# How are Sobolev spaces on compact Riemannian manifolds defined?

For an open subset $$\Omega\subset \mathbb R^n$$ one can define the Sobolev space

$$H^1(\Omega):=W^{1,2}(\Omega)=\{u \in L^2(\Omega) \, \vert \, \partial u \in L^2(\Omega)\}.$$

Is there a "simple" way to introduce the Sobolev space $$H^1$$ on a compact, Riemannian manifold? Is it equivalent tot he Euclidean case?

Sure. If $$M$$ is a compact manifold, and $$k$$ is a non-negative integer, then $$H^k(M)$$ is the space of function $$u\in L^2(M)$$ with the property that for any $$\ell$$ smooth vector fields $$X_1,\cdots, X_\ell$$ on $$M$$, with $$\ell\leq k$$, we have $$X_1\cdots X_\ell u\in L^2(M)$$. For a reference, you can see Michael Taylor's PDE I text.
• Thank you for the reference. So in our case, $H^1(M)$ is the space of functions $u \in L^2(M)$ such that for a vector field $X$ on $M$ we have $Xu \in L^2(M)$? since $k=1$? What does $Xu$ mean? – Tesla Jun 5 at 14:39
• Correct. As per your question, vector fields act on functions, right? So, if $X=a^j\frac{\partial}{\partial x^j}$ in a coordinate neighborhood, then $Xu=a^j\frac{\partial u}{\partial x^j}$ (note that I'm using the Einstein summation convention). – cmk Jun 5 at 14:44
• Yes but a vector field maps from $M$ to $\mathbb R^n$ right? So shouldnt it be $Xu \in (L^2(M))^n$? – Tesla Jun 5 at 15:34