Mixing
Space of Riemannian metrics as a topological groupoid
$begingroup$
I'm reading these notes on groupoids and I'm struggling with example 1.4. I recall the relevant definitions below.
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?
general-topology differential-geometry riemannian-geometry groupoids
asked Jan 28 at 11:58
Rhys SteeleRhys Steele
7,4251930
$endgroup$
add a comment |
$begingroup$
I'm reading these notes on groupoids and I'm struggling with example 1.4. I recall the relevant definitions below.
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?
general-topology differential-geometry riemannian-geometry groupoids
asked Jan 28 at 11:58
Rhys SteeleRhys Steele
7,4251930
$endgroup$
$begingroup$
Indeed, it is not immediate that the composition is continuous with respect to the c-o topology. It is not hard though and holds for general Hausdorff locally compact topological spaces. One place where it is proven is Ratcliffe's book "Foundations of Hyperbolic Manifolds". I think he also proves that the inversion is continuous as well but I am not sure.
$endgroup$
– Moishe Kohan
Jan 28 at 13:49
$begingroup$
@MoisheCohen Do you happen to know where in Ratcliffe's book this is written? The closest I can see is the claim that this a "basic property of the compact open topology ... when $X$ is locally compact" (theorem 5.2.2. in the second edition). Also as far I understand, I'm not actually dealing with the compact open topology on the space of diffeomorphisms directly since the space of morphisms contains more than one arrow for each diffeomorphism. Is this wrong?
$endgroup$
– Rhys Steele
Jan 28 at 14:00
$begingroup$
I've found a satisfactory reference for continuity of composition on this site here. Unfortunately it is still unclear exactly how I am viewing my space of morphisms as having the compact-open topology.
$endgroup$
– Rhys Steele
Jan 28 at 14:05
$begingroup$
Local compactness is required for the underlying topological space, not for its group of self-homeomorphisms. Unless you are thinking about infinite-dimensional manifolds $N$, it will work for you.
$endgroup$
– Moishe Kohan
Jan 28 at 16:07
$begingroup$
@MoisheCohen I think you misunderstand my problem. I can see that composition is continuous say from $operatorname{Diff}(N) times operatorname{Diff}(N) to operatorname{Diff}(N)$ when everything is given the c-o topology, this is no issue. The problem is that my space of arrows is not $operatorname{Diff}(N)$ and it is not clear to me how the c-o topology on $operatorname{Diff}(N)$ is inducing a topology on my space of arrows, as claimed in the notes I link to.
$endgroup$
– Rhys Steele
Jan 28 at 17:28
add a comment |
$begingroup$
I'm reading these notes on groupoids and I'm struggling with example 1.4. I recall the relevant definitions below.
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?
general-topology differential-geometry riemannian-geometry groupoids
asked Jan 28 at 11:58
Rhys SteeleRhys Steele
7,4251930
$endgroup$
I'm reading these notes on groupoids and I'm struggling with example 1.4. I recall the relevant definitions below.
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?
general-topology differential-geometry riemannian-geometry groupoids
general-topology differential-geometry riemannian-geometry groupoids
asked Jan 28 at 11:58
Rhys SteeleRhys Steele
7,4251930
asked Jan 28 at 11:58
Rhys SteeleRhys Steele
7,4251930
asked Jan 28 at 11:58
Rhys SteeleRhys Steele
7,4251930
asked Jan 28 at 11:58
Rhys SteeleRhys Steele
7,4251930
asked Jan 28 at 11:58
Rhys SteeleRhys Steele
7,4251930
7,4251930
$begingroup$
Indeed, it is not immediate that the composition is continuous with respect to the c-o topology. It is not hard though and holds for general Hausdorff locally compact topological spaces. One place where it is proven is Ratcliffe's book "Foundations of Hyperbolic Manifolds". I think he also proves that the inversion is continuous as well but I am not sure.
$endgroup$
– Moishe Kohan
Jan 28 at 13:49
$begingroup$
@MoisheCohen Do you happen to know where in Ratcliffe's book this is written? The closest I can see is the claim that this a "basic property of the compact open topology ... when $X$ is locally compact" (theorem 5.2.2. in the second edition). Also as far I understand, I'm not actually dealing with the compact open topology on the space of diffeomorphisms directly since the space of morphisms contains more than one arrow for each diffeomorphism. Is this wrong?
$endgroup$
– Rhys Steele
Jan 28 at 14:00
$begingroup$
I've found a satisfactory reference for continuity of composition on this site here. Unfortunately it is still unclear exactly how I am viewing my space of morphisms as having the compact-open topology.
$endgroup$
– Rhys Steele
Jan 28 at 14:05
$begingroup$
Local compactness is required for the underlying topological space, not for its group of self-homeomorphisms. Unless you are thinking about infinite-dimensional manifolds $N$, it will work for you.
$endgroup$
– Moishe Kohan
Jan 28 at 16:07
$begingroup$
@MoisheCohen I think you misunderstand my problem. I can see that composition is continuous say from $operatorname{Diff}(N) times operatorname{Diff}(N) to operatorname{Diff}(N)$ when everything is given the c-o topology, this is no issue. The problem is that my space of arrows is not $operatorname{Diff}(N)$ and it is not clear to me how the c-o topology on $operatorname{Diff}(N)$ is inducing a topology on my space of arrows, as claimed in the notes I link to.
$endgroup$
– Rhys Steele
Jan 28 at 17:28
add a comment |
$begingroup$
Indeed, it is not immediate that the composition is continuous with respect to the c-o topology. It is not hard though and holds for general Hausdorff locally compact topological spaces. One place where it is proven is Ratcliffe's book "Foundations of Hyperbolic Manifolds". I think he also proves that the inversion is continuous as well but I am not sure.
$endgroup$
– Moishe Kohan
Jan 28 at 13:49
$begingroup$
@MoisheCohen Do you happen to know where in Ratcliffe's book this is written? The closest I can see is the claim that this a "basic property of the compact open topology ... when $X$ is locally compact" (theorem 5.2.2. in the second edition). Also as far I understand, I'm not actually dealing with the compact open topology on the space of diffeomorphisms directly since the space of morphisms contains more than one arrow for each diffeomorphism. Is this wrong?
$endgroup$
– Rhys Steele
Jan 28 at 14:00
$begingroup$
I've found a satisfactory reference for continuity of composition on this site here. Unfortunately it is still unclear exactly how I am viewing my space of morphisms as having the compact-open topology.
$endgroup$
– Rhys Steele
Jan 28 at 14:05
$begingroup$
Local compactness is required for the underlying topological space, not for its group of self-homeomorphisms. Unless you are thinking about infinite-dimensional manifolds $N$, it will work for you.
$endgroup$
– Moishe Kohan
Jan 28 at 16:07
$begingroup$
@MoisheCohen I think you misunderstand my problem. I can see that composition is continuous say from $operatorname{Diff}(N) times operatorname{Diff}(N) to operatorname{Diff}(N)$ when everything is given the c-o topology, this is no issue. The problem is that my space of arrows is not $operatorname{Diff}(N)$ and it is not clear to me how the c-o topology on $operatorname{Diff}(N)$ is inducing a topology on my space of arrows, as claimed in the notes I link to.
$endgroup$
– Rhys Steele
Jan 28 at 17:28
$begingroup$
Indeed, it is not immediate that the composition is continuous with respect to the c-o topology. It is not hard though and holds for general Hausdorff locally compact topological spaces. One place where it is proven is Ratcliffe's book "Foundations of Hyperbolic Manifolds". I think he also proves that the inversion is continuous as well but I am not sure.
$endgroup$
– Moishe Kohan
Jan 28 at 13:49
$begingroup$
Indeed, it is not immediate that the composition is continuous with respect to the c-o topology. It is not hard though and holds for general Hausdorff locally compact topological spaces. One place where it is proven is Ratcliffe's book "Foundations of Hyperbolic Manifolds". I think he also proves that the inversion is continuous as well but I am not sure.
$endgroup$
– Moishe Kohan
Jan 28 at 13:49
$begingroup$
@MoisheCohen Do you happen to know where in Ratcliffe's book this is written? The closest I can see is the claim that this a "basic property of the compact open topology ... when $X$ is locally compact" (theorem 5.2.2. in the second edition). Also as far I understand, I'm not actually dealing with the compact open topology on the space of diffeomorphisms directly since the space of morphisms contains more than one arrow for each diffeomorphism. Is this wrong?
$endgroup$
– Rhys Steele
Jan 28 at 14:00
$begingroup$
@MoisheCohen Do you happen to know where in Ratcliffe's book this is written? The closest I can see is the claim that this a "basic property of the compact open topology ... when $X$ is locally compact" (theorem 5.2.2. in the second edition). Also as far I understand, I'm not actually dealing with the compact open topology on the space of diffeomorphisms directly since the space of morphisms contains more than one arrow for each diffeomorphism. Is this wrong?
$endgroup$
– Rhys Steele
Jan 28 at 14:00
$begingroup$
I've found a satisfactory reference for continuity of composition on this site here. Unfortunately it is still unclear exactly how I am viewing my space of morphisms as having the compact-open topology.
$endgroup$
– Rhys Steele
Jan 28 at 14:05
$begingroup$
I've found a satisfactory reference for continuity of composition on this site here. Unfortunately it is still unclear exactly how I am viewing my space of morphisms as having the compact-open topology.
$endgroup$
– Rhys Steele
Jan 28 at 14:05
$begingroup$
Local compactness is required for the underlying topological space, not for its group of self-homeomorphisms. Unless you are thinking about infinite-dimensional manifolds $N$, it will work for you.
$endgroup$
– Moishe Kohan
Jan 28 at 16:07
$begingroup$
Local compactness is required for the underlying topological space, not for its group of self-homeomorphisms. Unless you are thinking about infinite-dimensional manifolds $N$, it will work for you.
$endgroup$
– Moishe Kohan
Jan 28 at 16:07
$begingroup$
@MoisheCohen I think you misunderstand my problem. I can see that composition is continuous say from $operatorname{Diff}(N) times operatorname{Diff}(N) to operatorname{Diff}(N)$ when everything is given the c-o topology, this is no issue. The problem is that my space of arrows is not $operatorname{Diff}(N)$ and it is not clear to me how the c-o topology on $operatorname{Diff}(N)$ is inducing a topology on my space of arrows, as claimed in the notes I link to.
$endgroup$
– Rhys Steele
Jan 28 at 17:28
$begingroup$
@MoisheCohen I think you misunderstand my problem. I can see that composition is continuous say from $operatorname{Diff}(N) times operatorname{Diff}(N) to operatorname{Diff}(N)$ when everything is given the c-o topology, this is no issue. The problem is that my space of arrows is not $operatorname{Diff}(N)$ and it is not clear to me how the c-o topology on $operatorname{Diff}(N)$ is inducing a topology on my space of arrows, as claimed in the notes I link to.
$endgroup$
– Rhys Steele
Jan 28 at 17:28
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $phi_0in Diff(N)$ and use the fact that there is a continuous section of $t$,
$$
sigma_{phi_0}: g_2mapsto (phi_0^*(g_2), phi_0, g_2), Riem(N)to Mor({mathcal G}),
$$
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).
answered Jan 28 at 21:50
Moishe KohanMoishe Kohan
48.1k344110
$endgroup$
add a comment |
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1 Answer
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active
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votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $phi_0in Diff(N)$ and use the fact that there is a continuous section of $t$,
$$
sigma_{phi_0}: g_2mapsto (phi_0^*(g_2), phi_0, g_2), Riem(N)to Mor({mathcal G}),
$$
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).
answered Jan 28 at 21:50
Moishe KohanMoishe Kohan
48.1k344110
$endgroup$
add a comment |
$begingroup$
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 $phi_0in Diff(N)$ and use the fact that there is a continuous section of $t$,
$$
sigma_{phi_0}: g_2mapsto (phi_0^*(g_2), phi_0, g_2), Riem(N)to Mor({mathcal G}),
$$
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).
answered Jan 28 at 21:50
Moishe KohanMoishe Kohan
48.1k344110
$endgroup$
add a comment |
$begingroup$
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 $phi_0in Diff(N)$ and use the fact that there is a continuous section of $t$,
$$
sigma_{phi_0}: g_2mapsto (phi_0^*(g_2), phi_0, g_2), Riem(N)to Mor({mathcal G}),
$$
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).
answered Jan 28 at 21:50
Moishe KohanMoishe Kohan
48.1k344110
$endgroup$
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 $phi_0in Diff(N)$ and use the fact that there is a continuous section of $t$,
$$
sigma_{phi_0}: g_2mapsto (phi_0^*(g_2), phi_0, g_2), Riem(N)to Mor({mathcal G}),
$$
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).
answered Jan 28 at 21:50
Moishe KohanMoishe Kohan
48.1k344110
answered Jan 28 at 21:50
Moishe KohanMoishe Kohan
48.1k344110
answered Jan 28 at 21:50
Moishe KohanMoishe Kohan
48.1k344110
answered Jan 28 at 21:50
Moishe KohanMoishe Kohan
48.1k344110
48.1k344110
add a comment |
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$begingroup$
Indeed, it is not immediate that the composition is continuous with respect to the c-o topology. It is not hard though and holds for general Hausdorff locally compact topological spaces. One place where it is proven is Ratcliffe's book "Foundations of Hyperbolic Manifolds". I think he also proves that the inversion is continuous as well but I am not sure.
$endgroup$
– Moishe Kohan
Jan 28 at 13:49
$begingroup$
@MoisheCohen Do you happen to know where in Ratcliffe's book this is written? The closest I can see is the claim that this a "basic property of the compact open topology ... when $X$ is locally compact" (theorem 5.2.2. in the second edition). Also as far I understand, I'm not actually dealing with the compact open topology on the space of diffeomorphisms directly since the space of morphisms contains more than one arrow for each diffeomorphism. Is this wrong?
$endgroup$
– Rhys Steele
Jan 28 at 14:00
$begingroup$
I've found a satisfactory reference for continuity of composition on this site here. Unfortunately it is still unclear exactly how I am viewing my space of morphisms as having the compact-open topology.
$endgroup$
– Rhys Steele
Jan 28 at 14:05
$begingroup$
Local compactness is required for the underlying topological space, not for its group of self-homeomorphisms. Unless you are thinking about infinite-dimensional manifolds $N$, it will work for you.
$endgroup$
– Moishe Kohan
Jan 28 at 16:07
$begingroup$
@MoisheCohen I think you misunderstand my problem. I can see that composition is continuous say from $operatorname{Diff}(N) times operatorname{Diff}(N) to operatorname{Diff}(N)$ when everything is given the c-o topology, this is no issue. The problem is that my space of arrows is not $operatorname{Diff}(N)$ and it is not clear to me how the c-o topology on $operatorname{Diff}(N)$ is inducing a topology on my space of arrows, as claimed in the notes I link to.
$endgroup$
– Rhys Steele
Jan 28 at 17:28