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Question as in title (Diff = category of smooth manifolds and smooth maps)

I thought I'd convinced myself this is true, so this is just a sanity check.

Also, what about for settings other than smooth manifolds? (like analytic manifolds, complex manifolds, or less differentiable - say, $C^2$ manifolds)

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By "local-section-admitting" what do you mean, exactly? That locally at each point there is a section? is the map $(x,y)\in\mathbb R^2\mapsto xy\in\mathbb R$ local-section-admitting? –  Mariano Suárez-Alvarez Sep 28 '10 at 12:59
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Given points $x$ in the source and $y$ in the target of $f$, you can find a local section $s$ such that $s(y)=x$ iff $f(x)=y$ and $D_x f$ is onto. (Use the chain rule and the implicit function theorem.) –  Tim Perutz Sep 28 '10 at 13:27
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That's funnny, without thinking twice, I would have said that $f:X \to Y$ admits local sections if for every $y\in Y$ there is a neighbourhood V of y and a morphism $g:V \to X$ with $f \circ g=Id_V$. Surprisingly there is no consensus, judging from the reactions by Tim, Mariano and André. Where is Bourbaki when you need him to establish the definitive terminology? [I know he wrote a "Fascicule des Résultats" on manifolds, but nobody seems to ever have used it] –  Georges Elencwajg Sep 28 '10 at 14:15
    
@David, in summary, once you settle on a precise definition the answer to your questions is going to follow from the implicit function theorem. –  Ryan Budney Sep 28 '10 at 14:18
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@Georges: Lang's book Fundamentals of differential geometry (which is expanded from his earlier DG book) was partially an attempt to expand the Fascicule des Résultats out to a full book. –  Harry Gindi Sep 28 '10 at 14:43
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2 Answers

up vote 11 down vote accepted

There are two possible meanings for the sentence "f : MN admits local sections", so let's first disambiguate.

Meaning 1: For every point of N, there exists a neighborhood of that points and a section from that neighborhood back to M.

That's what people typically check in order to verify that, say, a map is a $G$-principal bundle.

Meaning 2: For every point mM, there exists a neighborhood of $f(m)$, and a section s from that neighborhood back to M, subject to the extra condition that $s(f(m))=m$.

Clearly, you care about the second meaning of that sentence.


It is correct that a map is a submersion (not necessarily surjective!) iff it admits local sections.

If a map has local sections, then the maps on tangent spaces are sujective: that's just obvious.

Conversely, if a map is surjective at the level of tangent spaces, you first pick a local section of the maps of tangent spaces. Then, to finish the argument, you use the fact that any subspace of the tangent space $T_mM$ is the tangent space of a submanifold of M, and apply the implicit function theorem.

Note: if you care about infinite dimensional Banach manifolds, then the existence of a section for the map to tangent spaces needs to be assumed a separate condition. Indeed, it's not enough to assume that the map of tangent spaces is surjective, since it's not true that any surjective map of Banach spaces has a section.

Note: For complex varieties, you don't have the implicit function theorem, so it doens't work. Counterexample: the map $z\mapsto z^2$ from ℂ* to itself. The fix is to pass the the "étale topology"... but that's another story.

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In fact, a map with property "Meaning 2", in the setting of topological spaces, is said to be a "topological submersion". –  David Carchedi Sep 28 '10 at 14:12
    
Thanks, Andre. That disambiguated things nicely. I don't think there would be a nice characterisation of maps in Diff with local sections in the sense of Meaning 1 after all. –  David Roberts Sep 28 '10 at 21:21
    
On the first Note: indeed, a linear bounded operator between Banach spaces is a linear section iff it is surjective and its kernel splits. Linear sections are an open set in the space of linear operators between two Banach, also. So the precise criterium for "$f:M\to N$ is a submersion locally at $m\in M$" (this is how I express the Meaning 2) is just: $Df(m)$ is a linear section. –  Pietro Majer Sep 28 '10 at 22:32
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Dear David: yes!

In one direction this is just the functoriality of tangent maps. Let $f:X\to Y$ be the morphism, $x$ a point in $X$ with image $y\in Y$ and $g:V\to X$ a local section. From $f \circ g=Id_V$ you get $f_{\ast x} \circ g_{\ast y}=Id_{\ast y}$ and this implies that $f_{\ast x}$ is surjective i.e. that f is a submersion at $x$.

The other direction is not formal and depends on a theorem giving the local form of a submersion: this is much harder and is equivalent to the implicit function theorem or the local diffeomorphism theorm. It is true in the category of $C^k-$ manifolds, $k\geq 1$, and in that of real or complex analytic manifolds.

However it is not true in an algebraic geometry context. For example the squaring map $\mathbb C\to \mathbb C:z\mapsto z^2$ is a surjective submersion but has no local (in the Zariski sense) algebraic (= rational) section.To remedy this, Grothendieck introduced a new branch in Algebraic Geometry called Etale Topology, and more generally Grothendieck Topologies.

Edited (later): I hadn't defined "admitting local sections". Just as Tim observes in his comment, the answer "yes" is only correct with the understanding that through every $x\in X$ there passes a section defined in a neighbourhood of $y=f(x)$. This is also the " Meaning 2" in André's post, who quite judiciously chooses it as the relevant one.

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+1 for the extension to the other categories, but yes, I was confused about how I was defining 'local-section-admitting'. –  David Roberts Sep 28 '10 at 21:20
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