# Decomposition of linear partial differential operators

I was wondering about the following:

Let $M$ be a smooth, second-countable (possibly noncompact) manifold and let $E$ and $F$ be smooth vector bundles over $M$.

1. Can every smooth linear partial differential operator $P$ from $E$ to $F$ of finite order be written as $\sum_{i=0}^n (T_i)_* ∘ P_i$ for certain smooth linear partial differential operators $P_0, \dots, P_n$ from $E$ to $E$ and certain vector bundle homomorphisms $T_0, \dots, T_n$ from $E$ to $F$?

2. Can every smooth linear partial differential operator $P$ from $E$ to $E$ of finite order be written as a finite sum of compositions of smooth linear partial differential operators from $E$ to $E$ of order at most 1?

Of course, locally (i.e., on a chart domain over which the vector bundles trivialize) the answer to both questions is yes, but this does not seem to be of much help when trying to answer the questions globally.

• Why not just use a partition of unity to make the local results global? Aug 6, 2011 at 17:48
• Because, as far as I can see, on noncompact manifolds using a partition of unity will not give finite sums. Aug 7, 2011 at 7:50
• @Marcel: You can always cover a connected manifold with a finite set of coordinate charts, see this question: mathoverflow.net/questions/58988/… Aug 7, 2011 at 8:51
• @Dmitri: Thanks a lot for pointing this out! This looks really promising. Of course, here we need a finite cover by coordinate charts such that $E$ trivializes over these charts and $M$ is real, but this does not seem to be a problem (for example, the proof of Proposition 4.1 in Wells' Differential analysis on complex manifolds looks fine). Using a smooth partition of unity subordinate to such a finite cover to patch the local expressions indeed seems to solve the problem, but I will come back to this after I have checked the details (probably tomorrow; I am quite busy right now). Aug 7, 2011 at 11:55

If the (say, $d$-dimensional) base manifold $M$ is parallelizable (i.e. $TM\to M$ is trivial), then the answer to both questions is yes even globally, provided we choose a (say, torsion-free) covariant derivative $\nabla^M$ on $TM$ and a covariant derivative $\nabla^E$ on $E\to M$. We may then combine these connections by means of the Leibniz rule to define a covariant derivative on $\otimes^k T^*M\otimes E\to M$ for all $k$. Let us denote this common extension of $\nabla^M$ and $\nabla^E$ simply by $\nabla$. One may then define iterated covariant derivatives $\nabla^k$ of any order $k\geq 0$ on $E\to M$ recursively as follows: $$\nabla^0\phi=\phi\ ,\,\nabla^1\phi=\nabla^E\phi\ ,\,\nabla^{k+1}\phi=\nabla(\nabla^k\phi)\ ,\quad\phi\in\Gamma(M,E)\ ,$$ so that $\nabla^k\phi\in\Gamma(M,\otimes^kT^*M\otimes E)$ for every $k\geq 0$. Let us denote the contraction of $\nabla^k\phi$ with $k$ smooth vector fields $Y_1,\ldots,Y_k\in\mathfrak{X}(M)$ by $\nabla^k_{Y_1,\ldots,Y_k}\phi\in\Gamma(M,E)$. The key fact we need (apparently due to R. Palais, but I am not sure), holding even when $M$ is not parallelizable, is:
Any linear partial differential operator of order $r$ $P:\Gamma(M,J^rE)\to\Gamma(M,F)$ may be written uniquely as $$\tag{1}\label{e1}P\phi=\sum^r_{k=0}A_k\nabla^k\phi\ ,\quad\phi\in\Gamma(M,E)\ ,$$ where $A_k\in\Gamma(M,S^k TM\otimes E^*\otimes F)$ for each $k=0,1,\ldots,r$, $S^kTM\to M$ is the symmetric, rank-$k$ contravariant tensor sub-bundle of $\otimes^kTM\to M$ and $E^*\to M$ is the dual bundle to $E\to M$. Particularly, one may replace $\nabla^k$ in \eqref{e1} by its symmetrized part.
The coefficients $A_k$ of the representation \eqref{e1} of $P$ may of course be seen as vector bundle morphisms from $\otimes^kT^*M\otimes E\to M$ to $F\to M$ covering $\mathrm{id}_M$. This is "almost" what you want, what gets in the way is precisely the potential non-triviality of $TM\to M$ - that is where the hypothesis of parallelizability of $M$ enters.
Recall now that the latter amounts to being able to choose $d$ smooth vector fields $X_1,\ldots,X_d\in\mathfrak{X}(M)$ such that $\{X_1(p),\ldots,X_d(p)\}$ is a basis of $T_pM$ for every $p\in M$ - that is, a (global) linear frame in $TM$. Let $\omega^1,\ldots,\omega^d\in\Omega^1(M)$ be the corresponding linear coframe (i.e. $\omega^i(X_j)=\delta^i_j$) - from \eqref{e1} one may write $$\tag{2}\label{e2}P\phi=\sum^r_{k=0}\sum^n_{j_1,\ldots,j_k=1}A_k(\omega^{j_1}\otimes\cdots\otimes\omega^{j_k})\nabla^k_{X_{j_1},\ldots,X_{j_k}}\phi\ ,\quad\phi\in\Gamma(M,E)\ ,$$ where $A_k(\omega^{j_1}\otimes\cdots\otimes\omega^{j_k})\in\Gamma(M,E^*\otimes F)$ is the contraction of $A_k$ with $\omega^{j_1}\otimes\cdots\otimes\omega^{j_k}\in\Gamma(M,\otimes^kT^*M)$. Formula \eqref{e2} is precisely the affirmative answer to your first question. From there to answering affirmatively your second question is just a(n essentially combinatorial) matter of rewriting $\nabla^k_{X_{j_1},\ldots,X_{j_k}}\phi$ in \eqref{e2} in terms of $\nabla^E_{X_{j_1}}\cdots\nabla^E_{X_{j_{m+1}}}\phi$ and $\nabla^M_{X_{l_1}}\cdots\nabla^M_{X_{l_m}}X_i$ for $i,j_1,\ldots,j_{m+1},l_1,\ldots,l_m=1,\ldots,d$, $m=1,\ldots,k-1$ using Leibniz's rule (the first-order operators you want are, of course, $\nabla^E_{X_j}$ with $j=1,\ldots,d$).
For most applications to global analysis, however, \eqref{e1} suffices, with the advantage that it holds in complete generality. There is, of course, the possibility of using a partition of unity subordinated to a finite atlas (an atlas with $d+1$ charts always exists whenever $M$ is second countable by topological dimension theory, see e.g. the Corollary to Theorem 1.2.I, pp. 20-21 of the book of W. Greub, S. Halperin and R. Vanstone, Connections, Curvature, and Cohomology, Volume I (Academic Press, 1972)) and write $P$ in each chart using coordinates, as suggested by you in your question and Dmitri Pavlov in his comment above, but another advantage of \eqref{e1} and \eqref{e2} is that they are both coordinate-free representations of $P$.
There is a middle course to circumvent the hypothesis of parallelizability which relies on the following remark: if $U\subset M$ is the (open) domain of a chart $x:U\to\mathbb{R}^d$, then $TU$ is trivial (just pick the coordinate vector fields $X_j(p)=(T_px)^{-1}\frac{\partial}{\partial x^j}$, $j=1,\ldots,d$). Let now $\{(U_\alpha,x_\alpha)\ |\ \alpha=1,\ldots,d+1\}$ be a finite atlas of $M$ as in the previous paragraph, $\{f_\alpha\ |\ \alpha=1,\ldots,d+1\}$ a partition of unity subordinate to the open cover $\{U_\alpha\ |\ \alpha=1,\ldots,d+1\}$ of $M$ and $X_{1,\alpha},\ldots,X_{d,\alpha}$ be the linear frame on $U_\alpha$ obtained from $x_\alpha$ as above for each $\alpha$, with corresponding linear coframe $\omega^{1,\alpha},\ldots,\omega^{d,\alpha}$. We can then write $$P\phi=\sum^{d+1}_{\alpha=1}\sum^r_{k=0}\sum^n_{j_1,\ldots,j_k=1}A_k(\omega^{j_1,\alpha}\otimes\cdots\otimes\omega^{j_k,\alpha})\nabla^k_{X_{j_1,\alpha},\ldots,X_{j_k,\alpha}}(f_\alpha\phi)\ ,$$ recalling that $$\nabla^k\phi=\sum^{d+1}_{\alpha=1}\nabla^k(f_\alpha\phi)$$ and, by Leibniz's rule, $$\nabla^k_{Y_1,\ldots,Y_k}(f_\alpha\phi)=\sum_{I\subset\{1,\ldots,k\}}\nabla^{M|I|}_{Y_I}f_\alpha\nabla^{n-|I|}_{Y_{I^c}}\phi\ ,\quad Y_I=(Y_j)_{j\in I}\ .$$ This leads us to the following, not-so-appetizing formula $$\tag{3}\label{e3}P\phi=\sum^{d+1}_{\alpha,\beta=1}\sum^r_{k=0}\sum^n_{j_1,\ldots,j_k=1}\sum_{I\subset\{1,\ldots,k\}}(\nabla^{M|I|}_{X_{j_I,\alpha}}f_\alpha)A_k(\omega^{j_1,\alpha}\otimes\cdots\otimes\omega^{j_k,\alpha})(f_\beta\nabla^{n-|I|}_{X_{j_{I^c},\alpha}}\phi)\ ,$$ where $Y_{j_I,\alpha}=(Y_{j_l,\alpha})_{l\in I}$. Since $\mathrm{supp}\nabla^kf_\alpha\subset\mathrm{supp} f_\alpha$ for all $k,\alpha$, formula \eqref{e3} is globally well defined and yields a positive answer to your first question even in the absence of parallelizability. To answer positively your second question, one proceeds as in the parallelizable case with each term $\nabla^{n-|I|}_{Y_{j_{I^c}}}\phi$ in \eqref{e3}. At this point, yet another advantage of working with \eqref{e1} instead should become clear - namely, economy and simplicity.