I get a little confused by the details of the following theorem from John Lee's book Riemannian Manifold page 100. It basically about construction of vector field along a piecewise smooth curve using smooth bump function. Let $\gamma : [a,b] \rightarrow M$ be piecewise smooth curve (i.e there are subdivision $a_0=a<a_1<...<a_{n-1}<a_n=b$ s.t $\gamma_{[a_i,a_{i+1}]}$ is smooth). Now we want a vector field $V : [a,b] \rightarrow M$ along $\gamma$ such that at a point in the "corner", say at $\gamma(a_i)$ for $0<i<n$, $V$ having a particular value, say $V(a_i)=V_i$, but at any other corner $a_j (j\neq i)$, $V(a_j)=0$. This particular construction arise from the following proof.
$\textbf{Theorem 6.6.}$ Every minimizing curve is a geodesic when it is given a unit speed parametrization The idea is to use First Variation Formula \begin{equation} \frac{d}{ds}\Bigg|_{s=0} L(\Gamma_s) = -\int_a^b \langle V,D_t\dot{\gamma} \rangle dt - \sum_{i=1}^{k-1} \langle V(a_i), \Delta_i\dot{\gamma} \rangle \end{equation} to show that the initial piecewise curve $\gamma$ satisfying the hypothesis (minimizing and unit speed) is a broken geodesic, that is $D_t \dot{\gamma} \equiv 0$ on each subdivision where its smooth, and to show that it has no corners, that is $\Delta_i \dot{\gamma} = 0$ for all $i$.
My trouble is to follow the second step. It is says that to show $\Delta_i \dot{\gamma} = 0$, we can use a smooth bump function in a coordinate chart to construct a vector field $V$ along $\gamma$ such that $V(a_i)=\Delta_i \dot{\gamma}$ and $V(a_j)=0$ for all $j\neq i$. After a lot of thinking i already did such construction but not sure it is right. I worried because the similar proof for the above theorem in another text like do Carmo (Riemannian Geometry, Ch.9 prop 2.5 p.196) does not show either how to construct such vector field.
Here is my ideas :
Choose a small neighbourhood $U$ of $\gamma (a_i)$ such that it does not contain any other points $\gamma(a_j)$ except for $j=i$. Build a local vector field $V$ such that $V$ is constant with the constant coefficient equal to $\Delta_i \dot{\gamma}$. Choose a bump function $\varphi$ supported in $U$ such that $\varphi(\gamma(a_i)) = 1$. Restrict $\varphi$ to $U$ and multiplied it by $V$. Therefore we have a vector field $\varphi V$ defined in $U$. By Gluing lemma for smooth map we can extend $\varphi V$ to the whole $M$ such that $\varphi V$ vanish outside $U$ (details for extending $\varphi V$ is in here : Using Gluing Lemma for Smooth Functions). Finally defined vector field along $\gamma$ by restrict the global vector field above to the image of $\gamma$ in $M$.
I've been thinking some easier way (i'm not sure its correct or not). First we choose a appropriate small neighbourhood of $a_i$ in the domain of $\gamma : I \rightarrow M$, say $U \subset I$ such that $U$ does not contain $a_j$ for $j \neq i$. Choose a bump function $\varphi$ supported in $U$. Parallel translate the vector $\Delta_i \dot{\gamma}$ at $\gamma(a_i)$ along all portion of $\gamma(I)$, called this vector field $V$. Multiplied the resulting vector field by $\varphi$. The result is the vector field along $\gamma$, $\varphi V$, with the property that we needed, which is vanish for all $\gamma(a_j)$ for $j\neq i$
Any one can suggest another way (or correct way) to construct such vector field ? Thank you.
