Skip to main content

Discussion Board for the Lecture Series on Bing topology and Casson handles

Welcome to the discussion board for Mike Freedman's lectures on Bing topology and Casson handles. To add a new question or comment, press "Add new comment" below. For questions please add "Question" to the subject line, or "Answer" for answers. To reply to a question or add another answer, press "reply" at the lower-right corner of the question.

For all editing you need to login first, see the login button in the top menu bar. People at MPIM can use their usual login data, external users please send a short email to operator$@$ to get a login.

LaTeX expressions can be entered with the usual dollar notation.

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:

The ABG is simply connected but removing the 2-sphere at the frontier leads to a non-simply connected space, see question 3. Therefore, the AGB cannot have a collar. In particular, it's not a manifold! Note that the other side of the AHS is a 3-ball and hence there is a collar, just not a bicollar.

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.

© MPI f. Mathematik, Bonn Impressum & Datenschutz
-A A +A