18
$\begingroup$

The title is the question. Sorry, this isn't quite research level. I imagine the answer is well-known, just not to me. Thanks for any help!

$\endgroup$
  • $\begingroup$ some tag with "topology" would be natural (I'm not sure whether this is considered as algebraic or geometric topology, or both) $\endgroup$ – YCor May 7 '15 at 22:28
  • 1
    $\begingroup$ btw there was no comment about the meaning of the question: does it mean continuous immersions modulo continuous isotopy, smooth immersion modulo smooth isotopy, and are these two point of views equivalent in any dimension? $\endgroup$ – YCor May 7 '15 at 22:30
  • $\begingroup$ @YCor, I think "immersion" is inherently a smooth notion, so it means smooth immersions modulo smooth isotopy. But the Smale-Hirsch theorem turns it into a topological problem about continuous maps. $\endgroup$ – Dylan Thurston May 8 '15 at 0:40
  • $\begingroup$ @DylanThurston: an immersion can be characterized as a map between manifolds $f:X\to Y$ such that for every $x\in X$ there are neighborhoods $V,V'$ of $x$ and $f(x)$ such that $f(V)\subset V'$ and the restriction of $V$ can be described in coordinates as the inclusion of a subspace of $\mathbf{R}^d$ into $\mathbf{R}^n$. This makes sense in the continuous setting. But OK, I understand there's consensus that the setting is the smooth setting. $\endgroup$ – YCor May 8 '15 at 8:01
  • $\begingroup$ @YCor: this is smooth immersions up to 1-parameter families of immersions. Sometimes this is called "regular isotopy" to distinguish it from "isotopy". We're talking about the homotopy-type of the space of immersions. Immersions up to plain isotopy have an enormous amount of components, primarily indexed by the image (as a stratified space). $\endgroup$ – Ryan Budney May 8 '15 at 15:13
15
$\begingroup$

I'm not a professional topologist by any means, but let me give this a shot. There's some discussion in the first few lectures of John Francis from this course. Please correct me if I've made mistakes below...

By Smale and Hirsch, the space of immersions of $S^n$ into $R^{n+1}$ is homotopy equivalent to the space of unbased maps from $S^n$ into $V_{n}(\mathbb{R}^{n+1})$, $Map(S^n,V_n(\mathbb{R}^{n+1})$. Here $V_n(\mathbb{R}^{n+1})$ is the Stiefel manifold of $n$ frames in $n+1$-space. $V_{n}(\mathbb{R}^{n+1})$ is homeomorphic to $SO(n+1)$.

See Ryan's answer for some more details.

Your question is then basically equivalent to characterizing $\pi_1(Map(S^{n},SO(n+1)))$, the fundamental group of that mapping space.

As B.S. points out in his answer, since $SO(n+1)$ is a group, $Map(S^n,SO(n+1))$ is homotopy equivalent to $SO(n+1)\times\Omega^n(SO(n+1))$, therefore the above fundamental group is $\pi_1(SO(n+1))\times\pi_1(\Omega^n(SO(n+1)))\cong\pi_1(SO(n+1))\times\pi_{n+1}(SO(n+1))$.

Though B.S. has pointed out that we can see that it's always nontrivial just from the first factor, we can in fact compute the group from known results on homotopy groups of $SO(n+1)$ (see e.g. this table compiled from the literature by Klaus Johannson).

The result for all $n$ is (scroll right in the grey box below):

n| 1   | 2      | 3              | 4         | 5    | 6      | 8s-1                | 8s             | 8s+1        | 8s+2      | 8s+3           | 8s+4      | 8s+5      | 8s+6   |
-| --- | ------ | -------------- | --------- | ---- | ------ | ------------------- | -------------- | ----------- | --------- | -------------- | --------- | --------- | ------ |
 | Z   | Z_2+Z  | Z_2+Z_2+Z_2    | Z_2+Z_2   | Z_2  | Z_2+Z  | Z_2+Z_2+Z_2+Z_2     | Z_2+Z_2+Z_2    | Z_2+Z+Z_2   | Z_2+Z_2   | Z_2+Z_2+Z_2    | Z_2+Z_2   | Z_2+Z_4   | Z_2+Z  |

In the rightmost columns, $s$ is any integer greater than or equal to 1, and the plus signs denote direct sum. Apologies for the ugly formatting of the table.

$\endgroup$
  • 1
    $\begingroup$ I think that is not what Hirsch-Smale says here. Hirsch-Smale says that the space of immersions is homotopy equivalent to the space of sections of the fiber bundle over $S^n$ with fiber $V_n(\mathbb{R}^{n+1})$ associated to the tangent bundle, which is trivializable only for $n = 1, 3, 7$. You also need to be more careful about the distinction between spaces of based vs. unbased maps; e.g. based maps from $S^1$ is the based loop space but free maps from $S^1$ is the free loop space, and these have different homotopy groups in general. $\endgroup$ – Qiaochu Yuan May 7 '15 at 18:10
  • 1
    $\begingroup$ @Qiaochu: literally you're right but there is a reduction in this particular case. Smale-Hirsch says immersions (via the derivative) are homotopy-equivalent to the bundle mono-morphisms $TS^n \to T \mathbb R^{n+1}$, but using the almost-trivializability of $TS^n$ you can check this mapping space has the same homotopy-type as $\Omega^n SO_{n+1}$. $\endgroup$ – Ryan Budney May 7 '15 at 18:14
  • 4
    $\begingroup$ @Ryan: huh. Okay, I'm willing to believe that it has the same homotopy type as the space of maps from $S^n$ to $SO(n+1)$, but surely this doesn't in turn have the same homotopy type as the space of based maps...? $\endgroup$ – Qiaochu Yuan May 7 '15 at 18:16
  • 2
    $\begingroup$ Oh, yes, you're right there. It's the unbased loop space. The based loop space would be immersions with a fixed behaviour at a point. I'll supply the details in a short partial-answer. $\endgroup$ – Ryan Budney May 7 '15 at 18:17
  • 2
    $\begingroup$ I think there's a fibration $\Omega^n(SO(n+1))\rightarrow X \rightarrow SO(n+1)$, where X is the space of unbased maps from $S^n$ to $SO(n+1)$. So you can look at the long exact sequence attached to this fibration to try to compute $\pi_1(X)$. Since $\pi_2(SO(n+1))$ vanishes, this only has five nontrivial terms. $\star \rightarrow \pi_{n+1}(SO(n+1)) \rightarrow \pi_1(X) \rightarrow \mathbb{Z}_2 \rightarrow \pi_n(SO(n+1)) \rightarrow \pi_0(X) \rightarrow \star$. By Kervaire's table for $\pi_{n+1}(SO(n+1))$ you basically have the answer up to knowing how $\pi_1$ acts on $\pi_n$ for $SO(n+1)$ $\endgroup$ – Noah Snyder May 7 '15 at 20:12
13
$\begingroup$

Your space is never simply connected (for $n\geq 1$). As already answered, it is (weakly) homotopy equivalent by Smale-Hirsch h-principle to the space of unbased maps from $S^n$ to $SO(n+1)$, which is itself $SO(n+1)\times \Omega^n SO(n+1)$ ($\Omega^n$ = based maps from $S^n$). So $\pi_1$ is at least $\pi_1(SO(n+1))$ (and more in general). In fact it is not even connected as soon as $\pi_n SO(n+1)$ isn't trivial, which occurs for $n=1,3,4,5$ and many more (maybe all except $n=2$?). But it is connected for $n=2$, which awarded its celebrity to Smale (sphere eversion), even if his advisor Raoul Bott didn't believed it at first. EDIT: according to j.c. $\pi_n SO(n+1)$ is trivial only for $n=2,6$. Quite an interesting fact !

$\endgroup$
  • 1
    $\begingroup$ Your answer shows that for any choice of base-point there is a canonical surjective homomorphism from this fundamental group to the cyclic group of order two (induced by the projection to $SO(n+1)$). This should correspond to a natural 2-fold covering (of the space of immersions). What is it? $\endgroup$ – YCor May 7 '15 at 22:26
  • $\begingroup$ Could you say a little more about why the space of unbased maps from $S^n$ to $SO(n+1)$ the same as $SO(n+1)\times\Omega^nSO(n+1)$? By the way, $\pi_n SO(n+1)$ is given in the table cited in my answer, it's nontrivial except for $n=2$ and $n=6$. $\endgroup$ – j.c. May 7 '15 at 22:31
  • 2
    $\begingroup$ @j.c.: Since $SO(n+1)$ is a group, the space of maps from $S^n$ to $SO(n+1)$ is the same as the product of $SO(n+1)$ (image of the base point of $S^n$) and $\Omega^n SO(n+1)$ (based maps $\infty\mapsto I$). By the way, I didn't know of the case $n=6$. Thanks ! $\endgroup$ – BS. May 7 '15 at 23:26
12
$\begingroup$

Let me just fill-in the gap in j.c.'s exposition. Smale-Hirsch states that the derivative from the space of immersions $Imm(S^n, \mathbb R^{n+1})$ to the space of bundle monomorphisms $Mono(TS^n, T\mathbb R^{n+1})$ is a homotopy-equivalence.

There is a cute observation that allows you to nail-down the space of fibrewise one-to-one maps $TS^n \to T\mathbb R^{n+1}$.

Notice that $TS^n$ is virtually trivial, i.e. $TS^n \oplus \epsilon^1 = S^n \times \mathbb R^{n+1}$, where $\epsilon^1$ is a line bundle over $S^n$.

So the idea is to extend any bundle monomorphism to an orientation-preserving bundle monomorphism $S^n \times \mathbb R^{n+1} \to T \mathbb R^{n+1}$. You can do this continuously, over the entire space using the a cross-product type construction. This is a homotopy-equivalence between $Mono(TS^n, T\mathbb R^{n+1})$ and the $n$-fold free loop space of $SO_{n+1}$.

$\endgroup$
  • 1
    $\begingroup$ The space of unbased maps from $S^n$ to a space is not the $n$-fold free loop space; that would be the space of unbased maps from $T^n$. $\endgroup$ – Qiaochu Yuan May 7 '15 at 18:45
  • 2
    $\begingroup$ Qiaochu you're too much a grad student. That's needlessly pedantic. $\endgroup$ – Ryan Budney May 7 '15 at 19:38
  • 5
    $\begingroup$ @QiaochuYuan If it helps, I suspect Ryan's comment lies in the space of unbased critiques from Professors to Grad Students. $\endgroup$ – Vidit Nanda May 7 '15 at 21:11
  • 1
    $\begingroup$ Unbased, only up to a controlled homotopy. $\endgroup$ – Ryan Budney May 7 '15 at 22:05
  • 2
    $\begingroup$ @RyanBudney Observe: the mapping space $\text{map}(S^n,G)$ of a Lie group $G$ splits as a product $\Omega^n G \times G$. $\endgroup$ – John Klein May 8 '15 at 2:02
7
$\begingroup$

For $n=1$, the space $Imm(S^1,\mathbb R^2)$ has $\mathbb Z$ many connected components described by the rotation index. In each case the fundamental group is $\mathbb Z$. See Thm 2.10 of here for the components with rotation index $\ne 0$, and see this paper for rotation index 0.

$\endgroup$

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.