Let $\pi \colon M\to N$ be a smooth map between real smooth manifolds. Then $C^\infty(M)$ forms a module over $C^\infty(N)$ (via pullback). Is this module flat when $\pi$ is a submersion?
Recall that the usual definition of flatness is equivalent to the following equational condition:
whenever $ h_1 \ldots h_k\in C^\infty(N) $ and $g_1 \ldots g_k\in C^\infty(M)$ are such that:
$$h_1 g_1 + \ldots + h_k g_k = 0$$ (as functions on $M$)
then there are functions $G_1 \ldots G_r\in C^\infty(M)$ and $a_{i,j}\in C^\infty(N)$ such that:
$$g_i= \sum_j a_{i,j}G_j \; \forall i $$
and
$$\sum_i h_i a_{i,j}= 0 \; \forall j$$
Some remarks:
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It's known that the inclusion of an open subset $U\subset N$ is a flat morphism since smooth functions on $U$ are obtained from smooth functions on $N$ by localizing w.r.t. functions vanishing nowhere on $U$.
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It's also known that smooth flat maps have to be open. Proofs of both of these facts can be found for example in the book: Gonzales, Salas, $C^\infty$-differentiable spaces, Lecture notes in Mathematics, Springer 2000.
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I've asked some of the experts including Malgrange and the above authors and it seems that the answer is not known.
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I gave the equational condition of flatness since it seems like the most reasonable thing to use here. But considering already the simplest situation here's what gets me stuck: suppose you want to check flatness of the standard projection $\mathbb{R}^2 \to \mathbb{R}, (x,y)\mapsto x$, and take the case of just one $h\in C^\infty(\mathbb{R})$ and one $g\in C^\infty(\mathbb{R}^2)$ with $hg=0$. If you pick $h(x)$ to be strictly positive for $x<0$ and $0$ for $x\geq 0$, then the flatness condition translates into:
Any smooth function $g(x,y) \in C^\infty (\mathbb{R}^2)$ that vanishes on the half plane $x\leq 0 $ admits a "factorization":
$$g(x,y)= \sum_j a_j (x)G_j (x,y)$$
where the $a_j\in C^\infty(\mathbb{R})$ all vanish on $x\leq 0$ and the $G_j\in C^\infty(\mathbb{R}^2)$ are arbitrary.
Anyone has an idea how to prove this "simple" case, or sees a counter example?
(Edit: George Lowther beautifully proves this "simple" case, and also comes closer to the full result in his second answer. If you also think he deserves some credit consider up-voting his second answer since the first one turned community wiki.)
Motivation
My personal interest is that a positive answer would allow me to finish a certain proof, which trying to explain here would take this too far afield. But I may try to put the question into context as follows: the notion of flat morphism plays an important role in algebraic geometry where it is basically the right way to formalize the notion of parametrized families of varieties (fibers of such a morphism being these families). One may also say that it is the right "technical" notion allowing one to do all the things one expects to do with such parametrized families (correct me if I'm wrong).
Now I've been taught that differential topology may also be seen as a part of commutative algebra (and that taking such a point of view might even be useful at times). For example: a manifold itself may be recovered completely from the algebra of smooth functions on it, and any smooth map between manifolds is completely encoded by the corresponding algebra morphism. Other examples: vector fields are just derivations of the algebra, vector bundles are just finitely generated projective modules over the algebra etc. Good places to learn this point of view are: Jet Nestruev, Smooth manifolds and observables, as well as the above mentioned book.
Now in differential topology there is a well know notion of smooth parametrized families of manifolds, namely smooth fiber bundles. Hence from this algebraic point of view it would be natural to expect that fiber bundles are flat morphisms.
Best Answer
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.
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. █
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. █
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). █
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. █
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?