Identifying the space of geodesics on the hyperbolic plane as a topological space.
$begingroup$
On pg. 2 of this PDF, the author defines $mathcal G=(-infty, 0)times(0, 1)$ and mentions that $mathcal G$ can be thought of as "a space of geodesics on the hyperbolic $2$-space $mathbb H^2$."
One of the things that is confusing me is the use of the indefinite article "a" in the phrase "a space of geodesics..."
The other thing is that how are we thinking of the collection of geodesics as a topological space?
Can somebody please spell out the details here, and please feel free to offer an insights.
hyperbolic-geometry continued-fractions
asked Dec 22 '18 at 7:10
caffeinemachinecaffeinemachine
6,62721354
$endgroup$
add a comment |
$begingroup$
On pg. 2 of this PDF, the author defines $mathcal G=(-infty, 0)times(0, 1)$ and mentions that $mathcal G$ can be thought of as "a space of geodesics on the hyperbolic $2$-space $mathbb H^2$."
One of the things that is confusing me is the use of the indefinite article "a" in the phrase "a space of geodesics..."
The other thing is that how are we thinking of the collection of geodesics as a topological space?
Can somebody please spell out the details here, and please feel free to offer an insights.
hyperbolic-geometry continued-fractions
asked Dec 22 '18 at 7:10
caffeinemachinecaffeinemachine
6,62721354
$endgroup$
add a comment |
$begingroup$
On pg. 2 of this PDF, the author defines $mathcal G=(-infty, 0)times(0, 1)$ and mentions that $mathcal G$ can be thought of as "a space of geodesics on the hyperbolic $2$-space $mathbb H^2$."
One of the things that is confusing me is the use of the indefinite article "a" in the phrase "a space of geodesics..."
The other thing is that how are we thinking of the collection of geodesics as a topological space?
Can somebody please spell out the details here, and please feel free to offer an insights.
hyperbolic-geometry continued-fractions
asked Dec 22 '18 at 7:10
caffeinemachinecaffeinemachine
6,62721354
$endgroup$
On pg. 2 of this PDF, the author defines $mathcal G=(-infty, 0)times(0, 1)$ and mentions that $mathcal G$ can be thought of as "a space of geodesics on the hyperbolic $2$-space $mathbb H^2$."
One of the things that is confusing me is the use of the indefinite article "a" in the phrase "a space of geodesics..."
The other thing is that how are we thinking of the collection of geodesics as a topological space?
Can somebody please spell out the details here, and please feel free to offer an insights.
hyperbolic-geometry continued-fractions
hyperbolic-geometry continued-fractions
asked Dec 22 '18 at 7:10
caffeinemachinecaffeinemachine
6,62721354
asked Dec 22 '18 at 7:10
caffeinemachinecaffeinemachine
6,62721354
asked Dec 22 '18 at 7:10
caffeinemachinecaffeinemachine
6,62721354
asked Dec 22 '18 at 7:10
caffeinemachinecaffeinemachine
6,62721354
asked Dec 22 '18 at 7:10
caffeinemachinecaffeinemachine
6,62721354
6,62721354
add a comment |
add a comment |
1 Answer
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$begingroup$
What this means, I believe, is that it is a subspace of the full space of geodesics. In the upper half plane model $mathbb H$, the boundary is
$$partial mathbb H = mathbb R cup {infty}
$$
which is topologized as the one point compactification of $mathbb R$, and is therefore homeomorphic to the circle $S^1$.
The full space of (oriented) geodesics can be identified with the set of ordered pairs
$${(x,y) in partial mathbb H times partial mathbb H mid x ne y }
$$
where $x$ is the initial ideal endpoint of the geodesic and $y$ is the terminal ideal endpoint. That space can be topologized as a subset of the torus $partial mathbb H times partial mathbb H approx S^1 times S^1$ with the product topology.
In that passage, the author seems to be thinking of $(-infty,0) times (0,1)$ very literally as a subset of $partial mathbb H times partial mathbb H$, with the subspace topology --- which, of course, is also the ordinary product topology on $(-infty,0) times (0,1)$.
So, the reason that it is "a" space of geodesics is simply because it is "a" subset of the full space of geodesics.
answered Dec 26 '18 at 5:08
Lee MosherLee Mosher
50.1k33787
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
What this means, I believe, is that it is a subspace of the full space of geodesics. In the upper half plane model $mathbb H$, the boundary is
$$partial mathbb H = mathbb R cup {infty}
$$
which is topologized as the one point compactification of $mathbb R$, and is therefore homeomorphic to the circle $S^1$.
The full space of (oriented) geodesics can be identified with the set of ordered pairs
$${(x,y) in partial mathbb H times partial mathbb H mid x ne y }
$$
where $x$ is the initial ideal endpoint of the geodesic and $y$ is the terminal ideal endpoint. That space can be topologized as a subset of the torus $partial mathbb H times partial mathbb H approx S^1 times S^1$ with the product topology.
In that passage, the author seems to be thinking of $(-infty,0) times (0,1)$ very literally as a subset of $partial mathbb H times partial mathbb H$, with the subspace topology --- which, of course, is also the ordinary product topology on $(-infty,0) times (0,1)$.
So, the reason that it is "a" space of geodesics is simply because it is "a" subset of the full space of geodesics.
answered Dec 26 '18 at 5:08
Lee MosherLee Mosher
50.1k33787
$endgroup$
add a comment |
$begingroup$
What this means, I believe, is that it is a subspace of the full space of geodesics. In the upper half plane model $mathbb H$, the boundary is
$$partial mathbb H = mathbb R cup {infty}
$$
which is topologized as the one point compactification of $mathbb R$, and is therefore homeomorphic to the circle $S^1$.
The full space of (oriented) geodesics can be identified with the set of ordered pairs
$${(x,y) in partial mathbb H times partial mathbb H mid x ne y }
$$
where $x$ is the initial ideal endpoint of the geodesic and $y$ is the terminal ideal endpoint. That space can be topologized as a subset of the torus $partial mathbb H times partial mathbb H approx S^1 times S^1$ with the product topology.
In that passage, the author seems to be thinking of $(-infty,0) times (0,1)$ very literally as a subset of $partial mathbb H times partial mathbb H$, with the subspace topology --- which, of course, is also the ordinary product topology on $(-infty,0) times (0,1)$.
So, the reason that it is "a" space of geodesics is simply because it is "a" subset of the full space of geodesics.
answered Dec 26 '18 at 5:08
Lee MosherLee Mosher
50.1k33787
$endgroup$
add a comment |
$begingroup$
What this means, I believe, is that it is a subspace of the full space of geodesics. In the upper half plane model $mathbb H$, the boundary is
$$partial mathbb H = mathbb R cup {infty}
$$
which is topologized as the one point compactification of $mathbb R$, and is therefore homeomorphic to the circle $S^1$.
The full space of (oriented) geodesics can be identified with the set of ordered pairs
$${(x,y) in partial mathbb H times partial mathbb H mid x ne y }
$$
where $x$ is the initial ideal endpoint of the geodesic and $y$ is the terminal ideal endpoint. That space can be topologized as a subset of the torus $partial mathbb H times partial mathbb H approx S^1 times S^1$ with the product topology.
In that passage, the author seems to be thinking of $(-infty,0) times (0,1)$ very literally as a subset of $partial mathbb H times partial mathbb H$, with the subspace topology --- which, of course, is also the ordinary product topology on $(-infty,0) times (0,1)$.
So, the reason that it is "a" space of geodesics is simply because it is "a" subset of the full space of geodesics.
answered Dec 26 '18 at 5:08
Lee MosherLee Mosher
50.1k33787
$endgroup$
What this means, I believe, is that it is a subspace of the full space of geodesics. In the upper half plane model $mathbb H$, the boundary is
$$partial mathbb H = mathbb R cup {infty}
$$
which is topologized as the one point compactification of $mathbb R$, and is therefore homeomorphic to the circle $S^1$.
The full space of (oriented) geodesics can be identified with the set of ordered pairs
$${(x,y) in partial mathbb H times partial mathbb H mid x ne y }
$$
where $x$ is the initial ideal endpoint of the geodesic and $y$ is the terminal ideal endpoint. That space can be topologized as a subset of the torus $partial mathbb H times partial mathbb H approx S^1 times S^1$ with the product topology.
In that passage, the author seems to be thinking of $(-infty,0) times (0,1)$ very literally as a subset of $partial mathbb H times partial mathbb H$, with the subspace topology --- which, of course, is also the ordinary product topology on $(-infty,0) times (0,1)$.
So, the reason that it is "a" space of geodesics is simply because it is "a" subset of the full space of geodesics.
answered Dec 26 '18 at 5:08
Lee MosherLee Mosher
50.1k33787
answered Dec 26 '18 at 5:08
Lee MosherLee Mosher
50.1k33787
answered Dec 26 '18 at 5:08
Lee MosherLee Mosher
50.1k33787
answered Dec 26 '18 at 5:08
Lee MosherLee Mosher
50.1k33787
50.1k33787
add a comment |
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