# A very basic question about projections in formal PDE theory

I am learning formal PDE theory for my research and I am currently struggling to have a basic understanding of the operations involved in completing a (say, linear) PDE system to an involutive one (Cartan-Kuranishi procedure) in terms of the partial differential operators involved.

First, let us put some context in order to make the question a bit more interesting to a larger audience and to fix notation. In what follows, all manifolds are smooth, Hausdorff, paracompact, connected and oriented, and all maps between any two of them are assumed to be smooth. Given two vector bundles $\pi:E\rightarrow M$, $\pi':E'\rightarrow M$ over the base manifold $M$, a (linear) partial differential operator of type $E\rightarrow E'$ and order $k\geq 0$ is a linear map $$P:\Gamma(\pi)\rightarrow\Gamma(\pi')$$ ($\Gamma(\pi^{(}{}'{}^{)})=\{\phi\in C^\infty(M,E^{(}{}'{}^{)})\ |\ \pi^{(}{}'{}^{)}\circ\phi=\text{id}_M\}$ is the space of sections of $\pi^{(}{}'{}^{)}$) of the form $$P=\Phi_P\circ j^k\ ,$$ where $j^k$ stands for the $k$-th order jet prolongation of sections of vector bundles and $$\Phi_P:J^kE\rightarrow E'$$ is a vector bundle map covering $\text{id}_M$, i.e. $\pi'\circ\Phi_P=\pi\circ\pi^k$, where $\pi^k:J^k E\rightarrow E$ is the projection map of the $k$-th order jet bundle $J^k E$ of $\pi$ onto the base $E$ (with the convention $J^0 E=E$, $\pi^0=\text{id}_E$, $j^0=\text{id}$). The (homogeneous linear) PDE system associated to $P$ is given by $$\mathcal{R}_k=\ker\Phi_P\ .$$ The $q$-th order prolongation of $P$ ($q\geq 0$) is the partial differential operator $\rho_q P$ of type $E\rightarrow J^q F$ and order $k+q$ given by $$\rho_q P\doteq j^q\circ P=\rho_q(\Phi_P)\circ j^{k+q}\ ,$$ so that $\rho_0 P=P$. The vector bundle map $\rho_q(\Phi_P):J^{k+q}E\rightarrow J^q E'$ covering $\text{id}_M$ is uniquely determined by the second equality above (particularly, $\rho_0(\Phi_P)=\Phi_P$). The PDE system associated to $\rho_q P$, called the $q$-th order prolongation of the PDE system $\mathcal{R}_k$, is then denoted by $$\mathcal{R}_{k+q}=\ker\rho_q(\Phi_P)\ .$$ Therefore, the operation of prolongation of PDE systems has a clear, global interpretation in terms of prolongation of the associated partial differential operators. In what follows, we assume in addition that $P$ is regular, that is, $\rho_q(\Phi_P)$ has constant rank for all $q\geq 0$ - this is demonstrably equivalent to assuming that $\mathcal{R}_{q+k}$ is a vector sub-bundle of $J^{k+q}E$ for all $q\geq 0$. Conversely, if $\mathcal{R}_k$ is a vector sub-bundle of $J^kE$, it is locally the kernel of $\Phi_P$ for some partial differential operator $P$ or order $k$.

Finally, let $$\pi^{r+s}_r:J^{r+s}E\rightarrow J^r E\ ,\quad r,s\geq 0$$ be the natural jet projection maps, so that $$\pi^r_0=\pi^r\ ,\quad\pi^{r+s}_r\circ\pi^{r+s+t}_{r+s}=\pi^{r+s+t}_r$$ for all $r,s,t\geq 0$. It can be shown that $\pi^{k+q}_{k+r}(\mathcal{R}_{k+q})\subset\mathcal{R}_{k+r}$ for all $0\leq r\leq q$. If $P$ is regular and $\pi^{k+q}_{k+r}:\mathcal{R}_{k+q}\rightarrow\mathcal{R}_{k+r}$ have constant rank for all such $r,q$, we say that $P$ is sufficiently regular. Particularly, if $P$ is regular and the above maps are surjective for all such $r,q$, we say that $P$ is formally integrable.

The other key operation employed in the Cartan-Kuranishi completion algorithm besides prolongation is projection. Both operations together unveil the hidden integrability conditions of $P$. In terms of the PDE system $\mathcal{R}_k$ associated to $P$ and its prolongations $\mathcal{R}_{k+q}$ of order $q>0$, projection is a simple operation to describe (unlike prolongation, which is more straightforwardly described in terms of $P$ as done above). Namely, the projection $\mathcal{R}^{(r)}_{k+q}$ of order $r\geq 0$ of $\mathcal{R}_{k+q+r}$ is given by $$\mathcal{R}^{(r)}_{k+q}=\pi^{k+q+r}_{k+q}(\mathcal{R}_{k+q+r})\subset\mathcal{R}_{k+q}\ .$$ For the purposes of the Cartan-Kuranishi algorithm, it suffices to consider $r$-th order projections of PDE systems with $r=1$.

Now we have reached a point where I can ask my

Question: Suppose that $P$ is sufficiently regular. Is there a simple, global formula for a partial differential operator to which the projected PDE system $\mathcal{R}^{(1)}_{k+q}=\pi^{k+q+1}_{k+q}(\mathcal{R}_{k+q+1})$ is associated? (I understand that such an operator should not be unique)

I simply could not find such a formula in the standard literature on the subject (e.g. the 1964 thesis of Quillen and the books of Pommaret, Bryant et al. and Seiler). My conjecture is that composing $\rho_{q+1}P=\rho_{q+1}(\Phi_P)\circ j^{k+q+1}=j^{q+1}\circ\Phi_P\circ j^k$ with the natural projection $\psi_{q+1}$ onto the cokernel of its principal symbol $\sigma(\rho_{q+1} P)$ would provide such an operator and hence a positive answer to my question. In other words,

Subquestion: Is there a vector bundle map $$\rho^{(1)}_q(\Phi_P):J^{k+q}E\rightarrow\text{coker}(\sigma(\rho_{q+1} P))$$ covering $\text{id}_M$ such that $$\rho^{(1)}_q(\Phi_P)\circ\pi^{k+q+1}_{k+q}=\psi_{q+1}\circ\rho_{q+1}(\Phi_P)\ ?$$ If that is the case, is it true that $\mathcal{R}^{(1)}_{k+q}=\ker(\rho^{(1)}_q(\Phi_P))$?

(Remark: if $P$ is sufficiently regular, then $\sigma(\rho_{q+1} P)$ has constant rank and therefore both $\ker(\sigma(\rho_{q+1} P))$ and $\text{coker}(\sigma(\rho_{q+1} P))$ are vector bundles)

If true, this would provide a nice formula for the partial differential operator $D$ produced by the Cartan-Kuranishi algorithm such that $DP$ is involutive and equivalent to $P$, in the same spirit as the compatibility complex associated to an involutive (or, more generally, a formally integrable) partial differential operator, construced by Quillen in his thesis. Moreover, a variation of this procedure would also possibly work with quasilinear partial differential operators.

Your conjecture is almost correct. But consider the extreme case when your equation has no integrability conditions, so that $\operatorname{coker}(\sigma(\rho_{q+1} P)) = 0$, whence $\rho^{(1)}_{q+1}(\Phi_P) = 0$, which definitely doesn't give you the the equation $\mathcal{R}^{(1)}_{k+q}$ that you wanted. What you need to do instead is to augment the $q$-prolonged equation to include the integrability conditions obtained from the projection: $$\rho^{(1)}_q(\Phi_P)\colon J^{k+q}E \to J^qE' \oplus_M \operatorname{coker}(\sigma(\rho_{q+1} P)) ,$$ so that $\rho^{(1)}_q(\Phi_P) = \rho_q(\Phi_P) \oplus \Phi_{Q\circ P}$, where $\Phi_Q\colon J^{k+q}E \to \operatorname{coker}(\sigma(\rho_{q+1} P))$ is some differential operator whose principal symbol is precisely the projection $$S^qT^*\otimes_M E' \xrightarrow{\sigma(Q)} \operatorname{coker}(\sigma(\rho_{q+1} P)) .$$ Of course, the lift of a symbol $\sigma(Q)$ to a differential operator is not unique. But it can be chosen globally, for instance, using an auxiliary connection like you did in your earlier question here.
• Sorry for not getting back to you sooner Igor, only now have I got time to go through your answer. First of all, there is something I don't understand in the reasoning of your first paragraph: as far as I know, no integrability conditions means that $\pi^{k+q+1}_{k+q}(R_{k+q+1})=R_{k+q}$, but not necessarily that the principal symbol of $\rho_{q+1}P$ is surjective (which is the same as $\text{coker}(\sigma(\rho_{q+1}P))=0$). – Pedro Lauridsen Ribeiro Jul 12 '17 at 5:19
• What seems clear is that killing the highest-order derivatives of $\rho_{q+1}P$ by linear operations amounts to finding a vector bundle map $\psi_{q+1}:J^{q+1}E'\rightarrow F$ that sends the image of the principal symbol $\sigma(\rho_{q+1}P)$ of $\rho_{q+1}P$ to zero. The natural projection onto the cokernel of $\sigma(\rho_{q+1}P)$ seems the simplest choice to do so (I've rephrased the subquestion to make this clearer). – Pedro Lauridsen Ribeiro Jul 12 '17 at 5:29
• Nonetheless, if $\sigma(\rho_{q+1}P)$ does happen to be surjective, it is clear that the condition $R^{(1)}_{k+q}\subset R_{k+q}$ can no longer be satisfied with my conjectured choice. However, it seems that going along your suggestion and choosing instead $\rho^{(1)}_q(\Phi_P)=\rho_q(\Phi_P)\oplus\rho'_q(\Phi_P)$ with $\rho'_q\circ\pi^{k+q+1}_{k+q}=\psi_{q+1}\circ\rho_{q+1}(\Phi_P)$, where $\psi_{q+1}$ is the natural projection onto $\text{coker}(\sigma(\rho_{q+1}P))$, does the trick (recall that $\psi_{q+1}\circ\rho_{q+1}(\Phi_P)$ has vanishing principal symbol by construction). – Pedro Lauridsen Ribeiro Jul 12 '17 at 6:04
• @PedroLauridsenRibeiro, I should have been a bit more precise in my first paragraph, but I think you got the idea. The question of whether symbol $\rho_{q+1} P$ is surjective is decided by the non-existence of compatibility conditions for $P$ (though sometimes they are also called "integrability conditions"). As I'm sure you know, a compatibility condition is any operator $P'$ such that $P'\circ P = 0$. And there certainly are operators $P$ for which you can only have $P' = 0$. – Igor Khavkine Jul 13 '17 at 8:04