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For concreteness, I will work in the category of smooth manifolds, but my question makes sense in topological and PL category as well. Recall that a manifold $M$ is called open if every connected component of $M$ is non-compact.

Question. Is it true that for every $n$ there exists a compact $n$-dimensional manifold $N^n$ so that every open $n$-dimensional manifold $M^n$ admits an immersion in $N^n$? (In this context an immersion is just a local diffeomorphism.)

I think that the answer is positive and that the manifolds $N$ are connected sums of products of projective spaces of various dimensions.

Edit: Igor noted that real-projective spaces are not enough, one should include complex-projective spaces as factors in the products. (The reason I think that products real and complex projective spaces are the right thing to use is that real and complex projective spaces generate rings of unoriented and oriented cobordisms.)

Some background: As we know very well, not every manifold admits an open embedding in a compact manifold. For instance, the infinite connected sum of 2-dimensional tori does not. However, it is easy to prove that every open surface admits an immersion in ${\mathbb R}P^2$. Whitehead proved that every open oriented 3-dimensional manifold admits an immersion in ${\mathbb R}^3$ (and, hence, to any 3-dimensional manifold). I also convinced myself (although I do not have a complete proof) that every open non-orinentable 3-manifold admits an immersion in ${\mathbb R}P^2\times S^1$. More generally, every open paralelizable $n$-manifold admits an immersion in ${\mathbb R}^n$. This is a special case of the Hirsch-Smale immersion theory, which reduces existence of immersion from an open manifold to a homotopy-theoretic question about maps of tangent bundles.

John Pardon
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Misha
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    Doesn't the connected sum of products of projective spaces have zero rational Pontryagin classes? If so, the same would be true for any immersed submanifold of the same dimension. – Igor Belegradek Jun 19 '12 at 23:21
  • What does the torus trick, as given in Kirby Siebenmann do for you in regards to this question? – Scott Carter Jun 19 '12 at 23:51
  • @Igor: You are right, I have to include complex-projective spaces as possible factors. – Misha Jun 20 '12 at 03:49
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    @Scott: Torus trick is based on the fact that if $M$ is a closed $n$-manifold with trivial tangent bundle (say, a torus), then $M-p$ admits an immersion in ${\mathbb R}^n$, which is a direct corollary of Hirsch's theorem. This does not really help (at least in an obvious way) with general smooth manifolds which have nontrivial tangent bundles. They will not immerse in ${\mathbb R}^n$, so one needs more general targets, which you can see already in dimension $2$. – Misha Jun 20 '12 at 03:57
  • @Daniele: I know this paper and used it in the past in a work which was a partial motivation for my question. Philips' result is a variation on H-S theory and gives the same answer: Reduction of the problem to a homotopy-theoretic question. – Misha Jun 20 '12 at 15:15
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    May be this paper can help you: A. Phillips, "Submersions of open manifolds", Topology 6 (1967), 171-206. The main result is that, for $M$ open and $\dim M \geq \dim N$, there is a submersion $M \to N$ iff there is a bundle epimorphism $TM \to TN$. From this you get the following: every oriented 4-manifold with the homotopy type of a (not necessarily finite) 2-complex can be immersed in $CP^2$. – Daniele Zuddas Jun 20 '12 at 15:22
  • I modified my comment with a further remark. – Daniele Zuddas Jun 20 '12 at 15:25
  • @Daniele: This is nice and goes in the right direction. How do you construct a bundle map in the case of an open 4-manifold like this? – Misha Jun 20 '12 at 16:52
  • @Daniele: Could you provide details as an "answer" since there is probably not enough room for a proof in the comments? – Misha Jun 20 '12 at 18:53
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    Roughly, you are trying to approximate the $n-1$-skeleton of $BO(n)$ by a closed (compact?) $n$-manifold and its canonical map. I think that the finiteness of the skeleton means that there is a best $n$-manifold, so that all open $n$-manifolds that immerse in a closed $n$-manifold immerse in this one. The Wu formula restricts tangent bundles, so we cannot fully approximate the $n-1$-skeleton. Key question: do all restrictions on tangent bundles of closed manifolds apply to open manifolds? (Brown and Peterson may be relevant. They probably built something more relevant than BO(n).) – Ben Wieland Jun 20 '12 at 19:24
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    Note that an obvious relative version is false. You might hope to construct an immersion $M\to N$ by induction over a handle structure of $M$ (without any index $n$ handles, since $M$ is open)). But there can be no $N$ such that immersions of $n$-manifolds in $N$ can always be extended across a handle of index less than $n$: by Smale-Hirsch, that would mean that the relative homotopy groups of the tangent bundle map $N\to BO_n$ vanish up to $\pi_{n-1}$. But that would force the mod $2$ Betti numbers of $N$ to violate Poincare duality. – Tom Goodwillie Jun 21 '12 at 21:52
  • @Ben: Thanks for the comments, I will take a look at the B-P papers. – Misha Jun 22 '12 at 00:38
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    @Tom: Could you write more details as an "answer": I tried, unsuccessfully, inductive arguments and it would be nice to know why exactly they do no work. What is the least dimension where the relative version fails? – Misha Jun 22 '12 at 21:55
  • I don't know what I was thinking! – Tom Goodwillie May 13 '16 at 17:15

1 Answers1

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This is the special case announced in the comment. We assume $M$ to be an oriented 4-manifold with the homotopy type of a 2-complex. We prove that $M$ immerses in $\mathbb{C}P^2$.

In the case of compact 4-manifolds with boundary I have a more direct and elementary proof in my paper which does not depend on Phillips' theorem.

Anyway, a bundle map $TM \to T\mathbb{C}P^2$ (needed to apply the theorem of Phillips) can be constructed in this way. Firstly, endow $M$ with an almost-complex structure (which exists by our assumptions). There is a nowhere vanishing vector field on $M$, hence $TM$ splits as a Whitney sum of complex rank 1 bundles $TM = \xi \oplus \varepsilon^1$ with $\varepsilon^1$ trivial and $\xi$ a pullback of the complex universal bundle. By cellular approximation, $\xi$ is a pullback of the canonical bundle $\gamma^1_1$ on $\mathbb{C}P^1$, so $TM$ is the pullback of $\gamma^1_1 \oplus \varepsilon^1$. Now, rank 4 real oriented vector bundles over $\mathbb{C}P^1$ are classified by $\Bbb Z_2$ (via the second Stiefel-Whitney class), hence $T\mathbb{CP}^2_{|CP^1} = \gamma^1_1 \oplus \varepsilon^1$ (as real vector bundles). So $TM$ is a pullback of $T\mathbb{C}P^2$ by a suitable map $M \to \mathbb{C}P^2$ (with values in $\mathbb{C}P^1$).

David White
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Daniele Zuddas
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