Metrics on symmetric spaces II

April 18, 2026

Metrics on symmetric spaces II

Following on from the previous post, we discuss a second result of Avramidi that makes precise the statement 'the locally symmetric metric on a locally symmetric manifold is the most symmetric metric on that manifold’.

Our setting is this: 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 \[\rho: \rm{Isom}(M, g) \to \rm{Out}(\pi_1(M))\] Last time we proved
Theorem 1: $\rho$ is injective.

We will use this, or more accurately we will quote something we showed along the way, to deduce

Theorem 2: Let $(M, \text{sym})$ be a finite volume, irreducible, locally symmetric manifold of dimension $\geq 3$. Suppose that $M$ has no local torus factors. If $g$ is any complete, finite volume Riemannian metric on $M$, then either:

$g$ is a constant multiple of the locally symmetric metric, or
the isometry group of the universal cover $\text{Isom}(\widetilde{M}, \widetilde{g})$ is discrete and contains $\pi_1 M$ as a subgroup of index $\leq \text{vol}(M, h_{\text{sym}})/\varepsilon(h_{\text{sym}})$.

The constant $\varepsilon(h_{\text{sym}})$ is the volume of the smallest locally symmetric orbifold covered by $(M, h_{\text{sym}})$. It does not depend on the metric $g$. We will give an example to show that the assumption that $g$ has finite volume is necessary.

We first fix some background and notation: throughout this post

$M$ = finite volume, irreducible, locally symmetric manifold of dimension $\geq 3$.
$\Gamma = \pi_{1}(M)$
$I = \operatorname{Isom}(\widetilde{M})$ = the group of isometries of $\widetilde{M}$
$I_{0}$ = the connected component of $I$ containing the identity
$\Gamma_{0} = \Gamma \cap I_{0}$

Here $\widetilde{M}$ is endowed with the unique Riemannian metric for which the covering map $\widetilde{M}\to M$ is a Riemannian covering. Hence $\Gamma$ acts on $\widetilde{M}$ isometrically by deck transformations, giving a natural inclusion $\Gamma\to I$, where $I = \operatorname{Isom}(\widetilde{M})$ is the isometry group of $\widetilde{M}$.

By Myers-Steenrod, $I$ is a Lie group, possibly with infinitely many components. Let $I_{0}$ denote the connected component of the identity of $I$; note that $I_{0}$ is normal in $I$. If $I$ is discrete, then we are done, so suppose that $I$ is not discrete. Myers-Steenrod again then gives that the dimension of $I$ is positive, and so $I_{0}$ is a connected, positive-dimensional Lie group.

We have the following exact sequences:

\[1\longrightarrow I_{0}\longrightarrow I\longrightarrow I/I_{0} \longrightarrow 1 \]

and

\[1\longrightarrow\Gamma_{0}\longrightarrow\Gamma\longrightarrow\Gamma/\Gamma_{0} \longrightarrow 1 \]

We will also use the following standard result

Proposition 3: 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 $A$.

Proof of Theorem 2

The proof breaks up into 4 steps, analogously to that of the Riemannian orbibundle theorem

Step 1: If $\Gamma$ commutes with the compact Lie group $K$, then $K=1$

To show that $K$ is trivial, it suffices to show that it has no elements $\phi$ of prime order $p$. Since $\phi$ commutes with $\Gamma$, it descends to a homotopically trivial $\mathbb{Z}/p$-action on the locally symmetric space $M$. We showed in the proof of Theorem 1 last time that there are no such actions.

Step 2: If $\Gamma$ normalizes the compact Lie group $K$, then $K=1$

First recall that Bestvina-Feighn proved

Theorem 4: Let $G$ be a connected semisimple Lie group, $K \subset G$ a maximal compact subgroup, $G/K$ the associated contractible manifold, and $\Gamma$ a lattice in $G$. If $\Gamma$ acts properly discontinuously on a contractible manifold $W$, then $\rm{dim}W \geq \rm{dim}G/K$.

We claim that there is a contractible manifold $ Z $ of dimension $\dim(\text{Fix}(K)) + 1 <\rm{dim}G/K $ on which $\Gamma$ acts properly discontinuously. This would contradict Bestvina-Feighn, thus $ T $ would be trivial.

We now prove the claim. Note that by Smith theory (see the previous post), Fix$(T)$ is acyclic. We want to replace this $\Gamma$-manifold by another one that is contractible, i.e. we want to kill the fundamental group. If $\text{Fix}(K)/\Gamma$ is compact, then the kernel that we are trying to kill is finitely normally generated, and the construction is standard: one kills the elements of this kernel by surgering the circles, giving rise to new homology $=$ homotopy in dimension 2, which can then be killed by surgering the 2-spheres as well. Taking the product of $\text{Fix}(K)$ with $\mathbf{R}$ if $\dim(\text{Fix}(K)) = 4$, the 2-spheres needed at this stage can be embedded by general position since $\dim(\text{Fix}(K) \times \mathbf{R}) > 4$.

If $\text{Fix}(K)/\Gamma$ is not compact, then one wants to do the same argument, but one has to be careful to make sure that the circles and 2-spheres that one wants to surger do not accumulate. However, by replacing $\text{Fix}(K)$ by $\text{Fix}(K) \times \mathbf{R}$, this problem disappears: one simply does the $i^{\text{th}}$ surgery at the ``height'' $\text{Fix}(K) \times \{i\}$, producing at height $i$ a 2-dimensional homology class. This class can be represented by a sphere which lies in a compact region that is above level $i - 1/2$ (by the Hurewicz theorem) that is embedded (by general position) with trivial normal bundle (by appropriate framing). The infinite collection of constructed 2-spheres do not accumulate, since at most $i$ of them pass through level $i$. Surgering this free basis for the homology of the previous stage gives us back an acyclic manifold, which is now, in addition, simply connected, and, thus, contractible.

Remark: Avramidi suggests two other ways of proving this step: one via Thurston's smearing method, and one which is a bare hands computation using the various homology theories mentioned last time.

Step 3: $\Gamma$ is semisimple with trivial centre and no compact factors

We give a brief indication of how to show this. The connected Lie group $I_{0}$ can be expressed as an extension

\begin{equation}
1\to I_{0}^{\rm sol}\to I_{0}\to I_{0}^{\rm ss}\to 1
\end{equation}

of the maximal connected solvable normal subgroup of $I_{0}$ by a semisimple group $I_{0}^{\rm ss}$. The group $\Gamma_{0}:=\Gamma\cap I_{0}$ also has a unique maximal normal solvable subgroup, denoted $\Gamma_{0}^{\rm sol}$.

Claim: $\Gamma_{0}^{\rm sol}$ is trivial.

Proof: The solvable group $\Gamma_{0}^{\rm sol}$ is a characteristic subgroup of $\Gamma_{0}$, so it is a normal subgroup of $\Gamma$. On the other hand, recall that the fundamental group $\Gamma$ of a locally symmetric space with no local torus factors has no non-trivial abelian normal subgroups, hence no non-trivial normal solvable subgroups. This proves the claim.

Now one uses an argument due to Farb and Weinberger to show that $\Gamma_{0}$ is a lattice in $I_{0}$, and iteratively applies step 2 to kill off compact factors.

Step 4: The group generated by $I_{0}$ and $\Gamma$ is an extension:

\begin{equation}
1\to I_{0}\to\langle I_{0},\Gamma\rangle\xrightarrow{\pi}\Gamma/\Gamma_{0} \to 1
\end{equation}

Since $I_{0}$ is semisimple with trivial centre, its outer automorphism group is finite. Thus, the kernel of the conjugation homomorphism $\Gamma\to\text{Out}(I_{0})$ is a finite index subgroup $\Gamma^{\prime}<\Gamma$. Let $\Gamma^{\prime}_{0}=\Gamma^{\prime}\cap I_{0}$. Proposition 3 gives a projection $\varphi\colon\langle I_{0},\Gamma^{\prime}\rangle\to I_{0}$, which restricts to the identity map on $I_{0}$. One easily checks that the product map

\begin{equation}
\langle I_{0},\Gamma^{\prime}\rangle\xrightarrow{\varphi\times\pi}I_{0}\times \Gamma^{\prime}/\Gamma^{\prime}_{0}
\end{equation}

is an isomorphism that maps $\Gamma^{\prime}$ onto the product $\Gamma^{\prime}_{0}\times\Gamma^{\prime}/\Gamma^{\prime}_{0}$. Now, we use the hypothesis that $\Gamma^{\prime}$ is irreducible to conclude that either $\Gamma^{\prime}_{0}=1$ or $\Gamma^{\prime}_{0}$ is a finite index subgroup of $\Gamma^{\prime}$. We look at the two possibilities.

Case 1: $\Gamma^{\prime}_{0}<\Gamma^{\prime}$ \textit{is a finite index subgroup}. We will show that in this case $(M,g)$ is isometric to a locally symmetric space. Since $\Gamma_{0}$ is a lattice in the Lie group $I_{0}$, the finite index subgroup $\Gamma^{\prime}_{0}$ is also a lattice in $I_{0}$. Denote by $K<I_{0}$ a maximal compact subgroup. Then, the quotient $I_{0}/K$ is a symmetric space with no compact or Euclidean factors. Since $\Gamma^{\prime}_{0}$ is a finite index subgroup of $\Gamma$, Margulis-Mostow-Prasad rigidity implies the locally symmetric space $\Gamma^{\prime}_{0}\setminus I_{0}/K$ is isometric to a finite cover of $(M,h_{\rm sym})$. In particular, the two spaces have the same dimension. On the other hand, an orbit of the group $I_{0}$ acting on $\widetilde{M}$ has the form $I_{0}\cdot x=I_{0}/K_{x}$ for a compact subgroup $K_{x}<I_{0}$. We get

\[\dim\widetilde{M}=\dim I_{0}/K\leq\dim I_{0}/K_{x}\leq\dim\widetilde{M}\]

where all the inequalities are actually equalities. This means the isometry group $I_{0}$ is acting transitively on $\widetilde{M}$, so that $(\widetilde{M},\widetilde{g})$ is a symmetric space and the quotient $(M,g)$ is locally symmetric.

Case 2: $\Gamma^{\prime}_{0}=1$. The group $I_{0}$ is compact, since it contains $\Gamma^{\prime}_{0}$ as a lattice. On the other hand, we have shown that $I_{0}$ has no compact factors, so $I_{0}=1$, ie, the isometry group $I$ is discrete. Since $\Gamma$ is a lattice in $I$, we conclude that it is a finite index subgroup of $I$. Thus, there is a further finite index subgroup $\Gamma^{\prime\prime}<\Gamma$ that is normalized by $I$. Conjugation by $I$ gives a group homomorphism:

\[\rho\colon I/\,\Gamma^{\prime\prime}\to{\rm Out}(\Gamma^{\prime\prime})\]

An element in the kernel of this homomorphism is a homotopically trivial isometry of $(\widetilde{M}/\,\Gamma^{\prime\prime},\widetilde{g})$. By Theorem 1, there are no such homotopically trivial isometries, so the map $\rho$ is injective. By Margulis-Mostow-Prasad rigidity, the image $\rho(I/\,\Gamma^{\prime\prime})$ can be represented by isometries of the locally symmetric metric $h_{\rm sym}$. Thus, the group $L$ of lifts

\begin{equation}
1\to\Gamma^{\prime\prime}\to L\to\rho(I/\,\Gamma^{\prime\prime})\to 1
\end{equation}

can be represented by a discrete group of isometries of the symmetric metric. This extension is determined by $\rho$ since $\Gamma^{\prime\prime}$ has trivial centre (Proposition 3) so the group of lifts $L$ is isomorphic to $I$. We have shown that $I$ is a group of isometries of the symmetric metric. Let $\Gamma$ act on the universal cover $\widetilde{M}$ by isometries via:

\[\Gamma<I\cong L<{\rm Isom}(\widetilde{M},\widetilde{h}_{\rm sym})\]

By Margulis--Mostow--Prasad rigidity, the quotient $ \tilde{M} / \Gamma, \tilde{h}_{\text{sym}} $ is isometric to $ (M, h_{\text{sym}}) $. Consequently,

\[|I / \Gamma| = \frac{\text{vol}(\tilde{M} / \Gamma, \tilde{h}_{\text{sym}})}{\text{vol}(\tilde{M} / I, \tilde{h}_{\text{sym}})} \leq \frac{\text{vol}(M, h_{\text{sym}})}{\epsilon(h_{\text{sym}})}\]

where $ \epsilon(h_{\text{sym}}) $ is the volume of the smallest locally symmetric orbifold covered by $ (M, h_{\text{sym}}) $. This completes the proof of Theorem 2. $\blacksquare$

Remark: If $ (\tilde{M}, \tilde{h}_{\text{sym}}) $ is a symmetric space with no compact or Euclidean factors, then the volume of the smallest locally symmetric orbifold isometrically covered by $ (\tilde{M}, \tilde{h}_{\text{sym}}) $ is bounded below by a positive constant $ \epsilon(\tilde{h}_{\text{sym}}) $ that depends only on the symmetric space. This is a theorem of Kazhdan and Margulis. (I hope to discuss a generalisation due to Gelander soon.) By contrast, the real line covers circles of arbitrarily small volume. Similarly, any symmetric space with a compact or Euclidean factor will support compact quotients of arbitrarily small volume.

Infinite volume case

In this section, we give an example which shows that Theorem 2 does not extend to metrics $ g $ of infinite volume.

Let $ M $ and $ N $ be non-compact aspherical manifolds whose universal cover is diffeomorphic to $ \mathbb{R}^m $. Pick a proper embedded rays $ r_1 : [0, \infty) \to M $ and $ r_2 : [0, \infty) \to N $, cut out tubular neighborhoods of these and glue to get a new manifold $ M \#_r N $ which is a connect sum along the rays. This manifold is homotopy equivalent to the wedge $ M \lor N $, so it is aspherical, and its fundamental group is $ \pi_1 M \star \pi_1 N $. The quotient $ M' := (M \#_r N)/\pi_1 M $ is homotopy equivalent to $ M $ and, in fact, one can check that it is diffeomorphic to $ M $. Thus, $ M $ is an infinite cover of the manifold $ M \#_r N $. Now, pick a complete Riemannian metric $ \widetilde{g} $ on $ M \#_r N $. It lifts to an infinite volume complete Riemannian metric $ g $ on $ M $, which further lifts to a metric $ \widetilde{g} $ on the universal cover whose isometry group Isom$(M, \widetilde{g})$ contains the free product $ \pi_1 M \star \pi_1 N $.

Further results

In fact, Avramidi manages to show more with these ideas. It turns out that for many nice enough manifolds, not just locally-symmetric ones, similar results hold. Before stating the next theorem, where we make this precise, we need a definition.

Definition: A Riemannian manifold is said to have bounded geometry if the sectional curvatures are bounded above and below, and there is some $\epsilon>0$ such that the injectivity radius is $\geq \epsilon$. A manifold $\widetilde{M}$ with a properly discontinuous $\Gamma$-action is $\Gamma$-tameable or $\Gamma$-tame if there is a complete Riemannian bounded geometry metric on $\widetilde{M}$ which is $\Gamma$-invariant and has finite $\Gamma$-covolume.

Theorem 5: Let $ M $ be an aspherical manifold with $\pi_1M$-tame universal cover. Suppose all three of the following conditions hold.

The fundamental group $\pi_1M$ is residually finite.
$M$ is the interior of a compact manifold with boundary $\partial M$.
The Euler characteristic $\chi(M)$ is non-zero.

Then,

Any homotopically trivial periodic diffeomorphism of $M$ is the identity.
$\pi_1M$ does not commute with any non-trivial compact Lie group $K < \text{Diff} \widetilde{M}$. More generally, $\pi_1M$ does not normalize any such $K$.

From this, one can deduce the following:

Theorem 6: Let $ M $ be an aspherical manifold with $\pi_1M$-tame universal cover. Suppose all three of the following conditions hold.

The fundamental group $\pi_1M$ is residually finite.
$M$ is the interior of a compact manifold with boundary $\partial M$.
The Euler characteristic $\chi(M)$ is non-zero.

Pick a complete Riemannian (or more generally Finsler) metric $g$ on $M$. Then the isometry group $\operatorname{Isom}(M,g)$ is isomorphic to a finite subgroup of $\operatorname{Out}(\pi_1M)$.

Proof: The main point is to show that the group Isom$(M,g)$ is compact, even if the metric $g$ has infinite volume. Then we will show that there are no homotopically trivial isometries.

Compactness of the isometry group: Let $M_0 \subset M$ be the complement of an open collar neighborhood of the boundary. We first show that this subset cannot be moved off itself by a homeomorphism because it `contains all of the $L^2$-cohomology'. If a homeomorphism $\phi$ moves $M_0$ completely off itself ($\phi(M_0) \cap M_0 = \emptyset$), then the induced isomorphism $\phi^*$ on $L^2$-homology factors through $M \setminus M_0 \sim \partial M$. That is, we have maps

\[H_{(2)}^*(M; \Gamma) \to H_{(2)}^*(\partial M; \Gamma) \to H_{(2)}^*(\widetilde{M}; \Gamma)\]

and the composition is the isomorphism $\phi^*$. This cannot happen because $\partial M$ is $L^2$-acyclic while $M$ is not.

We've shown that the subset $ M_0 \subset M $ cannot be moved off itself. Thus, a point $ x \in M_0 $ gets moved at most a distance 2 · diameter$ (M_0) =: 2D $ by any isometry of $ (M, g) $. The isometry group Isom$ (M, g) $ is a Lie group acting smoothly on $ M $, and a theorem of Deng and Hou says that the action

\[A : \text{Isom}(M, g) \to M \times M\]
\[(\phi, x) \mapsto (\phi(x), x)\]

is a proper map for any Finsler metric. Thus $ A^{-1}(B_x(2D) \times \{x\}) $ is compact. It contains the closed subset Isom$ (M, g) \times \{x\} $ (because the orbit Isom$ (M, g) \cdot x $ is contained in the closed ball $ B_x(2D) $) which shows that Isom$ (M, g) $ is compact.

No homotopically trivial isometries: Let $ K := \ker(\text{Isom}(M, g) \to \text{Out}(\pi_1 M)) $ be the group of homotopically trivial isometries. It is a closed subgroup of the isometry group so it is also a compact Lie group. Since $ M $ has no periodic homotopically trivial isometries previous theorem, this group is trivial. Thus, Isom$ (M, g) < \text{Out}(\pi_1 M)$ is a compact subgroup of a discrete group. Consequently, it is finite. $\blacksquare$

Similar techniques to those mentioned in this post also prove:

Theorem 7: Let $\widetilde{M}$ be a contractible $\Gamma$-tame manifold. Suppose the action of $\Gamma$ on $\widetilde{M}$ is effective and there is a finite index torsionfree subgroup $\Gamma' < \Gamma$. Suppose, in addition, that the following three conditions hold.

$\Gamma$ is residually finite and irreducible, i.e. all finite index subgroups of the form $A \times B$ satisfy that $A$ or $B$ is finite.
$\widetilde{M}/\Gamma'$ is the interior of a compact manifold with boundary.
The Euler characteristic $\chi(\widetilde{M}/\Gamma')$ is non-zero.

If $\widetilde{g}$ is a complete Finsler metric on $\widetilde{M}$ which is $\Gamma$-invariant and has finite $\Gamma$-covolume then either

$(\widetilde{M},\widetilde{g})$ is isometric to a symmetric space, or
The isometry group Isom$(\widetilde{M},\widetilde{g})$ is discrete and $\Gamma$ is a finite index subgroup. Moreover, there exist inclusions \begin{equation}\Gamma < \text{Isom}(\widetilde{M},\widetilde{g}) < \text{Comm}(\Gamma). \end{equation}

The substance of the last equation is that the isometry group injects into the abstract commensurator. This is a group theoretic analogue of the geometric statement that there are no homotopically trivial isometries in any finite cover of $\widetilde{M}/\Gamma'$. It means that information about the commensurator can give quantitative bounds on the size of the isometry group. In general, the abstract commensurator is difficult to compute, but much is known about it for lattices in symmetric spaces, and it is completely determined for (finite index subgroups of) mapping class groups.

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