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There are two classes of maps $S^1\to S^1$ for which I know how to define the winding number:

Continuous maps:
Using the unique path lifting property of the universal covering map $\mathbb R\to S^1$, any continuous map $\gamma:S^1\to S^1$ can be lifted to a path $\tilde\gamma:[0,1]\to \mathbb R$. The winding number of $\gamma$ is then defined by the formula $$W(\gamma)=\tilde \gamma(1)-\tilde \gamma(0).$$

Sobolev-1/2 maps:
If we view a map $\gamma:S^1 \to S^1$ as a special case of a function $S^1\to \mathbb C$, then the winding number is the algebraic area (I'm omitting all factors of $\pi$ here) enclosed by $\gamma$. By Stokes' theorem, this is the integral over $S^1$ of the 1-form $\gamma^*(-ydx+xdy)$. The latter is clearly quadratic in $\gamma$. If $\gamma=\sum_{n\in \mathbb Z}\gamma_nz^n$ is the Fourier series of $\gamma$, then each loop $z\mapsto \gamma_nz^n$ encloses an area of $n|\gamma_n|^2$ and the cross-terms don't contribute anything. Therefore, we get the formula $$ W(\gamma)=\sum_{n\in \mathbb Z}n|\gamma_n|^2. $$ At this point, recall that $\sum_{n\in \mathbb Z}|n+1||\gamma_n|^2$ is the definition of (the square of) the Sobolev-1/2 norm of $\gamma$. Therefore, $\gamma$ having finite Sobolev-1/2 norm is the obvious condition for the above sum to converge.


Now, it is well known that $$ \{\text{Continuous}\} \quad\not\subseteq\quad \{\text{Sobolev-}1/2\} $$ and $$ \{\text{Sobolev-}1/2\} \quad\not\subseteq\quad \{\text{Continuous}\}. $$ Thus my question:

Is there a reasonable class of maps $S^1\to S^1$ that contains both the continuous maps and Sobolev-1/2 maps, and on which the winding number makes sense?

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A class of maps including both continuous and $H^{1/2}$, where an extension is available, is Vanishing Mean Oscillation, VMO. This has been treated by several authors starting I think with Haïm Brezis. You can find quite a lot googling "degree theory for VMO maps".

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  • $\begingroup$ Does BMO with BMO-norm smaller than the radius of my circle also work? $\endgroup$ – André Henriques Apr 13 '14 at 0:34
  • $\begingroup$ Any measurable function with $\infty$-norm strictly less than $\text{radius}/2$ is in BMO and has BMO-norm strictly less than the radius. It'd be surprising that such a thing has a winding number! $\endgroup$ – Mariano Suárez-Álvarez Apr 13 '14 at 5:16
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    $\begingroup$ @Mariano Suárez-Alvarez: There are no $S^1$-valued functions with $\infty$-norm strictly less than radius/2. By definition, an $S^1$-valued function has its $\infty$-norm equal to the radius of the circle. $\endgroup$ – André Henriques Apr 13 '14 at 6:25

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