Why do we use two constants in the definition of quasi-isometric embedding?Quasi-isometric embedding and Quasi-isometryIf $H$ is a subgroup of $G$, then the inclusion is a quasi-isometry if and only if $H$ has finite index in $G$.Is every metric space quasi-isometric to a graph?Show: $(X,d_X)$ is complete $Leftrightarrow $ $f(X)$ is closed in $(Y, d_Y)$ ($f: X to Y$ is an isometric embedding)Nearest points to image of isometric embeddingPreimage of convergent sequence under isometric embeddingImage of a complete space under an isometric embeddingIs There any quasi-isomorphism between $mathbbR$ and $mathbbR^2$?On a characterization of quasi-isometric embedding for groupsEvery quasiisometry is a quasiisometric embedding
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Why do we use two constants in the definition of quasi-isometric embedding?
Quasi-isometric embedding and Quasi-isometryIf $H$ is a subgroup of $G$, then the inclusion is a quasi-isometry if and only if $H$ has finite index in $G$.Is every metric space quasi-isometric to a graph?Show: $(X,d_X)$ is complete $Leftrightarrow $ $f(X)$ is closed in $(Y, d_Y)$ ($f: X to Y$ is an isometric embedding)Nearest points to image of isometric embeddingPreimage of convergent sequence under isometric embeddingImage of a complete space under an isometric embeddingIs There any quasi-isomorphism between $mathbbR$ and $mathbbR^2$?On a characterization of quasi-isometric embedding for groupsEvery quasiisometry is a quasiisometric embedding
$begingroup$
A map $f: X to Y$ is a quasi-isometric embedding if there exists constants $K ge 1$ and $Cge 0$ so that for all $x, x' in X$ we have
$$frac1Kd_X(x,x') -C le d_Y(f(x), f(x')) le K d_X(x,x') +C.$$
But we could also let $R=maxK,C$ and the quasi-isometric embedding still holds.
Why is this not used as the definition?
abstract-algebra group-theory isometry geometric-group-theory
asked Mar 30 at 2:19
Al JebrAl Jebr
4,42143478
$endgroup$
add a comment |
$begingroup$
A map $f: X to Y$ is a quasi-isometric embedding if there exists constants $K ge 1$ and $Cge 0$ so that for all $x, x' in X$ we have
$$frac1Kd_X(x,x') -C le d_Y(f(x), f(x')) le K d_X(x,x') +C.$$
But we could also let $R=maxK,C$ and the quasi-isometric embedding still holds.
Why is this not used as the definition?
abstract-algebra group-theory isometry geometric-group-theory
asked Mar 30 at 2:19
Al JebrAl Jebr
4,42143478
$endgroup$
$begingroup$
In part, this is because there are special cases when quasi-isometries are particularly nice: when $C=0$ and when $K=1$. Otherwise, you are right, one constant is enough.
$endgroup$
– Moishe Kohan
Mar 30 at 2:54
$begingroup$
I have to admit that my immediate reaction to this question was to wonder why you might think it was preferable to use one constant rather than two. As Moishe Kohan commented, that makes the definition less precise.
$endgroup$
– Derek Holt
Mar 30 at 5:39
$begingroup$
We can even use 4 constants: $K_1d_Y^*f-C_1le d_Yle K_2d_Y^*f+C_2$. For instance, when we pass to the asymptotic cone, the additive constant disappear while the multiplicative one are preserved.
$endgroup$
– YCor
Mar 30 at 8:41
1
$begingroup$
The answer is that is not a single way to formulate the definition, but several more or less obviously equivalent formulations. When making quantitative statements it is useful to specify the definition, and talk, for instance, of a $(K,C)$-QI-embedding, and such a specification is sensitive on the precise formulation, and is chosen according to your particular needs.
$endgroup$
– YCor
Mar 30 at 8:45
$begingroup$
I really HATE to throw away information. Discarding mathematical information is something I will resist. I will protest if I can no longer formulate statements like: "An isometry is a $(1,0)$ quasi-isometry"; or "A $K$-bilipshitz homeomorphism is a $(K,0)$ quasi-isometry"; or "If $f : X to Y$ is a $(K,C)$ quasi-isometry, and if $X' < X$ is a net (i.e. each point of $X$ is uniformly close to some point of $X'$), then the restriction $f : X' to Y$ is a $(K,C')$ quasi-isometry for some $C'$ ".
$endgroup$
– Lee Mosher
Apr 4 at 21:43
add a comment |
$begingroup$
A map $f: X to Y$ is a quasi-isometric embedding if there exists constants $K ge 1$ and $Cge 0$ so that for all $x, x' in X$ we have
$$frac1Kd_X(x,x') -C le d_Y(f(x), f(x')) le K d_X(x,x') +C.$$
But we could also let $R=maxK,C$ and the quasi-isometric embedding still holds.
Why is this not used as the definition?
abstract-algebra group-theory isometry geometric-group-theory
asked Mar 30 at 2:19
Al JebrAl Jebr
4,42143478
$endgroup$
A map $f: X to Y$ is a quasi-isometric embedding if there exists constants $K ge 1$ and $Cge 0$ so that for all $x, x' in X$ we have
$$frac1Kd_X(x,x') -C le d_Y(f(x), f(x')) le K d_X(x,x') +C.$$
But we could also let $R=maxK,C$ and the quasi-isometric embedding still holds.
Why is this not used as the definition?
abstract-algebra group-theory isometry geometric-group-theory
abstract-algebra group-theory isometry geometric-group-theory
asked Mar 30 at 2:19
Al JebrAl Jebr
4,42143478
asked Mar 30 at 2:19
Al JebrAl Jebr
4,42143478
asked Mar 30 at 2:19
Al JebrAl Jebr
4,42143478
asked Mar 30 at 2:19
Al JebrAl Jebr
4,42143478
asked Mar 30 at 2:19
Al JebrAl Jebr
4,42143478
4,42143478
$begingroup$
In part, this is because there are special cases when quasi-isometries are particularly nice: when $C=0$ and when $K=1$. Otherwise, you are right, one constant is enough.
$endgroup$
– Moishe Kohan
Mar 30 at 2:54
$begingroup$
I have to admit that my immediate reaction to this question was to wonder why you might think it was preferable to use one constant rather than two. As Moishe Kohan commented, that makes the definition less precise.
$endgroup$
– Derek Holt
Mar 30 at 5:39
$begingroup$
We can even use 4 constants: $K_1d_Y^*f-C_1le d_Yle K_2d_Y^*f+C_2$. For instance, when we pass to the asymptotic cone, the additive constant disappear while the multiplicative one are preserved.
$endgroup$
– YCor
Mar 30 at 8:41
1
$begingroup$
The answer is that is not a single way to formulate the definition, but several more or less obviously equivalent formulations. When making quantitative statements it is useful to specify the definition, and talk, for instance, of a $(K,C)$-QI-embedding, and such a specification is sensitive on the precise formulation, and is chosen according to your particular needs.
$endgroup$
– YCor
Mar 30 at 8:45
$begingroup$
I really HATE to throw away information. Discarding mathematical information is something I will resist. I will protest if I can no longer formulate statements like: "An isometry is a $(1,0)$ quasi-isometry"; or "A $K$-bilipshitz homeomorphism is a $(K,0)$ quasi-isometry"; or "If $f : X to Y$ is a $(K,C)$ quasi-isometry, and if $X' < X$ is a net (i.e. each point of $X$ is uniformly close to some point of $X'$), then the restriction $f : X' to Y$ is a $(K,C')$ quasi-isometry for some $C'$ ".
$endgroup$
– Lee Mosher
Apr 4 at 21:43
add a comment |
$begingroup$
In part, this is because there are special cases when quasi-isometries are particularly nice: when $C=0$ and when $K=1$. Otherwise, you are right, one constant is enough.
$endgroup$
– Moishe Kohan
Mar 30 at 2:54
$begingroup$
I have to admit that my immediate reaction to this question was to wonder why you might think it was preferable to use one constant rather than two. As Moishe Kohan commented, that makes the definition less precise.
$endgroup$
– Derek Holt
Mar 30 at 5:39
$begingroup$
We can even use 4 constants: $K_1d_Y^*f-C_1le d_Yle K_2d_Y^*f+C_2$. For instance, when we pass to the asymptotic cone, the additive constant disappear while the multiplicative one are preserved.
$endgroup$
– YCor
Mar 30 at 8:41
1
$begingroup$
The answer is that is not a single way to formulate the definition, but several more or less obviously equivalent formulations. When making quantitative statements it is useful to specify the definition, and talk, for instance, of a $(K,C)$-QI-embedding, and such a specification is sensitive on the precise formulation, and is chosen according to your particular needs.
$endgroup$
– YCor
Mar 30 at 8:45
$begingroup$
I really HATE to throw away information. Discarding mathematical information is something I will resist. I will protest if I can no longer formulate statements like: "An isometry is a $(1,0)$ quasi-isometry"; or "A $K$-bilipshitz homeomorphism is a $(K,0)$ quasi-isometry"; or "If $f : X to Y$ is a $(K,C)$ quasi-isometry, and if $X' < X$ is a net (i.e. each point of $X$ is uniformly close to some point of $X'$), then the restriction $f : X' to Y$ is a $(K,C')$ quasi-isometry for some $C'$ ".
$endgroup$
– Lee Mosher
Apr 4 at 21:43
$begingroup$
In part, this is because there are special cases when quasi-isometries are particularly nice: when $C=0$ and when $K=1$. Otherwise, you are right, one constant is enough.
$endgroup$
– Moishe Kohan
Mar 30 at 2:54
$begingroup$
In part, this is because there are special cases when quasi-isometries are particularly nice: when $C=0$ and when $K=1$. Otherwise, you are right, one constant is enough.
$endgroup$
– Moishe Kohan
Mar 30 at 2:54
$begingroup$
I have to admit that my immediate reaction to this question was to wonder why you might think it was preferable to use one constant rather than two. As Moishe Kohan commented, that makes the definition less precise.
$endgroup$
– Derek Holt
Mar 30 at 5:39
$begingroup$
I have to admit that my immediate reaction to this question was to wonder why you might think it was preferable to use one constant rather than two. As Moishe Kohan commented, that makes the definition less precise.
$endgroup$
– Derek Holt
Mar 30 at 5:39
$begingroup$
We can even use 4 constants: $K_1d_Y^*f-C_1le d_Yle K_2d_Y^*f+C_2$. For instance, when we pass to the asymptotic cone, the additive constant disappear while the multiplicative one are preserved.
$endgroup$
– YCor
Mar 30 at 8:41
$begingroup$
We can even use 4 constants: $K_1d_Y^*f-C_1le d_Yle K_2d_Y^*f+C_2$. For instance, when we pass to the asymptotic cone, the additive constant disappear while the multiplicative one are preserved.
$endgroup$
– YCor
Mar 30 at 8:41
1
1
$begingroup$
The answer is that is not a single way to formulate the definition, but several more or less obviously equivalent formulations. When making quantitative statements it is useful to specify the definition, and talk, for instance, of a $(K,C)$-QI-embedding, and such a specification is sensitive on the precise formulation, and is chosen according to your particular needs.
$endgroup$
– YCor
Mar 30 at 8:45
$begingroup$
The answer is that is not a single way to formulate the definition, but several more or less obviously equivalent formulations. When making quantitative statements it is useful to specify the definition, and talk, for instance, of a $(K,C)$-QI-embedding, and such a specification is sensitive on the precise formulation, and is chosen according to your particular needs.
$endgroup$
– YCor
Mar 30 at 8:45
$begingroup$
I really HATE to throw away information. Discarding mathematical information is something I will resist. I will protest if I can no longer formulate statements like: "An isometry is a $(1,0)$ quasi-isometry"; or "A $K$-bilipshitz homeomorphism is a $(K,0)$ quasi-isometry"; or "If $f : X to Y$ is a $(K,C)$ quasi-isometry, and if $X' < X$ is a net (i.e. each point of $X$ is uniformly close to some point of $X'$), then the restriction $f : X' to Y$ is a $(K,C')$ quasi-isometry for some $C'$ ".
$endgroup$
– Lee Mosher
Apr 4 at 21:43
$begingroup$
I really HATE to throw away information. Discarding mathematical information is something I will resist. I will protest if I can no longer formulate statements like: "An isometry is a $(1,0)$ quasi-isometry"; or "A $K$-bilipshitz homeomorphism is a $(K,0)$ quasi-isometry"; or "If $f : X to Y$ is a $(K,C)$ quasi-isometry, and if $X' < X$ is a net (i.e. each point of $X$ is uniformly close to some point of $X'$), then the restriction $f : X' to Y$ is a $(K,C')$ quasi-isometry for some $C'$ ".
$endgroup$
– Lee Mosher
Apr 4 at 21:43
add a comment |
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$begingroup$
In part, this is because there are special cases when quasi-isometries are particularly nice: when $C=0$ and when $K=1$. Otherwise, you are right, one constant is enough.
$endgroup$
– Moishe Kohan
Mar 30 at 2:54
$begingroup$
I have to admit that my immediate reaction to this question was to wonder why you might think it was preferable to use one constant rather than two. As Moishe Kohan commented, that makes the definition less precise.
$endgroup$
– Derek Holt
Mar 30 at 5:39
$begingroup$
We can even use 4 constants: $K_1d_Y^*f-C_1le d_Yle K_2d_Y^*f+C_2$. For instance, when we pass to the asymptotic cone, the additive constant disappear while the multiplicative one are preserved.
$endgroup$
– YCor
Mar 30 at 8:41
1
$begingroup$
The answer is that is not a single way to formulate the definition, but several more or less obviously equivalent formulations. When making quantitative statements it is useful to specify the definition, and talk, for instance, of a $(K,C)$-QI-embedding, and such a specification is sensitive on the precise formulation, and is chosen according to your particular needs.
$endgroup$
– YCor
Mar 30 at 8:45
$begingroup$
I really HATE to throw away information. Discarding mathematical information is something I will resist. I will protest if I can no longer formulate statements like: "An isometry is a $(1,0)$ quasi-isometry"; or "A $K$-bilipshitz homeomorphism is a $(K,0)$ quasi-isometry"; or "If $f : X to Y$ is a $(K,C)$ quasi-isometry, and if $X' < X$ is a net (i.e. each point of $X$ is uniformly close to some point of $X'$), then the restriction $f : X' to Y$ is a $(K,C')$ quasi-isometry for some $C'$ ".
$endgroup$
– Lee Mosher
Apr 4 at 21:43