# $H^1(\Omega)$ in Euclidean space vs $H^1(\Sigma)$ on a compact surface

For an open subset $$\Omega \subset \mathbb R^n$$ the Sobolev space $$H^1(\Omega)=W^{1,2}(\Omega)$$ is defined as $$$$H^1(\Omega)=\{ u \in L^2(\Omega) \, \vert \, \partial^{\alpha}u \in L^2(\Omega) \}$$$$ in distributional sense. I stumbled upon some literature on the construction of Sobolev spaces on manifolds that seem too technical for me to understand. Is there an "easy" derivation of the Sobolev space on a compact two-dimensional surface $$\Sigma \subset \mathbb R^3$$?

• Let $X = \Bbb{C}/L,L=\Bbb{Z}+i\Bbb{Z}$ a complex torus, for $(x,y) \in U= (0,3) \times (0,3) \subset \Bbb{R}^2$ let $\phi(x,y) = x+iy+L \in X$ then a function $f: X \to \Bbb{C}$ is in $H^k(X)$ iff $F = f \circ \phi : U \to \Bbb{C}$ and its partial derivatives $\partial_x^a \partial_y^b F,a+b \le k$ are in $L^2( U)$. The point is that this doesn't depend on the chosen smooth covering $\phi$, ie. with $\phi_2(x,y) = (x+1)^2+i (y+1)^2 +L$ then $H^k(X)$ would stay the same. Next $\partial_x F = \lim_{h \to 0} \frac{F(.+(h,0))-F}{h}$ limit taken in $L^2(U)$ (if it converges) – reuns Jun 1 at 16:10
• Thanks for your effort, but I don't understand connection between your given construction of a complex torus and my question due to my lack of knowledge.. – Tesla Jun 1 at 16:15
• So in "layman" terms, when does $H^k(X)$ stay the same as in Euclidean space, $X$ being a two dimensional surface? – Tesla Jun 1 at 16:16
• $\Bbb{C}/(\Bbb{Z}+i\Bbb{Z})$ is the simplest example of compact Riemann surface. It is a set together with a few "charts" from which we can define continuous, smooth, holomorphic, meromorphic, $L^2$ functions, exactly as I did. For a surface $S \subset \Bbb{R}^3$ it works the same way : you need a few charts (or cover) $\phi : U \subset \Bbb{R}^2\to S$ to make it a manifold and say $f : S\to \Bbb{C}$ is smooth iff $f \circ \phi : U \to \Bbb{C}$ is smooth – reuns Jun 1 at 16:20
• Thanks, will have a look into it – Tesla Jun 1 at 17:50