Mixing
Do harmonic frames always exist locally?
$begingroup$
Let $M$ be a smooth Riemannian manifold of dimension $d$. Let $1 < k <d$ be an integer. Consider the exterior power bundle $Lambda_{k}(T^*M)$.
Do closed and co-closed frames for $Lambda_{k}(T^*M)$ always exist locally?
In other words, let $p in M$. Does there exist a neighbourhood $U$ of $p$, which admits a frame of $k$-forms, all of which are closed and co-closed?
For $k=1$, the answer is positive, due to the existence of harmonic coordinates. Note that I do not require the frame to come "from coordinates" in any way; In Euclidean space, however, we can use coordinates of course; Take $dx^{i_1} wedge dots wedge dx^{i_k}$, where $x_i$ are the standard coordinates.
Edit:
Let us specialize to even dimension $d$, and let $k=frac{d}{2}$.
Then, for a generic metric $g$, there are no coordinate systems where even one wedge is harmonic: $delta(mathrm{d}x^{i_1}wedgecdotswedgemathrm{d}x^{i_n})
=0$. This implies harmonic frames, if exist generically, cannot be induced in general by coordinates. Moreover, if such a frame exists, then each member in it cannot be decomposable, since a closed decomposable form can always be expressed locally as the wedge of coordinate differentials.
In fact, such a harmonic frame $omega^i$ must have the following property:
There are no non-zero decomposable elements, spanned by $omega^i$ using constant coefficients. (i.e. there do not exist real numbers $a_i$ such that $sum a_i omega^i neq 0$ is decomposable). Indeed, if such numbers existed, then $sum a_i omega^i$ would be decomposable and harmonic, which is generically impossible as mentioned above.
For a "generic" frame of $k$-forms, this property probably holds. (Here is an example for such "strongly non-decomposable frames").
pde riemannian-geometry differential-forms vector-bundles
asked May 30 '18 at 11:38
Asaf ShacharAsaf Shachar
5,77531141
$endgroup$
add a comment |
$begingroup$
Let $M$ be a smooth Riemannian manifold of dimension $d$. Let $1 < k <d$ be an integer. Consider the exterior power bundle $Lambda_{k}(T^*M)$.
Do closed and co-closed frames for $Lambda_{k}(T^*M)$ always exist locally?
In other words, let $p in M$. Does there exist a neighbourhood $U$ of $p$, which admits a frame of $k$-forms, all of which are closed and co-closed?
For $k=1$, the answer is positive, due to the existence of harmonic coordinates. Note that I do not require the frame to come "from coordinates" in any way; In Euclidean space, however, we can use coordinates of course; Take $dx^{i_1} wedge dots wedge dx^{i_k}$, where $x_i$ are the standard coordinates.
Edit:
Let us specialize to even dimension $d$, and let $k=frac{d}{2}$.
Then, for a generic metric $g$, there are no coordinate systems where even one wedge is harmonic: $delta(mathrm{d}x^{i_1}wedgecdotswedgemathrm{d}x^{i_n})
=0$. This implies harmonic frames, if exist generically, cannot be induced in general by coordinates. Moreover, if such a frame exists, then each member in it cannot be decomposable, since a closed decomposable form can always be expressed locally as the wedge of coordinate differentials.
In fact, such a harmonic frame $omega^i$ must have the following property:
There are no non-zero decomposable elements, spanned by $omega^i$ using constant coefficients. (i.e. there do not exist real numbers $a_i$ such that $sum a_i omega^i neq 0$ is decomposable). Indeed, if such numbers existed, then $sum a_i omega^i$ would be decomposable and harmonic, which is generically impossible as mentioned above.
For a "generic" frame of $k$-forms, this property probably holds. (Here is an example for such "strongly non-decomposable frames").
pde riemannian-geometry differential-forms vector-bundles
asked May 30 '18 at 11:38
Asaf ShacharAsaf Shachar
5,77531141
$endgroup$
add a comment |
$begingroup$
Let $M$ be a smooth Riemannian manifold of dimension $d$. Let $1 < k <d$ be an integer. Consider the exterior power bundle $Lambda_{k}(T^*M)$.
Do closed and co-closed frames for $Lambda_{k}(T^*M)$ always exist locally?
In other words, let $p in M$. Does there exist a neighbourhood $U$ of $p$, which admits a frame of $k$-forms, all of which are closed and co-closed?
For $k=1$, the answer is positive, due to the existence of harmonic coordinates. Note that I do not require the frame to come "from coordinates" in any way; In Euclidean space, however, we can use coordinates of course; Take $dx^{i_1} wedge dots wedge dx^{i_k}$, where $x_i$ are the standard coordinates.
Edit:
Let us specialize to even dimension $d$, and let $k=frac{d}{2}$.
Then, for a generic metric $g$, there are no coordinate systems where even one wedge is harmonic: $delta(mathrm{d}x^{i_1}wedgecdotswedgemathrm{d}x^{i_n})
=0$. This implies harmonic frames, if exist generically, cannot be induced in general by coordinates. Moreover, if such a frame exists, then each member in it cannot be decomposable, since a closed decomposable form can always be expressed locally as the wedge of coordinate differentials.
In fact, such a harmonic frame $omega^i$ must have the following property:
There are no non-zero decomposable elements, spanned by $omega^i$ using constant coefficients. (i.e. there do not exist real numbers $a_i$ such that $sum a_i omega^i neq 0$ is decomposable). Indeed, if such numbers existed, then $sum a_i omega^i$ would be decomposable and harmonic, which is generically impossible as mentioned above.
For a "generic" frame of $k$-forms, this property probably holds. (Here is an example for such "strongly non-decomposable frames").
pde riemannian-geometry differential-forms vector-bundles
asked May 30 '18 at 11:38
Asaf ShacharAsaf Shachar
5,77531141
$endgroup$
Let $M$ be a smooth Riemannian manifold of dimension $d$. Let $1 < k <d$ be an integer. Consider the exterior power bundle $Lambda_{k}(T^*M)$.
Do closed and co-closed frames for $Lambda_{k}(T^*M)$ always exist locally?
In other words, let $p in M$. Does there exist a neighbourhood $U$ of $p$, which admits a frame of $k$-forms, all of which are closed and co-closed?
For $k=1$, the answer is positive, due to the existence of harmonic coordinates. Note that I do not require the frame to come "from coordinates" in any way; In Euclidean space, however, we can use coordinates of course; Take $dx^{i_1} wedge dots wedge dx^{i_k}$, where $x_i$ are the standard coordinates.
Edit:
Let us specialize to even dimension $d$, and let $k=frac{d}{2}$.
Then, for a generic metric $g$, there are no coordinate systems where even one wedge is harmonic: $delta(mathrm{d}x^{i_1}wedgecdotswedgemathrm{d}x^{i_n})
=0$. This implies harmonic frames, if exist generically, cannot be induced in general by coordinates. Moreover, if such a frame exists, then each member in it cannot be decomposable, since a closed decomposable form can always be expressed locally as the wedge of coordinate differentials.
In fact, such a harmonic frame $omega^i$ must have the following property:
There are no non-zero decomposable elements, spanned by $omega^i$ using constant coefficients. (i.e. there do not exist real numbers $a_i$ such that $sum a_i omega^i neq 0$ is decomposable). Indeed, if such numbers existed, then $sum a_i omega^i$ would be decomposable and harmonic, which is generically impossible as mentioned above.
For a "generic" frame of $k$-forms, this property probably holds. (Here is an example for such "strongly non-decomposable frames").
pde riemannian-geometry differential-forms vector-bundles
pde riemannian-geometry differential-forms vector-bundles
asked May 30 '18 at 11:38
Asaf ShacharAsaf Shachar
5,77531141
asked May 30 '18 at 11:38
Asaf ShacharAsaf Shachar
5,77531141
edited Jan 22 at 13:56
Asaf Shachar
asked May 30 '18 at 11:38
Asaf ShacharAsaf Shachar
5,77531141
asked May 30 '18 at 11:38
Asaf ShacharAsaf Shachar
5,77531141
asked May 30 '18 at 11:38
Asaf ShacharAsaf Shachar
5,77531141
5,77531141
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$newcommand{M}{mathcal{M}}$
$newcommand{ep}{epsilon}$
The answer is positive. Closed and co-closed frames always exist locally.
I will sketch here one proof: (For other approaches, see here and here). We want to prove that around every point $p in M$ there exist a local frame for $bigwedge^k(T^*M)$ whose elements are closed and co-closed.
For the Euclidean metric this is immediate: We have the standard (constant) frame $dx^I=dx^{i_1} wedge ldots dx^{i_k} $. Since every metric is locally close to being Euclidean on small neighbourhoods, the idea is to use an approximation argument:
Given a Riemannian metric $g$, we denote the space of $g$-harmonic forms of degree $k$ by $H^k_{g}$.
We view $H^k_{g}$, as a subspace of $Omega^k(M)$ which is "changing continuously" with the metric $g$. Suppose $g_{ep} to g_0$ in the $C^1$-norm where $g_0$ is the Euclidean metric; Then $H^k_{g_{ep}} to H^k_{g_0}$ in the following sense: there exist a family of bases of $H^k_{g_{ep}}$, which converges to a basis of $H^k_{g_{0}} $ in $C^1$; this basis of $H^k_{g_{0}}$ forms a local frame for $bigwedge^k(T^*M)$. Since being a frame is an open condition, those bases for $H^k_{g_{ep}}$ are local frames for sufficiently small $ep$.
For the full details, see Appendix A in my paper here.
Some more details:
Even though the claim is local, and the approximation scheme is also inspired by a local phenomena, the implementation of the proof is based on a combination of local and global arguments. The reason is that on a closed manifold, being closed and co-closed is equivalent to being harmonic, and the dimension of the space of harmonic forms is a finite number which is a topological invariant of the manifold; it does not depend on the chosen metric.
Thus, given a family of metrics $g_{ep} to g_0$ on a closed manifold $M$, we consider the behaviour of the finite-dimensional subspaces $H^k_{g_{ep}}$ (all of the same dimension) as $ep to 0$.
That is, we look at the map $g to H^k_{g}=ker Delta_g$. It turns out that this map is continuous in some appropriate sense; this relies on a certain "stability property of kernels of linear operators". It turns out that a crucial factor in the existence of such a stability phenomenon is the assumption that all the kernels have the same finite dimension. The convergence of kernels does not always hold when the dimensions are not equal or infinite.
answered Jan 22 at 13:55
Asaf ShacharAsaf Shachar
5,77531141
$endgroup$
add a comment |
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$begingroup$
$newcommand{M}{mathcal{M}}$
$newcommand{ep}{epsilon}$
The answer is positive. Closed and co-closed frames always exist locally.
I will sketch here one proof: (For other approaches, see here and here). We want to prove that around every point $p in M$ there exist a local frame for $bigwedge^k(T^*M)$ whose elements are closed and co-closed.
For the Euclidean metric this is immediate: We have the standard (constant) frame $dx^I=dx^{i_1} wedge ldots dx^{i_k} $. Since every metric is locally close to being Euclidean on small neighbourhoods, the idea is to use an approximation argument:
Given a Riemannian metric $g$, we denote the space of $g$-harmonic forms of degree $k$ by $H^k_{g}$.
We view $H^k_{g}$, as a subspace of $Omega^k(M)$ which is "changing continuously" with the metric $g$. Suppose $g_{ep} to g_0$ in the $C^1$-norm where $g_0$ is the Euclidean metric; Then $H^k_{g_{ep}} to H^k_{g_0}$ in the following sense: there exist a family of bases of $H^k_{g_{ep}}$, which converges to a basis of $H^k_{g_{0}} $ in $C^1$; this basis of $H^k_{g_{0}}$ forms a local frame for $bigwedge^k(T^*M)$. Since being a frame is an open condition, those bases for $H^k_{g_{ep}}$ are local frames for sufficiently small $ep$.
For the full details, see Appendix A in my paper here.
Some more details:
Even though the claim is local, and the approximation scheme is also inspired by a local phenomena, the implementation of the proof is based on a combination of local and global arguments. The reason is that on a closed manifold, being closed and co-closed is equivalent to being harmonic, and the dimension of the space of harmonic forms is a finite number which is a topological invariant of the manifold; it does not depend on the chosen metric.
Thus, given a family of metrics $g_{ep} to g_0$ on a closed manifold $M$, we consider the behaviour of the finite-dimensional subspaces $H^k_{g_{ep}}$ (all of the same dimension) as $ep to 0$.
That is, we look at the map $g to H^k_{g}=ker Delta_g$. It turns out that this map is continuous in some appropriate sense; this relies on a certain "stability property of kernels of linear operators". It turns out that a crucial factor in the existence of such a stability phenomenon is the assumption that all the kernels have the same finite dimension. The convergence of kernels does not always hold when the dimensions are not equal or infinite.
answered Jan 22 at 13:55
Asaf ShacharAsaf Shachar
5,77531141
$endgroup$
add a comment |
$begingroup$
$newcommand{M}{mathcal{M}}$
$newcommand{ep}{epsilon}$
The answer is positive. Closed and co-closed frames always exist locally.
I will sketch here one proof: (For other approaches, see here and here). We want to prove that around every point $p in M$ there exist a local frame for $bigwedge^k(T^*M)$ whose elements are closed and co-closed.
For the Euclidean metric this is immediate: We have the standard (constant) frame $dx^I=dx^{i_1} wedge ldots dx^{i_k} $. Since every metric is locally close to being Euclidean on small neighbourhoods, the idea is to use an approximation argument:
Given a Riemannian metric $g$, we denote the space of $g$-harmonic forms of degree $k$ by $H^k_{g}$.
We view $H^k_{g}$, as a subspace of $Omega^k(M)$ which is "changing continuously" with the metric $g$. Suppose $g_{ep} to g_0$ in the $C^1$-norm where $g_0$ is the Euclidean metric; Then $H^k_{g_{ep}} to H^k_{g_0}$ in the following sense: there exist a family of bases of $H^k_{g_{ep}}$, which converges to a basis of $H^k_{g_{0}} $ in $C^1$; this basis of $H^k_{g_{0}}$ forms a local frame for $bigwedge^k(T^*M)$. Since being a frame is an open condition, those bases for $H^k_{g_{ep}}$ are local frames for sufficiently small $ep$.
For the full details, see Appendix A in my paper here.
Some more details:
Even though the claim is local, and the approximation scheme is also inspired by a local phenomena, the implementation of the proof is based on a combination of local and global arguments. The reason is that on a closed manifold, being closed and co-closed is equivalent to being harmonic, and the dimension of the space of harmonic forms is a finite number which is a topological invariant of the manifold; it does not depend on the chosen metric.
Thus, given a family of metrics $g_{ep} to g_0$ on a closed manifold $M$, we consider the behaviour of the finite-dimensional subspaces $H^k_{g_{ep}}$ (all of the same dimension) as $ep to 0$.
That is, we look at the map $g to H^k_{g}=ker Delta_g$. It turns out that this map is continuous in some appropriate sense; this relies on a certain "stability property of kernels of linear operators". It turns out that a crucial factor in the existence of such a stability phenomenon is the assumption that all the kernels have the same finite dimension. The convergence of kernels does not always hold when the dimensions are not equal or infinite.
answered Jan 22 at 13:55
Asaf ShacharAsaf Shachar
5,77531141
$endgroup$
add a comment |
$begingroup$
$newcommand{M}{mathcal{M}}$
$newcommand{ep}{epsilon}$
The answer is positive. Closed and co-closed frames always exist locally.
I will sketch here one proof: (For other approaches, see here and here). We want to prove that around every point $p in M$ there exist a local frame for $bigwedge^k(T^*M)$ whose elements are closed and co-closed.
For the Euclidean metric this is immediate: We have the standard (constant) frame $dx^I=dx^{i_1} wedge ldots dx^{i_k} $. Since every metric is locally close to being Euclidean on small neighbourhoods, the idea is to use an approximation argument:
Given a Riemannian metric $g$, we denote the space of $g$-harmonic forms of degree $k$ by $H^k_{g}$.
We view $H^k_{g}$, as a subspace of $Omega^k(M)$ which is "changing continuously" with the metric $g$. Suppose $g_{ep} to g_0$ in the $C^1$-norm where $g_0$ is the Euclidean metric; Then $H^k_{g_{ep}} to H^k_{g_0}$ in the following sense: there exist a family of bases of $H^k_{g_{ep}}$, which converges to a basis of $H^k_{g_{0}} $ in $C^1$; this basis of $H^k_{g_{0}}$ forms a local frame for $bigwedge^k(T^*M)$. Since being a frame is an open condition, those bases for $H^k_{g_{ep}}$ are local frames for sufficiently small $ep$.
For the full details, see Appendix A in my paper here.
Some more details:
Even though the claim is local, and the approximation scheme is also inspired by a local phenomena, the implementation of the proof is based on a combination of local and global arguments. The reason is that on a closed manifold, being closed and co-closed is equivalent to being harmonic, and the dimension of the space of harmonic forms is a finite number which is a topological invariant of the manifold; it does not depend on the chosen metric.
Thus, given a family of metrics $g_{ep} to g_0$ on a closed manifold $M$, we consider the behaviour of the finite-dimensional subspaces $H^k_{g_{ep}}$ (all of the same dimension) as $ep to 0$.
That is, we look at the map $g to H^k_{g}=ker Delta_g$. It turns out that this map is continuous in some appropriate sense; this relies on a certain "stability property of kernels of linear operators". It turns out that a crucial factor in the existence of such a stability phenomenon is the assumption that all the kernels have the same finite dimension. The convergence of kernels does not always hold when the dimensions are not equal or infinite.
answered Jan 22 at 13:55
Asaf ShacharAsaf Shachar
5,77531141
$endgroup$
$newcommand{M}{mathcal{M}}$
$newcommand{ep}{epsilon}$
The answer is positive. Closed and co-closed frames always exist locally.
I will sketch here one proof: (For other approaches, see here and here). We want to prove that around every point $p in M$ there exist a local frame for $bigwedge^k(T^*M)$ whose elements are closed and co-closed.
For the Euclidean metric this is immediate: We have the standard (constant) frame $dx^I=dx^{i_1} wedge ldots dx^{i_k} $. Since every metric is locally close to being Euclidean on small neighbourhoods, the idea is to use an approximation argument:
Given a Riemannian metric $g$, we denote the space of $g$-harmonic forms of degree $k$ by $H^k_{g}$.
We view $H^k_{g}$, as a subspace of $Omega^k(M)$ which is "changing continuously" with the metric $g$. Suppose $g_{ep} to g_0$ in the $C^1$-norm where $g_0$ is the Euclidean metric; Then $H^k_{g_{ep}} to H^k_{g_0}$ in the following sense: there exist a family of bases of $H^k_{g_{ep}}$, which converges to a basis of $H^k_{g_{0}} $ in $C^1$; this basis of $H^k_{g_{0}}$ forms a local frame for $bigwedge^k(T^*M)$. Since being a frame is an open condition, those bases for $H^k_{g_{ep}}$ are local frames for sufficiently small $ep$.
For the full details, see Appendix A in my paper here.
Some more details:
Even though the claim is local, and the approximation scheme is also inspired by a local phenomena, the implementation of the proof is based on a combination of local and global arguments. The reason is that on a closed manifold, being closed and co-closed is equivalent to being harmonic, and the dimension of the space of harmonic forms is a finite number which is a topological invariant of the manifold; it does not depend on the chosen metric.
Thus, given a family of metrics $g_{ep} to g_0$ on a closed manifold $M$, we consider the behaviour of the finite-dimensional subspaces $H^k_{g_{ep}}$ (all of the same dimension) as $ep to 0$.
That is, we look at the map $g to H^k_{g}=ker Delta_g$. It turns out that this map is continuous in some appropriate sense; this relies on a certain "stability property of kernels of linear operators". It turns out that a crucial factor in the existence of such a stability phenomenon is the assumption that all the kernels have the same finite dimension. The convergence of kernels does not always hold when the dimensions are not equal or infinite.
answered Jan 22 at 13:55
Asaf ShacharAsaf Shachar
5,77531141
answered Jan 22 at 13:55
Asaf ShacharAsaf Shachar
5,77531141
answered Jan 22 at 13:55
Asaf ShacharAsaf Shachar
5,77531141
answered Jan 22 at 13:55
Asaf ShacharAsaf Shachar
5,77531141
5,77531141
add a comment |
add a comment |
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