Graduate Quotient maps, group action, open maps

[PsychonautQQ]

This is not homework, it's self study material. I would rather post it here than where questions are usually posted (homework help section) because i think it's much more likely to be seen here by somebody with knowledge on the subject.

Let G be a topological group acting continuously on a topological space X. Show that the quotient map p: X--> X/G is open.

So G is acting continuously on X... so let's say m : G x X ---> X is a map given by (g,x) = gx.

Since m is continuous, this means if that U is open in X then m^-1(U) is open in GxX, this means in particular that for each g in G, the map x --> gx is continuous from x to itself. (is this all true?)

Anyway, the question here is to show that the quotient map p: X ---> X/G is open. X/G is the orbit space of the action of G on X, where x~y iff there is some g s.t. m(g,x)=y.

Let R be an open neighborhood of X. Then I'm trying to think of what p(R) would look like... Am i on the right track here?

 

[fresh_42]

To me this reduces itself to the question whether $g.U \subseteq X$ is open for $U \subseteq X$ open.

 

[PsychonautQQ]

To me this reduces itself to the question whether $g.U \subseteq X$ is open for $U \subseteq X$ open.

Right. What does g.U look like? Sorry posting from phone

 

[fresh_42]

$g.U$ looks like $\{g.x\,\vert \,x \in U\}$ and my intuition says it doesn't have a chance not to be open as it is basically a continuous copy of $U$, but I don't have the right argument at hand. Probably something with $g^{-1}.[(g.x)]=[(g^{-1} \cdot g)].x = x$ together with the fact that group multiplication and inversion are also continuous.

 

[Infrared]

$g.U$ looks like $\{g.x\,\vert \,x \in U\}$ and my intuition says it doesn't have a chance not to be open as it is basically a continuous copy of $U$, but I don't have the right argument at hand. Probably something with $g^{-1}.[(g.x)]=[(g^{-1} \cdot g)].x = x$ together with the fact that group multiplication and inversion are also continuous.

I'm not sure what to think of this- you basically have the proof and yet say that you can't think of it!

The map $l_g: U\to gU$ given by left multiplication by $g$ is continuous with a continuous inverse given by $l_{g^{-1}}$ and hence is a homeomorphism.

 

[fresh_42]

I'm not sure what to think of this- you basically have the proof and yet say that you can't think of it!

The map $l_g: U\to gU$ given by left multiplication by $g$ is continuous with a continuous inverse given by $l_{g^{-1}}$ and hence is a homeomorphism.

Yes, I recognized it while writing. It was my last thought about it. Before that trick with the inverse I was caught in the mantra: open and continuous are not the same. I'm notoriously cautious when it comes to topology and as long as I haven't checked whether all steps are waterproof ... There are too many absurdities around in topology and I generally avoid intuition in this (mine) field.

 

[WWGD]

This is not homework, it's self study material. I would rather post it here than where questions are usually posted (homework help section) because i think it's much more likely to be seen here by somebody with knowledge on the subject.
So G is acting continuously on X... so let's say m : G x X ---> X is a map given by (g,x) = gx.

Since m is continuous, this means if that U is open in X then m^-1(U) is open in GxX, this means in particular that for each g in G, the map x --> gx is continuous from x to itself. (is this all true?)

I am not sure I understand your quantification, but, yes, a continuous function is continuous at each point in the most general point-set-topological sense: $f: X \rightarrow Y$ is continuous at $x \in X ; f(x)=y$ if for every open 'hood $V_y$ of $y$ there is an open 'hood $W_x$ of $x$ with $f(W_x) \subset V_y$. And, yes, if $f: G \times X \rightarrow X$ is continuous , then for every $U$ open in $Y$, $f^{-1}(U)$ is open in $G \times X$.

 

[WWGD]

Right. What does g.U look like? Sorry posting from phone

It depends on the definition of the particular group action you are working with. Otherwise, the question is too general; I don't see how to give a description that would cover all, or some (non-trivial) cases.

 

[WWGD]

$g.U$ looks like $\{g.x\,\vert \,x \in U\}$ and my intuition says it doesn't have a chance not to be open as it is basically a continuous copy of $U$, but I don't have the right argument at hand. Probably something with $g^{-1}.[(g.x)]=[(g^{-1} \cdot g)].x = x$ together with the fact that group multiplication and inversion are also continuous.

Let $f(x)=x^2 : \mathbb R \rightarrow \mathbb R$ . Then $(-1,1)$ is open in the Reals; $f(-1,1)=[0,1)$ is a continuous copy of $(-1,1)$...

EDIT: As Infrared points out, given $f:G \times X \rightarrow X$ is continuous, inversion is continuous ( I am not sure if we need G to be a Lie group to guarantee this ) then $f^{-1}:=g^{-1}x=g^{-1}(g.x):= (g^{-1}g)x=x$. But we also need the action to be bijective, I don't know if this is guaranteed by the conditions of the problem. EDIT2: Per the definition of Topological Group I have seen, yes, product and inversion are _defined_ to be continuous.

 

[fresh_42]

Let $f(x)=x^2 : \mathbb R \rightarrow \mathbb R$ . Then $(-1,1)$ is open in the Reals; $f(-1,1)=[0,1)$ is a continuous copy of $(-1,1)$...

I know that open and continuous are two different properties. That's why I didn't say $g.U$ is automatically open, but the group action saves the day.

EDIT: As Infrared points out, given $f:G \times X \rightarrow X$ is continuous, inversion is continuous ( I am not sure if we need G to be a Lie group to guarantee this ) then $f^{-1}:=g^{-1}x=g^{-1}(g.x):= (g^{-1}g)x=x$. But we also need the action to be bijective, I don't know if this is guaranteed by the conditions of the problem.

We only need the action to be a group action, i.e. $g.(h.x) = (gh).x$ and $1.x=x$ and we don't need a Lie group, but a topological group, for otherwise we could lose continuity in some of the steps. Bijective is the wrong term here, as $G$ and $X$ are different sets and the operations as elements of $GL(X)$ are automatically bijective. I don't see that a free operation (only $1$ fixes points) would be needed either, as the inversion takes places in the group. E.g. if we define the trivial operation $G.X = \operatorname{id}_X$ with $g.x=x$ for all $g\in G, x\in X$ we would still have $g.U$ open.

 

[WWGD]

I know that open and continuous are two different properties. That's why I didn't say $g.U$ is automatically open, but the group action saves the day.

We only need the action to be a group action, i.e. $g.(h.x) = (gh).x$ and $1.x=x$ and we don't need a Lie group, but a topological group, for otherwise we could lose continuity in some of the steps. Bijective is the wrong term here, as $G$ and $X$ are different sets and the operations as elements of $GL(X)$ are automatically bijective. I don't see that a free operation (only $1$ fixes points) would be needed either, as the inversion takes places in the group. E.g. if we define the trivial operation $G.X = \operatorname{id}_X$ with $g.x=x$ for all $g\in G, x\in X$ we would still have $g.U$ open.

Ah, yes, I meant injective into X EDIT Is a group action necessarily an element of GL(X)? EDIT for the map to be a homeomorphism. Although this seems to require for X to be (homeomorphic to a ) product space, which cannot happen, e.g., for $X=\mathbb R^{2n+1}$. But maybe I am confusing something here since group actions are not my specialty, EDIT2: Isn't every Lie group a topological group? My confusion lies in that , e.g., $G \times X$ is topologically a product space, while $X$ may not be. This is why I have trouble seeing how this is a homeomorphism.

 

[fresh_42]

Ah, yes, I meant injective into $X$, i.e., $gx \neq g'x ; g \neq g'$ Is a group action necessarily an element of GL(X)?

Yes, it is by definition (unless explicitly ruled out and otherwise stated) a representation (on $X\,$), aka a group homomorphism to $GL(X)$, aka an operation $G \times X \longrightarrow X$, aka an action $x \mapsto g.x$ with $g.h.x = (gh).x$ and $1.x=x$. Four names for the same thing, only different in which property is emphasized. Even injectivity isn't needed. The argument still holds in the case $g.x = x$ for all $g,x$. The group property guarantees that we can reverse the action, not the action itself. Only the categorial conditions of $\operatorname{Top}$ are needed, continuity, and that $G$ is a topological group. But as I said before, in topology I'm never certain until I've seen all the small steps: $L_g$ and $L_{g^{-1}}$ are continuous bijections on $X$ (always), ergo homeomorphisms and thus open: $(L_{g^{-1}})^{-1}(U)=L_g(U)=g.U\,.$

__________
Shhh... We have a little corner here on PF where we are allowed to consume opium, i.e. no smoothness required, no Lie groups present, no manifolds around. Don't tell the others!

 

[WWGD]

Yes, it is by definition (unless explicitly ruled out and otherwise stated) a representation (on $X\,$), aka a group homomorphism to $GL(X)$, aka an operation $G \times X \longrightarrow X$, aka an action $x \mapsto g.x$ with $g.h.x = (gh).x$ and $1.x=x$. Four names for the same thing, only different in which property is emphasized. Even injectivity isn't needed. The argument still holds in the case $g.x = x$ for all $g,x$. The group property guarantees that we can reverse the action, not the action itself. Only the categorial conditions of $\operatorname{Top}$ are needed, continuity, and that $G$ is a topological group. But as I said before, in topology I'm never certain until I've seen all the small steps: $L_g$ and $L_{g^{-1}}$ are continuous bijections on $X$ (always), ergo homeomorphisms and thus open: $(L_{g^{-1}})^{-1}(U)=L_g(U)=g.U\,.$__________
Shhh... We have a little corner here on PF where we are allowed to consume opium, i.e. no smoothness required, no Lie groups present, no manifolds around. Don't tell the others!

Yes, you see, but what confuses me is that the homeomorphism is not, AFAIK, from X to itself, but from $G \times X \rightarrow X$. But X may not always be a product space, so it cannot be homeo. to $G \times X$. I don't see what I am missing. Since G is a topological group, $G \times X$ is given the product topology, right? So, if we were to select , say $X= \mathbb R$, we know it is not a product space, so I don't see how the map can be a homeomorphism. I just don't see what I am missing here.

 

[WWGD]

Ah, Duh myself, the map I was referring to, I just realized the map is from G to itself using multiplication by a _fixed_ element. I was imagine a bizarre map no one ever brought up, only in my mind. Sorry, I made up a whole problem in my mind. Also, my apologies to PsychonautQQ for trashing his post. Maybe we can move it or delete my previous. I would do it but then maybe other posts look weird/awkward.

 

[fresh_42]

As I understand a continuous action, all $L_g \, : \, X \longrightarrow X$ with $L_g(x)=g.x$ have to be continuous. $g \mapsto L_g$ is probably also continuous. See, if it were Zariski, we wouldn't have to bother. But I think it's true anyways. I haven't thought about the product case, though. Homeomorphisms are certainly not around here. Interesting question. Maybe it has an easy answer, too, but it's not needed.

Edit: I don't think we should delete them. It's a vital example on how scientific discussions are. Remember your struggle, when you tried to convince me that Banach-Tarski cannot be done by a homeomorphism? Oh dear, 90% of the thread would have had to be deleted for my stubbornness. I still can't believe how long it took me to see, that the second ball couldn't be sucked through a single point ... This little detour is nothing in comparison. I'm sure @PsychonautQQ understands this.

 

[Infrared]

$L_g$ and $L_{g^{-1}}$ are continuous bijections on $X$ (always), ergo homeomorphisms

I'm sure you're aware of this, but I don't want the OP to read this as "bijective continuous maps are homeomorphisms", which is false.

$g \mapsto L_g$ is probably also continuous.

What topology are you putting on the space of functions from $G$ to itself?

 

[fresh_42]

I'm sure you're aware of this, but I don't want the OP to read this as "bijective continuous maps are homeomorphisms", which is false.

In this case, my little topology book would be wrong: $f$ is homeomorph, if it is bijective and $f$ and $f^{-1}$ are continuous. Makes sense, as it has to be a bijection as being an isomorphism in the category of topological spaces, and both, the function and its inverse must be morphisms, i.e. continuous.

What do you understand under a homeomorphism?

What topology are you putting on the space of functions from $G$ to itself?

I have no functions from $G$ to itself, except the group operations which are continuous by definition of $G$. The question is, what topology has $\operatorname{Iso}(X)$, the isomorphisms of $X$. How about the product topology defined via the projections? Of course if the example allowed it, I would prefer a Zariski topology for functions. I even think that it is a topological group via a duality argument, but I'm not certain.

 

[Infrared]

In this case, my little topology book would be wrong: $f$ is homeomorph, if it is bijective and $f$ and $f^{-1}$ are continuous. Makes sense, as it has to be a bijection as being an isomorphism in the category of topological spaces, and both, the function and its inverse must be morphisms, i.e. continuous.

My point was just that continuity+bijectivity doesn't imply that the inverse is continuous.

I have no functions from $G$ to itself, except the group operations which are continuous by definition of $G$. The question is, what topology has $\operatorname{Iso}(X)$, the isomorphisms of $X$. How about the product topology defined via the projections? Of course if the example allows it, I would prefer a Zariski topology for functions. I even think that it is a topological group via a duality argument, but I'm not certain.

Okay, if I understand what you're saying, you're viewing $Iso(X)$ as a subset of $Fun(X,X)$ which you identify with $\prod_{x\in X} X$ and getting a topology this way. Is this correct?

 

[fresh_42]

My point was just that continuity+bijectivity doesn't imply that the inverse is continuous.

Yes, of course, but in a case where $(L_g)^{-1}=L_{g^{-1}}$ I thought it's quite obvious that the continuity of all $L_g$ automatically provides the continuity of all inverses.

 

[lavinia]

By the definition of a quotient topology, a set in the quotient space $X/G$ is open if and only if its inverse image in $X$ is open in $X$.

Let $U$ be open in $X$. The inverse image of its projection into $X/G$ is its orbit under the action of the group $G$. But $G$ acts by homeomorphisms so the orbit of $U$ is open. Therefore the projection of $U$ into $X/G$ is open.

 

[Andres316]

I'm not sure what to think of this- you basically have the proof and yet say that you can't think of it!

The map $l_g: U\to gU$ given by left multiplication by $g$ is continuous with a continuous inverse given by $l_{g^{-1}}$ and hence is a homeomorphism.

Why does the last thing you say happen?

 

[fresh_42]

Why does the last thing you say happen?

Which part do you mean?

The map $l_g: U\to gU$ given by left multiplication by $g$ is continuous ...

by definition of an action, other mappings aren't considered an action in this context

... with a continuous inverse given by $l_{g^{-1}}$...

by definition of an action: $g.(h.(u))=(g \cdot h).u$ and $1.u=u$ ...

... and hence is a homeomorphism.

Homeomorphisms $\varphi$ are the isomorphisms in the category of topological spaces, i.e. $\varphi$ is bijective, and $\varphi\, , \,\varphi^{-1}$ are both morphisms in this category, i.e. continuous. Homeomorphisms map open sets one-to-one on open sets.

 

[Andres316]

Could you help me with other questions I have in algebraic topology?

 

[fresh_42]

Could you help me with other questions I have in algebraic topology?

Just create a new thread, either in the linear algebra section or the topology section, probably the latter, i.e. in the forum here. I'm not sure whether I can help you, because things here can quickly become non trivial, however, we have members who are really specialists in algebraic topology.

 

[Andres316]

What are these specialist members? Are they willing to help?

 

[WWGD]

Yes, please go ahead and pose the question in that forum.

 

[Andres316]

In the topology and analysis forum? I already raised it and deleted it because it did not belong to that topic, I do not understand why.

 

[WWGD]

You deleted it yourself or a mod deleted it? If you deleted it, re-post it and see what happens.

 

[Andres316]

A mod deleted it :(

 

[WWGD]

A mod deleted it :(

How about trying the general math forum and see what happens?