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Question: Freedman JDG paper
I have a few questions about what seem to me to be typos in "The topology of four-dimensional manifolds" that I haven't been able to resolve. Sometimes there is a letter and I can't find where that letter has been introduced, so apologies if I have just not looked hard enough.
Page 389.
When I see $T^1_5$ should I replace it with $T^0_5$? In my understanding $T^1_5$ has yet to be defined? Is it correct that the subscript on a torus $\tau_j$ corresponds to that torus' association with the jth application of Theorem 4.2, but they all lie in the fourth stage $T^1_4 - T^1_3$ of the new Casson handle? This contradicts the preamble to Theorems 4.1 - 4, where it says that the subscript to a $\tau$ will indicate which stage of $T_n$ a distinguished torus lies, however, so I'm confused.
Page 390
Line 11: What is the index $r$ ? I don't understand what $\overline{\delta^i}$ is being connect summed with, nor how we know that this $x$ (or some number of copies of it) is able to make the framing zero.
Page 391
Line 8. Defn of $V^1_5$, should it be $T^{w+1}_4$?
Line 12: Should it be $\partial (\tilde{T}^w_4 \cap V^1_5)$ ?
Page 394 last line: What are the $T_n$ here?
Page 395 line 14-5: I don't understand this sentence.
Page 397 line 21: Should it say $T^{k+1}_{6k+1 - 6(k+1)}$ instead of $T^k_{6k+1 - 6(k+1)}$ ? Otherwise disjointness with $\beta_k$ doesn't seem to hold?
Page 403, line 1: Should it be the infinite intersection of the $\overline{X_n}$ ? Line 17: what does the + mean? Should it be a minus or another union? Line 25: there doesn't seem to be a diagram 5.6.
Last two paragraphs of page 403: I get rather confused here. What is $D$, what is $f^-$, what is the curly L? What theorem is the q.e.d for?
Page 406 last line, what is S?
Question 4: The converse of Push-Pull
Are there two non-homotopy equivalent spaces $X,Y$ such that $X\times S^1$ is homeomorphic to $Y\times S^1$?
Comment on 4)
Question 4 may be misleading. The question "which compact manifolds \(X, Y\) become homeomorphic after \(\times R\)" is equivalent to \(X\times S^1\simeq Y\times S^1\) so that the projections to \(S^1\) are homotopic". (The homeomorphism lifts to cyclic covers). The answer: if they are h-cobordant. The answer above lifts to a homeomorphism of \(X\times R\) to a cover of \(Y\times S^1\) that is not the obvious one.
Answer 4
(the groups can be found in "Hirshon, On cancellation in groups," "Some cancellation theorems with applications to nilpotent groups") Yes, $X$ and $Y$ can even be compact, aspherical manifolds. There exist two nonisomorphic torsionfree polycyclic groups $G,H$ such that $G\times \mathbb{Z}$ and $H\times \mathbb{Z}$ are isomorphic. Both groups act freely on $\mathbb{R}^n$ and thus we obtain closed aspehrical manifolds as a model for $BH$ and $BG$. If we cross with $S^1$ we get two closed aspherical manifolds as a model for $B(G\times S^1)$. Since both manifolds are topologically rigid (homotopy equivalence => homeomorphism), they are homeomorphic.
Let me give some details. Let us first construct a group that can be written as a direct product of $\mathbb{Z}$ and another group in different ways. Let us try a group of the form $G\rtimes \mathbb{Z}^2$, where the first generator $x_1$ acts through some automorphism $\varphi$ of order $m$ and the second generator $x_2$ acts trivially. What happens if we picked different generators, say $x_1'= ax_1+bx_2,x_2'=mx_1+dx_2$. for $a$ coprime to $m$ and some suitable numbers $c,d$. Now $x_1'$ acts by $\varphi^a$ and $x_2$ still acts trivially. Thus we get $(G\rtimes_\varphi \mathbb{Z})\times \mathbb{Z} \cong (G\rtimes_{\varphi^a} \mathbb{Z})\times \mathbb{Z}$.
It remains to find a group $G$ with an automorphism of finite order $m$, a number $a$ such that $(a,m)=1$ and $G\rtimes_\varphi\mathbb{Z}$ and $G\rtimes_{\varphi^a}\mathbb{Z}$ are not isomorphic. Hirshon took $G=\mathbb{Z}/1023,\varphi(x)=4x$ and $a=3$. Of course there might be plenty of choices possible. I think at this point it is already reasonable to think one can build a fiber bundle over a torus with similar properties.
I wanted to get a more explicit (torsionfree, virtually Abelian) example than the one given in the beginning, so I tried $G=\mathbb{Z^5}\rtimes_\Psi \mathbb{Z}$ where the automorphism is given by $x_5\mapsto -x_1$ and $x_i\mapsto x_{i+1}$. The automorphism $\varphi$ of $G$ should be given by $\Psi$ on $\mathbb{Z}^5$ and it should send a generator $t$ of the acting group to $x_1t$. The order of $\varphi$ is $10$. However I could not show that then the groups $G\rtimes_\varphi \mathbb{Z}$ and $G\rtimes_{\varphi^3} \mathbb{Z}$ are not isomorphic. (Maybe someone has an idea...)
$\Psi$ can be realized by an homeomorphism of the torus $T^5$. Its mapping torus is a compact, aspherical manifold $M$. This should be the fiber of a fiber bundle over $T^2$. The automorphism $\varphi$ of $\pi_1(M)$ can also be realized by a homeomorphism $h:T^5\times [0,1] /\sim \rightarrow T^5\times [0,1] \qquad (x_1,\ldots, x_5,h)\mapsto (x_1+h,x_2,\ldots, x_5,h)$. Now let us consider $M\times [0,1]\rightarrow [0,1]$ and identify the fibers over the left and right sides by $h$. Crossing with $S^1$ gives a fiber bundle over the torus. A similar base change as in the world of groups then gives two ways of expressing it as a product with $S^1$. The fundamental groups of the other factors are then $G\rtimes_\varphi \mathbb{Z}$ and $G\rtimes_{\varphi^3} \mathbb{Z}$. To argue that both factors are really different, I would need now that those groups are not isomorphic, which I do not know...
Question 3
One of the two components of the complement of the Alexander horned sphere is not simply connected. What is its fundamental group?
Answer to 3:
It's the direct limit of free groups on 2^n generators, as n goes to infinity. More precisely, it's the fundamental group of the infinite union of height n-gropes (of genus one), one sitting inside the next as usual. By construction, each n-th level generator becomes a commutator of a pair of generators at the (n+1)-st level. In particular, the abelianization of this group is trivial and it is therefore NOT a free group.
Answer 3
The construction shows that the complement deformation retracts to a (infinite) 1-complex, so the fundamental group is free of infinite rank.
Response to answer 3
No, see answer above.
Question 2
Why is the Alexander horned sphere not bicollared, ie. why is it impossible to extend the embedding to S2×[0,1]?.
Answer to 2:
Answer 2
It would then have a simply-connected neighborhood, but it doesn't.
Question 1b
Are there also examples where the Quotient is Hausdorff, but not homeomorphic to the $n$-ball?
Answer 1b
Plenty. First published ones by Bing 1959 (Dogbone space). Worst currently known in Daverman, Robert J.; Walsh, John J. "A ghastly generalized n-manifold", 1981. Any such quotient becomes a manifold when crossed with $R^2$, and so far all examples become manifolds when crossed with $R$.
Question 1
Find an example for a collection of disjoint, cellular subsets of the n-ball such that if we collapse each of those sets to a point, the quotient is not homeomorphic to an n-ball. I tried to look at the collection of closed intervals in $[0,1]$ that appear in the construction of the Cantor set, but funnily the quotients seems to be homeomorphic to an interval.
Answer 1
For example if we take $\{1/n\}\times [1/n,1-1/n]\subset B_2(0)\subset \mathbb{R}^2$. Then the quotient would not be Hausdorff, any two points in $\{0\} \times [0,1]$ cannot be saparated from each other.