Proof: Let $K$ be a triangulation of the given PL-manifold. Let $\widehat{K}$ denote $K \times [0, 1]$ with the cone on $K$ glued on to $K \times 0$. Applying the functor $\mathcal{F}_X$, there is a unique vertex $v_0 = \mathcal{F}_X(x_0)$ of $\mathcal{F}_X(K_X)$ corresponding to the cone point $x_0$ of $\widehat{K}$. The link of $v_0$ in $\mathcal{F}_X(K_X)$ is PL-homeomorphic to the link of $x_0$ in $\widehat{K}$, namely $K \times 0$. Thus, removing a neighborhood of $v_0$ gives the desired cobordism from $K \times 0$ to $\mathcal{F}_X(K \times 1)$. $\blacksquare$

Further directions

This construction is an important one, and lots of relevant subsequent work has been done. We attempt to list a few important results and brief indications of proofs:

• In general, any hyperbolization procedure $\mathcal{H}$ admits a relative version, which allows to work relatively to a subcomplex, i.e., keep it unaltered under the hyperbolization. More precisely, if $L\subseteq K$ is a subcomplex, then one can attach to $K$ the cone over $L$, apply the hyperbolization procedure to the coned-off complex, and the remove a small neighborhood of the cone point. Thanks to axiom (2) stated above, the link of the cone point is a copy of $L$, so removing a small neighborhood of the cone point results in a boundary component homeomorphic to $L$.
• If ${\mathcal {H}}$ is the strict hyperbolization of Charney-Davis, then Belegradek showed that the relative version of $\displaystyle {\mathcal {H}}$ results in a space whose fundamental group is hyperbolic relative to $\displaystyle \pi _{1}(K)$.
• Lafont and Ruffoni showed that when the input manifold has hyperbolic and cubulated fundamental group, so does the output. In particular, the construction preserves linearity, residual finiteness, and a whole host of other good properties, since closed arithmetic manifolds of simplest type, including the ones used above, are cubulated.

In an impressive paper Ontaneda showed that if K is a smooth triangulation of a smooth manifold, then the Charney-Davis hyperbolisation procedure can be refined to ensure that $\mathcal {H}(K)$ is a smooth manifold and that it admits a Riemannian metric of negative sectional curvature. Moreover, it is possible to pinch the curvature arbitrarily close to $-1$. We can deduce the follow corollaries (most of which were known before but without the pinching result in the conclusion).

Corollary 16: Every closed smooth manifold is smoothly cobordant to a closed Riemannian manifold with sectional curvatures in the interval \([-1-\varepsilon,-1]\), for every \(\varepsilon>0\).

The proof is completely analogous to Theorem 15.

Corollary 17: The cohomology ring of any finite CW-complex embeds in the cohomology ring of a closed Riemannian manifold with sectional curvatures in the interval \([-1-\varepsilon,-1]\), for every \(\varepsilon>0\).


Proof: Let \(X\) be a finite CW-complex. Embed \(X\) in some \(\mathbb{R}^n\) and let \(P\) be a compact neighborhood of \(X\) that retracts to \(X\). Let \(M\) be the double of \(P\). Then there is a retraction \(M \rightarrow X\), and Corollary 17 follows from the retraction property (the surjection on homology with all coefficients means that the map on cohomology is injective by naturality of the Poincare duality isomorphism). $\blacksquare$

Since degree one maps between closed orientable manifolds are \(\pi_1\)-surjective we obtain the following result.

Corollary 18: For every finite CW-complex \(X\) there is a closed Riemannian manifold \(N\) and a map \(f\colon N\rightarrow X\) such that:

1. \(N\) has sectional curvatures in the interval \([-1-\varepsilon,-1]\),
2.  \(f\) is \(\pi_1\)-surjective,
3.  \(f\) is homology surjective.

All previously known examples of closed negatively curved Riemannian manifolds with less than \(\frac{1}{4}\)-pinched curvature have zero rational Pontryagin classes. The next corollary gives examples of such manifolds with nonzero rational Pontryagin classes.

Corollary 19: For every \(\varepsilon>0\) and \(n\geq 4\) there is a closed Riemannian \(n\)-manifold with sectional curvatures in the interval \([-1-\varepsilon,-1]\) and nonzero rational Pontryagin classes.

To show this, input an orientable manifold $M$ with nonzero Pontryagin classes.

All manifolds given in Corollary 19 were at the time new examples of closed negatively curved manifolds, provided \(\varepsilon<3\). 

Corollary 20: For any \(\varepsilon>0\) and \(n\geq 4\) there are closed Riemannian \(n\)-manifolds with sectional curvatures in the interval \([-1-\varepsilon,-1]\) that are not homeomorphic to a hyperbolic manifold (real, complex, quaternionic, or Cayley), or the Gromov-Thurston branched cover of a real hyperbolic manifold, or one of the Mostow-Siu or Deraux examples.

Proof Sketch: Let \(N\) be as in Corollary 19, with \(\varepsilon<3\). So, \(N\) is less than quarter-pinched negatively curved. Then \(N\) is not homeomorphic to a real hyperbolic manifold or the Gromov-Thurston branched cover of a real hyperbolic manifold. This follows from Novikov's topological invariance of the rational Pontryagin classes, and Ardanza's result that the branched covers have vanishing Pontryagin classes mentioned. Also the quarter-pinched rigidity results given in (or implied by) the work of Corlette, Gromov, Hernandez, and Mok-Siu-Yeung imply that \(N\) is not homeomorphic to a quaternionic or Cayley hyperbolic manifold.  Finally, it follows from work of Hernandez that a closed K\"ahler manifold of dimension \(\geq 4\) cannot be homeomorphic to a less than \(\frac{1}{4}\)-pinched negatively curved manifold, so \(N\) cannot be homeomorphic to a complex hyperbolic manifold, or any of the Mostow-Siu or Deraux examples. This is because Mostow-Siu and Deraux examples are all K\"ahler. $\blacksquare$