I am working on a question regarding a smooth section, which seems quite intuitive in a Euclidian space.

Let $M$ be a smooth manifold an $c:(0,1) \to M$ a smooth curve. Show, that $\dot c:(0,1) \to TM, \, t \mapsto dc_t(1)$ is a smooth section along (?) c, which means, that $\pi \circ \dot c = c$.

  • $\pi \circ \dot c = c$ ist obviously true, as $\dot c$ maps $t$ to a vector $(c(t), dc_t(1))$ in $TM$
  • Now the smoothness, which is the difficult part for me. In general: Is there a suitable way for daily life to calculate the differential? I have come across different definitions (with equivalence classes of curves, the algebraic definition, as the "Jacobian matrix" of the charts), but nevertheless I still don't really grasp the meaning and the connection to the Euclidian case (despite sending tangential vectors of one tangent space to another), particularily when it comes to questions like the one above. So what I have got so far is, that $\dot c$ is smooth, when $\tilde \varphi \circ \dot c$ is smooth as a function from $(0,1) \to \mathbb{R}^{2m}$ where $m$ is the dimension of $M$ and $\tilde \varphi$ the to $\varphi$ corresponding chart of $TM$. $$(\tilde \varphi \circ \dot c)(t) = (\varphi(c(t)), \xi)$$ with $\xi$ given by the representation $dc_t(1) = \xi^i \frac{\partial}{\partial x^I}\big|_t$ What is $\xi$ exactly? $\phi(c(t))$ is smooth, as $c(t)$ is smooth and I guess, that $\xi$ is smooth, because it is given by the partial derivatives of $c(t)$, but the partial derivatives of $c$. Is that correct? I there a general approach to questions like this with the differential

Thanks in advance for any hints and advice


I know this feeling pretty well, it's really confusing, until it starts to make sense! I am not an expert, but maybe my student-like point of view can help you a bit.

Notation: I use $T$ to denote the tangential (or differential) of a map, i.e. for $f: M \to N$, we get the tangential as $Tf: TM \to TN$. The restriction to a point $x$ is then given by $T_x f: T_xM \to T_{f(x)}N$. For tangent spaces, I use the directional derivatives as the definition. The Jacobian of a real valued function at point $x$ is denoted by $Df(x)$.

About the smoothness:

Let me define $U = (0,1)$ to avoid some confusion. My approach would be to write $\dot{c}$ as a composition of smooth maps $U \rightarrow TU \rightarrow TM$, where the first map is given by $t \mapsto (t,1)$ and the second map is given by the tangential of $c$.

Smoothness of the first map: The first map is smooth. To check this, we compute the chart representation. If we take $\mathrm{id}_U: U \to (0,1)$ as a chart, then the chart for $TU$ is given by $T \mathrm{id}_U: TU \to (0,1) \times \mathbb{R}$. With resprect to these charts the map is then given by $$ (0,1) \to (0,1) \times \mathbb{R} : t \mapsto (t,1),$$ which is a smooth map.

Smoothness of the second map: If $c : U \rightarrow M$ is smooth, then also it tangential $Tc : TU \rightarrow TM$ is a smooth map between the two manifolds $TU$ and $TM$. (Is this already known at the point of this excercise?)

What is $\xi$ exactly?

Let us fix $t$, then $\dot{c}(t) =: (c(t),v(t)) \in T_{c(t)}M$ is the tangent vector we are interested in.

To get some coordinates of $v(t)$, we first need a basis for the vector space $T_{c(t)}M$. Like in your question, we take a chart around $c(t)$, say $$\varphi: U \to \mathbb{R}^n: p \mapsto (x^1(p), \dots, x^n(p)),$$ where $U$ is an open set containing $c(t)$.

Now each individual component $x^i$ of the chart is related to a tangent vector, denoted by $\frac{\partial}{\partial x^i}\big|_p$. Its defining property is being the corresponding directional derivative after applying the chart:

$$\frac{\partial}{\partial x^i}\big|_p(f) = (\partial_i (f \circ \varphi^{-1}) ) ( \varphi(p) ),\quad \text{for all } f\in C^{\infty}(M).$$

In this way the chart $\tilde{\varphi}$ is constructed. An explicit formula for the coefficients is then given by

$$\xi^i(t) = D( x^i \circ c ) ( t ) = \frac{\partial (x^i \circ c)}{\partial t}\big|_t.$$

Hence you are right with the intuition, that the coefficients are smooth in $t$, as being partial derivatives $$\dot{c}(t) = \sum_i \frac{\partial (x^i \circ c)}{\partial t}\big|_t \frac{\partial}{\partial x^i} \big|_{c(t)}$$

The more general situation is summarized in this diagram, where $h$ and $k$ are some charts and $\eta$, $\xi$ are charts for the associated tangent spaces. (The domains of these maps may need to be restricted, but I don't want to make it more complicated. So I assume that $h,k$ are global charts and I don't restrict the map $k \circ f \circ h^{-1}$ onto its real domain.) $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{lll} TM & \ra{Tf} & TN \\ \da{h \times \eta} & & \da{k \times \xi}\\ \mathbb{R}^m \times \mathbb{R}^m & \ra{k \circ f \circ h^{-1} \times D(k \circ f \circ h^{-1})} & \mathbb{R}^n \times \mathbb{R}^n \\ (x,\eta) & \mapsto ((k \circ f \circ h^{-1}) (x), D(k \circ f \circ h^{-1})(x)[\eta]) & = (y,\xi) \end{array} $$

For $x = h(p)$ and $v = \sum_i \eta^i(v) \frac{\partial}{\partial x^i} \big|_p \in T_p M$ we get the formula for the coefficients of

$$T f( v ) = \sum_j \xi^j( Tf(v) ) \frac{\partial}{\partial y^j} \big|_{f(x)},$$ given by

$$ \xi( Tf(v) ) = D(k \circ f \circ h^{-1})(x)[\eta(v)] = \left( \sum_{j=1}^m \frac{\partial (k^i \circ f \circ h^{-1})}{\partial x^j}\big|_{x} \cdot \eta^j(v) \right)_{i=1,\dots,n} \in \mathbb{R}^n$$

Like you see, this is not really an equation you want to use often, but on the other hand it is just the normal derivative and some chain rules. In my experience, you can usually take advantage of the special kind of manifold you are working on. Since $T_x f$ is a linear map, you 'just' need to find out how it acts on a basis of $T_x M$. For example if you know how it acts on the vectors of a special charts, say polar coordinates, then you can find the coefficients of $T_x f$ directly, without the need to compute the derivative.

Anyway, quite often the questions are of general nature and the numeric values are of less interest. And if you are lucky, you maybe can use some theorems and the calculus related to differential forms, Lie derivatives, Riemannian metrics, Stokes etc... I guess that is the point where all these abstract notations really show off their power.

Is there a general approach?

This question is beyond my knowledge. Quite often you can use diagrams to proof certain statements in a nice way. For example the tangential $T$ acts like a functor on the category of smooth manifolds and there exists many 'canonical' chart independent constructions and objects, which can be used.

Proving smoothness is often no really needed nor interesting, since almost all objects are smooth by construction and you just combine them. But in the beginning you spend a lot of time proving all these basics, which will later be used without any notice.

  • $\begingroup$ Thanks a lot for the effort. It really helps. Do you have an example for polar coordinates as you mentioned above, because I haven't seen any example at all yet. $\endgroup$ – Jonas W. May 27 '18 at 17:49
  • $\begingroup$ A constructed example: $c:U=(0,1)\to \mathbb{R}^2: \lambda \mapsto (\sin(\lambda),\cos(\lambda))^T$, then $T_{\lambda} c\big( v \frac{\partial}{\partial t}\big|_\lambda \big) = v \cdot \cos(\lambda) \frac{\partial}{\partial x^1}\big|_{c(\lambda)} - v \cdot \sin(\lambda) \frac{\partial}{\partial x^2}\big|_{c(\lambda)}.$ But in polar coordinates $\varphi \circ c~(\lambda) = (0,\lambda)$ (with $\varphi$ denoting the chart), hence $T_{\lambda} c\big( v \frac{\partial}{\partial t}\big|_\lambda \big) = v \frac{\partial}{\partial \phi}\big|_{c(\lambda)}$. $\endgroup$ – Steffen Plunder May 28 '18 at 2:21
  • $\begingroup$ Note that, in contrast to the Jacobian matrices, both expressions describe the same point in $T\mathbb{R}^2$. The chain rule, which is usually needed to translate tangent vector between different coordinate systems, is already included in the definition of the tangent vectors. In this way using the notions of differential geometry may help to clarify in settings of many different coordinates. This is for example useful in rigid body motion or elasticity. $\endgroup$ – Steffen Plunder May 28 '18 at 2:25
  • $\begingroup$ Alright, thanks a lot again! $\endgroup$ – Jonas W. May 28 '18 at 17:05

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