I've been working through Jost's Riemannian Geometry and Geometric Analysis and I've been having some trouble understanding the definition of the Sobolev space $H^{1,2}(M,N)$. In particular, I'm not sure how he defines the space $L^2(M,N)$.

Let $M$ and $N$ be Riemannian manifolds. Then the Sobolev space $H^{1,2}(M,N)$ is defined to be the set of functions $f\in L^2(M,N)$ with finite energy, $E_2(f)<+\infty$.

The first mention of the space $L^2(M,N)$ is in $\S 8.3$, but it's provided without definition. Jost refers to convergence in this space $L^2(M,N)$ in several statements of theorems (e.g. Lemma 8.3.2). It would make more sense to me if $N$ was a vector bundle over $M$ since we would have a norm in $N$, and then we could look at integrability of functions $M\to N$ in a way that is similar to Bochner integrability. But this doesn't seem to be the case.

One could alternatively use Nash's isometric embedding theorem to find an isometry $i:N\to\mathbb{R}^k$ for some $k\in\mathbb{N}$ and then say that $f\in L^2(M,N)$ if $f\in L^2(M,i(N))$, but this doesn't seem to be well-defined or canonical in any way (if $M$ is non-compact) since we can just shift $i(N)$ arbitrarily around in $\mathbb{R}^k$.

So my question really is: how are the $L^p(M,N)$ spaces defined for maps between Riemannian manifolds?


A function is in $L^p(M,N)$ if it is measurable (the preimage of every open set in $N$ is measurable in $M$) and the integral $$ \int_M d(f(x),Q)^p\,dx $$ is finite for some $Q\in N$. Here $d$ is the intrinsic distance function of the manifold $N$.

Observe that this makes sense for every metric space $N$, the manifold structure is not really relevant.

If $M$ is has finite measure, then "for some $Q\in N$" is equivalent to "for all $Q\in N$", via the triangle inequality.

This definition is explicit in "Sobolev spaces and harmonic maps for metric space targets" by Korevaar and Schoen. Jost somehow neglects to state it either in the book or his paper "Equilibrium maps between metric spaces" from which the book borrows.

(To be precise, Korevaar and Schoen require $f$ to be Borel measurable.)