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Conformal mappings that preserve angles and areas but not perimeters?
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Conformal mappings from $U$ to $V$, both subsets of $mathbbC$, locally preserve angles.
But, in general, such mappings neither preserve areas nor preserve perimeters.
Q. Are there clear examples of analytic
conformal mappings that preserve areas but not perimeters? And vice versa?
    Â
    Â
The conformal mapping $w=z^2$ in rectangular coordinates:
(John Mathews.)
cv.complex-variables complex-geometry conformal-geometry analytic-geometry
asked 3 hours ago
Joseph O'Rourke
83.4k15220684
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up vote
1
down vote
favorite
Conformal mappings from $U$ to $V$, both subsets of $mathbbC$, locally preserve angles.
But, in general, such mappings neither preserve areas nor preserve perimeters.
Q. Are there clear examples of analytic
conformal mappings that preserve areas but not perimeters? And vice versa?
    Â
    Â
The conformal mapping $w=z^2$ in rectangular coordinates:
(John Mathews.)
cv.complex-variables complex-geometry conformal-geometry analytic-geometry
asked 3 hours ago
Joseph O'Rourke
83.4k15220684
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Conformal mappings from $U$ to $V$, both subsets of $mathbbC$, locally preserve angles.
But, in general, such mappings neither preserve areas nor preserve perimeters.
Q. Are there clear examples of analytic
conformal mappings that preserve areas but not perimeters? And vice versa?
    Â
    Â
The conformal mapping $w=z^2$ in rectangular coordinates:
(John Mathews.)
cv.complex-variables complex-geometry conformal-geometry analytic-geometry
asked 3 hours ago
Joseph O'Rourke
83.4k15220684
Conformal mappings from $U$ to $V$, both subsets of $mathbbC$, locally preserve angles.
But, in general, such mappings neither preserve areas nor preserve perimeters.
Q. Are there clear examples of analytic
conformal mappings that preserve areas but not perimeters? And vice versa?
    Â
    Â
The conformal mapping $w=z^2$ in rectangular coordinates:
(John Mathews.)
cv.complex-variables complex-geometry conformal-geometry analytic-geometry
cv.complex-variables complex-geometry conformal-geometry analytic-geometry
asked 3 hours ago
Joseph O'Rourke
83.4k15220684
asked 3 hours ago
Joseph O'Rourke
83.4k15220684
edited 3 hours ago
asked 3 hours ago
Joseph O'Rourke
83.4k15220684
asked 3 hours ago
Joseph O'Rourke
83.4k15220684
asked 3 hours ago
Joseph O'Rourke
83.4k15220684
83.4k15220684
add a comment |Â
add a comment |Â
2 Answers
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No.
Either condition implies $|,f'(z)| = 1$ for all $z$ in the domain of $,f$,
which in turn implies that $,f'(z)$ is locally constant
(for instance using the open mapping theorem) and thus that
$,f$ is a Euclidean isometry on each connected component of its domain.
answered 1 hour ago
Noam D. Elkies
54.5k9192278
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4
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See this handout where Brian Conrad talks about the mapmaker's paradox. A mapmaker would like to draw a map that is area preserving and conformal, however a $C^infty$ isomorphism between Riemannian manifolds with corners is conformal and volume preserving if and only if it is an isometry (theorem 2.4)! The case of perimeter preserving conformal maps was discussed in an older question MO172764. Therefore cartographers make do with maps that are conformal but not area preserving, or nonconformal and area preserving, like examples given in the handout.
answered 1 hour ago
Gjergji Zaimi
59.6k3154296
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
No.
Either condition implies $|,f'(z)| = 1$ for all $z$ in the domain of $,f$,
which in turn implies that $,f'(z)$ is locally constant
(for instance using the open mapping theorem) and thus that
$,f$ is a Euclidean isometry on each connected component of its domain.
answered 1 hour ago
Noam D. Elkies
54.5k9192278
add a comment |Â
up vote
6
down vote
No.
Either condition implies $|,f'(z)| = 1$ for all $z$ in the domain of $,f$,
which in turn implies that $,f'(z)$ is locally constant
(for instance using the open mapping theorem) and thus that
$,f$ is a Euclidean isometry on each connected component of its domain.
answered 1 hour ago
Noam D. Elkies
54.5k9192278
add a comment |Â
up vote
6
down vote
up vote
6
down vote
No.
Either condition implies $|,f'(z)| = 1$ for all $z$ in the domain of $,f$,
which in turn implies that $,f'(z)$ is locally constant
(for instance using the open mapping theorem) and thus that
$,f$ is a Euclidean isometry on each connected component of its domain.
answered 1 hour ago
Noam D. Elkies
54.5k9192278
No.
Either condition implies $|,f'(z)| = 1$ for all $z$ in the domain of $,f$,
which in turn implies that $,f'(z)$ is locally constant
(for instance using the open mapping theorem) and thus that
$,f$ is a Euclidean isometry on each connected component of its domain.
answered 1 hour ago
Noam D. Elkies
54.5k9192278
answered 1 hour ago
Noam D. Elkies
54.5k9192278
answered 1 hour ago
Noam D. Elkies
54.5k9192278
answered 1 hour ago
Noam D. Elkies
54.5k9192278
54.5k9192278
add a comment |Â
add a comment |Â
up vote
4
down vote
See this handout where Brian Conrad talks about the mapmaker's paradox. A mapmaker would like to draw a map that is area preserving and conformal, however a $C^infty$ isomorphism between Riemannian manifolds with corners is conformal and volume preserving if and only if it is an isometry (theorem 2.4)! The case of perimeter preserving conformal maps was discussed in an older question MO172764. Therefore cartographers make do with maps that are conformal but not area preserving, or nonconformal and area preserving, like examples given in the handout.
answered 1 hour ago
Gjergji Zaimi
59.6k3154296
add a comment |Â
up vote
4
down vote
See this handout where Brian Conrad talks about the mapmaker's paradox. A mapmaker would like to draw a map that is area preserving and conformal, however a $C^infty$ isomorphism between Riemannian manifolds with corners is conformal and volume preserving if and only if it is an isometry (theorem 2.4)! The case of perimeter preserving conformal maps was discussed in an older question MO172764. Therefore cartographers make do with maps that are conformal but not area preserving, or nonconformal and area preserving, like examples given in the handout.
answered 1 hour ago
Gjergji Zaimi
59.6k3154296
add a comment |Â
up vote
4
down vote
up vote
4
down vote
See this handout where Brian Conrad talks about the mapmaker's paradox. A mapmaker would like to draw a map that is area preserving and conformal, however a $C^infty$ isomorphism between Riemannian manifolds with corners is conformal and volume preserving if and only if it is an isometry (theorem 2.4)! The case of perimeter preserving conformal maps was discussed in an older question MO172764. Therefore cartographers make do with maps that are conformal but not area preserving, or nonconformal and area preserving, like examples given in the handout.
answered 1 hour ago
Gjergji Zaimi
59.6k3154296
See this handout where Brian Conrad talks about the mapmaker's paradox. A mapmaker would like to draw a map that is area preserving and conformal, however a $C^infty$ isomorphism between Riemannian manifolds with corners is conformal and volume preserving if and only if it is an isometry (theorem 2.4)! The case of perimeter preserving conformal maps was discussed in an older question MO172764. Therefore cartographers make do with maps that are conformal but not area preserving, or nonconformal and area preserving, like examples given in the handout.
answered 1 hour ago
Gjergji Zaimi
59.6k3154296
answered 1 hour ago
Gjergji Zaimi
59.6k3154296
answered 1 hour ago
Gjergji Zaimi
59.6k3154296
answered 1 hour ago
Gjergji Zaimi
59.6k3154296
59.6k3154296
add a comment |Â
add a comment |Â
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