3
$\begingroup$

R. Cohen proved the immersion conjecture in a 1985 Annals paper:

Cohen, Ralph L., The immersion conjecture for differentiable manifolds, Ann. Math. (2) 122, 237-328 (1985). ZBL0592.57022.

Any smooth compact n-dimensional manifold admits an immersion into Euclidean space of dimension 2n-a(n), where a(n) is the number of 1's in the binary decomposition of n.

Is there any result of this kind for (the total space of) a vector bundle E over compact manifold? Notice that the sphere bundle of E is compact. Maybe there is a silly argument...

$\endgroup$
13
  • $\begingroup$ What do you mean by immersion of a vector bundle? Do you mean an immersion of the total space? $\endgroup$
    – Misha
    May 20, 2019 at 20:51
  • $\begingroup$ @Misha: yes, sorry. $\endgroup$ May 20, 2019 at 21:08
  • 1
    $\begingroup$ The total space of a vector bundle is a smooth submanifold of the corresponding fiberwise one-point compactification, aka $S(E \oplus \Bbb R)$, which is a closed manifold of the same dimension $n$ as $E$ so long as the base is closed. Since the larger space immerses in $\Bbb R^{2n-a(n)}$, so does the smaller space. $\endgroup$
    – mme
    May 21, 2019 at 1:00
  • 1
    $\begingroup$ I suspect that you can do better using the h-principle. As vague evidence for this, if you have an immersion of the base $M$ in some Euclidean space, then its normal bundle $E$ immerses in the same Euclidean space (as an immersed tubular neighbourhood). $\endgroup$
    – Mark Grant
    May 21, 2019 at 8:59
  • 2
    $\begingroup$ @MarkGrant of course, that also works --- it was only obvious to me in retrospect that Cohen's theorem was also true for compact manifolds with boundary; but after all, I was just taking the double of what you said to get rid of the boundary, and whence a proof for the boundary case follows from the proof in the closed case. $\endgroup$
    – mme
    May 21, 2019 at 16:47

2 Answers 2

4
$\begingroup$

The answer is no.

Perhaps the simplest counter-example is for vector bundles over $0$-manifolds. They all immerse in $\mathbb R^n$ where $n$ is the dimension of the bundle. This is a considerably better number than $2n-a(n)$.

You might say, "but what about for non-discrete manifolds?" Such manifolds do immerse in $\mathbb R^{2n-a(n)}$ (according to Cohen). The issue is whether or not you can do better for this restricted subclass of manifolds.

Consider for example the line bundles over surfaces. The immersion conjecture would say they all immerse in $\mathbb R^{2\cdot 3 - 1} = \mathbb R^5$ but I think we can argue they immerse in $\mathbb R^4$.

Said another way, we are asking what line bundles are sub-bundles of normal bundles of immersed surfaces in $\mathbb R^4$. For orientable surfaces you can quickly generate any isomorphism type. For non-orientable surfaces I believe you can, as well. You immerse the surface in $\mathbb R^3$ and use the $1$-cocycle as a way to guide how you "flip" the bundle, when thought of as inside the larger $\mathbb R^4$.

$\endgroup$
6
  • $\begingroup$ An obnoxiously fancy way to phrase your final paragraph: rank 2 vector bundles are classified by their first Stiefel-Whitney class and their twisted Euler class. Given any immersion $i: S \to \Bbb R^3$, the normal bundle of $i': S \to \Bbb R^4$ is isomorphic to $\det(S) \oplus \Bbb R$, and hence has trivial twisted Euler class; but given any line bundle $\lambda$ over $S$, $\lambda \oplus (\det(S) \otimes \lambda^{-1})$ has $w_1 = w_1(S)$ and has $w_2 = w_1(\lambda)^2 - w_1(S) w_1(\lambda) = 0$, using the Wu relations or an explicit calculation of $w_1(S)$ and the cohomology of a surface $S$. $\endgroup$
    – mme
    May 21, 2019 at 17:18
  • $\begingroup$ Therefore $di'$ extends to a bundle monomorphism of $\lambda \to S$ and hence $i'$ extends to an immersion $\lambda \to \Bbb R^4$; of course because we can make the underlying manifold embedded, we can make this line bundle embedded. $\endgroup$
    – mme
    May 21, 2019 at 17:20
  • $\begingroup$ Sorry, maybe I should have stated my question more explicitly. By "this kind" I did not meant to say that "2n-a(n)" is sharp. Any bound, as long as it is better than Whitney's. $\endgroup$ May 21, 2019 at 20:14
  • 1
    $\begingroup$ I don't think anyone has written anything up. The Massey conjecture bound is "any bound" but as noted above, it's not sharp. I suppose if you were looking for sharp bounds, the place to start would be to do what Massey did: enumerate bundles over all cobordism class representatives, and find the best immersion dimension for those bundles. $\endgroup$ May 22, 2019 at 10:53
  • $\begingroup$ @Ryan: I don't see how to do what you propose, since the base is any compact manifold of any dimension... $\endgroup$ May 22, 2019 at 14:14
1
$\begingroup$

I don't know if the following remarks can be considered "of this kind".

Let $E$ be the total space of a rank $r$ vector bundle over the compact $n$-manifold $M$. Then, as you said, by Cohen $M\subseteq \mathbb{R}^{2n-a(n)}$.

Also (see Atiyah, K-theory, Corollary 1.4.14) you can realize $E$ as a sub-vector-bundle of a trivial bundle $E\subseteq \underline{\mathbb{R}}^m$ for a suitable $m$. In fact, looking at the proof of the preceding Lemma 1.4.12 it seems* to me that you can take $m$ to be $\leq r\times t(E)$ where $t(E)$ is the minimum number of elements of an open cover of $M$ which is trivializing for $E$. And from this, it seems that $t(E)\leq n+1$.

So, you could embed $E$ in $\mathbb{R}^{2n-a(n)+r\cdot (n+1)}$.

$^*$ You have maps of vector bundles $\theta_\alpha:{\underline{\mathbb{R}}}_{U_\alpha}^r\to E|_{U_\alpha}$ which you glue by a partition of unity $\{p_\alpha\}$, to obtain that $E$ is generated by global sections, by the map $\theta=\sum_\alpha p_\alpha \cdot\theta_\alpha:\prod^{t(E)}\underline{\mathbb{R}}^r\to E\to 0$. Since $E$ is isomorphic to its dual, you can dualize the map and obtain an inclusion of $E$ into a trivial v.b.

Remark: I haven't thought if this would give, in the case of a total space of a v.b., a better result than Cohen's embedding theorem itself applied to the projectivization $\mathbb{P}(E\oplus\mathbb{R})$ (or the one-point compactification of $E$ along the fibers) which has dimension $N=n+r$. One should ask if $2n-a(n)+r(n+1)< 2N-a(N)$ for some $n,r$.

$\endgroup$
3
  • $\begingroup$ Because $a(n) \leq \log_2(n)$, roughly your bounds are $(2+r)n - O(\log_2 n)$; the factor of $r$ is pretty noxious. Call the optimal bound $D(n,r)$; Cohen gives $D(n,0) = 2n - a(n)$. A slight improvement on the Whitney embedding theorem (see here) gives $D(n,r) \leq 2n+r$, which is a stronger bound for essentially every $n$. I don't see any clear way to improve the linked argument, even knowing Cohen's result. $\endgroup$
    – mme
    May 21, 2019 at 16:45
  • $\begingroup$ I'm not sure if I understood you correctly, but since a(n)<n, then 2n−a(n)+r(n+1) > 2(n+r) (except maybe for r=1), which is already Whitney's bound for E. So no better. (Sorry, I just saw Mike's answer to yours). $\endgroup$ May 21, 2019 at 20:33
  • 1
    $\begingroup$ Yes, the only cases that escape are $r=1$, which doesn't give anything useful; and $n=1$, which formally is better than Whitney by $1$, but equal to the improvement by Haefliger-Hirsch that I can see in wikipedia; and anyway not better than Cohen for any $r$ (And, by the way, there aren't many interesting vector bundles on the circle, are there?...). $\endgroup$
    – Qfwfq
    May 21, 2019 at 22:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.