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The group $\mathbb{Z}_2$ acts on $S^7\times S^7$ by switching the coordinates with fixed point set $\Delta(S^7\times S^7)\cong S^7$. I want to know whether there are some other $\mathbb{Z}_2$ actions on $S^7\times S^7$ with fixed point set homeomorphic to $S^7$. We regard two actions as the same if they are conjugate in $\text{Aut}(S^7\times S^7)$.

I guess this question could be discussed in different categories, say TOP or SMOOTH, so i put $\text{Aut}$ instead of $\text{Homeo}$ or $\text{Diff}$.

The space $S^n×S^n$ is topologically rigid for odd $n$ (i.e. every homotopy equivalence from $M$ to $S^n×S^n$ is homotopic to a homeomorphism). You may replace $7$ by any other odd number. That would still be interesting to me.

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    $\begingroup$ Is there some reason $7$ is important here? $\endgroup$ – PVAL Oct 6 '14 at 4:42
  • $\begingroup$ $\ S^7\ $ has a rich algebraic structure. $\endgroup$ – Włodzimierz Holsztyński Oct 6 '14 at 9:01
  • $\begingroup$ One reason that i put $S^7\times S^7$ there:$S^n\times S^n$ is topologically rigid for odd $n$ (i.e. every homotopy equivalence from $M$ to $S^n\times S^n$ is homotopic to a homeomorphism).You may replace 7 by any other odd number.That would still be interesting to me. $\endgroup$ – sara Oct 6 '14 at 12:57
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    $\begingroup$ Well ... I'm not sure this is what you want, but you can sure obtain an exotic sphere as fixed point set ... $\endgroup$ – few_reps Oct 8 '14 at 9:13
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First, a warning. This kind of question, classifying structures up to the automorphisms of a manifold, is usually quite difficult. It is easier to classify them up to the connected component of the automorphism group, that is, up to isotopy.

It is possible to classify involutions of $S^n\times S^n$ whose fixed points are the diagonal, up to conjugacy by automorphisms in the connected component of the identity. There are finitely many equivalence classes. If $n$ is sufficiently large (7 is probably adequate), there are multiple classes.

A strategy for classifying actions by $F$ on $M$ that are free away from fixed set $M^F=X$: (1) Classify homotopy classes of maps $X\to M$; immersions up to isotopy; embeddings up to isotopy. (2) Classify appropriate actions on the normal bundle. (3) Remove a tubular neighborhood of $X$ to produce a manifold with boundary, where the putative action is already known on the boundary. Classify (free) homotopy actions on the space, compatible with the boundary, ie, free $F$-spaces that are homotopy equivalent, but may not be manifolds. (4) Apply surgery theory to the quotient to produce a manifold with fundamental group $F$ and boundary the quotient of the sphere bundle with the prescribed action. Taking the universal cover gives a manifold with boundary with a free $F$-action that can be glued to $X$ to produce an action of $F$ on a manifold homotopy equivalent to $M$ with prescribed fixed set. Surgery can be applied again to determine whether it is $M$.

All of these steps are individually computable, at least if both $M$ and the complement of $X$ are simply connected, so it's useful for $X$ to be codimension at least 3. The surgery step requires dimension at least 5. However, a step will often produce infinitely many possibilities; for any particular one, the next step is computable, but there is no guarantee that the infinite family has a nice description that can be dealt with simultaneously. For example, there are infinitely many homotopy classes of maps from $S^7$ to $S^7\times S^7$, each with their own embeddings.

For the smooth case, I believe that this procedure classifies actions up to conjugacy by the connected component of the diffeomorphism group. In the PL case, I am not sure how to do the bundle step. In the topological category, the fixed set can have a wild embedding and the procedure breaks down. But if you restrict to group actions with tamely embedded fixed point set, it is no worse than the PL case.

Let us pursue step 4. This makes sense in any category. For sufficiently large $n$, there are smooth involutions with fixed set the diagonal that are not conjugate to the usual by a diffeomorphism in the connected component of the identity, or even by a homeomorphism in the connected component of the identity. They are detected by deleting a tubular neighborhood of the diagonal to get a manifold with boundary and a free involution. The diffeomorphism or homeomorphism type of the quotient is an invariant of the involution. Recall the warning from the beginning: such classification is difficult. What is accessible to surgery theory is automorphisms within a fixed class, marked by its cover being homotopy equivalent to a subset of $M$, and other data from earlier steps. Classification up to this marking can be computed explicitly, but I will not do the calculations. I will wave my hands and assert that a group is large, hence nontrivial. It grows with $n$, but is probably nontrivial already at $n=7$.

Surgery theory says that the choice of manifold structures on a Poincaré duality space $Y$ is first approximated by a choice of a stable tangent bundle. More precisely, it a reduction of the structure group of the homotopy-invariant Spivak normal fibration (with structure "group" $G$) to the structure group $CAT$ for bundles for the appropriate category. If such a reduction exists (eg, if $Y$ is already a manifold), the choice of reduction is a torsor for the space of maps into the coset space $[Y,G/CAT]$. If $Y$ is already a manifold with boundary and we do not want to change the manifold structure on the boundary, the new bundle must restrict to the boundary to be the original, so we consider maps $[Y/\partial,G/CAT]$. The deviation of this first approximation from the answer is controlled by $L$-theory, which depends only on the dimension modulo 4 and the fundamental group. This is a bounded deviation, while $[Y,G/CAT]$ can become arbitrarily large for complicated $Y$. In our example, $[Y,G/CAT]$ becomes large as the dimension increases.

We apply surgery theory to two manifolds. One is $Y_1$, the complement of a tubular neighborhood of the diagonal, a deformation retract of the complement of the diagonal. It is homotopy equivalent to $S^n$ and has boundary the bundle of unit tangent vectors of $S^n$. The other is $Y_2$, the quotient by the involution, with boundary a $\mathbb RP^{n-1}$ bundle over $S^n$. Surgery yields manifolds homotopy equivalent to $Y_2$, with the same boundary. Taking the double cover and gluing back in the tubular neighborhood yields an involution on a manifold homotopy equivalent to $S^n\times S^n$. If $n$ is odd and we are in the PL or topological categories, there is only one such manifold, as you said, so this yields exotic involutions on $S^n\times S^n$.

So exotic involutions in the PL or topological case are close to $[Y_2/\partial,G/CAT]$. Maps into $G/CAT$ are a generalized cohomology theory, computable from ordinary cohomology by the Atiyah-Hirzebruch spectral sequence. $Y_1$ is homotopy equivalent to $S^n$ and $Y_2$ is homotopy equivalent to $\mathbb RP^n$. So the larger the dimension, the more total cohomology $Y_2$ has, and the larger $[Y_2/\partial,G/CAT]$, and eventually it overwhelms the bounded obstructions and exotic examples appear. In a little more detail: for 2-torsion, $G/TOP$ is basically just ordinary cohomology in specific dimensions, with period 4. So it does detect much of the cohomology of $\mathbb RP^n$. (Actually, we should work not with $Y_2$, but with $Y_2/\partial$, but by Poincaré duality, its cohomology is close to that of $Y_2$.)

If we want to deal with the smooth category or even $n$, we must only use exotic $Y_2$ whose double cover is standard as a $Y_1$. That is, we must restrict to the kernel $[Y_2/\partial,G/CAT]\to [Y_1/\partial,G/CAT]$. (Also, we must worry about the bounded indeterminacy coming from $L$-theory, but that is zero in our case of simply connected even dimensional manifolds.) $Y_1$ is homotopy equivalent to $S^n$, so it does not become more complicated as $n$ increases. In particular, $[Y_1/\partial,G/TOP]$ is just two homotopy groups of $G/TOP$, which are periodic and do not become more complicated with $n$. Whereas, $[Y_2/\partial,G/TOP]$ has increasing amounts of 2-torsion in the kernel. In the smooth category, the homotopy groups of $G/O$ reflect the homotopy groups of spheres and do become more complicated with $n$, but this affects both groups and is independent of $Y_2$ becoming complicated. The computation of the kernel is more complicated, but it grows and thus there are exotic smooth involutions. Indeed, that the group is large because $Y_2$ is complicated means that there are exotic smooth involutions represented by elements of $[Y_2/\partial,G/O]$ that are still nontrivial in $[Y_2/\partial,G/TOP]$, and thus remain exotic as continuous involutions.

These exotic involutions are not isotopic to the original, that is, not conjugate by automorphsims in the connected component of the identity, but it is not immediately clear that they are exotic in the way you asked, that they are not conjugate by arbitrary automorphisms. The invariant used to distinguish them is the tangent bundle, but this is relative to a homotopy equivalence. There might be a homotopy self-equivalence of $Y_2$ that switches the two tangent bundles; and such a homotopy equivalence might be induced by an automorphism of $M$. Ruling out this possibility requires understanding the group of homotopy equivalences of $Y_2$ and its action on the set of bundles. It is probably tractable in this case.

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  • $\begingroup$ Very nice answer ! Could you give some more detail about " The invariant used to distinguish them is the tangent bundle, but this is relative to a homotopy equivalence…". I guess this is similar to what happened in surgery theory: to yield the topological classification within a homotopy type,we need quotient the structure set by the action of the group of self-homotopy equivalences. $\endgroup$ – student Oct 27 '14 at 0:40
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    $\begingroup$ Yes, the method is surgery, so it has the drawbacks of surgery. We'd like to distinguish involutions by the isomorphism type of the quotient of the complement of a regular neighborhood of the fixed set, but that's hard so we mark so that we can apply surgery. And even if there are isomorphism once we drop the marking, that doesn't mean that there are automorphisms of $M$ that induce them. $\endgroup$ – Ben Wieland Oct 27 '14 at 21:49
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The octonions are a division algebra; so that any two of the three maps $$(x,y)\mapsto x \\ (x,y)\mapsto y\\ (x,y)\mapsto xy $$ exhibit a pullback square $$ \begin{array}{c} \mathbb{S}^7 \times \mathbb{S}^7 & \to & \mathbb{S}^7 \\ \downarrow & \ulcorner & \downarrow \\ \mathbb{S}^7 & \to & * \end{array} $$ and so any pair of these corresponds to such an involution, e.g. as the map $$ (x,y) \mapsto (x^{-1},xy) $$ interchanges product and second projection.

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  • $\begingroup$ Right, the algebraic structure of $\ S^7\ $ is rich. It'd be nice if you could show that your homeomorphism is not conjugate to the coordinate swap. I guess that $\ S^7\times S^7 \setminus\ $ fixed points is not homeomorphic in the two cases. $\endgroup$ – Włodzimierz Holsztyński Oct 10 '14 at 4:02
  • $\begingroup$ This involution is conjugate to the swap by $(x,y)\mapsto (xy,y)$. $\endgroup$ – Ben Wieland Oct 11 '14 at 14:06
  • $\begingroup$ @sara Is the answer for $n=1$ known? In particular is there an order 2 oriention preserving homeo of $\mathbb{T}^{2}$ which fixed points is a circle? $\endgroup$ – Ali Taghavi Oct 18 '14 at 19:27
  • $\begingroup$ @Ali mathoverflow.net/questions/21177/… $\endgroup$ – sara Oct 18 '14 at 23:39

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