Theorem about smooth functions on manifolds
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Here is a theorem as it appears in Spivak's Differential Geometry Vol. 1.
THEOREM. (1) If $f:M^nto N^m$ has rank $k$ at $p$, then there is some coordinate system $(x,U)$ around $p$ and some coordinate system $(y,V)$ around $f(p)$ with $$ycirc fcirc x^{-1}(a^1,ldots, a^n)=(a^1,ldots, a^k,psi^{k+1}(a),ldots,psi^m(a)).$$ Moreover, given any coordinate system $y,$ appropriate coordinate system on $N$ can be obtained merely by permuting the component functions of $y.$
(2) If $f$ has rank $k$ in a neighborhood of $p,$ then there are coordinate systems $(x,U)$ and $(y,V)$ such that $$ycirc fcirc x^{-1}(a^1,ldots, a^n)=(a^1,ldots, a^k,0,ldots,0).$$
My question is: how am I supposed to interpret this this result? The only way I can memorize theorems and proofs is by packing the result into a short mnemonic (usually just an easy to remember or intuitive fact), but I have no idea what the point of this theorem is. Is there anyone who can shed some intuitive understanding of what this theorem is saying.
differential-geometry differential-topology smooth-manifolds smooth-functions
asked Dec 16 '18 at 1:58
D. BroganD. Brogan
638513
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add a comment |
$begingroup$
Here is a theorem as it appears in Spivak's Differential Geometry Vol. 1.
THEOREM. (1) If $f:M^nto N^m$ has rank $k$ at $p$, then there is some coordinate system $(x,U)$ around $p$ and some coordinate system $(y,V)$ around $f(p)$ with $$ycirc fcirc x^{-1}(a^1,ldots, a^n)=(a^1,ldots, a^k,psi^{k+1}(a),ldots,psi^m(a)).$$ Moreover, given any coordinate system $y,$ appropriate coordinate system on $N$ can be obtained merely by permuting the component functions of $y.$
(2) If $f$ has rank $k$ in a neighborhood of $p,$ then there are coordinate systems $(x,U)$ and $(y,V)$ such that $$ycirc fcirc x^{-1}(a^1,ldots, a^n)=(a^1,ldots, a^k,0,ldots,0).$$
My question is: how am I supposed to interpret this this result? The only way I can memorize theorems and proofs is by packing the result into a short mnemonic (usually just an easy to remember or intuitive fact), but I have no idea what the point of this theorem is. Is there anyone who can shed some intuitive understanding of what this theorem is saying.
differential-geometry differential-topology smooth-manifolds smooth-functions
asked Dec 16 '18 at 1:58
D. BroganD. Brogan
638513
$endgroup$
4
$begingroup$
The point is that if a function has constant rank in a neighborhood, then in suitable coordinate patches it "looks like" the standard projection $mathbb{R}^n to mathbb{R}^k subset mathbb{R}^m$. Similarly, even if the function just has rank $k$ at a single point, the first $k$ can be chosen to look "standard" on a suitable coordinate patch. You can think of it as an analogue to choosing a nice basis for a linear transformation which puts it in a standard form (say, upper triangular, or diagonal if possible).
$endgroup$
– Max
Dec 16 '18 at 2:12
add a comment |
$begingroup$
Here is a theorem as it appears in Spivak's Differential Geometry Vol. 1.
THEOREM. (1) If $f:M^nto N^m$ has rank $k$ at $p$, then there is some coordinate system $(x,U)$ around $p$ and some coordinate system $(y,V)$ around $f(p)$ with $$ycirc fcirc x^{-1}(a^1,ldots, a^n)=(a^1,ldots, a^k,psi^{k+1}(a),ldots,psi^m(a)).$$ Moreover, given any coordinate system $y,$ appropriate coordinate system on $N$ can be obtained merely by permuting the component functions of $y.$
(2) If $f$ has rank $k$ in a neighborhood of $p,$ then there are coordinate systems $(x,U)$ and $(y,V)$ such that $$ycirc fcirc x^{-1}(a^1,ldots, a^n)=(a^1,ldots, a^k,0,ldots,0).$$
My question is: how am I supposed to interpret this this result? The only way I can memorize theorems and proofs is by packing the result into a short mnemonic (usually just an easy to remember or intuitive fact), but I have no idea what the point of this theorem is. Is there anyone who can shed some intuitive understanding of what this theorem is saying.
differential-geometry differential-topology smooth-manifolds smooth-functions
asked Dec 16 '18 at 1:58
D. BroganD. Brogan
638513
$endgroup$
Here is a theorem as it appears in Spivak's Differential Geometry Vol. 1.
THEOREM. (1) If $f:M^nto N^m$ has rank $k$ at $p$, then there is some coordinate system $(x,U)$ around $p$ and some coordinate system $(y,V)$ around $f(p)$ with $$ycirc fcirc x^{-1}(a^1,ldots, a^n)=(a^1,ldots, a^k,psi^{k+1}(a),ldots,psi^m(a)).$$ Moreover, given any coordinate system $y,$ appropriate coordinate system on $N$ can be obtained merely by permuting the component functions of $y.$
(2) If $f$ has rank $k$ in a neighborhood of $p,$ then there are coordinate systems $(x,U)$ and $(y,V)$ such that $$ycirc fcirc x^{-1}(a^1,ldots, a^n)=(a^1,ldots, a^k,0,ldots,0).$$
My question is: how am I supposed to interpret this this result? The only way I can memorize theorems and proofs is by packing the result into a short mnemonic (usually just an easy to remember or intuitive fact), but I have no idea what the point of this theorem is. Is there anyone who can shed some intuitive understanding of what this theorem is saying.
differential-geometry differential-topology smooth-manifolds smooth-functions
differential-geometry differential-topology smooth-manifolds smooth-functions
asked Dec 16 '18 at 1:58
D. BroganD. Brogan
638513
asked Dec 16 '18 at 1:58
D. BroganD. Brogan
638513
asked Dec 16 '18 at 1:58
D. BroganD. Brogan
638513
asked Dec 16 '18 at 1:58
D. BroganD. Brogan
638513
asked Dec 16 '18 at 1:58
D. BroganD. Brogan
638513
638513
4
$begingroup$
The point is that if a function has constant rank in a neighborhood, then in suitable coordinate patches it "looks like" the standard projection $mathbb{R}^n to mathbb{R}^k subset mathbb{R}^m$. Similarly, even if the function just has rank $k$ at a single point, the first $k$ can be chosen to look "standard" on a suitable coordinate patch. You can think of it as an analogue to choosing a nice basis for a linear transformation which puts it in a standard form (say, upper triangular, or diagonal if possible).
$endgroup$
– Max
Dec 16 '18 at 2:12
add a comment |
4
$begingroup$
The point is that if a function has constant rank in a neighborhood, then in suitable coordinate patches it "looks like" the standard projection $mathbb{R}^n to mathbb{R}^k subset mathbb{R}^m$. Similarly, even if the function just has rank $k$ at a single point, the first $k$ can be chosen to look "standard" on a suitable coordinate patch. You can think of it as an analogue to choosing a nice basis for a linear transformation which puts it in a standard form (say, upper triangular, or diagonal if possible).
$endgroup$
– Max
Dec 16 '18 at 2:12
4
4
$begingroup$
The point is that if a function has constant rank in a neighborhood, then in suitable coordinate patches it "looks like" the standard projection $mathbb{R}^n to mathbb{R}^k subset mathbb{R}^m$. Similarly, even if the function just has rank $k$ at a single point, the first $k$ can be chosen to look "standard" on a suitable coordinate patch. You can think of it as an analogue to choosing a nice basis for a linear transformation which puts it in a standard form (say, upper triangular, or diagonal if possible).
$endgroup$
– Max
Dec 16 '18 at 2:12
$begingroup$
The point is that if a function has constant rank in a neighborhood, then in suitable coordinate patches it "looks like" the standard projection $mathbb{R}^n to mathbb{R}^k subset mathbb{R}^m$. Similarly, even if the function just has rank $k$ at a single point, the first $k$ can be chosen to look "standard" on a suitable coordinate patch. You can think of it as an analogue to choosing a nice basis for a linear transformation which puts it in a standard form (say, upper triangular, or diagonal if possible).
$endgroup$
– Max
Dec 16 '18 at 2:12
add a comment |
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The point is that if a function has constant rank in a neighborhood, then in suitable coordinate patches it "looks like" the standard projection $mathbb{R}^n to mathbb{R}^k subset mathbb{R}^m$. Similarly, even if the function just has rank $k$ at a single point, the first $k$ can be chosen to look "standard" on a suitable coordinate patch. You can think of it as an analogue to choosing a nice basis for a linear transformation which puts it in a standard form (say, upper triangular, or diagonal if possible).
$endgroup$
– Max
Dec 16 '18 at 2:12