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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

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2 Answers 2

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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 $\mathbb Z/2\mathbb Z$.

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    $\begingroup$ 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
    May 15, 2016 at 17:20
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    $\begingroup$ 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 $k$. $\endgroup$ 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
    May 17, 2016 at 9:58
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    $\begingroup$ rank means dimension of the fibers $\endgroup$ May 18, 2016 at 11:12
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    $\begingroup$ My guess above is simply wrong. $\endgroup$ Jul 10, 2016 at 15:02
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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.

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