Mixing
Understanding some definitions of smooth manifolds and the inverse function theorem
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
I am an electrical engineering grad student trying to learn some of the basics of differential geometry to better understand some work on signal processing and learning on Grassmanian manifolds. Pretty much I want to understand this paper pretty completely. My relevant background includes a formal undergraduate analysis course, a graduate measure theoretic probability course, and of course a bunch of random engineering mathematics/mathematical physics courses. I'm using the book "Introduction to Differential Geometry for Engineers" by Doolin/Martin to try and get up to speed quickly.
The first "definition" given is the following:
A subset $M$ of $mathbbR^n$ is a k-dimensional manifold if for each $mathbfx in M$ there are: open subsets $U$ and $V$ of $mathbbR^n$ with $x in U$ and a diffeomorphism $f$ from $U$ to $V$ such that $f(U cap M) = mathbfy in V : y^k+1=y^k+2=...= y^n = 0 $. Thus, each point $y$ in the image of $f$ has a representation like $y = (y^1(x), y^2(x), ... y^k(x),0,...,0)$.
The book also quotes an implicit function theorem, which I'll summarize as:
If $F:mathbbR^n-ktimesmathbbR^krightarrowmathbbR^k$ is continuously differentiable in an open set containing $(mathbfa,mathbfb)$ and $F(mathbfa,mathbfb) = mathbf0$ then if the Jacobian of F is of rank $k$ then there is an open set $Ain mathbbR^n-k$ containing $mathbfa$ and an open set $B in mathbbR^k$ containing $mathbfb$ such that for each $mathbfxin A$ there is a unique $g(mathbfx)in B$ such that $F(mathbfx, g(mathbfx))=0$ Furthermore the function $g$ is differentiable. Furthermore, there is a change of coordinates such that assigns $g(mathbfx) = mathbf0$.
My question is in regards to the following example from the text:
Consider a $C^infty$ function $F$ with the domain $A in
mathbbR^n$ and a range in $mathbbR^k$, $k<n$. Consider the set
$M = mathbfxin mathbbR^n:F(mathbfx) =mathbf0$ Let the
rank of the Jacobian of $F$ be equal to $k$ $forall$ $mathbfxin M$.
$M$ is an $n-k$ dimensional manifold.
They indicate that this is true precisely because in the inverse function theorem there exists the change of coordinates that allows us to set $g(mathbfx) = 0$ above. I don't see why that fact leads to the conclusion that $M$ is a manifold. Am I missing some obvious construction or something?
I tried to construct an alternative argument:
If I consider the function $f:mathbbR^n rightarrow mathbbR^n$ such that $f( mathbfx ) = [x^1,x^2,x^3,...,x^n, F(mathbfx)]$. I have $fin C^infty$ and defined on $A$, the domain of $F$, and I think its easy to see that for an open subset of $A$ we have the diffeomorphism as in the "definition" above".
Does this argument work?
multivariable-calculus differential-geometry manifolds smooth-manifolds
edited Aug 6 at 22:29
Arnaud Mortier
19.2k22159
asked Aug 6 at 22:05
Travis C Cuvelier
484
add a comment |Â
up vote
2
down vote
favorite
I am an electrical engineering grad student trying to learn some of the basics of differential geometry to better understand some work on signal processing and learning on Grassmanian manifolds. Pretty much I want to understand this paper pretty completely. My relevant background includes a formal undergraduate analysis course, a graduate measure theoretic probability course, and of course a bunch of random engineering mathematics/mathematical physics courses. I'm using the book "Introduction to Differential Geometry for Engineers" by Doolin/Martin to try and get up to speed quickly.
The first "definition" given is the following:
A subset $M$ of $mathbbR^n$ is a k-dimensional manifold if for each $mathbfx in M$ there are: open subsets $U$ and $V$ of $mathbbR^n$ with $x in U$ and a diffeomorphism $f$ from $U$ to $V$ such that $f(U cap M) = mathbfy in V : y^k+1=y^k+2=...= y^n = 0 $. Thus, each point $y$ in the image of $f$ has a representation like $y = (y^1(x), y^2(x), ... y^k(x),0,...,0)$.
The book also quotes an implicit function theorem, which I'll summarize as:
If $F:mathbbR^n-ktimesmathbbR^krightarrowmathbbR^k$ is continuously differentiable in an open set containing $(mathbfa,mathbfb)$ and $F(mathbfa,mathbfb) = mathbf0$ then if the Jacobian of F is of rank $k$ then there is an open set $Ain mathbbR^n-k$ containing $mathbfa$ and an open set $B in mathbbR^k$ containing $mathbfb$ such that for each $mathbfxin A$ there is a unique $g(mathbfx)in B$ such that $F(mathbfx, g(mathbfx))=0$ Furthermore the function $g$ is differentiable. Furthermore, there is a change of coordinates such that assigns $g(mathbfx) = mathbf0$.
My question is in regards to the following example from the text:
Consider a $C^infty$ function $F$ with the domain $A in
mathbbR^n$ and a range in $mathbbR^k$, $k<n$. Consider the set
$M = mathbfxin mathbbR^n:F(mathbfx) =mathbf0$ Let the
rank of the Jacobian of $F$ be equal to $k$ $forall$ $mathbfxin M$.
$M$ is an $n-k$ dimensional manifold.
They indicate that this is true precisely because in the inverse function theorem there exists the change of coordinates that allows us to set $g(mathbfx) = 0$ above. I don't see why that fact leads to the conclusion that $M$ is a manifold. Am I missing some obvious construction or something?
I tried to construct an alternative argument:
If I consider the function $f:mathbbR^n rightarrow mathbbR^n$ such that $f( mathbfx ) = [x^1,x^2,x^3,...,x^n, F(mathbfx)]$. I have $fin C^infty$ and defined on $A$, the domain of $F$, and I think its easy to see that for an open subset of $A$ we have the diffeomorphism as in the "definition" above".
Does this argument work?
multivariable-calculus differential-geometry manifolds smooth-manifolds
edited Aug 6 at 22:29
Arnaud Mortier
19.2k22159
asked Aug 6 at 22:05
Travis C Cuvelier
484
Are you sure of your transcription of the implicit function theorem? There is nothing in the assumptions that implies that the level $0$ set of $F$ can be seen as the graph of a function of the first $n-k$ coordinates. Think of the function $F:Bbb Rtimes Bbb RtoBbb R$ defined by $f(x,y)=x$.
– Arnaud Mortier
Aug 6 at 22:43
You're definitely right, I was not thinking when I was transcribing. The assumption should definitely be that the last k columns of the Jacobian are invertible.
– Travis C Cuvelier
Aug 7 at 3:19
If you have a second, would you mind answering this clarification re the statement of the implicit function theorem. While the version you posted below is definitely what's stated in the back of the book, they give the example x^2+y^2-1 =0 as implicitly defining the unit circle manifold in $mathbbR^2$. The Jacobian of F(x,y) = x^2+y^2 doesn't vanish on this set, but certainly at y = 0 the last column isn't full rank. They give the fact that the Jabobian is never the zero vector as the reason the implicit function theorem applies. Is it just a change of coordinates again? Seems alright...
– Travis C Cuvelier
Aug 7 at 3:52
Yes, while the theorem is stated with the last $k$ columns, this point is not particularly crucial, if some $k$ columns give an invertible matrix, then you can move them to the right (which does correspond to a change of variables in a way, as you permute the variables).
– Arnaud Mortier
Aug 7 at 9:02
Thanks so much! Owe you one!
– Travis C Cuvelier
Aug 7 at 15:54
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am an electrical engineering grad student trying to learn some of the basics of differential geometry to better understand some work on signal processing and learning on Grassmanian manifolds. Pretty much I want to understand this paper pretty completely. My relevant background includes a formal undergraduate analysis course, a graduate measure theoretic probability course, and of course a bunch of random engineering mathematics/mathematical physics courses. I'm using the book "Introduction to Differential Geometry for Engineers" by Doolin/Martin to try and get up to speed quickly.
The first "definition" given is the following:
A subset $M$ of $mathbbR^n$ is a k-dimensional manifold if for each $mathbfx in M$ there are: open subsets $U$ and $V$ of $mathbbR^n$ with $x in U$ and a diffeomorphism $f$ from $U$ to $V$ such that $f(U cap M) = mathbfy in V : y^k+1=y^k+2=...= y^n = 0 $. Thus, each point $y$ in the image of $f$ has a representation like $y = (y^1(x), y^2(x), ... y^k(x),0,...,0)$.
The book also quotes an implicit function theorem, which I'll summarize as:
If $F:mathbbR^n-ktimesmathbbR^krightarrowmathbbR^k$ is continuously differentiable in an open set containing $(mathbfa,mathbfb)$ and $F(mathbfa,mathbfb) = mathbf0$ then if the Jacobian of F is of rank $k$ then there is an open set $Ain mathbbR^n-k$ containing $mathbfa$ and an open set $B in mathbbR^k$ containing $mathbfb$ such that for each $mathbfxin A$ there is a unique $g(mathbfx)in B$ such that $F(mathbfx, g(mathbfx))=0$ Furthermore the function $g$ is differentiable. Furthermore, there is a change of coordinates such that assigns $g(mathbfx) = mathbf0$.
My question is in regards to the following example from the text:
Consider a $C^infty$ function $F$ with the domain $A in
mathbbR^n$ and a range in $mathbbR^k$, $k<n$. Consider the set
$M = mathbfxin mathbbR^n:F(mathbfx) =mathbf0$ Let the
rank of the Jacobian of $F$ be equal to $k$ $forall$ $mathbfxin M$.
$M$ is an $n-k$ dimensional manifold.
They indicate that this is true precisely because in the inverse function theorem there exists the change of coordinates that allows us to set $g(mathbfx) = 0$ above. I don't see why that fact leads to the conclusion that $M$ is a manifold. Am I missing some obvious construction or something?
I tried to construct an alternative argument:
If I consider the function $f:mathbbR^n rightarrow mathbbR^n$ such that $f( mathbfx ) = [x^1,x^2,x^3,...,x^n, F(mathbfx)]$. I have $fin C^infty$ and defined on $A$, the domain of $F$, and I think its easy to see that for an open subset of $A$ we have the diffeomorphism as in the "definition" above".
Does this argument work?
multivariable-calculus differential-geometry manifolds smooth-manifolds
edited Aug 6 at 22:29
Arnaud Mortier
19.2k22159
asked Aug 6 at 22:05
Travis C Cuvelier
484
I am an electrical engineering grad student trying to learn some of the basics of differential geometry to better understand some work on signal processing and learning on Grassmanian manifolds. Pretty much I want to understand this paper pretty completely. My relevant background includes a formal undergraduate analysis course, a graduate measure theoretic probability course, and of course a bunch of random engineering mathematics/mathematical physics courses. I'm using the book "Introduction to Differential Geometry for Engineers" by Doolin/Martin to try and get up to speed quickly.
The first "definition" given is the following:
A subset $M$ of $mathbbR^n$ is a k-dimensional manifold if for each $mathbfx in M$ there are: open subsets $U$ and $V$ of $mathbbR^n$ with $x in U$ and a diffeomorphism $f$ from $U$ to $V$ such that $f(U cap M) = mathbfy in V : y^k+1=y^k+2=...= y^n = 0 $. Thus, each point $y$ in the image of $f$ has a representation like $y = (y^1(x), y^2(x), ... y^k(x),0,...,0)$.
The book also quotes an implicit function theorem, which I'll summarize as:
If $F:mathbbR^n-ktimesmathbbR^krightarrowmathbbR^k$ is continuously differentiable in an open set containing $(mathbfa,mathbfb)$ and $F(mathbfa,mathbfb) = mathbf0$ then if the Jacobian of F is of rank $k$ then there is an open set $Ain mathbbR^n-k$ containing $mathbfa$ and an open set $B in mathbbR^k$ containing $mathbfb$ such that for each $mathbfxin A$ there is a unique $g(mathbfx)in B$ such that $F(mathbfx, g(mathbfx))=0$ Furthermore the function $g$ is differentiable. Furthermore, there is a change of coordinates such that assigns $g(mathbfx) = mathbf0$.
My question is in regards to the following example from the text:
Consider a $C^infty$ function $F$ with the domain $A in
mathbbR^n$ and a range in $mathbbR^k$, $k<n$. Consider the set
$M = mathbfxin mathbbR^n:F(mathbfx) =mathbf0$ Let the
rank of the Jacobian of $F$ be equal to $k$ $forall$ $mathbfxin M$.
$M$ is an $n-k$ dimensional manifold.
They indicate that this is true precisely because in the inverse function theorem there exists the change of coordinates that allows us to set $g(mathbfx) = 0$ above. I don't see why that fact leads to the conclusion that $M$ is a manifold. Am I missing some obvious construction or something?
I tried to construct an alternative argument:
If I consider the function $f:mathbbR^n rightarrow mathbbR^n$ such that $f( mathbfx ) = [x^1,x^2,x^3,...,x^n, F(mathbfx)]$. I have $fin C^infty$ and defined on $A$, the domain of $F$, and I think its easy to see that for an open subset of $A$ we have the diffeomorphism as in the "definition" above".
Does this argument work?
multivariable-calculus differential-geometry manifolds smooth-manifolds
edited Aug 6 at 22:29
Arnaud Mortier
19.2k22159
asked Aug 6 at 22:05
Travis C Cuvelier
484
edited Aug 6 at 22:29
Arnaud Mortier
19.2k22159
edited Aug 6 at 22:29
Arnaud Mortier
19.2k22159
edited Aug 6 at 22:29
Arnaud Mortier
19.2k22159
19.2k22159
asked Aug 6 at 22:05
Travis C Cuvelier
484
asked Aug 6 at 22:05
Travis C Cuvelier
484
asked Aug 6 at 22:05
Travis C Cuvelier
484
484
Are you sure of your transcription of the implicit function theorem? There is nothing in the assumptions that implies that the level $0$ set of $F$ can be seen as the graph of a function of the first $n-k$ coordinates. Think of the function $F:Bbb Rtimes Bbb RtoBbb R$ defined by $f(x,y)=x$.
– Arnaud Mortier
Aug 6 at 22:43
You're definitely right, I was not thinking when I was transcribing. The assumption should definitely be that the last k columns of the Jacobian are invertible.
– Travis C Cuvelier
Aug 7 at 3:19
If you have a second, would you mind answering this clarification re the statement of the implicit function theorem. While the version you posted below is definitely what's stated in the back of the book, they give the example x^2+y^2-1 =0 as implicitly defining the unit circle manifold in $mathbbR^2$. The Jacobian of F(x,y) = x^2+y^2 doesn't vanish on this set, but certainly at y = 0 the last column isn't full rank. They give the fact that the Jabobian is never the zero vector as the reason the implicit function theorem applies. Is it just a change of coordinates again? Seems alright...
– Travis C Cuvelier
Aug 7 at 3:52
Yes, while the theorem is stated with the last $k$ columns, this point is not particularly crucial, if some $k$ columns give an invertible matrix, then you can move them to the right (which does correspond to a change of variables in a way, as you permute the variables).
– Arnaud Mortier
Aug 7 at 9:02
Thanks so much! Owe you one!
– Travis C Cuvelier
Aug 7 at 15:54
add a comment |Â
Are you sure of your transcription of the implicit function theorem? There is nothing in the assumptions that implies that the level $0$ set of $F$ can be seen as the graph of a function of the first $n-k$ coordinates. Think of the function $F:Bbb Rtimes Bbb RtoBbb R$ defined by $f(x,y)=x$.
– Arnaud Mortier
Aug 6 at 22:43
You're definitely right, I was not thinking when I was transcribing. The assumption should definitely be that the last k columns of the Jacobian are invertible.
– Travis C Cuvelier
Aug 7 at 3:19
If you have a second, would you mind answering this clarification re the statement of the implicit function theorem. While the version you posted below is definitely what's stated in the back of the book, they give the example x^2+y^2-1 =0 as implicitly defining the unit circle manifold in $mathbbR^2$. The Jacobian of F(x,y) = x^2+y^2 doesn't vanish on this set, but certainly at y = 0 the last column isn't full rank. They give the fact that the Jabobian is never the zero vector as the reason the implicit function theorem applies. Is it just a change of coordinates again? Seems alright...
– Travis C Cuvelier
Aug 7 at 3:52
Yes, while the theorem is stated with the last $k$ columns, this point is not particularly crucial, if some $k$ columns give an invertible matrix, then you can move them to the right (which does correspond to a change of variables in a way, as you permute the variables).
– Arnaud Mortier
Aug 7 at 9:02
Thanks so much! Owe you one!
– Travis C Cuvelier
Aug 7 at 15:54
Are you sure of your transcription of the implicit function theorem? There is nothing in the assumptions that implies that the level $0$ set of $F$ can be seen as the graph of a function of the first $n-k$ coordinates. Think of the function $F:Bbb Rtimes Bbb RtoBbb R$ defined by $f(x,y)=x$.
– Arnaud Mortier
Aug 6 at 22:43
Are you sure of your transcription of the implicit function theorem? There is nothing in the assumptions that implies that the level $0$ set of $F$ can be seen as the graph of a function of the first $n-k$ coordinates. Think of the function $F:Bbb Rtimes Bbb RtoBbb R$ defined by $f(x,y)=x$.
– Arnaud Mortier
Aug 6 at 22:43
You're definitely right, I was not thinking when I was transcribing. The assumption should definitely be that the last k columns of the Jacobian are invertible.
– Travis C Cuvelier
Aug 7 at 3:19
You're definitely right, I was not thinking when I was transcribing. The assumption should definitely be that the last k columns of the Jacobian are invertible.
– Travis C Cuvelier
Aug 7 at 3:19
If you have a second, would you mind answering this clarification re the statement of the implicit function theorem. While the version you posted below is definitely what's stated in the back of the book, they give the example x^2+y^2-1 =0 as implicitly defining the unit circle manifold in $mathbbR^2$. The Jacobian of F(x,y) = x^2+y^2 doesn't vanish on this set, but certainly at y = 0 the last column isn't full rank. They give the fact that the Jabobian is never the zero vector as the reason the implicit function theorem applies. Is it just a change of coordinates again? Seems alright...
– Travis C Cuvelier
Aug 7 at 3:52
If you have a second, would you mind answering this clarification re the statement of the implicit function theorem. While the version you posted below is definitely what's stated in the back of the book, they give the example x^2+y^2-1 =0 as implicitly defining the unit circle manifold in $mathbbR^2$. The Jacobian of F(x,y) = x^2+y^2 doesn't vanish on this set, but certainly at y = 0 the last column isn't full rank. They give the fact that the Jabobian is never the zero vector as the reason the implicit function theorem applies. Is it just a change of coordinates again? Seems alright...
– Travis C Cuvelier
Aug 7 at 3:52
Yes, while the theorem is stated with the last $k$ columns, this point is not particularly crucial, if some $k$ columns give an invertible matrix, then you can move them to the right (which does correspond to a change of variables in a way, as you permute the variables).
– Arnaud Mortier
Aug 7 at 9:02
Yes, while the theorem is stated with the last $k$ columns, this point is not particularly crucial, if some $k$ columns give an invertible matrix, then you can move them to the right (which does correspond to a change of variables in a way, as you permute the variables).
– Arnaud Mortier
Aug 7 at 9:02
Thanks so much! Owe you one!
– Travis C Cuvelier
Aug 7 at 15:54
Thanks so much! Owe you one!
– Travis C Cuvelier
Aug 7 at 15:54
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
There is nothing hidden there, all you need is to put the pieces together:
First, fix the implicit function theorem: the one you stated is incomplete, it misses the fact that the partial Jacobian of $F$ that consists of the last $k$ columns of the full Jacobian should be invertible. This condition is stronger than just asking for the full Jacobian to have rank $k$.
If $M = mathbfxin mathbbR^n:F(mathbfx) =mathbf0$ where $operatornamerkoperatornameJac_mathbfx(F)=k$ for all $mathbfx$, then by the implicit function theorem, for every $(mathbfa, mathbfb)in M$ there are open subsets $Asubset Bbb R^n-k$ and $Bsubset Bbb R^k$ such that $mathbfain A$ and $mathbfbin B$ such that:
$Mcap (Atimes B)$ is the graph of a differentiable function $g:Ato
B$.
Furthermore, there is a change of coordinates that allows for $g$ to be simply the constant map $forall xin A,g(x)=0$. What this really means is that for each $xin A$ you change the coordinates of $xtimes B$ in such a way that the point of $M$ whose $A$-value is $x$ is now $(x,0)$. You do so smoothly over $Atimes B$, in other words
There is a diffeomorphism $f:Atimes Bto Atimes B'$ (for some open
subset $B'subset Bbb R^k$ which has to contain $0$), such that $f(Mcap (Atimes B))=Atimes0$.
Now to translate this into the notations used for submanifolds:
Let $U=Atimes B$, $V=Atimes B'$, and let $f:Atimes Bto Atimes B'$ represent the change of coordinates above. The condition for being a submanifold is obviously met: $f(Ucap M)$ is the set of elements of $V$ whose last $k$ coordinates are $0$. All the difficulty was to show the existence of the change of variables.
answered Aug 6 at 23:45
Arnaud Mortier
19.2k22159
Thanks so much for this answer. My whole confusion was in the construction of the diffeomorphism. I'm dutifully writing this explicitly in the margins of my book. Thanks again.
– Travis C Cuvelier
Aug 7 at 3:22
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
There is nothing hidden there, all you need is to put the pieces together:
First, fix the implicit function theorem: the one you stated is incomplete, it misses the fact that the partial Jacobian of $F$ that consists of the last $k$ columns of the full Jacobian should be invertible. This condition is stronger than just asking for the full Jacobian to have rank $k$.
If $M = mathbfxin mathbbR^n:F(mathbfx) =mathbf0$ where $operatornamerkoperatornameJac_mathbfx(F)=k$ for all $mathbfx$, then by the implicit function theorem, for every $(mathbfa, mathbfb)in M$ there are open subsets $Asubset Bbb R^n-k$ and $Bsubset Bbb R^k$ such that $mathbfain A$ and $mathbfbin B$ such that:
$Mcap (Atimes B)$ is the graph of a differentiable function $g:Ato
B$.
Furthermore, there is a change of coordinates that allows for $g$ to be simply the constant map $forall xin A,g(x)=0$. What this really means is that for each $xin A$ you change the coordinates of $xtimes B$ in such a way that the point of $M$ whose $A$-value is $x$ is now $(x,0)$. You do so smoothly over $Atimes B$, in other words
There is a diffeomorphism $f:Atimes Bto Atimes B'$ (for some open
subset $B'subset Bbb R^k$ which has to contain $0$), such that $f(Mcap (Atimes B))=Atimes0$.
Now to translate this into the notations used for submanifolds:
Let $U=Atimes B$, $V=Atimes B'$, and let $f:Atimes Bto Atimes B'$ represent the change of coordinates above. The condition for being a submanifold is obviously met: $f(Ucap M)$ is the set of elements of $V$ whose last $k$ coordinates are $0$. All the difficulty was to show the existence of the change of variables.
answered Aug 6 at 23:45
Arnaud Mortier
19.2k22159
Thanks so much for this answer. My whole confusion was in the construction of the diffeomorphism. I'm dutifully writing this explicitly in the margins of my book. Thanks again.
– Travis C Cuvelier
Aug 7 at 3:22
add a comment |Â
up vote
0
down vote
accepted
There is nothing hidden there, all you need is to put the pieces together:
First, fix the implicit function theorem: the one you stated is incomplete, it misses the fact that the partial Jacobian of $F$ that consists of the last $k$ columns of the full Jacobian should be invertible. This condition is stronger than just asking for the full Jacobian to have rank $k$.
If $M = mathbfxin mathbbR^n:F(mathbfx) =mathbf0$ where $operatornamerkoperatornameJac_mathbfx(F)=k$ for all $mathbfx$, then by the implicit function theorem, for every $(mathbfa, mathbfb)in M$ there are open subsets $Asubset Bbb R^n-k$ and $Bsubset Bbb R^k$ such that $mathbfain A$ and $mathbfbin B$ such that:
$Mcap (Atimes B)$ is the graph of a differentiable function $g:Ato
B$.
Furthermore, there is a change of coordinates that allows for $g$ to be simply the constant map $forall xin A,g(x)=0$. What this really means is that for each $xin A$ you change the coordinates of $xtimes B$ in such a way that the point of $M$ whose $A$-value is $x$ is now $(x,0)$. You do so smoothly over $Atimes B$, in other words
There is a diffeomorphism $f:Atimes Bto Atimes B'$ (for some open
subset $B'subset Bbb R^k$ which has to contain $0$), such that $f(Mcap (Atimes B))=Atimes0$.
Now to translate this into the notations used for submanifolds:
Let $U=Atimes B$, $V=Atimes B'$, and let $f:Atimes Bto Atimes B'$ represent the change of coordinates above. The condition for being a submanifold is obviously met: $f(Ucap M)$ is the set of elements of $V$ whose last $k$ coordinates are $0$. All the difficulty was to show the existence of the change of variables.
answered Aug 6 at 23:45
Arnaud Mortier
19.2k22159
Thanks so much for this answer. My whole confusion was in the construction of the diffeomorphism. I'm dutifully writing this explicitly in the margins of my book. Thanks again.
– Travis C Cuvelier
Aug 7 at 3:22
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
There is nothing hidden there, all you need is to put the pieces together:
First, fix the implicit function theorem: the one you stated is incomplete, it misses the fact that the partial Jacobian of $F$ that consists of the last $k$ columns of the full Jacobian should be invertible. This condition is stronger than just asking for the full Jacobian to have rank $k$.
If $M = mathbfxin mathbbR^n:F(mathbfx) =mathbf0$ where $operatornamerkoperatornameJac_mathbfx(F)=k$ for all $mathbfx$, then by the implicit function theorem, for every $(mathbfa, mathbfb)in M$ there are open subsets $Asubset Bbb R^n-k$ and $Bsubset Bbb R^k$ such that $mathbfain A$ and $mathbfbin B$ such that:
$Mcap (Atimes B)$ is the graph of a differentiable function $g:Ato
B$.
Furthermore, there is a change of coordinates that allows for $g$ to be simply the constant map $forall xin A,g(x)=0$. What this really means is that for each $xin A$ you change the coordinates of $xtimes B$ in such a way that the point of $M$ whose $A$-value is $x$ is now $(x,0)$. You do so smoothly over $Atimes B$, in other words
There is a diffeomorphism $f:Atimes Bto Atimes B'$ (for some open
subset $B'subset Bbb R^k$ which has to contain $0$), such that $f(Mcap (Atimes B))=Atimes0$.
Now to translate this into the notations used for submanifolds:
Let $U=Atimes B$, $V=Atimes B'$, and let $f:Atimes Bto Atimes B'$ represent the change of coordinates above. The condition for being a submanifold is obviously met: $f(Ucap M)$ is the set of elements of $V$ whose last $k$ coordinates are $0$. All the difficulty was to show the existence of the change of variables.
answered Aug 6 at 23:45
Arnaud Mortier
19.2k22159
There is nothing hidden there, all you need is to put the pieces together:
First, fix the implicit function theorem: the one you stated is incomplete, it misses the fact that the partial Jacobian of $F$ that consists of the last $k$ columns of the full Jacobian should be invertible. This condition is stronger than just asking for the full Jacobian to have rank $k$.
If $M = mathbfxin mathbbR^n:F(mathbfx) =mathbf0$ where $operatornamerkoperatornameJac_mathbfx(F)=k$ for all $mathbfx$, then by the implicit function theorem, for every $(mathbfa, mathbfb)in M$ there are open subsets $Asubset Bbb R^n-k$ and $Bsubset Bbb R^k$ such that $mathbfain A$ and $mathbfbin B$ such that:
$Mcap (Atimes B)$ is the graph of a differentiable function $g:Ato
B$.
Furthermore, there is a change of coordinates that allows for $g$ to be simply the constant map $forall xin A,g(x)=0$. What this really means is that for each $xin A$ you change the coordinates of $xtimes B$ in such a way that the point of $M$ whose $A$-value is $x$ is now $(x,0)$. You do so smoothly over $Atimes B$, in other words
There is a diffeomorphism $f:Atimes Bto Atimes B'$ (for some open
subset $B'subset Bbb R^k$ which has to contain $0$), such that $f(Mcap (Atimes B))=Atimes0$.
Now to translate this into the notations used for submanifolds:
Let $U=Atimes B$, $V=Atimes B'$, and let $f:Atimes Bto Atimes B'$ represent the change of coordinates above. The condition for being a submanifold is obviously met: $f(Ucap M)$ is the set of elements of $V$ whose last $k$ coordinates are $0$. All the difficulty was to show the existence of the change of variables.
answered Aug 6 at 23:45
Arnaud Mortier
19.2k22159
answered Aug 6 at 23:45
Arnaud Mortier
19.2k22159
answered Aug 6 at 23:45
Arnaud Mortier
19.2k22159
answered Aug 6 at 23:45
Arnaud Mortier
19.2k22159
19.2k22159
Thanks so much for this answer. My whole confusion was in the construction of the diffeomorphism. I'm dutifully writing this explicitly in the margins of my book. Thanks again.
– Travis C Cuvelier
Aug 7 at 3:22
add a comment |Â
Thanks so much for this answer. My whole confusion was in the construction of the diffeomorphism. I'm dutifully writing this explicitly in the margins of my book. Thanks again.
– Travis C Cuvelier
Aug 7 at 3:22
Thanks so much for this answer. My whole confusion was in the construction of the diffeomorphism. I'm dutifully writing this explicitly in the margins of my book. Thanks again.
– Travis C Cuvelier
Aug 7 at 3:22
Thanks so much for this answer. My whole confusion was in the construction of the diffeomorphism. I'm dutifully writing this explicitly in the margins of my book. Thanks again.
– Travis C Cuvelier
Aug 7 at 3:22
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2874360%2funderstanding-some-definitions-of-smooth-manifolds-and-the-inverse-function-theo%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Are you sure of your transcription of the implicit function theorem? There is nothing in the assumptions that implies that the level $0$ set of $F$ can be seen as the graph of a function of the first $n-k$ coordinates. Think of the function $F:Bbb Rtimes Bbb RtoBbb R$ defined by $f(x,y)=x$.
– Arnaud Mortier
Aug 6 at 22:43
You're definitely right, I was not thinking when I was transcribing. The assumption should definitely be that the last k columns of the Jacobian are invertible.
– Travis C Cuvelier
Aug 7 at 3:19
If you have a second, would you mind answering this clarification re the statement of the implicit function theorem. While the version you posted below is definitely what's stated in the back of the book, they give the example x^2+y^2-1 =0 as implicitly defining the unit circle manifold in $mathbbR^2$. The Jacobian of F(x,y) = x^2+y^2 doesn't vanish on this set, but certainly at y = 0 the last column isn't full rank. They give the fact that the Jabobian is never the zero vector as the reason the implicit function theorem applies. Is it just a change of coordinates again? Seems alright...
– Travis C Cuvelier
Aug 7 at 3:52
Yes, while the theorem is stated with the last $k$ columns, this point is not particularly crucial, if some $k$ columns give an invertible matrix, then you can move them to the right (which does correspond to a change of variables in a way, as you permute the variables).
– Arnaud Mortier
Aug 7 at 9:02
Thanks so much! Owe you one!
– Travis C Cuvelier
Aug 7 at 15:54