The image of the exterior derivative is closed: Hodge theory
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If we have an operator $T$ in a Hilbert space $H$ then one can decmpose $H$ as an orthogonal sum $ker(T^*) oplus overline{ran(T)}$. This is purely operator theoretic context. On the other hand there is an important theory which is called Hodge theory: one possible application of this theory is to find a special (unique) representative of cohomology class (as a harmonic differential form). My question is the following:
Does the Hodge theory imply that the range of the exterior derivative is closed?
differential-geometry operator-theory hodge-theory
$endgroup$
add a comment |
$begingroup$
If we have an operator $T$ in a Hilbert space $H$ then one can decmpose $H$ as an orthogonal sum $ker(T^*) oplus overline{ran(T)}$. This is purely operator theoretic context. On the other hand there is an important theory which is called Hodge theory: one possible application of this theory is to find a special (unique) representative of cohomology class (as a harmonic differential form). My question is the following:
Does the Hodge theory imply that the range of the exterior derivative is closed?
differential-geometry operator-theory hodge-theory
$endgroup$
3
$begingroup$
The range of the exterior derivative as a map from where to where?
$endgroup$
– Eric Wofsey
Dec 26 '18 at 1:06
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Probably one has to take a closure of this operator which initially is defined on all smooth forms (it is unbounded) which produces something like Sobolev space
$endgroup$
– truebaran
Dec 26 '18 at 1:08
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Are you sure you haven't missed a star (or assuming $T$ is self-adjoint, for example)? The correct formula should be (variant of) $ker T^*=(operatorname{im} T)^perp$.
$endgroup$
– user10354138
Dec 27 '18 at 19:10
$begingroup$
Of course, corrected
$endgroup$
– truebaran
Dec 30 '18 at 22:14
add a comment |
$begingroup$
If we have an operator $T$ in a Hilbert space $H$ then one can decmpose $H$ as an orthogonal sum $ker(T^*) oplus overline{ran(T)}$. This is purely operator theoretic context. On the other hand there is an important theory which is called Hodge theory: one possible application of this theory is to find a special (unique) representative of cohomology class (as a harmonic differential form). My question is the following:
Does the Hodge theory imply that the range of the exterior derivative is closed?
differential-geometry operator-theory hodge-theory
$endgroup$
If we have an operator $T$ in a Hilbert space $H$ then one can decmpose $H$ as an orthogonal sum $ker(T^*) oplus overline{ran(T)}$. This is purely operator theoretic context. On the other hand there is an important theory which is called Hodge theory: one possible application of this theory is to find a special (unique) representative of cohomology class (as a harmonic differential form). My question is the following:
Does the Hodge theory imply that the range of the exterior derivative is closed?
differential-geometry operator-theory hodge-theory
differential-geometry operator-theory hodge-theory
edited Dec 30 '18 at 22:14
truebaran
asked Dec 26 '18 at 0:54
truebarantruebaran
2,2002723
2,2002723
3
$begingroup$
The range of the exterior derivative as a map from where to where?
$endgroup$
– Eric Wofsey
Dec 26 '18 at 1:06
$begingroup$
Probably one has to take a closure of this operator which initially is defined on all smooth forms (it is unbounded) which produces something like Sobolev space
$endgroup$
– truebaran
Dec 26 '18 at 1:08
$begingroup$
Are you sure you haven't missed a star (or assuming $T$ is self-adjoint, for example)? The correct formula should be (variant of) $ker T^*=(operatorname{im} T)^perp$.
$endgroup$
– user10354138
Dec 27 '18 at 19:10
$begingroup$
Of course, corrected
$endgroup$
– truebaran
Dec 30 '18 at 22:14
add a comment |
3
$begingroup$
The range of the exterior derivative as a map from where to where?
$endgroup$
– Eric Wofsey
Dec 26 '18 at 1:06
$begingroup$
Probably one has to take a closure of this operator which initially is defined on all smooth forms (it is unbounded) which produces something like Sobolev space
$endgroup$
– truebaran
Dec 26 '18 at 1:08
$begingroup$
Are you sure you haven't missed a star (or assuming $T$ is self-adjoint, for example)? The correct formula should be (variant of) $ker T^*=(operatorname{im} T)^perp$.
$endgroup$
– user10354138
Dec 27 '18 at 19:10
$begingroup$
Of course, corrected
$endgroup$
– truebaran
Dec 30 '18 at 22:14
3
3
$begingroup$
The range of the exterior derivative as a map from where to where?
$endgroup$
– Eric Wofsey
Dec 26 '18 at 1:06
$begingroup$
The range of the exterior derivative as a map from where to where?
$endgroup$
– Eric Wofsey
Dec 26 '18 at 1:06
$begingroup$
Probably one has to take a closure of this operator which initially is defined on all smooth forms (it is unbounded) which produces something like Sobolev space
$endgroup$
– truebaran
Dec 26 '18 at 1:08
$begingroup$
Probably one has to take a closure of this operator which initially is defined on all smooth forms (it is unbounded) which produces something like Sobolev space
$endgroup$
– truebaran
Dec 26 '18 at 1:08
$begingroup$
Are you sure you haven't missed a star (or assuming $T$ is self-adjoint, for example)? The correct formula should be (variant of) $ker T^*=(operatorname{im} T)^perp$.
$endgroup$
– user10354138
Dec 27 '18 at 19:10
$begingroup$
Are you sure you haven't missed a star (or assuming $T$ is self-adjoint, for example)? The correct formula should be (variant of) $ker T^*=(operatorname{im} T)^perp$.
$endgroup$
– user10354138
Dec 27 '18 at 19:10
$begingroup$
Of course, corrected
$endgroup$
– truebaran
Dec 30 '18 at 22:14
$begingroup$
Of course, corrected
$endgroup$
– truebaran
Dec 30 '18 at 22:14
add a comment |
1 Answer
1
active
oldest
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$begingroup$
Sure. Given a degree $d$ differential operator with smooth coefficients $D: Gamma^infty(E) to Gamma^infty(F)$ between spaces of smooth sections of some vector bundles, there is a corresponding map $D: L^2_k(E) to L^2_{k-d}(F)$, given by extension to the closure. And there is also a corresponding map $D^*: L^2_k(E) to L^2_{k-d}(F)$, the "formal adjoint", which satisfies $langle Df, grangle_{L^2} = langle f, D^* grangle_{L^2}$.
You are now in a purely functional-analytic context: you have a "closed densely defined unbounded operator" $D: L^2(E) to L^2(F)$, with natural domain $L^2_d(E)$. The adjoint of this operator in the sense of densely defined operators is a map $D^*: L^2(F) to L^2(E)$, with natural domain $L^2_d(F)$. And in this context, the formula $text{Im}(D) = text{ker}(D^*)^perp$ as a subset of $L^2(F)$ remains true.
If you want this formula to hold for any Sobolev space, you can bootstrap this to an $L^2$-orthogonal direct sum decomposition of $L^2_k(F)$ formally, so long as $text{Im}(D)$ is closed. But this is what you were after! So I don't think this strategy is very helpful.
For what remains you need the power of 'elliptic regularity'. I will state this here for the level of generality of "Fredholm complexes", because this is what you are looking at in your post.
Let $$0 to Gamma(E_0) xrightarrow{D_1} cdots xrightarrow{D_n} Gamma(E_n) to 0$$ be a complex of differential operators of common degree $d$ over a single manifold $M$, meaning $D_i D_{i-1} = 0$. If this is an elliptic complex, meaning the corresponding linear maps of vector bundles $$0 to T^*(E_0 setminus 0) xrightarrow{sigma(D_1)} cdots xrightarrow{sigma(D_n)} T^*(E_n setminus 0) to 0$$ form, fiberwise, an exact complex.
This implies that $Delta_D = oplus_i (D_i + D_i^*)^2$ is an elliptic operator. In particular, observe that we have the ellieptic regularity theorem: if $Delta_D omega in L^2_{k-d}$, then $omega in L^2_{k+d};$ in fact we have a continuous right inverse $F: text{Im}(Delta_D) cap L^2_{k-d} to L^2_{k+d}$. So if $Delta_D omega_n to psi in L^2_k$ for some $k$, then $Delta_D F Delta_D omega_n to Delta_D F psi = psi$; because $FDelta_D omega_n in L^2_{k+d}$, we see that we have constructed a sequence in $L^2_{k+d}$ so that after applying $Delta_D$, it converges to $psi$, as desired.
That is, $text{Im}(Delta_D)$ is always a closed subspace (of the appropriate Sobolev space) when $D$ is elliptic; its $L^2$-orthogonal complement is $text{ker(Delta_D)}$.
As for the rest, observe that $text{Im}(D^*D)$ and $text{Im}(DD^*)$ are $L^2$-orthogonal, and hence never intersect. Because they are further $L^2$ orthogonal with $text{ker(Delta_D)}$, we may write eg $DD^* omega = Delta_D psi$ for any $omega$ and some appropriate $psi$; thus $DD^* omega = DD^* psi + D^* D psi$, and hence $DD^* (omega - psi) = D^* D psi$; thus $D^*D psi = 0$. So what we see is we can write any element of $text{Im}(Delta_D)$ as a sum of two unique elements of $text{Im}(D^*D)$ and $text{Im}(DD^*)$. You can use this fact to argue as above that these subspaces are closed.
From here, I claim that $text{Im}(D D^* D) = text{Im}(D^* D) = text{Im}(D^*)$ and similarly for $text{Im}(DD^*)$. This is just as straightforward; write $omega = DD^* omega_0 + omega_h + D^*D omega_1$ where $omega_h$ is harmonic. Then $Domega = DD^*Domega_1$. (You appeal to the regularity results for $Delta_D$ to ensure that $D^* D omega_1$ is of the same regularity.)
Thus we always have the direct sum decomposition $L^2_k(E_i) = text{Im}(D) oplus text{Im}(D^*) oplus text{ker}(Delta_D)$, where all terms are $L^2$-orthogonal, whenever the $D_i$ form an elliptic complex. This is certainly the case for the Hodge complex. It implies the following elliptic regularity statement for Fredholm complexes:
Given a Fredholm complex $(D_i)$ of differential operators of degree $d$ over some closed manifold, if one has $D_i omega in L^2_k$, then one may write $omega = D_{i-1} omega_0 + omega_1$, where $omega_1 in L^2_{k+d}$.
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Sure. Given a degree $d$ differential operator with smooth coefficients $D: Gamma^infty(E) to Gamma^infty(F)$ between spaces of smooth sections of some vector bundles, there is a corresponding map $D: L^2_k(E) to L^2_{k-d}(F)$, given by extension to the closure. And there is also a corresponding map $D^*: L^2_k(E) to L^2_{k-d}(F)$, the "formal adjoint", which satisfies $langle Df, grangle_{L^2} = langle f, D^* grangle_{L^2}$.
You are now in a purely functional-analytic context: you have a "closed densely defined unbounded operator" $D: L^2(E) to L^2(F)$, with natural domain $L^2_d(E)$. The adjoint of this operator in the sense of densely defined operators is a map $D^*: L^2(F) to L^2(E)$, with natural domain $L^2_d(F)$. And in this context, the formula $text{Im}(D) = text{ker}(D^*)^perp$ as a subset of $L^2(F)$ remains true.
If you want this formula to hold for any Sobolev space, you can bootstrap this to an $L^2$-orthogonal direct sum decomposition of $L^2_k(F)$ formally, so long as $text{Im}(D)$ is closed. But this is what you were after! So I don't think this strategy is very helpful.
For what remains you need the power of 'elliptic regularity'. I will state this here for the level of generality of "Fredholm complexes", because this is what you are looking at in your post.
Let $$0 to Gamma(E_0) xrightarrow{D_1} cdots xrightarrow{D_n} Gamma(E_n) to 0$$ be a complex of differential operators of common degree $d$ over a single manifold $M$, meaning $D_i D_{i-1} = 0$. If this is an elliptic complex, meaning the corresponding linear maps of vector bundles $$0 to T^*(E_0 setminus 0) xrightarrow{sigma(D_1)} cdots xrightarrow{sigma(D_n)} T^*(E_n setminus 0) to 0$$ form, fiberwise, an exact complex.
This implies that $Delta_D = oplus_i (D_i + D_i^*)^2$ is an elliptic operator. In particular, observe that we have the ellieptic regularity theorem: if $Delta_D omega in L^2_{k-d}$, then $omega in L^2_{k+d};$ in fact we have a continuous right inverse $F: text{Im}(Delta_D) cap L^2_{k-d} to L^2_{k+d}$. So if $Delta_D omega_n to psi in L^2_k$ for some $k$, then $Delta_D F Delta_D omega_n to Delta_D F psi = psi$; because $FDelta_D omega_n in L^2_{k+d}$, we see that we have constructed a sequence in $L^2_{k+d}$ so that after applying $Delta_D$, it converges to $psi$, as desired.
That is, $text{Im}(Delta_D)$ is always a closed subspace (of the appropriate Sobolev space) when $D$ is elliptic; its $L^2$-orthogonal complement is $text{ker(Delta_D)}$.
As for the rest, observe that $text{Im}(D^*D)$ and $text{Im}(DD^*)$ are $L^2$-orthogonal, and hence never intersect. Because they are further $L^2$ orthogonal with $text{ker(Delta_D)}$, we may write eg $DD^* omega = Delta_D psi$ for any $omega$ and some appropriate $psi$; thus $DD^* omega = DD^* psi + D^* D psi$, and hence $DD^* (omega - psi) = D^* D psi$; thus $D^*D psi = 0$. So what we see is we can write any element of $text{Im}(Delta_D)$ as a sum of two unique elements of $text{Im}(D^*D)$ and $text{Im}(DD^*)$. You can use this fact to argue as above that these subspaces are closed.
From here, I claim that $text{Im}(D D^* D) = text{Im}(D^* D) = text{Im}(D^*)$ and similarly for $text{Im}(DD^*)$. This is just as straightforward; write $omega = DD^* omega_0 + omega_h + D^*D omega_1$ where $omega_h$ is harmonic. Then $Domega = DD^*Domega_1$. (You appeal to the regularity results for $Delta_D$ to ensure that $D^* D omega_1$ is of the same regularity.)
Thus we always have the direct sum decomposition $L^2_k(E_i) = text{Im}(D) oplus text{Im}(D^*) oplus text{ker}(Delta_D)$, where all terms are $L^2$-orthogonal, whenever the $D_i$ form an elliptic complex. This is certainly the case for the Hodge complex. It implies the following elliptic regularity statement for Fredholm complexes:
Given a Fredholm complex $(D_i)$ of differential operators of degree $d$ over some closed manifold, if one has $D_i omega in L^2_k$, then one may write $omega = D_{i-1} omega_0 + omega_1$, where $omega_1 in L^2_{k+d}$.
$endgroup$
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$begingroup$
Sure. Given a degree $d$ differential operator with smooth coefficients $D: Gamma^infty(E) to Gamma^infty(F)$ between spaces of smooth sections of some vector bundles, there is a corresponding map $D: L^2_k(E) to L^2_{k-d}(F)$, given by extension to the closure. And there is also a corresponding map $D^*: L^2_k(E) to L^2_{k-d}(F)$, the "formal adjoint", which satisfies $langle Df, grangle_{L^2} = langle f, D^* grangle_{L^2}$.
You are now in a purely functional-analytic context: you have a "closed densely defined unbounded operator" $D: L^2(E) to L^2(F)$, with natural domain $L^2_d(E)$. The adjoint of this operator in the sense of densely defined operators is a map $D^*: L^2(F) to L^2(E)$, with natural domain $L^2_d(F)$. And in this context, the formula $text{Im}(D) = text{ker}(D^*)^perp$ as a subset of $L^2(F)$ remains true.
If you want this formula to hold for any Sobolev space, you can bootstrap this to an $L^2$-orthogonal direct sum decomposition of $L^2_k(F)$ formally, so long as $text{Im}(D)$ is closed. But this is what you were after! So I don't think this strategy is very helpful.
For what remains you need the power of 'elliptic regularity'. I will state this here for the level of generality of "Fredholm complexes", because this is what you are looking at in your post.
Let $$0 to Gamma(E_0) xrightarrow{D_1} cdots xrightarrow{D_n} Gamma(E_n) to 0$$ be a complex of differential operators of common degree $d$ over a single manifold $M$, meaning $D_i D_{i-1} = 0$. If this is an elliptic complex, meaning the corresponding linear maps of vector bundles $$0 to T^*(E_0 setminus 0) xrightarrow{sigma(D_1)} cdots xrightarrow{sigma(D_n)} T^*(E_n setminus 0) to 0$$ form, fiberwise, an exact complex.
This implies that $Delta_D = oplus_i (D_i + D_i^*)^2$ is an elliptic operator. In particular, observe that we have the ellieptic regularity theorem: if $Delta_D omega in L^2_{k-d}$, then $omega in L^2_{k+d};$ in fact we have a continuous right inverse $F: text{Im}(Delta_D) cap L^2_{k-d} to L^2_{k+d}$. So if $Delta_D omega_n to psi in L^2_k$ for some $k$, then $Delta_D F Delta_D omega_n to Delta_D F psi = psi$; because $FDelta_D omega_n in L^2_{k+d}$, we see that we have constructed a sequence in $L^2_{k+d}$ so that after applying $Delta_D$, it converges to $psi$, as desired.
That is, $text{Im}(Delta_D)$ is always a closed subspace (of the appropriate Sobolev space) when $D$ is elliptic; its $L^2$-orthogonal complement is $text{ker(Delta_D)}$.
As for the rest, observe that $text{Im}(D^*D)$ and $text{Im}(DD^*)$ are $L^2$-orthogonal, and hence never intersect. Because they are further $L^2$ orthogonal with $text{ker(Delta_D)}$, we may write eg $DD^* omega = Delta_D psi$ for any $omega$ and some appropriate $psi$; thus $DD^* omega = DD^* psi + D^* D psi$, and hence $DD^* (omega - psi) = D^* D psi$; thus $D^*D psi = 0$. So what we see is we can write any element of $text{Im}(Delta_D)$ as a sum of two unique elements of $text{Im}(D^*D)$ and $text{Im}(DD^*)$. You can use this fact to argue as above that these subspaces are closed.
From here, I claim that $text{Im}(D D^* D) = text{Im}(D^* D) = text{Im}(D^*)$ and similarly for $text{Im}(DD^*)$. This is just as straightforward; write $omega = DD^* omega_0 + omega_h + D^*D omega_1$ where $omega_h$ is harmonic. Then $Domega = DD^*Domega_1$. (You appeal to the regularity results for $Delta_D$ to ensure that $D^* D omega_1$ is of the same regularity.)
Thus we always have the direct sum decomposition $L^2_k(E_i) = text{Im}(D) oplus text{Im}(D^*) oplus text{ker}(Delta_D)$, where all terms are $L^2$-orthogonal, whenever the $D_i$ form an elliptic complex. This is certainly the case for the Hodge complex. It implies the following elliptic regularity statement for Fredholm complexes:
Given a Fredholm complex $(D_i)$ of differential operators of degree $d$ over some closed manifold, if one has $D_i omega in L^2_k$, then one may write $omega = D_{i-1} omega_0 + omega_1$, where $omega_1 in L^2_{k+d}$.
$endgroup$
add a comment |
$begingroup$
Sure. Given a degree $d$ differential operator with smooth coefficients $D: Gamma^infty(E) to Gamma^infty(F)$ between spaces of smooth sections of some vector bundles, there is a corresponding map $D: L^2_k(E) to L^2_{k-d}(F)$, given by extension to the closure. And there is also a corresponding map $D^*: L^2_k(E) to L^2_{k-d}(F)$, the "formal adjoint", which satisfies $langle Df, grangle_{L^2} = langle f, D^* grangle_{L^2}$.
You are now in a purely functional-analytic context: you have a "closed densely defined unbounded operator" $D: L^2(E) to L^2(F)$, with natural domain $L^2_d(E)$. The adjoint of this operator in the sense of densely defined operators is a map $D^*: L^2(F) to L^2(E)$, with natural domain $L^2_d(F)$. And in this context, the formula $text{Im}(D) = text{ker}(D^*)^perp$ as a subset of $L^2(F)$ remains true.
If you want this formula to hold for any Sobolev space, you can bootstrap this to an $L^2$-orthogonal direct sum decomposition of $L^2_k(F)$ formally, so long as $text{Im}(D)$ is closed. But this is what you were after! So I don't think this strategy is very helpful.
For what remains you need the power of 'elliptic regularity'. I will state this here for the level of generality of "Fredholm complexes", because this is what you are looking at in your post.
Let $$0 to Gamma(E_0) xrightarrow{D_1} cdots xrightarrow{D_n} Gamma(E_n) to 0$$ be a complex of differential operators of common degree $d$ over a single manifold $M$, meaning $D_i D_{i-1} = 0$. If this is an elliptic complex, meaning the corresponding linear maps of vector bundles $$0 to T^*(E_0 setminus 0) xrightarrow{sigma(D_1)} cdots xrightarrow{sigma(D_n)} T^*(E_n setminus 0) to 0$$ form, fiberwise, an exact complex.
This implies that $Delta_D = oplus_i (D_i + D_i^*)^2$ is an elliptic operator. In particular, observe that we have the ellieptic regularity theorem: if $Delta_D omega in L^2_{k-d}$, then $omega in L^2_{k+d};$ in fact we have a continuous right inverse $F: text{Im}(Delta_D) cap L^2_{k-d} to L^2_{k+d}$. So if $Delta_D omega_n to psi in L^2_k$ for some $k$, then $Delta_D F Delta_D omega_n to Delta_D F psi = psi$; because $FDelta_D omega_n in L^2_{k+d}$, we see that we have constructed a sequence in $L^2_{k+d}$ so that after applying $Delta_D$, it converges to $psi$, as desired.
That is, $text{Im}(Delta_D)$ is always a closed subspace (of the appropriate Sobolev space) when $D$ is elliptic; its $L^2$-orthogonal complement is $text{ker(Delta_D)}$.
As for the rest, observe that $text{Im}(D^*D)$ and $text{Im}(DD^*)$ are $L^2$-orthogonal, and hence never intersect. Because they are further $L^2$ orthogonal with $text{ker(Delta_D)}$, we may write eg $DD^* omega = Delta_D psi$ for any $omega$ and some appropriate $psi$; thus $DD^* omega = DD^* psi + D^* D psi$, and hence $DD^* (omega - psi) = D^* D psi$; thus $D^*D psi = 0$. So what we see is we can write any element of $text{Im}(Delta_D)$ as a sum of two unique elements of $text{Im}(D^*D)$ and $text{Im}(DD^*)$. You can use this fact to argue as above that these subspaces are closed.
From here, I claim that $text{Im}(D D^* D) = text{Im}(D^* D) = text{Im}(D^*)$ and similarly for $text{Im}(DD^*)$. This is just as straightforward; write $omega = DD^* omega_0 + omega_h + D^*D omega_1$ where $omega_h$ is harmonic. Then $Domega = DD^*Domega_1$. (You appeal to the regularity results for $Delta_D$ to ensure that $D^* D omega_1$ is of the same regularity.)
Thus we always have the direct sum decomposition $L^2_k(E_i) = text{Im}(D) oplus text{Im}(D^*) oplus text{ker}(Delta_D)$, where all terms are $L^2$-orthogonal, whenever the $D_i$ form an elliptic complex. This is certainly the case for the Hodge complex. It implies the following elliptic regularity statement for Fredholm complexes:
Given a Fredholm complex $(D_i)$ of differential operators of degree $d$ over some closed manifold, if one has $D_i omega in L^2_k$, then one may write $omega = D_{i-1} omega_0 + omega_1$, where $omega_1 in L^2_{k+d}$.
$endgroup$
Sure. Given a degree $d$ differential operator with smooth coefficients $D: Gamma^infty(E) to Gamma^infty(F)$ between spaces of smooth sections of some vector bundles, there is a corresponding map $D: L^2_k(E) to L^2_{k-d}(F)$, given by extension to the closure. And there is also a corresponding map $D^*: L^2_k(E) to L^2_{k-d}(F)$, the "formal adjoint", which satisfies $langle Df, grangle_{L^2} = langle f, D^* grangle_{L^2}$.
You are now in a purely functional-analytic context: you have a "closed densely defined unbounded operator" $D: L^2(E) to L^2(F)$, with natural domain $L^2_d(E)$. The adjoint of this operator in the sense of densely defined operators is a map $D^*: L^2(F) to L^2(E)$, with natural domain $L^2_d(F)$. And in this context, the formula $text{Im}(D) = text{ker}(D^*)^perp$ as a subset of $L^2(F)$ remains true.
If you want this formula to hold for any Sobolev space, you can bootstrap this to an $L^2$-orthogonal direct sum decomposition of $L^2_k(F)$ formally, so long as $text{Im}(D)$ is closed. But this is what you were after! So I don't think this strategy is very helpful.
For what remains you need the power of 'elliptic regularity'. I will state this here for the level of generality of "Fredholm complexes", because this is what you are looking at in your post.
Let $$0 to Gamma(E_0) xrightarrow{D_1} cdots xrightarrow{D_n} Gamma(E_n) to 0$$ be a complex of differential operators of common degree $d$ over a single manifold $M$, meaning $D_i D_{i-1} = 0$. If this is an elliptic complex, meaning the corresponding linear maps of vector bundles $$0 to T^*(E_0 setminus 0) xrightarrow{sigma(D_1)} cdots xrightarrow{sigma(D_n)} T^*(E_n setminus 0) to 0$$ form, fiberwise, an exact complex.
This implies that $Delta_D = oplus_i (D_i + D_i^*)^2$ is an elliptic operator. In particular, observe that we have the ellieptic regularity theorem: if $Delta_D omega in L^2_{k-d}$, then $omega in L^2_{k+d};$ in fact we have a continuous right inverse $F: text{Im}(Delta_D) cap L^2_{k-d} to L^2_{k+d}$. So if $Delta_D omega_n to psi in L^2_k$ for some $k$, then $Delta_D F Delta_D omega_n to Delta_D F psi = psi$; because $FDelta_D omega_n in L^2_{k+d}$, we see that we have constructed a sequence in $L^2_{k+d}$ so that after applying $Delta_D$, it converges to $psi$, as desired.
That is, $text{Im}(Delta_D)$ is always a closed subspace (of the appropriate Sobolev space) when $D$ is elliptic; its $L^2$-orthogonal complement is $text{ker(Delta_D)}$.
As for the rest, observe that $text{Im}(D^*D)$ and $text{Im}(DD^*)$ are $L^2$-orthogonal, and hence never intersect. Because they are further $L^2$ orthogonal with $text{ker(Delta_D)}$, we may write eg $DD^* omega = Delta_D psi$ for any $omega$ and some appropriate $psi$; thus $DD^* omega = DD^* psi + D^* D psi$, and hence $DD^* (omega - psi) = D^* D psi$; thus $D^*D psi = 0$. So what we see is we can write any element of $text{Im}(Delta_D)$ as a sum of two unique elements of $text{Im}(D^*D)$ and $text{Im}(DD^*)$. You can use this fact to argue as above that these subspaces are closed.
From here, I claim that $text{Im}(D D^* D) = text{Im}(D^* D) = text{Im}(D^*)$ and similarly for $text{Im}(DD^*)$. This is just as straightforward; write $omega = DD^* omega_0 + omega_h + D^*D omega_1$ where $omega_h$ is harmonic. Then $Domega = DD^*Domega_1$. (You appeal to the regularity results for $Delta_D$ to ensure that $D^* D omega_1$ is of the same regularity.)
Thus we always have the direct sum decomposition $L^2_k(E_i) = text{Im}(D) oplus text{Im}(D^*) oplus text{ker}(Delta_D)$, where all terms are $L^2$-orthogonal, whenever the $D_i$ form an elliptic complex. This is certainly the case for the Hodge complex. It implies the following elliptic regularity statement for Fredholm complexes:
Given a Fredholm complex $(D_i)$ of differential operators of degree $d$ over some closed manifold, if one has $D_i omega in L^2_k$, then one may write $omega = D_{i-1} omega_0 + omega_1$, where $omega_1 in L^2_{k+d}$.
answered Dec 31 '18 at 0:36
Mike MillerMike Miller
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The range of the exterior derivative as a map from where to where?
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– Eric Wofsey
Dec 26 '18 at 1:06
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Probably one has to take a closure of this operator which initially is defined on all smooth forms (it is unbounded) which produces something like Sobolev space
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– truebaran
Dec 26 '18 at 1:08
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Are you sure you haven't missed a star (or assuming $T$ is self-adjoint, for example)? The correct formula should be (variant of) $ker T^*=(operatorname{im} T)^perp$.
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– user10354138
Dec 27 '18 at 19:10
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Of course, corrected
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– truebaran
Dec 30 '18 at 22:14