Complex multiplication and symetry
The complex multiplication is a similar map, i.e. for all $w,z_1,z_2in mathbb{C}$ we have
$$|w||z_1 - z_2|= |wz_1-wz_2|.$$
Therefore $f:mathbb{C}rightarrow mathbb{C}, f(z):=zetabarzeta^{-1}bar{z}$ for any nonzero $zetainBbb{C}$ must also be a isometry and since the only fixed points of $f$ are ${azeta:ainmathbb{R}}$, $f$ must be the the map which provides the point $z'$ which is symmetrical to $z$ on the $azeta$-axis.
Can somebody please explain me the last implication? That given a isometry on a plane and a set of fixed points implicates a symetry-map? I have tried to understand this Statement by drawing upon elementar-geometry books, but my linear Algebra skills are not sufficient enough. That's why I would prefer a intuitive explaination over a maybe rigorous proof in this case.
Thanks in advance.
complex-numbers euclidean-geometry
edited Nov 29 at 15:18
Servaes
22.3k33793
asked Nov 29 at 13:32
RM777
1808
|
show 3 more comments
The complex multiplication is a similar map, i.e. for all $w,z_1,z_2in mathbb{C}$ we have
$$|w||z_1 - z_2|= |wz_1-wz_2|.$$
Therefore $f:mathbb{C}rightarrow mathbb{C}, f(z):=zetabarzeta^{-1}bar{z}$ for any nonzero $zetainBbb{C}$ must also be a isometry and since the only fixed points of $f$ are ${azeta:ainmathbb{R}}$, $f$ must be the the map which provides the point $z'$ which is symmetrical to $z$ on the $azeta$-axis.
Can somebody please explain me the last implication? That given a isometry on a plane and a set of fixed points implicates a symetry-map? I have tried to understand this Statement by drawing upon elementar-geometry books, but my linear Algebra skills are not sufficient enough. That's why I would prefer a intuitive explaination over a maybe rigorous proof in this case.
Thanks in advance.
complex-numbers euclidean-geometry
edited Nov 29 at 15:18
Servaes
22.3k33793
asked Nov 29 at 13:32
RM777
1808
What is $zeta$ here? If $azeta,ainBbb{R}$ and $aneq0$ then $zetainBbb{R}$ and so $f$ is simply complex conjugation.
– Servaes
Nov 29 at 13:46
$zeta$ is a complex number
– RM777
Nov 29 at 14:10
So then either $zetainBbb{R}$ or $a=0$. Your question is very unclear.
– Servaes
Nov 29 at 14:13
Given the complex plane, you choose a $zeta in mathbb{C}$, you draw a line intersecting $0$ to $zeta$ and define the line as $zeta$-axis. You have the function f and you put in a complex number z. You observe that if the complex number is $zeta$ , $f(zeta)=zetabarzeta^{-1}bar{zeta}= zeta$, so if you choose $zeta = z, f$ behaves like the identity function. If your Input is $azeta$ then $f(azeta) = (azeta)$ thus for values of $azeta,ain mathbb{R}$ you get the identity back otherwise the result is different and due to f being isometric, $f$ must be the symetrical projection.
– RM777
Nov 29 at 14:30
Maybe my notes of the lecture are also false but that was just the context of my Question. My Question was "Does given a isometry on a plane and a set of fixed points implicates a symetry-map?"
– RM777
Nov 29 at 14:34
|
show 3 more comments
The complex multiplication is a similar map, i.e. for all $w,z_1,z_2in mathbb{C}$ we have
$$|w||z_1 - z_2|= |wz_1-wz_2|.$$
Therefore $f:mathbb{C}rightarrow mathbb{C}, f(z):=zetabarzeta^{-1}bar{z}$ for any nonzero $zetainBbb{C}$ must also be a isometry and since the only fixed points of $f$ are ${azeta:ainmathbb{R}}$, $f$ must be the the map which provides the point $z'$ which is symmetrical to $z$ on the $azeta$-axis.
Can somebody please explain me the last implication? That given a isometry on a plane and a set of fixed points implicates a symetry-map? I have tried to understand this Statement by drawing upon elementar-geometry books, but my linear Algebra skills are not sufficient enough. That's why I would prefer a intuitive explaination over a maybe rigorous proof in this case.
Thanks in advance.
complex-numbers euclidean-geometry
edited Nov 29 at 15:18
Servaes
22.3k33793
asked Nov 29 at 13:32
RM777
1808
The complex multiplication is a similar map, i.e. for all $w,z_1,z_2in mathbb{C}$ we have
$$|w||z_1 - z_2|= |wz_1-wz_2|.$$
Therefore $f:mathbb{C}rightarrow mathbb{C}, f(z):=zetabarzeta^{-1}bar{z}$ for any nonzero $zetainBbb{C}$ must also be a isometry and since the only fixed points of $f$ are ${azeta:ainmathbb{R}}$, $f$ must be the the map which provides the point $z'$ which is symmetrical to $z$ on the $azeta$-axis.
Can somebody please explain me the last implication? That given a isometry on a plane and a set of fixed points implicates a symetry-map? I have tried to understand this Statement by drawing upon elementar-geometry books, but my linear Algebra skills are not sufficient enough. That's why I would prefer a intuitive explaination over a maybe rigorous proof in this case.
Thanks in advance.
complex-numbers euclidean-geometry
complex-numbers euclidean-geometry
edited Nov 29 at 15:18
Servaes
22.3k33793
asked Nov 29 at 13:32
RM777
1808
edited Nov 29 at 15:18
Servaes
22.3k33793
asked Nov 29 at 13:32
RM777
1808
edited Nov 29 at 15:18
Servaes
22.3k33793
edited Nov 29 at 15:18
Servaes
22.3k33793
edited Nov 29 at 15:18
Servaes
22.3k33793
22.3k33793
asked Nov 29 at 13:32
RM777
1808
asked Nov 29 at 13:32
RM777
1808
asked Nov 29 at 13:32
RM777
1808
1808
What is $zeta$ here? If $azeta,ainBbb{R}$ and $aneq0$ then $zetainBbb{R}$ and so $f$ is simply complex conjugation.
– Servaes
Nov 29 at 13:46
$zeta$ is a complex number
– RM777
Nov 29 at 14:10
So then either $zetainBbb{R}$ or $a=0$. Your question is very unclear.
– Servaes
Nov 29 at 14:13
Given the complex plane, you choose a $zeta in mathbb{C}$, you draw a line intersecting $0$ to $zeta$ and define the line as $zeta$-axis. You have the function f and you put in a complex number z. You observe that if the complex number is $zeta$ , $f(zeta)=zetabarzeta^{-1}bar{zeta}= zeta$, so if you choose $zeta = z, f$ behaves like the identity function. If your Input is $azeta$ then $f(azeta) = (azeta)$ thus for values of $azeta,ain mathbb{R}$ you get the identity back otherwise the result is different and due to f being isometric, $f$ must be the symetrical projection.
– RM777
Nov 29 at 14:30
Maybe my notes of the lecture are also false but that was just the context of my Question. My Question was "Does given a isometry on a plane and a set of fixed points implicates a symetry-map?"
– RM777
Nov 29 at 14:34
|
show 3 more comments
What is $zeta$ here? If $azeta,ainBbb{R}$ and $aneq0$ then $zetainBbb{R}$ and so $f$ is simply complex conjugation.
– Servaes
Nov 29 at 13:46
$zeta$ is a complex number
– RM777
Nov 29 at 14:10
So then either $zetainBbb{R}$ or $a=0$. Your question is very unclear.
– Servaes
Nov 29 at 14:13
Given the complex plane, you choose a $zeta in mathbb{C}$, you draw a line intersecting $0$ to $zeta$ and define the line as $zeta$-axis. You have the function f and you put in a complex number z. You observe that if the complex number is $zeta$ , $f(zeta)=zetabarzeta^{-1}bar{zeta}= zeta$, so if you choose $zeta = z, f$ behaves like the identity function. If your Input is $azeta$ then $f(azeta) = (azeta)$ thus for values of $azeta,ain mathbb{R}$ you get the identity back otherwise the result is different and due to f being isometric, $f$ must be the symetrical projection.
– RM777
Nov 29 at 14:30
Maybe my notes of the lecture are also false but that was just the context of my Question. My Question was "Does given a isometry on a plane and a set of fixed points implicates a symetry-map?"
– RM777
Nov 29 at 14:34
What is $zeta$ here? If $azeta,ainBbb{R}$ and $aneq0$ then $zetainBbb{R}$ and so $f$ is simply complex conjugation.
– Servaes
Nov 29 at 13:46
What is $zeta$ here? If $azeta,ainBbb{R}$ and $aneq0$ then $zetainBbb{R}$ and so $f$ is simply complex conjugation.
– Servaes
Nov 29 at 13:46
$zeta$ is a complex number
– RM777
Nov 29 at 14:10
$zeta$ is a complex number
– RM777
Nov 29 at 14:10
So then either $zetainBbb{R}$ or $a=0$. Your question is very unclear.
– Servaes
Nov 29 at 14:13
So then either $zetainBbb{R}$ or $a=0$. Your question is very unclear.
– Servaes
Nov 29 at 14:13
Given the complex plane, you choose a $zeta in mathbb{C}$, you draw a line intersecting $0$ to $zeta$ and define the line as $zeta$-axis. You have the function f and you put in a complex number z. You observe that if the complex number is $zeta$ , $f(zeta)=zetabarzeta^{-1}bar{zeta}= zeta$, so if you choose $zeta = z, f$ behaves like the identity function. If your Input is $azeta$ then $f(azeta) = (azeta)$ thus for values of $azeta,ain mathbb{R}$ you get the identity back otherwise the result is different and due to f being isometric, $f$ must be the symetrical projection.
– RM777
Nov 29 at 14:30
Given the complex plane, you choose a $zeta in mathbb{C}$, you draw a line intersecting $0$ to $zeta$ and define the line as $zeta$-axis. You have the function f and you put in a complex number z. You observe that if the complex number is $zeta$ , $f(zeta)=zetabarzeta^{-1}bar{zeta}= zeta$, so if you choose $zeta = z, f$ behaves like the identity function. If your Input is $azeta$ then $f(azeta) = (azeta)$ thus for values of $azeta,ain mathbb{R}$ you get the identity back otherwise the result is different and due to f being isometric, $f$ must be the symetrical projection.
– RM777
Nov 29 at 14:30
Maybe my notes of the lecture are also false but that was just the context of my Question. My Question was "Does given a isometry on a plane and a set of fixed points implicates a symetry-map?"
– RM777
Nov 29 at 14:34
Maybe my notes of the lecture are also false but that was just the context of my Question. My Question was "Does given a isometry on a plane and a set of fixed points implicates a symetry-map?"
– RM777
Nov 29 at 14:34
|
show 3 more comments
1 Answer
1
active
oldest
votes
Suppose a plane isometry $f$ fixes a line through the origin $L$ pointwise. Let $p,qin L$ be two distinct points. For every point $z$ in the plane we have
$$|p-z|=|f(p)-f(z)|=|p-f(z)|
qquadtext{ and }qquad
|q-z|=|f(q)-f(z)|=|q-f(z)|.$$
Geometrically this means that if we draw two circles centered at $p$ and $q$ and both passing through $z$, they also both pass through $f(z)$, as in the picture:
![[Picture in the making]](https://i.stack.imgur.com/vPwKC.png)
This tells us that either $f(z)=z$, or $f(z)$ is the reflection of $z$ in the line $L$. As this is true for every $z$, and $f$ is an isometry, it follows (prove this!) that $f$ is either the identity or the reflection in the line $L$.
answered Nov 29 at 15:10
Servaes
22.3k33793
I don't know how to prove it properly but because (I use the bar Notation to describe a track) $|overline{zp}| = |overline{f(z)p}|$ we have an isosceles triangle $zpf(z)$ therefor the hight of the triangle must be perpendicular to $overline{zf(z)}$ showing that f(z) is the reflection or the triangle is deformed because $|overline{zf(z)}|=0$ and therefore f must be the identity then.
– RM777
Nov 30 at 20:19
add a comment |
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1 Answer
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Suppose a plane isometry $f$ fixes a line through the origin $L$ pointwise. Let $p,qin L$ be two distinct points. For every point $z$ in the plane we have
$$|p-z|=|f(p)-f(z)|=|p-f(z)|
qquadtext{ and }qquad
|q-z|=|f(q)-f(z)|=|q-f(z)|.$$
Geometrically this means that if we draw two circles centered at $p$ and $q$ and both passing through $z$, they also both pass through $f(z)$, as in the picture:
This tells us that either $f(z)=z$, or $f(z)$ is the reflection of $z$ in the line $L$. As this is true for every $z$, and $f$ is an isometry, it follows (prove this!) that $f$ is either the identity or the reflection in the line $L$.
answered Nov 29 at 15:10
Servaes
22.3k33793
I don't know how to prove it properly but because (I use the bar Notation to describe a track) $|overline{zp}| = |overline{f(z)p}|$ we have an isosceles triangle $zpf(z)$ therefor the hight of the triangle must be perpendicular to $overline{zf(z)}$ showing that f(z) is the reflection or the triangle is deformed because $|overline{zf(z)}|=0$ and therefore f must be the identity then.
– RM777
Nov 30 at 20:19
add a comment |
Suppose a plane isometry $f$ fixes a line through the origin $L$ pointwise. Let $p,qin L$ be two distinct points. For every point $z$ in the plane we have
$$|p-z|=|f(p)-f(z)|=|p-f(z)|
qquadtext{ and }qquad
|q-z|=|f(q)-f(z)|=|q-f(z)|.$$
Geometrically this means that if we draw two circles centered at $p$ and $q$ and both passing through $z$, they also both pass through $f(z)$, as in the picture:
This tells us that either $f(z)=z$, or $f(z)$ is the reflection of $z$ in the line $L$. As this is true for every $z$, and $f$ is an isometry, it follows (prove this!) that $f$ is either the identity or the reflection in the line $L$.
answered Nov 29 at 15:10
Servaes
22.3k33793
I don't know how to prove it properly but because (I use the bar Notation to describe a track) $|overline{zp}| = |overline{f(z)p}|$ we have an isosceles triangle $zpf(z)$ therefor the hight of the triangle must be perpendicular to $overline{zf(z)}$ showing that f(z) is the reflection or the triangle is deformed because $|overline{zf(z)}|=0$ and therefore f must be the identity then.
– RM777
Nov 30 at 20:19
add a comment |
Suppose a plane isometry $f$ fixes a line through the origin $L$ pointwise. Let $p,qin L$ be two distinct points. For every point $z$ in the plane we have
$$|p-z|=|f(p)-f(z)|=|p-f(z)|
qquadtext{ and }qquad
|q-z|=|f(q)-f(z)|=|q-f(z)|.$$
Geometrically this means that if we draw two circles centered at $p$ and $q$ and both passing through $z$, they also both pass through $f(z)$, as in the picture:
This tells us that either $f(z)=z$, or $f(z)$ is the reflection of $z$ in the line $L$. As this is true for every $z$, and $f$ is an isometry, it follows (prove this!) that $f$ is either the identity or the reflection in the line $L$.
answered Nov 29 at 15:10
Servaes
22.3k33793
Suppose a plane isometry $f$ fixes a line through the origin $L$ pointwise. Let $p,qin L$ be two distinct points. For every point $z$ in the plane we have
$$|p-z|=|f(p)-f(z)|=|p-f(z)|
qquadtext{ and }qquad
|q-z|=|f(q)-f(z)|=|q-f(z)|.$$
Geometrically this means that if we draw two circles centered at $p$ and $q$ and both passing through $z$, they also both pass through $f(z)$, as in the picture:
This tells us that either $f(z)=z$, or $f(z)$ is the reflection of $z$ in the line $L$. As this is true for every $z$, and $f$ is an isometry, it follows (prove this!) that $f$ is either the identity or the reflection in the line $L$.
answered Nov 29 at 15:10
Servaes
22.3k33793
answered Nov 29 at 15:10
Servaes
22.3k33793
answered Nov 29 at 15:10
Servaes
22.3k33793
answered Nov 29 at 15:10
Servaes
22.3k33793
22.3k33793
I don't know how to prove it properly but because (I use the bar Notation to describe a track) $|overline{zp}| = |overline{f(z)p}|$ we have an isosceles triangle $zpf(z)$ therefor the hight of the triangle must be perpendicular to $overline{zf(z)}$ showing that f(z) is the reflection or the triangle is deformed because $|overline{zf(z)}|=0$ and therefore f must be the identity then.
– RM777
Nov 30 at 20:19
add a comment |
I don't know how to prove it properly but because (I use the bar Notation to describe a track) $|overline{zp}| = |overline{f(z)p}|$ we have an isosceles triangle $zpf(z)$ therefor the hight of the triangle must be perpendicular to $overline{zf(z)}$ showing that f(z) is the reflection or the triangle is deformed because $|overline{zf(z)}|=0$ and therefore f must be the identity then.
– RM777
Nov 30 at 20:19
I don't know how to prove it properly but because (I use the bar Notation to describe a track) $|overline{zp}| = |overline{f(z)p}|$ we have an isosceles triangle $zpf(z)$ therefor the hight of the triangle must be perpendicular to $overline{zf(z)}$ showing that f(z) is the reflection or the triangle is deformed because $|overline{zf(z)}|=0$ and therefore f must be the identity then.
– RM777
Nov 30 at 20:19
I don't know how to prove it properly but because (I use the bar Notation to describe a track) $|overline{zp}| = |overline{f(z)p}|$ we have an isosceles triangle $zpf(z)$ therefor the hight of the triangle must be perpendicular to $overline{zf(z)}$ showing that f(z) is the reflection or the triangle is deformed because $|overline{zf(z)}|=0$ and therefore f must be the identity then.
– RM777
Nov 30 at 20:19
add a comment |
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What is $zeta$ here? If $azeta,ainBbb{R}$ and $aneq0$ then $zetainBbb{R}$ and so $f$ is simply complex conjugation.
– Servaes
Nov 29 at 13:46
$zeta$ is a complex number
– RM777
Nov 29 at 14:10
So then either $zetainBbb{R}$ or $a=0$. Your question is very unclear.
– Servaes
Nov 29 at 14:13
Given the complex plane, you choose a $zeta in mathbb{C}$, you draw a line intersecting $0$ to $zeta$ and define the line as $zeta$-axis. You have the function f and you put in a complex number z. You observe that if the complex number is $zeta$ , $f(zeta)=zetabarzeta^{-1}bar{zeta}= zeta$, so if you choose $zeta = z, f$ behaves like the identity function. If your Input is $azeta$ then $f(azeta) = (azeta)$ thus for values of $azeta,ain mathbb{R}$ you get the identity back otherwise the result is different and due to f being isometric, $f$ must be the symetrical projection.
– RM777
Nov 29 at 14:30
Maybe my notes of the lecture are also false but that was just the context of my Question. My Question was "Does given a isometry on a plane and a set of fixed points implicates a symetry-map?"
– RM777
Nov 29 at 14:34