Understanding exterior differential systems

Question

Let $M$ be an $n$-dimensional smooth manifold. An exterior differential system on $M$ is by definition a graded ideal $\mathcal{I}\subset \Omega^{\bullet}(M)$ in the ring $\Omega^{\bullet}(M)$ of differential forms on $M$ that is closed under exterior differentiation. The goal is then to find submanifolds $\iota \colon N \to M$ on which $\mathcal{I}$ vanishes, namely submanifolds for which $\iota^{\ast}\mathcal{I} = 0$. Now, suppose you want to study the existence of a global coframe $(e^1,\dots , e^n)$ on $M$, namely a section of its frame bundle, that satisfies the following differential equations: $$ d e^i = T^i_{jk} \, e^j\wedge e^k\, , \quad d ( F_k e^k) = 0 $$ where $T^i_{jk}, F_k \in C^{\infty}(M)$ are given smooth functions on $M$ and summation over repeated indices is assumed. If we interpret the $T^i_{jk}$ as map: $$ T \colon M \to \mathbb{R}^n\otimes \bigwedge\nolimits^2 (\mathbb{R}^n)^{\ast} $$ then typically (in the cases I want to study), the image of $T$ will belong to a given explicit subspace of $C\subset \mathbb{R}^n\otimes \bigwedge\nolimits^2 (\mathbb{R}^n)^{\ast}$. Intuitively speaking, I think this should define an exterior differential system, perhaps even the most natural one. Using the previous equations as "generators", I can define an ideal $\mathcal{I}$. However, I do not want to find submanifolds on $M$ where this ideal is zero, I want it to be zero on the whole manifold $M$! What am I missing in order to properly interpret this problem as an exterior differential system? Do I have to somehow construct a larger auxiliary manifold for which solutions will span appropriate integral submanifolds? If that is the case, is there a canonical way of doing so? I have thought of lifting the problem to the frame bundle, but that seems to be a sort of a random choice.

Answer 1

Here's an expansion of my comment that the natural formulation of this problem as an EDS is on the coframe bundle $\pi: P\to M$, which, I hope, will be helpful. Also, because it will match my usual notation better, I'm going to relabel the $T^i_{jk}$ so that the equation to be solved is $\mathrm{d}e^i = -\tfrac12\, T^i_{jk}\,e^j\wedge e^k$. I have to do this, or I will get confused.

I'm going to regard $\mathbb{R}^n$ as the vector space of columns of height $n$, so a differential form $\phi$ that takes values in $\mathbb{R}^n$ can be written as $\phi = (\phi^i)$ where each $\phi^i$ an ordinary differential form. Recall $\pi:P\to M$ is the principal right $\mathrm{GL}(n,\mathbb{R})$-bundle over $M$ (an $n$-manifold), for which an element $u\in P$ that satisfies $\pi(u)=p\in M$ is a linear isomorphism $u:T_pM\to\mathbb{R}^n$. The right action by $\mathrm{GL}(n,\mathbb{R})$ is given by $u\cdot A = A^{-1} u$ for $A\in \mathrm{GL}(n,\mathbb{R})$. There is a canonical $\mathbb{R}^n$-valued $1$-form $\omega$ on $P$ defined by $\omega(v) = u\bigl(\pi'(v)\bigr)$ for $v\in T_uP$. It satisfies $R_A^*(\omega) = A^{-1}\omega$. I'll write $\omega = (\omega^i)$. Note that $\omega$ is $\pi$-semi-basic, i.e., that the kernel of $\pi': TP\to TM$ is the kernel $\omega$. By construction, a (smooth) section $e = (e^i):M\to P$ of $\pi:P\to M$ necessarily satisfies $e^*(\omega^i) = e^i$.

Now, the functions $T^i_{jk}=-T^i_{kj}$ and $F_i$ are defined on $M$, but I want to use them on $P$, so I'll just pull them up to $P$ via $\pi$, but, as is standard in EDS, I won't notate the pullback. Thus, when I regard $F_i$ as a function on $P$, I'll really mean $F_i\circ\pi:P\to \mathbb{R}$. That understood, I am going to write down $(n{+}1)$ $2$-forms that are defined on $P$: $$ \Omega^i = \mathrm{d}\omega^i + \tfrac12\,T^i_{jk}\,\omega^j\wedge\omega^k \quad\text{and}\quad \Phi = \mathrm{d}\bigl( F_i\,\omega^i \bigr). $$ If $e:M\to P$ is a coframing of $M$, then $e = (e^i)$ satisfies $e^*(\Omega^i) = e^*(\Phi) = 0$ if and only if $\mathrm{d}e^i = - \tfrac12\,T^i_{jk}\,e^j\wedge e^k$ and $\mathrm{d}\bigl( F_i\,e^i \bigr)=0$. Thus, if $\mathcal{I}$ is the differential ideal generated by the $\Omega^i$ and $\Phi$, then coframings satisfying the desired equations are exactly the integral manifolds of $\mathcal{I}$ that are sections of $\pi:P\to M$.

To see what this buys us, we need to see the algebraic generators of $\mathcal{I}$. For this, we need to observe that $\mathrm{d}F_i = F_{ij}\,\omega^k$ and $\mathrm{d}T^i_{jk} = T^i_{jkl}\,\omega^l$ for unique functions $F_{ij}$ and $T^i_{jkl}$ on $P$. (These new functions will not be functions on $M$ unless they are identically zero, i.e., the functions $F_i$ and $T^i_{jk}$ on $M$ are constants.). Using these functions, we compute $$ \Phi - F_i\Omega^i = \tfrac12\bigl(F_{kj}-F_{jk} - F_iT^i_{jk}\bigr)\,\omega^j\wedge\omega^k := \tfrac12 A_{jk}\,\omega^j\wedge\omega^k $$ and $$ \mathrm{d}\Omega^i - \tfrac12 T^i_{jk}\,\Omega^j\wedge\omega^k + \tfrac12 T^i_{jk}\,\omega^j\wedge\Omega^k := \tfrac16 B^i_{jkl}\,\omega^j\wedge\omega^k\wedge\omega^l $$ where $$ B^i_{jkl} = T^i_{jkl} + T^i_{klj} + T^i_{ljk} - T^i_{mj}T^m_{kl} - T^i_{mk}T^m_{lj} - T^i_{ml}T^m_{jk}\,. $$ Thus, $\mathcal{I}$ is generated algebraically by the $\Omega^i$, the $\pi$-semi-basic $2$-form $\alpha = \tfrac12 A_{jk}\,\omega^j\wedge\omega^k$, and the $\pi$-semi-basic $3$-forms $\beta^i = \tfrac16 B^i_{jkl}\,\omega^j\wedge\omega^k\wedge\omega^l$.

Now, pulled back to any section $e:M\to P$, the $e^i$ are linearly independent $1$-forms, so it follows that any $e:P\to M$ that is an integral manifold of $\mathcal{I}$ must pull back the functions $A_{jk}$ and $B^i_{jkl}$ to be zero. In other words, $e(M)$ must lie in the zero locus $Z\subset P$ of the functions $A_{jk}$ and $B^i_{jkl}$.

For any $n>1$, it's easy to construct examples for which $Z$ is empty. Just take the $F_i$ and $T^i_{jk}$ to be constants for which some $A_{jk}$ (which is also constant in this case) is nonzero. When $n\ge 4$ and the functions $T^i_{jk}$ and $F_i$ are chosen generically, the set $Z$ will be empty, so there will be no solutions to the problem in such a case.

Finally, one nice thing that may not be immediately apparent is that the equations $A=B=0$ are actually more tractable than one might expect: If one chooses a 'reference section' $\eta:M\to P$, then we can write $\omega = g^{-1}\,\eta$ where $(\pi,g):P\to M\times \mathrm{GL}(n,\mathbb{R})$ defines a trivialization of $P$ as a $\mathrm{GL}(n,\mathbb{R})$-bundle. Then $F_{ij}$ and $T^i_{jkl}$ are actually linear in the coefficients of the matrix-valued function $g$, so it follows that $A_{jk}$ and $B^i_{jkl}$ are linear in the coefficients of $g$ as well, i.e., the 'fiber variables'. Hence the locus $Z\subset P$, when nonempty, is actually not very bad. It could even happen that $\pi:Z\to M$ is a smooth submersion, and you can now simplify your problem by looking for sections $e:M\to Z$ that are integral manifolds of the ideal $\mathcal{I}$ pulled back to $Z$ (which, by the calculation above, is generated algebraically by the $2$-forms $\Omega^i$). When $n=2$ or $3$, this is actually a reasonable thing to do; you get an exterior differential system that has a reasonable structure.