# How is orthonormal basis $\{e^{2πikt}\}_{k=-\infty}^\infty$ with index going to $-\infty$ constructed in $L^2(\Bbb S^1)$?

Orthogonal basises can be constructed in any Hilbert space $L^2(Q)$ in any interval $Q$ by using any countable dense family of function (the Gram-Schmidt recipe), and they are indexed from $1$ to $\infty$. Depending on the Lie group, we have them as special functions:

But how does the basis $\{e^{2πikt}\}_{k=-\infty}^\infty$ with index going to $-\infty$ constructed in $L^2(\Bbb S^1)$? This is given at the beginning of any text as an example of orthonormality, but none of them explains how the basis is constructed. I suspect it's related to the fact that the eigenvalues of the generator of the group $SO(2)$ must be in $\Bbb Z$ due to the restriction of the rotation group in $\Bbb R^2$ ($J|m\rangle=|m\rangle m, m\in\mathbb Z$). The irreducible representation of $SO(2)$ is $U^m(\phi)=e^{-iφm}$ and from the Schur's lemma we can deduce the orthogonality and completeness, which is the Fourier series. However the translation group in the same space does not have this restriction, and in both case I don't see how $-\infty$ comes up.

• It doesn’t actually start at $-\infty$. $k$ just runs over all integers. – Qiaochu Yuan Nov 27 '17 at 5:41
• If $G$ is any compact abelian group, $L^2(G)$ has a basis consisting of the characters of $G$, which are the continuous homomorphisms $G \to S^1$. The characters form an abelian group called the Pontryagin dual of $G$ (en.wikipedia.org/wiki/Pontryagin_duality), and every abelian group arises this way. For the generalization to compact nonabelian groups see en.wikipedia.org/wiki/Peter%E2%80%93Weyl_theorem . – Qiaochu Yuan Nov 27 '17 at 6:12
• To add on what Qiaochu said : $\{e^{2i \pi n x}\}$ is the orthonormal basis diagonalizing (eigenfunctions) the convolution operators $f \mapsto T[f](x) = \int_0^1 f(x-y) h(y)dy$, those operators satistying $T[f(.+a)](x) = T[f](x+a)$. This generalizes in an obvious way in any compact abelian group. In non-abelian groups, the convolution is non-commutative and things get more complicated. In locally-compact groups, the eigenfunctions aren't in $L^2(G)$ (they are distributions) and things get more complicated (this is the topic of the spectral theorem and representation theory) – reuns Nov 27 '17 at 9:52
• Sure $\partial_{x_i}$ and $\Delta$, on $\mathbb{R}^n$ or $\mathbb{R}^n/\mathbb{Z}^n$, are convolution operators. Orthogonal basis is common to bounded normal/self-adjoint operators on separable Hilbert spaces (spectral theorem) – reuns Dec 2 '17 at 5:42
• @Ooker: you mean finite abelian groups, and no. – Qiaochu Yuan Dec 3 '17 at 2:48