# How is the differential of a Sobolev function on a manifold regarded as an a.e. defined section of $T^*M$?

Let $$(M,g)$$ be a smooth compact Riemannian manifold, and let $$f \in W^{1,p}(M)$$ for $$p\ge 1$$. (I don't assume $$p>\dim M$$).

I have seen in various sources that people refer to the weak derivative of $$f$$ as a linear functional $$T_pM \to \mathbb{R}$$, which is defined for almost every $$p \in M$$. (an a.e.defined section of $$T^*M$$).

I define $$W^{1,p}(M)$$ to be the completion of the space of compactly supported smooth functions $$C_c^{\infty}(M)$$ w.r.t the $$\|\cdot\|_{1,p}$$ norm.

Optional: I suggest below $$2$$ approaches; I would like to know if they are compatible, i.e. if they both produce the same element in $$(T_pM)^*$$.

(Regarding the second approach, I am not even sure if it produces a well-defined functional).

Approach 1:

Given $$f \in W^{1,p}(M)$$, there exist $$f_n \in C_c^{\infty}(M)$$, $$f_n \to f$$ in $$W^{1,p}$$.

$$df_n \in \Gamma(T^*M)$$ is a Cauchy sequence in $$L^p(M,T^*M)$$, where $$L^p(M,T^*M)$$ is the completion of the space of smooth sections $$\Gamma(T^*M)$$ w.r.t the natural $$p$$-norm. By completeness, $$df_n$$ converges to an element in $$L^p(M,T^*M)$$, which we can realize as a measurable section $$T^*M$$. We set $$df=\lim_{n \to \infty} df_n$$.

Approach 2 ("Local picture"):

Let $$\phi:U\subseteq M \to \mathbb{R}^n$$ be a surjective coordinate chart around $$p \in M$$, and $$\phi(p)=0$$. Set $$f_{\phi}=f|_U \circ \phi^{-1} :\mathbb{R}^n \to \mathbb{R}$$. Then $$f_{\phi} \in W^{1,p}(\mathbb{R}^n)$$ (we might need to shrink $$U$$ to ensure nothing will explode). We define $$df_p$$ by the equation

$$df_p \circ d(\phi^{-1})_0(e_i):= d(f_{\phi})_0(e_i)=(\partial_i f_{\phi})(0). \tag{1}$$

Does equation $$(1)$$ well-defines an element in $$T_p^*M$$ independently of the coordinate chart? Does it coincide with $$\lim_{n \to \infty} df_n$$ from the previous approach?

• Concerning point 2: its well-definiteness relies on the validity of the chain rule for weak derivatives. This is, as far as I can tell, a bit subtle: mathoverflow.net/q/291992/13042 – Giuseppe Negro May 17 at 7:45

You have a notion of weak derivative in this case: An almost everywhere section of $$T^*M$$, denoted $$df$$, is a weak derivative of $$f$$ if for every smooth, compactly supported 1-form $$\phi$$ you have $$\int_M g(df,\phi) \,\text{dVol} = \int_M f\,\delta\phi \,\text{dVol}.$$ where the pairing is with respect to the metric $$g$$. A function is in $$W^{1,p}(M)$$ if it is in $$L^p$$, and has a weak derivative in $$L^p(T^*M)$$. Since this definition of weak derivative is consistent with taking limits in $$L^p$$, it is consistent with your approach 1. I believe that taking $$\phi$$ to be supported in a coordinate patch and writing the definition above in coordinates would just yield your approach 2. The question why $$W^{1,p}$$ as defined here is actually the same as the completion should be very similar to the same question in Euclidean spaces (the classic $$H=W$$ question).