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edited Sep 22 2011 at 8:26 |
Now i I see that in order for the first-order terms to vanish for all I have two questions at this point: Why is $df =-K\phi$ necessary for $E$ to vanish? More important (since that leads to the restrictions on the metric): Why is $*\phi$ dual to a Killing field? Maybe this is obvious but i don't see it.
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edited Sep 21 2011 at 10:19 |
$\alpha $\alpha \mapsto a \alpha + b (*\alpha) \qquad a,b \in \mathbb{R}$mathbb{R}$$
$\Delta^1 $\Delta^1 \alpha = \widehat{\Delta^1} \alpha + K\cdot \alpha$alpha$$
$\widehat{d^*} $\widehat{d^*} \ \widehat{\Delta^1} = \widehat{\Delta^0} \ \widehat{ d^* }$$$
Update: (in reply to the answer of Robert Bryant below):below) I have tried to spell out the calculations leading to some of the results in the answer below in some more detail in order to understand them by myself and for future reference. Unfortunately I am not familiar with the principal symbol calculus, so I apologize for my low-level approach. My goal is to calculate the eigenvalues for the Bochner Laplacian numerically. From the implementation point of view it is easier to deal with scalar valued second order equations than with vector (or in this case 1-form valued equations). I think the special case below is instructive. I wonder if it is possible to avoid the need of putting restrictions on the metric. Of course, that would be great. But it would be also interesting to have an argument that says that it is impossible in the general case. Anyway, In the special case$\widehat{\Delta^1} $\widehat{\Delta^1} \alpha := \nabla^*\nabla \alpha + L \alpha = \Delta^1 \alpha + (L-K) \alpha$,$\widehat{\Delta^0f} $\widehat{\Delta^0f} := \Delta^0 f + H f$, and$\widehat{d^*}\alpha $\widehat{d^*}\alpha :=d^* \alpha + \langle \phi,\alpha \rangle $$$\widehat{d^*} $\begin{aligned} E \alpha &:= \widehat{d^*} \widehat{\Delta^1}\alpha - \widehat{\Delta^0} \widehat{d^*} \alpha \\& = d^*(\Delta^1\alpha + (L-K)\alpha) \widehat{d^*}(\Delta^1\alpha+(L-K)\alpha)-\widehat{\Delta^0}(d^*\alpha + \langle \phi, phi,\alpha\rangle) \Delta^1\alpha \& =d^*\Delta^1\alpha+d^*((L-K)\alpha)+\langle \phi,\Delta^1\alpha \rangle + \langle \phi, (L-K)\alpha L-K) \alpha \rangle \\ & -\Delta^0 d^*\alpha -Hd^*\alpha -\Delta^0\langle \phi,\alpha \rangle -H \langle \phi,\alpha \rangle\end{aligned}$$ Because of the identities $ \widehat{\Delta^0} \widehat{d^*} begin{aligned}d^*( (L-K) \alpha ) &= (L-K)d^*\alpha -\langle d(L-K) ,\alpha \rangle \\\Delta^0 \langle \phi,\alpha \rangle & = \Delta^0(d^*\alpha +\langle langle \nabla^*\nabla \phi,\alpha \rangle) rangle + H (d^* \langle \phi,\nabla^*\nabla \alpha + \rangle - 2\langle \phi,\alpha nabla \rangle)$ Since we require $\widehat{d^*} phi, \widehat{\Delta^1}\alpha nabla \alpha \rangle \\\nabla^*\nabla &= \Delta^1 - K\\d^*\alpha &= -\nabla^i \alpha_i = -g_{ab} g^{ak} g^{bi} \widehat{\Delta^1} nabla_k\alpha_i = \widehat{d^*} langle -g, \alpha$nabla \alpha \rangle \\d^*\Delta^1&=\Delta^0 d^* \end{aligned} $$that the third and second order terms in $\alpha$ cancel, yielding thelast two equations implyfollowing first-order operator:$d^*((L-K)\alpha) + $E\alpha = \langle -g(L-K-H) + 2\nabla \phi, \Delta^1+(L-K)\alpha \rangle =\Delta^0\langle nabla \phi,\alpha alpha \rangle +H \langle (d^* L-K-H)\phi - d(L-K)- \alpha nabla^*\nabla \phi + K \langle phi, \phi,\alpha alpha \rangle)$ From here onwards it is not clear to me how to obtain the condition that rangle $K\phi$ has to be exact and deduce the other relationships $ Now i see that in order for the answer below. I assume it has first-order terms to follow from some manipulations of this formula involving vanish for all$\alpha$ we need $\nabla\phi = f g$ for $f := \frac{1}{2}(L-K-H)$.Taking the derivatives covariant derivative of the scalar products in it. As an alternative, applying this equation yields $*$ \nabla \nabla \phi = df \otimes g + df \otimes 0$ and integrating I arrived at taking the trace yields $\Delta^1 -\nabla^*\nabla \phi + (L-K-H)\phi =df$. Therefore, if we set $df = -dH$. @Robert: Could you please provide some hints in order K\phi$, the zeroth-order part reducesto point my thinking into the right direction?operator $$E\alpha = \langle 2f\phi -d(L-K),\alpha \rangle$$
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edited Sep 5 2011 at 15:21 |
Update: (in reply to answer of Robert Bryant below): In the special case$\widehat{\Delta^1} \alpha := \nabla^*\nabla \alpha + L \alpha = \Delta^1 \alpha + (L-K) \alpha$,$\widehat{\Delta^0f} := \Delta^0 f + H f$, and$\widehat{d^*}\alpha :=d^* \alpha + \langle \phi,\alpha \rangle $ we get$\widehat{d^*} \widehat{\Delta^1}\alpha = d^*(\Delta^1\alpha + (L-K)\alpha) + \langle \phi, \Delta^1\alpha + (L-K)\alpha \rangle $$\widehat{\Delta^0} \widehat{d^*} \alpha = \Delta^0(d^*\alpha +\langle \phi,\alpha \rangle) + H (d^* \alpha + \langle \phi,\alpha \rangle)$ Since we require $\widehat{d^*} \widehat{\Delta^1}\alpha = \widehat{\Delta^1} \widehat{d^*} \alpha$, the last two equations imply: $d^*((L-K)\alpha) + \langle \phi, \Delta^1+(L-K)\alpha \rangle =\Delta^0\langle \phi,\alpha \rangle + H (d^* \alpha + \langle \phi,\alpha \rangle)$
From here onwards it is not clear to me how to obtain the condition that $K\phi$ has to be exact and deduce the other relationships in the answer below. I assume it has to follow from some manipulations of this formula involving the derivatives of the scalar products in it. As an alternative, applying $*$ and integrating I arrived at $\Delta^1 \phi + (L-K-H)\phi = -dH$. @Robert: Could you please provide some hints in order to point my thinking into the right direction?
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edited Sep 5 2011 at 0:46 |
How to construct a scalar differntial differential operator having the same spectrum as a non-scalar differential operator exploiting symmetries?
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edited Aug 30 2011 at 18:05 |
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edited Aug 30 2011 at 15:34 |
Eigenvalue spectra of 1-Form and How to construct a scalar differntial operator having the same spectrum as a non-scalar differential operators on manifolds operator exploiting symmetries?
I am interested in eigenvalue problems for differential operators acting on one forms on closed two-dimensional manifolds and how they relate to eigenvalue problems of associated operators acting on scalar functions.
For the discussion I assume $M$ to be a compact two-dimensional Riemannian manifold without boundary. Let $\Delta^k=dd^*+d^*d$ be the De Rham Laplacian on the space $\Omega^k(M)$ of real-valued $k$-Forms. Here $d:\Omega^k(M) \to \Omega^{k+1}(M)$ is the exterior derivative and $d^*: \Omega^{k+1}(M) \to \Omega^{k}$ is its adjoint.
For $k=0$ we get the Laplace Beltrami Operator acting on functions: $\Delta^0=d^*d$.
Consider the eigenvalue problem: Find $\lambda \in \mathbb{R}$ and an $1$-Form $\alpha$ such that $\Delta^1\alpha = \lambda \alpha$. Because of the identity
$d^* \Delta^1 = \Delta^0 d^* $
any eigenform $\alpha$ of $\Delta^1$ yields an eigenfunction $f:= d^*\alpha$ of $\Delta^0$ for the same eigenvalue, provided that $\alpha$ is not co-closed.
Also, since $\Delta^1$ commutes with the Hodge star we have that with any eigenform $\alpha$, the $1$-form $*\alpha$ (imagine $\alpha$ being rotated pointwise by 90 degrees) is also a linearly independent eigenform for the same eigenvalue. Therefore all eigenspaces of $\Delta^1$ have even dimension and are invariant under the symmetry
$\alpha \mapsto (\cos \varphi) a \alpha + b (\sin *\alpha) \varphi) (*\alpha)$qquad a,b \in \mathbb{R}$
To sum up, the spectra of $\Delta^1$ and $\Delta^0$ are closely related: They are essentially the same exccept except for the multiplicities of the eigenvalues. More precisely, any non-zero eigenvalue of the scalar operator $\Delta^0$ with multiplicity $m$ becomes an eigenvalue of the $1$-form operator $\Delta^1$ with multiplicity $2m$.
Now for the question: Does a similar relation hold for other differential operators? For example, I consider the Bochner Laplacian $\widehat{\Delta^1} = \nabla^* \nabla$ on $T^*M$. The Weizenböck identity
$\Delta^1 \alpha = \widehat{\Delta^1} \alpha + K\cdot \alpha$
shows that the Bochner Laplacian and the De Rham Laplacian on $1$-Forms differ only by a an operator that is a pointwise multiplication with the Gaussian curvature $K$. Moreover, the Bochner Laplacian also commutes with the Hodge star $*$, therefore all eigenspaces of $\widehat{\Delta^1}$ have even dimension and are invariant under the symmetry mentioned above. So far, the situation looks similar. Now, is there an operator $\widehat{\Delta^0}$ acting on $\Omega^0(M)$ and an operator
$\widehat{d^*}: \Omega^1(M) \to \Omega^0(M)$ such that the following intertwining relation holds?
$\widehat{d^*} \ \widehat{\Delta^1} = \widehat{\Delta^0} \ \widehat{ d^* }$
In that case the the spectrum of the Bochner Laplacian on $1$-Forms would be essentially equal (up to multiplicity as discussed above) to the spectrum of the unknown scalar operator $\widehat{\Delta^0}$. Is this possible? More generally, for what other
operators (apart from the De Rham Laplacian) is such a "reduction" possible?
Any references would be appreciated.
Thanks.
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edited Aug 30 2011 at 14:57 |
I am interested in eigenvalue problems for differential operators acting on one forms on closed two-dimensional manifolds and how they relate to eigenvalue problems of associated operators acting on scalar functions.
For the discussion I assume $M$ to be a compact two-dimensional Riemannian manifold without boundary. Let $\Delta^k=dd^*+d^*d$ be the De Rham Laplacian on the space $\Omega^k(M)$ of real-valued $k$-Forms. Here $d:\Omega^k(M) \to \Omega^{k+1}(M)$ is the exterior derivative and $d^*: \Omega^{k+1}(M) \to \Omega^{k}$ is its adjoint.
For $k=0$ we get the Laplace Beltrami Operator acting on functions: $\Delta^0=d^*d$.
Consider the eigenvalue problem: Find $\lambda \in \mathbb{R}$ and an $1$-Form $\alpha$ such that $\Delta^1\alpha = \lambda \alpha$. Because of the identity
$d^* \Delta^1 = \Delta^0 d^* $
any eigenform $\alpha$ of $\Delta^1$ yields an eigenfunction $f:= d^*\alpha$ of $\Delta^0$ for the same eigenvalue, provided that $\alpha$ is not co-closed.
Also, since $\Delta^1$ commutes with the Hodge star we have that with any eigenform $\alpha$, the $1$-form $*\alpha$ (imagine $\alpha$ being rotated pointwise by 90 degrees) is also a linearly independent eigenform for the same eigenvalue. Therefore all eigenspaces of $\Delta^1$ have even dimension and are invariant under the symmetry
$\alpha \mapsto (\cos \varphi) \alpha + (\sin \varphi) (*\alpha)$
To sum up, the spectra of $\Delta^1$ and $\Delta^0$ are closely related: They are essentially the same exccept for the multiplicities of the eigenvalues. More precisely, any non-zero eigenvalue of the scalar operator $\Delta^0$ with multiplicity $m$ becomes an eigenvalue of the $1$-form operator $\Delta^1$ with multiplicity $2m$.
Now for the question: Does a similar relation hold for other differential operators? For example, I consider the Bochner Laplacian $\widehat{\Delta^1} = \nabla^* \nabla$ on $T^*M$. The Weizenboeck Weizenböck identity
$\Delta^1 \alpha = \widehat{\Delta^1} \alpha + K\cdot \alpha$
shows that the Bochner Laplacian and the De Rham Laplacian on $1$-Forms differ only by a an operator that is a pointwise multiplication with the Gaussian curvature $K$. Moreover, the Laplace Bochner operator Laplacian also commutes with the Hodge star $*$, therefore all eigenspaces of $\widehat{\Delta^1}$ have even dimension and are invariant under the symmetry mentioned above. So far, the situation looks similar. Now, is there an operator $\widehat{\Delta^0}$ acting on $\Omega^0(M)$ and an operator
$\widehat{d^*}: \Omega^1(M) \to \Omega^0(M)$ such that the following intertwining relation holds?
$\widehat{d^*} \ \widehat{\Delta^1} = \widehat{\Delta^0} \ \widehat{ d^* }$
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edited Aug 30 2011 at 14:47 |
I am interested in eigenvalue problems for differential operators acting on one forms on closed two-dimensional manifolds and how they relate to eigenvalue problems of associated operators acting on scalar functions.
For the discussion I assume $M$ to be a compact two-dimensional Riemannian manifold without boundary. Let $\Delta^k=dd^*+d^*d$ be the De Rham Laplacian on the space $\Omega^k(M)$of \Omega^k(M)$ of real-valued $k$-Forms. Here $d:\Omega^k(M) \to \Omega^{k+1}(M)$ is the exterior derivative and $d^*: \Omega^{k+1}(M) \to \Omega^{k}$ is its adjoint.
For $k=0$ we get the Laplace Beltrami Operator acting on functions: $\Delta^0=d^*d$.
Consider the eigenvalue problem: Find $\lambda \in \mathbb{R}$ and an $1$-Form $\alpha$ such that $\Delta^1\alpha = \lambda \alpha$. alpha$. Because of the identity
$\widehat{d^*}$$d^* \Delta^1 = \Delta^0 d^* $
any eigenform $\alpha$ of $\Delta^1$ yields an eigenfunction $f:= d^*\alpha$ of $\Delta^0$ for the same eigenvalue, provided that $\alpha$ is not co-closed.
Also, since $\Delta^1$ commutes with the Hodge star we have that with any eigenform $\alpha$, the $1$-form $\alpha$ *\alpha$ (imagine $\alpha$ being rotated pointwise by 90 degrees)is degrees) is also a linearly independent eigenform for the same eigenvalue. Therefore all eigenspaces of $\Delta^1$ have even dimension and are invariant under the symmetry
$\alpha \mapsto (\cos \varphi) \alpha + (\sin \varphi) (\alpha)$.*\alpha)$
To sum up, the spectra of $\Delta^1$ and $\Delta^0$ are closely related: They are essentially the same exccept for the multiplicities of the eigenvalues. More precisely, any non-zero eigenvalue of the scalar operator $\Delta^0$ with multiplicity $m$ becomes an eigenvalue of the $1$-form operator $\Delta^1$ with multiplicity $2m$.
Now for the question: Does a similar relation hold for other differential operators? For example, I consider the Bochner Laplacian $\widehat{\Delta^1} = \nabla^* \nabla$ on $T^*M$. The Weizenboeck identity
$\Delta^1 \alpha = \widehat{\Delta^1} \alpha + K\cdot \alpha$
shows that the Bochner Laplacian and the De Rham Laplacian on $1$-Forms differ only by a an operator that is a pointwise multiplication with the Gaussian curvature $K$. Moreover, the Laplace Bochner operator also commutes with the Hodge star $*$, therefore all eigenspaces of $\widehat{\Delta^1}$ have even dimension. So far, the situation looks similar. Now, is there an operator $\widehat{\Delta^0}$ acting on $\Omega^0(M)$ and an operator
$\widehat{d^*}: \Omega^1(M) \to \Omega^0(M)$ such that the following intertwining relation holds?
$\widehat{d^*} \ \widehat{\Delta^1} = \widehat{\Delta^0} \ \widehat{ d^* }$
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edited Aug 30 2011 at 14:40 |
I am interested in eigenvalue problems for differential operators acting on one forms on closed two-dimensional manifolds and how they relate to eigenvalue problems of associated operators acting on scalar functions.
For the discussion I assume $M$ to be a compact two-dimensional Riemannian manifold without boundary. Let $\Delta^k=dd^*+d^*d$ be the De Rham Laplacian on the space $\Omega^k(M)$of real-valued $k$-Forms.
Here $d:\Omega^k(M) \to \Omega^{k+1}(M)$ is the exterior derivative and
$d^*: \Omega^{k+1}(M) \to \Omega^{k}$ is its adjoint.
For $k=0$ we get the Laplace Beltrami Operator acting on functions: $\Delta^0=d^*d$.
Consider the eigenvalue problem: Find $\lambda \in \mathbb{R}$ and an $1$-Form $\alpha$ such that $\Delta^1\alpha = \lambda \alpha$. Because of the identity
$\widehat{d^*}$
$d^* \Delta^1 = \Delta^0 d^* $
any eigenform $\alpha$ of $\Delta^1$ yields an eigenfunction $f:= d^*\alpha$ of $\Delta^0$ for the same eigenvalue, provided that $\alpha$ is not co-closed.
Also, since $\Delta^1$ commutes with the Hodge star we have that with any eigenform $\alpha$, the $1$-form $\alpha$ (imagine $\alpha$ being rotated pointwise by 90 degrees)is also a linearly independent eigenform for the same eigenvalue. Therefore all eigenspaces of $\Delta^1$ have even dimension and are invariant under the symmetry
$\alpha \mapsto (\cos \varphi) \alpha + (\sin \varphi) (\alpha)$.
To sum up, the spectra of $\Delta^1$ and $\Delta^0$ are closely related: They are essentially the same exccept for the multiplicities of the eigenvalues. More precisely, any non-zero eigenvalue of the scalar operator $\Delta^0$ with multiplicity $m$ becomes an eigenvalue of the $1$-form operator $\Delta^1$ with multiplicity $2m$.
Now for the question: Does a similar relation hold for other differential operators? For example, I consider the Bochner Laplacian $\widehat{\Delta^1} = \nabla^* \nabla$ on $T^*M$. T^*M$. The Weizenboeck identity
$\Delta^1 \alpha = \widehat{\Delta^1} \alpha + K\cdot \alpha$
shows that the Bochner Laplacian and the De Rham Laplacian on $1$-Forms differ only by a an operator that is a pointwise multiplication with the Gaussian curvature $K$. Moreover, the Laplace Bochner operator also commutes with the Hodge star $*$, therefore all eigenspaces of $\widehat{\Delta^1}$ have even dimension. So far, the situation looks similar.
Now, is there an operator $\widehat{\Delta^0}$ acting on $\Omega^0(M)$ and an operator
$\widehat{d^*}: \Omega^1(M) \to \Omega^0(M)$ such that the following intertwining relation holds?
$\widehat{d^}\widehat{d^*} \ \widehat{\Delta^1} = \widehat{\Delta^1}\widehat{d^widehat{\Delta^0} \ \widehat{ d^* }$
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edited Aug 30 2011 at 14:35 |
I am interested in eigenvalue problems for differential operators acting on one forms on closed two-dimensional manifolds and how they relate to eigenvalue problems of associated operators acting on scalar functions.
For the discussion I assume $M$ to be a compact two-dimensional Riemannian manifold without boundary. Let $\Delta^k=dd^*+d^*d$ be the De Rham Laplacian on the space $\Omega^k(M)$of real-valued $k$-Forms.
Here $d:\Omega^k(M) \to \Omega^{k+1}(M)$ is the exterior derivative and
$d^*: \Omega^{k+1}(M) \to \Omega^{k}$ is its adjoint.
For $k=0$ we get the Laplace Beltrami Operator acting on functions: $\Delta^0=d^*d$.
Consider the eigenvalue problem: Find $\lambda \in \mathbb{R}$ and an $1$-Form $\alpha$ such that $\Delta^1\alpha = \lambda \alpha$. Because of the identity
$\widehat{d^*}$
$d^* \Delta^1 = \Delta^0 d^* $
any eigenform $\alpha$ of $\Delta^1$ yields an eigenfunction $f:= d^*\alpha$ of $\Delta^0$ for the same eigenvalue, provided that $\alpha$ is not co-closed.
Also, since $\Delta^1$ commutes with the Hodge star $$ we have that with any eigenform $\alpha$, the $1$-form $\alpha$ (imagine $\alpha$ being rotated pointwise by $90$ 90 degrees)is also a linearly independent eigenform for the same eigenvalue. Therefore all eigenspaces of $\Delta^1$ have even dimension and are invariant under the symmetry
$\alpha \mapsto (\cos \varphi) \alpha + (\sin \varphi) (*\alpha)$.\alpha)$.
To sum up, the spectra of $\Delta^1$ and $\Delta^0$ are closely related: They are essentially the same exccept for the multiplicities of the eigenvalues. More precisely, any non-zero eigenvalue of the scalar operator $\Delta^0$ with multiplicity $m$ becomes an eigenvalue of the $1$-form operator $\Delta^1$ with multiplicity $2m$.
Now for the question: Does a similar relation hold for other differential operators? For example, I consider the Bochner Laplacian $\widehat{\Delta^1} = \nabla^* \nabla$ on $T^*M$.
The Weizenboeck identity
$\Delta^1 \alpha = \widehat{\Delta^1} \alpha + K\cdot \alpha$
shows that the Bochner Laplacian and the De Rham Laplacian on $1$-Forms differ only by a an operator that is a pointwise multiplication with the Gaussian curvature $K$. Moreover, the Laplace Bochner operator also commutes with the Hodge star $*$, therefore all eigenspaces of $\widehat{\Delta^1}$ have even dimension. So far, the situation looks similar.
Now, is there an operator $\widehat{\Delta^0}$ acting on $\Omega^0(M)$ and an operator
$\widehat{d^*}: \Omega^1(M) \to \Omega^0(M)$ such that the following intertwining relation holds?
$\widehat{d^}\ \widehat{\Delta^1} = \widehat{\Delta^1}\widehat{d^}$
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edited Aug 30 2011 at 14:29 |
I am interested in eigenvalue problems for differential operators acting on one forms on closed two-dimensional manifolds and how they relate to eigenvalue problems of associated operators acting on scalar functions.
For the discussion I assume $M$ to be a compact two-dimensional Riemannian manifold without boundary. Let $\Delta^k=dd^*+d^*d$ be the De Rham Laplacian on the space $\Omega^k(M)$of real-valued $k$-Forms.
Here $d:\Omega^k(M) \to \Omega^{k+1}(M)$ is the exterior derivative and
$d^*: \Omega^{k+1}(M) \to \Omega^{k}$ is its adjoint.
For $k=0$ we get the Laplace Beltrami Operator acting on functions: $\Delta^0=d^*d$.
Consider the eigenvalue problem: Find $\lambda \in \mathbb{R}$ and an $1$-Form $\alpha$ such that $\Delta^1\alpha = \lambda \alpha$. Because of the identity
$\widehat{d^*}$
$d^* \Delta^1 = \Delta^0 d^* $
any eigenform $\alpha$ of $\Delta^1$ yields an eigenfunction $f:= d^*\alpha$ of $\Delta^0$ for the same eigenvalue, provided that $\alpha$ is not co-closed.
Also, since $\Delta^1$ commutes with the Hodge star $$ we have that with any eigenform $\alpha$, the $1$-form $\alpha$ (imagine $\alpha$ being rotated pointwise by $90$ degrees)is also a linearly independent eigenform for the same eigenvalue. Therefore all eigenspaces of $\Delta^1$ have even dimension and are invariant under the symmetry
$\alpha \mapsto (\cos \varphi) \alpha + (\sin \varphi) (*\alpha)$.
To sum up, the spectra of $\Delta^1$ and $\Delta^0$ are closely related: They are essentially the same exccept for the multiplicities of the eigenvalues. More precisely, any non-zero eigenvalue of the scalar operator $\Delta^0$ with multiplicity $m$ becomes an eigenvalue of the $1$-form operator $\Delta^1$ with multiplicity $2m$.
Now for the question: Does a similar relation hold for other differential operators? For example, I consider the Bochner Laplacian $\widehat{\Delta^1} = \nabla^* \nabla$ on $T^*M$.
The Weizenboeck identity :
$\Delta^1 \alpha = \widehat{\Delta^1} \alpha + K\cdot \alpha$
shows that the Bochner Laplacian and the De Rham Laplacian on $1$-Forms differ only by a an operator that is a pointwise multiplication with the Gaussian curvature $K$. Moreover, the Laplace Bochner operator also commutes with the Hodge star $*$, therefore all eigenspaces of $\widehat{\Delta^1}$ have even dimension. So far, the situation looks similar.
Now, is there an operator $\widehat{\Delta^0}$ acting on $\Omega^0(M)$ and an operator
$\widehat{d^*}: \Omega^1(M) \to \Omega^0(M)$ such that the following intertwining relation holds?
$\widehat{d^}\ widehat{\Delta^1}=\widehat{\Delta^0} \widehat{d^widehat{\Delta^1} = \widehat{\Delta^1}\widehat{d^}$
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edited Aug 30 2011 at 14:19 |
I am interested in eigenvalue problems for differential operators acting on one forms on closed two-dimensional manifolds and how they relate to eigenvalue problems of associated operators acting on scalar functions.
For the discussion I assume $M$ to be a compact two-dimensional Riemannian Manifold manifold without boundary. Let ${\Delta^k = dd^* + d^d}$
\Delta^k=dd^*+d^*d$
be the De Rham Laplacian on the space
$\Omega^k(M)$ of \Omega^k(M)$of real-valued $k$-Forms.
Here $d:\Omega^k(M) \to \Omega^{k+1}(M)$ is the exterior derivative and
$d^: d^*: \Omega^{k+1}(M) \to \Omega^{k}$, $d^*=-d$ Omega^{k}$ is its adjoint, $*$ being the Hodge-Star.
For $k=0$ we get the Laplace Beltrami Operator acting on functions: $\Delta^0=d^*d$.
Consider the eigenvalue problem: Find $\lambda \in \mathbb{R}$ and an $1$-Form $\alpha$ such that $\Delta^1\alpha = \lambda \alpha$. Because of the identity\begin{equation} \label{eq:intertwine1}
d^{}
$d^* \Delta^1 = \Delta^0 d^{}
\end{equation}
* $
any eigenform $\alpha$ of $\Delta^1$ yields an eigenfunction $f:= d^\alpha$ d^*\alpha$ of $\Delta^0$ for the same eigenvalue, provided that $\alpha$ is not co-closed.
Also, since $\Delta^1$ commutes with the Hodge star $$ we have that with any eigenform $\alpha$, the $1$-form $\alpha$ (imagine $\alpha$ being rotated pointwise by $90$ degrees)is also a linearly independent eigenform for the same eigenvalue. Therefore all eigenspaces of $\Delta^1$ have even dimension and are invariant under the symmetry
$\alpha \mapsto (\cos \varphi) \alpha + (\sin \varphi) (\alpha)$.*\alpha)$.
To sum up, the spectra of $\Delta^1$ and $\Delta^0$ are closely related: They are essentially the same exccept for the multiplicities of the eigenvalues. More precisely, any non-zero eigenvalue of the scalar operator $\Delta^0$ with multiplicity $m$ becomes an eigenvalue of the $1$-form operator $\Delta^1$ with multiplicity $2m$.
Now for the \textbf{question}: question: Does a similar relation hold for other differential operators? For example, I consider the Bochner Laplacian $\widehat{\Delta^1} = \nabla^* \nabla$
where $\nabla$ is connection on $T^M$ induced by the Levi-Civita connection of $M$. T^*M$.
The Weizenböck
Weizenboeck identity\begin{equation}
\Delta^1 :
$\Delta^1 \alpha = \widehat{\Delta^1} \alpha + K\cdot \alpha
\end{equation*}
alpha$
shows that the Bochner Laplacian and the De Rham Laplacian on $1$-Forms differ only by a zeroth order an operator that is a pointwise multiplication with the Gaussian curvature $K$. Moreover, the Laplace Bochner operator also commutes with the Hodge star $*$, therefore all eigenspaces of $\widehat{\Delta^1}$ have even dimension. So far, the situation looks similar.
Now, is there an operator $\widehat{\Delta^0}$ acting on $\Omega^0(M)$ and an operator
$\widehat{d^}: \widehat{d^*}: \Omega^1(M) \to \Omega^0(M)$
such that the following intertwining relation similar to eq.~\eqref{eq:intertwine1}
holds?\begin{equation}
\widehat{d^
$\widehat{d^} \widehat{\Delta^1} = \widehat{\Delta^0} widehat{\Delta^1}=\widehat{\Delta^0} \widehat{d^}\end{equation*}$
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asked Aug 30 2011 at 14:00 |
Eigenvalue spectra of 1-Form and scalar differential operators on manifolds exploiting symmetries
I am interested in eigenvalue problems for differential operators
acting on one forms on closed two-dimensional manifolds and how they
relate to eigenvalue problems of associated operators acting on scalar
functions.
For the discussion I assume $M$ to be a compact two-dimensional
Riemannian Manifold without boundary. Let
${\Delta^k = dd^* + d^d}$
be the De Rham Laplacian on the space $\Omega^k(M)$ of real-valued
$k$-Forms. Here $d:\Omega^k(M) \to \Omega^{k+1}(M)$ is the exterior
derivative and $d^: \Omega^{k+1}(M) \to \Omega^{k}$, $d^*=-d$ is
its adjoint, $*$ being the Hodge-Star. For $k=0$ we get the Laplace
Beltrami Operator acting on functions: $\Delta^0=d^*d$.
Consider the eigenvalue problem: Find $\lambda \in \mathbb{R}$ and an
$1$-Form $\alpha$ such that $\Delta^1\alpha = \lambda \alpha$.
Because of the identity
\begin{equation} \label{eq:intertwine1}
d^{} \Delta^1 = \Delta^0 d^{}
\end{equation}
any eigenform $\alpha$ of $\Delta^1$ yields an eigenfunction $f:=
d^\alpha$ of $\Delta^0$ for the same eigenvalue, provided that
$\alpha$ is not co-closed. Also, since $\Delta^1$ commutes with the
Hodge star $$ we have that with any eigenform $\alpha$, the $1$-form
$\alpha$ (imagine $\alpha$ being rotated pointwise by $90$ degrees)is
also a linearly independent eigenform for the same eigenvalue.
Therefore all eigenspaces of $\Delta^1$ have even dimension and are
invariant under the symmetry $\alpha \mapsto (\cos \varphi) \alpha + (\sin \varphi) (\alpha)$.
To sum up, the spectra of $\Delta^1$ and $\Delta^0$ are closely
related: They are essentially the same exccept for the multiplicities
of the eigenvalues. More precisely, any non-zero eigenvalue of the
scalar operator $\Delta^0$ with multiplicity $m$ becomes an eigenvalue
of the $1$-form operator $\Delta^1$ with multiplicity $2m$.
Now for the \textbf{question}: Does a similar relation hold for other
differential operators? For example, I consider the Bochner Laplacian
$\widehat{\Delta^1} = \nabla^* \nabla$ where $\nabla$ is connection on
$T^M$ induced by the Levi-Civita connection of $M$. The Weizenböck
identity
\begin{equation}
\Delta^1 \alpha = \widehat{\Delta^1} \alpha + K\cdot \alpha
\end{equation*}
shows that the Bochner Laplacian and the De Rham Laplacian on
$1$-Forms differ only by a zeroth order operator that is a pointwise
multiplication with the Gaussian curvature $K$. Moreover, the Laplace
Bochner operator also commutes with the Hodge star $*$, therefore all
eigenspaces of $\widehat{\Delta^1}$ have even dimension. So far, the
situation looks similar.
Now, is there an operator $\widehat{\Delta^0}$ acting on $\Omega^0(M)$
and an operator $\widehat{d^}: \Omega^1(M) \to \Omega^0(M)$ such that
the following intertwining relation similar to eq.~\eqref{eq:intertwine1}
holds?
\begin{equation}
\widehat{d^} \widehat{\Delta^1} = \widehat{\Delta^0} \widehat{d^}
\end{equation*}
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