Why do I need partitions of unity to define Sobolev norms on compact manifolds?

Question

I am considering the following possible definition of Sobolev spaces.

Let $M$ be a compact manifold and $H^k(M)$ be the space of measurable functions $f\colon M\to\Bbb R$ with the following property: every point $p\in M$ has a local chart $(U,\phi)$ via which $f\circ\phi^{-1}\in H^k(\phi(U))$. Now, clearly $C^\infty(M)\subset H^k(M)$.

Fix $f\in H^k(M)$. I will briefly show that if $(V,\varphi)$ is another local chart on $M$, then $f\circ\varphi^{-1}\in H^k(\varphi(V))$. Cover $M$ by finitely many $\{U_i,\phi_i\}$ such that $f\circ\phi_i^{-1}\in H^k(\phi_i(U_i))$. Choose a partition of unity $\rho_i$ subordinated in $\{\varphi(V\cap U_i)\}$, where $D_i=\mathrm{supp}\rho_i$. Therefore, on $\varphi(V)$, $$f\circ\varphi^{-1}=\sum_i\rho_i(f\circ\varphi^{-1})$$

If $|\alpha|\leq k$, $$\begin{align} \int_{\varphi(V)}|D^\alpha (f\circ\varphi^{-1})|^2 &\leq C\sum_i\sum_{\beta\leq\alpha} \int_{\varphi(V)}|D^{\alpha-\beta}\rho_i|^2|D^\beta(f\circ\varphi^{-1})|^2\\ &=C\sum_i\sum_{\beta\leq\alpha} \int_{\varphi(V)\cap D_i}|D^{\alpha-\beta}\rho_i|^2|D^\beta(f\circ\varphi^{-1})|^2\\ &=C\sum_i\sum_{\beta\leq\alpha} \int_{\overline{\varphi(V)\cap D_i}}|D^{\alpha-\beta}\rho_i|^2|D^\beta(f\circ\varphi^{-1})|^2 \end{align} $$ Note that $\overline{\varphi(V)\cap D_i}\subset\varphi(V\cap U_i)$ is compact. Hence, it is easy to show that the last integral is finite, since change of coordinates have bounded derivatives on this compact set.

Therefore, $H^k(M)$ can be equipped with a norm without partitions of unity: choose a finite local charts $\{U_i,\varphi_i\}$ covering $M$. Define $$|f|_k^2=\sum_i|f\circ\varphi_i^{-1}|_k^2$$ The above argument show that each $|f\circ\varphi_i^{-1}|_k^2$ is finite. Moreover, different choices of local charts $\{U_i,\varphi_i\}$ yield equivalent norms.

My question: What is wrong with this definition? Since partitions of unity are used in every literature I know on Sobolev spaces on manifolds, I am very confused. Any insights?

(I am aware of this answer. The accepted answer says that the reason we need partitions of unity is to control the derivatives of change of coordinates. However, it seems that we don't need it.)

Answer 1

Choose a partition of unity $\rho_i$ subordinated in $\{\varphi(V\cap U_i)\}$

You can do that but the support of $\rho_i$ will not be a compact subset of $\varphi(V\cap U_i)$. Indeed, if it was then, since the finite sum of functions with compact support still has compact support, $\sum_i \rho_i$ could not be identically equal to $1$ in $\varphi(V)$.

Thus, you don't get to have both

$$f\circ\varphi^{-1}=\sum_i\rho_i(f\circ\varphi^{-1})$$

and

$\overline{\varphi(V)\cap D_i}\subset\varphi(V\cap U_i)$ is compact.

The two forms of partition of unity, using same indexing set and compactly supported, can be combined only if we are covering a compact set, which $\varphi(V)$ is not.

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A concrete example to show why we need the partition of unity. Let $M=S^2$, realized as the compactification of $\mathbb{R}^2$. Cover it by two open sets: $$U_1 = \{x\in \mathbb{R}^2 : \|x\|< 2\}, \quad U_2 = \{x\in \mathbb{R}^2 : \|x\| > 1\}\cup\{\infty\}$$ On $U_1$, consider the coordinate chart $\phi : U_1\to U_1$ defined in polar coordinates as $$ (r, \theta) \mapsto (r, \theta + f(r)) $$ where $f:[0, 2)\to \mathbb{R}$ is $C^\infty$ smooth, identically zero on $[0, 1]$, and rapidly tends to infinity as $r\to 2$. This gives us a $C^\infty$ function $\phi $. It should be clear that $u\circ \phi^{-1}$ can fail to have integrable derivatives even if $u$ is a very nice function (linear on $U_1$, for example).