Stationary phase, Laplace’s method, and the Fourier transform for Gaussian integrals
1 Critical points
Let be a nonempty open subset of and let be smooth. Then . For each , is the unique element of satisfying11 1 http://individual.utoronto.ca/jordanbell/notes/gradienthilbert.pdf
and is itself smooth. is the derivative of . One checks that
We call a critical point of when , and we denote the set of critical points of by . For and let denote the dimension of the kernel of , and we then define the Morse index of to be
In other words, is the number of negative eigenvalues of counted according to geometric multiplicity. We say that is nondegenerate when is invertible.
For self-adjoint and for , let be the dimension of the kernel of . Let , let , and let . Because is self-adjoint, . We define the signature of as . In other words, is the number of positive eigenvalues of counted according to geometric multiplicity minus the number of negative eigenvalues of counted according to geometric multiplicity.22 2 cf. Sylvester’s law of inertia, http://individual.utoronto.ca/jordanbell/notes/principalaxis.pdf
We can connect the notions of Morse index and signature. For , write . For to be a nondegenerate critical point means that is invertible and because is finite-dimensional this is equivalent to . Then which yields .
The Morse lemma33 3 Serge Lang, Differential and Riemannian Manifolds, p. 182, chapter VII, Theorem 5.1. states that if is a nondegenerate critical point of then there is an open subset of with and a -diffeomorphism , , such that
2 Stationary phase
Let be a nonempty connected open subset of , and let be smooth functions such that has compact support. Suppose that each is nondegenerate.44 4 In particular, is called a Morse function if it has no degenerate critical points, and in this case of course each is nondegenerate. The stationary phase approximation states that
as .55 5 Liviu Nicolaescu, An Invitation to Morse Theory, second ed., p. 183, Proposition 3.88.
Let be self-adjoint and invertible and define
We calculate , so . The Hessian of is , and because is invertible, is indeed a nondegenerate critical point of . Thus we have the following.
For a nonempty connected open subset of and for smooth functions such that has compact support and such that each is nondegenerate,
3 The Fourier transform
For self-adjoint, the spectral theorem tells us that are and an orthonormal basis for such that .
We call positive when it is self-adjoint and satisfies for all . In this case, the eigenvalues of are nonnegative, thus the signature of is . Suppose furthermore that is invertible, and let and . Then
For and , using the change of variables formula with the fact that and then using Fubini’s theorem,
and using we have
and so, as is invertible we get the following.
When is positive and invertible, for and we have
4 Gaussian integrals
Let be positive and invertible and let . As above,
which gives the following.77 7 cf. Gaussian measures on : http://individual.utoronto.ca/jordanbell/notes/gaussian.pdf
If is positive and invertible, then for ,
5 Laplace’s method
Let be the open ball in with center and radius and let be smooth, attain its minimum value only at , and satisfy for all . Let be smooth and for let
Laplace’s method88 8 Peter D. Miller, Applied Asymptotic Analysis, p. 92, Exercise 3.16 and R. Wong, Asymptotic Approximations of Integrals, p. 495, Theorem 3. tells us
Let be positive and invertible. Define by
Then as above , with which . We get the following from according Laplace’s method.
Let be positive and invertible and let be smooth. Then