Show Circle Group $ mathbb{T} $ isomorphic to $mathbb { C } ^ { * } / mathbb { R } ^ { * } $
$begingroup$
I'm trying to think of an extension to this question, which asks to show whether $mathbb { C } ^ { * } / mathbb { R } ^ { + } simeq mathbb { T } $. They do it using the First Isomorphism Theorem.
I think my postulate should be true because $ mathbb { R } ^ { + } simeq { R } ^ { * } $, (the positive reals under multiplication and the reals under multiplication ), but I'm having difficulty coming up with a homomorphism
$$ phi : mathbb { C } ^ { * } mapsto mathbb{T} $$
Any suggestions?
abstract-algebra group-theory infinite-groups
asked Dec 2 '18 at 15:58
Andrew HardyAndrew Hardy
125
$endgroup$
add a comment |
$begingroup$
I'm trying to think of an extension to this question, which asks to show whether $mathbb { C } ^ { * } / mathbb { R } ^ { + } simeq mathbb { T } $. They do it using the First Isomorphism Theorem.
I think my postulate should be true because $ mathbb { R } ^ { + } simeq { R } ^ { * } $, (the positive reals under multiplication and the reals under multiplication ), but I'm having difficulty coming up with a homomorphism
$$ phi : mathbb { C } ^ { * } mapsto mathbb{T} $$
Any suggestions?
abstract-algebra group-theory infinite-groups
asked Dec 2 '18 at 15:58
Andrew HardyAndrew Hardy
125
$endgroup$
add a comment |
$begingroup$
I'm trying to think of an extension to this question, which asks to show whether $mathbb { C } ^ { * } / mathbb { R } ^ { + } simeq mathbb { T } $. They do it using the First Isomorphism Theorem.
I think my postulate should be true because $ mathbb { R } ^ { + } simeq { R } ^ { * } $, (the positive reals under multiplication and the reals under multiplication ), but I'm having difficulty coming up with a homomorphism
$$ phi : mathbb { C } ^ { * } mapsto mathbb{T} $$
Any suggestions?
abstract-algebra group-theory infinite-groups
asked Dec 2 '18 at 15:58
Andrew HardyAndrew Hardy
125
$endgroup$
I'm trying to think of an extension to this question, which asks to show whether $mathbb { C } ^ { * } / mathbb { R } ^ { + } simeq mathbb { T } $. They do it using the First Isomorphism Theorem.
I think my postulate should be true because $ mathbb { R } ^ { + } simeq { R } ^ { * } $, (the positive reals under multiplication and the reals under multiplication ), but I'm having difficulty coming up with a homomorphism
$$ phi : mathbb { C } ^ { * } mapsto mathbb{T} $$
Any suggestions?
abstract-algebra group-theory infinite-groups
abstract-algebra group-theory infinite-groups
asked Dec 2 '18 at 15:58
Andrew HardyAndrew Hardy
125
asked Dec 2 '18 at 15:58
Andrew HardyAndrew Hardy
125
asked Dec 2 '18 at 15:58
Andrew HardyAndrew Hardy
125
asked Dec 2 '18 at 15:58
Andrew HardyAndrew Hardy
125
asked Dec 2 '18 at 15:58
Andrew HardyAndrew Hardy
125
125
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Just use the same homomorphism, and then double the angle. Dividing by negatives means that we identify numbers on opposite sides of the origin. Double the angle, and that's not a problem anymore. After all, we can map the circle 2-1 to the circle.
(Or, as Tsemo Aristide's answer put it, going from the projective line to the circle is an angle-doubling)
answered Dec 2 '18 at 16:15
jmerryjmerry
5,097514
$endgroup$
add a comment |
$begingroup$
$mathbb{R}^*$ is not isomorphic to $mathbb{R}^+$ as topological groups because the first set is not connected and the second is connected. But the quotient of $mathbb{C}^*$ by $mathbb{R}^*$ is the projective line which is the circle.
answered Dec 2 '18 at 16:04
Tsemo AristideTsemo Aristide
57.2k11444
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Just use the same homomorphism, and then double the angle. Dividing by negatives means that we identify numbers on opposite sides of the origin. Double the angle, and that's not a problem anymore. After all, we can map the circle 2-1 to the circle.
(Or, as Tsemo Aristide's answer put it, going from the projective line to the circle is an angle-doubling)
answered Dec 2 '18 at 16:15
jmerryjmerry
5,097514
$endgroup$
add a comment |
$begingroup$
Just use the same homomorphism, and then double the angle. Dividing by negatives means that we identify numbers on opposite sides of the origin. Double the angle, and that's not a problem anymore. After all, we can map the circle 2-1 to the circle.
(Or, as Tsemo Aristide's answer put it, going from the projective line to the circle is an angle-doubling)
answered Dec 2 '18 at 16:15
jmerryjmerry
5,097514
$endgroup$
add a comment |
$begingroup$
Just use the same homomorphism, and then double the angle. Dividing by negatives means that we identify numbers on opposite sides of the origin. Double the angle, and that's not a problem anymore. After all, we can map the circle 2-1 to the circle.
(Or, as Tsemo Aristide's answer put it, going from the projective line to the circle is an angle-doubling)
answered Dec 2 '18 at 16:15
jmerryjmerry
5,097514
$endgroup$
Just use the same homomorphism, and then double the angle. Dividing by negatives means that we identify numbers on opposite sides of the origin. Double the angle, and that's not a problem anymore. After all, we can map the circle 2-1 to the circle.
(Or, as Tsemo Aristide's answer put it, going from the projective line to the circle is an angle-doubling)
answered Dec 2 '18 at 16:15
jmerryjmerry
5,097514
answered Dec 2 '18 at 16:15
jmerryjmerry
5,097514
answered Dec 2 '18 at 16:15
jmerryjmerry
5,097514
answered Dec 2 '18 at 16:15
jmerryjmerry
5,097514
5,097514
add a comment |
add a comment |
$begingroup$
$mathbb{R}^*$ is not isomorphic to $mathbb{R}^+$ as topological groups because the first set is not connected and the second is connected. But the quotient of $mathbb{C}^*$ by $mathbb{R}^*$ is the projective line which is the circle.
answered Dec 2 '18 at 16:04
Tsemo AristideTsemo Aristide
57.2k11444
$endgroup$
add a comment |
$begingroup$
$mathbb{R}^*$ is not isomorphic to $mathbb{R}^+$ as topological groups because the first set is not connected and the second is connected. But the quotient of $mathbb{C}^*$ by $mathbb{R}^*$ is the projective line which is the circle.
answered Dec 2 '18 at 16:04
Tsemo AristideTsemo Aristide
57.2k11444
$endgroup$
add a comment |
$begingroup$
$mathbb{R}^*$ is not isomorphic to $mathbb{R}^+$ as topological groups because the first set is not connected and the second is connected. But the quotient of $mathbb{C}^*$ by $mathbb{R}^*$ is the projective line which is the circle.
answered Dec 2 '18 at 16:04
Tsemo AristideTsemo Aristide
57.2k11444
$endgroup$
$mathbb{R}^*$ is not isomorphic to $mathbb{R}^+$ as topological groups because the first set is not connected and the second is connected. But the quotient of $mathbb{C}^*$ by $mathbb{R}^*$ is the projective line which is the circle.
answered Dec 2 '18 at 16:04
Tsemo AristideTsemo Aristide
57.2k11444
answered Dec 2 '18 at 16:04
Tsemo AristideTsemo Aristide
57.2k11444
answered Dec 2 '18 at 16:04
Tsemo AristideTsemo Aristide
57.2k11444
answered Dec 2 '18 at 16:04
Tsemo AristideTsemo Aristide
57.2k11444
57.2k11444
add a comment |
add a comment |
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