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I am decently experienced on Sobolev spaces on Euclidean spaces, but I just know basic ideas on Riemannian manifolds and want to understand something on Sobolev spaces on them.

Let $(M,g)$ be smooth Riemannian manifold of dimension $m$ and let $\phi:M \to \mathbb{R}^l$. For $x \in M$ and $v \in \mathbb{S}^{m}$ we write $\frac{d}{dt}\phi(\exp_x(tv))|_{t=0}=D\phi(x)v$ and $\frac{d^2}{dt^2}\phi(\exp_x(tv))|_{t=0}=Hess(\phi)(x)(v,v)$ and observe functionals $\int_M \int_{\mathbb{S}^m} |D\phi(x)v|^2 dv dx$ and $\int_M \int_{\mathbb{S}^m} |Hess(\phi)(x)(v,v)|^2 dv dx$. These functionals should be norms equivalent to $L^2(M)$-norms: $\int_M |D\phi(x)|^2 dx$ and $\int_M |Hess(\phi)(x)|^2 dx$, respectively.

Question. How do these $L^2(M)$-norms come into play in the definition of $\phi \in H^2(M,\mathbb{R}^l)$? More precise, can one define Sobolev norm as $\|\phi\|^2_{H^2(M)}=\|\phi\|^2_{L^2(M)} + \|D\phi\|^2_{L^2(M)}+\|Hess(\phi)\|^2_{L^2(M)}$ and $H^2(M)$ as closure of $C^\infty(M)$ with respect to that norm?

I have seen several different (yet equivalent) ways to define Sobolev spaces on Riemannian manifold, but it is not fully clear to me what kind of weak derivatives one observes.

UPDATE: A way to tackle this question is Proposition 1 in our paper.

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  • $\begingroup$ I don't understand your precise question. You seem to be asking about potential differences between different definitions of Sobolev spaces. Can you give precise descriptions of the different definitions you are referring to? You've given one particular one here. (Other approaches include: do it locally with coordinate charts. Or as a subspace of the space of distributions with some norm property.) If $(M,g)$ is compact, then all definitions I am aware of agree. There can be subtleties when the manifold is open. $\endgroup$ Commented Feb 18, 2021 at 19:29
  • $\begingroup$ In other words, how exactly are you defining $H^2(M)$? Your "can one define sobolev norms..." question seems to suggest that you have a different definition in mind. What is it? $\endgroup$ Commented Feb 18, 2021 at 19:31
  • $\begingroup$ From definitions I want to compare the one I gave with, I was just able to find one where you have norms of $\|\nabla^k \phi\|_{L^2(M)}$, these being iterated covariant derivatives. Since I am new in this, I am not quite familiar in relations among the notions of derivatives I gave and covariant derivatives. $\endgroup$ Commented Feb 18, 2021 at 20:35
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    $\begingroup$ Ah: in that case, you should just remark that $Hess = \nabla^2$ acting on scalars. $\endgroup$ Commented Feb 18, 2021 at 20:57
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    $\begingroup$ Yes. Hessian with respect to any connection $\nabla$ is just $\nabla^2$; you are using Levi-Civita connection from the Riemannian structure. I don't know what you mean by $D$ precisely; but if it is a derivation (which your post suggest it is), then note that all derivations act the same on scalars. $\endgroup$ Commented Feb 18, 2021 at 21:57

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Indeed, those norms are equivalent to the classical integrals of $|D\phi|^2$ and $|\mathrm{Hess} \phi|^2$. Then you can define $H^2(M, \mathbb{R}^l)$ as usual as the completion of the space of smoothly compactly supported fonctions. For nice manifolds (e.g. compact or asymptotically Euclidean ones) this is equivalent to the classical definition of functions whose weak first and second derivatives (same definition as on $\mathbb{R}^n$) belong to $L^2$. And, in practice (for me at least), there is no difference. For more recent results please check this paper: https://arxiv.org/abs/2011.14630

To answer your question in the comments, the Sobolev spaces are independent of the metric you choose as long as the metrics are uniformly equivalent (i.e. there exists a constant $C > 1$ such that $C^{-1} g \leq h \leq C g$), leaving aside regularity questions. The main difficulty when learning Sobolev spaces on manifolds is that there is no definite reference to read and this property among others is something which you have to figure out yourself. The proof is by no means complicated, it is based on the fact that if you have two connections (say Levi-Civita connections), e.g. $\nabla$ and $D$ (associated to $g$ and $h$ respectively), then they differ by a first order term:

  • For functions the difference is zero as we have $\nabla f = Df = df$ (the usual differential).
  • For 1-forms (e.g. $df$) the difference can be expressed in terms of the "Christoffel symbols": $$ \nabla_i X_j = D_i X_j - \Gamma_{ij}^k X_k $$ where $$ \Gamma_{ij}^k = \frac{1}{2} g^{kl} \left(D_i h_{lj} + D_j h_{il} - D_l h_{ij}\right) $$ (this corresponds to the usual Christoffel symbols if $h$ is Euclidean). I can explain better later if you need me to provide the details.
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  • $\begingroup$ Thanks for the answer. I am aware of results in defining Sobolev spaces on manifolds as closure of smooth functions under Sobolev norm. My concern is more into direction that in the Sobolev $H^2(M)$-norm one uses covariant derivative of first and second degree and, possibly, weak derivatives in sense of $\mathbb{R}^n$ (I guess one assumes there is an extension of the function to some ball in the surrounding space). Thus, I am not sure if one can use first and second derivatives as I defined them. $\endgroup$ Commented Feb 18, 2021 at 14:43
  • $\begingroup$ If you're asking whether $\int_M \int_{\mathbb{S}^m} |D\phi(x)v|^2 dv dx$ and $\int_M \int_{\mathbb{S}^m} |Hess(\phi)(x)(v,v)|^2 dv dx$ are equivalent to $\int_M \int_{\mathbb{S}^m} |D\phi(x)|^2 dv dx$ and $\int_M \int_{\mathbb{S}^m} |Hess(\phi)(x)|^2 dv dx$, it suffices to show that, given $x \in M$, $\int_{\mathbb{S}^m} |D\phi(x)v|^2 dv = c(n)|D\phi(x)|^2$ and $\int_{\mathbb{S}^m} |Hess(\phi)(x)(v,v)|^2 dv$ = c(n) |Hess(\phi)(x)|^2$. This is a straightforward calculation. $\endgroup$
    – Deane Yang
    Commented Feb 18, 2021 at 15:06
  • $\begingroup$ Thanks for the comment, but that is not what I wanted. My question regards connection between $\|\nabla^2 \phi\|_{L^2(M)}$ and $\|Hess \phi\|_{L^2(M)}$ and same for the first derivative(s). I made it more clear in the question now. $\endgroup$ Commented Feb 18, 2021 at 15:54
  • $\begingroup$ I updated my answer accordingly. If you need more details, feel free to ask. $\endgroup$ Commented Feb 18, 2021 at 17:00
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    $\begingroup$ Normaly yes. Unless you work on weird non-compact manifolds. Can you tell me which manifold you are considering? $\endgroup$ Commented Feb 18, 2021 at 20:42

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