Constructing an $L^2$ space on the unit ring $\mathcal{S^1}$

Question

Revised Question:

Starting with $L^2[0,2\pi]$, does the canonical map $$[0,2\pi)\ni\theta\mapsto e^{i\theta}\in\mathcal{S^1}$$(with functions going across in the obvious way) turn $L^2[\mathcal{S^1}]$ into a bona fide Hilbert space?

In particular, does the difference in topology between $[0,2\pi]$ and ${S^1}$ have any nasty implications?

Original Question:

Is the Hilbert Space of $L^2$ functions on $[0,2\pi]$ with $f(0)=f(2\pi)$ equivalent to a Hilbert Space of $L^2$ functions defined on the unit ring $\mathcal{S^1}$? Can I even construct the latter?

The reason I ask is that I'm uncertain whether the difference in the measure and topologies between $[0,2\pi]$ and $\mathcal{S^1}$ 'bubbles' its way up into functional analytic results.

Note: The boundary condition given is shorthand for: "The set of equivalence classes of Lebesgue square measurable functions (modulo sets of measure zero) containing a continuous member satisfying the given boundary condition".

Answer 1

$L^2$ spaces should not be sensitive to the topology or shape of whatever underlying space you're working with. Indeed, given a "manifold" (a generalization of circles and surfaces), one way of defining an $L^2$ space on it is to pick a chart $D^n \to M$, where $D$ is the unit disc, that is injective except at a set of measure zero, and then pull back functions to $D^n$; one says functions on $M$ are measurable if the pullback is measurable, $L^2$ if the pullback is $L^2$, etc.

Here is my favorite way of defining functions on a manifold $M$. The things you can integrate on manifolds aren't functions, they're differential forms; but we will hide this difficulty by picking a top-dimensional nonvanishing form $\omega$, and defining "$\int_M f$" $= \int_M f\omega$. For the circle, we're going to pick $\omega = d\theta$ (so that the integral looks like something you've seen before). Now I define $L^2(M)$ to be the completion of $C(M)$, the set of continuous functions on $M$, with respect to the $L^2$ norm $\|f\|^2 = \int_M f^2 \omega$. This does not exhibit $L^2(M)$ as an actual set of functions (modulo measure zero etc), but it does give a perfectly serviceable definition; I would say an $L^2$ function is just an element of this completion!

From this perspective, it is completely clear that $L^2(S^1)$ is the completion of $C_p([0,2\pi])$, where the subscript means I'm considering functions with $f(0) = f(2\pi)$. This is because there is a canonical isometry $C_p([0,2\pi]) \to C(S^1)$, the point being that $\int_{S^1} f^2 d\theta$ is the exact same thing as $\int_{[0,2\pi]} f^2 dx$ (just think of how the first term is defined in any way you've ever seen it defined). A similar argument shows that both of the definitions of $L^2(M)$ I gave are equivalent.

You could also go a straightforward route and say that a function on $M$ is measureable iff it's measureable in every chart, and if it's measurable, define $\int f$ by choosing a finite atlas of charts and a partition of unity with respect to this atlas and summing up the integrals of $\rho_\alpha f$ over each chart. But I find this aesthetically displeasing.

Answer 2

Of course as Hilbert spaces $L^2(S^1)$ and $L^2\bigl([0,1]\bigr)$ are isomorphic, and you could also say that $L^2\bigl([0,1]\bigr)$ is the prime example of a Hilbert space arising from Lebesgue theory.

But note that $L^2(S^1)$ is one of the most important Hilbert spaces in the world, and there definitively is an essential difference between $L^2\bigl([0,1]\bigr)$ and $L^2(S^1)$ which has not been exposed so far: While $[0,1]$ is just a measure space the ground set $S^1$ carries additional structure, namely a transitive group of translations. It is not necessary to go into details here. We all have heard of Fourier theory. E.g., there is a famous theorem saying that any function $f\in L^2(S^1)$ is represented (meaning "equal to") almost everywhere by its Fourier series.