Mixing
In what sense is an ultralimit a “limit”
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Recently, I ended up having to use the notion of an ultralimit of metric spaces and realized that I do not have a good intuition for the "limit object" this construction creates. Given a sequence of metric spaces $(X_n,d_n)$, in what way does the ultralimit resemble $X_n$'s?
Let me explain what kind of answers I am looking for by pointing to a specific example. A special case of the ultralimit construction is the construction of the asymptotic cone of a metric spaces, in which one takes the ultralimit of the metric spaces $(X,d(cdot,cdot)/n)$ for a fixed metric space $(X,d(cdot,cdot))$ and some fixed non-principal ultrafilter on $mathbb{N}$.
One may think of the asymptotic cone as the space which codes the recurring configurations of the space $(X,d(cdot,cdot))$ if an observer was to "zoom out" from this space. As $n$ gets bigger, points that are not close will be seen closer in $(X,d(cdot,cdot)/n)$ and, in the limit case, various "local" configurations will disappear. In a sense, the asymptotic cone somehow codes information regarding the "global" geometry of the underlying space.
Is there some intuitive way to describe what the ultralimit does in general? Let me ask a very concrete question. What is an ultralimit of the sequence of spaces $(C_n,d/n)$ where $C_n$ is the cyclic graph with $n$ vertices and $d$ gives the length of the shortest path between two vertices? In what sense can such ultralimits be "obtained" from these cyclic graphs?
I also added the reference request tag to the post since I would also appreciate being directed to a survey or book that covers basic facts about these ultralimits.
reference-request metric-spaces metric-geometry
asked Jan 28 at 16:31
BurakBurak
1,84431019
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add a comment |
$begingroup$
Recently, I ended up having to use the notion of an ultralimit of metric spaces and realized that I do not have a good intuition for the "limit object" this construction creates. Given a sequence of metric spaces $(X_n,d_n)$, in what way does the ultralimit resemble $X_n$'s?
Let me explain what kind of answers I am looking for by pointing to a specific example. A special case of the ultralimit construction is the construction of the asymptotic cone of a metric spaces, in which one takes the ultralimit of the metric spaces $(X,d(cdot,cdot)/n)$ for a fixed metric space $(X,d(cdot,cdot))$ and some fixed non-principal ultrafilter on $mathbb{N}$.
One may think of the asymptotic cone as the space which codes the recurring configurations of the space $(X,d(cdot,cdot))$ if an observer was to "zoom out" from this space. As $n$ gets bigger, points that are not close will be seen closer in $(X,d(cdot,cdot)/n)$ and, in the limit case, various "local" configurations will disappear. In a sense, the asymptotic cone somehow codes information regarding the "global" geometry of the underlying space.
Is there some intuitive way to describe what the ultralimit does in general? Let me ask a very concrete question. What is an ultralimit of the sequence of spaces $(C_n,d/n)$ where $C_n$ is the cyclic graph with $n$ vertices and $d$ gives the length of the shortest path between two vertices? In what sense can such ultralimits be "obtained" from these cyclic graphs?
I also added the reference request tag to the post since I would also appreciate being directed to a survey or book that covers basic facts about these ultralimits.
reference-request metric-spaces metric-geometry
asked Jan 28 at 16:31
BurakBurak
1,84431019
$endgroup$
1
$begingroup$
Your question regarding $(C_n,d/n)$ doesn't make sense until you tell us what the metric is on $C_n$. Perhaps your intention is that each edge of $C_n$ has length $1$? And so each edge of $(C_n,d/n)$ has length $1/n$? If that's so, then each $(C_n,d/n)$ is isometric to the length metric on the circle of circumference $1$ (radius $1/2pi$). From a metric space point of view, your sequence $(C_n,d/n)$ is therefore a constant sequence. The ultralimit of a constant sequence of homogeneous metric spaces is isometric to that same metric space.
$endgroup$
– Lee Mosher
Jan 29 at 3:10
2
$begingroup$
As for your question in general, you've described some of the intuition pretty well in your own question. I would suggest that you proceed by building intuition through an understanding of examples and theorems. Try to understand why $n$-dimensional Euclidean space is its own asymptotic cone. Try to understand why the asymptotic cone of the hyperbolic plane is the $mathbb R$-tree of uniform valence $2^{aleph_0}$. Try to understand why the asymptotic cone of the 3-dimensional Heisenberg group (with its left invariant Riemannian metric) is the 3-dimensional Carnot-Caratheodory space.
$endgroup$
– Lee Mosher
Jan 29 at 3:13
add a comment |
$begingroup$
Recently, I ended up having to use the notion of an ultralimit of metric spaces and realized that I do not have a good intuition for the "limit object" this construction creates. Given a sequence of metric spaces $(X_n,d_n)$, in what way does the ultralimit resemble $X_n$'s?
Let me explain what kind of answers I am looking for by pointing to a specific example. A special case of the ultralimit construction is the construction of the asymptotic cone of a metric spaces, in which one takes the ultralimit of the metric spaces $(X,d(cdot,cdot)/n)$ for a fixed metric space $(X,d(cdot,cdot))$ and some fixed non-principal ultrafilter on $mathbb{N}$.
One may think of the asymptotic cone as the space which codes the recurring configurations of the space $(X,d(cdot,cdot))$ if an observer was to "zoom out" from this space. As $n$ gets bigger, points that are not close will be seen closer in $(X,d(cdot,cdot)/n)$ and, in the limit case, various "local" configurations will disappear. In a sense, the asymptotic cone somehow codes information regarding the "global" geometry of the underlying space.
Is there some intuitive way to describe what the ultralimit does in general? Let me ask a very concrete question. What is an ultralimit of the sequence of spaces $(C_n,d/n)$ where $C_n$ is the cyclic graph with $n$ vertices and $d$ gives the length of the shortest path between two vertices? In what sense can such ultralimits be "obtained" from these cyclic graphs?
I also added the reference request tag to the post since I would also appreciate being directed to a survey or book that covers basic facts about these ultralimits.
reference-request metric-spaces metric-geometry
asked Jan 28 at 16:31
BurakBurak
1,84431019
$endgroup$
Recently, I ended up having to use the notion of an ultralimit of metric spaces and realized that I do not have a good intuition for the "limit object" this construction creates. Given a sequence of metric spaces $(X_n,d_n)$, in what way does the ultralimit resemble $X_n$'s?
Let me explain what kind of answers I am looking for by pointing to a specific example. A special case of the ultralimit construction is the construction of the asymptotic cone of a metric spaces, in which one takes the ultralimit of the metric spaces $(X,d(cdot,cdot)/n)$ for a fixed metric space $(X,d(cdot,cdot))$ and some fixed non-principal ultrafilter on $mathbb{N}$.
One may think of the asymptotic cone as the space which codes the recurring configurations of the space $(X,d(cdot,cdot))$ if an observer was to "zoom out" from this space. As $n$ gets bigger, points that are not close will be seen closer in $(X,d(cdot,cdot)/n)$ and, in the limit case, various "local" configurations will disappear. In a sense, the asymptotic cone somehow codes information regarding the "global" geometry of the underlying space.
Is there some intuitive way to describe what the ultralimit does in general? Let me ask a very concrete question. What is an ultralimit of the sequence of spaces $(C_n,d/n)$ where $C_n$ is the cyclic graph with $n$ vertices and $d$ gives the length of the shortest path between two vertices? In what sense can such ultralimits be "obtained" from these cyclic graphs?
I also added the reference request tag to the post since I would also appreciate being directed to a survey or book that covers basic facts about these ultralimits.
reference-request metric-spaces metric-geometry
reference-request metric-spaces metric-geometry
asked Jan 28 at 16:31
BurakBurak
1,84431019
asked Jan 28 at 16:31
BurakBurak
1,84431019
edited Jan 28 at 16:57
Burak
asked Jan 28 at 16:31
BurakBurak
1,84431019
asked Jan 28 at 16:31
BurakBurak
1,84431019
asked Jan 28 at 16:31
BurakBurak
1,84431019
1,84431019
1
$begingroup$
Your question regarding $(C_n,d/n)$ doesn't make sense until you tell us what the metric is on $C_n$. Perhaps your intention is that each edge of $C_n$ has length $1$? And so each edge of $(C_n,d/n)$ has length $1/n$? If that's so, then each $(C_n,d/n)$ is isometric to the length metric on the circle of circumference $1$ (radius $1/2pi$). From a metric space point of view, your sequence $(C_n,d/n)$ is therefore a constant sequence. The ultralimit of a constant sequence of homogeneous metric spaces is isometric to that same metric space.
$endgroup$
– Lee Mosher
Jan 29 at 3:10
2
$begingroup$
As for your question in general, you've described some of the intuition pretty well in your own question. I would suggest that you proceed by building intuition through an understanding of examples and theorems. Try to understand why $n$-dimensional Euclidean space is its own asymptotic cone. Try to understand why the asymptotic cone of the hyperbolic plane is the $mathbb R$-tree of uniform valence $2^{aleph_0}$. Try to understand why the asymptotic cone of the 3-dimensional Heisenberg group (with its left invariant Riemannian metric) is the 3-dimensional Carnot-Caratheodory space.
$endgroup$
– Lee Mosher
Jan 29 at 3:13
add a comment |
1
$begingroup$
Your question regarding $(C_n,d/n)$ doesn't make sense until you tell us what the metric is on $C_n$. Perhaps your intention is that each edge of $C_n$ has length $1$? And so each edge of $(C_n,d/n)$ has length $1/n$? If that's so, then each $(C_n,d/n)$ is isometric to the length metric on the circle of circumference $1$ (radius $1/2pi$). From a metric space point of view, your sequence $(C_n,d/n)$ is therefore a constant sequence. The ultralimit of a constant sequence of homogeneous metric spaces is isometric to that same metric space.
$endgroup$
– Lee Mosher
Jan 29 at 3:10
2
$begingroup$
As for your question in general, you've described some of the intuition pretty well in your own question. I would suggest that you proceed by building intuition through an understanding of examples and theorems. Try to understand why $n$-dimensional Euclidean space is its own asymptotic cone. Try to understand why the asymptotic cone of the hyperbolic plane is the $mathbb R$-tree of uniform valence $2^{aleph_0}$. Try to understand why the asymptotic cone of the 3-dimensional Heisenberg group (with its left invariant Riemannian metric) is the 3-dimensional Carnot-Caratheodory space.
$endgroup$
– Lee Mosher
Jan 29 at 3:13
1
1
$begingroup$
Your question regarding $(C_n,d/n)$ doesn't make sense until you tell us what the metric is on $C_n$. Perhaps your intention is that each edge of $C_n$ has length $1$? And so each edge of $(C_n,d/n)$ has length $1/n$? If that's so, then each $(C_n,d/n)$ is isometric to the length metric on the circle of circumference $1$ (radius $1/2pi$). From a metric space point of view, your sequence $(C_n,d/n)$ is therefore a constant sequence. The ultralimit of a constant sequence of homogeneous metric spaces is isometric to that same metric space.
$endgroup$
– Lee Mosher
Jan 29 at 3:10
$begingroup$
Your question regarding $(C_n,d/n)$ doesn't make sense until you tell us what the metric is on $C_n$. Perhaps your intention is that each edge of $C_n$ has length $1$? And so each edge of $(C_n,d/n)$ has length $1/n$? If that's so, then each $(C_n,d/n)$ is isometric to the length metric on the circle of circumference $1$ (radius $1/2pi$). From a metric space point of view, your sequence $(C_n,d/n)$ is therefore a constant sequence. The ultralimit of a constant sequence of homogeneous metric spaces is isometric to that same metric space.
$endgroup$
– Lee Mosher
Jan 29 at 3:10
2
2
$begingroup$
As for your question in general, you've described some of the intuition pretty well in your own question. I would suggest that you proceed by building intuition through an understanding of examples and theorems. Try to understand why $n$-dimensional Euclidean space is its own asymptotic cone. Try to understand why the asymptotic cone of the hyperbolic plane is the $mathbb R$-tree of uniform valence $2^{aleph_0}$. Try to understand why the asymptotic cone of the 3-dimensional Heisenberg group (with its left invariant Riemannian metric) is the 3-dimensional Carnot-Caratheodory space.
$endgroup$
– Lee Mosher
Jan 29 at 3:13
$begingroup$
As for your question in general, you've described some of the intuition pretty well in your own question. I would suggest that you proceed by building intuition through an understanding of examples and theorems. Try to understand why $n$-dimensional Euclidean space is its own asymptotic cone. Try to understand why the asymptotic cone of the hyperbolic plane is the $mathbb R$-tree of uniform valence $2^{aleph_0}$. Try to understand why the asymptotic cone of the 3-dimensional Heisenberg group (with its left invariant Riemannian metric) is the 3-dimensional Carnot-Caratheodory space.
$endgroup$
– Lee Mosher
Jan 29 at 3:13
add a comment |
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$begingroup$
Your question regarding $(C_n,d/n)$ doesn't make sense until you tell us what the metric is on $C_n$. Perhaps your intention is that each edge of $C_n$ has length $1$? And so each edge of $(C_n,d/n)$ has length $1/n$? If that's so, then each $(C_n,d/n)$ is isometric to the length metric on the circle of circumference $1$ (radius $1/2pi$). From a metric space point of view, your sequence $(C_n,d/n)$ is therefore a constant sequence. The ultralimit of a constant sequence of homogeneous metric spaces is isometric to that same metric space.
$endgroup$
– Lee Mosher
Jan 29 at 3:10
2
$begingroup$
As for your question in general, you've described some of the intuition pretty well in your own question. I would suggest that you proceed by building intuition through an understanding of examples and theorems. Try to understand why $n$-dimensional Euclidean space is its own asymptotic cone. Try to understand why the asymptotic cone of the hyperbolic plane is the $mathbb R$-tree of uniform valence $2^{aleph_0}$. Try to understand why the asymptotic cone of the 3-dimensional Heisenberg group (with its left invariant Riemannian metric) is the 3-dimensional Carnot-Caratheodory space.
$endgroup$
– Lee Mosher
Jan 29 at 3:13