What is the analogous of $\|\nabla X\|^2=\sum_{i=1}^ng(\nabla_i X,\nabla_i X)$ for two forms? For a vector field $X$ it is well-known that $\|\nabla X\|^2=\langle\nabla X,\nabla X\rangle=\sum_{i=1}^ng(\nabla_{e_i} X,\nabla_{e_i} X)$ where $\{e_i\}$ is an ONB and $\langle \cdot,\cdot \rangle$ is the inner product on tensors induced by Riemannian metric $g$.

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*First how to see the preceding formula? I know that in local coordinate it is $g_{si}g^{tj}(\nabla X)^{s}_t(\nabla X)^i_j.$

*second, I want to know is there an analogous  for two forms? i.e. $\|\nabla \omega\|^2=\langle\nabla \omega ,\nabla \omega \rangle$ in terms of ONB. Is that equal to $ \sum_{i=1}^n\sum_{j=1}^ng((\nabla_{e_i} \omega)e_j,(\nabla_{e_i} \omega)e_j)$ or  $\sum_{i=1}^n\langle \nabla_{e_i} \omega,\nabla_{e_i} \omega\rangle$?

 A: If $T$ and $S$ are type $(r,s)$ tensor fields on a pseudo-Riemannian manifold $(M,g)$, then $$\langle T,S\rangle = \sum g_{i_1k_1}\cdots g_{i_rk_r} g^{j_1\ell_1}\cdots g^{j_s\ell_s}T^{i_1\cdots i_r}_{\phantom{i_1\cdots i_r}j_1\cdots j_s}S^{k_1\cdots k_r}_{\phantom{k_1\cdots k_r}\ell_1\cdots \ell_s}.$$This is a definition (easily seen to be coordinate-independent). If the metric is positive and we have fixed a local orthonormal frame $(e_i)_{i=1}^n$ (and hence a metrically equivalent coframe $(\theta^i)_{i=1}^n$), then:

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*If you have an orthonormal frame, then $$\langle \nabla X, \nabla X\rangle = \sum \delta_{si}\delta^{tj} (\nabla X)^s_t(\nabla X)^i_j = \sum [(\nabla X)^i_j]^2,$$but what is $(\nabla X)^i_j$? By orthonormal expansion, we have that $\nabla_{e_j}X = \sum \langle \nabla_{e_j}X, e_i\rangle e_i$, so $(\nabla X)^i_j = \langle \nabla_{e_j}X,e_i\rangle$, and we can plug this into the above formula to get $$\begin{align} \langle \nabla X, \nabla X\rangle &= \sum [(\nabla X)^i_j]^2 = \sum \langle \nabla_{e_i}X,e_j\rangle \langle \nabla_{e_j}X,e_i\rangle \\ &= \sum \left\langle \nabla_{e_j}X, \sum\langle \nabla_{e_j}X,e_i\rangle e_i\right\rangle = \sum \langle \nabla_{e_j}X,\nabla_{e_j}X\rangle\end{align}$$


*A $2$-form $\omega$ is just a very particular type $(0,2)$ tensor field, so $$\langle \omega,\omega\rangle = \sum g^{ik}g^{j\ell}\omega_{ij}\omega_{k\ell}$$and that's it. The covariant derivative $\nabla\omega$ is a type $(0,3)$ tensor field, so all of the above applies, and we have $$\langle \nabla \omega, \nabla \omega\rangle = \sum g^{ir}g^{js}g^{kt}\omega_{ij;k}\omega_{rs;t}.$$Particularizing this, we have $$\langle \nabla \omega,\nabla\omega\rangle = \sum \delta^{ir} \delta^{js}\delta^{kt} \omega_{ij;k}\omega_{rs;t} = \sum [\omega_{ij;k}]^2,$$but what is $\omega_{ij;k}$? By orthonormal expansion again, we have that $$\nabla_{e_k}\omega = \sum \langle \nabla_{e_k}\omega, \theta^i \otimes \theta^j\rangle \theta^i\otimes \theta^j,$$so $\omega_{ij;k} = \langle \nabla_{e_k}\omega, \theta^i\otimes \theta^j\rangle$, leading to $$\begin{align} 
 \langle \nabla \omega, \nabla \omega\rangle &= \sum [\omega_{ij;k}]^2 = \sum \langle \nabla_{e_k}\omega, \theta^i\otimes \theta^j\rangle \langle \nabla_{e_k}\omega, \theta^i \otimes \theta^j\rangle \\ &= \sum \left\langle \nabla_{e_k}\omega, \sum\langle \nabla_{e_k}\omega, \theta^i\otimes \theta^j\rangle \theta^i\otimes \theta^j\right\rangle =\sum \langle \nabla_{e_k}\omega, \nabla_{e_k}\omega\rangle.\end{align}$$
Let me know if you want me to be more careful with the bounds of summation, I was just using Einstein's convention in my mind (which apparently you're familiar with). If you want to consider the inner product induced in the exterior algebra as opposed to the tensor algebra, all of this should work the same (but you pay the price either with a $(\deg \omega)!$ somewhere or restricting the sum to increasing indices --- I'll let you pick your poison there).
