# how does the basis of an inner product space change when the domain is deformed

Assume we have a complete orthogonal system on a domain $D$, given by the eigenfunctions of the Laplacian on $D$. For example, the set $\{e^{int}\}$ on $[-\pi, \pi]$, or the spherical harmonics on the unit ball. Now consider a domain $D'$, which is "close" to D in some sense (the boundary of $D$ is close to the boundary of $D'$ in some suitable norm).

Are the eigenfunctions of the Laplacian on $D$ close, in some sense, to the eigenfunctions of the Laplacian on $D'$? Does knowing the basis of $D$ help approximate the basis of $D'$ ? Any known results along these or similar lines appreciated.

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This is touching on something interesting, but I think you need to be more precise, because there is no such thing as the basis for $L^2(D')$. It seems like you have some preferred basis in mind, but if so you need to pose your question more precisely – Yemon Choi Jan 24 '10 at 22:53
What do you mean by L^2 distances between boundaries of domains? – Jonas Meyer Jan 24 '10 at 22:55
I think fredjalves is thinking about the preferred basis for L^2(D) and L^2(D') from eigenfunctions of the Laplacian. At least that's what the two examples suggest. – j.c. Jan 24 '10 at 22:55
good point Yemon and jc. Yes, I was thinking of the basis from eigenfunctions of the Laplacian, not just any basis. Jonas: assuming $D$ and $D'$ are subsets of $R^n$, then their boundaries are functions on $R^{n-1}$. I'd like to say that those boundaries are close, with respect to some suitable norm (different from the the norm induced by the inner product on $R^n$) – fredjalves Jan 24 '10 at 23:30

## 1 Answer

The study of eigenfunctions and eigenvalues of the Laplacian is a well-developed field called spectral geometry. You might start with the first few lectures in the course notes here or the book by Craioveanu et al. "Old and new aspects in spectral geometry".

Perhaps some experts can point you to better introductory references.

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