No, it should be pretty quick. In my case, I did this computation first using abstract indices, following the conventions of Section 4.2 in my recent preprint (arxiv.org/pdf/2403.16710#page16) on renormalized areas of minimal surfaces. After that, I translated the notation to your conventions.

As written, $IV$ is not symmetric, but the other three are. My guess is that Schimming means $dt^i dt^j$ to mean the symmetric product, though, in which case symmetry is automatic. For the original question, $Q^{ij}(Q_{ij} + R_{ikj}^k) = \frac{1}{2}C_{\mu nn \nu}R^{\mu \nu}$. Let me know if you can't verify this (or the other part of the claim) on your own.

There is a transcription error in your formula for $B_{nn}$ from Schimming's paper. The $R_{ij}$ is supposed to be $R_{ikj}^k$ (the contraction is only over intrinsic indices, not extrinsic ones). Can you complete the derivation with this observation?

@Nikolai Yes, his argument is incorrect, though it is true that if $\xi$ is harmonic, then $\xi \wedge \omega^{m-1}$ is harmonic (Proof: $\xi \wedge \omega^{m-1} = L^{m-1}\xi$ and $\Delta$ commutes with $L$). For an example of two harmonic forms whose product is not harmonic, consider a nonzero harmonic $(0,1)$-form $\xi$ on a compact Riemann surface $\Sigma_g$ of genus $g \geq 2$. Then $\star\xi$ is also harmonic, but $\xi \wedge \star\xi$ is necessarily nonzero but vanishes somewhere, and hence (being nonconstant) cannot be harmonic.

@Nikolai Yes, that was what I intended to write in my previous comment. In Siu’s lecture notes he assumes that the complex manifold is compact. In this setting, harmonic is equivalent to being $\bar\partial$- and $\smash{\bar\partial}^*$-closed, and so one gets the statement “if $\eta$ is a closed $(1,1)$-form, then $\eta$ is harmonic if and only if $\Lambda\eta$ is locally constant”. The converse of your original statement is false without compactness: On $\mathbb{C}$, take $\eta = zL1$. Then $z=\Lambda\eta$ is harmonic but not constant, yet $\eta$ is harmonic.

Sorry, I made a mistake in my last comment. The wedge product of two harmonic forms need not be harmonic, even if one of the factors is harmonic. I’ve also edited my answer to address this and to make sure my answer isn’t implicitly assuming $X$ is compact and connected, as in David Speyer’s comment to your question.

No, the wedge product of harmonic forms need not be harmonic. But it is if one of those forms is parallel.

@JMK I have added a description of how $h_2$ corresponds to a choice of Möbius structure and how it can be computed using the limits of the second fundamental forms of the sets $\{ x = \varepsilon \}$ as $\varepsilon to 0$.