The image of the exterior derivative is closed: Hodge theory
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
asked Dec 26 at 0:54
truebaran
2,1952723
add a comment |
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
asked Dec 26 at 0:54
truebaran
2,1952723
2
The range of the exterior derivative as a map from where to where?
– Eric Wofsey
Dec 26 at 1:06
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
– truebaran
Dec 26 at 1:08
Are you sure you haven't missed a star (or assuming $T$ is self-adjoint, for example)? The correct formula should be (variant of) $ker T^*=(operatorname{im} T)^perp$.
– user10354138
yesterday
add a comment |
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
asked Dec 26 at 0:54
truebaran
2,1952723
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
asked Dec 26 at 0:54
truebaran
2,1952723
asked Dec 26 at 0:54
truebaran
2,1952723
asked Dec 26 at 0:54
truebaran
2,1952723
asked Dec 26 at 0:54
truebaran
2,1952723
asked Dec 26 at 0:54
truebaran
2,1952723
2,1952723
2
The range of the exterior derivative as a map from where to where?
– Eric Wofsey
Dec 26 at 1:06
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
– truebaran
Dec 26 at 1:08
Are you sure you haven't missed a star (or assuming $T$ is self-adjoint, for example)? The correct formula should be (variant of) $ker T^*=(operatorname{im} T)^perp$.
– user10354138
yesterday
add a comment |
2
The range of the exterior derivative as a map from where to where?
– Eric Wofsey
Dec 26 at 1:06
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
– truebaran
Dec 26 at 1:08
Are you sure you haven't missed a star (or assuming $T$ is self-adjoint, for example)? The correct formula should be (variant of) $ker T^*=(operatorname{im} T)^perp$.
– user10354138
yesterday
2
2
The range of the exterior derivative as a map from where to where?
– Eric Wofsey
Dec 26 at 1:06
The range of the exterior derivative as a map from where to where?
– Eric Wofsey
Dec 26 at 1:06
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
– truebaran
Dec 26 at 1:08
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
– truebaran
Dec 26 at 1:08
Are you sure you haven't missed a star (or assuming $T$ is self-adjoint, for example)? The correct formula should be (variant of) $ker T^*=(operatorname{im} T)^perp$.
– user10354138
yesterday
Are you sure you haven't missed a star (or assuming $T$ is self-adjoint, for example)? The correct formula should be (variant of) $ker T^*=(operatorname{im} T)^perp$.
– user10354138
yesterday
add a comment |
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2
The range of the exterior derivative as a map from where to where?
– Eric Wofsey
Dec 26 at 1:06
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
– truebaran
Dec 26 at 1:08
Are you sure you haven't missed a star (or assuming $T$ is self-adjoint, for example)? The correct formula should be (variant of) $ker T^*=(operatorname{im} T)^perp$.
– user10354138
yesterday