Metrics on symmetric spaces I

April 11, 2026

Metrics on symmetric spaces I

We have previously mentioned the approach of studying the topology of manifolds by asking what sort of metrics they admit. This has proved rather fruitful. For locally symmetric spaces, it is plausible, given the terminology, that a statement like 'the locally symmetric metric on a locally symmetric manifold is the most symmetric metric on that manifold’. This is indeed true in many senses, and the goal of this and the next post is to discuss two ways, due to Avramidi, that one can make this notion precise.

Let $(M; g)$ be a complete Riemannian manifold homeomorphic to a finite volume, nonpositively curved locally symmetric space with no local torus factors. The isometry group $\rm{Isom}(M, g)$ acts on free homotopy classes of loops, which gives an action homomorphism \[\rho: \rm{Isom}(M, g) \to \rm{Out}(\pi_1(M))\] The result we will show today is

Theorem 1: $\rho$ is injective.

I should say a couple of words about how this ties in with the philosophy that we are trying to show. For the fundamental group of a closed hyperbolic surface, the Dehn-Nielsen-Baer theorem says that this is equivalent to the mapping class group of the surface, which is the group of all \emph{topological} isomorphisms, hence in some sense the maximum amount of symmetry seen by the topology. If, in addition, the locally symmetric space is irreducible and of dimension at least 3, then Margulis–Mostow–Prasad rigidity shows that the group $\rm{Out}(\pi_1(M))$ is represented by isometries of the locally symmetric metric $h_{sym}$. Thus, $\rm{Isom}(M, g) < \rm{Isom}(M, h_{sym})$ so that, in a sense, the locally symmetric metric has the most symmetry.

Background:

We recall some classical facts from group theory and Riemannian geometry

Proposition 2: A group extension
\[1 \rightarrow A < B \xrightarrow{\pi} C \rightarrow 1\]
is determined by:

the conjugation representation $\rho: C \rightarrow \text{Out}(A)$, and
a cohomology class in the group $H^2(C; Z(A))$.

In particular, if both the centre $Z(A)$ and the representation $\rho$ are trivial, then the extension splits as a product. In this case:

there is a projection $B \xrightarrow{\varphi} A$ which restricts to the identity map on $A$, and
a section $C \xrightarrow{s} B$ whose image $s(C)$ commutes with $A$.

I have previously mentioned the Myers-Steenrod theorem(s) on this blog, but we will need one of them later and it's pretty so I may as well state it again.

Myers-Steenrod Theorem: Let $M$ be a Riemannian manifold. Then $\operatorname{Isom}(M)$ is a Lie group, and acts properly on $M$. If $M$ is compact then $\operatorname{Isom}(M)$ is compact.

Note that the Lie group $\operatorname{Isom}(M)$ may have infinitely many components. For example, let $M$ be the universal cover of a bumpy metric on the torus. An exposition of the proof can be found in these notes.

Flats

In this section, homology can have coefficients in any of $\mathbb{Z}$, $\mathbb{R}$ and $\mathbb{F}_p$. As is often the case when thinking about symmetric spaces, understanding the behaviour of flats is a key ingredient. Our proof of Theorem 1 is based on a result of Pettet-Souto, which says that the fundamental group of a locally symmetric space contains a free abelian subgroup that ``moves the boundary as much as possible''.

Suppose that $ G $ is a connected semisimple Lie group with no compact or Euclidean factors. Let $ K < G $ be a maximal compact and $ \Gamma < G $ a torsion-free lattice. Suppose that $ r := \text{rank}_{\mathbb{R}}G $ is the real rank of $ G $. By a maximal $\Gamma$-periodic flat we will mean a totally geodesic $ r $-dimensional flat subspace $ \mathbb{R}^r \subset G/K $ which is invariant under some free abelian subgroup $ \mathbb{Z}^r < \Gamma $.

We will use the following facts, which are proved in the paper of Pettet-Souto.

The symmetric space $ G/K $ has a maximal $\Gamma$-periodic flat.
There is a number $ q = \text{rank}_{\mathbb{Q}}(M) $ called the $\mathbb{Q}$-rank of the locally symmetric space $ M := \Gamma \setminus G/K $ and a $(q-1)$-dimensional simplicial complex $ \Delta_{\mathbb{Q}}(M) $ on which the group $ \Gamma $ acts simplicially. The complex is called the rational Tits building of the locally symmetric space $ M $. It is homotopy equivalent to an infinite wedge of $(q-1)$-spheres, i.e. $ \Delta_{\mathbb{Q}}(M) \cong \bigvee S^{q-1} $.
The locally symmetric space $ M $ is the interior of a compact manifold $ \overline{M} $ with boundary $ \partial \overline{M} $. There is a $\Gamma$-equivariant homotopy equivalence \[ \partial \widetilde{\overline{M}} \to \Delta_{\mathbb{Q}}(M).\]

From now we will use $\mathbb{R}^r$ to denote a fixed maximal $\Gamma$-periodic flat and $\mathbb{Z}^r < \Gamma$ a free abelian subgroup of rank $r$ which leaves that flat invariant. We quote the following result of Pettet-Souto.

Theorem 3: \emph{The stabilizer $\operatorname{Stab}_{\mathbb{Z}^r}(\sigma_k)$ of a $k$-simplex in the building $\Delta_{\mathbb{Q}}(M)$ is a free abelian group of rank $\leq r - k - 1$.}

Remark: Heuristically, the stabilizer spans a torus which is complementary to the $(k + 1)$-dimensional $\mathbb{Q}$-split torus corresponding to $\sigma_k$, and together they span a torus of rank $\operatorname{rank}_{\mathbb{Z}}(\operatorname{Stab}_{\mathbb{Z}^r}(\sigma_k)) + k + 1 \leq r$.

Theorem 4: Let $\mathbb{Z}^r < \Gamma$ be a free abelian subgroup leaving invariant a maximal $\Gamma$-periodic flat $\mathbb{R}^r \subset M$. Then, the bundle \[\Delta_{\mathbb{Q}}(M) \to \frac{\mathbb{R}^r \times \Delta_{\mathbb{Q}}(M)}{\mathbb{Z}^r} \to T^r\] does not have a section.

Proof: We will show that $(\mathbb{R}^r \times \Delta_{\mathbb{Q}}(M))/\mathbb{Z}^r$ has no homology in dimension $r$, which implies there can't be a section. (If there were a section $s$, then $s(T^r)$ represents a non-trivial homology class.) The double complex
\[E_{s,t}^0 := C_s(\mathbb{R}^r) \otimes_{\mathbb{Z}^r} C_t(\Delta_{\mathbb{Q}}(M))\]
has the spectral sequence
\[E_{s,t}^1 := H_s(\mathbb{Z}^r; C_t(\Delta_{\mathbb{Q}}(M))) \implies H_{s+t} \left( \frac{\mathbb{R}^r \times \Delta_{\mathbb{Q}}(M)}{\mathbb{Z}^r} \right)\]
associated to it. We will use the fact that $\mathbb{Z}^r$ acts with small stabilizers on the rational Tits building $\Delta_{\mathbb{Q}}(M)$. Note that
\begin{align*}
E_{s,t}^1 &:= H_s(\mathbb{Z}^r; C_t(\Delta_{\mathbb{Q}}(M))) \\
&= H_s(\mathbb{Z}^r; \bigoplus_{\sigma_t} \mathbb{Z}^r \cdot \sigma_t) \\
&= \bigoplus_{\sigma_t} H_s(\mathbb{Z}^r; \mathbb{Z}^r \cdot \sigma_t) \\
&= \bigoplus_{\sigma_t} H_s(\text{Stab}_{\mathbb{Z}^r}(\sigma_t)),
\end{align*}
where the sum is taken over all $\mathbb{Z}^r$-orbits of $t$-simplices. By Theorem 3, $\text{Stab}_{\mathbb{Z}^r}(\sigma_t)$ is a free abelian group of rank $\leq r-t-1$, so we find that $E_{s,t}^1 = 0$ for $s \geq r-t$, i.e. for $s+t \geq r$. Since this spectral sequence converges to the homology of $(\mathbb{R}^r \times \Delta_{\mathbb{Q}}(M))/\mathbb{Z}^r$, we find that
\[H_k \left( \frac{\mathbb{R}^r \times \Delta_{\mathbb{Q}}(M)}{\mathbb{Z}^r} \right) = 0 \quad \text{for} k \geq r. \tag{1} \blacksquare\]

Theorem 5: With the notations above, let $\mathbb{R}^r \subset G/K = \widetilde{M}$ be a maximal $\Gamma$-periodic flat invariant under a free abelian subgroup $\mathbb{Z}^r < \Gamma$. Then, $\mathbb{R}^r$ cannot be $\mathbb{Z}^r$-equivariantly homotoped into the boundary $\partial M$.

Proof: Notice that if the flat $\mathbb{R}^r$ can be $\mathbb{Z}^r$-equivariantly homotoped to the boundary, then we have a $\mathbb{Z}^r$-equivariant map $s : \mathbb{R}^r \to \partial M \to \Delta_{\mathbb{Q}}(M)$. Give the product $\mathbb{R}^r \times \Delta_{\mathbb{Q}}(M)$ the diagonal $\mathbb{Z}^r$-action. By projecting onto the first factor, we get a bundle $(\mathbb{R}^r \times \Delta_{\mathbb{Q}}(M))/\mathbb{Z}^r \to T^r$ over the torus $T^r$ with fibre $\Delta_{\mathbb{Q}}(M)$. The $\mathbb{Z}^r$-equivariant map
\[\mathbb{R}^r \xrightarrow{v \mapsto (v,v)} \mathbb{R}^r \times \mathbb{R}^r \xrightarrow{\text{id} \times s} \mathbb{R}^r \times \Delta_{\mathbb{Q}}(M)\]
gives a section of this bundle. We are done by theorem 4.

Corollary 6: If $(M,g)$ is a complete, Riemannian manifold homeomorphic to a finite volume, aspherical locally symmetric space with no local torus factors, then the group of homotopically trivial isometries $K := \ker(\rho : \operatorname{Isom}(M,g) \to \operatorname{Out}(\pi_1 M))$ is a compact Lie group.

Proof: The group $K$ is a Lie group by the Myers-Steenrod theorem. Let $N \subset \widetilde{M}$ be a collar neighborhood of the boundary. If the group $K$ is not compact, then there is a homotopically trivial isometry $\phi$ which sends $\widetilde{M} \setminus N$ into the collar neighborhood of the boundary $N$. This isometry defines a $\mathbb{Z}^r$-equivariant homotopy of a maximal periodic flat into a neighborhood of the boundary, contradicting Theorem 5. $\blacksquare$

The Borel-Serre boundary

In many settings, it is useful to have compact objects. For manifolds, it is natural to want to know whether they can be compactified by adding a boundary, and often there are different ways of doing this: one can compactify an open disc to a closed disc or a sphere, for instance. Hence, when there are different possibilities, one naturally asks what the different boundaries are good for. For locally symmetric spaces, Borel-Serre found one which is extremely useful, which we briefly describe now.

Fix $ \hat{G}/\mathbb{Q} $ a connected reductive algebraic group, $ K \subset G(\mathbb{R}) $ a maximal compact subgroup, and $\Gamma \subset G(Q)$ an arithmetic subgroup. A nice illustrative example to keep in mind is $ G = SL_2 $, with $ K = SO_2(\mathbb{R}) \subset SL_2(\mathbb{R}) $ and $\Gamma \subset SL_2 Q$ a congruence subgroup.

The centre of $ G $ is an algebraic torus over $ Q $, which contains a greatest $ Q $-split subtorus $ A_G $. Define
\[^0G = \bigcap_{\chi} \ker(\chi^2)\]
where $\chi : G \to G_m$ runs over rational characters of $ G $. This is again a connected reductive group over $ Q $, and the group of real points of $ G $ decomposes as
\[G(\mathbb{R}) = ^0G(\mathbb{R}) \times A_G(\mathbb{R})^+\]
(where the $+$ superscript denotes the topological identity component).

Our symmetric space for $ G $ is
\[D = G(\mathbb{R})/K \cdot A_G(\mathbb{R})^+\]
and our locally symmetric space is
\[X = \Gamma \backslash D = \Gamma \backslash G(\mathbb{R})/K \cdot A_G(\mathbb{R})^+\]

For convenience, we assume $ A_G $ is trivial; this doesn't affect the formation of our locally symmetric space because $ G(\mathbb{R})/K \cdot A_G(\mathbb{R})^+ = ^0G(\mathbb{R})/K $ (and we can replace $ G $ by $ ^0G $).
In our example of $ SL_2 $, the centre is $\pm 1$ so contains no non-trivial torus, and our symmetric spaces are
\[D = SL_2(\mathbb{R})/SO_2(\mathbb{R}) = \mathfrak{H} \quad \text{and} \quad X = \Gamma \backslash SL_2(\mathbb{R})/SO_2(\mathbb{R}),\] the classical upper half-plane and modular curves.

The Borel-Serre compactification of $ X $ is obtained by adjoining boundary components to $ D $ to get a ``partial compactification'', and then quotienting by $ \Gamma $ to get a true compactification of $ X $. The boundary components we add to $ D $ correspond to parabolic subgroups of $ G $.

Let $ P $ be a proper rational parabolic subgroup of $ G $. Then we have a Levi decomposition
\[P = U_P \rtimes L_P\]
where $ U_P $ is the unipotent radical of $ P $ and $ L_P $ is a Levi subgroup. As above we have $ L_P(\mathbb{R}) = ^0L_P(\mathbb{R}) \times A_P(\mathbb{R})^+ $ where $ A_P $ is the largest $ Q $-split torus in the centre of $ L_P $. Thus
\[P(\mathbb{R}) = U_P(\mathbb{R}) \cdot ^0L_P(\mathbb{R}) \cdot A_P(\mathbb{R})^+\]

From the Iwasawa decomposition of $ G $ one can see that $ P $ acts transitively on $ D $, which is to say
\[D = G(\mathbb{R})/K = P(\mathbb{R})/K_P\]
(where $ K_P = K \cap P(\mathbb{R}) $). Now we can define a right action of $ A_P(\mathbb{R})^+ $ on $ D $ by $(g \cdot K_P)a = ga \cdot K_P $ ($ g \in P(\mathbb{R}) $ and $ a \in A_P(\mathbb{R})^+ $), which is well defined because $ A_P(\mathbb{R})^+ $ commutes with $ K_P \subset ^0L(\mathbb{R}) $.

Define the Borel-Serre boundary component associated to $ P $ to be $ e_P = D / A_P(\mathbb{R})^+ $, and the Borel-Serre compactification $ D^{\text{BS}} $ to be (as a set) the disjoint union of $ D $ and $ e_P $ for each proper rational parabolic $ P $.

Borel-Serre manage to topologise $D^{\text{BS}}$ such that the action of $ G(\mathbb{Q}) $ on $ D $ extends to an action by homeomorphisms on $ D^{\text{BS}} $.

Homology of $\partial \widetilde{\overline{M}}/\mathbb{Z}^r$

For future use, we rephrase the above computation in terms of the homology of the Borel-Serre boundary. We have homotopy equivalences
\[\partial \widetilde{\overline{M}}/\mathbb{Z}^r \leftarrow (\mathbb{R}^r \times \partial \widetilde{\overline{M}})/\mathbb{Z}^r \rightarrow (\mathbb{R}^r \times \Delta_{\mathbb{Q}}(M))/\mathbb{Z}^r\]
The left map is a homotopy equivalence because it is the projection map of a bundle with fibre $\mathbb{R}^r$. The right map is the obvious homotopy equivalence obtained from the $\mathbb{Z}^r$-equivariant homotopy equivalence $\partial \widetilde{\overline{M}} \rightarrow \Delta_{\mathbb{Q}}(M)$. Thus, equation (1) shows that the homology of the Borel-Serre boundary of the $\mathbb{Z}^r$ cover vanishes in dimensions $\geq r$, i.e.:
\[H_k(\partial \widetilde{\overline{M}}/\mathbb{Z}^r) = 0 \quad \text{for } k \geq r.\]
This equation, the long exact homology sequence
\[\cdots \rightarrow H_k(\widetilde{\overline{M}}/\mathbb{Z}^r) \rightarrow H_k(\widetilde{\overline{M}}/\mathbb{Z}^r, \partial \widetilde{\overline{M}}/\mathbb{Z}^r) \rightarrow H_{k-1}(\partial \widetilde{\overline{M}}/\mathbb{Z}^r) \rightarrow \cdots\]
and the fact that $\widetilde{\overline{M}}/\mathbb{Z}^r$ has no homology above dimension $r$ (it is homotopy equivalent to the $r$-torus) implies that:
\[H_k(\widetilde{\overline{M}}/\mathbb{Z}^r, \partial \widetilde{\overline{M}}/\mathbb{Z}^r) = 0 \quad \text{for } k > r.\]
It is useful to rewrite this in terms of homology with local coefficients in the $\Gamma$-module $\Gamma/\mathbb{Z}^r$:
\[H_k(\partial \overline{M}; [\Gamma/\mathbb{Z}^r]) = 0 \quad \text{for } k \geq r, \tag{2}\]
\[H_k(\overline{M}, \partial \overline{M}; [\Gamma/\mathbb{Z}^r]) = 0 \quad \text{for } k > r. \tag{3}\]

Homology theories

Here we briefly recall different homology theories and defer the explanation of why they are needed to later.

Homology with closed supports

Given a $\Gamma$-cover $\widetilde{X} \to X$, denote by $C^{\text{cl}}_{*}(X; V)$ the complex of chains with \emph{closed} support on $X$ and coefficients in the $[\Gamma]$-module $V$. More precisely, first define $C^{\text{cl}}_{*}(X; [\Gamma])$ as the complex of those chains on the cover $\widetilde{X}$ that for every compact set $K \subseteq \widetilde{X}$ meet only finitely many $\Gamma$-translates of $K$. Then, define $C^{\text{cl}}_{*}(X; V) := C^{\text{cl}}_{*}(X; [\Gamma]) \otimes_{[\Gamma]} V$.

If $M$ is the interior of a compact manifold with boundary, $\partial M \times (0, \infty)$ is an open neighborhood of the boundary, and $M_0 := M \setminus (\partial M \times (0, \infty))$ its complement, then the relative homology of the pair $(M_0, \partial M_0)$ is isomorphic to homology with closed supports on $M$ via:
\begin{align}
H_{*}(M_0, \partial M_0; V) &\cong H^{\text{cl}}_{*}(M; V), \tag{4} \\
(c, \partial c) &\mapsto c \cup_{\partial c} \partial c \times [0, \infty).
\end{align}
One way to see this is to note that the map is Poincaré dual to the cohomology isomorphism $H^{m-*}(M; O \otimes V) \cong H^{m-*}(M_0, O \otimes V)$, where $O$ is the orientation module.

Homology of ends

The complex $C^{\text{end}}_{*}(X; V)$ of \emph{chains on the end of $X$} is defined to be the quotient:
\[0 \to C_{*+1}(X; V) \to C^{\text{cl}}_{*+1}(X; V) \to C^{\text{end}}_{*}(X; V) \to 0\]
Denote the homology of this complex by $H^{\text{end}}_{*}(X; V) := H_{*}(C^{\text{end}}_{*}(X; V))$. Associated to the short exact sequence of chain complexes above there is a long exact homology sequence:
\[\cdots \to H^{\text{end}}_{*+1}(X; V) \to H^{\text{end}}_{*}(X; V) \to H_{*}(X; V) \to \cdots.\]
Putting together the long exact homology sequence of the pair $(M_0, \partial M_0)$ with the long exact homology sequence we have just written above, we get a commutative diagram:
\[\begin{array}{ccccccccc}
\cdots & \to & H_{*+1}(M_0, \partial M_0; V) & \to & H_{*}(\partial M_0; V) & \to & H_{*}(M_0; V) & \to & \cdots \\
& & \downarrow & & \downarrow & & \downarrow & & \\
\cdots & \to & H^{\text{cl}}_{*+1}(M; V) & \to & H^{\text{end}}_{*}(M; V) & \to & H_{*}(M; V) & \to & \cdots
\end{array}\]
The left vertical arrow is an isomorphism by (4) and the right vertical arrow is an isomorphism because (ordinary) homology is homotopy invariant, so the middle arrow is an isomorphism:
\begin{align}
H_{*}(\partial M_0; V) &\cong H^{\text{end}}_{*}(M; V), \tag{5} \\
a &\mapsto a \times [0, \infty).
\end{align}

Putting together (2), (3), (4) and (5) we get:
\begin{align}
H_k^{\text{end}}(M; [\Gamma/\mathbb{Z}^r]) &= 0 \quad \text{for } k \geq r, \tag{6} \\
H_k^{\text{cl}}(M; [\Gamma/\mathbb{Z}^r]) &= 0 \quad \text{for } k > r. \tag{7}
\end{align}

Smith theory, homology, and cohomology with coefficients in a $\mathbb{Z}/p[\pi_1 M]$-module

In this subsection we recall the Smith theory we need to count the fixed set of a homotopically trivial $\mathbb{Z}/p$ (or $S^1$)-action on an ambient space.

Smith theory: Let $G$ be a $p$-group, and $X$ be a finite $G$-CW complex that is a mod $p$ cohomology $n$-sphere. Then, $X^G$ is either empty or a cohomology $m$-sphere for some $m \leq n$. If $p$ odd, then we have that $n - m$ is even, and if further $n$ is even, $X^G$ is non-empty.

This has been generalised to Bredon cohomology. For an exposition see these notes.

Next, we strengthen this to show that the inclusion of the fixed point set $F' \subset M'$ is a $\mathbb{Z}/p$-homology equivalence in every cover $M' \to M$.

Theorem 7: Let $M$ be an $m$-dimensional smooth aspherical manifold and $\widetilde{M}$ its universal cover. Suppose we have a smooth $(\mathbb{Z}/p)^n$-action on $\widetilde{M}$ that commutes with the action of $\pi_1 M$ by covering translations. Let $F \subset M$ be the fixed point set of the projected $(\mathbb{Z}/p)^n$-action on $M$. Let $V$ be a $\mathbb{Z}/p[\pi_1 M]$-module. Then, the inclusion of the fixed point set induces isomorphisms on homology and cohomology with coefficients in $V$, i.e. \[H_*(F; V) \cong H_*(M; V), \tag{8} \] \[H^*(M; V) \cong H^*(F; V) \tag{9}.\]

If there is a smooth $(S^1)^n$-action on $\widetilde{M}$ that commutes with $\Gamma$ then the same isomorphisms are true with coefficients in any $\mathbb{R}[\Gamma]$-module $V$.

We prove this for $\mathbb{Z}/p$ coefficients only. The proof for $\mathbb{R}$ coefficients is analogous, since the fact we use that the fixed set of a $(\mathbb{Z}/p)^n$-action ($(S^1)^n$-action) on the contractible manifold $M$ is $\mathbb{Z}/p$–acyclic ($\mathbb{R}$-acyclic).

Proof: The proof uses the following exact sequence of chain complexes:
\[0 \rightarrow C_*(\widetilde{F}; \mathbb{Z}/p) \rightarrow C_*(\widetilde{M}; \mathbb{Z}/p) \rightarrow C_*(\widetilde{M}, \widetilde{F}; \mathbb{Z}/p) \rightarrow 0\]
is an exact sequence of complexes of free $\mathbb{Z}/p[\pi_1 M]$-modules, and the complex $C_*(\widetilde{M}, \widetilde{F}; \mathbb{Z}/p)$ is acyclic.

The sequence is exact by definition.
It is a sequence of complexes of $\mathbb{Z}/p[\pi_1 M]$-modules because the $\mathbb{Z}/p$-action commutes with the $\pi_1 M$-action.
For each $k$-simplex $\sigma_k$ in the base $M$, pick a lift $\tilde{\sigma}_k$ in the universal cover $\widetilde{M}$. Then,
\begin{align*}C_k(\widetilde{F}; \mathbb{Z}/p) &\cong \oplus_{\sigma_k \in F} \mathbb{Z}/p[\pi_1 M]\tilde{\sigma}_k, \\ C_k(\widetilde{M}; \mathbb{Z}/p) &\cong \oplus_{\sigma_k} \mathbb{Z}/p[\pi_1 M]\tilde{\sigma}_k, \\C_k(\widetilde{M}, \widetilde{F}; \mathbb{Z}/p) &\cong \oplus_{\sigma_k \notin F} \mathbb{Z}/p[\pi_1 M]\tilde{\sigma}_k, \end{align*} show that the $\mathbb{Z}/p[\pi_1 M]$-modules in the short exact sequence are all free.
Smith theory shows that the complex $C_*(\widetilde{M}, \widetilde{F}; \mathbb{Z}/p)$ is acyclic.

Recall that homology and cohomology with coefficients in $V$ are computed by
\begin{align}
H_*(-; V) &= H_*(C_*(-; \mathbb{Z}/p) \otimes_{\mathbb{Z}/p[\pi_1 M]} V), \\
H^*(-; V) &= H_*(\text{Hom}_{\mathbb{Z}/p[\pi_1 M]}(C_*(-; \mathbb{Z}/p), V)).
\end{align}

Since the $\mathbb{Z}/p[\pi_1 M]$-modules appearing in the first short exact sequence are all free, the functor $- \otimes_{\mathbb{Z}/p[\pi_1 M]} V$ preserves exactness of the sequence and acyclicity of the relative complex $C_*(\widetilde{M}, \widetilde{F}; \mathbb{Z}/p)$, so the long exact homology sequence shows that $H_*(F; V) \rightarrow H_*(M; V)$ is an isomorphism. The same remarks for the functor $\text{Hom}_{\mathbb{Z}/p[\pi_1 M]}(-, V)$ give the cohomology isomorphism. $\square$

Corollary 8: If the fixed point set has dimension $f$, then we have an isomorphism of homology with closed supports:\[H^{\text{cl}}_{*+m-f}(M;V) \cong H^{\text{cl}}_{*}(F;V). \tag{10}\]

Geometrically, the isomorphism is obtained by sending a cycle $c$ on $M$ to the transverse intersection $c \cap F$ on $F$.

Proof: Denote by $\mathcal{O}_{M}$ and $\mathcal{O}_{F}$ the orientation $[\Gamma]$-modules on $M$ and $F$, respectively. If $\kappa = \mathbb{F}_{2}$, then these modules are trivial, so they are equal. Else, then the normal bundle of the fixed point set (to an $(S^{1})^{n}$-action or $(\mathbb{Z}/p)^{n}$-action for an odd prime $p$) is orientable, so again the orientation modules are equal. Combining this with the Poincaré duality isomorphisms,
\begin{align}
H^{\text{cl}}_{*}(M;V) &\cong H^{m-*}(M;\mathcal{O}_{M} \otimes V), \\
H^{\text{cl}}_{*}(F;V) &\cong H^{f-*}(F;\mathcal{O}_{F} \otimes V),
\end{align}
and the cohomology isomorphism (9) proves (10).

Proof of Theorem 1

We have seen that the group $K = \ker(\rho)$ of homotopically trivial isometries is a compact Lie group (Corollary 6). Thus, to show this group is trivial we only need to check that there are no elements of prime order $p$. In other words, we need to show that the locally symmetric space $M$ has no non-trivial, homotopically trivial $\mathbb{Z}/p$-actions. We lift the $\mathbb{Z}/p$-action to the cover $\widetilde{M}/\mathbb{Z}^r$, look at the fixed point set ``near the boundary'' of this cover, and use (2) and (3) (with $\kappa = \mathbb{F}_p$) to show the fixed point set is everything.

The $\mathbb{Z}/p$-action may not extend to the Borel–Serre boundary of $M$, so we need to replace homology of the boundary and homology relative to the boundary by homology of the end and homology with closed supports, respectively.

Proof: Eberlein proved that the fundamental groups of locally symmetric spaces with no local torus factors are centreless. Thus, by Proposition 2, the homotopically trivial $\mathbb{Z}/p$-action on $M$ lifts to a $\mathbb{Z}/p$-action on $\widetilde{M}$ that commutes with the covering action of the fundamental group $\Gamma$. This lets one apply the Smith theory mentioned above.

Let $F \subset M$ be the fixed set of $\mathbb{Z}/p$ acting on $M$. Denote by $f$ and $m$ the dimensions of $F$ and $M$, respectively. Our goal is to show that $f = m$, i.e., that the fixed point set has the same dimension as the manifold $M$. Since the fixed point set is a closed submanifold of $M$, and $M$ is connected, this will show that it is everything, i.e., that the $\mathbb{Z}/p$-action is trivial.

In this proof we will use $\kappa = \mathbb{F}_{p}$ coefficients. The inclusion $F \hookrightarrow M$ induces maps on homology, homology with closed supports and homology of the end with coefficients in $V := \mathbb{F}_{p}[\Gamma/\mathbb{Z}^r]$. These fit together in the commutative diagram

whose rows are long exact homology sequences. The top row is computed using (6) and the fact that $H_r(M; \mathbb{F}_p[\Gamma/\mathbb{Z}^r]) = H_r(\widetilde{M}/\mathbb{Z}^r; \mathbb{F}_p) = \mathbb{F}_p$. The vertical isomorphisms are the Smith theory isomorphisms (8) and (10). If $m > f$ then equation (7) shows that the last term is 0, hence $\psi$ is onto. This is a contradiction, since the commutative diagram shows that $\psi$ is the zero map. Thus, $m = f$, which means the fixed point set is the entire manifold $M$, i.e., the homotopically trivial $\mathbb{Z}/p$-action is actually trivial. This completes the proof of Theorem 1.

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