# Are submersions of differentiable manifolds flat morphisms?

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:

• 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 $U$.

• 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.

• I've asked some of the experts including Malgrange and the above authors and it seems that the answer is not known.

• 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.

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Is your definition of flatness the same as saying that O_{M,x} is a flat O_{N,f(x)} module for every x in M? Or that p^* is exact on O_N modules? – David Zureick-Brown Oct 19 '09 at 16:29
The flatness condition I gave is equivalent to saying that tensoring exact sequences of C<sup>&\infin;</sup>(N) modules with C<sup>&\infin;</sup>(M) gives exact sequences (a proof of this equivalence may be found in Eisenbud). I guess that is what you meant with your second question. I think this condition implies the local flatness you mention. I don't know if it is equivalent to it, but it might be useful having such an equivalence. – Michael Bächtold Oct 19 '09 at 21:03
Oh i guess html doesn't work in comments. What i meant was C^\infty(N) etc. – Michael Bächtold Oct 19 '09 at 21:05
Could someone explain why this is an interesting or useful thing to know? – Deane Yang Feb 17 '10 at 1:55
I've added a small section trying to motivate the question – Michael Bächtold Feb 17 '10 at 18:23

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. █

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?

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Great, that looks very promising, and the type of argument I would have expected to see. I could fill in most details, though I still need to figure out your estimate in lemma 1 for the n-th derivative of g (my estimate toolbox is somewhat rusty). But thanks! – Michael Bächtold Feb 17 '10 at 17:43
I have an easier to follow argument for lemma 1, which I'll add when I have some time to log on my pc. Also, the simple case generalizes to arbitrary submersions, which I'll also add. – George Lowther Feb 17 '10 at 18:34
hmm, how come my answer is community wiki? Did I check some box without noticing, or did someone change it to community? – George Lowther Feb 17 '10 at 22:55
@George, you made 12 revisions, so it's community wiki now. You should probably repost this in a new answer so you can get reputation for it. – Harry Gindi Feb 17 '10 at 22:55
ok, I see. It's because I edited it too many times. Guess I should avoid edits if I want credit... – George Lowther Feb 17 '10 at 22:57

It is a consequence of Malgrange's preparation theorem for differentiable functions that $C^{\infty}(M)$ is a faithfully flat $C^{\omega}(M)$-module ($C^{\omega}(M)$ is the sheaf of analytic functions on $M$). See Corollary 1.12, Chapter VI of his book Ideals of differentiable functions.

On the other hand $C^{\omega}(M)$ is a flat $C^{\omega}(N)$-module as the argument pointed out by Greg Stevenson shows.

I believe, but don't know how, these two facts can be put together to give a positive answer to the question.

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I can get quite close to proving this. That doesn't mean that the result is true but it does at least seem to be very nearly true. We can also see what any counterexamples must look like if it does fail. Basically, if the details of my argument below go through as expected, the only problem can occur at points where h=(h1,h2,...) vanishes along with all its derivatives, but where the derivatives (to some order) of h/|h| explode.

I hope that this result is true, as it is a simple algebraic statement which would encode some rather deep properties of smooth functions. However, I am not yet convinced that this is the case.

Let me restrict to the simple case where $p\colon M\to N$ is the projection from $\mathbb{R}^2$ to $\mathbb{R}$, p(x,y)=x. Slightly reorganizing your definition of flatness, let n be a positive integer. Let $g\colon M\to\mathbb{R}^n,\ g=(g_1,\ldots,g_n)$ and $h\colon N\to\mathbb{R}^n,\ h=(h_1,\ldots,h_n)$ be smooth maps such that their inner product is zero, $\langle h,g\rangle=\sum_k h_k g_k=0$. Then, flatness of the projection p says that there is a finite set of smooth functions $a_k\colon N\to\mathbb{R}^n,\ a_k=(a_{k1},\ldots,a_{kn})$ and $G_k\in C^\infty(M)$ satisfying $\langle h,a_k\rangle=0$ and $g=\sum_k a_kG_k$. That is, g is spanned by ak as a module over $C^\infty(M)$.

The subset of N on which h is nonzero is easy to deal with, simply by projecting orthogonal to h. If $e_i\colon N\to\mathbb{R}^n,\ e_{ij}\equiv\delta_{ij}$ then $g=\sum_ig_ie_i$ and orthogonal projection gives $$g=g-|h|^{-2}\langle h,g\rangle h=\sum_i(e_i-|h|^{-2}h_ih)g_i$$ so we can take $a_k=e_k-|h|^{-2}h_kh$ and $G_k=g_k$. On the other hand, if h vanishes on an open set then it doesn't put any restriction on either g or ak and the problem is easy. The only issues occur on neighborhoods of the boundary of the set $\lbrace h=0\rbrace\subset N$.

Let me denote the space of smooth functions $a\colon N\to\mathbb{R}^n$ such that $\langle h,a\rangle=0$ by $V$, which is to be considered as a $C^\infty(N)$-module. Also, as I am restricting to $M=\mathbb{R}^2$, then $g_y\equiv g(\cdot,y)\in V$ can be considered as a smoothly parametrized family in V. If the flatness property is true then it says that $g_y$ is contained in a finitely generated submodule of V, generated by the ak. Conversely, if V is finitely generated, then a modified version of the standard proof of flatness for algebraic varieties should also work here (for varieties, the Noetherian property guarantees that V is finitely generated). More generally, it needs to be shown that the family $g_y\in V$ is contained in a finitely generated submodule of V, then flatness should follow.

Let's try to show that V is finitely generated in some neighborhood of a point $P\in N$. Suppose that there exists an $a\in V$ satisfying $a(P)\not=0$. Then, on the set $\lbrace a\not=0 \rbrace$, any $b\in V$ can be decomposed uniquely into a multiple of a and its component orthogonal to a. By projecting onto the subspace of $\mathbb{R}^n$ orthogonal to a at each point, this allows us to reduce the problem to $\mathbb{R}^{n-1}$. Continuing in this way, by induction, we can reduce to the case where every $a\in V$ has $a(P)=0$ (at which point, h(P) must also be zero).

So, now consider the case of a point $P\in N$ at which all $a\in V$ vanish. Let r be the smallest positive integer at which the rth-order derivatives of some $a\in V$ are nonzero (if it exists). Then, recalling that I am only considering the case of $N=\mathbb{R}$, we can define $b(x)= a(x)/(x-P)^r\in V$ which satisfies $b(P)=a^{(r)}(P)/r!\not=0$. This is a contradiction.

So, this reduces the problem to the points $P\in N$ on the boundary of the set $\lbrace h=0\rbrace$ at which both h and every $a\in V$ along with all their derivatives vanish. I will denote this (necessarily closed) set by S. We need to show that the smooth family $g_y\in V$ is contained in a finitely generated submodule of V, in some neighborhood of S.

The case for n=1 can be completed, as I did in my other response proving your "simple" case. Then, as I explained, there exists a $G\in V$ which is strictly positive outside of {h=0} and which goes to zero much slower than any of the gy at S, so that $g_y/G\in C^\infty(N)$ and $g_y = (g_y/G)G$ are all in the singly generated submodule of V generated by G.

We can attempt to solve the case n>1 similarly. First, let $a_i = e_i - |h|^{-2} h_i h$ be the projections of $e_{ij}=\delta_{ij}$ orthogonal to h, which is defined, bounded, and smooth on $\lbrace h\not= 0\rbrace$, and gy decomposes as $g_y=\sum_k a_{k}(g_{y})_k$, as I did above. The coordinates $(g_y)_k\in C^{\infty}(N)$ and all their derivatives vanish at S. Then, as in the n=1 case, there will be a smooth function G vanishing on S such that $(g_y)_k/G$ are all smooth, and we can decompose $g_y=\sum_k (G a_{k})((g_y)_k/G)$. If $G a_k$ are smooth at S then they generate a submodule of V containing $g_y$. The problem is that, even if G and all its derivatives go to zero faster than polynomially in the distance d from S, the same cannot be said for $G a_k$. It is possible that, even though ak is bounded, its derivatives can explode faster than polynomially in 1/d.

If any counterexample exists to disprove flatness, the problem must occur at a point P on the manifold where h and all its derivatives vanish, but the derivatives of $\hat h=h/|h|$ explode faster than exponentially in the reciprocal of the distance from S. That is, $\hat h=h/|h|$ moves around very wildly at P.

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Wouldn't something like h=f(x)(sin(k(x)),cos(k(x))) where k(x)=exp(1/x) and f(x) is positive on x>0 and goes sufficiently fast to zero at 0 provide such an example? – Michael Bächtold Feb 19 '10 at 19:02
In that case, I think a(x) in V has to be of the form a(x)=b(x)(cos(k),-sin(k)) where b(x)k(x) is smooth, and it just reduces to the n=1 case. But that is the kind of thing I was thinking of. – George Lowther Feb 19 '10 at 22:47

I have an idea for the case of a submersion - maybe it is nonsense but maybe not.

In the algebraic case of nonsingular varieties X,Y over an algebraically closed field one can check smoothness (which implies flatness) of a morphism X -> Y in terms of the induced maps of the Zariski tangent spaces at closed points being surjective. So in particular, identifying the Zariski tangent spaces with the fibres of the tangent bundles the "algebraic version of submersion checked on closed points" implies flatness.

If we take the viewpoint of differentiable spaces and work with injectivity of the cotangent sheaf tensored with residue fields rather than surjectivity of the tangent bundle maybe it is possible to transfer the proof to the case you are interested in? All one really needs is to translate the statement that we have an injective morphism on the fibres of the cotangent bundle into one about regular sequences in the rings of germs and then hopefully one could use a version of the local criterion of flatness to conclude.

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Thanks i'll have a look at that. Could you give me a reference for that argument of translating injectivity into regular sequences? – Michael Bächtold Oct 25 '09 at 17:00
A reference is Hartshorne Ch III Prop 10.4 (the proof of (iii) implies (i)). The basic idea is that identifying the stalks of the cotangent sheaf with m/m^2 one can lift a basis to a regular sequence at x and f(x) and injectivity says the sequence at f(x) maps injectively into the one at x. One can take the quotient at f(x) by the max ideal and by the corresponding image at x which is flat since the quotient at f(x) is the residue field. Then one uses the local criterion to finish by induction. – Greg Stevenson Oct 25 '09 at 20:21
So everything would go trough, except for the "local criterion of flatness" (Eisenbud Ch 6.4) which assumes notherianity. I've had a look at two proofs of it and they both rely on Krulls theorem applied to the maximal ideal in the germ of functions. That doesn't work here because of "flat smooth functions". For the moment I don't see an easy way to adapt the proof. – Michael Bächtold Oct 27 '09 at 14:04
@Greg: In general, the stalks of the sheaf of a manifold are not noetherian, so Idon't believe it will work. – Harry Gindi Feb 17 '10 at 0:47
Oh, haha, that's what the comment above mine says! – Harry Gindi Feb 17 '10 at 0:48