Let $M$ be a smooth manifold. My question is simple:
Given a point $p\in M$ and a tangent vector $v \in T_p M$, is there a smooth vector field $X$ on $M$ such that $X(p)=v$?
I tried to answer this as follows:
Choose a coordinate chart $(U,\phi)=(U,x^1,…,x^n)$ about $p$. Then for each $q \in U$, $\frac{\partial}{\partial x^1}|_q,…, \frac{\partial}{\partial x^n}|_q$ is a basis for the tangent space $T_qM$. In particular, we can write $v= \sum _i a^i \frac{\partial}{\partial x^i}|_p$ where $a^i$ are constants. Define $Y$ on $U$ by $Y=\sum _i a^i \frac{\partial}{\partial x^i}$. Then $Y$ is smooth on $U$ (since the coefficient functions $a^i$'s are smooth, being constant functions), and we have $Y(p)=v$. Now using the bump function, we may obtain a smooth vector field $X$, defined on $M$, which agrees with $Y$ in a neighborhood (possibly smaller than $U$) of $p$.
Is my argument fine? Also, is there a more simple approach?
Best Answer
This proof is correct, and I think you can't simplify it any further.
By the way, the choice of bump function is pretty arbitrary, and not unique. So, maybe you should not use the article the for it.