Laplacian on Calabi-Yau manifold Is the (Hodge) Laplacian of a $(p,q)$-form on a Calabi-Yau manifold equal to the covariant Laplacian? I.e. is $(\Delta \omega)_{\mu_1 \dots \mu_p \bar{\nu}_1 \dots \bar{\nu}_q} = - \nabla^a \nabla_a \omega_{\mu_1 \dots \mu_p \bar{\nu}_1 \dots \bar{\nu}_q}$? I believe the answer is negative -- there should be additional terms involving the Riemann tensor.
 A: An attempt at an answer: on a Kähler manifold, the various Laplacians are identical up to a factor of 2. Furthermore, we can view a complex $k$-dimensional manifold as a real $2k$-dimensional one and trivially extend any $(p,q)$-form to a $(p+q)$-form by setting the appropriate coefficients to zero. The Laplacian of a $(p,q)$-form on a Kähler manifold is equal to the Laplacian of the "realified" $(p+q)$-form. This lets us adapt a Weitzenböck identity from the differential geometry literature (e.g. eq. (9.13) of http://www.math.wisc.edu/~jeffv/courses/865_Fall_2007.pdf) for $(p+q = n)$-forms $\omega$ on real Riemannian manifolds of dimension $2k$:
\begin{equation} \label{weitzenbock}
\Delta \omega = - \nabla^2 \omega + \rho_\omega,
\end{equation}
where $\nabla^2 = \nabla^a \nabla_a$ is the "rough'' or covariant Laplacian and $\rho_\omega$ is a "correction" term involving the curvature tensors:
\begin{align}
\left( \rho_\omega \right)_{a_1 a_2 \dots a_n} = - \sum_{i=1}^{n-1} & \sum_{j > i}^n \mbox{$R^{lm}$}_{a_i a_j} \omega_{a_1 \dots a_{i-1} l a_{i+1} \dots a_{j-1} m a_{j+1} \dots a_n} \notag \\
& + \sum_{i=1}^n \mbox{$R^l$}_{a_i} \omega_{a_1 \dots a_{i-1} l a_{i+1} \dots a_n}.
\end{align}
To get the result for Kähler manifolds we simply plug in $(a_1 a_2 \dots a_n) = (\mu_1 \dots \mu_p \bar{\nu}_1 \dots \bar{\nu}_q)$ and possibly simplify the expression using the properties of the Riemann tensor on Kähler manifolds. For example, we could note that the indices $l,m$ in the $\mbox{$R^{lm}$}_{a_i a_j}$ terms have to be of different holomorphy in order for that contribution to be non-vanishing, and that the indices $l, a_i$ in $\mbox{$R^l$}_{a_i}$ have to be of the same type to make it non-vanishing. We get, using the antisymmetry of $\omega$,
\begin{align}
\left( \rho_\omega \right)_{\mu_1 \dots \mu_p \bar{\nu}_1 \dots \bar{\nu}_q} = - 2 \sum_{i=1}^p & \sum_{j = 1}^q \mbox{$R^{\alpha \bar{\beta}}$}_{\mu_i \bar{\nu}_j} \omega_{\mu_1 \dots \mu_{i-1} \alpha \mu_{i+1} \dots \mu_p \bar{\nu}_1 \dots \bar{\nu}_{j-1} \bar{\beta} \bar{\nu}_{j+1} \dots \bar{\nu}_q} \notag \\
& + \sum_{i=1}^p \mbox{$R^\alpha$}_{\mu_i} \omega_{\mu_1 \dots \mu_{i-1} \alpha \mu_{i+1} \dots \mu_p \bar{\nu}_1 \dots \bar{\nu}_q} + \sum_{j=1}^q \mbox{$R^\bar{\beta}$}_{\bar{\nu}_j} \omega_{\mu_1 \dots \mu_p \bar{\nu}_1 \dots \bar{\nu}_{j-1} \bar{\beta} \bar{\nu}_{j+1} \dots \bar{\nu}_q}.
\end{align}
If correct, the Wikipedia article https://en.wikipedia.org/wiki/Weitzenb%C3%B6ck_identity states an incorrect result in the section "Complex differential geometry" and we should change it. On a Calabi-Yau manifold, the last two sums do not contribute but the Riemann tensor terms generically remain present.
