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Isn't this a trivial corollary?


Calculus on surfaces and chain ruleConstructing submanifolds. Did I understand this right?Some trivial questions about Tangent SpacesHow to see this is a normal vector field?Finding tangent plane to $2$ dimensional submanifold of $mathbbR^4$Generic condition for vector fields/normal sectionsHow to recover the tangent space from the metricInequality for gradients under different metricsShow that if $f$ is a smooth function, $M$ is a manifold and $x$ is a local extremum of $f$ on $M$, then $D_f(x)(v) = 0$ in the tangent space.Why $dim(ker T_z f)=dim(T_z(f^-1(c)))$?













3












$begingroup$


Let $U subseteq mathbb R^n$ be an open subset and let $M subseteq U$ be a $k$-dimensional submanifold of $mathbb R^n$. Consider a differentiable function $f: U longrightarrow mathbb R$. There's a corollary that states, that if $f big|_M$ takes on a local extremum at a point $p in M$, then the gradient $nabla f(p)$ is normal to $M$ at $p$, i.e. $nabla f(p) perp T_pM$, where $T_pM$ is the tangent space. I've studied and understood the (short) proof, but isn't this statement absolutely trivial or have I got things mixed up? Let me explain: Assuming $p$ is an extremum, it follows that the differential vanishes, i.e. $nabla f(p) = 0$, since this is a necessary condition. But the zero vector is orthogonal to everything. So what's the point of this corollary?










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  • 2




    $begingroup$
    Please make the title more informative.
    $endgroup$
    – YuiTo Cheng
    Jun 22 at 4:07















3












$begingroup$


Let $U subseteq mathbb R^n$ be an open subset and let $M subseteq U$ be a $k$-dimensional submanifold of $mathbb R^n$. Consider a differentiable function $f: U longrightarrow mathbb R$. There's a corollary that states, that if $f big|_M$ takes on a local extremum at a point $p in M$, then the gradient $nabla f(p)$ is normal to $M$ at $p$, i.e. $nabla f(p) perp T_pM$, where $T_pM$ is the tangent space. I've studied and understood the (short) proof, but isn't this statement absolutely trivial or have I got things mixed up? Let me explain: Assuming $p$ is an extremum, it follows that the differential vanishes, i.e. $nabla f(p) = 0$, since this is a necessary condition. But the zero vector is orthogonal to everything. So what's the point of this corollary?










share|cite|improve this question









$endgroup$







  • 2




    $begingroup$
    Please make the title more informative.
    $endgroup$
    – YuiTo Cheng
    Jun 22 at 4:07













3












3








3





$begingroup$


Let $U subseteq mathbb R^n$ be an open subset and let $M subseteq U$ be a $k$-dimensional submanifold of $mathbb R^n$. Consider a differentiable function $f: U longrightarrow mathbb R$. There's a corollary that states, that if $f big|_M$ takes on a local extremum at a point $p in M$, then the gradient $nabla f(p)$ is normal to $M$ at $p$, i.e. $nabla f(p) perp T_pM$, where $T_pM$ is the tangent space. I've studied and understood the (short) proof, but isn't this statement absolutely trivial or have I got things mixed up? Let me explain: Assuming $p$ is an extremum, it follows that the differential vanishes, i.e. $nabla f(p) = 0$, since this is a necessary condition. But the zero vector is orthogonal to everything. So what's the point of this corollary?










share|cite|improve this question









$endgroup$




Let $U subseteq mathbb R^n$ be an open subset and let $M subseteq U$ be a $k$-dimensional submanifold of $mathbb R^n$. Consider a differentiable function $f: U longrightarrow mathbb R$. There's a corollary that states, that if $f big|_M$ takes on a local extremum at a point $p in M$, then the gradient $nabla f(p)$ is normal to $M$ at $p$, i.e. $nabla f(p) perp T_pM$, where $T_pM$ is the tangent space. I've studied and understood the (short) proof, but isn't this statement absolutely trivial or have I got things mixed up? Let me explain: Assuming $p$ is an extremum, it follows that the differential vanishes, i.e. $nabla f(p) = 0$, since this is a necessary condition. But the zero vector is orthogonal to everything. So what's the point of this corollary?







derivatives differential-geometry orthogonality submanifold tangent-spaces






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asked Jun 21 at 17:32









playdisplaydis

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444 bronze badges







  • 2




    $begingroup$
    Please make the title more informative.
    $endgroup$
    – YuiTo Cheng
    Jun 22 at 4:07












  • 2




    $begingroup$
    Please make the title more informative.
    $endgroup$
    – YuiTo Cheng
    Jun 22 at 4:07







2




2




$begingroup$
Please make the title more informative.
$endgroup$
– YuiTo Cheng
Jun 22 at 4:07




$begingroup$
Please make the title more informative.
$endgroup$
– YuiTo Cheng
Jun 22 at 4:07










2 Answers
2






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5












$begingroup$

Take$$M=(x,y)inmathbb R^2mid x^2+y^2=1$$and let $f(x,y)=x^2+y^2$. Then $f|_M$ is constant, and therefore it has an extreme point at every point of $M$. However, $nabla f$ is never equal to $0$, at any point of $M$ (although, as the corollory states, it is orthogonal to $M$ at each point).



The gradient has to be $0$ indeed, if the point at which we are computing it is both an extreme point of $f$ and an interior point of $M$ (if you see $f$ as a subset of $mathbb R^n$), but points of submanifolds usually are not interior points.






share|cite|improve this answer









$endgroup$




















    4












    $begingroup$

    $p$ is not necessarily a local extremum on $U$, it's a local extremum on $M$.



    Consider for instance $f$ defined on $mathbbR^2$ by the square of the euclidean norm; then on the circle $S^1$, it reaches a local extremum at each point, but its gradient is not zero on $S^1$ : $nabla f (p) = 2p$ which is indeed orthogonal to $T_pS^1= p^bot$, but is not $0$






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      2 Answers
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      active

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      2 Answers
      2






      active

      oldest

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      active

      oldest

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      active

      oldest

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      5












      $begingroup$

      Take$$M=(x,y)inmathbb R^2mid x^2+y^2=1$$and let $f(x,y)=x^2+y^2$. Then $f|_M$ is constant, and therefore it has an extreme point at every point of $M$. However, $nabla f$ is never equal to $0$, at any point of $M$ (although, as the corollory states, it is orthogonal to $M$ at each point).



      The gradient has to be $0$ indeed, if the point at which we are computing it is both an extreme point of $f$ and an interior point of $M$ (if you see $f$ as a subset of $mathbb R^n$), but points of submanifolds usually are not interior points.






      share|cite|improve this answer









      $endgroup$

















        5












        $begingroup$

        Take$$M=(x,y)inmathbb R^2mid x^2+y^2=1$$and let $f(x,y)=x^2+y^2$. Then $f|_M$ is constant, and therefore it has an extreme point at every point of $M$. However, $nabla f$ is never equal to $0$, at any point of $M$ (although, as the corollory states, it is orthogonal to $M$ at each point).



        The gradient has to be $0$ indeed, if the point at which we are computing it is both an extreme point of $f$ and an interior point of $M$ (if you see $f$ as a subset of $mathbb R^n$), but points of submanifolds usually are not interior points.






        share|cite|improve this answer









        $endgroup$















          5












          5








          5





          $begingroup$

          Take$$M=(x,y)inmathbb R^2mid x^2+y^2=1$$and let $f(x,y)=x^2+y^2$. Then $f|_M$ is constant, and therefore it has an extreme point at every point of $M$. However, $nabla f$ is never equal to $0$, at any point of $M$ (although, as the corollory states, it is orthogonal to $M$ at each point).



          The gradient has to be $0$ indeed, if the point at which we are computing it is both an extreme point of $f$ and an interior point of $M$ (if you see $f$ as a subset of $mathbb R^n$), but points of submanifolds usually are not interior points.






          share|cite|improve this answer









          $endgroup$



          Take$$M=(x,y)inmathbb R^2mid x^2+y^2=1$$and let $f(x,y)=x^2+y^2$. Then $f|_M$ is constant, and therefore it has an extreme point at every point of $M$. However, $nabla f$ is never equal to $0$, at any point of $M$ (although, as the corollory states, it is orthogonal to $M$ at each point).



          The gradient has to be $0$ indeed, if the point at which we are computing it is both an extreme point of $f$ and an interior point of $M$ (if you see $f$ as a subset of $mathbb R^n$), but points of submanifolds usually are not interior points.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jun 21 at 17:36









          José Carlos SantosJosé Carlos Santos

          196k24 gold badges153 silver badges272 bronze badges




          196k24 gold badges153 silver badges272 bronze badges





















              4












              $begingroup$

              $p$ is not necessarily a local extremum on $U$, it's a local extremum on $M$.



              Consider for instance $f$ defined on $mathbbR^2$ by the square of the euclidean norm; then on the circle $S^1$, it reaches a local extremum at each point, but its gradient is not zero on $S^1$ : $nabla f (p) = 2p$ which is indeed orthogonal to $T_pS^1= p^bot$, but is not $0$






              share|cite|improve this answer









              $endgroup$

















                4












                $begingroup$

                $p$ is not necessarily a local extremum on $U$, it's a local extremum on $M$.



                Consider for instance $f$ defined on $mathbbR^2$ by the square of the euclidean norm; then on the circle $S^1$, it reaches a local extremum at each point, but its gradient is not zero on $S^1$ : $nabla f (p) = 2p$ which is indeed orthogonal to $T_pS^1= p^bot$, but is not $0$






                share|cite|improve this answer









                $endgroup$















                  4












                  4








                  4





                  $begingroup$

                  $p$ is not necessarily a local extremum on $U$, it's a local extremum on $M$.



                  Consider for instance $f$ defined on $mathbbR^2$ by the square of the euclidean norm; then on the circle $S^1$, it reaches a local extremum at each point, but its gradient is not zero on $S^1$ : $nabla f (p) = 2p$ which is indeed orthogonal to $T_pS^1= p^bot$, but is not $0$






                  share|cite|improve this answer









                  $endgroup$



                  $p$ is not necessarily a local extremum on $U$, it's a local extremum on $M$.



                  Consider for instance $f$ defined on $mathbbR^2$ by the square of the euclidean norm; then on the circle $S^1$, it reaches a local extremum at each point, but its gradient is not zero on $S^1$ : $nabla f (p) = 2p$ which is indeed orthogonal to $T_pS^1= p^bot$, but is not $0$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jun 21 at 17:37









                  MaxMax

                  19.9k1 gold badge12 silver badges46 bronze badges




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