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  Riemannian generalization/adaption of the Hubbard–Stratonovich transformation

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I'd like to write the Hubbard–Stratonovich (HS) transformation of a scalar function on a Riemannian manifold. This transformation is quite simple in Euclidean space. One can consider it as a Fourier transformation of a Gaussian function: $$ \exp\left(-\frac{a}{2}\frac{d^2}{dx^2}\right) f(x)~=~\int_{\mathbb{R}} \!\frac{dy}{\sqrt{2\pi a}}\exp\left(-\frac{y^2}{2a}\right) f(x+y), \qquad a~>~0. $$ However, my question is can whether one apply HS transformation to a scalar function on a compact Riemannian manifold? If yes, then how?

I couldn't understand how to define the Fourier transform on a compact Riemannian manifold. It seems to me the Fourier transform is defined by the Pontryagin duality but it goes a little technical and I'm not familiar with.. Nevertheless, I guess one can write the Hubbard-Stratonovich transform: $$ \exp\left(-\frac{a}{2}\Delta\right) f({x})~=~ (2\pi a)^{-\frac{n}{2}}\int d^n{y}\sqrt {g(y)}\exp\left(-\frac{{y}^2}{2a}\right) f({x+y})~, $$ where $\Delta$ is the Laplace–Beltrami operator. In addition, I read that the Weierstrass transform can be defined on any Riemannian manifold, but it doesn't make sense to me, I couldn't find any proper references and don't know how to write this transform.

Context:

My real problem involves the free energy of a harmonic oscillator on a Riemannian manifold which leads to an trasformation similar to the HS transformation. Indeed, potential energy of harmonic oscillator is $U=\frac{1}{2}kd(y,y_0)^2$ which $k$ is a constant, $d(y,y_0)^2$ is distance. I simply assumed that $d(y, 0)=y^2$. Under what conditions on the Riemannian manifold $M$ can I write a transformation similar to to the one mentioned above? As far as I understand, one can define the FT on a Riemannian manifold when it is either a Lie group or a symmetric space.

This post imported from StackExchange Physics at 2015-09-06 15:05 (UTC), posted by SE-user AFZQ
asked Sep 5, 2015 in Mathematics by AFZQ (10 points) [ no revision ]

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