Is the metric completion of a Riemannian manifold always a geodesic space?
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A length space is a metric space $X$, where the distance between two points is the infimum of the lengths of curves joining them. The length of a curve $c: [0,1] rightarrow X$ is the sup of
$$ d(c(0), c(t_1)) + d(c(t_1), d(t_2)) + cdots + d(c(t_{N-1}), c(1)) $$
over all $0 < t_1 < t_2cdots < t_{N-1} < 1$ and $N > 0$.
A geodesic space is a length space, where for each $x,y in X$, there is a curve $c$ connecting $x$ to $y$ whose length is equal to $d(x,y)$.
A Riemannian manifold $M$ and its metric completion $overline{M}$ are length spaces. If the Riemannian manifold is complete, then it is a geodesic space.
But is $overline{M}$ necessarily a geodesic space? If not, what is a counterexample?
This was motivated by my flawed answer to Minimizing geodesics in incomplete Riemannian manifolds
Also, note that if $overline{M}$ is locally compact, then it is a geodesic space by the usual proof. One example of $M$, where $overline{M}$ is not locally compact is the universal cover of the punctured flat plane. However, this is still a geodesic space.
dg.differential-geometry mg.metric-geometry riemannian-geometry
asked 9 hours ago
Deane YangDeane Yang
20.1k562141
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add a comment |
$begingroup$
A length space is a metric space $X$, where the distance between two points is the infimum of the lengths of curves joining them. The length of a curve $c: [0,1] rightarrow X$ is the sup of
$$ d(c(0), c(t_1)) + d(c(t_1), d(t_2)) + cdots + d(c(t_{N-1}), c(1)) $$
over all $0 < t_1 < t_2cdots < t_{N-1} < 1$ and $N > 0$.
A geodesic space is a length space, where for each $x,y in X$, there is a curve $c$ connecting $x$ to $y$ whose length is equal to $d(x,y)$.
A Riemannian manifold $M$ and its metric completion $overline{M}$ are length spaces. If the Riemannian manifold is complete, then it is a geodesic space.
But is $overline{M}$ necessarily a geodesic space? If not, what is a counterexample?
This was motivated by my flawed answer to Minimizing geodesics in incomplete Riemannian manifolds
Also, note that if $overline{M}$ is locally compact, then it is a geodesic space by the usual proof. One example of $M$, where $overline{M}$ is not locally compact is the universal cover of the punctured flat plane. However, this is still a geodesic space.
dg.differential-geometry mg.metric-geometry riemannian-geometry
asked 9 hours ago
Deane YangDeane Yang
20.1k562141
$endgroup$
add a comment |
$begingroup$
A length space is a metric space $X$, where the distance between two points is the infimum of the lengths of curves joining them. The length of a curve $c: [0,1] rightarrow X$ is the sup of
$$ d(c(0), c(t_1)) + d(c(t_1), d(t_2)) + cdots + d(c(t_{N-1}), c(1)) $$
over all $0 < t_1 < t_2cdots < t_{N-1} < 1$ and $N > 0$.
A geodesic space is a length space, where for each $x,y in X$, there is a curve $c$ connecting $x$ to $y$ whose length is equal to $d(x,y)$.
A Riemannian manifold $M$ and its metric completion $overline{M}$ are length spaces. If the Riemannian manifold is complete, then it is a geodesic space.
But is $overline{M}$ necessarily a geodesic space? If not, what is a counterexample?
This was motivated by my flawed answer to Minimizing geodesics in incomplete Riemannian manifolds
Also, note that if $overline{M}$ is locally compact, then it is a geodesic space by the usual proof. One example of $M$, where $overline{M}$ is not locally compact is the universal cover of the punctured flat plane. However, this is still a geodesic space.
dg.differential-geometry mg.metric-geometry riemannian-geometry
asked 9 hours ago
Deane YangDeane Yang
20.1k562141
$endgroup$
A length space is a metric space $X$, where the distance between two points is the infimum of the lengths of curves joining them. The length of a curve $c: [0,1] rightarrow X$ is the sup of
$$ d(c(0), c(t_1)) + d(c(t_1), d(t_2)) + cdots + d(c(t_{N-1}), c(1)) $$
over all $0 < t_1 < t_2cdots < t_{N-1} < 1$ and $N > 0$.
A geodesic space is a length space, where for each $x,y in X$, there is a curve $c$ connecting $x$ to $y$ whose length is equal to $d(x,y)$.
A Riemannian manifold $M$ and its metric completion $overline{M}$ are length spaces. If the Riemannian manifold is complete, then it is a geodesic space.
But is $overline{M}$ necessarily a geodesic space? If not, what is a counterexample?
This was motivated by my flawed answer to Minimizing geodesics in incomplete Riemannian manifolds
Also, note that if $overline{M}$ is locally compact, then it is a geodesic space by the usual proof. One example of $M$, where $overline{M}$ is not locally compact is the universal cover of the punctured flat plane. However, this is still a geodesic space.
dg.differential-geometry mg.metric-geometry riemannian-geometry
dg.differential-geometry mg.metric-geometry riemannian-geometry
asked 9 hours ago
Deane YangDeane Yang
20.1k562141
asked 9 hours ago
Deane YangDeane Yang
20.1k562141
asked 9 hours ago
Deane YangDeane Yang
20.1k562141
asked 9 hours ago
Deane YangDeane Yang
20.1k562141
asked 9 hours ago
Deane YangDeane Yang
20.1k562141
20.1k562141
add a comment |
add a comment |
1 Answer
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oldest
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I have been thinking about this since Deane and I discussed it this morning, and I came up with the following idea. Let $Sigma:={1,tfrac{1}{2},tfrac{1}{3},ldots}cup {-1,-tfrac{1}{2},-tfrac{1}{3},ldots}$. The set $Sigmacup {0}$ is closed in $mathbb{R}$.
Let $(M,g)$ be the complement of $[0,1]times (Sigmacup{0})$ in the Euclidean plane. Offhand, it seems to me that the metric completion $overline{M}$ of $(M,g)$ contains the following "extra points":
${0,1}times (Sigmacup{0})$
for each $(t,s)in(0,1)timesSigma$, two points $(t,s)_pm$, coming from the (two different) directional limits $lim_{yto s^pm}(t,y)$.
Importantly, as far as I can tell, there is nothing in $overline{M}$ corresponding to the points in the segment $(0,1)times{0}$.
If that's so, then the distance between the points $(0,0)$ and $(1,0)$ is 1, but there is no curve of distance 1 in $overline{M}$ connecting them.
answered 7 hours ago
macbethmacbeth
1,7781427
$endgroup$
1
$begingroup$
Yes, this does work. It is similar to Ballmann's example mentioned in Benoit's answer here: mathoverflow.net/questions/15592/…
$endgroup$
– Misha
6 hours ago
1
$begingroup$
Oh, that's interesting. Yes, this seems to be a "fattening" of the Ballman example which makes it a Riemannian manifold.
$endgroup$
– macbeth
6 hours ago
1
$begingroup$
Right, topologically speaking, you are taking a neighborhood of Ballmann's example as embedded in $R^2$.
$endgroup$
– Misha
6 hours ago
$begingroup$
@macbeth, thanks!
$endgroup$
– Deane Yang
4 hours ago
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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votes
$begingroup$
I have been thinking about this since Deane and I discussed it this morning, and I came up with the following idea. Let $Sigma:={1,tfrac{1}{2},tfrac{1}{3},ldots}cup {-1,-tfrac{1}{2},-tfrac{1}{3},ldots}$. The set $Sigmacup {0}$ is closed in $mathbb{R}$.
Let $(M,g)$ be the complement of $[0,1]times (Sigmacup{0})$ in the Euclidean plane. Offhand, it seems to me that the metric completion $overline{M}$ of $(M,g)$ contains the following "extra points":
${0,1}times (Sigmacup{0})$
for each $(t,s)in(0,1)timesSigma$, two points $(t,s)_pm$, coming from the (two different) directional limits $lim_{yto s^pm}(t,y)$.
Importantly, as far as I can tell, there is nothing in $overline{M}$ corresponding to the points in the segment $(0,1)times{0}$.
If that's so, then the distance between the points $(0,0)$ and $(1,0)$ is 1, but there is no curve of distance 1 in $overline{M}$ connecting them.
answered 7 hours ago
macbethmacbeth
1,7781427
$endgroup$
1
$begingroup$
Yes, this does work. It is similar to Ballmann's example mentioned in Benoit's answer here: mathoverflow.net/questions/15592/…
$endgroup$
– Misha
6 hours ago
1
$begingroup$
Oh, that's interesting. Yes, this seems to be a "fattening" of the Ballman example which makes it a Riemannian manifold.
$endgroup$
– macbeth
6 hours ago
1
$begingroup$
Right, topologically speaking, you are taking a neighborhood of Ballmann's example as embedded in $R^2$.
$endgroup$
– Misha
6 hours ago
$begingroup$
@macbeth, thanks!
$endgroup$
– Deane Yang
4 hours ago
add a comment |
$begingroup$
I have been thinking about this since Deane and I discussed it this morning, and I came up with the following idea. Let $Sigma:={1,tfrac{1}{2},tfrac{1}{3},ldots}cup {-1,-tfrac{1}{2},-tfrac{1}{3},ldots}$. The set $Sigmacup {0}$ is closed in $mathbb{R}$.
Let $(M,g)$ be the complement of $[0,1]times (Sigmacup{0})$ in the Euclidean plane. Offhand, it seems to me that the metric completion $overline{M}$ of $(M,g)$ contains the following "extra points":
${0,1}times (Sigmacup{0})$
for each $(t,s)in(0,1)timesSigma$, two points $(t,s)_pm$, coming from the (two different) directional limits $lim_{yto s^pm}(t,y)$.
Importantly, as far as I can tell, there is nothing in $overline{M}$ corresponding to the points in the segment $(0,1)times{0}$.
If that's so, then the distance between the points $(0,0)$ and $(1,0)$ is 1, but there is no curve of distance 1 in $overline{M}$ connecting them.
answered 7 hours ago
macbethmacbeth
1,7781427
$endgroup$
1
$begingroup$
Yes, this does work. It is similar to Ballmann's example mentioned in Benoit's answer here: mathoverflow.net/questions/15592/…
$endgroup$
– Misha
6 hours ago
1
$begingroup$
Oh, that's interesting. Yes, this seems to be a "fattening" of the Ballman example which makes it a Riemannian manifold.
$endgroup$
– macbeth
6 hours ago
1
$begingroup$
Right, topologically speaking, you are taking a neighborhood of Ballmann's example as embedded in $R^2$.
$endgroup$
– Misha
6 hours ago
$begingroup$
@macbeth, thanks!
$endgroup$
– Deane Yang
4 hours ago
add a comment |
$begingroup$
I have been thinking about this since Deane and I discussed it this morning, and I came up with the following idea. Let $Sigma:={1,tfrac{1}{2},tfrac{1}{3},ldots}cup {-1,-tfrac{1}{2},-tfrac{1}{3},ldots}$. The set $Sigmacup {0}$ is closed in $mathbb{R}$.
Let $(M,g)$ be the complement of $[0,1]times (Sigmacup{0})$ in the Euclidean plane. Offhand, it seems to me that the metric completion $overline{M}$ of $(M,g)$ contains the following "extra points":
${0,1}times (Sigmacup{0})$
for each $(t,s)in(0,1)timesSigma$, two points $(t,s)_pm$, coming from the (two different) directional limits $lim_{yto s^pm}(t,y)$.
Importantly, as far as I can tell, there is nothing in $overline{M}$ corresponding to the points in the segment $(0,1)times{0}$.
If that's so, then the distance between the points $(0,0)$ and $(1,0)$ is 1, but there is no curve of distance 1 in $overline{M}$ connecting them.
answered 7 hours ago
macbethmacbeth
1,7781427
$endgroup$
I have been thinking about this since Deane and I discussed it this morning, and I came up with the following idea. Let $Sigma:={1,tfrac{1}{2},tfrac{1}{3},ldots}cup {-1,-tfrac{1}{2},-tfrac{1}{3},ldots}$. The set $Sigmacup {0}$ is closed in $mathbb{R}$.
Let $(M,g)$ be the complement of $[0,1]times (Sigmacup{0})$ in the Euclidean plane. Offhand, it seems to me that the metric completion $overline{M}$ of $(M,g)$ contains the following "extra points":
${0,1}times (Sigmacup{0})$
for each $(t,s)in(0,1)timesSigma$, two points $(t,s)_pm$, coming from the (two different) directional limits $lim_{yto s^pm}(t,y)$.
Importantly, as far as I can tell, there is nothing in $overline{M}$ corresponding to the points in the segment $(0,1)times{0}$.
If that's so, then the distance between the points $(0,0)$ and $(1,0)$ is 1, but there is no curve of distance 1 in $overline{M}$ connecting them.
answered 7 hours ago
macbethmacbeth
1,7781427
answered 7 hours ago
macbethmacbeth
1,7781427
answered 7 hours ago
macbethmacbeth
1,7781427
answered 7 hours ago
macbethmacbeth
1,7781427
1,7781427
1
$begingroup$
Yes, this does work. It is similar to Ballmann's example mentioned in Benoit's answer here: mathoverflow.net/questions/15592/…
$endgroup$
– Misha
6 hours ago
1
$begingroup$
Oh, that's interesting. Yes, this seems to be a "fattening" of the Ballman example which makes it a Riemannian manifold.
$endgroup$
– macbeth
6 hours ago
1
$begingroup$
Right, topologically speaking, you are taking a neighborhood of Ballmann's example as embedded in $R^2$.
$endgroup$
– Misha
6 hours ago
$begingroup$
@macbeth, thanks!
$endgroup$
– Deane Yang
4 hours ago
add a comment |
1
$begingroup$
Yes, this does work. It is similar to Ballmann's example mentioned in Benoit's answer here: mathoverflow.net/questions/15592/…
$endgroup$
– Misha
6 hours ago
1
$begingroup$
Oh, that's interesting. Yes, this seems to be a "fattening" of the Ballman example which makes it a Riemannian manifold.
$endgroup$
– macbeth
6 hours ago
1
$begingroup$
Right, topologically speaking, you are taking a neighborhood of Ballmann's example as embedded in $R^2$.
$endgroup$
– Misha
6 hours ago
$begingroup$
@macbeth, thanks!
$endgroup$
– Deane Yang
4 hours ago
1
1
$begingroup$
Yes, this does work. It is similar to Ballmann's example mentioned in Benoit's answer here: mathoverflow.net/questions/15592/…
$endgroup$
– Misha
6 hours ago
$begingroup$
Yes, this does work. It is similar to Ballmann's example mentioned in Benoit's answer here: mathoverflow.net/questions/15592/…
$endgroup$
– Misha
6 hours ago
1
1
$begingroup$
Oh, that's interesting. Yes, this seems to be a "fattening" of the Ballman example which makes it a Riemannian manifold.
$endgroup$
– macbeth
6 hours ago
$begingroup$
Oh, that's interesting. Yes, this seems to be a "fattening" of the Ballman example which makes it a Riemannian manifold.
$endgroup$
– macbeth
6 hours ago
1
1
$begingroup$
Right, topologically speaking, you are taking a neighborhood of Ballmann's example as embedded in $R^2$.
$endgroup$
– Misha
6 hours ago
$begingroup$
Right, topologically speaking, you are taking a neighborhood of Ballmann's example as embedded in $R^2$.
$endgroup$
– Misha
6 hours ago
$begingroup$
@macbeth, thanks!
$endgroup$
– Deane Yang
4 hours ago
$begingroup$
@macbeth, thanks!
$endgroup$
– Deane Yang
4 hours ago
add a comment |
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