Mixing
Understanding submersions in differential topology
$begingroup$
I'm having trouble understanding an example in Guillemin and Pollack. If $f:R^krightarrow R$ be defined by
$f(x) = |x|^2 = x_1^2+...+x_k^2$
The derivative $df_x$ at the point $a = (a_1,..,a_k)$ has matrix $(2a_1,..,2a_k)$. Thus
$df_a:R^krightarrow R$ is surjective unless $f(a)=0$, so every nonzero real number is a regular value of $f$. I can't understand how to check $df_a$ is surjective and why it is obvious that every nonzero real number is a regular value of $f$? A hint is appreciated. Thanks.
multivariable-calculus differential-geometry differential-topology smooth-manifolds
edited Jan 31 at 9:10
Sou
3,3242923
asked Jan 31 at 8:38
manifoldedmanifolded
50219
$endgroup$
add a comment |
$begingroup$
I'm having trouble understanding an example in Guillemin and Pollack. If $f:R^krightarrow R$ be defined by
$f(x) = |x|^2 = x_1^2+...+x_k^2$
The derivative $df_x$ at the point $a = (a_1,..,a_k)$ has matrix $(2a_1,..,2a_k)$. Thus
$df_a:R^krightarrow R$ is surjective unless $f(a)=0$, so every nonzero real number is a regular value of $f$. I can't understand how to check $df_a$ is surjective and why it is obvious that every nonzero real number is a regular value of $f$? A hint is appreciated. Thanks.
multivariable-calculus differential-geometry differential-topology smooth-manifolds
edited Jan 31 at 9:10
Sou
3,3242923
asked Jan 31 at 8:38
manifoldedmanifolded
50219
$endgroup$
add a comment |
$begingroup$
I'm having trouble understanding an example in Guillemin and Pollack. If $f:R^krightarrow R$ be defined by
$f(x) = |x|^2 = x_1^2+...+x_k^2$
The derivative $df_x$ at the point $a = (a_1,..,a_k)$ has matrix $(2a_1,..,2a_k)$. Thus
$df_a:R^krightarrow R$ is surjective unless $f(a)=0$, so every nonzero real number is a regular value of $f$. I can't understand how to check $df_a$ is surjective and why it is obvious that every nonzero real number is a regular value of $f$? A hint is appreciated. Thanks.
multivariable-calculus differential-geometry differential-topology smooth-manifolds
edited Jan 31 at 9:10
Sou
3,3242923
asked Jan 31 at 8:38
manifoldedmanifolded
50219
$endgroup$
I'm having trouble understanding an example in Guillemin and Pollack. If $f:R^krightarrow R$ be defined by
$f(x) = |x|^2 = x_1^2+...+x_k^2$
The derivative $df_x$ at the point $a = (a_1,..,a_k)$ has matrix $(2a_1,..,2a_k)$. Thus
$df_a:R^krightarrow R$ is surjective unless $f(a)=0$, so every nonzero real number is a regular value of $f$. I can't understand how to check $df_a$ is surjective and why it is obvious that every nonzero real number is a regular value of $f$? A hint is appreciated. Thanks.
multivariable-calculus differential-geometry differential-topology smooth-manifolds
multivariable-calculus differential-geometry differential-topology smooth-manifolds
edited Jan 31 at 9:10
Sou
3,3242923
asked Jan 31 at 8:38
manifoldedmanifolded
50219
edited Jan 31 at 9:10
Sou
3,3242923
asked Jan 31 at 8:38
manifoldedmanifolded
50219
edited Jan 31 at 9:10
Sou
3,3242923
edited Jan 31 at 9:10
Sou
3,3242923
edited Jan 31 at 9:10
Sou
3,3242923
3,3242923
asked Jan 31 at 8:38
manifoldedmanifolded
50219
asked Jan 31 at 8:38
manifoldedmanifolded
50219
asked Jan 31 at 8:38
manifoldedmanifolded
50219
50219
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
It is clear that if $f(a)=a_1^2+cdots+a_k^2 = 0$, then $a=0$ which implies $df_a = (0,dots,0)$ is the zero map (obviously not surjective).
If $f(a)neq 0$, then at least there is one component of $a$ which is nonzero. Wlog, assume $a_1 neq 0$. Therefore the components of $df_a$ not all zero. For any nonzero $x in Bbb{R}$,
$$
df_a([x/2a_1, , 0, cdots ,, 0]) = 2a_1 frac{x}{2a_1} + 2a_2 cdot 0 + cdots + 2a_k cdot 0 = x.
$$
So $df_a$ is onto.
Since any nonzero real number is the image of a particular vector under all linear map $df_a$ (where $a$ is point such that $f(a) neq 0$), therefore they're all regular value.
answered Jan 31 at 8:54
SouSou
3,3242923
$endgroup$
add a comment |
$begingroup$
Hint: What is the rank of the linear map $2(a_1,dots, a_k):Bbb R^krightarrow Bbb R$? (And, what is the dimension of $Bbb R$?)
answered Jan 31 at 8:51
Chris CusterChris Custer
14.3k3827
$endgroup$
$begingroup$
$dim(R)=1$. I'm not sure about the rank of the linear map. It is a row matrix and can have $k$ linearly independent elements?
$endgroup$
– manifolded
Jan 31 at 8:56
$begingroup$
The rank always equals the dimension of the column space, which equals that of the row space. So...?
$endgroup$
– Chris Custer
Jan 31 at 9:01
add a comment |
$begingroup$
For $a neq 0$, the reason why $df_a$ is surjective because $df_a$ is a linear map from $T_p mathbb{R}^k$ to $T_{f(p)} mathbb{R}$ and since $operatorname{rank} df_a = 1 = operatorname{dim} T_{f(p)}mathbb{R}$ it follows that $df_a$ is surjective.
To see that $operatorname{rank} df_a = 1$, note that the matrix $[ 2a_1 dots 2a_k]$ will always have some entry to be non-zero (since we assumed that $a neq 0$ so some $a_i$ must be non-zero for some $0 leq i leq k$) and having one non-zero entry is all we need to conclude that the rank of this matrix is $1$.
Recall that the for a smooth map $F : M to N$ between smooth manifolds, a point $c in N$ is a regular value of $F$ if and only if either $c$ is not in the image of $F$ or at every point $p in F^{-1}(c)$ we have $dF_p : T_pM to T_{F(p)}M$ to be surjective.
So choose some non-zero real number $y$, if $y notin f[mathbb{R^k}] $ then we're done otherwise suppose that $y in f[mathbb{R^k}]$, then $f^{-1}(y)$ is nonempty. Choose any $a in f^{-1}(y)$, and then note that $a$ must clearly be non-zero in $mathbb{R}^k$, thus by the above we have $df_a$ to be surjective and since $a$ was chosen arbitrarily here, this holds for all $a in f^{-1}(y)$, and so it follows that every non-zero real number is a regular value for $f$.
answered Feb 1 at 14:44
PerturbativePerturbative
4,47621554
$endgroup$
add a comment |
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3 Answers
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3 Answers
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$begingroup$
It is clear that if $f(a)=a_1^2+cdots+a_k^2 = 0$, then $a=0$ which implies $df_a = (0,dots,0)$ is the zero map (obviously not surjective).
If $f(a)neq 0$, then at least there is one component of $a$ which is nonzero. Wlog, assume $a_1 neq 0$. Therefore the components of $df_a$ not all zero. For any nonzero $x in Bbb{R}$,
$$
df_a([x/2a_1, , 0, cdots ,, 0]) = 2a_1 frac{x}{2a_1} + 2a_2 cdot 0 + cdots + 2a_k cdot 0 = x.
$$
So $df_a$ is onto.
Since any nonzero real number is the image of a particular vector under all linear map $df_a$ (where $a$ is point such that $f(a) neq 0$), therefore they're all regular value.
answered Jan 31 at 8:54
SouSou
3,3242923
$endgroup$
add a comment |
$begingroup$
It is clear that if $f(a)=a_1^2+cdots+a_k^2 = 0$, then $a=0$ which implies $df_a = (0,dots,0)$ is the zero map (obviously not surjective).
If $f(a)neq 0$, then at least there is one component of $a$ which is nonzero. Wlog, assume $a_1 neq 0$. Therefore the components of $df_a$ not all zero. For any nonzero $x in Bbb{R}$,
$$
df_a([x/2a_1, , 0, cdots ,, 0]) = 2a_1 frac{x}{2a_1} + 2a_2 cdot 0 + cdots + 2a_k cdot 0 = x.
$$
So $df_a$ is onto.
Since any nonzero real number is the image of a particular vector under all linear map $df_a$ (where $a$ is point such that $f(a) neq 0$), therefore they're all regular value.
answered Jan 31 at 8:54
SouSou
3,3242923
$endgroup$
add a comment |
$begingroup$
It is clear that if $f(a)=a_1^2+cdots+a_k^2 = 0$, then $a=0$ which implies $df_a = (0,dots,0)$ is the zero map (obviously not surjective).
If $f(a)neq 0$, then at least there is one component of $a$ which is nonzero. Wlog, assume $a_1 neq 0$. Therefore the components of $df_a$ not all zero. For any nonzero $x in Bbb{R}$,
$$
df_a([x/2a_1, , 0, cdots ,, 0]) = 2a_1 frac{x}{2a_1} + 2a_2 cdot 0 + cdots + 2a_k cdot 0 = x.
$$
So $df_a$ is onto.
Since any nonzero real number is the image of a particular vector under all linear map $df_a$ (where $a$ is point such that $f(a) neq 0$), therefore they're all regular value.
answered Jan 31 at 8:54
SouSou
3,3242923
$endgroup$
It is clear that if $f(a)=a_1^2+cdots+a_k^2 = 0$, then $a=0$ which implies $df_a = (0,dots,0)$ is the zero map (obviously not surjective).
If $f(a)neq 0$, then at least there is one component of $a$ which is nonzero. Wlog, assume $a_1 neq 0$. Therefore the components of $df_a$ not all zero. For any nonzero $x in Bbb{R}$,
$$
df_a([x/2a_1, , 0, cdots ,, 0]) = 2a_1 frac{x}{2a_1} + 2a_2 cdot 0 + cdots + 2a_k cdot 0 = x.
$$
So $df_a$ is onto.
Since any nonzero real number is the image of a particular vector under all linear map $df_a$ (where $a$ is point such that $f(a) neq 0$), therefore they're all regular value.
answered Jan 31 at 8:54
SouSou
3,3242923
edited Feb 1 at 14:16
answered Jan 31 at 8:54
SouSou
3,3242923
answered Jan 31 at 8:54
SouSou
3,3242923
answered Jan 31 at 8:54
SouSou
3,3242923
3,3242923
add a comment |
add a comment |
$begingroup$
Hint: What is the rank of the linear map $2(a_1,dots, a_k):Bbb R^krightarrow Bbb R$? (And, what is the dimension of $Bbb R$?)
answered Jan 31 at 8:51
Chris CusterChris Custer
14.3k3827
$endgroup$
$begingroup$
$dim(R)=1$. I'm not sure about the rank of the linear map. It is a row matrix and can have $k$ linearly independent elements?
$endgroup$
– manifolded
Jan 31 at 8:56
$begingroup$
The rank always equals the dimension of the column space, which equals that of the row space. So...?
$endgroup$
– Chris Custer
Jan 31 at 9:01
add a comment |
$begingroup$
Hint: What is the rank of the linear map $2(a_1,dots, a_k):Bbb R^krightarrow Bbb R$? (And, what is the dimension of $Bbb R$?)
answered Jan 31 at 8:51
Chris CusterChris Custer
14.3k3827
$endgroup$
$begingroup$
$dim(R)=1$. I'm not sure about the rank of the linear map. It is a row matrix and can have $k$ linearly independent elements?
$endgroup$
– manifolded
Jan 31 at 8:56
$begingroup$
The rank always equals the dimension of the column space, which equals that of the row space. So...?
$endgroup$
– Chris Custer
Jan 31 at 9:01
add a comment |
$begingroup$
Hint: What is the rank of the linear map $2(a_1,dots, a_k):Bbb R^krightarrow Bbb R$? (And, what is the dimension of $Bbb R$?)
answered Jan 31 at 8:51
Chris CusterChris Custer
14.3k3827
$endgroup$
Hint: What is the rank of the linear map $2(a_1,dots, a_k):Bbb R^krightarrow Bbb R$? (And, what is the dimension of $Bbb R$?)
answered Jan 31 at 8:51
Chris CusterChris Custer
14.3k3827
answered Jan 31 at 8:51
Chris CusterChris Custer
14.3k3827
answered Jan 31 at 8:51
Chris CusterChris Custer
14.3k3827
answered Jan 31 at 8:51
Chris CusterChris Custer
14.3k3827
14.3k3827
$begingroup$
$dim(R)=1$. I'm not sure about the rank of the linear map. It is a row matrix and can have $k$ linearly independent elements?
$endgroup$
– manifolded
Jan 31 at 8:56
$begingroup$
The rank always equals the dimension of the column space, which equals that of the row space. So...?
$endgroup$
– Chris Custer
Jan 31 at 9:01
add a comment |
$begingroup$
$dim(R)=1$. I'm not sure about the rank of the linear map. It is a row matrix and can have $k$ linearly independent elements?
$endgroup$
– manifolded
Jan 31 at 8:56
$begingroup$
The rank always equals the dimension of the column space, which equals that of the row space. So...?
$endgroup$
– Chris Custer
Jan 31 at 9:01
$begingroup$
$dim(R)=1$. I'm not sure about the rank of the linear map. It is a row matrix and can have $k$ linearly independent elements?
$endgroup$
– manifolded
Jan 31 at 8:56
$begingroup$
$dim(R)=1$. I'm not sure about the rank of the linear map. It is a row matrix and can have $k$ linearly independent elements?
$endgroup$
– manifolded
Jan 31 at 8:56
$begingroup$
The rank always equals the dimension of the column space, which equals that of the row space. So...?
$endgroup$
– Chris Custer
Jan 31 at 9:01
$begingroup$
The rank always equals the dimension of the column space, which equals that of the row space. So...?
$endgroup$
– Chris Custer
Jan 31 at 9:01
add a comment |
$begingroup$
For $a neq 0$, the reason why $df_a$ is surjective because $df_a$ is a linear map from $T_p mathbb{R}^k$ to $T_{f(p)} mathbb{R}$ and since $operatorname{rank} df_a = 1 = operatorname{dim} T_{f(p)}mathbb{R}$ it follows that $df_a$ is surjective.
To see that $operatorname{rank} df_a = 1$, note that the matrix $[ 2a_1 dots 2a_k]$ will always have some entry to be non-zero (since we assumed that $a neq 0$ so some $a_i$ must be non-zero for some $0 leq i leq k$) and having one non-zero entry is all we need to conclude that the rank of this matrix is $1$.
Recall that the for a smooth map $F : M to N$ between smooth manifolds, a point $c in N$ is a regular value of $F$ if and only if either $c$ is not in the image of $F$ or at every point $p in F^{-1}(c)$ we have $dF_p : T_pM to T_{F(p)}M$ to be surjective.
So choose some non-zero real number $y$, if $y notin f[mathbb{R^k}] $ then we're done otherwise suppose that $y in f[mathbb{R^k}]$, then $f^{-1}(y)$ is nonempty. Choose any $a in f^{-1}(y)$, and then note that $a$ must clearly be non-zero in $mathbb{R}^k$, thus by the above we have $df_a$ to be surjective and since $a$ was chosen arbitrarily here, this holds for all $a in f^{-1}(y)$, and so it follows that every non-zero real number is a regular value for $f$.
answered Feb 1 at 14:44
PerturbativePerturbative
4,47621554
$endgroup$
add a comment |
$begingroup$
For $a neq 0$, the reason why $df_a$ is surjective because $df_a$ is a linear map from $T_p mathbb{R}^k$ to $T_{f(p)} mathbb{R}$ and since $operatorname{rank} df_a = 1 = operatorname{dim} T_{f(p)}mathbb{R}$ it follows that $df_a$ is surjective.
To see that $operatorname{rank} df_a = 1$, note that the matrix $[ 2a_1 dots 2a_k]$ will always have some entry to be non-zero (since we assumed that $a neq 0$ so some $a_i$ must be non-zero for some $0 leq i leq k$) and having one non-zero entry is all we need to conclude that the rank of this matrix is $1$.
Recall that the for a smooth map $F : M to N$ between smooth manifolds, a point $c in N$ is a regular value of $F$ if and only if either $c$ is not in the image of $F$ or at every point $p in F^{-1}(c)$ we have $dF_p : T_pM to T_{F(p)}M$ to be surjective.
So choose some non-zero real number $y$, if $y notin f[mathbb{R^k}] $ then we're done otherwise suppose that $y in f[mathbb{R^k}]$, then $f^{-1}(y)$ is nonempty. Choose any $a in f^{-1}(y)$, and then note that $a$ must clearly be non-zero in $mathbb{R}^k$, thus by the above we have $df_a$ to be surjective and since $a$ was chosen arbitrarily here, this holds for all $a in f^{-1}(y)$, and so it follows that every non-zero real number is a regular value for $f$.
answered Feb 1 at 14:44
PerturbativePerturbative
4,47621554
$endgroup$
add a comment |
$begingroup$
For $a neq 0$, the reason why $df_a$ is surjective because $df_a$ is a linear map from $T_p mathbb{R}^k$ to $T_{f(p)} mathbb{R}$ and since $operatorname{rank} df_a = 1 = operatorname{dim} T_{f(p)}mathbb{R}$ it follows that $df_a$ is surjective.
To see that $operatorname{rank} df_a = 1$, note that the matrix $[ 2a_1 dots 2a_k]$ will always have some entry to be non-zero (since we assumed that $a neq 0$ so some $a_i$ must be non-zero for some $0 leq i leq k$) and having one non-zero entry is all we need to conclude that the rank of this matrix is $1$.
Recall that the for a smooth map $F : M to N$ between smooth manifolds, a point $c in N$ is a regular value of $F$ if and only if either $c$ is not in the image of $F$ or at every point $p in F^{-1}(c)$ we have $dF_p : T_pM to T_{F(p)}M$ to be surjective.
So choose some non-zero real number $y$, if $y notin f[mathbb{R^k}] $ then we're done otherwise suppose that $y in f[mathbb{R^k}]$, then $f^{-1}(y)$ is nonempty. Choose any $a in f^{-1}(y)$, and then note that $a$ must clearly be non-zero in $mathbb{R}^k$, thus by the above we have $df_a$ to be surjective and since $a$ was chosen arbitrarily here, this holds for all $a in f^{-1}(y)$, and so it follows that every non-zero real number is a regular value for $f$.
answered Feb 1 at 14:44
PerturbativePerturbative
4,47621554
$endgroup$
For $a neq 0$, the reason why $df_a$ is surjective because $df_a$ is a linear map from $T_p mathbb{R}^k$ to $T_{f(p)} mathbb{R}$ and since $operatorname{rank} df_a = 1 = operatorname{dim} T_{f(p)}mathbb{R}$ it follows that $df_a$ is surjective.
To see that $operatorname{rank} df_a = 1$, note that the matrix $[ 2a_1 dots 2a_k]$ will always have some entry to be non-zero (since we assumed that $a neq 0$ so some $a_i$ must be non-zero for some $0 leq i leq k$) and having one non-zero entry is all we need to conclude that the rank of this matrix is $1$.
Recall that the for a smooth map $F : M to N$ between smooth manifolds, a point $c in N$ is a regular value of $F$ if and only if either $c$ is not in the image of $F$ or at every point $p in F^{-1}(c)$ we have $dF_p : T_pM to T_{F(p)}M$ to be surjective.
So choose some non-zero real number $y$, if $y notin f[mathbb{R^k}] $ then we're done otherwise suppose that $y in f[mathbb{R^k}]$, then $f^{-1}(y)$ is nonempty. Choose any $a in f^{-1}(y)$, and then note that $a$ must clearly be non-zero in $mathbb{R}^k$, thus by the above we have $df_a$ to be surjective and since $a$ was chosen arbitrarily here, this holds for all $a in f^{-1}(y)$, and so it follows that every non-zero real number is a regular value for $f$.
answered Feb 1 at 14:44
PerturbativePerturbative
4,47621554
answered Feb 1 at 14:44
PerturbativePerturbative
4,47621554
answered Feb 1 at 14:44
PerturbativePerturbative
4,47621554
answered Feb 1 at 14:44
PerturbativePerturbative
4,47621554
4,47621554
add a comment |
add a comment |
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