Sorry if this is a too basic question, but I didn't find an answer anywhere:

The connection Laplacian, or Bochner Laplacian, is the differential operator acting on $k$-tensor fields $T\in\Gamma^{\infty}(T^{\ast}M^{\otimes_{s}2})$ given by $$\Delta:=\mathrm{tr}(\nabla^{2}T)$$ Now, I am confused about the coordinate expression: Many references, for example this nice lecture notes by L. Nicolaescu, state that $$\Delta:=g^{ij}(\nabla_{i}\nabla_{j}-\Gamma_{ij}^{k}\nabla_{k})\tag{$\ast$}$$ where $\nabla$ denotes the Levi-Civita connection. However, I am a bit puzzled by the second term. For example, lets say I'll take the specific case $k=0$. Then, using formula ($\ast$) I get $$\Delta=g^{ij}(\underbrace{\nabla_{i}\nabla_{j}f}_{\partial_{i}\partial_{j}f-\Gamma_{ij}^{k}\nabla_{k}f}-\Gamma_{ij}^{k}\nabla_{k}f)=g^{ij}\partial_{i}\partial_{j}f-2\Gamma_{ij}^{k}\nabla_{k}f$$ which is wrong, since we should obtain the Laplace-Beltrami operator $$\Delta_{\mathrm{LB}}f=g^{ij}(\partial_{i}\partial_{j}f-\Gamma_{ij}^{k}\partial_{k}f)$$ In other words, the fact 2 is different.

I assume it is just a missunderstanding of notation or terminology. Any help appreciated.

Edit: I think I have an idea where the difference comes from, but I am not sure.

- When I write things like $\nabla_{i}\nabla_{j}f$, then I use the typical notation which is used in physics literature, i.e. $\nabla_{i}\omega_{j}$ for some 1-form $\omega$ are the coefficients of $\nabla_{\partial_{i}}\omega$ in coordinates, i.e. $\nabla_{\partial_{i}}\omega=:(\nabla_{i}\omega_{j})d x^{j}$.
- In the lecture notes above ($\ast$), the notation is meant to mean $\nabla_{i}:=\nabla_{\partial_{i}}$
- Now, you can see the difference. In my notation, I view $\nabla_{j}f$ is a 1-form $\omega_{j}$ yielding the formula $$\nabla_{i}\nabla_{j} f=\partial_{i}\partial_{j}f-\Gamma_{ij}^{k}\partial_{k}f$$ while in the lecture notes (equation ($\ast$)), you obtain $$\nabla_{\partial_{i}}\nabla_{\partial_{j}}f=\partial_{i}\partial_{j}f$$ since $\nabla_{\partial_{j}}f=\partial_{j}f$ is again a function. Hence, formula ($\ast$) viewed in this sense gives the correct result $$\Delta f=\nabla_{\partial_{i}}\nabla_{\partial_{j}}f-\Gamma_{ij}^{k}\nabla_{\partial_{k}}f=\partial_{i}\partial_{j}f-\Gamma_{ij}^{k}\partial_{k}f$$