Reference for the relation of the Casimir element to the Laplace Beltrami operator
$begingroup$
Wikipedia says,
"If $G$ is a Lie group with Lie algebra $mathfrak {g}$, the choice of an invariant bilinear form on $mathfrak {g}$ corresponds to a choice of bi-invariant Riemannian metric on $G$. Then under the identification of the universal enveloping algebra of $mathfrak {g}$ with the left invariant differential operators on $G$, the Casimir element of the bilinear form on $mathfrak {g}$ maps to the Laplacian of $G$ (with respect to the corresponding bi-invariant metric)."
Actually, a stronger claim is true for homogeneous spaces (and not only for the Lie group itself).
Do you have a reference to this claim, or even better, to a version of a stronger claim regarding homogeneous spaces?
reference-request lie-groups lie-algebras laplacian
asked Dec 3 '18 at 9:23
Racheli YovelRacheli Yovel
61
$endgroup$
add a comment |
$begingroup$
Wikipedia says,
"If $G$ is a Lie group with Lie algebra $mathfrak {g}$, the choice of an invariant bilinear form on $mathfrak {g}$ corresponds to a choice of bi-invariant Riemannian metric on $G$. Then under the identification of the universal enveloping algebra of $mathfrak {g}$ with the left invariant differential operators on $G$, the Casimir element of the bilinear form on $mathfrak {g}$ maps to the Laplacian of $G$ (with respect to the corresponding bi-invariant metric)."
Actually, a stronger claim is true for homogeneous spaces (and not only for the Lie group itself).
Do you have a reference to this claim, or even better, to a version of a stronger claim regarding homogeneous spaces?
reference-request lie-groups lie-algebras laplacian
asked Dec 3 '18 at 9:23
Racheli YovelRacheli Yovel
61
$endgroup$
$begingroup$
I suspect that this is, essentially, the familiar claim that the Laplacian on $mathbb R^n$ is the "only" differential operator that commutes with rototranslations. (The quotes are due to the fact that polynomials in the Laplacian also satisfy the property). Just a thought.
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 9:27
add a comment |
$begingroup$
Wikipedia says,
"If $G$ is a Lie group with Lie algebra $mathfrak {g}$, the choice of an invariant bilinear form on $mathfrak {g}$ corresponds to a choice of bi-invariant Riemannian metric on $G$. Then under the identification of the universal enveloping algebra of $mathfrak {g}$ with the left invariant differential operators on $G$, the Casimir element of the bilinear form on $mathfrak {g}$ maps to the Laplacian of $G$ (with respect to the corresponding bi-invariant metric)."
Actually, a stronger claim is true for homogeneous spaces (and not only for the Lie group itself).
Do you have a reference to this claim, or even better, to a version of a stronger claim regarding homogeneous spaces?
reference-request lie-groups lie-algebras laplacian
asked Dec 3 '18 at 9:23
Racheli YovelRacheli Yovel
61
$endgroup$
Wikipedia says,
"If $G$ is a Lie group with Lie algebra $mathfrak {g}$, the choice of an invariant bilinear form on $mathfrak {g}$ corresponds to a choice of bi-invariant Riemannian metric on $G$. Then under the identification of the universal enveloping algebra of $mathfrak {g}$ with the left invariant differential operators on $G$, the Casimir element of the bilinear form on $mathfrak {g}$ maps to the Laplacian of $G$ (with respect to the corresponding bi-invariant metric)."
Actually, a stronger claim is true for homogeneous spaces (and not only for the Lie group itself).
Do you have a reference to this claim, or even better, to a version of a stronger claim regarding homogeneous spaces?
reference-request lie-groups lie-algebras laplacian
reference-request lie-groups lie-algebras laplacian
asked Dec 3 '18 at 9:23
Racheli YovelRacheli Yovel
61
asked Dec 3 '18 at 9:23
Racheli YovelRacheli Yovel
61
asked Dec 3 '18 at 9:23
Racheli YovelRacheli Yovel
61
asked Dec 3 '18 at 9:23
Racheli YovelRacheli Yovel
61
asked Dec 3 '18 at 9:23
Racheli YovelRacheli Yovel
61
61
$begingroup$
I suspect that this is, essentially, the familiar claim that the Laplacian on $mathbb R^n$ is the "only" differential operator that commutes with rototranslations. (The quotes are due to the fact that polynomials in the Laplacian also satisfy the property). Just a thought.
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 9:27
add a comment |
$begingroup$
I suspect that this is, essentially, the familiar claim that the Laplacian on $mathbb R^n$ is the "only" differential operator that commutes with rototranslations. (The quotes are due to the fact that polynomials in the Laplacian also satisfy the property). Just a thought.
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 9:27
$begingroup$
I suspect that this is, essentially, the familiar claim that the Laplacian on $mathbb R^n$ is the "only" differential operator that commutes with rototranslations. (The quotes are due to the fact that polynomials in the Laplacian also satisfy the property). Just a thought.
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 9:27
$begingroup$
I suspect that this is, essentially, the familiar claim that the Laplacian on $mathbb R^n$ is the "only" differential operator that commutes with rototranslations. (The quotes are due to the fact that polynomials in the Laplacian also satisfy the property). Just a thought.
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 9:27
add a comment |
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$begingroup$
I suspect that this is, essentially, the familiar claim that the Laplacian on $mathbb R^n$ is the "only" differential operator that commutes with rototranslations. (The quotes are due to the fact that polynomials in the Laplacian also satisfy the property). Just a thought.
$endgroup$
– Giuseppe Negro
Dec 3 '18 at 9:27