Let $M$ be smooth manifold with boundary.Let $V,W$ be two smooth vector field,if we embed $M$ into double of $M$,then $V,W$ on $M$ is the same as $V,W$ on the closed subset $M\subset D(M)$(since diffeomorphism related vetor field on both side).After that We can extend $V,W$ from closed subset $M$ to $D(M)$.
Assume there are two smooth extension $X_1,X_2$ for $V$ and two smooth extension $Y_1,Y_2$ for $W$.
Prove $[X_1,Y_2]_p = [X_2,Y_2]_p$ for $p\in \partial M$
My attempt :
It seems no choice but using the definition $[X_1,Y_1]_p f = X_p(Yf) – Y_p(Xf) = V_p (Yf) – W_p (Xf)$ I have no idea how to preceed then.Intuitively Lie bracket gives the differential operator of order 2.So the change of vector field in a neiborhood is important.To express those information it's natural to write it under the coordinate chart.In side the coordiante chart we need to prove that $$\frac{\partial Y_1^j}{\partial x^i}(p) = \frac{\partial Y_2^j}{\partial x^i}(p)$$
If we subtract them $F = Y_1^j – Y_2^j$ then it's constant zero on$ \ M \subset D(M)$
Best Answer
You're practically already there with your last observation. Remember that a vector field $V$ is smooth iff the coordinate coefficient functions $V^i$ defined by $V = V^i \partial_i$ are smooth for every coordinate chart in your atlas.
Define the smooth vector field $Z:= [X_1,Y_1]-[X_2,Y_2]$, so that $Z|_M = 0$ (by passing to coordinates, if you like). Then in any coordinate chart neighborhood $U$ of $p \in \partial M \subset D(M)$, the coefficient functions $Z^i$ are smooth and identically zero on $M \cap U$. But the zero set of a continuous function is closed, so $Z^i$ must also be zero on $\partial M \cap U \ni p$, showing $Z_p = 0$.