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YCor
YCor
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Laplace Operatorsoperators that give $S^d$ Eigenvalueseigenvalues that are Perfect Squaresperfect squares

The Laplace-Beltrami operator on the sphere $S^d$ has eigenvalues $\{ k(k+d-1) : k \geq 0 \}$. Is there a geometrically natural Laplace operator / Laplace like operator (perhaps a Hodge Laplacian or a variant of these laplaciansLaplacians), defined on a general manifold that instead gives $S^d$ eigenvalues $\{ k^2 : k \geq 0 \}$?

Laplace Operators that give $S^d$ Eigenvalues that are Perfect Squares

The Laplace-Beltrami operator on the sphere $S^d$ has eigenvalues $\{ k(k+d-1) : k \geq 0 \}$. Is there a geometrically natural Laplace operator / Laplace like operator (perhaps a Hodge Laplacian or a variant of these laplacians), defined on a general manifold that instead gives $S^d$ eigenvalues $\{ k^2 : k \geq 0 \}$?

Laplace operators that give $S^d$ eigenvalues that are perfect squares

The Laplace-Beltrami operator on the sphere $S^d$ has eigenvalues $\{ k(k+d-1) : k \geq 0 \}$. Is there a geometrically natural Laplace operator / Laplace like operator (perhaps a Hodge Laplacian or a variant of these Laplacians), defined on a general manifold that instead gives $S^d$ eigenvalues $\{ k^2 : k \geq 0 \}$?

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Jacob Denson
Jacob Denson
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Laplace Operators that give $S^d$ Eigenvalues that are Perfect Squares

The Laplace-Beltrami operator on the sphere $S^d$ has eigenvalues $\{ k(k+d-1) : k \geq 0 \}$. Is there a geometrically natural Laplace operator / Laplace like operator (perhaps a Hodge Laplacian or a variant of these laplacians), defined on a general manifold that instead gives $S^d$ eigenvalues $\{ k^2 : k \geq 0 \}$?