If $f:S^1 to mathbb{R}$, defined by $f(x,y)=x-y$ and $X_p$ be a tangent vector at $p=(frac{1}{sqrt 2},...












1












$begingroup$


If $f:S^1 to mathbb{R}$, defined by $f(x,y)=x-y$ and $X_p$ be a tangent vector at $p=(frac{1}{sqrt 2}, frac{1}{sqrt 2})$ given by $X_p=4 (frac{partial}{partial t})_p$ where $t$ is the local coordinate on the chart $(U, phi)$, where $U=S^1-{(0,0)}$, and $phi(x,y)=frac{x}{1-y}$, then find $X_p(f)$.



Answer:



$X_p(f)=4(frac{partial}{partial t})_pf=4(frac{partial}{partial x})_p (x-y)+4(frac{partial}{partial y})_p(x-y)=4-4=0$



But where is the need of $ phi(x,y)=frac{x}{1-y}$ ?






differential-geometry







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$endgroup$












  • $begingroup$
    How did you replace $frac{partial}{partial t}$ with $frac{partial}{partial x} + frac{partial}{partial y}$? The point is that $t=phi(x,y)=frac x{1-y}$.
    $endgroup$
    – Ted Shifrin
    Dec 22 '18 at 23:52












  • $begingroup$
    @TedShifrin, so what would be the vector field $X_p=4(partial /partial t)$?
    $endgroup$
    – M. A. SARKAR
    Dec 23 '18 at 0:21
















1












$begingroup$


If $f:S^1 to mathbb{R}$, defined by $f(x,y)=x-y$ and $X_p$ be a tangent vector at $p=(frac{1}{sqrt 2}, frac{1}{sqrt 2})$ given by $X_p=4 (frac{partial}{partial t})_p$ where $t$ is the local coordinate on the chart $(U, phi)$, where $U=S^1-{(0,0)}$, and $phi(x,y)=frac{x}{1-y}$, then find $X_p(f)$.



Answer:



$X_p(f)=4(frac{partial}{partial t})_pf=4(frac{partial}{partial x})_p (x-y)+4(frac{partial}{partial y})_p(x-y)=4-4=0$



But where is the need of $ phi(x,y)=frac{x}{1-y}$ ?






differential-geometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    How did you replace $frac{partial}{partial t}$ with $frac{partial}{partial x} + frac{partial}{partial y}$? The point is that $t=phi(x,y)=frac x{1-y}$.
    $endgroup$
    – Ted Shifrin
    Dec 22 '18 at 23:52












  • $begingroup$
    @TedShifrin, so what would be the vector field $X_p=4(partial /partial t)$?
    $endgroup$
    – M. A. SARKAR
    Dec 23 '18 at 0:21














1












1








1


2



$begingroup$


If $f:S^1 to mathbb{R}$, defined by $f(x,y)=x-y$ and $X_p$ be a tangent vector at $p=(frac{1}{sqrt 2}, frac{1}{sqrt 2})$ given by $X_p=4 (frac{partial}{partial t})_p$ where $t$ is the local coordinate on the chart $(U, phi)$, where $U=S^1-{(0,0)}$, and $phi(x,y)=frac{x}{1-y}$, then find $X_p(f)$.



Answer:



$X_p(f)=4(frac{partial}{partial t})_pf=4(frac{partial}{partial x})_p (x-y)+4(frac{partial}{partial y})_p(x-y)=4-4=0$



But where is the need of $ phi(x,y)=frac{x}{1-y}$ ?






differential-geometry







share|cite|improve this question











$endgroup$




If $f:S^1 to mathbb{R}$, defined by $f(x,y)=x-y$ and $X_p$ be a tangent vector at $p=(frac{1}{sqrt 2}, frac{1}{sqrt 2})$ given by $X_p=4 (frac{partial}{partial t})_p$ where $t$ is the local coordinate on the chart $(U, phi)$, where $U=S^1-{(0,0)}$, and $phi(x,y)=frac{x}{1-y}$, then find $X_p(f)$.



Answer:



$X_p(f)=4(frac{partial}{partial t})_pf=4(frac{partial}{partial x})_p (x-y)+4(frac{partial}{partial y})_p(x-y)=4-4=0$



But where is the need of $ phi(x,y)=frac{x}{1-y}$ ?






differential-geometry




differential-geometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 22 '18 at 23:51









amWhy

1




1










asked Dec 22 '18 at 23:06









arifamatharifamath

1176




1176












  • $begingroup$
    How did you replace $frac{partial}{partial t}$ with $frac{partial}{partial x} + frac{partial}{partial y}$? The point is that $t=phi(x,y)=frac x{1-y}$.
    $endgroup$
    – Ted Shifrin
    Dec 22 '18 at 23:52












  • $begingroup$
    @TedShifrin, so what would be the vector field $X_p=4(partial /partial t)$?
    $endgroup$
    – M. A. SARKAR
    Dec 23 '18 at 0:21


















  • $begingroup$
    How did you replace $frac{partial}{partial t}$ with $frac{partial}{partial x} + frac{partial}{partial y}$? The point is that $t=phi(x,y)=frac x{1-y}$.
    $endgroup$
    – Ted Shifrin
    Dec 22 '18 at 23:52












  • $begingroup$
    @TedShifrin, so what would be the vector field $X_p=4(partial /partial t)$?
    $endgroup$
    – M. A. SARKAR
    Dec 23 '18 at 0:21
















$begingroup$
How did you replace $frac{partial}{partial t}$ with $frac{partial}{partial x} + frac{partial}{partial y}$? The point is that $t=phi(x,y)=frac x{1-y}$.
$endgroup$
– Ted Shifrin
Dec 22 '18 at 23:52






$begingroup$
How did you replace $frac{partial}{partial t}$ with $frac{partial}{partial x} + frac{partial}{partial y}$? The point is that $t=phi(x,y)=frac x{1-y}$.
$endgroup$
– Ted Shifrin
Dec 22 '18 at 23:52














$begingroup$
@TedShifrin, so what would be the vector field $X_p=4(partial /partial t)$?
$endgroup$
– M. A. SARKAR
Dec 23 '18 at 0:21




$begingroup$
@TedShifrin, so what would be the vector field $X_p=4(partial /partial t)$?
$endgroup$
– M. A. SARKAR
Dec 23 '18 at 0:21










2 Answers
2






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The easiest way to proceed here is to solve for $x(t)$ and $y(t)$ [Hint: Use $x^2+y^2=1$.] and then use
$$frac{partial}{partial t} = frac{partial x}{partial t}frac{partial}{partial x} + frac{partial y}{partial t}frac{partial}{partial y}.$$
In this case, at the point $(1/sqrt2,1/sqrt2)$, you should find that
$$frac{partial}{partial t} = frac{1-sqrt2}2,frac{partial}{partial x} + frac{sqrt2-1}2,frac{partial}{partial y}.$$
Note that this vector is a multiple of $-frac{partial}{partial x}+frac{partial}{partial y}$, as is appropriate at a point on the circle with $x=y$.



Even easier, once you know $x(t)$ and $y(t)$, then rewrite $f$ as $f(x(t),y(t))$ and take $partial/partial t$ !!






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    When you wrote $ f(x,y)=x-y $, you have already put $S^1$ into $R^2$ and used a chart of $ R^2 $. $(x,y)$ here is a point in $R^2$ chart. If you want to redefine a chart $ (U, phi) $, I think you have to rewrite $ f$ by $phi$.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

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      active

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      active

      oldest

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      2












      $begingroup$

      The easiest way to proceed here is to solve for $x(t)$ and $y(t)$ [Hint: Use $x^2+y^2=1$.] and then use
      $$frac{partial}{partial t} = frac{partial x}{partial t}frac{partial}{partial x} + frac{partial y}{partial t}frac{partial}{partial y}.$$
      In this case, at the point $(1/sqrt2,1/sqrt2)$, you should find that
      $$frac{partial}{partial t} = frac{1-sqrt2}2,frac{partial}{partial x} + frac{sqrt2-1}2,frac{partial}{partial y}.$$
      Note that this vector is a multiple of $-frac{partial}{partial x}+frac{partial}{partial y}$, as is appropriate at a point on the circle with $x=y$.



      Even easier, once you know $x(t)$ and $y(t)$, then rewrite $f$ as $f(x(t),y(t))$ and take $partial/partial t$ !!






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        The easiest way to proceed here is to solve for $x(t)$ and $y(t)$ [Hint: Use $x^2+y^2=1$.] and then use
        $$frac{partial}{partial t} = frac{partial x}{partial t}frac{partial}{partial x} + frac{partial y}{partial t}frac{partial}{partial y}.$$
        In this case, at the point $(1/sqrt2,1/sqrt2)$, you should find that
        $$frac{partial}{partial t} = frac{1-sqrt2}2,frac{partial}{partial x} + frac{sqrt2-1}2,frac{partial}{partial y}.$$
        Note that this vector is a multiple of $-frac{partial}{partial x}+frac{partial}{partial y}$, as is appropriate at a point on the circle with $x=y$.



        Even easier, once you know $x(t)$ and $y(t)$, then rewrite $f$ as $f(x(t),y(t))$ and take $partial/partial t$ !!






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          The easiest way to proceed here is to solve for $x(t)$ and $y(t)$ [Hint: Use $x^2+y^2=1$.] and then use
          $$frac{partial}{partial t} = frac{partial x}{partial t}frac{partial}{partial x} + frac{partial y}{partial t}frac{partial}{partial y}.$$
          In this case, at the point $(1/sqrt2,1/sqrt2)$, you should find that
          $$frac{partial}{partial t} = frac{1-sqrt2}2,frac{partial}{partial x} + frac{sqrt2-1}2,frac{partial}{partial y}.$$
          Note that this vector is a multiple of $-frac{partial}{partial x}+frac{partial}{partial y}$, as is appropriate at a point on the circle with $x=y$.



          Even easier, once you know $x(t)$ and $y(t)$, then rewrite $f$ as $f(x(t),y(t))$ and take $partial/partial t$ !!






          share|cite|improve this answer











          $endgroup$



          The easiest way to proceed here is to solve for $x(t)$ and $y(t)$ [Hint: Use $x^2+y^2=1$.] and then use
          $$frac{partial}{partial t} = frac{partial x}{partial t}frac{partial}{partial x} + frac{partial y}{partial t}frac{partial}{partial y}.$$
          In this case, at the point $(1/sqrt2,1/sqrt2)$, you should find that
          $$frac{partial}{partial t} = frac{1-sqrt2}2,frac{partial}{partial x} + frac{sqrt2-1}2,frac{partial}{partial y}.$$
          Note that this vector is a multiple of $-frac{partial}{partial x}+frac{partial}{partial y}$, as is appropriate at a point on the circle with $x=y$.



          Even easier, once you know $x(t)$ and $y(t)$, then rewrite $f$ as $f(x(t),y(t))$ and take $partial/partial t$ !!







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 23 '18 at 2:02

























          answered Dec 23 '18 at 1:56









          Ted ShifrinTed Shifrin

          63.9k44591




          63.9k44591























              0












              $begingroup$

              When you wrote $ f(x,y)=x-y $, you have already put $S^1$ into $R^2$ and used a chart of $ R^2 $. $(x,y)$ here is a point in $R^2$ chart. If you want to redefine a chart $ (U, phi) $, I think you have to rewrite $ f$ by $phi$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                When you wrote $ f(x,y)=x-y $, you have already put $S^1$ into $R^2$ and used a chart of $ R^2 $. $(x,y)$ here is a point in $R^2$ chart. If you want to redefine a chart $ (U, phi) $, I think you have to rewrite $ f$ by $phi$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  When you wrote $ f(x,y)=x-y $, you have already put $S^1$ into $R^2$ and used a chart of $ R^2 $. $(x,y)$ here is a point in $R^2$ chart. If you want to redefine a chart $ (U, phi) $, I think you have to rewrite $ f$ by $phi$.






                  share|cite|improve this answer









                  $endgroup$



                  When you wrote $ f(x,y)=x-y $, you have already put $S^1$ into $R^2$ and used a chart of $ R^2 $. $(x,y)$ here is a point in $R^2$ chart. If you want to redefine a chart $ (U, phi) $, I think you have to rewrite $ f$ by $phi$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 23 '18 at 0:18









                  GaoGao

                  514




                  514






























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