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Regular maps: What is it, what does it, who names it
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Consider the following definition of regular map: a $C^infty$ map $f:Urightarrow V$ between open subsets $Usubseteq mathbbR^m$ resp. $VsubseteqmathbbR^n$, such that for all $xin U$ the rank of $Df(x)$ be maximal, i.e., be equal to $minm,n$.
My questions are: 1) Does this mean that the rank of $Df(x)$ is constant as long as $x$ varies of a connected component of $U$? If yes, how can I prove this?
2) I could not find this definition on wikipedia. Can it perhaps be found under a different name?
analysis matrix-rank jacobian
asked Jul 25 at 14:47
temo
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Consider the following definition of regular map: a $C^infty$ map $f:Urightarrow V$ between open subsets $Usubseteq mathbbR^m$ resp. $VsubseteqmathbbR^n$, such that for all $xin U$ the rank of $Df(x)$ be maximal, i.e., be equal to $minm,n$.
My questions are: 1) Does this mean that the rank of $Df(x)$ is constant as long as $x$ varies of a connected component of $U$? If yes, how can I prove this?
2) I could not find this definition on wikipedia. Can it perhaps be found under a different name?
analysis matrix-rank jacobian
asked Jul 25 at 14:47
temo
1,7241444
Since the rank is maximal and $m, n$ are fixed, then yes the rank is constant, not only on each connected component, but overall. You can note that if the rank is maximal at one point, then it’s maximal in a neighbourhood around that point. It’s true that this implies that the rank on each connected component is constant, but moreso they’re all maximal so they’re all the same rank. Such maps would be called submersions or immersions, and you can find definitions of these on Wikipedia.
– Osama Ghani
Jul 25 at 19:23
@OsamaGhani Ah, thanks! Just a small question: If I only assume the rank is maximal in a single point, not on all of the domain, then it's also maximal in a neighborhood around that point; this is clear. But what argument do I then need to use to obtain that it's maximal on the whole connected component containing that point?
– temo
Jul 26 at 6:56
@temo I think this is not true. Look at $fcolon mathbbR to mathbbR^2$ with $t mapsto (t^2,t^3)$. So we have $m=1, n=2$. We obtain $Df(t) = beginpmatrix2t\3t^2endpmatrix$, which has $textrank=1$ iff $tneq0$. So it's maximal outside $0$, but not in the whole connected component. Note that still every point with maximal rank has a neighborhood with maximal rank.
– Babelfish
Jul 26 at 7:32
@temo You can use a connectedness argument. Let $S = p$ has an neighbourhood about it where $Df$ has maximal rank $$, then show that $S$ is both open and closed, and hence represents a connected component
– Osama Ghani
Jul 26 at 9:02
@osama Ghani, that's the way one would prove this. But in this case it's not possible, since S is not closed. See my comment for a counterexample.
– Babelfish
Jul 26 at 20:50
 |Â
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Consider the following definition of regular map: a $C^infty$ map $f:Urightarrow V$ between open subsets $Usubseteq mathbbR^m$ resp. $VsubseteqmathbbR^n$, such that for all $xin U$ the rank of $Df(x)$ be maximal, i.e., be equal to $minm,n$.
My questions are: 1) Does this mean that the rank of $Df(x)$ is constant as long as $x$ varies of a connected component of $U$? If yes, how can I prove this?
2) I could not find this definition on wikipedia. Can it perhaps be found under a different name?
analysis matrix-rank jacobian
asked Jul 25 at 14:47
temo
1,7241444
Consider the following definition of regular map: a $C^infty$ map $f:Urightarrow V$ between open subsets $Usubseteq mathbbR^m$ resp. $VsubseteqmathbbR^n$, such that for all $xin U$ the rank of $Df(x)$ be maximal, i.e., be equal to $minm,n$.
My questions are: 1) Does this mean that the rank of $Df(x)$ is constant as long as $x$ varies of a connected component of $U$? If yes, how can I prove this?
2) I could not find this definition on wikipedia. Can it perhaps be found under a different name?
analysis matrix-rank jacobian
asked Jul 25 at 14:47
temo
1,7241444
asked Jul 25 at 14:47
temo
1,7241444
asked Jul 25 at 14:47
temo
1,7241444
asked Jul 25 at 14:47
temo
1,7241444
1,7241444
Since the rank is maximal and $m, n$ are fixed, then yes the rank is constant, not only on each connected component, but overall. You can note that if the rank is maximal at one point, then it’s maximal in a neighbourhood around that point. It’s true that this implies that the rank on each connected component is constant, but moreso they’re all maximal so they’re all the same rank. Such maps would be called submersions or immersions, and you can find definitions of these on Wikipedia.
– Osama Ghani
Jul 25 at 19:23
@OsamaGhani Ah, thanks! Just a small question: If I only assume the rank is maximal in a single point, not on all of the domain, then it's also maximal in a neighborhood around that point; this is clear. But what argument do I then need to use to obtain that it's maximal on the whole connected component containing that point?
– temo
Jul 26 at 6:56
@temo I think this is not true. Look at $fcolon mathbbR to mathbbR^2$ with $t mapsto (t^2,t^3)$. So we have $m=1, n=2$. We obtain $Df(t) = beginpmatrix2t\3t^2endpmatrix$, which has $textrank=1$ iff $tneq0$. So it's maximal outside $0$, but not in the whole connected component. Note that still every point with maximal rank has a neighborhood with maximal rank.
– Babelfish
Jul 26 at 7:32
@temo You can use a connectedness argument. Let $S = p$ has an neighbourhood about it where $Df$ has maximal rank $$, then show that $S$ is both open and closed, and hence represents a connected component
– Osama Ghani
Jul 26 at 9:02
@osama Ghani, that's the way one would prove this. But in this case it's not possible, since S is not closed. See my comment for a counterexample.
– Babelfish
Jul 26 at 20:50
 |Â
show 1 more comment
Since the rank is maximal and $m, n$ are fixed, then yes the rank is constant, not only on each connected component, but overall. You can note that if the rank is maximal at one point, then it’s maximal in a neighbourhood around that point. It’s true that this implies that the rank on each connected component is constant, but moreso they’re all maximal so they’re all the same rank. Such maps would be called submersions or immersions, and you can find definitions of these on Wikipedia.
– Osama Ghani
Jul 25 at 19:23
@OsamaGhani Ah, thanks! Just a small question: If I only assume the rank is maximal in a single point, not on all of the domain, then it's also maximal in a neighborhood around that point; this is clear. But what argument do I then need to use to obtain that it's maximal on the whole connected component containing that point?
– temo
Jul 26 at 6:56
@temo I think this is not true. Look at $fcolon mathbbR to mathbbR^2$ with $t mapsto (t^2,t^3)$. So we have $m=1, n=2$. We obtain $Df(t) = beginpmatrix2t\3t^2endpmatrix$, which has $textrank=1$ iff $tneq0$. So it's maximal outside $0$, but not in the whole connected component. Note that still every point with maximal rank has a neighborhood with maximal rank.
– Babelfish
Jul 26 at 7:32
@temo You can use a connectedness argument. Let $S = p$ has an neighbourhood about it where $Df$ has maximal rank $$, then show that $S$ is both open and closed, and hence represents a connected component
– Osama Ghani
Jul 26 at 9:02
@osama Ghani, that's the way one would prove this. But in this case it's not possible, since S is not closed. See my comment for a counterexample.
– Babelfish
Jul 26 at 20:50
Since the rank is maximal and $m, n$ are fixed, then yes the rank is constant, not only on each connected component, but overall. You can note that if the rank is maximal at one point, then it’s maximal in a neighbourhood around that point. It’s true that this implies that the rank on each connected component is constant, but moreso they’re all maximal so they’re all the same rank. Such maps would be called submersions or immersions, and you can find definitions of these on Wikipedia.
– Osama Ghani
Jul 25 at 19:23
Since the rank is maximal and $m, n$ are fixed, then yes the rank is constant, not only on each connected component, but overall. You can note that if the rank is maximal at one point, then it’s maximal in a neighbourhood around that point. It’s true that this implies that the rank on each connected component is constant, but moreso they’re all maximal so they’re all the same rank. Such maps would be called submersions or immersions, and you can find definitions of these on Wikipedia.
– Osama Ghani
Jul 25 at 19:23
@OsamaGhani Ah, thanks! Just a small question: If I only assume the rank is maximal in a single point, not on all of the domain, then it's also maximal in a neighborhood around that point; this is clear. But what argument do I then need to use to obtain that it's maximal on the whole connected component containing that point?
– temo
Jul 26 at 6:56
@OsamaGhani Ah, thanks! Just a small question: If I only assume the rank is maximal in a single point, not on all of the domain, then it's also maximal in a neighborhood around that point; this is clear. But what argument do I then need to use to obtain that it's maximal on the whole connected component containing that point?
– temo
Jul 26 at 6:56
@temo I think this is not true. Look at $fcolon mathbbR to mathbbR^2$ with $t mapsto (t^2,t^3)$. So we have $m=1, n=2$. We obtain $Df(t) = beginpmatrix2t\3t^2endpmatrix$, which has $textrank=1$ iff $tneq0$. So it's maximal outside $0$, but not in the whole connected component. Note that still every point with maximal rank has a neighborhood with maximal rank.
– Babelfish
Jul 26 at 7:32
@temo I think this is not true. Look at $fcolon mathbbR to mathbbR^2$ with $t mapsto (t^2,t^3)$. So we have $m=1, n=2$. We obtain $Df(t) = beginpmatrix2t\3t^2endpmatrix$, which has $textrank=1$ iff $tneq0$. So it's maximal outside $0$, but not in the whole connected component. Note that still every point with maximal rank has a neighborhood with maximal rank.
– Babelfish
Jul 26 at 7:32
@temo You can use a connectedness argument. Let $S = p$ has an neighbourhood about it where $Df$ has maximal rank $$, then show that $S$ is both open and closed, and hence represents a connected component
– Osama Ghani
Jul 26 at 9:02
@temo You can use a connectedness argument. Let $S = p$ has an neighbourhood about it where $Df$ has maximal rank $$, then show that $S$ is both open and closed, and hence represents a connected component
– Osama Ghani
Jul 26 at 9:02
@osama Ghani, that's the way one would prove this. But in this case it's not possible, since S is not closed. See my comment for a counterexample.
– Babelfish
Jul 26 at 20:50
@osama Ghani, that's the way one would prove this. But in this case it's not possible, since S is not closed. See my comment for a counterexample.
– Babelfish
Jul 26 at 20:50
 |Â
show 1 more comment
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Since the rank is maximal and $m, n$ are fixed, then yes the rank is constant, not only on each connected component, but overall. You can note that if the rank is maximal at one point, then it’s maximal in a neighbourhood around that point. It’s true that this implies that the rank on each connected component is constant, but moreso they’re all maximal so they’re all the same rank. Such maps would be called submersions or immersions, and you can find definitions of these on Wikipedia.
– Osama Ghani
Jul 25 at 19:23
@OsamaGhani Ah, thanks! Just a small question: If I only assume the rank is maximal in a single point, not on all of the domain, then it's also maximal in a neighborhood around that point; this is clear. But what argument do I then need to use to obtain that it's maximal on the whole connected component containing that point?
– temo
Jul 26 at 6:56
@temo I think this is not true. Look at $fcolon mathbbR to mathbbR^2$ with $t mapsto (t^2,t^3)$. So we have $m=1, n=2$. We obtain $Df(t) = beginpmatrix2t\3t^2endpmatrix$, which has $textrank=1$ iff $tneq0$. So it's maximal outside $0$, but not in the whole connected component. Note that still every point with maximal rank has a neighborhood with maximal rank.
– Babelfish
Jul 26 at 7:32
@temo You can use a connectedness argument. Let $S = p$ has an neighbourhood about it where $Df$ has maximal rank $$, then show that $S$ is both open and closed, and hence represents a connected component
– Osama Ghani
Jul 26 at 9:02
@osama Ghani, that's the way one would prove this. But in this case it's not possible, since S is not closed. See my comment for a counterexample.
– Babelfish
Jul 26 at 20:50