Please forgive me for my inability to pose a mathematics question properly.

In one dimension the eigenfunctions of Laplacian (simply double derivative) are also eigenfunctions of its square root (single derivative). Is anything known regarding a possible generalisation to an arbitrary Riemannian manifold? Square root of the Laplacian is essentially a single derivative. Therefore, if such eigenfunctions exists or are known, then they will be exponential in some sense.

Even if it's not known in a fully general context, are there special situations where such eigenfunctions can be constructed?

Thanks in advance.

  • 2
    They have to be the same. For "functiona calculus" reasons. – András Bátkai Nov 14 at 13:25
  • 4
    The Dirac operator on a spin manifold is a far-reaching generalisation of the relationship between the one-dimensional laplacian and $d/dx$. – José Figueroa-O'Farrill Nov 14 at 13:27
  • 2
    There is a scalar square root of the Laplacian only it is no longer a differential operator, but a pseudodifferential operator. It has the same eigenfunctions as the Laplacian and the eigenvalues are the square roots of the eigenvalue s of the Laplacian. To get a differential operator you need to give up working with scalar functions and instead work with sections of (certain) vector bundles. – Liviu Nicolaescu Nov 14 at 21:12
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    Functional calculus is a good search term. – Neal Nov 14 at 21:25
  • Thank you all for your comments. I will look at it more carefully and then get back if needed. – Partha Mukhopadhyay Nov 19 at 8:10

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