11
$\begingroup$

If $M$ and $N$ are manifolds with $\dim M< \dim N$, we denote by $Imm\left(M,N\right)$ the space of immersions of $M$ in $N$.

Let $M$ and $M'$ be a simply connected manifolds of dimensions $m>0$. It is true that if $M$ is homotopic to $M'$, then for $k\geq m$, the spaces $Imm\left(M,\mathbb{R}^{m+k}\right)$ and $Imm\left(M',\mathbb{R}^{m+k}\right)$ are homotopic? i.e

if $k\geq m$, then $M\simeq M'\Rightarrow Imm\left(M,\mathbb{R}^{m+k}\right)\simeq Imm\left(M',\mathbb{R}^{m+k}\right)$?

Thanks

Abdoul

Improve this question
$\endgroup$
0

2 Answers 2

Reset to default
17
$\begingroup$

No. For example, if $M$ is a Moebius band then, at least for even $k$, $Imm(M,\mathbb R^{2+k})$ is not homotopy equivalent to $Imm(S^1\times \mathbb R,\mathbb R^{2+k})$.

The latter is equivalent to the space of all maps from $S^1$ to $O(2+k)/O(k)$. Its first non-trivial homotopy group is $\pi_{k-1}=\pi_k(O(2+k)/O(k))=\mathbb Z$.

The former is equivalent to the space of sections of a nontrivial bundle with base $S^1$ and fiber $O(2+k)/O(k)$, in which the action of the fundamental group of the base on $\pi_k(O(2+k)/O(k))$ is nontrivial. Its first non-trivial homotopy group is $\mathbb Z/2\mathbb Z$.

Improve this answer
$\endgroup$
7
  • 1
    $\begingroup$ Thank you very much for this example. It is possible to have an example for simply connected manifolds? Because in this case the action of the base on the fiber is always trivial. Thanks $\endgroup$
    – Abdoul
    Commented May 15, 2016 at 17:20
  • 1
    $\begingroup$ I'm pretty sure that if you take $M$ to be a rank two vector bundle over $S^2$ with odd Euler class then $Imm(M,\mathbb R^{4+k})$ is inequivalent to $Imm(S^2\times \mathbb R^2,\mathbb R^{4+k})$ for large $k$. $\endgroup$
    – Tom Goodwillie
    Commented May 16, 2016 at 12:27
  • $\begingroup$ Thank you, I will try to prove that. But, the mean of "M to be a rank two vector bundle over $S^2$" is not very clear for me. Could you explain it to me please? Thank you again. Abdoul $\endgroup$
    – Abdoul
    Commented May 17, 2016 at 9:58
  • 1
    $\begingroup$ rank means dimension of the fibers $\endgroup$
    – Tom Goodwillie
    Commented May 18, 2016 at 11:12
  • 1
    $\begingroup$ My guess above is simply wrong. $\endgroup$
    – Tom Goodwillie
    Commented Jul 10, 2016 at 15:02
2
$\begingroup$

Note that when $M$ and $M′$ are simply connected, I have a positive answer in the rational case. I.e

If $k\geq m$ is an odd integer, then $M\simeq M'\Rightarrow Imm\left(M,\mathbb{R}^{m+k}\right)\simeq_{\mathbb{Q}}Imm\left(M',\mathbb{R}^{m+k}\right).$

Here $"\simeq_{\mathbb{Q}}"$ denotes the rational homotopy equivalence.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .