critical points of a map












1












$begingroup$


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







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    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







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      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







      share|cite|improve this question









      $endgroup$




      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






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 24 '18 at 20:15









      PerelManPerelMan

      830414




      830414






















          1 Answer
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          1












          $begingroup$

          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:



          enter image description here






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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












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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:



          enter image description here






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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
















          1












          $begingroup$

          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:



          enter image description here






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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














          1












          1








          1





          $begingroup$

          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:



          enter image description here






          share|cite|improve this answer











          $endgroup$



          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:



          enter image description here







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 25 '18 at 17:03

























          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


















          • $begingroup$
            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


















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