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.

  • $\begingroup$ 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

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 $\mathbb{R}^n$.

  • $\begingroup$ 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

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