I am trying to prove the lemma from the Lee, Introduction to Smooth Manifolds book:

Lemma 3.16. Let $M$ be a smooth manifold. If $p \in M$ and $X \in T_pM$, there is a smooth vector field $\tilde X$ on $M$ such that $\tilde X_p=X$.

The proof in the book uses a bump function to construct the vector field $\tilde X$, however I made a proof without a bump function which is therefore suspiciously simpler and would like to verify if it is correct, otherwise what is wrong:

Let $X=X^i \frac{\partial}{\partial x^i}|_p$, then define the vector field $\tilde X:M \to TM$ by $r \mapsto X^i \frac{\partial}{\partial y^i}|_r$ where $r \in V$ for some chart $(V,\psi) \in \mathcal{A}_M$ with $\psi(V)$ having the local coordinates $y^i$. Then $\tilde X$ is a vector field since given the projection $\pi :TM \to M$ it is true that $\tilde X \circ \pi = id_M$. $\tilde X$ is smooth iff for any smooth charts $(U, \phi), (V, \psi) \in \mathcal{A}_M$ in an atlas for $M$ with the correspondent smooth chart $(\pi^{-1}(V), \Psi:\pi^{-1}(V) \to \mathbb{R}^{2n}) \in A_{TM}$ the function $\Psi \circ \tilde X \circ \phi^{-1}$ is smooth where $\Psi:v^i \frac{\partial}{\partial y^i}|_r \mapsto (y^1(r), ..., y^n(r), v^1, ..., v^n)$ with $y^i$ being a local coordinate of $\psi(V)$.

From the definition $\psi \circ \tilde X \circ \phi^{-1}:x \mapsto (y^1(\phi^{-1}(x)), ..., y^n(\phi^{-1}(x)), X^1, ..., X^n)$ where the $y^i$ are the local coordinates, i.e. $\psi(r)=(y^1(r), ..., y^n(r))$. Since the transition function $\psi \circ \phi^{-1}$ is smooth by the compatibility of the charts, all components of $\psi \circ \tilde X \circ \phi^{-1}$ are smooth, therefore $\psi \circ \tilde X \circ \phi^{-1}$ is smooth as required.

  • $\begingroup$ The field $\tilde{X}$ isn't defined on all of $M$, only on the open subset where the coordinates $x^{i}$ are defined. $\endgroup$ – Andrew D. Hwang Apr 5 '14 at 15:13
  • $\begingroup$ @user86418 I made an explicit distinction in an edit between the coordinates. Now $\tilde X$ is defined for all $p \in M$ as for every $p$ there is some chart $(V,\psi)$ such that $p \in V$. $\endgroup$ – Dávid Natingga Apr 5 '14 at 16:04
  • 1
    $\begingroup$ If I understand your strategy, your overlap maps do not respect the local definition of $\tilde{X}$; that is, two "patchwise" definitions probably do not agree where both are defined. (If that's not clear, consider the special case $X^{1} = 1$ and $X^{i} = 0$ for $i > 1$. Your construction purports to give a non-vanishing field, equal to $\partial/\partial y^{1}$ in each chart. But not every manifold admits a non-vanishing vector field.) $\endgroup$ – Andrew D. Hwang Apr 5 '14 at 16:33
  • $\begingroup$ @user86418 Thank you for the clarification, I think my confusion was that although $\frac{\partial}{\partial x^i}$ to $\frac{\partial}{\partial y^i}$ gives an isomorphism between the tangent spaces, the tangent vectors $\frac{\partial}{\partial x^i}$ and $\frac{\partial}{\partial y^i}$ are not necessarily the same ones. $\endgroup$ – Dávid Natingga Apr 12 '14 at 11:55

Your Answer

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

Browse other questions tagged or ask your own question.