Why $mathbb R^2/mathbb Z^2$ and $mathbb R/mathbb Ztimes mathbb R/mathbb Z$ are the same + $L^1(mathbb...
1
$begingroup$
Why $mathbb R^2/ mathbb Z^2$ is the Torus ? The Torus is indeed $mathbb R/mathbb Ztimes mathbb R/mathbb Z$. But why $mathbb R^2/mathbb Z^2$ also gives the torus ? For me if we glue all points of $mathbb Z^2$ together, then $mathbb R^2/mathbb Z^2$ should look as a sphere and not as a Torus ? Also, for me $L^1(mathbb S^2)$ is the set of integrable periodic functions on $mathbb R^2$. But in my course they wrote it as $L^1(mathbb T^2)$ (the torus in $mathbb R^3$). For me $L^1(mathbb T^2)$ is the set of function $f$ s.t. $xmapsto f(x,y )$ is $T_1$ periodic and $ymapsto f(x,y)$ is $T_2$ periodic, but there is no reason to have $f$ periodic.
real-analysis general-topology
asked Dec 26 '18 at 12:05
user623855user623855
1507
$endgroup$
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show 5 more comments
1
$begingroup$
Why $mathbb R^2/ mathbb Z^2$ is the Torus ? The Torus is indeed $mathbb R/mathbb Ztimes mathbb R/mathbb Z$. But why $mathbb R^2/mathbb Z^2$ also gives the torus ? For me if we glue all points of $mathbb Z^2$ together, then $mathbb R^2/mathbb Z^2$ should look as a sphere and not as a Torus ? Also, for me $L^1(mathbb S^2)$ is the set of integrable periodic functions on $mathbb R^2$. But in my course they wrote it as $L^1(mathbb T^2)$ (the torus in $mathbb R^3$). For me $L^1(mathbb T^2)$ is the set of function $f$ s.t. $xmapsto f(x,y )$ is $T_1$ periodic and $ymapsto f(x,y)$ is $T_2$ periodic, but there is no reason to have $f$ periodic.
real-analysis general-topology
asked Dec 26 '18 at 12:05
user623855user623855
1507
$endgroup$
$begingroup$
Think of the function $p:Bbb{R}^2toBbb{R}/Bbb{Z}timesBbb{R}/Bbb{Z}$ defined by $$p(x,y)=(x+Bbb{Z},y+Bbb{Z}).$$ You should convince yourself of the fact that $p(x_1,y_1)=p(x_2,y_2)$ if and only if $(x_1,y_1)$ and $(x_2,y_2)$ belong to the same coset of $Bbb{Z}^2$. In other words, by the first isomorphism theorem, we have an isomorphism of groups $$Bbb{R}^2/Bbb{Z}^2simeq Bbb{R}/Bbb{Z}timesBbb{R}/Bbb{Z}.$$ Because this bijection is so natural, it is bound to be a homeomorphism also.
$endgroup$
– Jyrki Lahtonen
Dec 26 '18 at 12:14
$begingroup$
Also, if for $x,yinBbb{R}$ and all $m,ninBbb{Z}$ you have $f(x,y)=f(x+n,y)$ as well as $f(x,y)=f(x,y+m)$, then surely you also have $f(x,y)=f(x+n,y+m)$. Doubly periodic.
$endgroup$
– Jyrki Lahtonen
Dec 26 '18 at 12:17
$begingroup$
A topological connection between the sphere and the plane comes from the stereographic projection (when you need to add a point at infinity not to miss the North pole). But, observe that in that projection there are no identifications between the points of the plane.
$endgroup$
– Jyrki Lahtonen
Dec 26 '18 at 12:19
$begingroup$
Or, if you start with a rectangle, you get the torus by identifying (=glueing) the pairs of points on opposite sides. You get the sphere only by identifying all the points of the border to form a single point on the sphere.
$endgroup$
– Jyrki Lahtonen
Dec 26 '18 at 12:28
$begingroup$
@JyrkiLahtonen: Thank you for your answer. So $mathbb R^2/mathbb Z^2$ can't be the sphere... strange, because gluing all points of $mathbb Z^2$ look like a sphere. So for periodic function, is really $L^1(mathbb R^2/mathbb Z^2)$ and not $L^1(mathbb S^2)$ ? Or both work ?
$endgroup$
– user623855
Dec 26 '18 at 12:28
|
show 5 more comments
1
1
1
$begingroup$
Why $mathbb R^2/ mathbb Z^2$ is the Torus ? The Torus is indeed $mathbb R/mathbb Ztimes mathbb R/mathbb Z$. But why $mathbb R^2/mathbb Z^2$ also gives the torus ? For me if we glue all points of $mathbb Z^2$ together, then $mathbb R^2/mathbb Z^2$ should look as a sphere and not as a Torus ? Also, for me $L^1(mathbb S^2)$ is the set of integrable periodic functions on $mathbb R^2$. But in my course they wrote it as $L^1(mathbb T^2)$ (the torus in $mathbb R^3$). For me $L^1(mathbb T^2)$ is the set of function $f$ s.t. $xmapsto f(x,y )$ is $T_1$ periodic and $ymapsto f(x,y)$ is $T_2$ periodic, but there is no reason to have $f$ periodic.
real-analysis general-topology
asked Dec 26 '18 at 12:05
user623855user623855
1507
$endgroup$
Why $mathbb R^2/ mathbb Z^2$ is the Torus ? The Torus is indeed $mathbb R/mathbb Ztimes mathbb R/mathbb Z$. But why $mathbb R^2/mathbb Z^2$ also gives the torus ? For me if we glue all points of $mathbb Z^2$ together, then $mathbb R^2/mathbb Z^2$ should look as a sphere and not as a Torus ? Also, for me $L^1(mathbb S^2)$ is the set of integrable periodic functions on $mathbb R^2$. But in my course they wrote it as $L^1(mathbb T^2)$ (the torus in $mathbb R^3$). For me $L^1(mathbb T^2)$ is the set of function $f$ s.t. $xmapsto f(x,y )$ is $T_1$ periodic and $ymapsto f(x,y)$ is $T_2$ periodic, but there is no reason to have $f$ periodic.
real-analysis general-topology
real-analysis general-topology
asked Dec 26 '18 at 12:05
user623855user623855
1507
asked Dec 26 '18 at 12:05
user623855user623855
1507
asked Dec 26 '18 at 12:05
user623855user623855
1507
asked Dec 26 '18 at 12:05
user623855user623855
1507
asked Dec 26 '18 at 12:05
user623855user623855
1507
1507
$begingroup$
Think of the function $p:Bbb{R}^2toBbb{R}/Bbb{Z}timesBbb{R}/Bbb{Z}$ defined by $$p(x,y)=(x+Bbb{Z},y+Bbb{Z}).$$ You should convince yourself of the fact that $p(x_1,y_1)=p(x_2,y_2)$ if and only if $(x_1,y_1)$ and $(x_2,y_2)$ belong to the same coset of $Bbb{Z}^2$. In other words, by the first isomorphism theorem, we have an isomorphism of groups $$Bbb{R}^2/Bbb{Z}^2simeq Bbb{R}/Bbb{Z}timesBbb{R}/Bbb{Z}.$$ Because this bijection is so natural, it is bound to be a homeomorphism also.
$endgroup$
– Jyrki Lahtonen
Dec 26 '18 at 12:14
$begingroup$
Also, if for $x,yinBbb{R}$ and all $m,ninBbb{Z}$ you have $f(x,y)=f(x+n,y)$ as well as $f(x,y)=f(x,y+m)$, then surely you also have $f(x,y)=f(x+n,y+m)$. Doubly periodic.
$endgroup$
– Jyrki Lahtonen
Dec 26 '18 at 12:17
$begingroup$
A topological connection between the sphere and the plane comes from the stereographic projection (when you need to add a point at infinity not to miss the North pole). But, observe that in that projection there are no identifications between the points of the plane.
$endgroup$
– Jyrki Lahtonen
Dec 26 '18 at 12:19
$begingroup$
Or, if you start with a rectangle, you get the torus by identifying (=glueing) the pairs of points on opposite sides. You get the sphere only by identifying all the points of the border to form a single point on the sphere.
$endgroup$
– Jyrki Lahtonen
Dec 26 '18 at 12:28
$begingroup$
@JyrkiLahtonen: Thank you for your answer. So $mathbb R^2/mathbb Z^2$ can't be the sphere... strange, because gluing all points of $mathbb Z^2$ look like a sphere. So for periodic function, is really $L^1(mathbb R^2/mathbb Z^2)$ and not $L^1(mathbb S^2)$ ? Or both work ?
$endgroup$
– user623855
Dec 26 '18 at 12:28
|
show 5 more comments
$begingroup$
Think of the function $p:Bbb{R}^2toBbb{R}/Bbb{Z}timesBbb{R}/Bbb{Z}$ defined by $$p(x,y)=(x+Bbb{Z},y+Bbb{Z}).$$ You should convince yourself of the fact that $p(x_1,y_1)=p(x_2,y_2)$ if and only if $(x_1,y_1)$ and $(x_2,y_2)$ belong to the same coset of $Bbb{Z}^2$. In other words, by the first isomorphism theorem, we have an isomorphism of groups $$Bbb{R}^2/Bbb{Z}^2simeq Bbb{R}/Bbb{Z}timesBbb{R}/Bbb{Z}.$$ Because this bijection is so natural, it is bound to be a homeomorphism also.
$endgroup$
– Jyrki Lahtonen
Dec 26 '18 at 12:14
$begingroup$
Also, if for $x,yinBbb{R}$ and all $m,ninBbb{Z}$ you have $f(x,y)=f(x+n,y)$ as well as $f(x,y)=f(x,y+m)$, then surely you also have $f(x,y)=f(x+n,y+m)$. Doubly periodic.
$endgroup$
– Jyrki Lahtonen
Dec 26 '18 at 12:17
$begingroup$
A topological connection between the sphere and the plane comes from the stereographic projection (when you need to add a point at infinity not to miss the North pole). But, observe that in that projection there are no identifications between the points of the plane.
$endgroup$
– Jyrki Lahtonen
Dec 26 '18 at 12:19
$begingroup$
Or, if you start with a rectangle, you get the torus by identifying (=glueing) the pairs of points on opposite sides. You get the sphere only by identifying all the points of the border to form a single point on the sphere.
$endgroup$
– Jyrki Lahtonen
Dec 26 '18 at 12:28
$begingroup$
@JyrkiLahtonen: Thank you for your answer. So $mathbb R^2/mathbb Z^2$ can't be the sphere... strange, because gluing all points of $mathbb Z^2$ look like a sphere. So for periodic function, is really $L^1(mathbb R^2/mathbb Z^2)$ and not $L^1(mathbb S^2)$ ? Or both work ?
$endgroup$
– user623855
Dec 26 '18 at 12:28
$begingroup$
Think of the function $p:Bbb{R}^2toBbb{R}/Bbb{Z}timesBbb{R}/Bbb{Z}$ defined by $$p(x,y)=(x+Bbb{Z},y+Bbb{Z}).$$ You should convince yourself of the fact that $p(x_1,y_1)=p(x_2,y_2)$ if and only if $(x_1,y_1)$ and $(x_2,y_2)$ belong to the same coset of $Bbb{Z}^2$. In other words, by the first isomorphism theorem, we have an isomorphism of groups $$Bbb{R}^2/Bbb{Z}^2simeq Bbb{R}/Bbb{Z}timesBbb{R}/Bbb{Z}.$$ Because this bijection is so natural, it is bound to be a homeomorphism also.
$endgroup$
– Jyrki Lahtonen
Dec 26 '18 at 12:14
$begingroup$
Think of the function $p:Bbb{R}^2toBbb{R}/Bbb{Z}timesBbb{R}/Bbb{Z}$ defined by $$p(x,y)=(x+Bbb{Z},y+Bbb{Z}).$$ You should convince yourself of the fact that $p(x_1,y_1)=p(x_2,y_2)$ if and only if $(x_1,y_1)$ and $(x_2,y_2)$ belong to the same coset of $Bbb{Z}^2$. In other words, by the first isomorphism theorem, we have an isomorphism of groups $$Bbb{R}^2/Bbb{Z}^2simeq Bbb{R}/Bbb{Z}timesBbb{R}/Bbb{Z}.$$ Because this bijection is so natural, it is bound to be a homeomorphism also.
$endgroup$
– Jyrki Lahtonen
Dec 26 '18 at 12:14
$begingroup$
Also, if for $x,yinBbb{R}$ and all $m,ninBbb{Z}$ you have $f(x,y)=f(x+n,y)$ as well as $f(x,y)=f(x,y+m)$, then surely you also have $f(x,y)=f(x+n,y+m)$. Doubly periodic.
$endgroup$
– Jyrki Lahtonen
Dec 26 '18 at 12:17
$begingroup$
Also, if for $x,yinBbb{R}$ and all $m,ninBbb{Z}$ you have $f(x,y)=f(x+n,y)$ as well as $f(x,y)=f(x,y+m)$, then surely you also have $f(x,y)=f(x+n,y+m)$. Doubly periodic.
$endgroup$
– Jyrki Lahtonen
Dec 26 '18 at 12:17
$begingroup$
A topological connection between the sphere and the plane comes from the stereographic projection (when you need to add a point at infinity not to miss the North pole). But, observe that in that projection there are no identifications between the points of the plane.
$endgroup$
– Jyrki Lahtonen
Dec 26 '18 at 12:19
$begingroup$
A topological connection between the sphere and the plane comes from the stereographic projection (when you need to add a point at infinity not to miss the North pole). But, observe that in that projection there are no identifications between the points of the plane.
$endgroup$
– Jyrki Lahtonen
Dec 26 '18 at 12:19
$begingroup$
Or, if you start with a rectangle, you get the torus by identifying (=glueing) the pairs of points on opposite sides. You get the sphere only by identifying all the points of the border to form a single point on the sphere.
$endgroup$
– Jyrki Lahtonen
Dec 26 '18 at 12:28
$begingroup$
Or, if you start with a rectangle, you get the torus by identifying (=glueing) the pairs of points on opposite sides. You get the sphere only by identifying all the points of the border to form a single point on the sphere.
$endgroup$
– Jyrki Lahtonen
Dec 26 '18 at 12:28
$begingroup$
@JyrkiLahtonen: Thank you for your answer. So $mathbb R^2/mathbb Z^2$ can't be the sphere... strange, because gluing all points of $mathbb Z^2$ look like a sphere. So for periodic function, is really $L^1(mathbb R^2/mathbb Z^2)$ and not $L^1(mathbb S^2)$ ? Or both work ?
$endgroup$
– user623855
Dec 26 '18 at 12:28
$begingroup$
@JyrkiLahtonen: Thank you for your answer. So $mathbb R^2/mathbb Z^2$ can't be the sphere... strange, because gluing all points of $mathbb Z^2$ look like a sphere. So for periodic function, is really $L^1(mathbb R^2/mathbb Z^2)$ and not $L^1(mathbb S^2)$ ? Or both work ?
$endgroup$
– user623855
Dec 26 '18 at 12:28
|
show 5 more comments
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$begingroup$
Think of the function $p:Bbb{R}^2toBbb{R}/Bbb{Z}timesBbb{R}/Bbb{Z}$ defined by $$p(x,y)=(x+Bbb{Z},y+Bbb{Z}).$$ You should convince yourself of the fact that $p(x_1,y_1)=p(x_2,y_2)$ if and only if $(x_1,y_1)$ and $(x_2,y_2)$ belong to the same coset of $Bbb{Z}^2$. In other words, by the first isomorphism theorem, we have an isomorphism of groups $$Bbb{R}^2/Bbb{Z}^2simeq Bbb{R}/Bbb{Z}timesBbb{R}/Bbb{Z}.$$ Because this bijection is so natural, it is bound to be a homeomorphism also.
$endgroup$
– Jyrki Lahtonen
Dec 26 '18 at 12:14
$begingroup$
Also, if for $x,yinBbb{R}$ and all $m,ninBbb{Z}$ you have $f(x,y)=f(x+n,y)$ as well as $f(x,y)=f(x,y+m)$, then surely you also have $f(x,y)=f(x+n,y+m)$. Doubly periodic.
$endgroup$
– Jyrki Lahtonen
Dec 26 '18 at 12:17
$begingroup$
A topological connection between the sphere and the plane comes from the stereographic projection (when you need to add a point at infinity not to miss the North pole). But, observe that in that projection there are no identifications between the points of the plane.
$endgroup$
– Jyrki Lahtonen
Dec 26 '18 at 12:19
$begingroup$
Or, if you start with a rectangle, you get the torus by identifying (=glueing) the pairs of points on opposite sides. You get the sphere only by identifying all the points of the border to form a single point on the sphere.
$endgroup$
– Jyrki Lahtonen
Dec 26 '18 at 12:28
$begingroup$
@JyrkiLahtonen: Thank you for your answer. So $mathbb R^2/mathbb Z^2$ can't be the sphere... strange, because gluing all points of $mathbb Z^2$ look like a sphere. So for periodic function, is really $L^1(mathbb R^2/mathbb Z^2)$ and not $L^1(mathbb S^2)$ ? Or both work ?
$endgroup$
– user623855
Dec 26 '18 at 12:28