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Is every homeomorphism $h$ of $S^n\times S^1$ with identity action in $\pi_1$ pseudo isotopic to a homeomorphism $g$ of $S^n\times S^1$ such that $g(S^n\times x)=S^n\times x$ for each $x\in S^1$? I would be satisfied with an answer for $n>3$.

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    $\begingroup$ Is this a homework question? $\endgroup$ – Anthony Quas Jan 31 '17 at 16:32
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    $\begingroup$ @AnthonyQuas: where do they teach courses on topological isotopies? I do not know the answer to the OP question. If you do, please give a hint. $\endgroup$ – Igor Belegradek Jan 31 '17 at 17:01
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    $\begingroup$ @olga: could you (edit the question to) clarify which $n$ you care about? $\endgroup$ – Igor Belegradek Jan 31 '17 at 17:24
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    $\begingroup$ While it may be tempting to be impatient about this question, I don't think it's deserved. Yes, there are a couple of basic obstructions to the existence of such an isotopy (n=1 and the orientation on $\pi_1$, as in the answers so far). Behind those, it seems to me that there is a mathematical question of some real substance, as indicated by the second case in Igor Rivin's answer. There is no need to be dismissive. $\endgroup$ – Tyler Lawson Jan 31 '17 at 18:09
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    $\begingroup$ @TylerLawson In fact, it is virtually certain that the case $n>3$ is completely open. $\endgroup$ – Igor Rivin Jan 31 '17 at 19:13
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This is false for $n=1.$ The mapping class group of the torus is $SL(2, Z),$ of which the homeomorphisms you describe are but a small part - the parabolic matrices $\begin{pmatrix}1 & n\\ 0 &1\end{pmatrix},$ unless I am very confused. I cautiously believe the statement is true for $n=2,$ by

Allen Hatcher, MR 420620 Homeomorphisms of sufficiently large $P^{2}$-irreducible $3$-manifolds, Topology 15 (1976), no. 4, 343--347.

UPDATE for $n=3,$ the best I can find is:

Richard Stong and Zhenghan Wang, MR 1769331 Self-homeomorphisms of 4-manifolds with fundamental group Z, Topology Appl. 106 (2000), no. 1, 49--56.

Which classifies up to pseudo-isotopy (NOT isotopy), and does give the expected result, as far as I can tell.

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  • $\begingroup$ I need only $n>3$ !!! $\endgroup$ – olga Jan 31 '17 at 18:59
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Some older results reduce the problem to a calculation. Browder (Diffeomorphisms of 1-Connected Manifolds, Transactions of the American Mathematical Society Vol. 128, No. 1 (Jul., 1967), pp. 155-163) looks at PL homeomorphisms of $S^1 \times S^n$, and shows that any such homeomorphism is pseudo-isotopic to one that preserves a copy of $S^n$. I would imagine that the same proof works for homeomorphisms. (I am assuming that you mean homeomorphism rather than eg diffeomorphisms.) In the dimensions you are considering, there is an obstruction due to Hatcher and Wagoner, to getting an isotopy from this pseudo-isotopy, that lives in a group called $Wh_2(\mathbb{Z})$. I had some notion that this group vanishes, but can't document this.

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    $\begingroup$ Hatcher-Wagoner probably already proved that $Wh_2(\mathbb Z)=0$, but it is also covered by Loday. It is just the fact that the higher $K$-theory of $\mathbb Z[\mathbb Z]=\mathbb Z[t^\pm]$ is easily described in terms of $K(\mathbb Z)$. $\endgroup$ – Ben Wieland Feb 1 '17 at 3:23
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    $\begingroup$ The difference between smooth and TOP pseudoisotopies is probably nontrivial. In any case all known computations are in the pseudoisotopy stable range, where the map on the homotopy groups induced by the inclusion from the smooth to TOP pseudoisotopies has finitely generated kernel and cokernel, see corollary 4.2 in arxiv.org/pdf/math/0607367v4 for a precise statement. $\endgroup$ – Igor Belegradek Feb 1 '17 at 4:20
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    $\begingroup$ Indeed I need the result that for $n>3$ there is no difference between smooth and TOP pseudoisotopies of homeomorphisms $S^n\times S^1$. $\endgroup$ – olga Feb 4 '17 at 6:28
  • $\begingroup$ The obstruction to isotopy is not just $Wh_2(\pi_1)$, as indicated by the title of Allen Hatcher's first paper. He mentions $S^n\times S^1$ as the very simplest example of the obstruction being realized here. $\endgroup$ – Ben Wieland Feb 20 '17 at 16:24
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I am happy to say that you help me to find the positive answer on my question. Due to our references I found what I wanted. Namely, on page 71-72 in "A SURVEY OF MULTIDIMENSIONAL KNOTS" by M. KERVAIRE and C. WEBERthere is an answer on my question, for me it was enough to have pseudo isotopy. Thank you!!!

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    $\begingroup$ Could you explain how what is written on pp.71-72 answers your question? The paper is at maths.ed.ac.uk/~aar/papers/kervsurv.pdf. $\endgroup$ – Igor Belegradek Feb 3 '17 at 13:47
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    $\begingroup$ With my pleasure. From pages 71-72 follows that the group of homeomorphisms under pseudo isotopy is isomorphic to $Z_2\times Z_2\times Z_2$. As I suppose that my homeomorphism preserve orientation in act identity in $\pi_1$ that their group is isomorphic to $Z_2$ and we now stamdart homeomorphism in each class: identity and with rotation (here arxiv.org/pdf/0903.1488v2.pdf it is possible to see some explanation for n=2 on page 2), both preserve spheres $S^n$. $\endgroup$ – olga Feb 4 '17 at 6:09
  • $\begingroup$ Isn't it true that all three standard generators fix $S^n \times \{0\}$? If so, then doesn't your argument show that there is no need for any assumptions on $h$? (And your argument also seems to show that if $h$ acts trivially on $\pi_1$, then it is possible to get the condition on $g$ for all $x$, not just some $x$, without assuming that $h$ is orientation preserving.) $\endgroup$ – Dave Witte Morris Feb 4 '17 at 22:43
  • $\begingroup$ I need that h is pseudo isotopic to a homeomorphism g which preserves all spheres. And it is true when h preserves orientation and acts identity in $\pi_1$. But, unfortunately, all results, ecxept n=3, are true not for homeomorphisms but for PL-homeomorphisms. Now I need some result in the direction that there is no difference between these groups. $\endgroup$ – olga Feb 5 '17 at 10:50
  • $\begingroup$ Then I think there is a mistake in the revised statement of your question: it says you want there to exist at least one sphere that is preserved, not all spheres, because it says "for some $x$", not "for all $x$". Please fix this if you really do mean "all" instead of "some". (Also, I don't see where you use the assumption that $h$ preserves orientation, because the homeomorphism that reverses the orientation on $S^n$ preserves all spheres.) $\endgroup$ – Dave Witte Morris Feb 5 '17 at 11:17
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For $n \neq 1$, the given homeomorphisms act trivially on the fundamental group, but applying an orientation-reversing homeomorphism to the $S^1$-factor yields a homeomorphism that is not trivial on the fundamental group.

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    $\begingroup$ Come on, the OP was slightly confused, but the question is interesting, and in her case completely open. No need to be mean. $\endgroup$ – Igor Rivin Jan 31 '17 at 19:49
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    $\begingroup$ In general, when someone posts an answer, the tacit implication should be that the question is worthy of consideration and not off-topic for this site. $\endgroup$ – Todd Trimble Jan 31 '17 at 21:52
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    $\begingroup$ @ToddTrimble I apologize for the error, but how should I have responded? In my mind, the answer to the stated question was easy, so the problem was not research level, but someone said they did not know the answer. I thought I was not supposed to answer a question in the comments, but is there an exception when I think the question does not belong on the site? And I did not realize that saying a question is not research level is being "mean", but I suppose I could have prefaced it with "In my opinion". $\endgroup$ – Dave Witte Morris Feb 1 '17 at 20:08
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    $\begingroup$ I understand that sometimes an urge to answer is almost irresistable; sometimes in the past I have commented that a question would belong better to Math.SE but also hint at an answer. But, I would avoid posting an answer and saying a question is off-topic or voting to close, because they really do pull in opposite directions. For people who answer who also help to get a question closed, nobody else can answer, but they got to answer (and net reputation points, etc.) -- that for example doesn't seem right. $\endgroup$ – Todd Trimble Feb 1 '17 at 21:20
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I attempt to answer the question:

What is the group of homeomorphisms of $S^n\times S^1$, up to pseudo-isotopy?

which seems of greater general interest. I will not completely answer it, but just do the easy part, producing an upper bound. I am not able to rule out the existence of a single pseudo-isotopy class of homeomorphisms homotopic to the identity but not pseudo-isotopic to the identity, which would be a counterexample to Olga’s question.

After writing this answer, I noticed that the paper of Browder cited by Danny Ruberman actually fully answers this question. And, in fact, it uses pretty much this method. But I think my answer is a little cleaner because it isolates surgery as an existing machine, while Browder was still working on the foundations of the theory. In particular, I have no restriction on the fundamental group of the manifold. He answers the whole question, ruling out the $\mathbb Z/2$ I was unable to address, but I don’t see how he handles it.


Absolute problems are difficult. Relative problems are easier. To attack homeomorphisms, put them in a long exact sequence involving homotopy equivalences and some kind of relative problem. Even after one computes the groups on either side in the long exact sequence, one is left with determining the image and quotient that contribute to the group of interest and the extension problem between them. But it gives pretty good upper and lower bounds.

Specifically, the group of homeomorphisms up to isotopy $\pi_0 Top(M)$ maps to the group of homotopy equivalences up to homotopy $\pi_0 hAut(M)$. The latter group is difficult to compute, let alone the image, though in your case it is easy. The kernel is the subgroup of homeomorphisms that are homotopic to the identity, up to isotopy. This receives a surjection from $\pi_1(hAut(M)/Top(M))$. I can’t compute that, but I mention it because of the exact sequence $$\pi_1hAut(M)\to \pi_1(hAut(M)/Top(M)) \to \pi_0 Top(M) \to \pi_0 hAut(M)$$ which suggests that $\pi_1hAut(M)$ will matter; indeed, it will reappear in an analogous role. Coarsening the equivalence relation from isotopy to pseudo-isotopy changes the group and makes it accessible. I will prove that the relative surgery structure set $S(M\times I;\partial)$ (which is a group under concatenation) surjects to the group of homeomorphisms homotopic to the identity, modulo pseudo-isotopy and the kernel receives a surjection from $\pi_1(hAut(M))$. And surgery structure sets are computable.

In your case it is easy to see that $$\pi_0hAut(S^n\times S^1) =(\mathbb Z/2)^3 =\pi_0O(2)\times\pi_0O(n+1)\times\pi_1O(n+1)$$ and thus that every homotopy equivalence lifts to a homeomorphism. I will compute $S(S^n\times S^1\times I;\partial)=\mathbb Z/2$. So that shows that the answer is either $(\mathbb Z/2)^3$ or an extension of that group by $\mathbb Z/2$.

Homotopy equivalences

First, compute the group of homotopy equivalences. Reduce from automorphisms to based automorphisms: in any geometric category the choice of a basepoint gives a map from the automorphism group to the space, with fiber are basepoint-preserving automorphisms: $Aut_*(M)\to Aut(M)\to M$. Based mapping spaces are easier to work with. Also, the coset space $hAut(M)/Top(M)$ doesn’t depend on whether there is a basepoint, so if we are interested in the image of $\pi_1hAut(M)$ we can replace it with the based version; $\pi_1hAut(M,*)=\mathbb Z/2$. Based mapping spaces are computed cell by cell. It is easy to compute $Map(S^n\vee S^1,S^n\times S^1)=\Omega S^1\times \Omega S^n\times \Omega^nS^n\times \Omega^nS^1=\mathbb Z\times\Omega S^n\times\Omega^n S^n$ because it is a map from a coproduct to a product. Adding a cell gives a fibration sequence: $Map((D^{n+1},\partial),S^n\times S^1)\to Map_*(S^n\times S^1,S^n\times S^1)\to Map_*(S^n\vee S^1,S^n\times S^1)$. And $Map((D^{n+1},\partial),X)$ is basically a torsor over $Map_*(S^{n+1},X)$. What makes this example easy, other than the small number of cells, is that two of the factors of the base split, so the homotopy groups in low degree are the sum of those of the base and the fiber.

Why Surgery?

Surgery computes the set of manifold structures on a given space. That’s something like the components of the homotopy fiber of the map from the category of manifolds and homeomorphisms to the category of spaces and homotopy equivalences. The components of the homotopy fiber have the homotopy type of the space $hAut(M)/Top(M)$ that I mentioned above. That isn’t the homotopy type of the space that surgery computes, but it gives a heuristic why surgery is relevant. A more precise, but more complicated reason is mapping tori. If $\varphi$ is a homeomorphism of $M$ then its mapping torus is the quotient $M\times I/(x,1)\sim(\varphi(x),0)$. If a homemorphism is homotopic to the identity, then its mapping torus is homotopy equivalent to the trivial bundle $M\times S^1$, and thus gives an element of the surgery structure set. If a homeomorphism is pseudo-isotopic to the identity, its mapping torus is homeomorphic to the trivial bundle. Thus homeomorphisms that are homotopic to the identity, up to pseudo-isotopy are relevant to $S(M\times S^1$). In fact, Shaneson’s splitting theorem says that $S(M\times S^1)=S(M\times I;\partial)\times S^h(M)$. The key ingredient is the relative surgery problem $S(M\times I;\partial)$ and this is also the fundamental group of the surgery space, connecting this to the prior heuristic.

Relative surgery

Leaving aside the motivation, I will show that $S(M\times I;\partial)$ is exactly what we want. It is a group under concatenation. It surjects to the group of homeomorphisms that are homotopic to the identity, up to pseudoisotopy; and the kernel of this map receives a surjection from $\pi_1hAut(M)$. Then I will compute $S(S^n\times S^1\times I;\partial)=\mathbb Z/2$.

For $W$ a manifold with boundary $S(W;\partial)$ is the set of maps $N\to W$ from a manifold to $W$ that are simple homotopy equivalences and such that they are homeomorphisms on the boundary, up to the equivalence relation of homeomorphisms $N\to N’$ that make triangle commute, up to homotopies which are constant on the boundary. Let us specialize to $W=M\times I$. By the $s$-cobordism theorem, any such $N$ is an $s$-cobordism, thus is homeomorphic to a cylinder $M\times I$. So this surgery set is really about the homotopy equivalence from $M\times I$ to itself. There is a homomorphism from $S(M\times I;\partial)$ to the group of homeomorphisms modulo pseudo-isotopy: given $N\in S(W;\partial)$ choose a cylinder structure $N=M\times I$ and compose with a constant homeomorphism of $M$ to make the structure map $N\to M\times I$ be the identity at time 0; and read off the homeomorphism at time 1. This is well-defined up to homeomorphisms of $M\times I$ that are the identity at time 0, ie, up to pseudo-isotopy. The image must be homotopic to the identity, because the structure map gives a homotopy. Conversely, every homeomorphism that is homotopic to the identity is in the image because a choice of homotopy gives a structure map. Finally, if we have $N\in S(M\times I)$ that yields the identity homeomorphism, its structure map gives a loop in $\pi_1(hAut(M))$. So we have an exact sequence $$\pi_1(hAut(M))\to S(M\times I;\partial)\to G \to \pi_0(hAut(M))$$ where $G$ is the sought group, the group of homeomorphisms up to pseudo-isotopy.

(This argument feels like I am identifying $S(M\times I;\partial)$ with $\pi_1(hAut(M)/H)$ for some group $H$, but I can’t see what it should be.)

Computation

Surgery theory computes $S(W;\partial)$ for a $w$-manifold $W$ by putting it in a long exact sequence $$[\Sigma W/\partial,G/Top]\to L_{w+1}(\pi_1W)\to S(W;\partial)\to [W/\partial,G/Top]\to L_w(\pi_1W)$$ Borel conjectured that a homotopy equivalence between aspherical closed manifolds is homotopic to the identity, ie, $S(B\pi)=*$. From the point of view of surgery theory, the natural generalization is that $L_*(\pi)=H_*(B\pi;L(e))$ for torsion-free groups $\pi$; and that $G/Top$ is the connected component of $L=L(e)$. This is known for our fundamental group $\mathbb Z$ and even though our space is not the circle, it means that there will be a lot of cancelation.

Specializing to $W=S^n\times S^1\times I$, we have $L_{n+2}(\mathbb Z)=L_{n+2}\oplus L_{n+1}$, while $S^n\times S^1$ is stably a wedge of spheres $S^{n+1}\vee S^n\vee S^1$. Thus $$[S^n\times S^1\times I/\partial,G/Top] =\pi_{n+2}G/Top\oplus \pi_{n+1}G/Top \oplus \pi_2G/Top\oplus \pi_1G/Top\\ =L_{n+2}\oplus L_{n+1}\oplus L_2\oplus L_1$$ which contains $L_{n+2}(\mathbb Z)$ as a summand. The Borel conjecture does not just say $L_{n+2}(\mathbb Z)$ is a summand, but that the map is compatible; and similarly for the terms a dimension up. Thus the structure set is the other summand: $S(S^n\times S^1\times I;\partial)=L_2\oplus L_1=\mathbb Z/2\oplus e=\mathbb Z/2$.

So $\mathbb Z/2$ is an upper bound to the size of the group of homeomorphisms homotopic to the identity, up to pseudo-isotopy. But my guess is that $\pi_1(hAut(S^n\times S^1,*))=\mathbb Z/2$ hits it and the actual group is trivial.

Addendum: Diffeomorphisms

There is a version of surgery theory for smooth manifolds. It replaces $G/Top$ with $G/O$. The space $G/O$ differs from $G/Top\sim L$ by the groups of exotic spheres. Thus there is less cancelation and it is generally more difficult to compute the smooth structure set. But in this case, with the stable splitting, it is fairly easy to compute, at least in terms of exotic spheres: $$S^{Diff}(S^n\times S^1\times I;\partial) =S^{Diff}(S^{n+2})\oplus S^{Diff}(S^{n+1})\oplus \mathbb Z/2$$ The $\mathbb Z/2$ is as before and I don’t know whether it gives exotic diffeomorphisms or comes from $\pi_1hAut(S^n\times S^1)$, but the others must yield diffeomorphisms not pseudo-isotopic to the identity. In fact, we can identify exotic spheres with exotic diffeomorphisms of lower dimensional spheres or disks: $S^{Diff}(S^{n+1})=\pi_0Diff^+(S^{n})=\pi_0Diff(D^{n};\partial)$. Thus $S^{Diff}(S^{n+1})$ gives diffeomorphisms of $S^n$ and thus $S^n\times S^1$; and $S^{Diff}(S^{n+2})$ gives diffeomorphisms of $D^{n+1}$ and thus any $n+1$-dimensional manifold, such as $S^{n}\times S^1$.

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