Why is partition of unity required in definition of Sobolev space on manfolds?

Why do we need to use $\phi_i u$ in the expression for the norm? Why not just $u$? The range of integration is over $R(x_i)$ anyway, so I don't understand why it is necessary.

If you check Kendall, Atkinson book, there they define Sobolev space on boundary without using partition of unity. am I missing something?

Solution 1:

Sobolev norms involve integrals. A perfectly smooth function on an open set can fail to have finite norm if the derivatives grow too fast near the boundary.

Chart maps are smooth, but we do not know anything about the "size" of their derivatives near the boundary. (For simplicity, pretend that $M$ is a Riemannian manifold, so that "size" of derivatives makes sense; otherwise I'd have to complicate things by looking at transition maps.) If we did not truncate $u$ by a compactly supported function, the composition $u\circ x_i^{-1}$ could fail to be in $W^{k,p}$ not because of what $u$ is, but because of how much $x_i$ contributes to derivatives via the chain rule. This is obviously not satisfactory; we want the definition of Sobolev space to be independent of a particular atlas used. Truncation by $\phi_i$ achieves this: all derivatives of the chart maps are bounded within a compact subset of the patch.

Another issue, pointed out by Ted Shifrin, is multiple counting of overlaps between charts; partition of unity mitigates this too.

By the way, in the above definition, the norm $\|u\|_{W^{k,p}(M)}$ depends on the particular charts and partition of unity we use. But this is something we can live with, as long as the space is independent of those choices.