Active questions tagged gt.geometric-topology - MathOverflowmost recent 30 from www.4124039.com2019-07-17T07:37:05Zhttp://www.4124039.com/feeds/tag?tagnames=gt.geometric-topology&sort=newesthttp://www.creativecommons.org/licenses/by-sa/3.0/rdfhttp://www.4124039.com/q/1110475Eilenberg-Mac Lane spaces for surface group extensions.Autumn Kenthttp://www.4124039.com/users/13352012-10-30T02:23:54Z2019-07-17T05:58:54Z
<p>(The question has been edited. It was pointed out in the comments that $\Gamma_G$ could be a surface group, thought of as a finite extension of another surface group, in which case $G$ is finite.)</p>
<p>Let $\pi$ be the fundamental group of a closed orientable surface, and let $G$ be a subgroup of $\mathrm{Out}(\pi)$.<br>
Associated to $G$ is an extension
\begin{equation}
1 \to \pi \to \Gamma_G \to G \to 1
\end{equation}
which sits inside the sequence
\begin{equation}
1 \to \mathrm{Inn}(\pi) \to \mathrm{Aut}(\pi) \to \mathrm{Out}(\pi) \to 1.
\end{equation}</p>
<p>If there is a finite $K(G,1)$, it is not difficult to see that there is a finite $K(\Gamma_G,1)$. A colleague and I are interested in a converse.</p>
<p>If $G$ is torsion-free and there is a finite $K(\Gamma_G, 1)$, is there a finite $K(G,1)$?</p>
<p>I am aware of Wall's theorem that a finitely presented group $\mathcal{G}$ has a finite $K(\mathcal{G},1)$ if and only if it is of type (FL), meaning that there is a finite resolution of $\mathbb{Z}$ by finitely generated free $\mathcal{G}$-modules. A group $\mathcal{G}$ is of type (FP) if there is a finite resolution of $\mathbb{Z}$ by finitely generated projective $\mathcal{G}$-modules. As far as I know, there is still no known group of type (FP) that is not of type (FL). </p>
<p>I imagine that careful examination of the Hochschild-Serre spectral sequence may tell you that $G$ is of type (FP), but I didn't pursue this as I'd like to get all the way to (FL). </p>
<p>Is there some geometric construction I'm missing?</p>
http://www.4124039.com/q/3350355Cobordism Theory of Topological Manifoldswonderichhttp://www.4124039.com/users/270042019-06-28T21:43:16Z2019-07-16T03:20:38Z
<p>Unfortunately, due to my ignorance, my present knowledge is limited to the cobordism Theory of Differentiable Manifolds.</p>
<ul>
<li>Cobordism Theory for DIFF/Differentiable/smooth manifolds</li>
</ul>
<p>However, there are Topological Manifolds which are not Differentiable Manifolds.</p>
<p><a href="https://i.stack.imgur.com/CJjAh.png" rel="noreferrer"><img src="https://i.stack.imgur.com/CJjAh.png" alt="enter image description here"></a></p>
<blockquote>
<p>So my question here for experts is that what do I need to beware and pay attention in order to master a cobordism theory of Topological Manifolds? What are the <strong>main differences of the computations of the bordism groups for the given following structures</strong>:</p>
<p>Say,</p>
<ol>
<li><p>Cobordism Theory of TOP/topological manifolds</p></li>
<li><p>Cobordism Theory for PDIFF/piecewise differentiable manifolds</p></li>
<li><p>Cobordism Theory for PL/piecewise-linear manifolds </p></li>
</ol>
<p>p.s. Are there Spin, Pin<span class="math-container">$^+$</span>, and Pin<span class="math-container">$^-$</span> versions of these cobordism theories of Topological Manifolds computed in the literature explicitly? </p>
</blockquote>
http://www.4124039.com/q/3360631Quadrilateral fundamental domainArun http://www.4124039.com/users/273302019-07-13T19:08:35Z2019-07-14T10:10:47Z
<p>Let P be a hyperbolic quadrilateral. Poincare polygon theorem provides sufficient condition for P to be a fundamental domain of some Fuchsian group in term of its inner angles. I find the angle sum condition in the theorem very restrictive. Is there less restrictive condition on angle sum for P to be a fundamental domain? For example, <span class="math-container">$P_1$</span> with angle sum <span class="math-container">$\pi/2+ \pi/2 + \pi/2+ \pi/3=11 \pi/6$</span> is not of the form <span class="math-container">$2\pi/n$</span> for any natural number <span class="math-container">$n$</span>, But <span class="math-container">$P_1$</span> is a fundamental domain for Fuchsian group generated by reflections on edges of <span class="math-container">$P_1$</span>. </p>
http://www.4124039.com/q/33586814Simplicial set of permutationsNikolai Mnevhttp://www.4124039.com/users/27022019-07-10T16:28:59Z2019-07-13T15:22:30Z
<p>Let <span class="math-container">$S_k$</span> be the set of all permutations of <span class="math-container">$k+1$</span> elements <span class="math-container">$0,1,...,k$</span>. introduce boundary maps <span class="math-container">$d_i : S_k \rightarrow S_{k-1}$</span> by deleting from permutation <span class="math-container">$\eta$</span> element <span class="math-container">$\eta(i)$</span> and monotone reordering and degeneracy<br>
<span class="math-container">$s_i :S_k \rightarrow S_{k+1} $</span> by adding 1 to all elements with <span class="math-container">$\eta(j)>\eta(i)$</span> and incerting into the result a new element <span class="math-container">$\eta(i)+1$</span> right after <span class="math-container">$\eta(i)$</span> on <span class="math-container">$i+1$</span> place. It is a simplicial set, contractible and classifies reorderings of simplicial sets. </p>
<p>Is it known? May be in higher symmetric something?</p>
<p>(Update) Boris Tsygan pointed the right direction in <a href="https://m.facebook.com/story.php?story_fbid=10157493966588914&id=811018913" rel="nofollow noreferrer">Facebook duscussion</a></p>
<p>The object is classical and it has a name
"Symmetric crossed simplicial group”.
It was introduced almost simultaneously
in </p>
<p>Appendix A10, page 191
“Symmetric objects”
B. L. Feigin and B. L. Tsygan
“Additive K-theory”
1987
K-theory, arithmetic and geometry, Semin., Moscow Univ. 1984-86
LNM 1289</p>
<p>Krasauskas, R.
"Skew-simplicial groups",
Lithuanian Mathematical Journal,
Jan 1987
vol 27 issue 1
p. 47--54</p>
<p>And independently</p>
<p>Zbigniew Fiedorowicz and Jean-Louis Loday “Crossed simplicial groups and their associated homology”
Trans. Amer. Math. Soc. 326 (1991), 57-87 </p>
<p>It has big value in everything symmetric. Geometric realization <span class="math-container">$|S_\bullet|$</span> is the topological group structure on infinite dimensional sphere. </p>
http://www.4124039.com/q/3359667Wildness of codimension 1 submanifolds of euclidean spaceLukashttp://www.4124039.com/users/1349672019-07-11T22:53:52Z2019-07-12T19:16:29Z
<p>This question arose out of <a href="https://math.stackexchange.com/questions/3287675/taming-mathbbr-diffeomorphisms">this stack exchange post</a>. I am wirting a thesis about the <span class="math-container">$s$</span>-cobordism theorem and Siebenmann's work about end obstructions. Combined they give a quick proof of the uniqueness of the smooth structure for <span class="math-container">$\mathbb{R}^n$</span>, <span class="math-container">$n \geq 6$</span>. Roughly Siebenmann's theorem says that for <span class="math-container">$n \geq 6$</span> a contractible <span class="math-container">$n$</span>-manifold <span class="math-container">$M$</span> that is simply connected at infinity embeds (smoothly) as the interior of a compact manifold. Since this compact manifold is contractible, by the <span class="math-container">$s$</span>-cobordism theorem, it is diffeomorphic to the standard <span class="math-container">$n$</span>-disk <span class="math-container">$D^n$</span> (see Minor's <em>Lectures on the <span class="math-container">$h$</span>-cobordism theorem</em> for example). It follows that <span class="math-container">$M = \text{int } D^n$</span> is diffeomorphic to <span class="math-container">$\mathbb{R}^n$</span>.</p>
<p>The problem is that the case <span class="math-container">$n = 5$</span> is not covered. I am aware of Stallings beautifully written <em>On the Piecewise-Linear Structure of Euclidean Space</em> but I am searching for a way to deal with the <span class="math-container">$n = 5$</span> case via Siebenmann's end theorem and the proper <span class="math-container">$s$</span>-cobordism theorem (see link to the mse question). This leads me to the following question, which is interesting in itself</p>
<blockquote>
<p>Given a smooth properly embedded codimension 1 submanifold <span class="math-container">$S \subset \mathbb{R}^{n+1}$</span>, is there a self diffeomorphism <span class="math-container">$\mathbb{R}^{n+1} \rightarrow \mathbb{R}^{n+1}$</span> which carries <span class="math-container">$S$</span> into a one dimensional bounded region <span class="math-container">$\mathbb{R}^n \times (-1, 1)$</span> ?</p>
</blockquote>
<p>Now if <span class="math-container">$M$</span> is a manifold that is homeomorphic to <span class="math-container">$\mathbb{R}^5$</span>, the product <span class="math-container">$M \times \mathbb{R}$</span> is homeomorphic to <span class="math-container">$\mathbb{R}^6$</span>, and hence also diffeomorphic. Granted the existence of the diffeomorphism in my question, we could find a diffeomorphism <span class="math-container">$f : M \times \mathbb{R} \rightarrow \mathbb{R}^6$</span> that maps <span class="math-container">$M \times 0$</span> into <span class="math-container">$\mathbb{R}^5 \times (-1, 1)$</span>. This would produce a proper <span class="math-container">$h$</span>-cobordism between <span class="math-container">$M$</span> and <span class="math-container">$\mathbb{R}^5$</span> by taking the region between <span class="math-container">$f(M \times 0)$</span> and <span class="math-container">$\mathbb{R}^5 \times 1$</span> in <span class="math-container">$\mathbb{R}^5 \times \mathbb{R}$</span>. Since <span class="math-container">$M$</span> is simply connected, the proper <span class="math-container">$s$</span>-cobordism theorem applies and shows that <span class="math-container">$M$</span> and <span class="math-container">$\mathbb{R}^5$</span> are actually diffeomorphic.</p>
http://www.4124039.com/q/3359026To find a point in Teichmüller space or measured foliation, how many lengths of curves do you need?Dylan Thurstonhttp://www.4124039.com/users/50102019-07-11T04:17:49Z2019-07-11T16:32:27Z
<p>To parametrize Teichmüller space, it suffices to measure the hyperbolic lengths of a finite number of curves. It is well-known that <span class="math-container">$9g-9$</span> curves suffice, by a standard pair-of-pants argument given in, for instance Fathi-Laudenbach-Poenaru.</p>
<p>I recall that you need exactly <span class="math-container">$6g-5$</span> curves: you cannot achieve it by <span class="math-container">$6g-6$</span>, because the character variety is not an algebraic subset of <span class="math-container">$\mathbb{C}^{6g-6}$</span>, but one extra curve suffices. Is this correct, and if so, who proved it?</p>
<p>Likewise for measured foliations, the cone over the boundary at infinity: can you parametrize measured foliations with <span class="math-container">$6g-5$</span> curves, and how to see that you cannot do it with <span class="math-container">$6g-6$</span>?</p>
http://www.4124039.com/q/3358393Diffeomorphism classification of Grassmannian manifoldsKafka91http://www.4124039.com/users/1429332019-07-10T07:27:02Z2019-07-11T07:33:21Z
<p>Is anything known about the diffeomorphism classification of Grassmannian manifolds? I know that there are some results on projective spaces (for example in Lopez de Medrano's "Involutions on manifolds"), </p>
<p><em>López de Medrano, S.</em>, Involutions on manifolds, Ergebnisse der Mathematik und ihrer Grenzgebiete. 59. Berlin-Heidelberg-New York: Springer-Verlag, ix, 102 p. (1971). <a href="https://zbmath.org/?q=an:0214.22501" rel="nofollow noreferrer">ZBL0214.22501</a>.</p>
<p>but I couldn't find anything on general Grassmannians.</p>
http://www.4124039.com/q/3358623Tiling of genus 2 surface by 8 pentagonsP Föhnhttp://www.4124039.com/users/1429462019-07-10T15:54:47Z2019-07-11T04:00:44Z
<p>In theses <a href="https://www.dpmms.cam.ac.uk/~hjrw2/Notes/cubenotes.pdf" rel="nofollow noreferrer">these notes</a>, Example 5.6, it is said that there is a "symmetric tiling of a genus 2 surface by 8 right-angled hyperbolic pentagons".</p>
<p>Question 1: What does this tiling look like?</p>
<p>Question 2: Is it always possible to tile a genus <span class="math-container">$n$</span> surface by <span class="math-container">$f$</span> regular <span class="math-container">$n$</span>-gons with interior angle <span class="math-container">$\pi/v$</span> (so that <span class="math-container">$v$</span> faces meet at every vertex) as long as the restriction given by the Euler characteristic
<span class="math-container">$$\chi=2-2n= f-nf/2+nf/v$$</span>
is satisfied?
Answering <a href="http://www.4124039.com/questions/198040/regular-tiling-of-a-surface-of-genus-2-by-heptagons?r=searchresults">this question</a>, Igor Rivin says yes, but it seems his argument only shows that regular hyperbolic <span class="math-container">$n$</span>-gons of interior angle smaller than the interior angle of a euclidean regular <span class="math-container">$n$</span>-gon exist.</p>
http://www.4124039.com/q/19804011regular tiling of a surface of genus 2 by heptagonsHaconhttp://www.4124039.com/users/193692015-02-20T15:05:26Z2019-07-10T17:36:21Z
<p>Let <span class="math-container">$S$</span> be a Riemann Surface of genus 2. Is there a picture in the literature for a regular tiling of <span class="math-container">$S$</span> by 12 heptagons (where three heptagons meet at each vertex)? Also, apart from the obvious restriction given by the Euler characteristic <span class="math-container">$2-2g=f-nf/2+nf/v$</span> (where <span class="math-container">$g$</span> is the genus, <span class="math-container">$f$</span> is the number of faces, <span class="math-container">$v$</span> is the number of faces meeting at each vertex) are there any obstructions for the existence of a tiling of a surface of genus <span class="math-container">$g$</span> by <span class="math-container">$n$</span>-gons (where the same number <span class="math-container">$v$</span> of <span class="math-container">$n$</span>-gons meet at each vertex).</p>
<p>I know that such a tiling exists (for a surface of genus 2 by heptagons), but I am unable to make a drawing. The construction goes like this (thanks to Mladen Bestvina): Start with a sphere. View this as a octahedron so that you have 8 triangles.
Pick a hexagon (start from the top go down, left, down come back up from the opposite side)
We fatten up these 6 sides to look like circles with one ray coming from the middle. You should think of these as the quotient of 2 triangles attached along one edge <|> via rotation by 180 degrees (so you get two edges meeting in a point A and 2 edges plus a ray meeting in the other point B). We call the center of the circle C so that the ray goes from C to B.<br>
You have to view the point <span class="math-container">$B_i$</span> (of the <span class="math-container">$i$</span>-th side, <span class="math-container">$i=1,2,3,4,5,6$</span>) as being attached to <span class="math-container">$A_{i+1}$</span> for <span class="math-container">$i=1,2,3,4,5,6$</span> (of course mod 6) so each of these vertices has valence <span class="math-container">$5+2=7$</span>.
Now you take the double cover with fixed points <span class="math-container">$C_1,...,C_6$</span> and you get a surface of genus 2 with 28 triangles (28=2x8+2x6 where 8 is the number of triangles and 6 are the circles with a ray that unwrap in to two triangles). The vertices still have valence 7 (because the cover is étale here). Take the dual tiling and you are done!</p>
<p>This is just a curiosity. I tried asking the first part of this on stackexchange with no luck.
I am also aware of the beautiful pictures of the famous tiling of a surface of genus 3 by 24 heptagons but you can not use that to obtain the tiling of a genus two surface in an obvious way.</p>
http://www.4124039.com/q/33574612Immersions of surfaces in $\mathbb{R}^3$Mariano Suárez-Álvarezhttp://www.4124039.com/users/14092019-07-08T23:01:15Z2019-07-10T04:21:38Z
<p>Stephen Smale famously proved in [<a href="https://doi.org/10.1090/S0002-9947-1959-0104227-9" rel="nofollow noreferrer">Trans. Amer. Math. Soc. 90 (1959), 281-290</a>] that any two <span class="math-container">$C^2$</span> immersions <span class="math-container">$S^2\to\mathbb R^3$</span> are regularly homotopic. This is how we knew that one can do a <a href="https://youtu.be/wO61D9x6lNY" rel="nofollow noreferrer">sphere eversion</a> before ever constructing such a thing, for example.</p>
<p>Later, Morris Hirsch wrote [<a href="https://doi.org/10.1090/S0002-9947-1959-0119214-4" rel="nofollow noreferrer">Trans. Amer. Math. Soc. 93 (1959), 242-276</a>], where he moves from the problem of immersing spheres to immersing general manifolds. His paper is based on Smale's, he says, in roughly the same way that obstruction theory is based on the theory of homotopy groups. </p>
<p>In his paper Hirsch uses his results to solve various problems in immersion theory, but I cannot find in it, nor after a google search, an answer to this:</p>
<blockquote>
<p>What is the classification up to regular homotopy of immersions <span class="math-container">$M\to\mathbb{R^3}$</span> from a closed oriented surface?</p>
</blockquote>
<p>In particular, is there just a single regular homotopy class of immersions of a closed orientable surface of positive genus? Can one everse a sphere with handles?</p>
<p><strong>Later.</strong> Looking at the explicit construction of representatives of regular homotopy classes of immersions given in the paper by Joel Hass and John Hughes that Danny Ruberman mentions in his answer, the following question asks itself. In some sense, it is closer to the eversion problem, as when turning spheres inside-out we start and end with embeddings.</p>
<blockquote>
<p>What is the classification up to regular homotopy of embeddings <span class="math-container">$M\to\mathbb{R^3}$</span> of a closed oriented surface?</p>
</blockquote>
<p>Of course, along the homotopy the embedding may degrade into just an immersion. Corollary 3.3 in the paper states that if <span class="math-container">$M$</span> has genus <span class="math-container">$g$</span>, then of the <span class="math-container">$4^g$</span> regular homotopy classes of immersions <span class="math-container">$M\to M\times[0,1]$</span> which are homotopic to the the obvious map <span class="math-container">$\iota:p\in M\mapsto (p,0)\in M\times [0,1]$</span> only one can be represented by an embedding, the one which contains <span class="math-container">$\iota$</span>. So the second question above is what happens when we replace <span class="math-container">$M\times[0,1]$</span> by <span class="math-container">$\mathbb{R^3}$</span>.</p>
http://www.4124039.com/q/3357882Is this a typo in Ihara's "On discrete subgroups of the two by two projective linear group over p-adic fields"?Ruperthttp://www.4124039.com/users/154822019-07-09T13:23:06Z2019-07-09T16:55:15Z
<p>In Eq. (9'') on p. 227 of Ihara's paper <i>"On discrete subgroups of the two by two projective linear group over p-adic fields"</i> (<a href="https://projecteuclid.org/euclid.jmsj/1260541107" rel="nofollow noreferrer">link</a>), where the second line says <span class="math-container">$$"\log Z_{\Gamma}(0,\chi)=1",$$</span> is this maybe a typo and he should have 0 instead of 1? Why does he even need to do two separate cases anyway?</p>
http://www.4124039.com/q/3357941Is every $(n-1)$-connected $n$-manifold embeddable in $\mathbb{R}^{n+1}$ homeomorphic to $\mathbb{S}^{n}$? [migrated]Ageloshttp://www.4124039.com/users/696812019-07-09T14:17:46Z2019-07-09T14:17:46Z
<p>Let <span class="math-container">$M^n$</span> be a compact, topological <span class="math-container">$n$</span>-manifold which is a subspace of <span class="math-container">$\mathbb{R}^{n+1}$</span>. If <span class="math-container">$M^n$</span> is <span class="math-container">$(n-1)$</span>-connected (i.e. <span class="math-container">$\pi_i$</span> vanishes for <span class="math-container">$i<n$</span>), does it have to be homeomorphic to the <span class="math-container">$n$</span>-sphere <span class="math-container">$\mathbb{S}^{n}$</span>?</p>
http://www.4124039.com/q/3355805Why does the longitude correspond to Frobenius in Arithmetic Topology, and other strange phenomenaFilippo Alberto Edoardohttp://www.4124039.com/users/182382019-07-06T15:38:55Z2019-07-06T16:20:38Z
<p>I am trying to adress Morishita's book <em>Knots and Primes</em> to discover a bit about Arithmetic Topology, but some analogies puzzle me. I know that the correspondence should be addressed with a grain of salt, but some parts of it look so fundamental that I would like to understand them better.</p>
<ol>
<li>In table (3.3) on page 50 of his book, Morishita writes that the longitude, called <span class="math-container">$\beta$</span>, should correspond to a lift of Frobenius and the meridian, called <span class="math-container">$\alpha$</span>, corresponds to a generator of tame inertia (both as elements in the maximal tame quotient of the absolute Galois group of a local field). He calls "longitude" a path going around one hole in the boundary of a tubular neighborhood of the knot and "meridian" the border of a disk which is a "cross-section" of the tube. Were the knot the unknot, this neighborhood would be the full torus <span class="math-container">$S^1\times D^2$</span>: in this case, <span class="math-container">$\alpha=\partial D^2$</span> and <span class="math-container">$\beta=S^1$</span>. This analogy does not agree with my idea that inertia acts as monodromy, which is "running around holes", but I tried to pursue. Then (page 63, after Theorem 5.1) he describes the analogue of decomposition groups for an unramified knot <span class="math-container">$K$</span>: he says that this group should be generated by "a loop going around <span class="math-container">$K$</span>", which I think is just the image of <span class="math-container">$\alpha$</span>. Then I am completely lost, as I would expect decomposition groups in unramified situations to be generated by Frobenius...</li>
<li>In Chapter 11, a tentative analogy with Iwasawa theory is suggested. Nevertheless, it seems to me that something is strange, as typical Iwasawa theory is concerned with very wild ramification, whereas the same table (3.3) on page 50 as before seems to indicate that there is <strong>no</strong> wild topological inertia. Thus, the Galois group of <span class="math-container">$X_\infty/X_K$</span>, where <span class="math-container">$X_K$</span> is a knot complement and <span class="math-container">$X_\infty$</span> is a <span class="math-container">$\mathbb{Z}$</span>-cover of <span class="math-container">$X_K$</span>, looks to me somehow similar to an "infinite tamely-ramified cover" (which has no arithmetic analogue) rather than a <span class="math-container">$\mathbb{Z}_p$</span>-extension. Am I missing something? Along the same lines, he has a small parenthesis between p. 144 and p. 145 where he writes that "[assuming that a <span class="math-container">$\mathbb{Z}_p$</span>-extension be ramified at one prime only, and that this prime be totally ramified] is an assumption analogous to the knot case". Why is it so? Neither the fact that a <span class="math-container">$\mathbb{Z}$</span>-cover must be ramified at only one knot nor the fact that there can't be a small unramified layer below look obvious to me (at least if the base manifold is not <span class="math-container">$S^3$</span>, otherwise <span class="math-container">$\pi_1(S^3)=0$</span> should say that no unramified extension exist).</li>
</ol>
http://www.4124039.com/q/3333404On a corollary of a paper by Colin and HondaPaulhttp://www.4124039.com/users/430972019-06-05T20:42:17Z2019-07-06T01:00:50Z
<p>The question is about the last sentence of the last corollary of <a href="https://arxiv.org/pdf/0706.0581.pdf" rel="nofollow noreferrer">Stabilizing the monodromy of an open book decomposition</a> by Vicent Colin and Ko Honda. This question is also related to this other question of mine <a href="http://www.4124039.com/posts/330596/edit">http://www.4124039.com/posts/330596/edit</a>. </p>
<p>The corollary of the aforementioned article says:</p>
<blockquote>
<p><span class="math-container">$\textbf{Corollary 3.1}$</span> Let <span class="math-container">$S$</span> be a surface with connected boundary <span class="math-container">$\partial S$</span> and let <span class="math-container">$h:S \to S$</span> be a pseudo-Anosov automorphism with fractional Dehn twist coefficient (at the only boundary component) greater than <span class="math-container">$2$</span>. Then, any elementary positive stabilization along a non-boundary-parallel arc <span class="math-container">$\gamma \subset S$</span> is pseudo-Anosov.</p>
</blockquote>
<p>The vast majority of the proof consists in showing that the stabilization <span class="math-container">$h'$</span> is not reducible. Then in the last sentence they conclude that it also cannot be periodic. Hence, by Nielsen-Thurston classification it must be pseudo-Anosov. However this last part is proven as follows in the last sentence of the paper:</p>
<blockquote>
<p>To show that <span class="math-container">$h'$</span> is not periodic, consider <span class="math-container">$\delta = \partial S$</span>. One easily verifies that the number of intersections with <span class="math-container">$a$</span> (the co-core of the handle used for stabilizing) increases with each iterate of <span class="math-container">$h'$</span>.</p>
</blockquote>
<p>They use all the hypothesis of the theorem to provide a proof for the first part and the way this last sentence is writen looks as if the last part was true in a more general setting. </p>
<p>The thing is that I cannot easily verify that. What I can easily verify is that the number of intersections of <span class="math-container">$h'(\delta)$</span> with <span class="math-container">$a$</span> is certainly <span class="math-container">$2$</span> (without using that <span class="math-container">$h$</span> is pseudo-anosov. I can also produce examples (see the edit below) of a periodic automorphism with connected boundary which, after a suitable positive stabilization, becomes a periodic automorphism aswell. So, in order to prove the last part we need to use that <span class="math-container">$h$</span> is pseudo-Anosov at some point, or the part on the fractional Dehn twist coefficient maybe. But it is not obvious to me why there are no relations in the surface that "un-wind" the twisting induced by the stabilization as in the examples that I can produce.</p>
<p>The rest of the question is an example that I claimed I could produce. It serves as a counter-example to Michael Lin's answer and it also serves as an example that the hypothesis on <span class="math-container">$h$</span> must be used. Also, since I have drawn some pictures, I hope this questions gets some more attention.</p>
<p><span class="math-container">$\textbf{Example}$</span></p>
<p>For more details on how to obtain a description of the monodromies that I use as examples, you may have a look at <a href="https://link.springer.com/article/10.1007/BF01883883" rel="nofollow noreferrer">Le groupe de monodromie du déploiement des singularités isolées de courbes planes I</a> by Norbert A'Campo for example. Or any textbook on plane curve singularities probably contains a description of these monodromies.</p>
<p>The monodromy of the <span class="math-container">$A_2$</span> singularity <span class="math-container">$x^2+y^3: \mathbb{C}^2 \to \mathbb{C}$</span> consists of the composition of two right-handed Dehn twists around a "parallel" and a "meridian" of the torus as in the picture (that I quickly made) below that I just made. The monodromy is the composition of the right-handed Dehn twist around the blue curve and then around the green curve. You can see this in the paper of A'Campo for example or in any introductory text to Picard-Lefschetz theory. That this monodromy is freely periodic of order <span class="math-container">$6$</span> follows from the observation that a representative of the monodromy is <span class="math-container">$(x,y) \mapsto (e^{\pi i}x, e^{2\pi i/3}y)$</span>. </p>
<p><a href="https://i.stack.imgur.com/qOQts.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qOQts.png" alt="Monodromy of the cusp"></a> </p>
<p>Losely speaking: in Singularity Theory "adding a Morse point" is more or less the same as "stabilizing". The singularity <span class="math-container">$x^2 +y^3$</span> "consists" of two Morse points. That is, after a generic perturbation it deforms to two Morse points. Also, its Dynkin diagram is the graph consisting of two vertices joint by a segment. Again, for more details on this, follow A'Campo's paper.</p>
<p>Now, the <span class="math-container">$A_3$</span> singularity <span class="math-container">$x^2+y^4$</span> has Milnor fiber the surface of genus <span class="math-container">$1$</span> and <span class="math-container">$2$</span> boundary components. And it can be obtain from the fiber of <span class="math-container">$A_2$</span> by attaching a <span class="math-container">$1$</span>-handle to its boundary. See the picture below.</p>
<p><a href="https://i.stack.imgur.com/tPeNn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tPeNn.png" alt="Monodromy of the stabilization of the cusp"></a></p>
<p>The description of the monodromy is the composition of a Dehn twist around the blue curve, then the green curve and then the yellow curve (observe that the last two Dehn twists commute since their support is disjoint) This part can also be deduced from A'Campo's article. That the monodromy is freely periodic follows from the same reasoning as before. For instance, a representative is <span class="math-container">$(x,y) \mapsto (e^{\pi i}x, e^{\pi i/2}y)$</span>. Which has order <span class="math-container">$4$</span>.</p>
<p>So by taking <span class="math-container">$h$</span> the first monodromy and <span class="math-container">$h'$</span> the second we can see how <span class="math-container">$h'^4(\delta) \cap a$</span> is empty. Also <span class="math-container">$h'(\delta) \cap a$</span> consists of two points.</p>
http://www.4124039.com/q/3354431Two multi-curves in a surface with the same transverse measureAdamhttp://www.4124039.com/users/239352019-07-04T16:29:20Z2019-07-04T20:18:31Z
<p>Let <span class="math-container">$(\cal F,\mu)$</span> be the stable measured foliation of a pseudo-Anosov on an oriented surface <span class="math-container">$S$</span>. Can there be two non-isotopic multi-loops (collections of disjoint simple loops) <span class="math-container">$L_1,L_2\subset S$</span>, both transverse to <span class="math-container">$\cal F$</span> with the same transverse measure, <span class="math-container">$\mu(L_1)=\mu(L_2)$</span>?</p>
<p>If yes, how one can construct such example?</p>
<p>PS. One can prove that "transversely isotopic"="isotopic" in this setting.</p>
http://www.4124039.com/q/3352885Curvature and asphericity of cube complexesPriyavrat Deshpandehttp://www.4124039.com/users/74942019-07-02T17:32:51Z2019-07-04T15:40:00Z
<p>Let <span class="math-container">$K$</span> be a connected cube complex (one may assume that its a cellulation of a smooth, closed manifold). Such a <span class="math-container">$K$</span> comes equipped with a length metric (one assumes that each edge is of unit length). It is a well known result of Gromov that if <span class="math-container">$K$</span> is non-positively curved (in the sense of Aleksandrov) then its universal cover is contractible; equivalently <span class="math-container">$K$</span> is aspherical.
Gromov also gave the following criterion to check the curvature: If the link of every vertex in <span class="math-container">$K$</span> is flag then <span class="math-container">$K$</span> is non-positively curved. </p>
<p>Recently I discovered a cube complex structure on some smooth, closed manifolds such that there are vertices whose link is the boundary of a simplex of appropriate dimension.
However, some of these manifolds are aspherical (since tori, positive genus surfaces are on that list).<br>
So I was wondering whether there are any other (combinatorial) criterion to decide the asphericity of a cube complex? </p>
<p>I would also like to know about the papers that prove a certain manifold is aspherical using the combinatorial properties of a regular CW structure on that manifold. </p>
http://www.4124039.com/q/3353862Category of Manifolds and Maps: TOP $\supseteq$ TRI $\supseteq$ PL $\supseteq$ DIFF? [closed]annie hearthttp://www.4124039.com/users/1064972019-07-04T00:02:34Z2019-07-04T00:02:34Z
<p>Please let me denote the following </p>
<ul>
<li>(TOP) topological manifolds <a href="https://en.wikipedia.org/wiki/Topological_manifold" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Topological_manifold</a></li>
<li>(PDIFF), for piecewise differentiable <a href="https://en.wikipedia.org/wiki/PDIFF" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/PDIFF</a></li>
<li>(PL) piecewise-linear manifolds <a href="https://en.wikipedia.org/wiki/Piecewise_linear_manifold#Smooth_manifolds" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Piecewise_linear_manifold#Smooth_manifolds</a></li>
<li>(DIFF) the smooth manifolds <a href="https://en.wikipedia.org/wiki/Differentiable_manifold#Definition" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Differentiable_manifold#Definition</a></li>
<li>(TRI) triangulable manifolds <a href="https://en.wikipedia.org/wiki/Triangulation_%28topology%29" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Triangulation_%28topology%29</a></li>
</ul>
<p>we consider the <strong>category</strong> of manifolds and their maps based on what I learned from Wikipedia above links.</p>
<p>Is it true that the above <strong>categories</strong> have the following relations:</p>
<blockquote>
<p>(1) TOP <span class="math-container">$\supseteq$</span> TRI ?</p>
<p>Namely, every TRI must be TOP manifolds?</p>
<p>(2) TRI <span class="math-container">$\supseteq$</span> PL ?</p>
<p>Namely, every PL must be TRI manifolds?</p>
<p>(3) TRI <span class="math-container">$\supseteq$</span> DIFF ?</p>
<p>Namely, every DIFF must be TRI manifolds?</p>
<p>(4) PL <span class="math-container">$\supseteq$</span> DIFF ?</p>
<p>Namely, every DIFF must be PL manifolds?</p>
<p>(5) So in a short summary, is it true that</p>
<p><span class="math-container">$$\text{ TOP $\supseteq$ TRI $\supseteq$ PL $\supseteq$ DIFF} ?$$</span></p>
</blockquote>
<p>(If what I said in (5) is false, what are their intersections, unions and complements of these categories?)</p>
<p>p.s. This is based on an improved unanswer question from MSE a week ago. I am sorry I hope more experts can fill in this question. Thanks! <3</p>
http://www.4124039.com/q/3353417Fibers of continuous maps of $\mathbb{R}^n$ which are injective at dense pointsShinpei Babahttp://www.4124039.com/users/1426162019-07-03T11:12:14Z2019-07-03T23:41:51Z
<p><strong>Question.</strong> Suppose that <span class="math-container">$f\colon\mathbb{R}^n \to \mathbb{R}^n$</span> is a continuous map and there is a dense subset <span class="math-container">$D \subset \mathbb{R}^n$</span> such that <span class="math-container">$f^{-1}(f(x)) = \{x\}$</span> for all <span class="math-container">$x \in D$</span>.
Is every fiber of <span class="math-container">$f$</span> connected?</p>
<p>When <span class="math-container">$n =1$</span>, it is easy to see that <span class="math-container">$f$</span> must moreover be a strictly increasing or decreasing function.
I think that it it true in general, but I am not sure. An idea to prove: Suppose, to the contrary, that <span class="math-container">$f(x)^{-1}$</span> has two connected components.
Then pick an arc, in the domain, with endpoints in different connected components fo <span class="math-container">$f^{-1}(x)$</span>. Then <span class="math-container">$f$</span> identifies the endpoints of the arc, and it seems unlikely that the resulting loop in the target is null homotopic, which would be a contradiction. </p>
http://www.4124039.com/q/3350309What do absolute neighborhood retracts look like?Tim Campionhttp://www.4124039.com/users/23622019-06-28T19:51:46Z2019-07-03T23:38:29Z
<p>In the course of filling in my <a href="http://www.4124039.com/questions/333363/a-map-of-non-pathological-topology">map of non-pathological topology</a>, I'd like to understand the class of <a href="https://en.wikipedia.org/wiki/Retract#Absolute_neighborhood_retract_(ANR)" rel="nofollow noreferrer">ANRs</a> (Absolute Neighborhood Retracts) as a sort of "neighborhood" of the class of CW complexes. This seems warranted by some of the nice properties of ANRs:</p>
<ul>
<li><p>Every ANR has the homotopy type of a CW complex.</p></li>
<li><p>Every ANR is locally contractible, and as a partial converse, any locally contractible finite-dimensional metric space is an ANR.</p></li>
</ul>
<p>But there are also important infinite-dimensional examples of ANR's:</p>
<ul>
<li><p>The Hilbert cube is an ANR.</p></li>
<li><p>Many function spaces are ANRs.</p></li>
</ul>
<p>This leaves me with some</p>
<p><strong>Questions:</strong></p>
<ol>
<li><p>What is a good example of a finite-dimensional ANR which is not a CW complex?</p></li>
<li><p>Are (finite-dimensional) ANRs an appropriate setting to study either (a) <a href="https://en.wikipedia.org/wiki/Fractal" rel="nofollow noreferrer">fractals</a> (wikipedia seems to define a fractal to be a subset of Euclidean space whose topological and Hausdorff dimensions differ) or (b) the limit sets of dynamical systems on CW complexes? I think my sense is that both (a) and (b) are generally <em>wilder</em> than ANRs, but I'm not really sure -- perhaps there's some overlap but no strict containments?</p></li>
<li><p>Do ANRs admit some kind of "generalized cell structure" like CW complexes do? Or is there some other sense in which ANRs can be "classified"? Is there at least a "classification" of what ANRs can look like locally?</p></li>
</ol>
<p>Less precisely, my feeling is that when somebody says "Let <span class="math-container">$X$</span> be a CW complex", I sort of know what they mean. But when somebody says "Let <span class="math-container">$X$</span> be an ANR", I don't -- I don't know what to think of as a "typical example", nor do I know what kinds of "typical pathologies" to watch out for. It would be nice if there were a book out there entirely devoted to the topology of ANRs <strike> but surprisingly I haven't been able to find one </strike>. I found a book <a href="https://books.google.com/books/about/Theory_of_retracts.html?id=GVTvAAAAMAAJ" rel="nofollow noreferrer">Theory of Retracts</a> by Sze-Tsen Hu, but I haven't yet found an example in it of a finite-dimensional ANR which is not a CW complex.</p>
<p><strong>EDIT:</strong> </p>
<ul>
<li><p>Another reference is Borsuk's <em>The Theory of Retracts</em>. This contains more examples in the later chapters, though I'm still struggling to piece together a coherent picture of the diversity of ANRs.</p></li>
<li><p>An important piece of context regarding (1): according to Thm V.10.1 of Borsuk, the (compact) finite-dimensional ANRs coincide with the retracts of (finite) polyhedra. Thus in finite dimensions, the question is: <em>How wild can an idempotent on a polyhedron be?</em>.</p></li>
</ul>
http://www.4124039.com/q/3347386What are orbifolds with corners?user142317http://www.4124039.com/users/1423172019-06-25T06:49:32Z2019-06-28T19:25:50Z
<p>What is the geometric definition of orbifolds with corners? Here “geometric" means that there is a definition in chapter 8 of the draft of Dominic Joyce's book <em>D-manifolds and d-orbifolds: a theory of derived differential geometry</em> (<a href="http://people.maths.ox.ac.uk/joyce/dmanifolds.html" rel="noreferrer">book website</a>, direct <a href="https://people.maths.ox.ac.uk/joyce/dmbook.pdf" rel="noreferrer">pdf link</a>), but that is too technical and abstract, and so I am trying to find a definition parallel to the definition of orbifolds in the normal way (locally it is an open set acted by a smooth group, etc.) or more intuitive than the definition in that book. If it is defined as an open set with boundary (its closure) acted by a group keeping the boundary invariant seems does not work, as the boundary point of a manifold with boundary is itself a point in an orbifold (a manifold with boundary is an ordinary orbifold).</p>
http://www.4124039.com/q/3350218Non-additivity of intersection formsuser101010http://www.4124039.com/users/994142019-06-28T18:28:39Z2019-06-28T18:28:39Z
<p>Given two oriented <span class="math-container">$4k$</span>-manifolds <span class="math-container">$X_1$</span> and <span class="math-container">$X_2$</span>, Novikov additivity tells us that
<span class="math-container">$$
\sigma(X_1 \sharp X_2) = \sigma(X_1) + \sigma(X_2).$$</span></p>
<p>More generally, if we glue the boundaries of two such manifolds <span class="math-container">$X_1$</span> and <span class="math-container">$X_2$</span> via a orientation-reversing diffeomorphism (or even just part of the boundaries - see <a href="http://www.4124039.com/questions/106237/additivity-of-signature">here</a>) then Novikov additivity holds. <a href="https://www.maths.ed.ac.uk/~v1ranick/papers/wall6.pdf" rel="noreferrer">Wall</a> has famously analyzed the failure of additivity when we glue along a submanifold of the boundary that itself has boundary. </p>
<p>I am interested in how this story extends to intersection forms. In particular:</p>
<blockquote>
<p>What can be said about the relationship between the intersection form <span class="math-container">$Q_{X}$</span>, where <span class="math-container">$X = X_1 \cup_\phi X_2$</span> where <span class="math-container">$\phi$</span> is an orientation-reversing diffeomorphism between closed submanifolds of the boundary, and the intersection forms <span class="math-container">$Q_{X_1}$</span> and <span class="math-container">$Q_{X_2}$</span>?</p>
</blockquote>
<p>I know that additivity of intersection forms holds in the case where the boundaries of <span class="math-container">$X_1$</span> and <span class="math-container">$X_2$</span> are both homology spheres and they are glued along their entire boundary - so in particular, <span class="math-container">$Q_{X_1 \sharp X_2} = Q_{X_1} + Q_{X_2}$</span>. I wonder if there is some error term in the general case that has been identified. For what it's worth, I am primarily interested in the case where <span class="math-container">$k=1$</span>. </p>
http://www.4124039.com/q/3347198Representation of the space of lattices in $\Bbb R^n$Mike Battagliahttp://www.4124039.com/users/246112019-06-24T20:13:02Z2019-06-25T05:16:17Z
<p>The space of 2D lattices in <span class="math-container">$\Bbb R^2$</span> can be represented with the two Eisenstein series <span class="math-container">$G_4$</span> and <span class="math-container">$G_6$</span>. Each lattice uniquely maps to a point in <span class="math-container">$\Bbb C^2$</span> using these two invariants, and the points <span class="math-container">$(1,0)$</span> and <span class="math-container">$(0,1)$</span> map to the unit square and triangular lattices.</p>
<p>This representation requires us to treat <span class="math-container">$\Bbb R^2$</span> as being isomorphic to <span class="math-container">$\Bbb C$</span>, so that we can evaluate, for some lattice <span class="math-container">$\Lambda(\omega_1, \omega_2)$</span> generated by <span class="math-container">$\omega_1$</span> and <span class="math-container">$\omega_2$</span></p>
<p><span class="math-container">$$
G_k(\Lambda) = \Bbb \sum_{0 \neq \omega \in \Lambda(\omega_1, \omega_2)} \frac{1}{\omega^k}
$$</span></p>
<p>where each element <span class="math-container">$\omega \in \Lambda$</span> is then treated as having the algebra structure from <span class="math-container">$\Bbb C$</span>.</p>
<p>The above is often treated in the setting of modular forms, where it is assumed that everything is in <span class="math-container">$\Bbb C$</span> from the start, and that one of the lattice vectors is <span class="math-container">$1$</span>. However, it is also given fairly often as an equivalent treatment in terms of real lattices.</p>
<p>So the basic question is, how do you develop a similar theory for <span class="math-container">$\Bbb R^n$</span>? Is there a similar algebra structure one can place on <span class="math-container">$\Bbb R^n$</span> taking the place of <span class="math-container">$\Bbb C$</span>?</p>
<p>Or, is there a simpler moduli space if we let there be different equivalent versions of the same lattice, e.g. "signed" lattices or something like that?</p>
<p><strong>My big-picture questions</strong>:</p>
<ol>
<li>What is a direct formula to represent the space of lattices in <span class="math-container">$\Bbb R^n$</span>, so that every lattice corresponds to a unique point?</li>
<li>If the above is too complicated, is there an easier representation if we allow for things like "signed lattices," or allow different automorphisms of the same lattice to be represented with different elements of the same space, etc? (As an example, in <span class="math-container">$\Bbb R^2$</span>, suppose we differentiated between two 90 degree rotations of the square lattice, or three 60 degree rotations of the triangular lattice, while equivocating between any 180 degree rotation of the same lattice.)</li>
</ol>
<p>Assuming some generalization of the Eisenstein series is relevant, then the follow-up questions would be</p>
<ol start="3">
<li>How do you generalize the Eisenstein series to lattices in <span class="math-container">$\Bbb R^3$</span>, or <span class="math-container">$\Bbb R^n$</span> in general?</li>
<li>Assuming you manage to do that, the space of lattices for <span class="math-container">$\Bbb R^n$</span> should have dimension <span class="math-container">$n^2$</span>. Which Eisenstein invariants are needed to uniquely represent this space of lattices?</li>
<li>The Eisenstein/Weierstrass invariants can be viewed as coefficients from the Laurent expansion of the Weierstrass elliptic function. Is the above simplified from looking at the Laurent expansions of higher-dimensional elliptic functions?</li>
</ol>
<p>I have been pointed in a few interesting directions - <a href="https://msp.org/pjm/1988/133-2/pjm-v133-n2-p08-s.pdf" rel="nofollow noreferrer">Eisenstein Series on Real, Complex, and Quaternionic Half-Spaces</a>, <a href="http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.493.9518&rep=rep1&type=pdf" rel="nofollow noreferrer">Eisenstein Series in Complexified Clifford Analysis</a>, <a href="https://arxiv.org/pdf/hep-th/9506060.pdf" rel="nofollow noreferrer">Generalization of Weierstrass Elliptic functions to <span class="math-container">$\Bbb R^n$</span></a>, <a href="http://www.ems-ph.org/journals/show_pdf.php?issn=0232-2064&vol=20&iss=4&rank=12" rel="nofollow noreferrer">On a New Type of Eisenstein Series in Clifford Analysis</a>, <a href="https://www.researchgate.net/publication/245495137_Clifford_Analysis_with_Generalized_Elliptic_and_Quasi_Elliptic_Functions" rel="nofollow noreferrer">Clifford analysis with generalized elliptic and quasi elliptic functions</a>. People have suggested the theory of automorphic forms and the Langlands' Eisenstein series.</p>
<p>However, I have somehow been unable to unpack all this to get a simple definition of the space of n-dimensional real lattices. I would assume that some generalized Eisenstein series would be necessary, but I don't know how to build them for arbitrary <span class="math-container">$\Bbb R^n$</span>. Some of the suggestions for replacing <span class="math-container">$\Bbb C$</span> with an arbitrary Clifford algebra can sometimes equivocate between different lattices. For instance, even in the simplest case of <span class="math-container">$\Bbb R^4$</span>, where we can treat the entire thing as the quaternion algebra, we tend to have that the lattices <span class="math-container">$\Lambda(1,2\mathbf e_1,\mathbf e_2,\mathbf e_3)$</span> and <span class="math-container">$\Lambda(1,\mathbf e_1,2\mathbf e_2,\mathbf e_3)$</span> produce the same Eisenstein series, if we naively use the formula <span class="math-container">$\sum_{0 \neq \omega \in \Lambda} \frac{1}{\omega^k}$</span>.</p>
<p>This was originally posted and bountied at MSE: <a href="https://math.stackexchange.com/questions/3260577/using-eisenstein-series-to-represent-the-space-of-lattices-generalization-to">https://math.stackexchange.com/questions/3260577/using-eisenstein-series-to-represent-the-space-of-lattices-generalization-to</a></p>
<p>I am hoping for some basic explication of how to do this that doesn't require extremely deep knowledge of elliptic curve geometry and etc - I just want a basic representation for the space of <span class="math-container">$n$</span>-d lattices.</p>
<p><strong>EDIT</strong>: some have also mentioned the <strong>Chabauty space</strong> <span class="math-container">$\cal C(\Bbb R^n)$</span> of <span class="math-container">$\Bbb R^n$</span>, which is the compactification of the space of lattices in that it is the space of all closed subgroups of <span class="math-container">$\Bbb R^n$</span>. Elements in the Chabauty space are isomorphic to <span class="math-container">$\Bbb R^p \times \Bbb Z^q$</span>, where <span class="math-container">$p+q \leq n$</span>. For any closed subgroup, the tuple <span class="math-container">$(p,q)$</span> is the <em>type</em> of the subgroup, and the set of all such <span class="math-container">$\cal C^{(p,q)}(\Bbb R^n)$</span> form a <strong>stratification</strong> of <span class="math-container">$\cal C(\Bbb R^n)$</span> (note that the set of all <span class="math-container">$\cal C^{(p,0)}$</span> is the Grassmannian). As some have mentioned, if <span class="math-container">$n>2$</span>, this is not a manifold, but it does seem to be a "pseudo-manifold" which is a union of manifolds that have been glued together, and the set <span class="math-container">$\cal C^{(p,q)}(\Bbb R^n)$</span> seems to be a manifold for any <span class="math-container">$p,q$</span>, as shown in <a href="https://arxiv.org/pdf/0812.2111.pdf" rel="nofollow noreferrer">this article</a>. Perhaps useful, although seems that it may be more general than necessary...</p>
http://www.4124039.com/q/878959Can every manifold be given an analytic structure?Theo Johnson-Freydhttp://www.4124039.com/users/782009-12-13T19:44:19Z2019-06-22T18:09:25Z
<p>Let $M$ be a (real) manifold. Recall that an <em>analytic structure</em> on $M$ is an atlas such that all transition maps are real-analytic (and maximal with respect to this property). (There's also a sheafy definition.) So in particular being analytic is a structure, not a property.</p>
<p>Q1: Is it true that any topological manifold can be equipped with an analytic structure?</p>
<p>Q2: Can any $C^\infty$ manifold (replace "analytic" by "$C^\infty$" in the first paragraph) be equipped with an analytic structure (consistent with the smooth structure)?</p>
http://www.4124039.com/q/3344326A coincidence between the Lambert cube, Lobell polyhedron, and hyperbolic 3-manifolds?Tito Piezas IIIhttp://www.4124039.com/users/129052019-06-20T14:45:46Z2019-06-21T13:24:38Z
<p><strong>I. Lambert cube</strong> <span class="math-container">$\mathfrak L(\alpha_1,\alpha_2,\alpha_3)$</span></p>
<p>In <a href="https://pdfs.semanticscholar.org/0de0/af4e7877fdebf45ad3be1af7faa04e787302.pdf?_ga=2.50967721.1479578514.1561033498-700242977.1518689157" rel="nofollow noreferrer">this paper</a> (p.8), we find the volume <span class="math-container">$V$</span> of the hyperbolic <em>Lambert cube</em> for the special case <span class="math-container">$\alpha=\alpha_1 = \alpha_2 = \alpha_3$</span> as</p>
<p><span class="math-container">$$V_\alpha = \int_x^\infty \ln\left(\frac1{t^2}\left(\frac{t^2-\tan^2\alpha}{1+\tan^2\alpha}\right)^3\right)\frac{dt}{t^2+1}$$</span></p>
<p>where <span class="math-container">$x$</span> is the positive root of <span class="math-container">$x^4-(1+3\tan^2\alpha)x^2-\tan^6\alpha =0$</span>. Hence,</p>
<p><span class="math-container">$$\begin{array}{|c|c|c|}
\hline
\alpha &x& V_{\alpha} & 8V_{\alpha}\\
\hline
\pi/3 &\left(\frac{1+\sqrt{13}}2\right)^{3/2}& 0.324423 & \color{brown}{2.59538}\\
\pi/4 &\left(\frac{1+\sqrt{5}}2\right)^{3/2}& 0.538275 & \color{blue}{4.30620}\\
\pi/5 &\left(\frac{5-\sqrt{5}}2\right)^{3/2}& 0.658081 & 5.26465\\
\hline
\end{array}$$</span></p>
<p><strong>II. Lobell polyhedron <span class="math-container">$L(n)$</span></strong></p>
<p>In <a href="https://arxiv.org/pdf/math/0603552.pdf" rel="nofollow noreferrer">this paper</a> (p.33), we find <a href="http://oeis.org/A321209" rel="nofollow noreferrer">the volume</a> (using a different formula) of <span class="math-container">$L(5)$</span> as</p>
<p><span class="math-container">$$\begin{array}{|c|c|c|}
\hline
n & L(n) & V_n\\
\hline
5 & L(5) & \color{blue}{4.30620}\\
6 & L(6) & 6.02304\\
\hline
\end{array}$$</span></p>
<p><strong>III. Closed hyperbolic 3-manifolds</strong></p>
<p>On a hunch, I checked those volumes in the <a href="http://plouffe.fr/simon/constants/table216.txt" rel="nofollow noreferrer">Hodgson & Weeks census</a> and found,</p>
<p><span class="math-container">$$\begin{array}{|c|c|c|}
\hline
\text{Dehn filling}& \text{Symmetry} & \text{Volume}\\
\hline
m160(-3, 2) & D6 & \color{brown}{2.59538}\\
\hline
s648(-5, 1) & D4 & \color{blue}{4.30620}\\
s921(-3, 1) & D2 & 4.30620\\
m400( 4, 1) & Z/2 & 4.30620\\
\hline
\end{array}$$</span></p>
<p><strong>IV. Questions</strong></p>
<ol>
<li>Why are the volumes for <span class="math-container">$\alpha =\pi/3, \pi/4$</span> and <span class="math-container">$L(5)$</span> found in the census?</li>
<li>Conversely, why for <span class="math-container">$\alpha =\pi/5$</span> and <span class="math-container">$L(6)$</span> are they NOT present?</li>
</ol>
http://www.4124039.com/q/3344504The handlebody decomposition of S^1 bundles over surfaces?Jake B.http://www.4124039.com/users/1013352019-06-20T19:26:07Z2019-06-21T12:03:41Z
<p>What is the most natural handlebody decomposition of <span class="math-container">$F_g \times S^1$</span>, if <span class="math-container">$F_g$</span> is an orientable closed surface of genus <span class="math-container">$g$</span>?</p>
http://www.4124039.com/q/33440411Which presentation of a 3-manifold group comes from a Heegaard splitting?Jiajun Wanghttp://www.4124039.com/users/993252019-06-20T06:41:00Z2019-06-20T07:19:12Z
<p>Given a balanced presentation of a three-manifold group G, is there any way to determine whether it comes from a Heegaard splitting? In other words, how to characterize such presentations? There exist Seifert fibered example (by Boileau and Zieschang) and hyperbolic example (by Tao Li) for which the rank of the fundamental group is strictly smaller than the Heegaard genus of the three-manifold. Clearly not every presentation comes from a Heegaard splitting. </p>
<p>For general groups, there exist obstructions for a balanced presentation to be the fundamental group of a three-manifold (for example, see <a href="http://www.4124039.com/questions/294655/obstructions-to-realizing-a-balanced-presentation-as-a-3-manifold-group">obstruction</a>). Berge's famous program <a href="https://www.math.uic.edu/t3m/" rel="noreferrer">heegaard</a> can construct a Heegaard diagram from a group presentation, provided that the presentation is realizable. However, it does not help to characterize such presentations.</p>
http://www.4124039.com/q/3342595Complete folds and one cutDirkhttp://www.4124039.com/users/96522019-06-18T11:10:50Z2019-06-20T00:36:58Z
<p>The <a href="https://en.wikipedia.org/wiki/Fold-and-cut_theorem" rel="noreferrer">fold-and-cut theorem</a> states that any shape with straight sides can be cut by a single complete straight cut if the paper is the folded flat in the right way. Here is an example from an <a href="http://www.4124039.com/a/143066/9652">answer on another MO question</a>:
<br /> <img src="https://i.stack.imgur.com/91eko.png" alt="Turtle"><br /></p>
<p>Note that the folds are not complete, i.e. they are not along lines that run through the whole paper but stop somewhere in the middle. This makes it a little cumbersome to fold the pattern.</p>
<p>My questions are:</p>
<blockquote>
<p>Is there any shape with straight edges that can't be obtained by complete folds and one complete cut? If yes, which kind of shapes can be obtained by complete folds and one complete cut?</p>
</blockquote>
http://www.4124039.com/q/3343476Can an exotic diffeomorphism of the 4-ball change the isotopy class of an embedded surface?Kevin Walkerhttp://www.4124039.com/users/2842019-06-19T14:26:52Z2019-06-19T22:02:34Z
<p>Let <span class="math-container">$W$</span> be <span class="math-container">$B^4$</span> or <span class="math-container">$S^3 \times I$</span>. Let <span class="math-container">$Y$</span> be a properly embedded surface in <span class="math-container">$W$</span>. Let <span class="math-container">$f : W \to W$</span> be a diffeomorphism which is the identity near <span class="math-container">$\partial W$</span>. Very little is known about <span class="math-container">$\pi_0(\mbox{Diff($W$) rel $\partial W$})$</span> (see <a href="http://pi.math.cornell.edu/~hatcher/Papers/Diff(M)2012.pdf" rel="noreferrer">page 4 of this survey by Hatcher</a>), so <span class="math-container">$f$</span> might not be isotopic to the identity (or, in the <span class="math-container">$S^3\times I$</span> case, isotopic to a standard twist diffeomorphism). Nevertheless, I believe it is true that <span class="math-container">$f(Y)$</span> is isotopic rel boundary to <span class="math-container">$Y$</span>. In other words, exotic diffeomorphisms cannot do anything exotic to embedded surfaces.</p>
<p>The proof I have in mind (based on conversations with an expert) uses a "push-pull" type argument (and, in the <span class="math-container">$S^3\times I$</span> case, straightening along an arc) to concentrate any potential exoticness of <span class="math-container">$f$</span> in the neighborhood of one or two points. We can take these points to be far from <span class="math-container">$Y$</span>, and it follows that <span class="math-container">$f(Y)$</span> is isotopic to <span class="math-container">$Y$</span>.</p>
<p>Rather than write down the details of this proof, I would prefer to just cite a published result. So my question is:</p>
<blockquote>
<p>Is there a citable reference for the above claim that <span class="math-container">$f(Y)$</span> is
isotopic rel boundary to <span class="math-container">$Y$</span>? Is there a citable reference for the claim that diffeomorphisms of simple 4-manifolds (like <span class="math-container">$B^4$</span> or <span class="math-container">$S^3 \times I$</span>) can be "tamed" in the complement of a finite number of points?</p>
</blockquote>
http://www.4124039.com/q/3343783What is topology of all Square matrices such that matrix times it transpose is diagonalDaryl Cooperhttp://www.4124039.com/users/425062019-06-19T20:15:26Z2019-06-19T20:15:26Z
<p>What is the topology of the subspace <span class="math-container">$X_n\subset M_n(\mathbb R)$</span>
consisting of all non-zero <span class="math-container">$n\times n$</span> matrices <span class="math-container">$A$</span> such that <span class="math-container">$A^t A$</span> is diagonal ? For example <span class="math-container">$X_2$</span> is the product of a torus and an interval. Is <span class="math-container">$X_n$</span> a manifold ? Reference sought.</p>
http://www.4124039.com/q/3342313Wall self-intersection invariant for odd-dimensional manifolds?user101010http://www.4124039.com/users/994142019-06-17T23:26:26Z2019-06-19T19:55:56Z
<p>I am trying to convince myself that a naïve definition of the Wall self intersection number should not work for odd-dimensional manifolds. Namely, let <span class="math-container">$X^{2n-1}$</span> be a smooth oriented closed manifold and let <span class="math-container">$f : S^n \to X$</span> be a map from a sphere. Also, let's assume that <span class="math-container">$X$</span> and <span class="math-container">$S^n$</span> are based and <span class="math-container">$f$</span> preserves base-points and <span class="math-container">$n \geq 2$</span>. </p>
<p>For now, suppose that <span class="math-container">$f$</span> is a smooth immersion with double point curves of intersection, and then by orienting the double point curves and attaching them to the basepoint along arcs contained in the image of <span class="math-container">$f$</span> to obtain <span class="math-container">$\mu(f) \in \mathbb{Z} \pi_1(X)$</span>. To remove the dependence on the orientations of the curves, I better mod out by <span class="math-container">$\{ g = g^{-1} : g \in \pi_1(X)\}$</span>. I would like for <span class="math-container">$\mu(f)$</span> to be invariant under changing <span class="math-container">$f$</span> by a homotopy, so I better also mod out by setting <span class="math-container">$1=0$</span>. </p>
<p>Is <span class="math-container">$\mu(f) \in \mathbb{Z}\pi_1(X) /\langle\{ g = g^{-1} \}, 1=0\rangle$</span> an invariant of the based homotopy class of <span class="math-container">$f$</span>?</p>
<p>I am guessing that the answer is no so ignoring everything above perhaps I should just ask: Is there a map <span class="math-container">$f : S^3 \to T^5$</span> that has exactly one double point curve in the image that is nontrivial in <span class="math-container">$\mathbb{Z}^5 = \pi_1(T^5)$</span>? </p>
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