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Is the diameter of a $S^2$ in arbitrary metric on $Bbb R^3$ just the distance between antipodal points?
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Let $S^2$ be a unit sphere in the Euclidean metric. Consider $S^2$ as a subset of $mathbbR^3$ with an arbitrary metric $d$. Now, we would like to compute the diameter of $S^2$ with respect to the metric $d$.
Is it always true that the diameter is the distance between antipodal points of $S^2$ or should we assume something about the metric (e.g. translational invariance etc.)?
This statement is obviously true for Euclidean metric, Manhattan metric, discrete metric and the maximum metric, but is true in general? Any hints?
general-topology metric-spaces
edited Aug 6 at 6:43
Asaf Karagila♦
292k31403733
asked Aug 5 at 19:13
mikis
1,3991721
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up vote
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Let $S^2$ be a unit sphere in the Euclidean metric. Consider $S^2$ as a subset of $mathbbR^3$ with an arbitrary metric $d$. Now, we would like to compute the diameter of $S^2$ with respect to the metric $d$.
Is it always true that the diameter is the distance between antipodal points of $S^2$ or should we assume something about the metric (e.g. translational invariance etc.)?
This statement is obviously true for Euclidean metric, Manhattan metric, discrete metric and the maximum metric, but is true in general? Any hints?
general-topology metric-spaces
edited Aug 6 at 6:43
Asaf Karagila♦
292k31403733
asked Aug 5 at 19:13
mikis
1,3991721
The distance of antipodal points is definitionally the diameter - you're simply giving it different values with different metrics. This is like asking, "Is the length of this line always the length of this line for an arbitrary metric?"
– Isky Mathews
Aug 5 at 19:23
2
The diameter of a set $A$ is define by $sup x,yin A$.
– mikis
Aug 5 at 19:26
add a comment |Â
up vote
3
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up vote
3
down vote
favorite
Let $S^2$ be a unit sphere in the Euclidean metric. Consider $S^2$ as a subset of $mathbbR^3$ with an arbitrary metric $d$. Now, we would like to compute the diameter of $S^2$ with respect to the metric $d$.
Is it always true that the diameter is the distance between antipodal points of $S^2$ or should we assume something about the metric (e.g. translational invariance etc.)?
This statement is obviously true for Euclidean metric, Manhattan metric, discrete metric and the maximum metric, but is true in general? Any hints?
general-topology metric-spaces
edited Aug 6 at 6:43
Asaf Karagila♦
292k31403733
asked Aug 5 at 19:13
mikis
1,3991721
Let $S^2$ be a unit sphere in the Euclidean metric. Consider $S^2$ as a subset of $mathbbR^3$ with an arbitrary metric $d$. Now, we would like to compute the diameter of $S^2$ with respect to the metric $d$.
Is it always true that the diameter is the distance between antipodal points of $S^2$ or should we assume something about the metric (e.g. translational invariance etc.)?
This statement is obviously true for Euclidean metric, Manhattan metric, discrete metric and the maximum metric, but is true in general? Any hints?
general-topology metric-spaces
edited Aug 6 at 6:43
Asaf Karagila♦
292k31403733
asked Aug 5 at 19:13
mikis
1,3991721
edited Aug 6 at 6:43
Asaf Karagila♦
292k31403733
edited Aug 6 at 6:43
Asaf Karagila♦
292k31403733
edited Aug 6 at 6:43
Asaf Karagila♦
292k31403733
292k31403733
asked Aug 5 at 19:13
mikis
1,3991721
asked Aug 5 at 19:13
mikis
1,3991721
asked Aug 5 at 19:13
mikis
1,3991721
1,3991721
The distance of antipodal points is definitionally the diameter - you're simply giving it different values with different metrics. This is like asking, "Is the length of this line always the length of this line for an arbitrary metric?"
– Isky Mathews
Aug 5 at 19:23
2
The diameter of a set $A$ is define by $sup x,yin A$.
– mikis
Aug 5 at 19:26
add a comment |Â
The distance of antipodal points is definitionally the diameter - you're simply giving it different values with different metrics. This is like asking, "Is the length of this line always the length of this line for an arbitrary metric?"
– Isky Mathews
Aug 5 at 19:23
2
The diameter of a set $A$ is define by $sup x,yin A$.
– mikis
Aug 5 at 19:26
The distance of antipodal points is definitionally the diameter - you're simply giving it different values with different metrics. This is like asking, "Is the length of this line always the length of this line for an arbitrary metric?"
– Isky Mathews
Aug 5 at 19:23
The distance of antipodal points is definitionally the diameter - you're simply giving it different values with different metrics. This is like asking, "Is the length of this line always the length of this line for an arbitrary metric?"
– Isky Mathews
Aug 5 at 19:23
2
2
The diameter of a set $A$ is define by $sup x,yin A$.
– mikis
Aug 5 at 19:26
The diameter of a set $A$ is define by $sup x,yin A$.
– mikis
Aug 5 at 19:26
add a comment |Â
3 Answers
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There is no reason for anything like this to be true for arbitrary metric. For example, consider the metric
$$
hatd(v,w):=sqrt(v_1-w_1)^2+2(v_2-w_2)^2+(v_3-w_3)^2.
$$
There is only one pair of points at which the diameter is attained, namely, $(0,1,0)$ and (0,-1,0). These happen to be antipodal, so $hatd$ does not quite qualify. However, we can slightly distort it by putting $d(v,w):=hatd(f(v),f(w))$, where $f:mathbbR^3mapstomathbbR^3$ is the bijection that reads, in spherical coordinates,
$$
f:(r,theta,varphi)mapsto (r,theta+frac12cos theta,varphi).
$$
(here $theta$ denotes the latitude). The diameter is now attained at the preimages of the above points, that both lie in the lower hemisphere, and thus are not antipodal.
In the positive direction, if your metric satisfies $2d(0,v)equiv d(v,-v)$, then the conclusion is true. Indeed, if the diameter is attained at $u,w$, then, by the triangle inequality, either $d(u,0)$ or $d(v,0)$ is at least half of the diameter. But then $d(u,-u)$ (resp., $d(v,-v)$) is at least diameter. In particular, this will be true for any norm-induced metric.
answered Aug 5 at 20:21
Kostya_I
408210
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In general, no. Let $M$ be some bijection $mathbbR^3to mathbbR^3$ which maps $S^2$ into ellipsoid $x^2+2y^2+2z^2=1$ and such that pre-images of points $(1,0,0)$ and $(-1,0,0)$ are not antipodal on $S^2$. Define $d$ by $d(a,b):= d_E(M(a), M(b))$, where $d_E$ is the standard Euclidean metric. Then the diameter of $S^2$ in this metric is $2$, and this distance is achieved only on pre-images of points $(1,0,0)$ and $(-1,0,0)$ (since the diameter of that ellipsoid in $d_E$ is $2$, and this distance is achieved only on $(1,0,0)$ and $(-1,0,0)$), but those two points are by construction not antipodal.
answered Aug 5 at 20:01
Litho
3,3421614
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up vote
5
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Not in general. First let's consider the problem in only two dimensions. Here is a counterexample.
Let $p$ be some point in the interior of the circle other than the usual center. Let the distance $$delta(x,y) = d(x,p)+d(p,y)$$ where $d$ is the usual metric. So for example the $delta$-distance between $a$ and $b$ follows the dotted path as shown. Clearly this distance is larger than the distance between any antipodal pair of points.
Extending this to three dimensions is straightforward.
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
There is no reason for anything like this to be true for arbitrary metric. For example, consider the metric
$$
hatd(v,w):=sqrt(v_1-w_1)^2+2(v_2-w_2)^2+(v_3-w_3)^2.
$$
There is only one pair of points at which the diameter is attained, namely, $(0,1,0)$ and (0,-1,0). These happen to be antipodal, so $hatd$ does not quite qualify. However, we can slightly distort it by putting $d(v,w):=hatd(f(v),f(w))$, where $f:mathbbR^3mapstomathbbR^3$ is the bijection that reads, in spherical coordinates,
$$
f:(r,theta,varphi)mapsto (r,theta+frac12cos theta,varphi).
$$
(here $theta$ denotes the latitude). The diameter is now attained at the preimages of the above points, that both lie in the lower hemisphere, and thus are not antipodal.
In the positive direction, if your metric satisfies $2d(0,v)equiv d(v,-v)$, then the conclusion is true. Indeed, if the diameter is attained at $u,w$, then, by the triangle inequality, either $d(u,0)$ or $d(v,0)$ is at least half of the diameter. But then $d(u,-u)$ (resp., $d(v,-v)$) is at least diameter. In particular, this will be true for any norm-induced metric.
answered Aug 5 at 20:21
Kostya_I
408210
add a comment |Â
up vote
3
down vote
accepted
There is no reason for anything like this to be true for arbitrary metric. For example, consider the metric
$$
hatd(v,w):=sqrt(v_1-w_1)^2+2(v_2-w_2)^2+(v_3-w_3)^2.
$$
There is only one pair of points at which the diameter is attained, namely, $(0,1,0)$ and (0,-1,0). These happen to be antipodal, so $hatd$ does not quite qualify. However, we can slightly distort it by putting $d(v,w):=hatd(f(v),f(w))$, where $f:mathbbR^3mapstomathbbR^3$ is the bijection that reads, in spherical coordinates,
$$
f:(r,theta,varphi)mapsto (r,theta+frac12cos theta,varphi).
$$
(here $theta$ denotes the latitude). The diameter is now attained at the preimages of the above points, that both lie in the lower hemisphere, and thus are not antipodal.
In the positive direction, if your metric satisfies $2d(0,v)equiv d(v,-v)$, then the conclusion is true. Indeed, if the diameter is attained at $u,w$, then, by the triangle inequality, either $d(u,0)$ or $d(v,0)$ is at least half of the diameter. But then $d(u,-u)$ (resp., $d(v,-v)$) is at least diameter. In particular, this will be true for any norm-induced metric.
answered Aug 5 at 20:21
Kostya_I
408210
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
There is no reason for anything like this to be true for arbitrary metric. For example, consider the metric
$$
hatd(v,w):=sqrt(v_1-w_1)^2+2(v_2-w_2)^2+(v_3-w_3)^2.
$$
There is only one pair of points at which the diameter is attained, namely, $(0,1,0)$ and (0,-1,0). These happen to be antipodal, so $hatd$ does not quite qualify. However, we can slightly distort it by putting $d(v,w):=hatd(f(v),f(w))$, where $f:mathbbR^3mapstomathbbR^3$ is the bijection that reads, in spherical coordinates,
$$
f:(r,theta,varphi)mapsto (r,theta+frac12cos theta,varphi).
$$
(here $theta$ denotes the latitude). The diameter is now attained at the preimages of the above points, that both lie in the lower hemisphere, and thus are not antipodal.
In the positive direction, if your metric satisfies $2d(0,v)equiv d(v,-v)$, then the conclusion is true. Indeed, if the diameter is attained at $u,w$, then, by the triangle inequality, either $d(u,0)$ or $d(v,0)$ is at least half of the diameter. But then $d(u,-u)$ (resp., $d(v,-v)$) is at least diameter. In particular, this will be true for any norm-induced metric.
answered Aug 5 at 20:21
Kostya_I
408210
There is no reason for anything like this to be true for arbitrary metric. For example, consider the metric
$$
hatd(v,w):=sqrt(v_1-w_1)^2+2(v_2-w_2)^2+(v_3-w_3)^2.
$$
There is only one pair of points at which the diameter is attained, namely, $(0,1,0)$ and (0,-1,0). These happen to be antipodal, so $hatd$ does not quite qualify. However, we can slightly distort it by putting $d(v,w):=hatd(f(v),f(w))$, where $f:mathbbR^3mapstomathbbR^3$ is the bijection that reads, in spherical coordinates,
$$
f:(r,theta,varphi)mapsto (r,theta+frac12cos theta,varphi).
$$
(here $theta$ denotes the latitude). The diameter is now attained at the preimages of the above points, that both lie in the lower hemisphere, and thus are not antipodal.
In the positive direction, if your metric satisfies $2d(0,v)equiv d(v,-v)$, then the conclusion is true. Indeed, if the diameter is attained at $u,w$, then, by the triangle inequality, either $d(u,0)$ or $d(v,0)$ is at least half of the diameter. But then $d(u,-u)$ (resp., $d(v,-v)$) is at least diameter. In particular, this will be true for any norm-induced metric.
answered Aug 5 at 20:21
Kostya_I
408210
answered Aug 5 at 20:21
Kostya_I
408210
answered Aug 5 at 20:21
Kostya_I
408210
answered Aug 5 at 20:21
Kostya_I
408210
408210
add a comment |Â
add a comment |Â
up vote
5
down vote
In general, no. Let $M$ be some bijection $mathbbR^3to mathbbR^3$ which maps $S^2$ into ellipsoid $x^2+2y^2+2z^2=1$ and such that pre-images of points $(1,0,0)$ and $(-1,0,0)$ are not antipodal on $S^2$. Define $d$ by $d(a,b):= d_E(M(a), M(b))$, where $d_E$ is the standard Euclidean metric. Then the diameter of $S^2$ in this metric is $2$, and this distance is achieved only on pre-images of points $(1,0,0)$ and $(-1,0,0)$ (since the diameter of that ellipsoid in $d_E$ is $2$, and this distance is achieved only on $(1,0,0)$ and $(-1,0,0)$), but those two points are by construction not antipodal.
answered Aug 5 at 20:01
Litho
3,3421614
add a comment |Â
up vote
5
down vote
In general, no. Let $M$ be some bijection $mathbbR^3to mathbbR^3$ which maps $S^2$ into ellipsoid $x^2+2y^2+2z^2=1$ and such that pre-images of points $(1,0,0)$ and $(-1,0,0)$ are not antipodal on $S^2$. Define $d$ by $d(a,b):= d_E(M(a), M(b))$, where $d_E$ is the standard Euclidean metric. Then the diameter of $S^2$ in this metric is $2$, and this distance is achieved only on pre-images of points $(1,0,0)$ and $(-1,0,0)$ (since the diameter of that ellipsoid in $d_E$ is $2$, and this distance is achieved only on $(1,0,0)$ and $(-1,0,0)$), but those two points are by construction not antipodal.
answered Aug 5 at 20:01
Litho
3,3421614
add a comment |Â
up vote
5
down vote
up vote
5
down vote
In general, no. Let $M$ be some bijection $mathbbR^3to mathbbR^3$ which maps $S^2$ into ellipsoid $x^2+2y^2+2z^2=1$ and such that pre-images of points $(1,0,0)$ and $(-1,0,0)$ are not antipodal on $S^2$. Define $d$ by $d(a,b):= d_E(M(a), M(b))$, where $d_E$ is the standard Euclidean metric. Then the diameter of $S^2$ in this metric is $2$, and this distance is achieved only on pre-images of points $(1,0,0)$ and $(-1,0,0)$ (since the diameter of that ellipsoid in $d_E$ is $2$, and this distance is achieved only on $(1,0,0)$ and $(-1,0,0)$), but those two points are by construction not antipodal.
answered Aug 5 at 20:01
Litho
3,3421614
In general, no. Let $M$ be some bijection $mathbbR^3to mathbbR^3$ which maps $S^2$ into ellipsoid $x^2+2y^2+2z^2=1$ and such that pre-images of points $(1,0,0)$ and $(-1,0,0)$ are not antipodal on $S^2$. Define $d$ by $d(a,b):= d_E(M(a), M(b))$, where $d_E$ is the standard Euclidean metric. Then the diameter of $S^2$ in this metric is $2$, and this distance is achieved only on pre-images of points $(1,0,0)$ and $(-1,0,0)$ (since the diameter of that ellipsoid in $d_E$ is $2$, and this distance is achieved only on $(1,0,0)$ and $(-1,0,0)$), but those two points are by construction not antipodal.
answered Aug 5 at 20:01
Litho
3,3421614
answered Aug 5 at 20:01
Litho
3,3421614
answered Aug 5 at 20:01
Litho
3,3421614
answered Aug 5 at 20:01
Litho
3,3421614
3,3421614
add a comment |Â
add a comment |Â
up vote
5
down vote
Not in general. First let's consider the problem in only two dimensions. Here is a counterexample.
Let $p$ be some point in the interior of the circle other than the usual center. Let the distance $$delta(x,y) = d(x,p)+d(p,y)$$ where $d$ is the usual metric. So for example the $delta$-distance between $a$ and $b$ follows the dotted path as shown. Clearly this distance is larger than the distance between any antipodal pair of points.
Extending this to three dimensions is straightforward.
add a comment |Â
up vote
5
down vote
Not in general. First let's consider the problem in only two dimensions. Here is a counterexample.
Let $p$ be some point in the interior of the circle other than the usual center. Let the distance $$delta(x,y) = d(x,p)+d(p,y)$$ where $d$ is the usual metric. So for example the $delta$-distance between $a$ and $b$ follows the dotted path as shown. Clearly this distance is larger than the distance between any antipodal pair of points.
Extending this to three dimensions is straightforward.
add a comment |Â
up vote
5
down vote
up vote
5
down vote
Not in general. First let's consider the problem in only two dimensions. Here is a counterexample.
Let $p$ be some point in the interior of the circle other than the usual center. Let the distance $$delta(x,y) = d(x,p)+d(p,y)$$ where $d$ is the usual metric. So for example the $delta$-distance between $a$ and $b$ follows the dotted path as shown. Clearly this distance is larger than the distance between any antipodal pair of points.
Extending this to three dimensions is straightforward.
Not in general. First let's consider the problem in only two dimensions. Here is a counterexample.
Let $p$ be some point in the interior of the circle other than the usual center. Let the distance $$delta(x,y) = d(x,p)+d(p,y)$$ where $d$ is the usual metric. So for example the $delta$-distance between $a$ and $b$ follows the dotted path as shown. Clearly this distance is larger than the distance between any antipodal pair of points.
Extending this to three dimensions is straightforward.
edited Aug 5 at 21:05
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MJD
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The distance of antipodal points is definitionally the diameter - you're simply giving it different values with different metrics. This is like asking, "Is the length of this line always the length of this line for an arbitrary metric?"
– Isky Mathews
Aug 5 at 19:23
2
The diameter of a set $A$ is define by $sup x,yin A$.
– mikis
Aug 5 at 19:26