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?


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".

  • $\begingroup$ 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
  • 4
    $\begingroup$ @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

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.