Why does the covering transformation group act properly discontinuously on the fiber?
$begingroup$
In Step 2, I don't understand the part "for otherwise, two points would belong to the same orbit and the restriction of $pi$ to $U_alpha$ would not be injective". What two points? If $gne e$ and $U_alphacap gU_alphane emptyset$, then $yin U_alphacap gU_alpha$. And we also know $xin U_alpha$. Either $xne y$ or $x=y$ (and it may happen that $x$ is the only point in the intersection). But in either case, I don't understand why the result follows. In the former case, why must $y$ lie in the same orbit as $x$? In the latter case, if $x$ is the only point in the intersection, where does the second point come from?
general-topology group-theory algebraic-geometry group-actions covering-spaces
asked Dec 5 '18 at 15:38
user531587user531587
261113
$endgroup$
add a comment |
$begingroup$
In Step 2, I don't understand the part "for otherwise, two points would belong to the same orbit and the restriction of $pi$ to $U_alpha$ would not be injective". What two points? If $gne e$ and $U_alphacap gU_alphane emptyset$, then $yin U_alphacap gU_alpha$. And we also know $xin U_alpha$. Either $xne y$ or $x=y$ (and it may happen that $x$ is the only point in the intersection). But in either case, I don't understand why the result follows. In the former case, why must $y$ lie in the same orbit as $x$? In the latter case, if $x$ is the only point in the intersection, where does the second point come from?
general-topology group-theory algebraic-geometry group-actions covering-spaces
asked Dec 5 '18 at 15:38
user531587user531587
261113
$endgroup$
$begingroup$
I've just found this: math.stackexchange.com/questions/1715615/…
$endgroup$
– user531587
Dec 5 '18 at 18:01
add a comment |
$begingroup$
In Step 2, I don't understand the part "for otherwise, two points would belong to the same orbit and the restriction of $pi$ to $U_alpha$ would not be injective". What two points? If $gne e$ and $U_alphacap gU_alphane emptyset$, then $yin U_alphacap gU_alpha$. And we also know $xin U_alpha$. Either $xne y$ or $x=y$ (and it may happen that $x$ is the only point in the intersection). But in either case, I don't understand why the result follows. In the former case, why must $y$ lie in the same orbit as $x$? In the latter case, if $x$ is the only point in the intersection, where does the second point come from?
general-topology group-theory algebraic-geometry group-actions covering-spaces
asked Dec 5 '18 at 15:38
user531587user531587
261113
$endgroup$
In Step 2, I don't understand the part "for otherwise, two points would belong to the same orbit and the restriction of $pi$ to $U_alpha$ would not be injective". What two points? If $gne e$ and $U_alphacap gU_alphane emptyset$, then $yin U_alphacap gU_alpha$. And we also know $xin U_alpha$. Either $xne y$ or $x=y$ (and it may happen that $x$ is the only point in the intersection). But in either case, I don't understand why the result follows. In the former case, why must $y$ lie in the same orbit as $x$? In the latter case, if $x$ is the only point in the intersection, where does the second point come from?
general-topology group-theory algebraic-geometry group-actions covering-spaces
general-topology group-theory algebraic-geometry group-actions covering-spaces
asked Dec 5 '18 at 15:38
user531587user531587
261113
asked Dec 5 '18 at 15:38
user531587user531587
261113
edited Dec 5 '18 at 15:50
user531587
asked Dec 5 '18 at 15:38
user531587user531587
261113
asked Dec 5 '18 at 15:38
user531587user531587
261113
asked Dec 5 '18 at 15:38
user531587user531587
261113
261113
$begingroup$
I've just found this: math.stackexchange.com/questions/1715615/…
$endgroup$
– user531587
Dec 5 '18 at 18:01
add a comment |
$begingroup$
I've just found this: math.stackexchange.com/questions/1715615/…
$endgroup$
– user531587
Dec 5 '18 at 18:01
$begingroup$
I've just found this: math.stackexchange.com/questions/1715615/…
$endgroup$
– user531587
Dec 5 '18 at 18:01
$begingroup$
I've just found this: math.stackexchange.com/questions/1715615/…
$endgroup$
– user531587
Dec 5 '18 at 18:01
add a comment |
1 Answer
1
active
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$begingroup$
Assume that $pi$ is a covering map.
(1) Each $g in G$ is a covering transformation because $g$ is a homeomorphism such that $pi circ g = pi$ (the latter is due to the definition of $X / G$ as the set of orbits under the action of $G$).
(2) No $g in G setminus { e }$ has a fixed point:
We know that a covering transformation is uniquely determined by its value on a single point because $X$ is path connected. Hence $e$ is the only covering transformation having a fixed point.
Now assume $g(U_alpha) cap U_alpha ne emptyset$, where $g ne e$.
So let $y in g(U_alpha) cap U_alpha$. Let $z in U_alpha$ be the unique point such that $g(z) = y$. We have $z ne y$ (if $z = y$, then $g$ would have a fixed point). But then $z,y$ belong to same orbit, i.e. we get $pi(z) = pi(y)$. This means that $pi : U_alpha to V$ is not injective, a contradiction.
Added on request:
What Munkres proves in Step 2 is actually the following:
Let $pi : X to B$ be a covering map with a path connected $X$. Then the group $mathcal{C}(X,pi,B)$ of covering transformations acts properly discontinuously on $X$.
Start with $x in X$ and do the same as above until you get $z, y in U_alpha$. Since $pi circ g = pi$, you get $pi(y) = pi g(z) = pi(z)$ which shows that $pi : U_alpha to V$ is not injective.
Munkres applies this to $pi : X to X/G$. At this point he only uses the fact $G subset mathcal{C}(X,pi,X/G)$ (see (1) above). That $G = mathcal{C}(X,pi,X/G)$ is shown in Step 3.
answered Dec 5 '18 at 17:50
Paul FrostPaul Frost
11.4k3934
$endgroup$
$begingroup$
Why is $z$ unique and is that used at all? Also, does the argument use that $yin U_alpha$?
$endgroup$
– user531587
Dec 5 '18 at 23:09
$begingroup$
It is unique because $g$ is a homeomorphism. But in fact uniqueness is not used, we only need $g(z) = y$. The essential part of the argument is that $z,y in U_alpha$ and $z ne y$.
$endgroup$
– Paul Frost
Dec 5 '18 at 23:15
$begingroup$
Is there a way to easily generalize this to the proof that the deck transformation group of a covering map acts properly discontinuously? In such a setting, the covering map is no longer the map that assigns to a point its orbit, so the given argument on injectivity doesn't work as far as I can see.
$endgroup$
– user531587
Dec 5 '18 at 23:32
$begingroup$
See my above edit.
$endgroup$
– Paul Frost
Dec 6 '18 at 13:54
add a comment |
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1 Answer
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$begingroup$
Assume that $pi$ is a covering map.
(1) Each $g in G$ is a covering transformation because $g$ is a homeomorphism such that $pi circ g = pi$ (the latter is due to the definition of $X / G$ as the set of orbits under the action of $G$).
(2) No $g in G setminus { e }$ has a fixed point:
We know that a covering transformation is uniquely determined by its value on a single point because $X$ is path connected. Hence $e$ is the only covering transformation having a fixed point.
Now assume $g(U_alpha) cap U_alpha ne emptyset$, where $g ne e$.
So let $y in g(U_alpha) cap U_alpha$. Let $z in U_alpha$ be the unique point such that $g(z) = y$. We have $z ne y$ (if $z = y$, then $g$ would have a fixed point). But then $z,y$ belong to same orbit, i.e. we get $pi(z) = pi(y)$. This means that $pi : U_alpha to V$ is not injective, a contradiction.
Added on request:
What Munkres proves in Step 2 is actually the following:
Let $pi : X to B$ be a covering map with a path connected $X$. Then the group $mathcal{C}(X,pi,B)$ of covering transformations acts properly discontinuously on $X$.
Start with $x in X$ and do the same as above until you get $z, y in U_alpha$. Since $pi circ g = pi$, you get $pi(y) = pi g(z) = pi(z)$ which shows that $pi : U_alpha to V$ is not injective.
Munkres applies this to $pi : X to X/G$. At this point he only uses the fact $G subset mathcal{C}(X,pi,X/G)$ (see (1) above). That $G = mathcal{C}(X,pi,X/G)$ is shown in Step 3.
answered Dec 5 '18 at 17:50
Paul FrostPaul Frost
11.4k3934
$endgroup$
$begingroup$
Why is $z$ unique and is that used at all? Also, does the argument use that $yin U_alpha$?
$endgroup$
– user531587
Dec 5 '18 at 23:09
$begingroup$
It is unique because $g$ is a homeomorphism. But in fact uniqueness is not used, we only need $g(z) = y$. The essential part of the argument is that $z,y in U_alpha$ and $z ne y$.
$endgroup$
– Paul Frost
Dec 5 '18 at 23:15
$begingroup$
Is there a way to easily generalize this to the proof that the deck transformation group of a covering map acts properly discontinuously? In such a setting, the covering map is no longer the map that assigns to a point its orbit, so the given argument on injectivity doesn't work as far as I can see.
$endgroup$
– user531587
Dec 5 '18 at 23:32
$begingroup$
See my above edit.
$endgroup$
– Paul Frost
Dec 6 '18 at 13:54
add a comment |
$begingroup$
Assume that $pi$ is a covering map.
(1) Each $g in G$ is a covering transformation because $g$ is a homeomorphism such that $pi circ g = pi$ (the latter is due to the definition of $X / G$ as the set of orbits under the action of $G$).
(2) No $g in G setminus { e }$ has a fixed point:
We know that a covering transformation is uniquely determined by its value on a single point because $X$ is path connected. Hence $e$ is the only covering transformation having a fixed point.
Now assume $g(U_alpha) cap U_alpha ne emptyset$, where $g ne e$.
So let $y in g(U_alpha) cap U_alpha$. Let $z in U_alpha$ be the unique point such that $g(z) = y$. We have $z ne y$ (if $z = y$, then $g$ would have a fixed point). But then $z,y$ belong to same orbit, i.e. we get $pi(z) = pi(y)$. This means that $pi : U_alpha to V$ is not injective, a contradiction.
Added on request:
What Munkres proves in Step 2 is actually the following:
Let $pi : X to B$ be a covering map with a path connected $X$. Then the group $mathcal{C}(X,pi,B)$ of covering transformations acts properly discontinuously on $X$.
Start with $x in X$ and do the same as above until you get $z, y in U_alpha$. Since $pi circ g = pi$, you get $pi(y) = pi g(z) = pi(z)$ which shows that $pi : U_alpha to V$ is not injective.
Munkres applies this to $pi : X to X/G$. At this point he only uses the fact $G subset mathcal{C}(X,pi,X/G)$ (see (1) above). That $G = mathcal{C}(X,pi,X/G)$ is shown in Step 3.
answered Dec 5 '18 at 17:50
Paul FrostPaul Frost
11.4k3934
$endgroup$
$begingroup$
Why is $z$ unique and is that used at all? Also, does the argument use that $yin U_alpha$?
$endgroup$
– user531587
Dec 5 '18 at 23:09
$begingroup$
It is unique because $g$ is a homeomorphism. But in fact uniqueness is not used, we only need $g(z) = y$. The essential part of the argument is that $z,y in U_alpha$ and $z ne y$.
$endgroup$
– Paul Frost
Dec 5 '18 at 23:15
$begingroup$
Is there a way to easily generalize this to the proof that the deck transformation group of a covering map acts properly discontinuously? In such a setting, the covering map is no longer the map that assigns to a point its orbit, so the given argument on injectivity doesn't work as far as I can see.
$endgroup$
– user531587
Dec 5 '18 at 23:32
$begingroup$
See my above edit.
$endgroup$
– Paul Frost
Dec 6 '18 at 13:54
add a comment |
$begingroup$
Assume that $pi$ is a covering map.
(1) Each $g in G$ is a covering transformation because $g$ is a homeomorphism such that $pi circ g = pi$ (the latter is due to the definition of $X / G$ as the set of orbits under the action of $G$).
(2) No $g in G setminus { e }$ has a fixed point:
We know that a covering transformation is uniquely determined by its value on a single point because $X$ is path connected. Hence $e$ is the only covering transformation having a fixed point.
Now assume $g(U_alpha) cap U_alpha ne emptyset$, where $g ne e$.
So let $y in g(U_alpha) cap U_alpha$. Let $z in U_alpha$ be the unique point such that $g(z) = y$. We have $z ne y$ (if $z = y$, then $g$ would have a fixed point). But then $z,y$ belong to same orbit, i.e. we get $pi(z) = pi(y)$. This means that $pi : U_alpha to V$ is not injective, a contradiction.
Added on request:
What Munkres proves in Step 2 is actually the following:
Let $pi : X to B$ be a covering map with a path connected $X$. Then the group $mathcal{C}(X,pi,B)$ of covering transformations acts properly discontinuously on $X$.
Start with $x in X$ and do the same as above until you get $z, y in U_alpha$. Since $pi circ g = pi$, you get $pi(y) = pi g(z) = pi(z)$ which shows that $pi : U_alpha to V$ is not injective.
Munkres applies this to $pi : X to X/G$. At this point he only uses the fact $G subset mathcal{C}(X,pi,X/G)$ (see (1) above). That $G = mathcal{C}(X,pi,X/G)$ is shown in Step 3.
answered Dec 5 '18 at 17:50
Paul FrostPaul Frost
11.4k3934
$endgroup$
Assume that $pi$ is a covering map.
(1) Each $g in G$ is a covering transformation because $g$ is a homeomorphism such that $pi circ g = pi$ (the latter is due to the definition of $X / G$ as the set of orbits under the action of $G$).
(2) No $g in G setminus { e }$ has a fixed point:
We know that a covering transformation is uniquely determined by its value on a single point because $X$ is path connected. Hence $e$ is the only covering transformation having a fixed point.
Now assume $g(U_alpha) cap U_alpha ne emptyset$, where $g ne e$.
So let $y in g(U_alpha) cap U_alpha$. Let $z in U_alpha$ be the unique point such that $g(z) = y$. We have $z ne y$ (if $z = y$, then $g$ would have a fixed point). But then $z,y$ belong to same orbit, i.e. we get $pi(z) = pi(y)$. This means that $pi : U_alpha to V$ is not injective, a contradiction.
Added on request:
What Munkres proves in Step 2 is actually the following:
Let $pi : X to B$ be a covering map with a path connected $X$. Then the group $mathcal{C}(X,pi,B)$ of covering transformations acts properly discontinuously on $X$.
Start with $x in X$ and do the same as above until you get $z, y in U_alpha$. Since $pi circ g = pi$, you get $pi(y) = pi g(z) = pi(z)$ which shows that $pi : U_alpha to V$ is not injective.
Munkres applies this to $pi : X to X/G$. At this point he only uses the fact $G subset mathcal{C}(X,pi,X/G)$ (see (1) above). That $G = mathcal{C}(X,pi,X/G)$ is shown in Step 3.
answered Dec 5 '18 at 17:50
Paul FrostPaul Frost
11.4k3934
edited Dec 6 '18 at 14:00
answered Dec 5 '18 at 17:50
Paul FrostPaul Frost
11.4k3934
answered Dec 5 '18 at 17:50
Paul FrostPaul Frost
11.4k3934
answered Dec 5 '18 at 17:50
Paul FrostPaul Frost
11.4k3934
11.4k3934
$begingroup$
Why is $z$ unique and is that used at all? Also, does the argument use that $yin U_alpha$?
$endgroup$
– user531587
Dec 5 '18 at 23:09
$begingroup$
It is unique because $g$ is a homeomorphism. But in fact uniqueness is not used, we only need $g(z) = y$. The essential part of the argument is that $z,y in U_alpha$ and $z ne y$.
$endgroup$
– Paul Frost
Dec 5 '18 at 23:15
$begingroup$
Is there a way to easily generalize this to the proof that the deck transformation group of a covering map acts properly discontinuously? In such a setting, the covering map is no longer the map that assigns to a point its orbit, so the given argument on injectivity doesn't work as far as I can see.
$endgroup$
– user531587
Dec 5 '18 at 23:32
$begingroup$
See my above edit.
$endgroup$
– Paul Frost
Dec 6 '18 at 13:54
add a comment |
$begingroup$
Why is $z$ unique and is that used at all? Also, does the argument use that $yin U_alpha$?
$endgroup$
– user531587
Dec 5 '18 at 23:09
$begingroup$
It is unique because $g$ is a homeomorphism. But in fact uniqueness is not used, we only need $g(z) = y$. The essential part of the argument is that $z,y in U_alpha$ and $z ne y$.
$endgroup$
– Paul Frost
Dec 5 '18 at 23:15
$begingroup$
Is there a way to easily generalize this to the proof that the deck transformation group of a covering map acts properly discontinuously? In such a setting, the covering map is no longer the map that assigns to a point its orbit, so the given argument on injectivity doesn't work as far as I can see.
$endgroup$
– user531587
Dec 5 '18 at 23:32
$begingroup$
See my above edit.
$endgroup$
– Paul Frost
Dec 6 '18 at 13:54
$begingroup$
Why is $z$ unique and is that used at all? Also, does the argument use that $yin U_alpha$?
$endgroup$
– user531587
Dec 5 '18 at 23:09
$begingroup$
Why is $z$ unique and is that used at all? Also, does the argument use that $yin U_alpha$?
$endgroup$
– user531587
Dec 5 '18 at 23:09
$begingroup$
It is unique because $g$ is a homeomorphism. But in fact uniqueness is not used, we only need $g(z) = y$. The essential part of the argument is that $z,y in U_alpha$ and $z ne y$.
$endgroup$
– Paul Frost
Dec 5 '18 at 23:15
$begingroup$
It is unique because $g$ is a homeomorphism. But in fact uniqueness is not used, we only need $g(z) = y$. The essential part of the argument is that $z,y in U_alpha$ and $z ne y$.
$endgroup$
– Paul Frost
Dec 5 '18 at 23:15
$begingroup$
Is there a way to easily generalize this to the proof that the deck transformation group of a covering map acts properly discontinuously? In such a setting, the covering map is no longer the map that assigns to a point its orbit, so the given argument on injectivity doesn't work as far as I can see.
$endgroup$
– user531587
Dec 5 '18 at 23:32
$begingroup$
Is there a way to easily generalize this to the proof that the deck transformation group of a covering map acts properly discontinuously? In such a setting, the covering map is no longer the map that assigns to a point its orbit, so the given argument on injectivity doesn't work as far as I can see.
$endgroup$
– user531587
Dec 5 '18 at 23:32
$begingroup$
See my above edit.
$endgroup$
– Paul Frost
Dec 6 '18 at 13:54
$begingroup$
See my above edit.
$endgroup$
– Paul Frost
Dec 6 '18 at 13:54
add a comment |
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$begingroup$
I've just found this: math.stackexchange.com/questions/1715615/…
$endgroup$
– user531587
Dec 5 '18 at 18:01