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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?

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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.

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  • $\begingroup$ 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? $\endgroup$ – Tesla Jun 5 at 14:39
  • $\begingroup$ 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). $\endgroup$ – cmk Jun 5 at 14:44
  • $\begingroup$ Yes but a vector field maps from $M$ to $\mathbb R^n$ right? So shouldnt it be $Xu \in (L^2(M))^n$? $\endgroup$ – Tesla Jun 5 at 15:34
  • $\begingroup$ Nope, vector fields map real-valued functions to real-valued functions. $\endgroup$ – cmk Jun 5 at 15:38
  • $\begingroup$ Ok thank you, I mixed it up with the definition of vector fields in the Euclidean case. I am a newbie on manifolds. Anyway, it seems more like a statement he makes, but where is the explanation to it? For example, why does the manifold have to be compact etc? $\endgroup$ – Tesla Jun 5 at 16:04

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