Laplacian on Projective plane The 2019 Stack Overflow Developer Survey Results Are InProjective Plane and Projective SpaceWhat is the volume of Complex Projective Space with Fubini-Study Metric?generalized warped productIsometries and geodesics in projective plane using coveringFinding the critical points of a quadratic form restricted to projective planeHermitian metric on $L=mathcalO_mathbbP^n(1)$ over the projective spaceComputing the Fubini-Study metric in affine chartUniformization of metrics vs. uniformization of Riemann surfacesOn hypersurfaces of the projective spaceConformal diffeomorphism of higher dimensional unit ball
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Laplacian on Projective plane
The 2019 Stack Overflow Developer Survey Results Are InProjective Plane and Projective SpaceWhat is the volume of Complex Projective Space with Fubini-Study Metric?generalized warped productIsometries and geodesics in projective plane using coveringFinding the critical points of a quadratic form restricted to projective planeHermitian metric on $L=mathcalO_mathbbP^n(1)$ over the projective spaceComputing the Fubini-Study metric in affine chartUniformization of metrics vs. uniformization of Riemann surfacesOn hypersurfaces of the projective spaceConformal diffeomorphism of higher dimensional unit ball
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The Laplacian of unit sphere with standard metric is well studied. Now I am wondering what is spectrum of Laplacian on Projective space. Since projection map is locally isometry, I guess they have same Laplacian operator, so do they have same Laplacian spectrum? This also sounds weird to me. Thanks for your help.
differential-geometry riemannian-geometry
asked Mar 25 at 18:16
H-HH-H
462210
$endgroup$
add a comment |
$begingroup$
The Laplacian of unit sphere with standard metric is well studied. Now I am wondering what is spectrum of Laplacian on Projective space. Since projection map is locally isometry, I guess they have same Laplacian operator, so do they have same Laplacian spectrum? This also sounds weird to me. Thanks for your help.
differential-geometry riemannian-geometry
asked Mar 25 at 18:16
H-HH-H
462210
$endgroup$
add a comment |
$begingroup$
The Laplacian of unit sphere with standard metric is well studied. Now I am wondering what is spectrum of Laplacian on Projective space. Since projection map is locally isometry, I guess they have same Laplacian operator, so do they have same Laplacian spectrum? This also sounds weird to me. Thanks for your help.
differential-geometry riemannian-geometry
asked Mar 25 at 18:16
H-HH-H
462210
$endgroup$
The Laplacian of unit sphere with standard metric is well studied. Now I am wondering what is spectrum of Laplacian on Projective space. Since projection map is locally isometry, I guess they have same Laplacian operator, so do they have same Laplacian spectrum? This also sounds weird to me. Thanks for your help.
differential-geometry riemannian-geometry
differential-geometry riemannian-geometry
asked Mar 25 at 18:16
H-HH-H
462210
asked Mar 25 at 18:16
H-HH-H
462210
asked Mar 25 at 18:16
H-HH-H
462210
asked Mar 25 at 18:16
H-HH-H
462210
asked Mar 25 at 18:16
H-HH-H
462210
462210
add a comment |
add a comment |
1 Answer
1
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Let $u$ be $Delta u=lambda u$ on the projective space $RP^n$, pull $u$ back to $S^n$
we get an eigenfunction $u^*$ with the same eigenvalue. One can find $alpha>0$ with
$alpha(alpha-n-1)=-lambda$, so that $p(x)=r^alpha u^*(theta)$ is harmonic on $mathbb R^n+1-0$, where $(r, theta)$ is the polar coordinate on $mathbb R^n+1$. One can check this using separation of variables.
$p$ must be a homogeneous polynomial of degree $alpha$ (first by removable singularity theorem we see $p$ is actually harmonic on the whole $mathbb R^n+1$, then one can use Gilbarg-Trudinger Theorem 2.10 to show $p$ is a polynomial) which turns out to be an integer. Now by the construction of $u^*$ we see $p$ is even, i.e. $p(-x)=p(x)$. Thus $alpha$ is an even integer.
So the spectrum of $RP^n$ is a part of the spectrum of $S^n$, missing those $lambda$ that correspond to odd $alpha$.
answered Mar 30 at 23:44
Yu DingYu Ding
7187
$endgroup$
add a comment |
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1 Answer
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active
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votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $u$ be $Delta u=lambda u$ on the projective space $RP^n$, pull $u$ back to $S^n$
we get an eigenfunction $u^*$ with the same eigenvalue. One can find $alpha>0$ with
$alpha(alpha-n-1)=-lambda$, so that $p(x)=r^alpha u^*(theta)$ is harmonic on $mathbb R^n+1-0$, where $(r, theta)$ is the polar coordinate on $mathbb R^n+1$. One can check this using separation of variables.
$p$ must be a homogeneous polynomial of degree $alpha$ (first by removable singularity theorem we see $p$ is actually harmonic on the whole $mathbb R^n+1$, then one can use Gilbarg-Trudinger Theorem 2.10 to show $p$ is a polynomial) which turns out to be an integer. Now by the construction of $u^*$ we see $p$ is even, i.e. $p(-x)=p(x)$. Thus $alpha$ is an even integer.
So the spectrum of $RP^n$ is a part of the spectrum of $S^n$, missing those $lambda$ that correspond to odd $alpha$.
answered Mar 30 at 23:44
Yu DingYu Ding
7187
$endgroup$
add a comment |
$begingroup$
Let $u$ be $Delta u=lambda u$ on the projective space $RP^n$, pull $u$ back to $S^n$
we get an eigenfunction $u^*$ with the same eigenvalue. One can find $alpha>0$ with
$alpha(alpha-n-1)=-lambda$, so that $p(x)=r^alpha u^*(theta)$ is harmonic on $mathbb R^n+1-0$, where $(r, theta)$ is the polar coordinate on $mathbb R^n+1$. One can check this using separation of variables.
$p$ must be a homogeneous polynomial of degree $alpha$ (first by removable singularity theorem we see $p$ is actually harmonic on the whole $mathbb R^n+1$, then one can use Gilbarg-Trudinger Theorem 2.10 to show $p$ is a polynomial) which turns out to be an integer. Now by the construction of $u^*$ we see $p$ is even, i.e. $p(-x)=p(x)$. Thus $alpha$ is an even integer.
So the spectrum of $RP^n$ is a part of the spectrum of $S^n$, missing those $lambda$ that correspond to odd $alpha$.
answered Mar 30 at 23:44
Yu DingYu Ding
7187
$endgroup$
add a comment |
$begingroup$
Let $u$ be $Delta u=lambda u$ on the projective space $RP^n$, pull $u$ back to $S^n$
we get an eigenfunction $u^*$ with the same eigenvalue. One can find $alpha>0$ with
$alpha(alpha-n-1)=-lambda$, so that $p(x)=r^alpha u^*(theta)$ is harmonic on $mathbb R^n+1-0$, where $(r, theta)$ is the polar coordinate on $mathbb R^n+1$. One can check this using separation of variables.
$p$ must be a homogeneous polynomial of degree $alpha$ (first by removable singularity theorem we see $p$ is actually harmonic on the whole $mathbb R^n+1$, then one can use Gilbarg-Trudinger Theorem 2.10 to show $p$ is a polynomial) which turns out to be an integer. Now by the construction of $u^*$ we see $p$ is even, i.e. $p(-x)=p(x)$. Thus $alpha$ is an even integer.
So the spectrum of $RP^n$ is a part of the spectrum of $S^n$, missing those $lambda$ that correspond to odd $alpha$.
answered Mar 30 at 23:44
Yu DingYu Ding
7187
$endgroup$
Let $u$ be $Delta u=lambda u$ on the projective space $RP^n$, pull $u$ back to $S^n$
we get an eigenfunction $u^*$ with the same eigenvalue. One can find $alpha>0$ with
$alpha(alpha-n-1)=-lambda$, so that $p(x)=r^alpha u^*(theta)$ is harmonic on $mathbb R^n+1-0$, where $(r, theta)$ is the polar coordinate on $mathbb R^n+1$. One can check this using separation of variables.
$p$ must be a homogeneous polynomial of degree $alpha$ (first by removable singularity theorem we see $p$ is actually harmonic on the whole $mathbb R^n+1$, then one can use Gilbarg-Trudinger Theorem 2.10 to show $p$ is a polynomial) which turns out to be an integer. Now by the construction of $u^*$ we see $p$ is even, i.e. $p(-x)=p(x)$. Thus $alpha$ is an even integer.
So the spectrum of $RP^n$ is a part of the spectrum of $S^n$, missing those $lambda$ that correspond to odd $alpha$.
answered Mar 30 at 23:44
Yu DingYu Ding
7187
edited Mar 30 at 23:52
answered Mar 30 at 23:44
Yu DingYu Ding
7187
answered Mar 30 at 23:44
Yu DingYu Ding
7187
answered Mar 30 at 23:44
Yu DingYu Ding
7187
7187
add a comment |
add a comment |
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