The image of the exterior derivative is closed: Hodge theory












2












$begingroup$


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?











share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    The range of the exterior derivative as a map from where to where?
    $endgroup$
    – Eric Wofsey
    Dec 26 '18 at 1:06










  • $begingroup$
    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
















2












$begingroup$


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?











share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    The range of the exterior derivative as a map from where to where?
    $endgroup$
    – Eric Wofsey
    Dec 26 '18 at 1:06










  • $begingroup$
    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














2












2








2





$begingroup$


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?











share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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














  • 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










1 Answer
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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 $omega_1 in L^2_{k+d}$.







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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 $omega_1 in L^2_{k+d}$.







    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      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 $omega_1 in L^2_{k+d}$.







      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        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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        $endgroup$



        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 $omega_1 in L^2_{k+d}$.








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        answered Dec 31 '18 at 0:36









        Mike MillerMike Miller

        37.4k472139




        37.4k472139






























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