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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.

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3  
Way too much abstraction here. You're talking about one point and a neighborhood of it. Do everything in local co-ordinate, and it's all very easy and clear. You can't answer this question without using the actual definition of a "jet". –  Deane Yang Feb 26 '12 at 12:19
    
Don't see how this will help. In local coordinates the jet equations are 'matrix equations' involving the Jacobi-matrices. But I still can't see how this lifts to the function equations. Can you give a little more details please? –  Nevermind Feb 26 '12 at 14:42
2  
diverietti is, of course, right. $j^1_mf$ is by definition an equivalence class, and you can't have an empty equivalence class. But you might as well write all the "matrix equations" out explicitly and write down explicit formulas for everything. By the way, presumably you know the answer when $M$ and $N$ are 1-dimensional, right? –  Deane Yang Feb 26 '12 at 15:35
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Aren't you done once you solve the matrix equation? By taking local coordinates you reduce to considering maps between Euclidean spaces and since you are only looking at $j^1$ you just need to consider linear maps, which are uniquely determined by their derivative at the origin. –  Willie Wong Feb 26 '12 at 16:14
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Ok.. Now I see whats the problem here. I made a mistake in writing the question! Sorry for that. Please read the second example again, I changed it. –  Nevermind Feb 26 '12 at 22:05
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2 Answers

up vote 2 down vote accepted

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.

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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.

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But isn't this question different from the original one? In the original one, you say that only the 1-jets at a given point of $f_1$, $f_2$, $g$, and $h$ are given and ask whether they can be extended to functions. Here, you say that the functions $f_1$, $f_2$, and $h$ are given and ask if the $1$-jet solution to $g$ can be extended to a function. The two questions are rather different. –  Deane Yang Feb 27 '12 at 8:29
    
No, I'm still looking for functions with given Jacobians. Here I'm looking for the functions $g$ with Jacobian $J$ at a point. Stating the the function $f_i$ and $h_i$ are known is not important. I just wrote it to point in the direction I'm looking at –  Nevermind Feb 27 '12 at 12:01
    
I just did this because it looked to me, that you don't get the idea. But in you last comment (@Deane) you are right: The question is wether or not the Jacobians could be 'extended' (better lifted or even integrated) to functions, having theses these Jacobians at a given point. The point here is, that comments about 'solving the matrix equation' doesn't makes sense to me at all, so I gave another example that is a little more simple but of the same kind. –  Nevermind Feb 27 '12 at 12:06
2  
The two questions (the original one and the one in the answer) and therefore their answers are quite different. Could you edit your original question to reflect what you really want? –  Deane Yang Feb 27 '12 at 13:37
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