Inverse Problem for jet equations

Question

The following is a well known fact and due to the functorial properties of the jet functor:

Suppose you have two smooth manifolds $M$ and $N$ and maps $f:M \rightarrow N$ as well as $g: M \rightarrow N$, then if there is a $m \in M$ and an open neighbourhood $U$ of $m$ in $M$ such that the equation

$$f(x) = g(x)$$

holds on $U$ then clearly $j^1_mf = j^1_mg$.

The inverse Problem is when you have jet equivalence classes $j^1_mf, j^1_mg \in J^1_m(M,N)$ with $$j_mf = j_mg$$

then is there an open neighbourhood $U$ of $m$ in $M$ and representatives $f \in j_mf$ as well as $g \in j_mg$ such that the equation $f(x) = g(x)$ holds on $U$ ?

Surely there are, but what is the argument in situations like this? (The functor properties of $J^1$ doesn't really fits here I guess, because the problem is inverse)

The point is, that I can't come up with a formal correct argument and when the situation gets more involved it is hard to say if the inverse question has a solution or not, due to the lack of an correct argument.

For example:

Suppose you have smooth manifolds and jet equivalence classes

$$j_{m_1}^1f \in J^1_{m_1}(M_1,N_1)$$ $$j_{m_2}^1h \in J^1_{m_2}(M_2,N_2)$$ $$j_{n_1}^1g_1, j_{n_1}^1g_2 \in J^1_{n_1}(N_1,N_2)$$ $$j_{m_1}^1k_1,j_{m_1}^1k_2 \in J^1_{m_1}(M_1,M_2)$$

(with $k_1(m_1) = m_2 = k_2(m_1)$, $f(m_1)=n_1$, $g_1(n_1)=h(m_2)=g_2(n_2)$ and a system of two jet equation of first order jets

$$j^1_{n_1}g_1 \circ j^1_{m_1}f = j^1_{m_2} h \circ j^1_{m_1}k_1$$ $$j^1_{n_1}g_2 \circ j^1_{m_1}f = j^1_{m_2} h \circ j^1_{m_1}k_2$$

and you know that this equation holds! So the question is not whether or not the jets satisfy the equation, but the equation is taken for granted.

But the inverse problem is the question whether or not there are appropriate reprensentatives $g_1' \in j^1_{n_1}$ $\ldots$ (and all others) and an open neighbourhood $U$ of $m_1$ in $M_1$ such that

$$g_1 \circ f(x) = h \circ k_1(x)$$ $$g_2 \circ f(x) = h \circ k_2(x)$$

holds on $U$.

THis last example is not obvious and a formal argument is really needed.

Answer 1

If you assume $U$ is a co-ordinate chart, the answer is yes. Simply choose local co-ordinates everywhere and assume each of your functions is an affine function of the co-ordinates. In that case, each function is equal to its own 1-jet map. So the desired equations involving the composition of functions follow directly from the 1-jet equations.

If you assume that, say, $f_1$, $f_2$, $h_1$, $h_2$ are given in advance and you want to solve for $g$, then this approach does not necessarily work. But, given suitable nondegeneracy assumptions on these functions, you can use the inverse function theorem to find local co-ordinates so that these functions become affine functions of the co-ordinates and therefore you can solve for $g$ as an affine function, too.

Answer 2

Restated in coordinates: Suppose you have (smooth) functions $f_1,f_2:\mathbb{R}^n \rightarrow \mathbb{R}^m$ and $h_1,h_2:\mathbb{R}^n \rightarrow \mathbb{R}^l$ AND YOU ALREADY KNOW that they satisfy the following system of (matrix)-equations:

$$\frac{\partial f_1^\mu}{\partial x_\nu}(x') = J_j^\mu \cdot \frac{\partial h_1^j}{\partial x_\nu}(x')$$ $$\frac{\partial f_2^\mu}{\partial x_\nu}(x') = J_j^\mu \cdot \frac{\partial h_2^j}{\partial x_\nu}(x')$$

at a point $x'$ with $x=h_1(x')=h_2(x')$ for a given $(l \times m)$ matrix $J$. Regarding $J$ as the coordinate representation of a jet at $x$, the question is, whether or not there is a (local) function $g: \mathbb{R}^l \rightarrow \mathbb{R}^m$ with Jacobi matrix $J$ at $x$ such that

$$f_1(t) = g\circ h_1(t)$$ $$f_2(t) = g\circ h_2(t)$$

holds.