Canonical (Multiplicative) maps between $m$--dimensional spaces to $n$--dimensional spaces, $n leq m$.
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
asked Jan 3 at 21:19
AmorFati
396529
add a comment |
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
asked Jan 3 at 21:19
AmorFati
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
add a comment |
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
asked Jan 3 at 21:19
AmorFati
396529
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
differential-geometry algebraic-geometry transformation canonical-transformation
asked Jan 3 at 21:19
AmorFati
396529
asked Jan 3 at 21:19
AmorFati
396529
asked Jan 3 at 21:19
AmorFati
396529
asked Jan 3 at 21:19
AmorFati
396529
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
add a comment |
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
add a comment |
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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