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We can define the degree of a function $f \in H^{\frac{1}{2}}(\mathbb{S}^1,\mathbb{S}^1)$ as $$\mathrm{deg} \hspace{1mm} f = \frac{1}{2\pi i} \int_{\mathbb{S}^1} f^{-1} \frac{\partial f}{\partial \theta} d\theta$$ What is the meaning of $f^{-1}$ and $\frac{\partial f}{\partial \theta}$ ?

I read somewhere that $f^{-1} = \bar{f} \in H^{\frac{1}{2}}(\mathbb{S}^1,\mathbb{S}^1)$ and $\frac{\partial f}{\partial \theta} \in H^{-\frac{1}{2}}(\mathbb{S}^1,\mathbb{S}^1)$. Why can we say that ?

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This survey of Brezis appears to comprehensively answer these sorts of questions:… (see in particular Section 3). – Terry Tao Sep 3 '13 at 4:58
Unfortunately, they do not explain why $f^{-1} \in H^{\frac{1}{2}}$ and $\frac{\partial f}{\partial \theta} \in H^{-\frac{1}{2}}$. I checked some of the references and they all say the same thing, arguing that the integral is well defined as a scalar product in the duality between $H^{\frac{1}{2}}$ and $H^{-\frac{1}{2}}$. I believe I missed something about that. – Gatz' Sep 3 '13 at 5:24
The functions $f$ and $\partial f/\partial\theta$ only need to lie in $H^{1/2}(S^1,C)$ and $H^{-1/2}(S^1,C)$ (rather than $H^{1/2}(S^1,S^1)$ and $H^{-1/2}(S^1,S^1)$) to define an inner product $\int_{S^1} \overline{f}\partial f/\partial \theta\ d\theta$, and this is immediate from the Fourier series characterisation of $H^s(S^1,C)$ (as given in the reference I gave above). – Terry Tao Sep 3 '13 at 15:52
Thank you for your help. But I still don't understand why a function $f \in H^{\frac{1}{2}}$ admits a weak derivative $\frac{\partial f}{\partial \theta}$ in $H^{-\frac{1}{2}}$. I believe it has something to do with the Fourier series characterisation of $H^s$ (and the duality between $H^{\frac{1}{2}}$ and $H^{-\frac{1}{2}}$), and I would be thankful for any good reference on the subject. – Gatz' Sep 4 '13 at 15:01

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