I can show that this is true for your "simple" case.
If g(x,y) ∈ C∞(ℝ2) vanishes on x ≤ 0 then it decomposes as g(x,y) = a(x)G(x,y) where a(x) ∈ C∞(ℝ) vanishes on x ≤ 0 and G(x,y) ∈ C∞(ℝ2).
This can be shown by proving the statements below. They could possibly be standard results, but I've never seen them before.
First, I'll refer to the following sets of functions.
- Let U be thet set of functions f(x) ∈ C∞(ℝ) which vanish on x ≤ 0 and are positive on x > 0.
- Let V be the set of functions f: ℝ+→ ℝ such that x-n f(x) → 0 as x → 0, for each positive integer n.
The statements I need to show the main result are as follows.
Lemma 1: For any f ∈ V, there is a g ∈ U such that f(x)/g(x) → 0 as x → 0.
Proof: Choose any smooth function r: ℝ+→ ℝ+ with r(0) = 1 and r(x) = 0 for x ≥ 1. For example, we can use r(x) = exp(1-1/(1-x)) for x < 1. Then, the idea is to choose a sequence of positive reals αk → 0 satisfying ∑k αk < ∞, and set
$$g(x) = x^{\theta(x)},\ \ \ \theta(x)=\sum_{k=1}^\infty r(x/\alpha_k)$$
for x > 0 and g(x) = 0 for x ≤ 0. Only finitely many terms in the summation will be nonzero outside any neighborhood of 0, so it is a well defined expression, and smooth on x > 0. Clearly, θ(x) → ∞ and, therefore, x-n g(x) → 0 as x → 0. It needs to be shown that all the derivatives of g vanish at 0 so that g ∈ U. As r and all its derivatives are bounded with compact support, r(n)(x) ≤ Knx-n-1 for some constants Kn. The nth derivative of θ is
$$\theta^{(n)}(x)=\sum_k\alpha_k^{-n}r^{(n)}(x/\alpha_k)\le K_nx^{-n-1}\sum_k\alpha_k$$
which has polynomially bounded growth in 1/x. The derivatives of log(g) satisfy
$$\frac{d^n}{dx^n}\log(g(x))=\frac{d^n}{dx^n}\left(\log(x)\theta(x)\right)$$
which also has polynomially bounded growth in 1/x. However, the derivative on the left hand side is g(n)(x)/g(x) plus a polynomial in g(i)(x)/g(x) for i < n. So, induction gives that g(n)(x)/g(x) has polynomially bounded growth in 1/x and, multiplying by g(x), g(n)(x) → 0 as x → 0.
By definition of f ∈ V, there is a decreasing sequence of positive reals εk such that f(x) ≤ xn for x ≤ εn. We just need to make sure that αk ≤ εn+1 for k ≥ n to ensure that g(x) ≥ xn-1 for εn+1 ≤ x ≤ min(εn,1). Then f(x)/g(x) goes to zero at rate x as x → 0.
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Lemma 2: For any sequence f1,f2,... ∈ V there is a g ∈ U such that fk(x)/g(x) → 0 as x → 0 for all k.
Proof: The idea is to apply Lemma 1 to f(x) = Σk λk|fk(x)| for positive reals λk. This works as long as f ∈ V, which is the case if Σk λksupx≤kmin(x,1)-k|fk(x)| is finite, and this condition is easy to ensure.
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Lemma 3: For any sequence f1,f2,... ∈ V there is a g ∈ U such that fk(x)/g(x)n → 0 as x → 0 for all positive integers k,n.
Proof: Apply Lemma 2 to the doubly indexed sequence fk,n = |fk|1/n.
The result follows from applying lemma 3 to the triply indexed sequence fi,j,k(x) = max{|(di+j/dxidyj)g(x,y)|: |y| ≤ k} ∈ V. Then, there is an a ∈ U such that fijk(x)/a(x)n → 0 as x → 0. Set G(x,y) = f(x,y)/a(x) for x > 0 and G(x,y) = 0 for x ≤ 0. On any bounded region for x > 0, the derivatives of G(x,y) to any order are bounded by a sum of terms, each of which is a product of fijk(x,y)/a(x)n with derivatives of a(x), so this vanishes as x → 0. Therefore, G ∈ C∞(ℝ2).
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In fact, using a similar method, the simple case can be generalized to arbitrary submersions.
Let p: M →N be a submersion. If h ∈ C∞(N) and g ∈ C∞(M) satisfy hg = 0 then, g = aG for some G ∈ C∞(M) and a ∈ C∞(N) satisfying ha = 0.
Very Rough Sketch:
If S ⊂ N is the open set {h≠0} then g and all its derivatives vanish on p-1(S).
The idea is to choose a smooth parameter u:N-S →ℝ+ which vanishes linearly with the distance to S. This can be done locally and then extended to the whole of N (I'm assuming manifolds satisfy the second countability property). As all the derivatives of g vanish on p-1(S), u-ng tends to zero at the boundary of S. This uses the fact that p is a submersion, so that u also goes to zero linearly with the distance from p-1(S) in M.
Then, following a similar argument as above, a can be expressed a function of u so that g/a and all its derivatives tend to zero at the boundary of S. Finally, G=0 on the closure of p-1(S) and G=g/a elsewhere.
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I suppose the next question is: does proving the special case above of a single g and h reduce the proof of flatness to algebraic manipulation?
The principal symbol of a differential operator
$\sum_{|\alpha| \leq m} a_\alpha(x) \partial_x^\alpha$
is by definition the function $\sum_{|\alpha| = m} a_\alpha(x) (i\xi)^\alpha$
Here $\alpha$ is a multi-index (so $\partial_x^\alpha$ denotes $\alpha_1$ derivatives
with respect to $x_1$, etc.)
At this point, the vector $\xi = (\xi_1, \ldots, \xi_n)$ is merely a formal variable.
The power of this definition is that if one interprets $(x,\xi)$ as variables in the
cotangent bundle in the usual way -- i.e. $x$ is any local coordinate chart, then
$\xi$ is the linear coordinate in each tangent space using the basis $dx^1, \ldots, dx^n$,
then the principal symbol is an invariantly defined function on $T^*X$, where $X$ is
the manifold on which the operator is initially defined, which is homogeneous of degree
$m$ in the cotangent variables.
Here is a more invariant way of defining it: fix $(x_0,\xi_0)$ to be any point in $T^*X$
and choose a function $\phi(x)$ so that $d\phi(x_0) = \xi_0$. If $L$ is the differential operator,
then $L( e^{i\lambda \phi})$ is some complicated sum of derivatives of $\phi$, multiplied together, but always with a common factor of $e^{i\lambda \phi}$. The `top order part' is the one which has a $\lambda^m$, and if we take only this, then its coefficient has only first derivatives of $\phi$ (lower order powers of $\lambda$ can be multiplied by higher derivatives of $\phi$).
Hence if we take the limit as $\lambda \to \infty$ of $\lambda^{-m} L( e^{i\lambda \phi})$
and evaluate at $x = x_0$, we get something which turns out to be exactly the principal symbol of $L$ at the point $(x_0, \xi_0)$.
There are many reasons the principal symbol is useful. There is indeed a `quantization map'
which takes a principal symbol to any operator of the correct order which has this as its principal symbol. This is not well defined, but is if we mod out by operators of one order lower. Hence the comment in a previous reply about this being an isomorphism between filtered algebras.
In special situations, e.g. on a Riemannian manifold where one has preferred coordinate
charts (Riemann normal coordinates), one can define a total symbol in an invariant fashion
(albeit depending on the metric). There are also other ways to take the symbol, e.g. corresponding to the Weyl quantization, but that's another story.
In microlocal analysis, the symbol captures some very strong properties of the operator $L$.
For example, $L$ is called elliptic if and only if the symbol is invertible (whenever $\xi \neq 0$). We can even talk about the operator being elliptic in certain directions if the principal symbol is nonvanishing in an open cone (in the $\xi$ variables) about those directions. Another interesting story is wave propagation: the characteristic set of the operator is the set of $(x,\xi)$ where the principal symbol $p(L)$ vanishes. If its differential (as a function on the cotangent bundle) is nonvanishing there, then the integral curves of the Hamiltonian flow associated to $p(L)$, i.e. for the Hamiltonian vector field determined by $p(L)$ using the standard symplectic structure on $T^*X$, ``carries'' the singularities of solutions of $Lu = 0$. This is the generalization of the classical fact that singularities of solutions of the wave equation propagate along light rays.
Best Answer
Given a paracompact smooth manifold, you have smooth partitions of unity (nLab), but on a real analytic manifold (e.g. a complex manifold viewed as a real manifold) one doesn't have analytic partitions of unity (much less holomorphic, if you are in the complex case). That is, given any open cover on a smooth manifold, one can find a partition of unity subordinate to that cover - this is a very topological property. Using partitions of unity you can paste together local functions as desired.
The existence of smooth partitions of unity comes down to the existence of a smooth (but not analytic!) bump function on $[-1,1]$. Edit: you can find details and formulas on Wikipedia.
A related fact is that on a paracompact smooth manifold, the sheaf of real-valued functions is fine (nLab,wikipedia).