2
$\begingroup$

Below is an exercise from Guillemin-Pollack:

A curve in a manifold $X$ is a smooth map $t\mapsto c(t)$ of an interval of $\mathbb R$ into $X$. The velocity vector of the curve $c$ at time $t_0$, denoted $dc/dt(t_0)$, is defined to be the vector $dc_{t_0}(1)\in T_{x_0}(X)$, where $x_0=c(t_0)$ and $dc_{t_0}: \mathbb R\rightarrow T_{x_0}(X)$. Prove that every vector in $T_x(X)$ is the velocity vector of some curve in $X$, and conversely.

For the first part, I was given the hint that says to define the curve $c$ as a composition of two maps, $(a,b)\rightarrow U\rightarrow X$ where $\phi: U\rightarrow X$ is a local parametrization around $x\in X$ with $\phi(y)=x$, and the map $\gamma: (a,b)\rightarrow U$ is given by $t\mapsto tu+y$ for a suitable $u$.

I have problems with filling the details into this hint. So given $v\in T_x(X)$, it lies, by definition, in the image of $d\phi_y$. So $v=d\phi_yu$. This $u$ can be chosen to lie in $u$ since $d\phi_y$ is an isomorphism (because $\phi$ is a diffeomorphism). But I don't understand why $tu$ will lie in $U$ when $t$ lies in a suitable interval $(a,b)$? $U$ need not be a vector subspace of the ambient space $\mathbb R^k.$ Modulo this fact, everything is clear, since $$d(\phi\circ \gamma)_0=d\phi_y\circ d\gamma_0$$ and $$d(\phi\circ \gamma)_0(1)=d\phi_y(d\gamma_0(1))=d\phi_yu=v.$$

Also, how to prove the converse? If $v=dc/dt(t_0)=dc_{t_0}(1)$, then $v$ certainly lies in the image of $dc_{t_0}$. But to show that $v\in T_x(X)$, we need to show that $v$ lies in the image of the differential of some local parametrization at some point.

$\endgroup$
  • $\begingroup$ I think the answer to my first question is that $U$ can be chosen to be $\mathbb R^k$ so $tu$ will lie in this $U$ for any real $t$. $\endgroup$ – user500094 Jan 23 '18 at 19:32
0
$\begingroup$

Remark that if $(U,f)$ is a chart which contains $x_0$, $f:U\rightarrow f(U)\subset \mathbb{R}^n$ is a diffeomorphism and $df_{x_0}^{-1}$ is a diffeomorphism, so it is enough to show that the result is true of an open subset of $\mathbb{R}^n$.

$\endgroup$
  • $\begingroup$ I can prove this for open subsets of $\mathbb R^n$ but what causes problems is the statement that "it is enough to show that the result is true of an open subset of $\mathbb R^n$". $\endgroup$ – user500094 Jan 23 '18 at 19:30

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.