For a given manifold $M$ and a point $x \in M$, we can define the tangent space at $x$, $T_xM$ in two ways (more, actually, but I am just concerned about these two for now):

1) Given a chart $(U, \phi)$, where $p \in U$, call two curves

$\gamma_1 : (-1,1) \to M $, $ \gamma_2 : (-1,1) \to M $, ($ \gamma_1(0)=\gamma_2(0)=p$)

$\textbf{equivalent}$ if

$ (\phi \circ \gamma_1)'(0)=(\phi \circ \gamma_2)'(0)$.

Call the equivalence classes $[\gamma]$ the $\textbf{tangent vectors of }$M$ \textbf{ at } x$. Define $T_xM$ as the collection of $[\gamma]$'s. These classes map to vectors in $\mathbb{R}^n$ via $[\gamma] \mapsto \frac{d}{dt}(\phi \circ \gamma)(0)$.

2) A $\textbf{derivation at } x$ is a linear map $D_x:C^\infty(M) \to \mathbb{R}$ such that, $\forall f,g \in C^\infty(M)$,


Call the vector space of all derviations at $x$ the tangent space at $x$, $T_xM$.

I understand that, given a $[\gamma]$, we can get a derivation $D_\gamma$ given by

$D_\gamma(f) := \frac{d}{dt}(f \circ \gamma)(0)$.

I am unclear as to how to go the other direction; given a derivation $D$, how can I get an equivalence class $[\gamma]$ of curves?


  • 2
    $\begingroup$ The other direction is considerably harder. In fact, it is not true for $C^k$ manifolds with $k<\infty$, only for smooth manifolds. A detailed proof may be found in Foundations of Differential Manifolds and Lie Groups by Frank Warner. $\endgroup$ – Amitai Yuval Feb 13 '15 at 20:02

Let $D$ be a derivation at $p$. Take local coordinates $(x^1, \dots, x^n)$ at $p$, with $p$ mapping to $0$. The goal is to show that $D$ can be written in these coordinates as $$ D = \sum_i c^i \frac{\partial}{\partial x^i} \tag{$*$} $$ for some constants $c^i$; having done this you can take your path to be the line whose coordinate representation is $$ t \mapsto t(c^1, \dots, c^n). $$ This gives a representative of the equivalence class you want.

For this to have a chance of being true, it's clear that $c^i$ must be the value $D(x^i)$ of the derivation on the smooth function $x^i$. Here's how this is proved:

We use the following lemma from calculus: If $f$ is a smooth function on a neighborhood $U$ of $0$ in $\mathbb{R}^n$ with $f(0) = 0$, then there exist smooth functions $g_i$ on $U$ such that $f(x) = \sum_i x^i g_i(x)$, and such that $g_i(0) = \tfrac{\partial f}{\partial x^i} (0)$. This lemma isn't hard to prove; one just writes $$ f(x) = \int_0^1 \frac{\partial}{\partial t} \Big( f(tx) \Big) dt $$ and uses the chain rule.

Now, note that the action of $D$ on $f$ is equal to the action of $D$ on $f - f(p)$, since $D$ is linear and $D(1) = 0$, and so it suffices to check $(*)$ for functions $f$ with $f(p) = 0$.

Let $f$ be such a function. Using the lemma, and noting that the $x^i$ are themselves smooth functions on $U$, $$ D(f) = \sum D(x^i g_i) = \sum D(x^i) g_i(p) = \sum D(x^i) \frac{\partial f}{\partial x^i} (p), $$ which establishes $(*)$ for $c^i = D(x^i)$. (The middle equality is just the Leibniz rule, using the fact that $x^i(p) = 0$.)


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