What is the difference between a chart and a tangent space?












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To my lay-person mind, a chart is a one-to-one function that maps an area on a manifold to a euclidean space of equal dimension.



Then I understand a tangent space to be the space of vectors that are parallel to the manifold at a specific point.



So does one use the appropriate charts to map points to the tangent space?



These seem like similar things to me (even though I am fairly certain they are not). Is the difference that any point on the manifold can be mapped to that specific tangent space whereas the chart only covers a subset of points in the manifold?



Or do I have two completely different ideas conflated?



Is the mapping of points from the manifold even related to the manifolds charts?






differential-geometry manifolds







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    To my lay-person mind, a chart is a one-to-one function that maps an area on a manifold to a euclidean space of equal dimension.



    Then I understand a tangent space to be the space of vectors that are parallel to the manifold at a specific point.



    So does one use the appropriate charts to map points to the tangent space?



    These seem like similar things to me (even though I am fairly certain they are not). Is the difference that any point on the manifold can be mapped to that specific tangent space whereas the chart only covers a subset of points in the manifold?



    Or do I have two completely different ideas conflated?



    Is the mapping of points from the manifold even related to the manifolds charts?






    differential-geometry manifolds







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      To my lay-person mind, a chart is a one-to-one function that maps an area on a manifold to a euclidean space of equal dimension.



      Then I understand a tangent space to be the space of vectors that are parallel to the manifold at a specific point.



      So does one use the appropriate charts to map points to the tangent space?



      These seem like similar things to me (even though I am fairly certain they are not). Is the difference that any point on the manifold can be mapped to that specific tangent space whereas the chart only covers a subset of points in the manifold?



      Or do I have two completely different ideas conflated?



      Is the mapping of points from the manifold even related to the manifolds charts?






      differential-geometry manifolds







      share|cite|improve this question









      $endgroup$




      To my lay-person mind, a chart is a one-to-one function that maps an area on a manifold to a euclidean space of equal dimension.



      Then I understand a tangent space to be the space of vectors that are parallel to the manifold at a specific point.



      So does one use the appropriate charts to map points to the tangent space?



      These seem like similar things to me (even though I am fairly certain they are not). Is the difference that any point on the manifold can be mapped to that specific tangent space whereas the chart only covers a subset of points in the manifold?



      Or do I have two completely different ideas conflated?



      Is the mapping of points from the manifold even related to the manifolds charts?






      differential-geometry manifolds




      differential-geometry manifolds






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 11 '18 at 10:36









      bidbybidby

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          $begingroup$

          Each point on a differentiable manifold has its own tangent space. Informally, you can think of the tangent space at a point as the space of tangent vectors to the surface at that point. But this informal picture only works if the manifold is embedded in some higher dimensional space.



          A more abstract and more formal point of view is that tangent space at a point $P$ of a differentiable manifold $M$ consists of the derivatives of all curves on $M$ that pass through $P$. The derivatives exist because each map between $M$ and a chart is differentiable and invertible. So we can map a curve on $M$ to a chart, carry out calculus in the chart and "lift" the result back to $M$. The result is independent of the choice of chart, so it is a property of $M$ itself (at $P$).



          It is not entirely obvious that we can create a vector space from the derivatives of the curves through $P$, but we can because the derivative is a linear operator. This vector space is then the tangent space at $P$, which we denote by $T_P$. The set of tangent spaces $T_P$ for all points $P$ on $M$ forms a vector bundle over $M$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            What is the difference between mapping the curve on M to a chart as you said, and mapping it to a tangent space at the same point?
            $endgroup$
            – bidby
            Dec 11 '18 at 11:56










          • $begingroup$
            Am I correct in my understanding if I say that one maps all the curves through a point P to Euclidean space using any chart that includes P. Once I am in this space I can differentiate all of these curves and I get a bunch of vectors, which form the tangent space/plane at P once I have mapped back to M using the inverse of the chart I originally used. While these tangent vectors may have looked different in the Euclidean space, depending on my choice of chart, the resulting tangent space/plane at P in M is independent of which chart I chose?
            $endgroup$
            – bidby
            Dec 11 '18 at 13:38






          • 1




            $begingroup$
            @bidby Yes, your second comment is correct. Going back to your first comment, each point in the tangent space $T_P$ represents the derivative at $P$ of a set of curves on $M$ (many curves may have the same derivative at $P$). A point in $T_P$ does not represent some other point on $M$. So you cannot map points on $M$ to $T_P$ - apart from $P$ itself, which you can say maps to the origin in $T_P$ (because the constant "curve" $f(t)=P space forall t$ has derivative $0$).
            $endgroup$
            – gandalf61
            Dec 11 '18 at 13:46










          • $begingroup$
            Ah ok! So is the exponential mapping of point A at point B to the tangent space $T_b$ really just the vector in the tangent space that is parallel with the geodesic between A and B?
            $endgroup$
            – bidby
            Dec 11 '18 at 14:05










          • $begingroup$
            @bidby Yes, but you have to add extra structure such as a Riemannian metric to the manifold in order to define geodesics and exponential maps.
            $endgroup$
            – gandalf61
            Dec 11 '18 at 14:20











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          1 Answer
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          $begingroup$

          Each point on a differentiable manifold has its own tangent space. Informally, you can think of the tangent space at a point as the space of tangent vectors to the surface at that point. But this informal picture only works if the manifold is embedded in some higher dimensional space.



          A more abstract and more formal point of view is that tangent space at a point $P$ of a differentiable manifold $M$ consists of the derivatives of all curves on $M$ that pass through $P$. The derivatives exist because each map between $M$ and a chart is differentiable and invertible. So we can map a curve on $M$ to a chart, carry out calculus in the chart and "lift" the result back to $M$. The result is independent of the choice of chart, so it is a property of $M$ itself (at $P$).



          It is not entirely obvious that we can create a vector space from the derivatives of the curves through $P$, but we can because the derivative is a linear operator. This vector space is then the tangent space at $P$, which we denote by $T_P$. The set of tangent spaces $T_P$ for all points $P$ on $M$ forms a vector bundle over $M$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            What is the difference between mapping the curve on M to a chart as you said, and mapping it to a tangent space at the same point?
            $endgroup$
            – bidby
            Dec 11 '18 at 11:56










          • $begingroup$
            Am I correct in my understanding if I say that one maps all the curves through a point P to Euclidean space using any chart that includes P. Once I am in this space I can differentiate all of these curves and I get a bunch of vectors, which form the tangent space/plane at P once I have mapped back to M using the inverse of the chart I originally used. While these tangent vectors may have looked different in the Euclidean space, depending on my choice of chart, the resulting tangent space/plane at P in M is independent of which chart I chose?
            $endgroup$
            – bidby
            Dec 11 '18 at 13:38






          • 1




            $begingroup$
            @bidby Yes, your second comment is correct. Going back to your first comment, each point in the tangent space $T_P$ represents the derivative at $P$ of a set of curves on $M$ (many curves may have the same derivative at $P$). A point in $T_P$ does not represent some other point on $M$. So you cannot map points on $M$ to $T_P$ - apart from $P$ itself, which you can say maps to the origin in $T_P$ (because the constant "curve" $f(t)=P space forall t$ has derivative $0$).
            $endgroup$
            – gandalf61
            Dec 11 '18 at 13:46










          • $begingroup$
            Ah ok! So is the exponential mapping of point A at point B to the tangent space $T_b$ really just the vector in the tangent space that is parallel with the geodesic between A and B?
            $endgroup$
            – bidby
            Dec 11 '18 at 14:05










          • $begingroup$
            @bidby Yes, but you have to add extra structure such as a Riemannian metric to the manifold in order to define geodesics and exponential maps.
            $endgroup$
            – gandalf61
            Dec 11 '18 at 14:20
















          1












          $begingroup$

          Each point on a differentiable manifold has its own tangent space. Informally, you can think of the tangent space at a point as the space of tangent vectors to the surface at that point. But this informal picture only works if the manifold is embedded in some higher dimensional space.



          A more abstract and more formal point of view is that tangent space at a point $P$ of a differentiable manifold $M$ consists of the derivatives of all curves on $M$ that pass through $P$. The derivatives exist because each map between $M$ and a chart is differentiable and invertible. So we can map a curve on $M$ to a chart, carry out calculus in the chart and "lift" the result back to $M$. The result is independent of the choice of chart, so it is a property of $M$ itself (at $P$).



          It is not entirely obvious that we can create a vector space from the derivatives of the curves through $P$, but we can because the derivative is a linear operator. This vector space is then the tangent space at $P$, which we denote by $T_P$. The set of tangent spaces $T_P$ for all points $P$ on $M$ forms a vector bundle over $M$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            What is the difference between mapping the curve on M to a chart as you said, and mapping it to a tangent space at the same point?
            $endgroup$
            – bidby
            Dec 11 '18 at 11:56










          • $begingroup$
            Am I correct in my understanding if I say that one maps all the curves through a point P to Euclidean space using any chart that includes P. Once I am in this space I can differentiate all of these curves and I get a bunch of vectors, which form the tangent space/plane at P once I have mapped back to M using the inverse of the chart I originally used. While these tangent vectors may have looked different in the Euclidean space, depending on my choice of chart, the resulting tangent space/plane at P in M is independent of which chart I chose?
            $endgroup$
            – bidby
            Dec 11 '18 at 13:38






          • 1




            $begingroup$
            @bidby Yes, your second comment is correct. Going back to your first comment, each point in the tangent space $T_P$ represents the derivative at $P$ of a set of curves on $M$ (many curves may have the same derivative at $P$). A point in $T_P$ does not represent some other point on $M$. So you cannot map points on $M$ to $T_P$ - apart from $P$ itself, which you can say maps to the origin in $T_P$ (because the constant "curve" $f(t)=P space forall t$ has derivative $0$).
            $endgroup$
            – gandalf61
            Dec 11 '18 at 13:46










          • $begingroup$
            Ah ok! So is the exponential mapping of point A at point B to the tangent space $T_b$ really just the vector in the tangent space that is parallel with the geodesic between A and B?
            $endgroup$
            – bidby
            Dec 11 '18 at 14:05










          • $begingroup$
            @bidby Yes, but you have to add extra structure such as a Riemannian metric to the manifold in order to define geodesics and exponential maps.
            $endgroup$
            – gandalf61
            Dec 11 '18 at 14:20














          1












          1








          1





          $begingroup$

          Each point on a differentiable manifold has its own tangent space. Informally, you can think of the tangent space at a point as the space of tangent vectors to the surface at that point. But this informal picture only works if the manifold is embedded in some higher dimensional space.



          A more abstract and more formal point of view is that tangent space at a point $P$ of a differentiable manifold $M$ consists of the derivatives of all curves on $M$ that pass through $P$. The derivatives exist because each map between $M$ and a chart is differentiable and invertible. So we can map a curve on $M$ to a chart, carry out calculus in the chart and "lift" the result back to $M$. The result is independent of the choice of chart, so it is a property of $M$ itself (at $P$).



          It is not entirely obvious that we can create a vector space from the derivatives of the curves through $P$, but we can because the derivative is a linear operator. This vector space is then the tangent space at $P$, which we denote by $T_P$. The set of tangent spaces $T_P$ for all points $P$ on $M$ forms a vector bundle over $M$.






          share|cite|improve this answer









          $endgroup$



          Each point on a differentiable manifold has its own tangent space. Informally, you can think of the tangent space at a point as the space of tangent vectors to the surface at that point. But this informal picture only works if the manifold is embedded in some higher dimensional space.



          A more abstract and more formal point of view is that tangent space at a point $P$ of a differentiable manifold $M$ consists of the derivatives of all curves on $M$ that pass through $P$. The derivatives exist because each map between $M$ and a chart is differentiable and invertible. So we can map a curve on $M$ to a chart, carry out calculus in the chart and "lift" the result back to $M$. The result is independent of the choice of chart, so it is a property of $M$ itself (at $P$).



          It is not entirely obvious that we can create a vector space from the derivatives of the curves through $P$, but we can because the derivative is a linear operator. This vector space is then the tangent space at $P$, which we denote by $T_P$. The set of tangent spaces $T_P$ for all points $P$ on $M$ forms a vector bundle over $M$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 11 '18 at 11:50









          gandalf61gandalf61

          8,801725




          8,801725












          • $begingroup$
            What is the difference between mapping the curve on M to a chart as you said, and mapping it to a tangent space at the same point?
            $endgroup$
            – bidby
            Dec 11 '18 at 11:56










          • $begingroup$
            Am I correct in my understanding if I say that one maps all the curves through a point P to Euclidean space using any chart that includes P. Once I am in this space I can differentiate all of these curves and I get a bunch of vectors, which form the tangent space/plane at P once I have mapped back to M using the inverse of the chart I originally used. While these tangent vectors may have looked different in the Euclidean space, depending on my choice of chart, the resulting tangent space/plane at P in M is independent of which chart I chose?
            $endgroup$
            – bidby
            Dec 11 '18 at 13:38






          • 1




            $begingroup$
            @bidby Yes, your second comment is correct. Going back to your first comment, each point in the tangent space $T_P$ represents the derivative at $P$ of a set of curves on $M$ (many curves may have the same derivative at $P$). A point in $T_P$ does not represent some other point on $M$. So you cannot map points on $M$ to $T_P$ - apart from $P$ itself, which you can say maps to the origin in $T_P$ (because the constant "curve" $f(t)=P space forall t$ has derivative $0$).
            $endgroup$
            – gandalf61
            Dec 11 '18 at 13:46










          • $begingroup$
            Ah ok! So is the exponential mapping of point A at point B to the tangent space $T_b$ really just the vector in the tangent space that is parallel with the geodesic between A and B?
            $endgroup$
            – bidby
            Dec 11 '18 at 14:05










          • $begingroup$
            @bidby Yes, but you have to add extra structure such as a Riemannian metric to the manifold in order to define geodesics and exponential maps.
            $endgroup$
            – gandalf61
            Dec 11 '18 at 14:20


















          • $begingroup$
            What is the difference between mapping the curve on M to a chart as you said, and mapping it to a tangent space at the same point?
            $endgroup$
            – bidby
            Dec 11 '18 at 11:56










          • $begingroup$
            Am I correct in my understanding if I say that one maps all the curves through a point P to Euclidean space using any chart that includes P. Once I am in this space I can differentiate all of these curves and I get a bunch of vectors, which form the tangent space/plane at P once I have mapped back to M using the inverse of the chart I originally used. While these tangent vectors may have looked different in the Euclidean space, depending on my choice of chart, the resulting tangent space/plane at P in M is independent of which chart I chose?
            $endgroup$
            – bidby
            Dec 11 '18 at 13:38






          • 1




            $begingroup$
            @bidby Yes, your second comment is correct. Going back to your first comment, each point in the tangent space $T_P$ represents the derivative at $P$ of a set of curves on $M$ (many curves may have the same derivative at $P$). A point in $T_P$ does not represent some other point on $M$. So you cannot map points on $M$ to $T_P$ - apart from $P$ itself, which you can say maps to the origin in $T_P$ (because the constant "curve" $f(t)=P space forall t$ has derivative $0$).
            $endgroup$
            – gandalf61
            Dec 11 '18 at 13:46










          • $begingroup$
            Ah ok! So is the exponential mapping of point A at point B to the tangent space $T_b$ really just the vector in the tangent space that is parallel with the geodesic between A and B?
            $endgroup$
            – bidby
            Dec 11 '18 at 14:05










          • $begingroup$
            @bidby Yes, but you have to add extra structure such as a Riemannian metric to the manifold in order to define geodesics and exponential maps.
            $endgroup$
            – gandalf61
            Dec 11 '18 at 14:20
















          $begingroup$
          What is the difference between mapping the curve on M to a chart as you said, and mapping it to a tangent space at the same point?
          $endgroup$
          – bidby
          Dec 11 '18 at 11:56




          $begingroup$
          What is the difference between mapping the curve on M to a chart as you said, and mapping it to a tangent space at the same point?
          $endgroup$
          – bidby
          Dec 11 '18 at 11:56












          $begingroup$
          Am I correct in my understanding if I say that one maps all the curves through a point P to Euclidean space using any chart that includes P. Once I am in this space I can differentiate all of these curves and I get a bunch of vectors, which form the tangent space/plane at P once I have mapped back to M using the inverse of the chart I originally used. While these tangent vectors may have looked different in the Euclidean space, depending on my choice of chart, the resulting tangent space/plane at P in M is independent of which chart I chose?
          $endgroup$
          – bidby
          Dec 11 '18 at 13:38




          $begingroup$
          Am I correct in my understanding if I say that one maps all the curves through a point P to Euclidean space using any chart that includes P. Once I am in this space I can differentiate all of these curves and I get a bunch of vectors, which form the tangent space/plane at P once I have mapped back to M using the inverse of the chart I originally used. While these tangent vectors may have looked different in the Euclidean space, depending on my choice of chart, the resulting tangent space/plane at P in M is independent of which chart I chose?
          $endgroup$
          – bidby
          Dec 11 '18 at 13:38




          1




          1




          $begingroup$
          @bidby Yes, your second comment is correct. Going back to your first comment, each point in the tangent space $T_P$ represents the derivative at $P$ of a set of curves on $M$ (many curves may have the same derivative at $P$). A point in $T_P$ does not represent some other point on $M$. So you cannot map points on $M$ to $T_P$ - apart from $P$ itself, which you can say maps to the origin in $T_P$ (because the constant "curve" $f(t)=P space forall t$ has derivative $0$).
          $endgroup$
          – gandalf61
          Dec 11 '18 at 13:46




          $begingroup$
          @bidby Yes, your second comment is correct. Going back to your first comment, each point in the tangent space $T_P$ represents the derivative at $P$ of a set of curves on $M$ (many curves may have the same derivative at $P$). A point in $T_P$ does not represent some other point on $M$. So you cannot map points on $M$ to $T_P$ - apart from $P$ itself, which you can say maps to the origin in $T_P$ (because the constant "curve" $f(t)=P space forall t$ has derivative $0$).
          $endgroup$
          – gandalf61
          Dec 11 '18 at 13:46












          $begingroup$
          Ah ok! So is the exponential mapping of point A at point B to the tangent space $T_b$ really just the vector in the tangent space that is parallel with the geodesic between A and B?
          $endgroup$
          – bidby
          Dec 11 '18 at 14:05




          $begingroup$
          Ah ok! So is the exponential mapping of point A at point B to the tangent space $T_b$ really just the vector in the tangent space that is parallel with the geodesic between A and B?
          $endgroup$
          – bidby
          Dec 11 '18 at 14:05












          $begingroup$
          @bidby Yes, but you have to add extra structure such as a Riemannian metric to the manifold in order to define geodesics and exponential maps.
          $endgroup$
          – gandalf61
          Dec 11 '18 at 14:20




          $begingroup$
          @bidby Yes, but you have to add extra structure such as a Riemannian metric to the manifold in order to define geodesics and exponential maps.
          $endgroup$
          – gandalf61
          Dec 11 '18 at 14:20


















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          Musa

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          Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
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