Let $(M,g)$ be a compact Riemannian manifold. Given a tensor bundle $\mathbb{E}$, let $\nabla:\Gamma(\mathbb{E}) \rightarrow \Gamma(T^{*}M\otimes \mathbb{E})$ be the canonical connection induced by the Levi-civita connection on vector fields, viewed as a mapping $E\mapsto \nabla E$. Let $L^{2}\Gamma(\mathbb{E})$ stand for the $L^{2}$-completion of the section space with the inner product $(V,W)_{L^{2}}=\int_{M}(V,W)_{g}dVol_{g}$ and let $H^{1}\Gamma(\mathbb{E})$ be the completion with the inner product $(V,W)_{H^{1}}=(V,W)_{L^{2}}+(\nabla V,\nabla W)_{L^{2}}$.
Denoting the vector fields by $\mathfrak{X}(M)$, it can be proven that $\nabla:H^{1}\mathfrak{X}(M) \rightarrow L^{2}\Gamma(T^{*}M\otimes TM)$ has a closed range, yielding a $L^{2}$-orthogonal splitting,
$L^{2}\Gamma(T^{*}M\otimes TM)=\mathrm{ker}{\nabla^{*}}\oplus\mathrm{Im}{\nabla}$
where $\nabla^{*}:H^{1}\Gamma(T^{*}M\otimes TM)\rightarrow H^{1}\mathfrak{X}(M)$ is the formal $L^{2}$-dual of $\nabla$. Now, the exterior covariant derivative is the operator $d:\Gamma(T^{*}M\otimes TM)\rightarrow\Gamma(\Lambda^{2}T^{*}M\otimes TM)$ given by $dE=\mathrm{Alt}(\nabla E)$, where the alternation is taken with respect to the first two entries of the tensor $\nabla E\in\Gamma(T^{*}M\otimes T^{*}M\otimes TM)$. (See p. 61 in Peterson's Riemannian Geomtery 3rd edition for a reference of this defintion.)
My question is this: does the continous linear operator $d:H^{1}\Gamma(T^{*}M\otimes TM)\rightarrow L^{2}\Gamma(\Lambda^{2}T^{*}M\otimes TM)$ have a closed range as well?
Here is my attempt: it can be established that there is the following a-priori estimate, $||{E}||_{H^{1}}\leq C ||dE||_{L^{2}}+C||\nabla^*E||_{L^{2}}+C||E||_{L^{2}}$. This implies that the range of $d$ when restricted to $\mathrm{ker}{\nabla^{*}}$ is closed. But unless $d\nabla=0$, which only happens when $g$ is locally flat, I cannot complete the argument for the closedenss of the range of $d$ when not restricted thus.
In the general case, $d\nabla=R$, where $R$ is the tensorial curvature endormoprhism of $g$. Hence, I would have expected that becuase $R$ is a continous operation in the $L^{2}$ norm, in which $H^{1}$ is compact, some closedness of the range of $d$ is due. Indeed, it does follow from this estimate, the fact that $d\nabla=R$, and the first decompostion that $d:L^{2}\Gamma(T^{*}M\otimes TM)\rightarrow L^{2}\Gamma(\Lambda^{2}T^{*}M\otimes TM)$ is a closed unbounded operator with domain $\mathscr{D}(d)=H^{1}\Gamma(T^{*}M\otimes TM)$, but I cannot complete the actual closedness of the range of $d$.
Any help, advice or reference for an answer will be appreciated!
