Vector bundle connection over complex manifold vs. over underlying real manifold

Let $(X,g)$ be an Hermitian manifold, and $(E,h)$ be an Hermitian vector bundle over $X$ of rank $r$. Denote by $(X^{\mathbb{R}},g^{\mathbb{R}})$ the underlying Riemannian manifold of $(X,g)$.

Question: Let $A$ be an integrable, unitary connection on $E$. Does $A$ induce an unitary connection $A^{\mathbb{R}}$ on $E$ as an Hermitian vector bundle over $X^{\mathbb{R}}$, such that under suitable definitions, $\|A\|_{W^{k,p}}$ and $\|A^{\mathbb{R}}\|_{W^{k,p}}$ are comparable? (i.e. they are equal, or differ by multiplying by an absolute constant)

Failure: A naïve way to define $A^{\mathbb{R}}$ will be taking the real part of the form. Specifically, we have locally $$A_{\alpha}\in\Lambda^1_{\mathbb{C}}(U_{\alpha})\otimes_{\mathbb{R}}\mathfrak{u}(r)$$ satisfying $$A_{\beta}=\phi_{\alpha\beta}^{-1}A_{\alpha}\phi_{\alpha\beta}+\phi_{\alpha\beta}^{-1}d\phi_{\alpha\beta},\;\;\phi_{\alpha\beta}\in\Gamma(U_{\alpha}\cap U_{\beta}, U(r))$$ Then there is the map of taking the real part: $$\Lambda_{\mathbb{C}}^1(U_{\alpha})\rightarrow\Lambda^1(U_{\alpha}^{\mathbb{R}})$$ Hence $A_{\alpha}$ becomes $A_{\alpha}^{\mathbb{R}}$ under this map, which still satisfies the relation $$A_{\beta}^{\mathbb{R}}=\phi_{\alpha\beta}^{-1}A_{\alpha}^{\mathbb{R}}\phi_{\alpha\beta}+\phi_{\alpha\beta}^{-1}d\phi_{\alpha\beta},\;\;\phi_{\alpha\beta}\in\Gamma(U_{\alpha}\cap U_{\beta}, U(r))$$ However, this makes any reasonable way of comparing the Sobolev norms impossible. For example, if the form part of $A_{\alpha}$ is purely imaginary, e.g. $dz^i-d\bar{z}^i$, then $A^{\mathbb{R}}_{\alpha}=0$.

Motivation: There are analytic results proved for connections on vector bundles over Riemannian manifolds, but I would like to use them on a complex manifold. Hence I need a dictionary between the two that preserves analytic properties as well as possible. What are the standard ways to do this, if the question simply fails?

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