Why euclidean topology on complex numbers?
This is a rather inconcrete question and i am hoping for different answers:
The topology on $mathbb{Q}$ and $mathbb{R}$ is natural in being the order topology of these ordered fields.
Endowing $mathbb{C}$ with the topology of $mathbb{R}^2$ is the key to all those nice results of classical complex analysis and the important fact, that a function $mathbb{C} to mathbb{C}$ is differentiable if and only if it is a conformal (locally angle-preserving) map. So the theory of holomorphic functions is more on geometry of $mathbb{R}^2$ than on the properties of $mathbb{C}$ as a field, I think.
Now suppose, the existence of an algebraic closure of $mathbb{Q}$ would have been discovered before the invention of $mathbb{C}$. It is hard to imagine that someone was like "lets endow $mathbb{Q}[sqrt{-1}]$ with the product metric of $mathbb{Q}^2$ and embed its algebraic closure into its metric completion!"
Because there are so many other choices: Choose $sqrt{2}$ instead of $sqrt{-1}$; endow $mathbb{Q}[sqrt{-1},sqrt{2}]$ with the product topology of $mathbb{Q}^3$; etc.
My question is: Why does $mathbb{C}$ deserve the euclidean topology of $mathbb{R}^2$ and are there other choices?
complex-analysis complex-numbers
asked Nov 25 at 0:10
Lucina
334
add a comment |
This is a rather inconcrete question and i am hoping for different answers:
The topology on $mathbb{Q}$ and $mathbb{R}$ is natural in being the order topology of these ordered fields.
Endowing $mathbb{C}$ with the topology of $mathbb{R}^2$ is the key to all those nice results of classical complex analysis and the important fact, that a function $mathbb{C} to mathbb{C}$ is differentiable if and only if it is a conformal (locally angle-preserving) map. So the theory of holomorphic functions is more on geometry of $mathbb{R}^2$ than on the properties of $mathbb{C}$ as a field, I think.
Now suppose, the existence of an algebraic closure of $mathbb{Q}$ would have been discovered before the invention of $mathbb{C}$. It is hard to imagine that someone was like "lets endow $mathbb{Q}[sqrt{-1}]$ with the product metric of $mathbb{Q}^2$ and embed its algebraic closure into its metric completion!"
Because there are so many other choices: Choose $sqrt{2}$ instead of $sqrt{-1}$; endow $mathbb{Q}[sqrt{-1},sqrt{2}]$ with the product topology of $mathbb{Q}^3$; etc.
My question is: Why does $mathbb{C}$ deserve the euclidean topology of $mathbb{R}^2$ and are there other choices?
complex-analysis complex-numbers
asked Nov 25 at 0:10
Lucina
334
In fact, there are other choices, depending on what you're interested about $mathbb{C}$. For instance, in Algebraic Geometry one does not endow $mathbb{C}$ with its usual topology but rather with the Zariski Topology, the minimal topology where polynomial functions are continuous. In this case, that topology is simply the co-finite topology, that is, closed sets are finite (or all $mathbb{C}$).
– Juan Diego Rojas
Nov 25 at 1:25
You answered your own question with "Endowing $C$ with the topology of $R^2$ is the key to all those nice results of classical complex analysis". Without it you don't get all those nice results. And yes, there are a lot of other choices.
– Somos
Nov 25 at 2:15
What is the field $mathbb{C}$ without its usual topology ? A non-countably infinite algebraically closed transcendental extension of $mathbb{Q}$ ? So all we can say is if a complex number $a$ is algebraic over $mathbb{Q}(b_1,b_2,ldots)$ or not ? The different topologies on algebraic numbers are interesting, for example the completion wrt. $sup_sigma |a^sigma|$.
– reuns
Nov 25 at 2:46
add a comment |
This is a rather inconcrete question and i am hoping for different answers:
The topology on $mathbb{Q}$ and $mathbb{R}$ is natural in being the order topology of these ordered fields.
Endowing $mathbb{C}$ with the topology of $mathbb{R}^2$ is the key to all those nice results of classical complex analysis and the important fact, that a function $mathbb{C} to mathbb{C}$ is differentiable if and only if it is a conformal (locally angle-preserving) map. So the theory of holomorphic functions is more on geometry of $mathbb{R}^2$ than on the properties of $mathbb{C}$ as a field, I think.
Now suppose, the existence of an algebraic closure of $mathbb{Q}$ would have been discovered before the invention of $mathbb{C}$. It is hard to imagine that someone was like "lets endow $mathbb{Q}[sqrt{-1}]$ with the product metric of $mathbb{Q}^2$ and embed its algebraic closure into its metric completion!"
Because there are so many other choices: Choose $sqrt{2}$ instead of $sqrt{-1}$; endow $mathbb{Q}[sqrt{-1},sqrt{2}]$ with the product topology of $mathbb{Q}^3$; etc.
My question is: Why does $mathbb{C}$ deserve the euclidean topology of $mathbb{R}^2$ and are there other choices?
complex-analysis complex-numbers
asked Nov 25 at 0:10
Lucina
334
This is a rather inconcrete question and i am hoping for different answers:
The topology on $mathbb{Q}$ and $mathbb{R}$ is natural in being the order topology of these ordered fields.
Endowing $mathbb{C}$ with the topology of $mathbb{R}^2$ is the key to all those nice results of classical complex analysis and the important fact, that a function $mathbb{C} to mathbb{C}$ is differentiable if and only if it is a conformal (locally angle-preserving) map. So the theory of holomorphic functions is more on geometry of $mathbb{R}^2$ than on the properties of $mathbb{C}$ as a field, I think.
Now suppose, the existence of an algebraic closure of $mathbb{Q}$ would have been discovered before the invention of $mathbb{C}$. It is hard to imagine that someone was like "lets endow $mathbb{Q}[sqrt{-1}]$ with the product metric of $mathbb{Q}^2$ and embed its algebraic closure into its metric completion!"
Because there are so many other choices: Choose $sqrt{2}$ instead of $sqrt{-1}$; endow $mathbb{Q}[sqrt{-1},sqrt{2}]$ with the product topology of $mathbb{Q}^3$; etc.
My question is: Why does $mathbb{C}$ deserve the euclidean topology of $mathbb{R}^2$ and are there other choices?
complex-analysis complex-numbers
complex-analysis complex-numbers
asked Nov 25 at 0:10
Lucina
334
asked Nov 25 at 0:10
Lucina
334
asked Nov 25 at 0:10
Lucina
334
asked Nov 25 at 0:10
Lucina
334
asked Nov 25 at 0:10
Lucina
334
334
In fact, there are other choices, depending on what you're interested about $mathbb{C}$. For instance, in Algebraic Geometry one does not endow $mathbb{C}$ with its usual topology but rather with the Zariski Topology, the minimal topology where polynomial functions are continuous. In this case, that topology is simply the co-finite topology, that is, closed sets are finite (or all $mathbb{C}$).
– Juan Diego Rojas
Nov 25 at 1:25
You answered your own question with "Endowing $C$ with the topology of $R^2$ is the key to all those nice results of classical complex analysis". Without it you don't get all those nice results. And yes, there are a lot of other choices.
– Somos
Nov 25 at 2:15
What is the field $mathbb{C}$ without its usual topology ? A non-countably infinite algebraically closed transcendental extension of $mathbb{Q}$ ? So all we can say is if a complex number $a$ is algebraic over $mathbb{Q}(b_1,b_2,ldots)$ or not ? The different topologies on algebraic numbers are interesting, for example the completion wrt. $sup_sigma |a^sigma|$.
– reuns
Nov 25 at 2:46
add a comment |
In fact, there are other choices, depending on what you're interested about $mathbb{C}$. For instance, in Algebraic Geometry one does not endow $mathbb{C}$ with its usual topology but rather with the Zariski Topology, the minimal topology where polynomial functions are continuous. In this case, that topology is simply the co-finite topology, that is, closed sets are finite (or all $mathbb{C}$).
– Juan Diego Rojas
Nov 25 at 1:25
You answered your own question with "Endowing $C$ with the topology of $R^2$ is the key to all those nice results of classical complex analysis". Without it you don't get all those nice results. And yes, there are a lot of other choices.
– Somos
Nov 25 at 2:15
What is the field $mathbb{C}$ without its usual topology ? A non-countably infinite algebraically closed transcendental extension of $mathbb{Q}$ ? So all we can say is if a complex number $a$ is algebraic over $mathbb{Q}(b_1,b_2,ldots)$ or not ? The different topologies on algebraic numbers are interesting, for example the completion wrt. $sup_sigma |a^sigma|$.
– reuns
Nov 25 at 2:46
In fact, there are other choices, depending on what you're interested about $mathbb{C}$. For instance, in Algebraic Geometry one does not endow $mathbb{C}$ with its usual topology but rather with the Zariski Topology, the minimal topology where polynomial functions are continuous. In this case, that topology is simply the co-finite topology, that is, closed sets are finite (or all $mathbb{C}$).
– Juan Diego Rojas
Nov 25 at 1:25
In fact, there are other choices, depending on what you're interested about $mathbb{C}$. For instance, in Algebraic Geometry one does not endow $mathbb{C}$ with its usual topology but rather with the Zariski Topology, the minimal topology where polynomial functions are continuous. In this case, that topology is simply the co-finite topology, that is, closed sets are finite (or all $mathbb{C}$).
– Juan Diego Rojas
Nov 25 at 1:25
You answered your own question with "Endowing $C$ with the topology of $R^2$ is the key to all those nice results of classical complex analysis". Without it you don't get all those nice results. And yes, there are a lot of other choices.
– Somos
Nov 25 at 2:15
You answered your own question with "Endowing $C$ with the topology of $R^2$ is the key to all those nice results of classical complex analysis". Without it you don't get all those nice results. And yes, there are a lot of other choices.
– Somos
Nov 25 at 2:15
What is the field $mathbb{C}$ without its usual topology ? A non-countably infinite algebraically closed transcendental extension of $mathbb{Q}$ ? So all we can say is if a complex number $a$ is algebraic over $mathbb{Q}(b_1,b_2,ldots)$ or not ? The different topologies on algebraic numbers are interesting, for example the completion wrt. $sup_sigma |a^sigma|$.
– reuns
Nov 25 at 2:46
What is the field $mathbb{C}$ without its usual topology ? A non-countably infinite algebraically closed transcendental extension of $mathbb{Q}$ ? So all we can say is if a complex number $a$ is algebraic over $mathbb{Q}(b_1,b_2,ldots)$ or not ? The different topologies on algebraic numbers are interesting, for example the completion wrt. $sup_sigma |a^sigma|$.
– reuns
Nov 25 at 2:46
add a comment |
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In fact, there are other choices, depending on what you're interested about $mathbb{C}$. For instance, in Algebraic Geometry one does not endow $mathbb{C}$ with its usual topology but rather with the Zariski Topology, the minimal topology where polynomial functions are continuous. In this case, that topology is simply the co-finite topology, that is, closed sets are finite (or all $mathbb{C}$).
– Juan Diego Rojas
Nov 25 at 1:25
You answered your own question with "Endowing $C$ with the topology of $R^2$ is the key to all those nice results of classical complex analysis". Without it you don't get all those nice results. And yes, there are a lot of other choices.
– Somos
Nov 25 at 2:15
What is the field $mathbb{C}$ without its usual topology ? A non-countably infinite algebraically closed transcendental extension of $mathbb{Q}$ ? So all we can say is if a complex number $a$ is algebraic over $mathbb{Q}(b_1,b_2,ldots)$ or not ? The different topologies on algebraic numbers are interesting, for example the completion wrt. $sup_sigma |a^sigma|$.
– reuns
Nov 25 at 2:46