User alex suciu - MathOverflow

Answer by Alex Suciu for Genus one fibered links

2013-04-11T12:59:26Z

This will not answer your questions explicitly, but perhaps give a hint on how to go about thinking about them. In his paper, How to construct all fibered knots and links (Topology 21 (1982), no. 3, 263–280, MR0649758), John Harer explained how to do just that: start from the unknot, and perform a finite sequence of Hopf plumbings/deplumbings and Stallings twists. Adding a Hopf band increases the genus of the fiber by 1, and multiplies the monodromy by the corresponding Dehh twist; performing a Stallings twist along a suitable curve leaves the genus unchanged, but again composes the monodromy with a certain Dehn twist. Starting from a known fibered $n$-component link (such as the one consisting of $n$ fibers of the Hopf map), and performing an arbitrarily long sequence of Stallings twists should give you at least some candidate to answer the first question, and perhaps a starting point towards answering the second one.

Answer by Alex Suciu for Do Nielsen transformations on a presentation preserve the homotopy type of the corresponding presentation complex?

2013-04-07T21:06:39Z

Yes. In fact, slightly more is true: the simple homotopy type of the presentation 2-complex is preserved under Nielsen transformations. For a proof of this fact, see Micheal N. Dyer and Allan J. Sieradski, Trees of homotopy types of two-dimensional CW-complexes. I., Comment. Math. Helv. 48 (1973), 31–44. MR0377905

Answer by Alex Suciu for Genus of Y^3 = X^4 - 1.

2013-02-26T14:36:58Z

The complex curve $X^n + Y^m = 1$ is the Milnor fiber (at the origin) of the weighted-homogeneous polynomial $f(X,Y)=X^n + Y^m$. Suppose $\gcd(n,m)=1$. Then the Milnor fiber deform-retracts onto a (minimal) Seifert fiber for the singularity link, which is an $(n,m)$-torus link. This Seifert fiber consists of $n$ stacked disks, each one joined to the one above by $m$ once-twisted bands. It is now a simple exercise to see that the genus of this surface (equal to the Milnor number of $f$) is $(n-1)(m-1)/2$.

More generally, if $f=f(z_1,\dots,z_m)$ is a weighted-homogeneous polynomial with weights $(w_1,\dots, w_m)$, then the Milnor fiber $f=1$ has the homotopy type of a wedge of $(m-1)$-dimensional spheres, and the (Milnor) number of these spheres is given by $\mu=(w_1−1)(w_2−1)\cdots (w_m−1)$, according to John Milnor and Peter Orlik, Isolated singularities defined by weighted homogeneous polynomials, Topology 9 (1970), 385-393.

Answer by Alex Suciu for The knot whose complement is the Hantzsche-Wendt manifold

2013-01-27T02:06:43Z

Supposing the question is, "Can the Hantsche-Wendt manifold be realized as a cyclic branched cover of a knot in the 3-sphere", then the answer is yes: it is a 3-fold cyclic branched cover along the figure-eight knot. Incidentally, it is also a 2-fold branched cover along the Borromean rings.

On the other hand, if the question is, "Can the Hantsche-Wendt manifold $M$ be realized by performing Dehn surgery along a knot in the 3-sphere" (as Misha seems to suggest), then the answer is no, since $H_1(M,\mathbb{Z})=\mathbb{Z}_2 \oplus \mathbb{Z}_2$ is not a cyclic group.

Answer by Alex Suciu for Poincare dodecahedron space

2012-03-21T05:17:34Z

The presentation for $\pi_1(X)$ that you write down simplifies to $\langle b, e \mid beb=eb^2e, ebe=be^2b\rangle$. As is well-known, the binary icosahedral group $I^*$ is isomorphic to the group of unimodular, $2\times 2$ matrices over $\mathbb{Z}_5$. An explicit isomorphism $\pi_1(M) \to {\rm SL}(2,5)$ is given by $$ b \mapsto \begin{pmatrix} 3 & 1 \\ -1 & 0 \end{pmatrix}, \quad e \mapsto \begin{pmatrix} -1 & -1 \\ 0 & -1 \end{pmatrix}. $$

Answer by Alex Suciu for 3-manifolds with solvable fundamental group

2012-01-18T04:14:56Z

Here are some references that may help answer your question:

Charles B. Thomas, Nilpotent groups and compact $3$-manifolds, Proc. Cambridge Philos. Soc. 64 (1968), 303-306; MR0233359.

Benny Evans and Louise Moser, Solvable fundamental groups of compact $3$-manifolds, Trans. Amer. Math. Soc. 168 (1972), 189–210; MR0301742.

Peter Teichner, Maximal nilpotent quotients of $3$-manifold groups, Math. Res. Lett. 4 (1997), no. 2-3, 283-293; MR1453060.

A little nugget (due to John Milnor): among the Brieskorn manifolds $\Sigma(p,q,r)$, the only nilmanifolds are $\Sigma(2,3,6)$, $\Sigma(2,4,4)$, and $\Sigma(3,3,3)$, which are circle bundles over the torus with Euler number $1$, $2$, and $3$, respectively.

Answer by Alex Suciu for Spaces of Finite Subsets

2011-12-25T01:55:47Z

The spaces $\exp_n(S^1)$, as well as the embeddings $\exp_n(S^1) \subset \exp_{n+2}(S^1)$ were studied by Christopher Tuffley in Finite subset spaces of $S^1$, Algebr. Geom. Topol. 2 (2002), 1119–1145, http://dx.doi.org/10.2140/agt.2002.2.1119; MR1998017 (2004f:54008), and, more recently, by Sadok Kallel and Denis Sjerve in Remarks on finite subset spaces, Homology, Homotopy Appl. 11 (2009), no. 2, 229–-250, http://www.intlpress.com/hha/v11/n2/a12/; MR2591920 (2011a:55019).

In particular, based on an argument from Clifford H. Wagner's thesis (Symmetric, cyclic, and permutation products of manifolds, Dissertationes Math. (Rozprawy Mat.) 182 (1980); MR0605369 (82h:55021)), Kallel and Sjerve show that $\exp_n(S^1)$ is a closed manifold if and only if $n=1$ or $n=3$. Furthermore, Tuffley shows that $$ \pi_1(\exp_{n+2}(S^1) \setminus \exp_{n}(S^1)) = \langle x, y \mid x^{n+2} = y^{n+1} \rangle. $$

Answer by Alex Suciu for Almost-direct product and 1-formality

2011-09-16T02:25:43Z

Yes, the direct product of two 1-formal groups is again 1-formal, and so is the free product of two 1-formal groups. A proof is given in arxiv:0902.1250, Proposition 9.2.

And no, the almost direct product of two 1-formal groups need not be 1-formal. A proof is given in the same paper, Example 8.2. The group in question is a semi-direct product of the form $G=F_4\rtimes F_1$, where $F_n$ is the free group of rank $n$, which is of course 1-formal. The action of $F_1$ on $F_4$ is given by a certain pure braid $\beta \in P_4$, acting via the Artin representation on $F_4$; thus, the action is trivial on $H_1(F_4)$. For this extension, the ``tangent cone formula" fails: the tangent cone to the characteristic variety $V_2(G)$ is strictly included in the resonance variety $R_2(G)$. In view of Theorem A from the cited paper, the group $G$ is not 1-formal.

It is worth noting that $G$ is the fundamental group of the complement of a certain link of 5 great circles in $S^3$. Alternatively, $G$ can be realized as the fundamental group of the complement of an arrangement of 5 planes in $\mathbb{R}^4$, meeting transversely at the origin (of course, this real arrangement cannot be isotoped to an arrangement of 5 complex lines in $\mathbb{C}^2$). For more details on the construction and properties of such arrangements, see arxiv:math.GT/9712251. In particular, the pure braid $\beta$ is described there in Propositions 4.4, 4.6, and 4.9.

Comment by Alex Suciu

2013-04-08T01:34:06Z

The 3-dimensional Brieskorn manifolds are discussed at length by John Milnor in his paper On the 3-dimensional Brieskorn manifolds M(p,q,r), in: Knots, groups, and 3-manifolds (Papers dedicated to the memory of R. H. Fox), pp. 175–225. Ann. of Math. Studies, No. 84, Princeton Univ. Press, Princeton, N. J., 1975, MR0418127.

Comment by Alex Suciu

2013-02-26T18:37:29Z

Yes, thanks for pointing that out: I was implicitly assuming $f$ is reduced. I added now the more general form of this result.

Comment by Alex Suciu

2013-01-27T02:51:03Z

Sorry, I saw your post after I finished writing my answer. And yes, that's the right reference.

Comment by Alex Suciu

2012-07-04T23:49:51Z

I guess I did make that claim, back in the day. Surely I must have had a reason, but I forgot by now what that reason was. Nevertheless, that paper of S.J. Lomonaco, cited by Asano, Marumoto, and Yanagawa as "*The homotopy groups of knots, II. A solution to Problem 36 of R.H. Fox*, to appear", is not listed on MathSciNet, and I could not find a reference to it on GoogleScholar, either.

Comment by Alex Suciu

2012-04-18T22:30:28Z

Does the question refer somehow to the Hilton-Milnor theorem (named after Peter Hilton and John Milnor) on homotopy groups of wedges of spheres?

Comment by Alex Suciu

2012-03-22T01:26:37Z

Or, one could cite <en.wikipedia.org/wiki/Binary_icosahedral_group>; or <groupprops.subwiki.org/wiki/…; (if url's would work nicely in this markup language, that is...)

Comment by Alex Suciu

2012-03-22T01:07:24Z

Well, I have to confess, I found this isomorphism with the help of GAP, which assures me this map (which can be checked "by hand" is a homomorphism) is indeed a bijection. I would need more motivation to verify this by hand, but surely it can be done. Now, as to why $I^*$ is isomorphic to ${\rm SL}(2,5)$. One could use the same kind argument: start with a known presentation of $I^*$, say, $\langle x, y \mid x^3=y^5=(xy)^2 \rangle$, write down a map to ${\rm SL}(2,5)$, and check it's an isomorphism. Or, one could use a more geometric approach (symmetries of the icosahedron, quaternions, etc).

Comment by Alex Suciu

2012-01-21T18:44:14Z

More generally, the total Stiefel-Whitney class of $E_{\mathbb R}$ is the reduction mod $2$ of the total Chern class of $E_{\mathbb C}$. This is Problem 14-B on page 171 of Characteristic Classes, by John Milnor and James Stasheff

Comment by Alex Suciu

2011-12-27T04:12:02Z

John: It's kind of like you say, up to some shift. The homotopy type of $\exp_n(S^1)$ for $n=1,2,\dots$ is $$ S^1, S^1, S^3, S^3, S^5, S^5, \dots $$ This is proved by Tuffley in his Ph.D. thesis (Berkeley, 2003).

Comment by Alex Suciu

2011-09-22T13:24:46Z

Ah, now we're talking. Off the top of my head, I don't know the answer to your modified question. Sounds worth thinking about.

Comment by Alex Suciu

2011-09-18T13:09:21Z

Good question. I asked that same question myself recently. It turns out the two definitions of (1-) formality are different: the one you mention just above is weaker.

Comment by Alex Suciu

2011-09-18T02:08:40Z

Careful: the surface pure braid groups $P_{g,n}$ are always 1-formal, except when $g=1$ and $n\ge 3$, when they are not. A proof that $P_{1,n}$ is not 1-formal for $n\ge 3$ is given in arxiv.org/abs/0902.1250, Example 10.1, again using the ``tangent cone formula".

Comment by Alex Suciu

2011-09-17T19:42:28Z

Actually, the group ${\rm SL}(2,\mathbb{Z})$ decomposes as a free product with amalgamation, of the form $\mathbb{Z}_4*_{\mathbb{Z}_2} \mathbb{Z}_6$. Thus, the complete list of finite orders of elements in this group is $\{1, 2, 3, 4, 6\}$.

Comment by Alex Suciu

2011-09-16T18:17:29Z

For instance, it is known that the full McCool group $P\Sigma_n$ (the group of basis-conjugating automorphisms of $F_n$) is 1-formal; see <a href="dx.doi.org/10.1142/S0218216509007257"; rel="nofollow">this</a> paper by Berceanu and Papadima. But, as far as I know, it is an open question whether the upper-triangular McCool group $P\Sigma_n^+$ is 1-formal (for $n\ge 4$). Note that this group can be realized as an iterated almost-direct product of free groups, $P\Sigma_n^+=F_{n-1}\rtimes \cdots \rtimes F_2\rtimes F_1$, much the same as the pure braid group $P_n$.

Comment by Alex Suciu

2011-09-16T14:13:56Z

There are not many general techniques: it all depends on what the automorphisms defining the almost-direct products are. Intuitively, I think, the deeper into the Johnson filtration your automorphism is (i.e., the more it looks like the identity through the eye of the successive nilpotent quotients), the more likely it is that the extension will be non-1-formal.