I'm having big troubles understanding the definition of Sobolev spaces on manifolds.

Ok, so we have a Riemannian manifold $(M, g)$, and then we can define a natural riemannian measure (which I will denote by $\nu_g$, and it is worth to say that it is a Radon measure, and its $\sigma$-algebra contains the Borel $\sigma$-algebra of $M$)[I have no troubles understanding this, it is explained very clearly in the books of Sakai and Chavel] In this way, we then have that $M$ is a measure space (in the usual sense), so it is very natural to define the spaces $L^p(M, g)$ as with any other measure space.

This is very cool, and then one might wonder if Sobolev spaces can be defined. In fact, they do exist, and in most books the following definition is usually given:

For $k\in \mathbb N$ and $1\leq p <\infty$, the Sobolev space $W^{k,p}(M, g)$ is the completion of $C^\infty(M)$ under the Sobolev norm $\|\cdot\|_{W^{k,p}(M, g)}$ given by $$\|u\|_{W^{k,p}(M, g)} := \left( \,\sum_{j=0}^k \int_M |\nabla^j u |^p \, d\nu_g\right)^{\frac 1 p},\qquad (*)$$ where $\nabla^j u$ is the $j$-th derivative of $u$.

$(*)$More precisely, one must take functions $u\in C^\infty(M)$ such that each one of $\displaystyle \int_M |\nabla^j u |^p\,d\nu_g$ is finite.

So, my problem is: what is exactly meant by $\nabla^j u$ as "the $j$-th derivative of $u$"? I can't really give a sensible meaning to this. Please give me some insight on this, I want a formal and clear definition (it seems that it is obvious for most autors, but I haven't seen a clear and formal explanation on this), my trouble is that I don't understand at all what is meant with $\nabla^j$ (I believe we can define it for general tensor fields). Can you elaborate a detailed answer on that construction? My problem is then that I don't understand iterated covariant derivatives (I have asked some of my friends, and they are as confused as I am) And please, I don't want a physicist's messy definition involving debauch of indices (if possible)

Any help or clear reference is very much appreciated. An elaborated example would be very useful too.

Edit: The original question: "what is $\nabla^j u$ ?" now has been settled (I'll post my own coordinate-free answer later). Now, the problem is that each one of $\nabla^j u$ is a $(0, j)$-tensor field on $M$, what is the correct definition for the integral of such thing?.

So, the question is: what is meant by $\displaystyle\int_M |\nabla^j u |^p \, d\nu_g$?

  • $\begingroup$ Roughly speaking, $\nabla^j u$ is a differential $j$-form, and you must assign meaning to $|\nabla^j u |^p$ in order to integrate it. Go for the books written respectively by Aubin and Hebey. Their books give an exposition of this thing. $\endgroup$
    – Boar
    Commented Sep 30, 2022 at 0:13
  • $\begingroup$ Sorry, I was wrong. $\nabla^j u$ is only a covariant $j$-tensor field. It may not be a differential form, but we still have to define what is meant by $|\nabla^j u|^p$ to carry out the integration. $\endgroup$
    – Boar
    Commented Sep 30, 2022 at 2:18

1 Answer 1


Let $(M,g)$ be a Riemannian manifold with associated Levi-Civita connection $\nabla$. Typically one introduces the covariant derivative on vector fields, i.e., $\nabla:\Gamma(TM)\times\Gamma(TM)\to\Gamma(TM)$, by $(X,Y)\mapsto\nabla_XY$ which satisfies the Leibniz property and some linearity conditions. Any connection on $TM$ uniquely determines a connection on each tensor bundle $T^{(k,l)}M$. In particular, if $u\in C^\infty(M)$, then $$\nabla u(X)=du(X)=X[u],$$ and in coordinates we have that $$\nabla u=\frac{\partial u}{\partial x^j}dx^j,$$ and hence $$g(\nabla u,\nabla u)=g^{ij}\frac{\partial u}{\partial x^i}\frac{\partial u}{\partial x^j}.$$

Given a $1$-form $\omega\in T^{(0,1)}M$, since our connection has generalized, we have that $\nabla\omega\in T^{(0,2)}M$, and in coordinates $$(\nabla\omega)_{ij}=\frac{\partial\omega_i}{\partial x^j}-\Gamma_{ij}^k\omega_k,$$ and so $$g(\nabla\omega,\nabla\omega)=g^{ij}g^{lm}(\nabla\omega)_{il}(\nabla\omega)_{jm}.$$ Letting $\omega=\nabla u$, we then see that $$(\nabla^2u)_{ij}=\frac{\partial ^2u}{\partial x^j\partial x^i}-\Gamma_{ij}^k\frac{\partial u}{\partial x^k}.$$

Let's do this one more time: Given a $(0,2)$-tensor $\omega\in T^{(0,2)}M$, we then have $\nabla\omega\in T^{(0,3)}M$, and $$(\nabla\omega)_{ijk}=\frac{\partial\omega_{ij}}{\partial x^k}-\Gamma_{ij}^l\omega_{lk}-\Gamma_{ij}^m\omega_{km}.$$ Then $$g(\nabla\omega,\nabla\omega)=g^{i_1j_1}g^{i_2j_2}g^{i_3j_3}(\nabla\omega)_{i_1i_2i_3}(\nabla\omega)_{j_1j_2j_3}.$$ Letting $\omega=\nabla^2u$, we then see that $$(\nabla^3 u)_{ijk}=\frac{\partial^3u}{\partial x^k\partial x^j\partial x^i}-\Gamma_{ij}^l\frac{\partial^2u}{\partial x^l\partial x^k}-\Gamma_{ij}^m\frac{\partial^2u}{\partial x^k\partial x^m}.$$

I believe the continued generalization to arbitrary $(k,l)$-tensor fields should be clear from here. Hopefully this helped any confusion.

Edit: For another exposition of the material, cf. Emmanuel Hebby's text "Sobolev Spaces on Riemannian Manifolds".

  • $\begingroup$ I avoid using coordinates, but it is good to see them in action. I now understand what is meant by $\nabla^j T$ for a general tensor field with or without coordinates. My problem now is that if $u\in C^\infty(M)$, then $\nabla^j u$ is a $(0, j)$-tensor field. What is then the meaning of $\displaystyle\int_M |\nabla^j u |^p \, d\nu_g$? $\endgroup$ Commented Jan 4, 2019 at 3:22
  • $\begingroup$ The formula for $(\nabla^3u)_{ijk}$ seems to be incorrect, the RHS should be $\frac{\partial}{\partial x^i}((\nabla^2u)_{jk})-\Gamma_{ij}^l(\nabla^2u)_{lk}-\Gamma_{ik}^l(\nabla^2u)_{jl}$ $\endgroup$ Commented Sep 3, 2020 at 8:10

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