Canonical (Multiplicative) maps between $m$--dimensional spaces to $n$--dimensional spaces, $n leq m$.












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Let $M$ and $N$ be two smooth manifolds which we may assume are $mathbb{R}^m$ and $mathbb{R}^n$ respectively, with $n leq m$. In fact, assume $M = mathbb{R}^m - { 0 }$ and $N = mathbb{R}^n - { 0 }$.



This is a very small part in a larger problem I am working on, but it is something that I am surprised I have never thought about.



I am wondering if there are canonical maps between $M $ and $N$ that are multiplicative. By this, I mean that I am looking for a generalisation of the map $mathbb{R}^n - { 0 } to mathbb{R} - { 0 }$ given by $$(x_1, ..., x_n) mapsto x_1 cdots x_n.$$ I am also happy with maps of this form: $$(x_1, ..., x_n) mapsto left( frac{x_1}{x_n}, ..., frac{x_{n-1}}{x_n} right).$$ But other than these simple examples from $dim(M) to dim(M)-1$ or $dim(M) to 1$, I cannot construct a map that does not require a choice of ordering.



The following concrete problem should suffice:



Concrete question: Construct a surjective map $mathbb{R}^4 - { 0 } longrightarrow mathbb{R}^2 - { 0 }$ that does not depend on the ordering of the coordinate functions on $mathbb{R}^4 - { 0 }$.



Note that this is possibly a trivial problem.



Thanks in advance, and please let me know if the question needs further clarification.










share|cite|improve this question






















  • The map $(x_1,cdots,x_n)mapsto x_1cdots x_n$ is not a map $Bbb R^nsetminus {0} to Bbb Rsetminus{0}$: consider $(0,1)$ as your input. Are you sure this is really the thing you're looking to generalize?
    – KReiser
    Jan 3 at 21:26










  • @KReiser Ahhh, Replace $mathbb{R}^n - { 0 }$ with the complement of all coordinate hyperplanes in $mathbb{R}^n$?
    – AmorFati
    Jan 3 at 21:28






  • 1




    If you do that, you should consider whether you'd like to adjust what you're asking for later in the question with the map $Bbb R^4setminus{0}toBbb R^2setminus{0}$. Your question seems a little confusing otherwise - you have a disconnect between your motivation and what you're asking for.
    – KReiser
    Jan 3 at 21:31


















0














Let $M$ and $N$ be two smooth manifolds which we may assume are $mathbb{R}^m$ and $mathbb{R}^n$ respectively, with $n leq m$. In fact, assume $M = mathbb{R}^m - { 0 }$ and $N = mathbb{R}^n - { 0 }$.



This is a very small part in a larger problem I am working on, but it is something that I am surprised I have never thought about.



I am wondering if there are canonical maps between $M $ and $N$ that are multiplicative. By this, I mean that I am looking for a generalisation of the map $mathbb{R}^n - { 0 } to mathbb{R} - { 0 }$ given by $$(x_1, ..., x_n) mapsto x_1 cdots x_n.$$ I am also happy with maps of this form: $$(x_1, ..., x_n) mapsto left( frac{x_1}{x_n}, ..., frac{x_{n-1}}{x_n} right).$$ But other than these simple examples from $dim(M) to dim(M)-1$ or $dim(M) to 1$, I cannot construct a map that does not require a choice of ordering.



The following concrete problem should suffice:



Concrete question: Construct a surjective map $mathbb{R}^4 - { 0 } longrightarrow mathbb{R}^2 - { 0 }$ that does not depend on the ordering of the coordinate functions on $mathbb{R}^4 - { 0 }$.



Note that this is possibly a trivial problem.



Thanks in advance, and please let me know if the question needs further clarification.










share|cite|improve this question






















  • The map $(x_1,cdots,x_n)mapsto x_1cdots x_n$ is not a map $Bbb R^nsetminus {0} to Bbb Rsetminus{0}$: consider $(0,1)$ as your input. Are you sure this is really the thing you're looking to generalize?
    – KReiser
    Jan 3 at 21:26










  • @KReiser Ahhh, Replace $mathbb{R}^n - { 0 }$ with the complement of all coordinate hyperplanes in $mathbb{R}^n$?
    – AmorFati
    Jan 3 at 21:28






  • 1




    If you do that, you should consider whether you'd like to adjust what you're asking for later in the question with the map $Bbb R^4setminus{0}toBbb R^2setminus{0}$. Your question seems a little confusing otherwise - you have a disconnect between your motivation and what you're asking for.
    – KReiser
    Jan 3 at 21:31
















0












0








0







Let $M$ and $N$ be two smooth manifolds which we may assume are $mathbb{R}^m$ and $mathbb{R}^n$ respectively, with $n leq m$. In fact, assume $M = mathbb{R}^m - { 0 }$ and $N = mathbb{R}^n - { 0 }$.



This is a very small part in a larger problem I am working on, but it is something that I am surprised I have never thought about.



I am wondering if there are canonical maps between $M $ and $N$ that are multiplicative. By this, I mean that I am looking for a generalisation of the map $mathbb{R}^n - { 0 } to mathbb{R} - { 0 }$ given by $$(x_1, ..., x_n) mapsto x_1 cdots x_n.$$ I am also happy with maps of this form: $$(x_1, ..., x_n) mapsto left( frac{x_1}{x_n}, ..., frac{x_{n-1}}{x_n} right).$$ But other than these simple examples from $dim(M) to dim(M)-1$ or $dim(M) to 1$, I cannot construct a map that does not require a choice of ordering.



The following concrete problem should suffice:



Concrete question: Construct a surjective map $mathbb{R}^4 - { 0 } longrightarrow mathbb{R}^2 - { 0 }$ that does not depend on the ordering of the coordinate functions on $mathbb{R}^4 - { 0 }$.



Note that this is possibly a trivial problem.



Thanks in advance, and please let me know if the question needs further clarification.










share|cite|improve this question













Let $M$ and $N$ be two smooth manifolds which we may assume are $mathbb{R}^m$ and $mathbb{R}^n$ respectively, with $n leq m$. In fact, assume $M = mathbb{R}^m - { 0 }$ and $N = mathbb{R}^n - { 0 }$.



This is a very small part in a larger problem I am working on, but it is something that I am surprised I have never thought about.



I am wondering if there are canonical maps between $M $ and $N$ that are multiplicative. By this, I mean that I am looking for a generalisation of the map $mathbb{R}^n - { 0 } to mathbb{R} - { 0 }$ given by $$(x_1, ..., x_n) mapsto x_1 cdots x_n.$$ I am also happy with maps of this form: $$(x_1, ..., x_n) mapsto left( frac{x_1}{x_n}, ..., frac{x_{n-1}}{x_n} right).$$ But other than these simple examples from $dim(M) to dim(M)-1$ or $dim(M) to 1$, I cannot construct a map that does not require a choice of ordering.



The following concrete problem should suffice:



Concrete question: Construct a surjective map $mathbb{R}^4 - { 0 } longrightarrow mathbb{R}^2 - { 0 }$ that does not depend on the ordering of the coordinate functions on $mathbb{R}^4 - { 0 }$.



Note that this is possibly a trivial problem.



Thanks in advance, and please let me know if the question needs further clarification.







differential-geometry algebraic-geometry transformation canonical-transformation






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asked Jan 3 at 21:19









AmorFati

396529




396529












  • The map $(x_1,cdots,x_n)mapsto x_1cdots x_n$ is not a map $Bbb R^nsetminus {0} to Bbb Rsetminus{0}$: consider $(0,1)$ as your input. Are you sure this is really the thing you're looking to generalize?
    – KReiser
    Jan 3 at 21:26










  • @KReiser Ahhh, Replace $mathbb{R}^n - { 0 }$ with the complement of all coordinate hyperplanes in $mathbb{R}^n$?
    – AmorFati
    Jan 3 at 21:28






  • 1




    If you do that, you should consider whether you'd like to adjust what you're asking for later in the question with the map $Bbb R^4setminus{0}toBbb R^2setminus{0}$. Your question seems a little confusing otherwise - you have a disconnect between your motivation and what you're asking for.
    – KReiser
    Jan 3 at 21:31




















  • The map $(x_1,cdots,x_n)mapsto x_1cdots x_n$ is not a map $Bbb R^nsetminus {0} to Bbb Rsetminus{0}$: consider $(0,1)$ as your input. Are you sure this is really the thing you're looking to generalize?
    – KReiser
    Jan 3 at 21:26










  • @KReiser Ahhh, Replace $mathbb{R}^n - { 0 }$ with the complement of all coordinate hyperplanes in $mathbb{R}^n$?
    – AmorFati
    Jan 3 at 21:28






  • 1




    If you do that, you should consider whether you'd like to adjust what you're asking for later in the question with the map $Bbb R^4setminus{0}toBbb R^2setminus{0}$. Your question seems a little confusing otherwise - you have a disconnect between your motivation and what you're asking for.
    – KReiser
    Jan 3 at 21:31


















The map $(x_1,cdots,x_n)mapsto x_1cdots x_n$ is not a map $Bbb R^nsetminus {0} to Bbb Rsetminus{0}$: consider $(0,1)$ as your input. Are you sure this is really the thing you're looking to generalize?
– KReiser
Jan 3 at 21:26




The map $(x_1,cdots,x_n)mapsto x_1cdots x_n$ is not a map $Bbb R^nsetminus {0} to Bbb Rsetminus{0}$: consider $(0,1)$ as your input. Are you sure this is really the thing you're looking to generalize?
– KReiser
Jan 3 at 21:26












@KReiser Ahhh, Replace $mathbb{R}^n - { 0 }$ with the complement of all coordinate hyperplanes in $mathbb{R}^n$?
– AmorFati
Jan 3 at 21:28




@KReiser Ahhh, Replace $mathbb{R}^n - { 0 }$ with the complement of all coordinate hyperplanes in $mathbb{R}^n$?
– AmorFati
Jan 3 at 21:28




1




1




If you do that, you should consider whether you'd like to adjust what you're asking for later in the question with the map $Bbb R^4setminus{0}toBbb R^2setminus{0}$. Your question seems a little confusing otherwise - you have a disconnect between your motivation and what you're asking for.
– KReiser
Jan 3 at 21:31






If you do that, you should consider whether you'd like to adjust what you're asking for later in the question with the map $Bbb R^4setminus{0}toBbb R^2setminus{0}$. Your question seems a little confusing otherwise - you have a disconnect between your motivation and what you're asking for.
– KReiser
Jan 3 at 21:31












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