Space of Riemannian metrics as a topological groupoid

Question

I'm reading these notes on groupoids and I'm struggling with example 1.4. I recall the relevant definitions below.

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Definition: A groupoid $\mathcal{G}$ is a small category in which every arrow is invertible.

I will write the same character $\mathcal{G}$ for the set of morphisms and write $M$ for the set of objects, which I will call the base space of the groupoid. Given a groupoid $\mathcal{G}$, we have natural maps $u,s,t,i,m$ where

• $u: M \to \mathcal{G}: x \mapsto 1_x$ where $1_x:x \to x$ is the identity morphism.
• $s,t : \mathcal{G} \to M$ where $s$ sends an arrow to its source and $t$ to its target.
• $i: \mathcal{G} \to \mathcal{G}$ which sends an arrow to its inverse.
• $m:G_2 \to \mathcal{G}$ is the composition of arrows, defined on $G_2 = \{(h,g) \in \mathcal{G}^2: s(h) = t(g)\}$ by $m(h,g) = hg$.

Definition: A topological groupoid is a groupoid $\mathcal{G}$ with base space $M$ such that $\mathcal{G}$ and $M$ are topological spaces, $s,t,u,i,m$ are continuous and additionally $s$ and $t$ are open.

Fix a smooth manifold $N$. We can consider the groupoid whose objects are Riemannian metrics on $N$ which has an arrow from $g_1$ to $g_2$ if and only if there is a diffeomorphism $\phi: N \to N$ such that $g_2 = \phi_* g_1$.

It is claimed in the notes that the compact-open topologies on the space of metrics and diffeomorphisms respectively induce a topology on $\mathcal{G}$ such that $\mathcal{G}$ is a topological groupoid.

Definition: Given topological spaces $X,Y$, the compact-open topology on the space of continuous maps $C(X,Y)$ has subbase given by sets of the form $$V(K,U) = \{ f \in C(X,Y): f(K) \subseteq U\}$$ where $K$ is compact and $U$ is open.

It is not clear to me how the compact-open topology on $\operatorname{Diff}(N)$ induces a topology on $\mathcal{G}$. It is clear that for every $\phi \in \operatorname{Diff}(N)$ and $g \in M$, we have an arrow $\phi_g \in \mathcal{G}$ from $g$ to $\phi_*g$ and in fact all arrows are of this form.

How does the compact-open topology induce a topology on $\mathcal{G}$ and how do we see that this makes $(\mathcal{G},M)$ a topological groupoid?

Answer 1

Here is one way to make this thing work. Consider the product space $$ P= Riem(N)\times Diff(N)\times Riem(N), $$ where $Riem(N)=M$ is the space of Riemannian metrics on $M$ equipped with the compact-open topology (we think of Riemannian metrics as maps from $M$ to $S^2TM$, the bundle of symmetric 2-tensors on $M$). An isometry $\phi: (N,g_1)\to (N,g_2)$ then is regarded as a triple $(g_1, \phi, g_2)\in P$. Hence, we obtain a topology on the space of morphisms $Mor({\mathcal G})$ in your category as a subspace topology of $P$. The projections $$ s: (g_1, \phi, g_2)\mapsto g_1, t: (g_1, \phi, g_2)\mapsto g_2 $$ are clearly continuous on $P$ and, hence, on $Mor({\mathcal G})$. Let's check that they are also open. For the map $t$:

Fix $\phi_0\in Diff(N)$ and use the fact that there is a continuous section of $t$,
$$ \sigma_{\phi_0}: g_2\mapsto (\phi_0^*(g_2), \phi_0, g_2), Riem(N)\to Mor({\mathcal G}), $$ Continuity of this map follows from continuity of the pull-back map $$ \phi_0^*: Riem(N)\to Riem(N), $$ which, in turn, is a local "vector calculus" computation.

Then, to verify that $t$ sends a neighborhood $U$ of $p=(g_1^0, \phi_0, g_2^0)$ to a neighborhood of $t(p)$ observe that $$ \sigma_{\phi_0}^{-1}(U) $$ is open and contains $g_2^0= t(p)$.

The proof for $s$ is similar, just use $\phi_0^{-1}$ instead of $\phi_0$. Continuity of $u$ is clear. Continuity of $m$ and $i$ follows from continuity of the composition and inversion maps on $C(N,N)$ (which is the only nontrivial part of the whole story).