critical points of a map
$begingroup$
Let $mathcal{C}$ be the $mathbb{R}^2$ submanifold defined by:
$y^2 = x(x − 1)(x − a)$ where $a>1$.
Let $f: mathcal{C} rightarrow mathbb{R}^2: (x,y) rightarrow x$
Find the critical point of $f$.
We have $partial_x f=1$ and $partial_y f=0$, which means that f has no critical points because its gradient is non zero.
On the other hand, I know that $f$ has 3 critical points in $(0,0), (1,0), (a,0)$. How to explain this contradiction?
Thank you for your help!
differential-geometry
asked Dec 24 '18 at 20:15
PerelManPerelMan
830414
$endgroup$
add a comment |
$begingroup$
Let $mathcal{C}$ be the $mathbb{R}^2$ submanifold defined by:
$y^2 = x(x − 1)(x − a)$ where $a>1$.
Let $f: mathcal{C} rightarrow mathbb{R}^2: (x,y) rightarrow x$
Find the critical point of $f$.
We have $partial_x f=1$ and $partial_y f=0$, which means that f has no critical points because its gradient is non zero.
On the other hand, I know that $f$ has 3 critical points in $(0,0), (1,0), (a,0)$. How to explain this contradiction?
Thank you for your help!
differential-geometry
asked Dec 24 '18 at 20:15
PerelManPerelMan
830414
$endgroup$
add a comment |
$begingroup$
Let $mathcal{C}$ be the $mathbb{R}^2$ submanifold defined by:
$y^2 = x(x − 1)(x − a)$ where $a>1$.
Let $f: mathcal{C} rightarrow mathbb{R}^2: (x,y) rightarrow x$
Find the critical point of $f$.
We have $partial_x f=1$ and $partial_y f=0$, which means that f has no critical points because its gradient is non zero.
On the other hand, I know that $f$ has 3 critical points in $(0,0), (1,0), (a,0)$. How to explain this contradiction?
Thank you for your help!
differential-geometry
asked Dec 24 '18 at 20:15
PerelManPerelMan
830414
$endgroup$
Let $mathcal{C}$ be the $mathbb{R}^2$ submanifold defined by:
$y^2 = x(x − 1)(x − a)$ where $a>1$.
Let $f: mathcal{C} rightarrow mathbb{R}^2: (x,y) rightarrow x$
Find the critical point of $f$.
We have $partial_x f=1$ and $partial_y f=0$, which means that f has no critical points because its gradient is non zero.
On the other hand, I know that $f$ has 3 critical points in $(0,0), (1,0), (a,0)$. How to explain this contradiction?
Thank you for your help!
differential-geometry
differential-geometry
asked Dec 24 '18 at 20:15
PerelManPerelMan
830414
asked Dec 24 '18 at 20:15
PerelManPerelMan
830414
asked Dec 24 '18 at 20:15
PerelManPerelMan
830414
asked Dec 24 '18 at 20:15
PerelManPerelMan
830414
asked Dec 24 '18 at 20:15
PerelManPerelMan
830414
830414
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $gcolonmathbb{R}^2tomathbb{R}$ defined by $g(x,y)=x$.
What you have shown is that $g$ has no critical point, however $f=g_{vert C}$ may have critical points!
Keep in mind that:
A regular map $hcolon Mrightarrow N$ restricted to a submanifold $S$ of $M$ may no longer be regular.
Example. The height function on $mathbb{R}^3$ has no critical point, however its restriction to the unit sphere $S^2$ has two critical points, the north pole and the south pole, at those points the tangent plane of $S^2$ has constant altitude.
The general phenomena is that at a point $pin M$, the linear map $T_phcolon T_pSto T_{f(p)}N$ may no longer be surjective, since $T_pS$ does not contain all the directions of $T_pM$.
When $N=mathbb{R}$, we are looking at points $p$, where $T_ph(T_pS)={0}$, since $T_ph$ is a linear form and it is surjective if and only if nonzero.
In your case, you are looking at point $(x,y)in C$ such that $T_{(x,y)}C$ is contained in a $x$-slice of $mathbb{R}^2$, this happens at the points $(0,0)$, $(1,0)$ and $(a,0)$.
Since $C$ is defined by the equation $varphi(x,y)=y^2-x(x-1)(x-a)=0$, the tangent line of $C$ at $(x,y)$ is given by $nabla_{(x,y)}varphi^perp$, so that $T_{(x,y)}C$ is contained in a $x$-slice if and only if $nabla_{(x,y)}varphi$ has no $y$-component.
Compute $nabla_{(x,y)}varphi=(-3x^2+2(a+1)x-a,2y)$, it has no $y$-component if and only if $y=0$.
Finally, the critical points of $f$ are the $(x,0)in C$, but $(x,0)in C$ if and only if $x=0,1,a$.
Here is a sketch, where $C$ is drawn in red:
answered Dec 24 '18 at 21:39
C. FalconC. Falcon
15.3k41951
$endgroup$
$begingroup$
Thank you for your thorough explanation. I have though a couple of questions: what is an x-slice of $mathbb{R}^2$. and what about $Msubset mathbb{R}^n and Nsubset mathbb{R}^p$? I think that in that case the regularity is preserved by the restriction to M and N ?
$endgroup$
– PerelMan
Dec 25 '18 at 0:08
1
$begingroup$
By a $x$-slice, I just mean a vertical line, the name is perhaps not that good! When I say that a map is regular, I really mean that it has no critical value, I am not talking about smoothness. A smooth map restricted to a submanifold is smooth and it is true for all manifolds, not only submanifolds of $mathbb{R}^n$. Is that clearer?
$endgroup$
– C. Falcon
Dec 25 '18 at 0:24
$begingroup$
Thank you! it is clear to me now.
$endgroup$
– PerelMan
Dec 25 '18 at 0:27
add a comment |
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1 Answer
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$begingroup$
Let $gcolonmathbb{R}^2tomathbb{R}$ defined by $g(x,y)=x$.
What you have shown is that $g$ has no critical point, however $f=g_{vert C}$ may have critical points!
Keep in mind that:
A regular map $hcolon Mrightarrow N$ restricted to a submanifold $S$ of $M$ may no longer be regular.
Example. The height function on $mathbb{R}^3$ has no critical point, however its restriction to the unit sphere $S^2$ has two critical points, the north pole and the south pole, at those points the tangent plane of $S^2$ has constant altitude.
The general phenomena is that at a point $pin M$, the linear map $T_phcolon T_pSto T_{f(p)}N$ may no longer be surjective, since $T_pS$ does not contain all the directions of $T_pM$.
When $N=mathbb{R}$, we are looking at points $p$, where $T_ph(T_pS)={0}$, since $T_ph$ is a linear form and it is surjective if and only if nonzero.
In your case, you are looking at point $(x,y)in C$ such that $T_{(x,y)}C$ is contained in a $x$-slice of $mathbb{R}^2$, this happens at the points $(0,0)$, $(1,0)$ and $(a,0)$.
Since $C$ is defined by the equation $varphi(x,y)=y^2-x(x-1)(x-a)=0$, the tangent line of $C$ at $(x,y)$ is given by $nabla_{(x,y)}varphi^perp$, so that $T_{(x,y)}C$ is contained in a $x$-slice if and only if $nabla_{(x,y)}varphi$ has no $y$-component.
Compute $nabla_{(x,y)}varphi=(-3x^2+2(a+1)x-a,2y)$, it has no $y$-component if and only if $y=0$.
Finally, the critical points of $f$ are the $(x,0)in C$, but $(x,0)in C$ if and only if $x=0,1,a$.
Here is a sketch, where $C$ is drawn in red:
answered Dec 24 '18 at 21:39
C. FalconC. Falcon
15.3k41951
$endgroup$
$begingroup$
Thank you for your thorough explanation. I have though a couple of questions: what is an x-slice of $mathbb{R}^2$. and what about $Msubset mathbb{R}^n and Nsubset mathbb{R}^p$? I think that in that case the regularity is preserved by the restriction to M and N ?
$endgroup$
– PerelMan
Dec 25 '18 at 0:08
1
$begingroup$
By a $x$-slice, I just mean a vertical line, the name is perhaps not that good! When I say that a map is regular, I really mean that it has no critical value, I am not talking about smoothness. A smooth map restricted to a submanifold is smooth and it is true for all manifolds, not only submanifolds of $mathbb{R}^n$. Is that clearer?
$endgroup$
– C. Falcon
Dec 25 '18 at 0:24
$begingroup$
Thank you! it is clear to me now.
$endgroup$
– PerelMan
Dec 25 '18 at 0:27
add a comment |
$begingroup$
Let $gcolonmathbb{R}^2tomathbb{R}$ defined by $g(x,y)=x$.
What you have shown is that $g$ has no critical point, however $f=g_{vert C}$ may have critical points!
Keep in mind that:
A regular map $hcolon Mrightarrow N$ restricted to a submanifold $S$ of $M$ may no longer be regular.
Example. The height function on $mathbb{R}^3$ has no critical point, however its restriction to the unit sphere $S^2$ has two critical points, the north pole and the south pole, at those points the tangent plane of $S^2$ has constant altitude.
The general phenomena is that at a point $pin M$, the linear map $T_phcolon T_pSto T_{f(p)}N$ may no longer be surjective, since $T_pS$ does not contain all the directions of $T_pM$.
When $N=mathbb{R}$, we are looking at points $p$, where $T_ph(T_pS)={0}$, since $T_ph$ is a linear form and it is surjective if and only if nonzero.
In your case, you are looking at point $(x,y)in C$ such that $T_{(x,y)}C$ is contained in a $x$-slice of $mathbb{R}^2$, this happens at the points $(0,0)$, $(1,0)$ and $(a,0)$.
Since $C$ is defined by the equation $varphi(x,y)=y^2-x(x-1)(x-a)=0$, the tangent line of $C$ at $(x,y)$ is given by $nabla_{(x,y)}varphi^perp$, so that $T_{(x,y)}C$ is contained in a $x$-slice if and only if $nabla_{(x,y)}varphi$ has no $y$-component.
Compute $nabla_{(x,y)}varphi=(-3x^2+2(a+1)x-a,2y)$, it has no $y$-component if and only if $y=0$.
Finally, the critical points of $f$ are the $(x,0)in C$, but $(x,0)in C$ if and only if $x=0,1,a$.
Here is a sketch, where $C$ is drawn in red:
answered Dec 24 '18 at 21:39
C. FalconC. Falcon
15.3k41951
$endgroup$
$begingroup$
Thank you for your thorough explanation. I have though a couple of questions: what is an x-slice of $mathbb{R}^2$. and what about $Msubset mathbb{R}^n and Nsubset mathbb{R}^p$? I think that in that case the regularity is preserved by the restriction to M and N ?
$endgroup$
– PerelMan
Dec 25 '18 at 0:08
1
$begingroup$
By a $x$-slice, I just mean a vertical line, the name is perhaps not that good! When I say that a map is regular, I really mean that it has no critical value, I am not talking about smoothness. A smooth map restricted to a submanifold is smooth and it is true for all manifolds, not only submanifolds of $mathbb{R}^n$. Is that clearer?
$endgroup$
– C. Falcon
Dec 25 '18 at 0:24
$begingroup$
Thank you! it is clear to me now.
$endgroup$
– PerelMan
Dec 25 '18 at 0:27
add a comment |
$begingroup$
Let $gcolonmathbb{R}^2tomathbb{R}$ defined by $g(x,y)=x$.
What you have shown is that $g$ has no critical point, however $f=g_{vert C}$ may have critical points!
Keep in mind that:
A regular map $hcolon Mrightarrow N$ restricted to a submanifold $S$ of $M$ may no longer be regular.
Example. The height function on $mathbb{R}^3$ has no critical point, however its restriction to the unit sphere $S^2$ has two critical points, the north pole and the south pole, at those points the tangent plane of $S^2$ has constant altitude.
The general phenomena is that at a point $pin M$, the linear map $T_phcolon T_pSto T_{f(p)}N$ may no longer be surjective, since $T_pS$ does not contain all the directions of $T_pM$.
When $N=mathbb{R}$, we are looking at points $p$, where $T_ph(T_pS)={0}$, since $T_ph$ is a linear form and it is surjective if and only if nonzero.
In your case, you are looking at point $(x,y)in C$ such that $T_{(x,y)}C$ is contained in a $x$-slice of $mathbb{R}^2$, this happens at the points $(0,0)$, $(1,0)$ and $(a,0)$.
Since $C$ is defined by the equation $varphi(x,y)=y^2-x(x-1)(x-a)=0$, the tangent line of $C$ at $(x,y)$ is given by $nabla_{(x,y)}varphi^perp$, so that $T_{(x,y)}C$ is contained in a $x$-slice if and only if $nabla_{(x,y)}varphi$ has no $y$-component.
Compute $nabla_{(x,y)}varphi=(-3x^2+2(a+1)x-a,2y)$, it has no $y$-component if and only if $y=0$.
Finally, the critical points of $f$ are the $(x,0)in C$, but $(x,0)in C$ if and only if $x=0,1,a$.
Here is a sketch, where $C$ is drawn in red:
answered Dec 24 '18 at 21:39
C. FalconC. Falcon
15.3k41951
$endgroup$
Let $gcolonmathbb{R}^2tomathbb{R}$ defined by $g(x,y)=x$.
What you have shown is that $g$ has no critical point, however $f=g_{vert C}$ may have critical points!
Keep in mind that:
A regular map $hcolon Mrightarrow N$ restricted to a submanifold $S$ of $M$ may no longer be regular.
Example. The height function on $mathbb{R}^3$ has no critical point, however its restriction to the unit sphere $S^2$ has two critical points, the north pole and the south pole, at those points the tangent plane of $S^2$ has constant altitude.
The general phenomena is that at a point $pin M$, the linear map $T_phcolon T_pSto T_{f(p)}N$ may no longer be surjective, since $T_pS$ does not contain all the directions of $T_pM$.
When $N=mathbb{R}$, we are looking at points $p$, where $T_ph(T_pS)={0}$, since $T_ph$ is a linear form and it is surjective if and only if nonzero.
In your case, you are looking at point $(x,y)in C$ such that $T_{(x,y)}C$ is contained in a $x$-slice of $mathbb{R}^2$, this happens at the points $(0,0)$, $(1,0)$ and $(a,0)$.
Since $C$ is defined by the equation $varphi(x,y)=y^2-x(x-1)(x-a)=0$, the tangent line of $C$ at $(x,y)$ is given by $nabla_{(x,y)}varphi^perp$, so that $T_{(x,y)}C$ is contained in a $x$-slice if and only if $nabla_{(x,y)}varphi$ has no $y$-component.
Compute $nabla_{(x,y)}varphi=(-3x^2+2(a+1)x-a,2y)$, it has no $y$-component if and only if $y=0$.
Finally, the critical points of $f$ are the $(x,0)in C$, but $(x,0)in C$ if and only if $x=0,1,a$.
Here is a sketch, where $C$ is drawn in red:
answered Dec 24 '18 at 21:39
C. FalconC. Falcon
15.3k41951
edited Dec 25 '18 at 17:03
answered Dec 24 '18 at 21:39
C. FalconC. Falcon
15.3k41951
answered Dec 24 '18 at 21:39
C. FalconC. Falcon
15.3k41951
answered Dec 24 '18 at 21:39
C. FalconC. Falcon
15.3k41951
15.3k41951
$begingroup$
Thank you for your thorough explanation. I have though a couple of questions: what is an x-slice of $mathbb{R}^2$. and what about $Msubset mathbb{R}^n and Nsubset mathbb{R}^p$? I think that in that case the regularity is preserved by the restriction to M and N ?
$endgroup$
– PerelMan
Dec 25 '18 at 0:08
1
$begingroup$
By a $x$-slice, I just mean a vertical line, the name is perhaps not that good! When I say that a map is regular, I really mean that it has no critical value, I am not talking about smoothness. A smooth map restricted to a submanifold is smooth and it is true for all manifolds, not only submanifolds of $mathbb{R}^n$. Is that clearer?
$endgroup$
– C. Falcon
Dec 25 '18 at 0:24
$begingroup$
Thank you! it is clear to me now.
$endgroup$
– PerelMan
Dec 25 '18 at 0:27
add a comment |
$begingroup$
Thank you for your thorough explanation. I have though a couple of questions: what is an x-slice of $mathbb{R}^2$. and what about $Msubset mathbb{R}^n and Nsubset mathbb{R}^p$? I think that in that case the regularity is preserved by the restriction to M and N ?
$endgroup$
– PerelMan
Dec 25 '18 at 0:08
1
$begingroup$
By a $x$-slice, I just mean a vertical line, the name is perhaps not that good! When I say that a map is regular, I really mean that it has no critical value, I am not talking about smoothness. A smooth map restricted to a submanifold is smooth and it is true for all manifolds, not only submanifolds of $mathbb{R}^n$. Is that clearer?
$endgroup$
– C. Falcon
Dec 25 '18 at 0:24
$begingroup$
Thank you! it is clear to me now.
$endgroup$
– PerelMan
Dec 25 '18 at 0:27
$begingroup$
Thank you for your thorough explanation. I have though a couple of questions: what is an x-slice of $mathbb{R}^2$. and what about $Msubset mathbb{R}^n and Nsubset mathbb{R}^p$? I think that in that case the regularity is preserved by the restriction to M and N ?
$endgroup$
– PerelMan
Dec 25 '18 at 0:08
$begingroup$
Thank you for your thorough explanation. I have though a couple of questions: what is an x-slice of $mathbb{R}^2$. and what about $Msubset mathbb{R}^n and Nsubset mathbb{R}^p$? I think that in that case the regularity is preserved by the restriction to M and N ?
$endgroup$
– PerelMan
Dec 25 '18 at 0:08
1
1
$begingroup$
By a $x$-slice, I just mean a vertical line, the name is perhaps not that good! When I say that a map is regular, I really mean that it has no critical value, I am not talking about smoothness. A smooth map restricted to a submanifold is smooth and it is true for all manifolds, not only submanifolds of $mathbb{R}^n$. Is that clearer?
$endgroup$
– C. Falcon
Dec 25 '18 at 0:24
$begingroup$
By a $x$-slice, I just mean a vertical line, the name is perhaps not that good! When I say that a map is regular, I really mean that it has no critical value, I am not talking about smoothness. A smooth map restricted to a submanifold is smooth and it is true for all manifolds, not only submanifolds of $mathbb{R}^n$. Is that clearer?
$endgroup$
– C. Falcon
Dec 25 '18 at 0:24
$begingroup$
Thank you! it is clear to me now.
$endgroup$
– PerelMan
Dec 25 '18 at 0:27
$begingroup$
Thank you! it is clear to me now.
$endgroup$
– PerelMan
Dec 25 '18 at 0:27
add a comment |
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