$\Delta = -d\delta - \delta d$ is the Laplace operator in the Riemann manifold $(M, g)$. I want to prove:
$\Delta\alpha (Y) = tr D^2\alpha(Y) - Ric(X, Y)$
where $\alpha \in A^1(M)$, $X$ its dual vector field, $Y \in \mathfrak{X}(M)$,.
Chosen local coordinate $(U, x)$, and $\alpha|_U = \alpha_i dx^i, X|_U = a^i\frac{\partial}{\partial x^i}$, and $tr D^2 = g^{ij}(D_{\frac{\partial}{\partial x^i}}D_{\frac{\partial}{\partial x^j}} - D_{D_{\frac{\partial}{\partial x^i}}\frac{\partial}{\partial x^j}})$.
By definition LHS
$\Delta\alpha = \alpha^k_{,ki}dx^i + g^{jk}(\alpha_{j,ik}-\alpha_{i,jk})dx^i $
$= g^{jk}\alpha_{i,jk}dx^i - \alpha^hR^k_{hki}dx^i $
$= g^{jk}\alpha_{i,jk}dx^i - \alpha^hR_{hi}dx^i$.
But when I compute the RHS, I confused about
$tr D^2\alpha = g^{ij}(D_{\frac{\partial}{\partial x^i}}D_{\frac{\partial}{\partial x^j}} - D_{D_{\frac{\partial}{\partial x^i}}\frac{\partial}{\partial x^j}})\alpha $
$= g^{ij}D_{\frac{\partial}{\partial x^i}}\alpha_{s,j}dx^s - g^{ij}D_{\Gamma^k_{ji}\frac{\partial}{\partial x^k}}\alpha $
$= g^{ij}(\alpha_{s,j})_{,i}dx^s - g^{ij}\Gamma^k_{ji}D_{\frac{\partial}{\partial x^k}}\alpha $
$= g^{ij}\alpha_{s,ji}dx^s-g^{ij}\Gamma^k_{ji}\alpha_{s,k}dx^s$
$Ric(X, \cdot) = R_{hi}\alpha^hdx^i$
There is one more term $-g^{ij}\Gamma^k_{ji}\alpha_{s,k}dx^s$ confuse me.
Did I make wrong or this term is zero? Any help will be appreciated.