Mixing
Submersions in differential topology
$begingroup$
If $f:R^3rightarrow R$ be a.t. $f(x) = x_1^2+x_2^2+x_3^2$, then I was explained how $df_a:R^3rightarrow R$ is surjective except except when $f(a)=0$ here Understanding submersions in differential topology
Now, for $f(x) = x_1^2+x_2^2-x_3^2$, $df_a$ is the matrix $(2a_1,2a_2,-2a_3)$. Why is it that for $f(a)=0$, $df_a$ is not surjective in this case also?
If $f(a) = a_1^2+a_2^2-a_3^2=0$, can we not take the derivative map to be the matrix $(2a_1,2a_2,-2sqrt{a_1^2+a_2^2})$ so that $forall xin R$, we can find a point whose image is $x$ under $df_a$ and the pre-image of $x$ is:
$df_a([1.5x/2a_1,1.5x/2a_2,x/sqrt{a_1^2+a_2^2}) = 2a_1.dfrac{1.5x}{2a_1}+2a_2.dfrac{1.5x}{2a_2}- 2sqrt{a_1^2+a_2^2}.dfrac{x}{sqrt{a_1^2+a_2^2}} = x?$ I know that I'm wrong. Can someone please point out where I went wrong? Thanks.
multivariable-calculus differential-geometry differential-topology smooth-manifolds
asked Jan 31 at 9:46
manifoldedmanifolded
50219
$endgroup$
add a comment |
$begingroup$
If $f:R^3rightarrow R$ be a.t. $f(x) = x_1^2+x_2^2+x_3^2$, then I was explained how $df_a:R^3rightarrow R$ is surjective except except when $f(a)=0$ here Understanding submersions in differential topology
Now, for $f(x) = x_1^2+x_2^2-x_3^2$, $df_a$ is the matrix $(2a_1,2a_2,-2a_3)$. Why is it that for $f(a)=0$, $df_a$ is not surjective in this case also?
If $f(a) = a_1^2+a_2^2-a_3^2=0$, can we not take the derivative map to be the matrix $(2a_1,2a_2,-2sqrt{a_1^2+a_2^2})$ so that $forall xin R$, we can find a point whose image is $x$ under $df_a$ and the pre-image of $x$ is:
$df_a([1.5x/2a_1,1.5x/2a_2,x/sqrt{a_1^2+a_2^2}) = 2a_1.dfrac{1.5x}{2a_1}+2a_2.dfrac{1.5x}{2a_2}- 2sqrt{a_1^2+a_2^2}.dfrac{x}{sqrt{a_1^2+a_2^2}} = x?$ I know that I'm wrong. Can someone please point out where I went wrong? Thanks.
multivariable-calculus differential-geometry differential-topology smooth-manifolds
asked Jan 31 at 9:46
manifoldedmanifolded
50219
$endgroup$
$begingroup$
A linear map $mathbb{R}^3to mathbb{R}$ is surjective if and only if it is nonzero. Hence, here, $df_a$ is surjective if and only if $a_1neq 0, a_2neq 0$ or $a_3 neq 0$. It could very well be that $a_1neq 0$ while $f(a)=0$, in this case $df_a$ is surjective. I think what was meant is that $f$ is not a submersion *on the whole of $f^{-1}(0)$ *
$endgroup$
– Max
Jan 31 at 10:06
$begingroup$
Understood! Thanks :)
$endgroup$
– manifolded
Jan 31 at 10:08
add a comment |
$begingroup$
If $f:R^3rightarrow R$ be a.t. $f(x) = x_1^2+x_2^2+x_3^2$, then I was explained how $df_a:R^3rightarrow R$ is surjective except except when $f(a)=0$ here Understanding submersions in differential topology
Now, for $f(x) = x_1^2+x_2^2-x_3^2$, $df_a$ is the matrix $(2a_1,2a_2,-2a_3)$. Why is it that for $f(a)=0$, $df_a$ is not surjective in this case also?
If $f(a) = a_1^2+a_2^2-a_3^2=0$, can we not take the derivative map to be the matrix $(2a_1,2a_2,-2sqrt{a_1^2+a_2^2})$ so that $forall xin R$, we can find a point whose image is $x$ under $df_a$ and the pre-image of $x$ is:
$df_a([1.5x/2a_1,1.5x/2a_2,x/sqrt{a_1^2+a_2^2}) = 2a_1.dfrac{1.5x}{2a_1}+2a_2.dfrac{1.5x}{2a_2}- 2sqrt{a_1^2+a_2^2}.dfrac{x}{sqrt{a_1^2+a_2^2}} = x?$ I know that I'm wrong. Can someone please point out where I went wrong? Thanks.
multivariable-calculus differential-geometry differential-topology smooth-manifolds
asked Jan 31 at 9:46
manifoldedmanifolded
50219
$endgroup$
If $f:R^3rightarrow R$ be a.t. $f(x) = x_1^2+x_2^2+x_3^2$, then I was explained how $df_a:R^3rightarrow R$ is surjective except except when $f(a)=0$ here Understanding submersions in differential topology
Now, for $f(x) = x_1^2+x_2^2-x_3^2$, $df_a$ is the matrix $(2a_1,2a_2,-2a_3)$. Why is it that for $f(a)=0$, $df_a$ is not surjective in this case also?
If $f(a) = a_1^2+a_2^2-a_3^2=0$, can we not take the derivative map to be the matrix $(2a_1,2a_2,-2sqrt{a_1^2+a_2^2})$ so that $forall xin R$, we can find a point whose image is $x$ under $df_a$ and the pre-image of $x$ is:
$df_a([1.5x/2a_1,1.5x/2a_2,x/sqrt{a_1^2+a_2^2}) = 2a_1.dfrac{1.5x}{2a_1}+2a_2.dfrac{1.5x}{2a_2}- 2sqrt{a_1^2+a_2^2}.dfrac{x}{sqrt{a_1^2+a_2^2}} = x?$ I know that I'm wrong. Can someone please point out where I went wrong? Thanks.
multivariable-calculus differential-geometry differential-topology smooth-manifolds
multivariable-calculus differential-geometry differential-topology smooth-manifolds
asked Jan 31 at 9:46
manifoldedmanifolded
50219
asked Jan 31 at 9:46
manifoldedmanifolded
50219
asked Jan 31 at 9:46
manifoldedmanifolded
50219
asked Jan 31 at 9:46
manifoldedmanifolded
50219
asked Jan 31 at 9:46
manifoldedmanifolded
50219
50219
$begingroup$
A linear map $mathbb{R}^3to mathbb{R}$ is surjective if and only if it is nonzero. Hence, here, $df_a$ is surjective if and only if $a_1neq 0, a_2neq 0$ or $a_3 neq 0$. It could very well be that $a_1neq 0$ while $f(a)=0$, in this case $df_a$ is surjective. I think what was meant is that $f$ is not a submersion *on the whole of $f^{-1}(0)$ *
$endgroup$
– Max
Jan 31 at 10:06
$begingroup$
Understood! Thanks :)
$endgroup$
– manifolded
Jan 31 at 10:08
add a comment |
$begingroup$
A linear map $mathbb{R}^3to mathbb{R}$ is surjective if and only if it is nonzero. Hence, here, $df_a$ is surjective if and only if $a_1neq 0, a_2neq 0$ or $a_3 neq 0$. It could very well be that $a_1neq 0$ while $f(a)=0$, in this case $df_a$ is surjective. I think what was meant is that $f$ is not a submersion *on the whole of $f^{-1}(0)$ *
$endgroup$
– Max
Jan 31 at 10:06
$begingroup$
Understood! Thanks :)
$endgroup$
– manifolded
Jan 31 at 10:08
$begingroup$
A linear map $mathbb{R}^3to mathbb{R}$ is surjective if and only if it is nonzero. Hence, here, $df_a$ is surjective if and only if $a_1neq 0, a_2neq 0$ or $a_3 neq 0$. It could very well be that $a_1neq 0$ while $f(a)=0$, in this case $df_a$ is surjective. I think what was meant is that $f$ is not a submersion *on the whole of $f^{-1}(0)$ *
$endgroup$
– Max
Jan 31 at 10:06
$begingroup$
A linear map $mathbb{R}^3to mathbb{R}$ is surjective if and only if it is nonzero. Hence, here, $df_a$ is surjective if and only if $a_1neq 0, a_2neq 0$ or $a_3 neq 0$. It could very well be that $a_1neq 0$ while $f(a)=0$, in this case $df_a$ is surjective. I think what was meant is that $f$ is not a submersion *on the whole of $f^{-1}(0)$ *
$endgroup$
– Max
Jan 31 at 10:06
$begingroup$
Understood! Thanks :)
$endgroup$
– manifolded
Jan 31 at 10:08
$begingroup$
Understood! Thanks :)
$endgroup$
– manifolded
Jan 31 at 10:08
add a comment |
1 Answer
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If $f(x_1,x_2,x_3)=x_1^2+x_2^2-x_3^2$, $df_a$ is surjective for every $ainmathbb{R}^3-{0}$. So it is not surjective on every point of ${a:f(a)=0}$.
answered Jan 31 at 9:54
Tsemo AristideTsemo Aristide
60.2k11446
$endgroup$
$begingroup$
Sorry didn't understand. I get $df_a$ is surjective for $ain R^3 - {0}$, can you please give me a $xin R$ which does not have a pre-image under $df_a$? Thanks.
$endgroup$
– manifolded
Jan 31 at 10:02
add a comment |
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$begingroup$
If $f(x_1,x_2,x_3)=x_1^2+x_2^2-x_3^2$, $df_a$ is surjective for every $ainmathbb{R}^3-{0}$. So it is not surjective on every point of ${a:f(a)=0}$.
answered Jan 31 at 9:54
Tsemo AristideTsemo Aristide
60.2k11446
$endgroup$
$begingroup$
Sorry didn't understand. I get $df_a$ is surjective for $ain R^3 - {0}$, can you please give me a $xin R$ which does not have a pre-image under $df_a$? Thanks.
$endgroup$
– manifolded
Jan 31 at 10:02
add a comment |
$begingroup$
If $f(x_1,x_2,x_3)=x_1^2+x_2^2-x_3^2$, $df_a$ is surjective for every $ainmathbb{R}^3-{0}$. So it is not surjective on every point of ${a:f(a)=0}$.
answered Jan 31 at 9:54
Tsemo AristideTsemo Aristide
60.2k11446
$endgroup$
$begingroup$
Sorry didn't understand. I get $df_a$ is surjective for $ain R^3 - {0}$, can you please give me a $xin R$ which does not have a pre-image under $df_a$? Thanks.
$endgroup$
– manifolded
Jan 31 at 10:02
add a comment |
$begingroup$
If $f(x_1,x_2,x_3)=x_1^2+x_2^2-x_3^2$, $df_a$ is surjective for every $ainmathbb{R}^3-{0}$. So it is not surjective on every point of ${a:f(a)=0}$.
answered Jan 31 at 9:54
Tsemo AristideTsemo Aristide
60.2k11446
$endgroup$
If $f(x_1,x_2,x_3)=x_1^2+x_2^2-x_3^2$, $df_a$ is surjective for every $ainmathbb{R}^3-{0}$. So it is not surjective on every point of ${a:f(a)=0}$.
answered Jan 31 at 9:54
Tsemo AristideTsemo Aristide
60.2k11446
answered Jan 31 at 9:54
Tsemo AristideTsemo Aristide
60.2k11446
answered Jan 31 at 9:54
Tsemo AristideTsemo Aristide
60.2k11446
answered Jan 31 at 9:54
Tsemo AristideTsemo Aristide
60.2k11446
60.2k11446
$begingroup$
Sorry didn't understand. I get $df_a$ is surjective for $ain R^3 - {0}$, can you please give me a $xin R$ which does not have a pre-image under $df_a$? Thanks.
$endgroup$
– manifolded
Jan 31 at 10:02
add a comment |
$begingroup$
Sorry didn't understand. I get $df_a$ is surjective for $ain R^3 - {0}$, can you please give me a $xin R$ which does not have a pre-image under $df_a$? Thanks.
$endgroup$
– manifolded
Jan 31 at 10:02
$begingroup$
Sorry didn't understand. I get $df_a$ is surjective for $ain R^3 - {0}$, can you please give me a $xin R$ which does not have a pre-image under $df_a$? Thanks.
$endgroup$
– manifolded
Jan 31 at 10:02
$begingroup$
Sorry didn't understand. I get $df_a$ is surjective for $ain R^3 - {0}$, can you please give me a $xin R$ which does not have a pre-image under $df_a$? Thanks.
$endgroup$
– manifolded
Jan 31 at 10:02
add a comment |
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$begingroup$
A linear map $mathbb{R}^3to mathbb{R}$ is surjective if and only if it is nonzero. Hence, here, $df_a$ is surjective if and only if $a_1neq 0, a_2neq 0$ or $a_3 neq 0$. It could very well be that $a_1neq 0$ while $f(a)=0$, in this case $df_a$ is surjective. I think what was meant is that $f$ is not a submersion *on the whole of $f^{-1}(0)$ *
$endgroup$
– Max
Jan 31 at 10:06
$begingroup$
Understood! Thanks :)
$endgroup$
– manifolded
Jan 31 at 10:08