4
$\begingroup$

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, $\forall f,g \in C^\infty(M)$,

$D_x(fg)=D_x(f)g(x)+D_x(g)f(x)$.

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?

Thanks!

$\endgroup$
  • 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
2
$\begingroup$

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 $x^i(p) = 0$.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.