Compelling evidence that two basepoints are better than one - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T17:42:13Z http://mathoverflow.net/feeds/question/40945 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/40945/compelling-evidence-that-two-basepoints-are-better-than-one Compelling evidence that two basepoints are better than one Daniel Moskovich 2010-10-03T18:26:20Z 2013-04-22T01:10:16Z <p>This question is inspired by <a href="http://mathoverflow.net/questions/40005/generalizing-a-problem-to-make-it-easier/40010#40010" rel="nofollow">an answer of Tim Porter</a>.<br><br> Ronnie Brown pioneered a framework for homotopy theory in which one may consider multiple basepoints. These ideas are accessibly presented in his book Topology and Groupoids. The idea of the fundamental groupoid, put forward as a multi-basepoint alternative to the fundamental group, is the highlight of the theory. The headline result seems to be that the van-Kampen Theorem looks more natural in the groupoid context.<br> I don't know whether I find this headline result compelling- the extra baggage of groupoids and pushouts makes me question whether the payoff is worth the effort, all the more so because I am a geometric topology person, rather than a homotopy theorist.</p> <blockquote> Do you have examples in geometric topology (3-manifolds, 4-manifolds, tangles, braids, knots and links...) where the concept of the fundamental groupoid has been useful, in the sense that it has led to new theorems or to substantially simplified treatment of known topics? </blockquote> <p>One place that I can imagine (but, for lack of evidence, only imagine) that fundamental groupoids might be useful (at least to simplify exposition) is in knot theory, where we're constantly switching between (at least) three different "natural" choices of basepoint- on the knot itself, on the boundary of a tubular neighbourhood, and in the knot complement. This change-of-basepoint adds a nasty bit of technical complexity which I have struggled with when writing papers. A recent proof (Proposition 8 of <a href="http://www.msp.warwick.ac.uk/agt/2009/09-03/p049.xhtml" rel="nofollow">my paper with Kricker</a>) which would have been a few lines if we hadn't had to worry about basepoints, became 3 pages. In another direction, what about fundamental groupoids of braids?<br> Have the ideas of fundamental groupoids been explored in geometric topological contexts? Conversely, if not, then why not? </p> http://mathoverflow.net/questions/40945/compelling-evidence-that-two-basepoints-are-better-than-one/40951#40951 Answer by Robert Bell for Compelling evidence that two basepoints are better than one Robert Bell 2010-10-03T19:22:45Z 2010-10-03T19:22:45Z <p>Crisp and Paris's proof of Tits's conjecture that the subgroup of an Artin group generated by the squares of the generators is itself a right-angled Artin group uses groupoids in an essential way. For each small type Artin group, they construct a surface with boundary by gluing annuli along squares and mark each square with a base point, and then then study the action of the Artin group on the fundamental groupoid (with respect to a basepoints which correspond to the gluing squares) of the obvious graph which is a deformation retract of this surface. In this way, they construct a representation of such an Artin group into automorphisms of the fundamental groupoid of a graph. Here is the reference:</p> <p>Crisp and Paris, The solution to a conjecture of Tits on the subgroup generated by the squares of the generators of an Artin group'', Invent. Math. 145 (2001).</p> http://mathoverflow.net/questions/40945/compelling-evidence-that-two-basepoints-are-better-than-one/40953#40953 Answer by André Henriques for Compelling evidence that two basepoints are better than one André Henriques 2010-10-03T19:32:13Z 2010-10-03T21:10:16Z <p>One situation in which it is essential to use groupoids is the study of orbifolds.</p> <p><b>Slogan:</b> <i>The set of points of an orbifold is a groupoid.</i> <Hr> Here's a concrete problem that is illuminated by the language of groupoids. Suppose I have an orbifold $M$ with a singuar stratum $X$. The stratum $X$ is isomorphic to $S^1$, and its isotropy group is some finite group $G$. Let's also assume that $X$ is oriented.</p> <p><b>Question:</b> What is the "monodromy" of going around that stratum?</p> <p>At first glance, one might guess that it's an element of $Aut(G)$.<br> That's wrong! The monodromy is an element of $Out(G)$.<br> So we have a somewhat paradoxical situation in front of us: there is a group associated to every point of $X$. Yet, the monodromy is not acting by automorphisms of that group.</p> <blockquote> <p>Here's an example of orbifold that nicely illustrates the kind if situation that can occur: $$M = (S^1\times V )/S_n,$$ where $S_n$ is the symmetric group and $V$ is a faithful representation. The group $S_n$ acts on the circle $S^1$ via the projection $S_n\twoheadrightarrow\mathbb Z_2$, and then the antipodal map. The representation $V$ of $S_n$ is just put there so that the orbifold isn't too degenerate (it can be omitted if you don't mind working with non-effective orbifolds). <br><br> In that example, the manifold $X$ is $S^1/\mathbb Z_2$. The isotropy group is the alternating group $A_n$. The monodromy is computed in the following way. Go half way around $S^1$, and then identifying "$A_n$ at point -1" with "$A_n$ at point +1" via any element of $S_n$ that sends -1 to +1. A choice of such an element yields an automorphism of $A_n$. But since there is no best way making such a choice, the only canonical thing is its class in $Out(A_n)$.</p> </blockquote> <p>Ok. Maybe now is good moment to try to remove some of the confusion.<br> It all becomes more clear once you realize that the thing that is associated to a point of $X$ is not a group. It's a groupoid:</p> <blockquote> <p>If $[M/G]$ is an orbifold and $x$ is a point in $M/G$, then the groupoid that lives above $x$ has objects given by points $m\in M$ mapping to $x$. An arrow from $m$ to $m'$ is given by a element of $G$ that sends $m$ to $m'$.</p> </blockquote> <p>The monodromy is then simply an automorphism of that groupoid (so now there's nothing weird any more). But this automorphism might fail to fix any of the objects of the groupoid. And so it can't be viewed as an automorphism of the corresponding group, unless you make some unnatural choices.</p> http://mathoverflow.net/questions/40945/compelling-evidence-that-two-basepoints-are-better-than-one/40965#40965 Answer by Jeffrey Giansiracusa for Compelling evidence that two basepoints are better than one Jeffrey Giansiracusa 2010-10-03T22:04:00Z 2010-10-03T22:04:00Z <p>Here is an interesting example where groupoids are useful. The mapping class group <code>$\Gamma_{g,n}$</code> is the group of isotopy classes of orientation preserving diffeomorphisms of a surface of genus $g$ with $n$ distinct marked points (labelled 1 through n). The classifying space <code>$B\Gamma_{g,n}$</code> is rational homology equivalent to the (coarse) moduli space <code>$\mathcal{M}_{g,n}$</code> of complex curves of genus $g$ with $n$ marked points (and if you are willing to talk about the moduli orbifold or stack, then it is actually a homotopy equivalence)</p> <p>The symmetric group <code>$\Sigma_n$</code> acts on <code>$\mathcal{M}_{g,n}$</code> by permuting the labels of the marked points.</p> <p><strong>Question:</strong> How do we describe the corresponding action of the symmetric group on the classifying space <code>$B\Gamma_{g,n}$</code>?</p> <p>It is possible to see <code>$\Sigma_n$</code> as acting by outer automorphisms on the mapping class group. I suppose that one could probably build an action on the classifying space directly from this, but here is a much nicer way to handle the problem.</p> <p>The group $\Gamma_{g,n}$ can be identified with the orbifold fundamental group of the moduli space. Let's replace it with a fundamental groupoid. Fix a surface $S$ with $n$ distinguished points, and take the groupoid where objects are labellings of the distinguished points by 1 through n, and morphisms are isotopy classes of diffeomorphisms that respect the labellings (i.e., sending the point labelled $i$ in the first labelling to the point labelled $i$ in the second labelling). </p> <p>Clearly this groupoid is equivalent to the original mapping class group, so its classifying space is homotopy equivalent. But now we have an honest action of the symmetric group by permuting the labels on the distinguished points of $S$.</p> http://mathoverflow.net/questions/40945/compelling-evidence-that-two-basepoints-are-better-than-one/40999#40999 Answer by James Griffin for Compelling evidence that two basepoints are better than one James Griffin 2010-10-04T09:22:50Z 2010-10-04T09:22:50Z <p>A short example.</p> <p>The family of pure braid groups does not possess a symmetric operad structure.</p> <p>But the fundamental groupoid of the little 2-discs operad is naturally a symmetric operad.</p> <p>Although the fundamental groups of the little 2-discs operad are the pure braid groups, there is no way to choose basepoints consistent with the operad structure.</p> <p>The moral is that groupoids are not naturally pointed, whilst groups are. If you're working with fundamental groups you should really be working with pointed spaces. Ofcourse you can ignore this and you'll only run into trouble if your mathematics doesn't work with pointed spaces, see the example above.</p> http://mathoverflow.net/questions/40945/compelling-evidence-that-two-basepoints-are-better-than-one/41000#41000 Answer by Greg Graviton for Compelling evidence that two basepoints are better than one Greg Graviton 2010-10-04T09:24:59Z 2010-10-04T09:24:59Z <p>My knowledge of algebraic topology is limited, but I have found one particular example where groupoids are almost imposing themselves: a purely algebraic proof that every subgroup of a free group is free.</p> <p>The usual geometric proof that a subgroup $H$ of the free group $F_2$ on two generators is free goes like this: one represents $F_2$ as the fundamental group $F_2 = \pi_1(S^1\vee S^1)$. Then, $H$ is the fundamental group of a covering of $p : X \to S^1\vee S^1$. But $X$ is a graph and the fundamental group of a graph is always free.</p> <p>Of course, it should be possible to translate the above into a purely algebraic proof, but this is very difficult if you don't introduce groupoids or a related notion. I haven't done it in detail, but it appears to me that one has to treat the group $F_2$ as a groupoid "inside" the subgroup $H$. The objects of this groupoid are equivalence classes</p> <p>$$g \sim g' \iff \exists h \in H. gh = g'$$</p> <p>and the morphisms are given by $\lbrace a : [g] \to [ga] \rbrace \cup \lbrace b : [g] \to [gb] \rbrace$ where $a$ and $b$ are the generators of $F_2$. This groupoid is a purely algebraic model of the covering space $X$. The group $H$ corresponds to the subgroupoid of morphisms of the form $\lbrace h : [e] \to [e] \rbrace$. It remains to be shown that $X$ can be contracted along a spanning tree while $H$ is left unchanged.</p> <p>I think the point is that this last contraction rips apart the group structure of $F_2$.</p> http://mathoverflow.net/questions/40945/compelling-evidence-that-two-basepoints-are-better-than-one/41009#41009 Answer by Laurence Taylor for Compelling evidence that two basepoints are better than one Laurence Taylor 2010-10-04T11:56:44Z 2010-10-04T11:56:44Z <p>Steenrod defined a local coefficient system as a functor from the path-groupoid of your space to a category. It is tricky to define homology/cohomology with local coefficients by just picking base points since the identifications given by the paths are needed as well. Calculations of induced maps are especially prone to error. Perhaps my favorite example is the following.</p> <p>Let $\tau\colon S^{2n} \to S^{2n}$ be the antipodal map with quotient $RP^{2n}$ and quotient map $\pi\colon S^{2n} \to RP^{2n}$. There is a twisted coefficient system $Z^w$ on $RP^{2n}$ so that $H_{2n}(RP^{2n};Z^w) \cong Z$ which is used in the non-orientable Poincare duality theorem.There is a natural notion of the pull-back coefficient system and the system $\pi^\ast Z^w$ is equivalent to the usual trivial system. Hence $H_{2n}(S^{2n};\pi^\ast Z^w) \cong Z$ but the equivalence involves a choice. </p> <p>Since $\pi\circ \tau = \pi$, there is a natural'' identification of $(\pi\circ\tau)^\ast Z^w$ with $\pi^\ast Z^w$. With these choices, the antipodal map has degree 1. If you make your choices so that the antipodal map has degree $-1$, then the two maps induced by $\pi$ have different degrees.</p> http://mathoverflow.net/questions/40945/compelling-evidence-that-two-basepoints-are-better-than-one/41517#41517 Answer by Bruno Martelli for Compelling evidence that two basepoints are better than one Bruno Martelli 2010-10-08T15:48:31Z 2010-10-08T15:48:31Z <p>If you study the set of ideal triangulations of a fixed punctured surface you find out that:</p> <ul> <li>("generators") by using sequences of flips you can relate any pair of triangulations,</li> <li>("relators") two such sequences are related by a well-understood set of moves (the most important one is the pentagon relation)</li> </ul> <p>The object you get with these "generators" and "relators" is not really a group, it is just a groupoid, called the Ptolemy groupoid. See for instance the <a href="http://arxiv.org/abs/math/9908165" rel="nofollow">paper of Chekhov and Fock</a> introducing quantum Teichmuller space.</p> http://mathoverflow.net/questions/40945/compelling-evidence-that-two-basepoints-are-better-than-one/46808#46808 Answer by Ronnie Brown for Compelling evidence that two basepoints are better than one Ronnie Brown 2010-11-21T10:32:57Z 2013-04-21T20:46:11Z <p>I came into groupoids by trying to find a new proof of the fundamental group of the circle. It turned out that one could do this using the fundamental groupoid on two base points. Writing the 1968 edition of my book now called Topology and Groupoids' (T&amp;G) (available on amazon.com and e-version from www.kagi.com) convinced me that all of 1-dimensional homotopy theory was better expressed in terms of groupoids rather than groups, in that one obtained more powerful theorems with simpler proofs. Later results on the fundamental groupoid of orbit spaces (Chapter 11 of T&amp;G) are more awkward to express in terms of groups; this elaborates on the point by Dustin Clausen. See further details below. </p> <p>Henry Whitehead answered the question of "Why not restrict to CW-complexes with just one vertex?" by considering covering spaces. Philip Higgins gave a considerable generalisation of Grusko's theorem by considering covering morphisms of groupoids, see his 1971 book Categories and groupoids' available as a TAC Reprint, 2005.</p> <p>In 1966 I thought about prospective uses of groupoids in higher homotopy theory, and this led over many years to higher dimensional Seifert-van Kampen Theorems, with a range of new nonabelian calculations of second relative homotopy groups and triad homotopy groups (for the latter, see the "nonabelian tensor product of groups"). That sounds relevant to geometric topology! </p> <p>So one answer to the original question is that the use of groupoids opens new worlds of possibilities. </p> <p>Actually the idea of `change of base point for the fundamental group' is a bit bizarre: one does not describe a railway timetable in terms of return journeys and change of starting point for these! Why is this still taught to students? </p> <p>In the end, an aesthetic viewpoint implies more power! </p> <p>Thanks to those above who give me additional examples. </p> <p>More information on my page <a href="http://www.bangor.ac.uk/r.brown/gpdsweb.html" rel="nofollow">From groups to groupoids</a>. </p> <p>September 2012: I forgot to add to this answer more information on <em>orbit spaces</em>, with particular reference to "two base points". </p> <p>Ross Geoghegan in his 1986 review (MR0760769) of two papers by M.A. Armstrong on the fundamental groups of orbit spaces wrote: "These two papers show which parts of elementary covering space theory carry over from the free to the nonfree case. This is the kind of basic material that ought to have been in standard textbooks on fundamental groups for the last fifty years." At present, to my knowledge, "Topology and Groupoids" is the only topology text to cover such results. </p> <p>Consider the action of the cyclic group of order 2, $Z_s$ on the unit circle $S$ by complex conjugation. Take $1$ as base point. The induced action of $Z_2$ on the fundamental group $\pi_1(S,1)$ is $n\mapsto -n$, and the quotient by this action is $Z_2$. But the quotient of $S$ by the action is a semicircle, which is contractible. What has gone wrong?</p> <p>The problem is there are <em>two</em> fixed points of the action. The quotient of the action of $Z_2$ on the groupoid $\pi_1(S, A)$, where $A$ consists of the points $\pm 1$, is indeed correct. </p> <p>The point is that a group acting on a space $X$ acts also on the fundamental groupoid $\pi_1 X$. If $X$ is Hausdorff, the action is properly discontinuous, and $X$ has a universal cover, then the fundamental groupoiud of the orbit space $X/G$ is the <em>orbit groupoid</em> of $\pi_1 X$. This is the groupoid expression of Armstrong's results. See Chapter 11 of <a href="http://pages.bangor.ac.uk/~mas010/topgpds.html" rel="nofollow">Topology and Groupoids</a>. </p> <p>April 21,2013: The book <a href="http://pages.bangor.ac.uk/~mas010/nonab-a-t.html" rel="nofollow">Nonabelian algebraic topology: filtered spaces, crossed complexes, cubical homotopy groupoids</a> gives an account of this new approach to basic algebraic topology at the border between homology and homotopy, <em>without</em> using singular homology theory, or simplicial approximation, but relying on the idea of multiple compositions of cubes. This also allows for results on second relative homotopy groups, results which, being essentially nonabelian, are not obtainable by traditional algebraic topology. It also avoids the "trick" of taking the free abelian group on ordered or oriented simplices in order to define chain groups, and the boundary map. </p> <p>All this comes from considering the question: if groupoids are useful in $1$-dimensional homotopy theory, how useful can they be in higher homotopy theory? One quickly notices that whereas group objects internal to groups are abelian groups, group objects internal to groupoids are in some sense "more nonabelian" than groups, as are groupoid objects internal to groupoids. So one looks to such objects to model higher homotopy properties: and this has been achieved. </p> http://mathoverflow.net/questions/40945/compelling-evidence-that-two-basepoints-are-better-than-one/128250#128250 Answer by Lee Mosher for Compelling evidence that two basepoints are better than one Lee Mosher 2013-04-21T14:31:34Z 2013-04-21T14:54:25Z <p>In my proof that mapping class groups are automatic, Ann. of Math. (2) 142 (1995), no. 2, 303–384, I used a theorem from ECHLPT "Word Processing in Groups" which says that if a groupoid is automatic then the corresponding group is automatic. </p> <p>That theorem was applied in the situation of a finite type surface $S$ with one or more punctures, using the groupoid mentioned in Bruno Martelli's answer which has come to be called the "Ptolemy groupoid" of $S$, due to connections with work of Robert Penner. That groupoid needs to be altered slightly for purposes of my proof, by adding data which breaks the finite symmetry group of an ideal triangulation. The data I added was an enumeration of the prongs of the triangulation, so the objects of the resulting groupoid are "ideal triangulations with enumerated prongs". The generating morphisms of this groupoid are of two types: permutations of the enumeration; and the flip relators mentioned by Bruno Martelli, called "elementary moves" in my paper, together with some rule for enumerating the prongs of the new ideal triangulation resulting from the elementary move.</p> <p>The group corresponding to this groupoid turns out to be the mapping class group of $S$, and hence the theorem from ECHLPT is applicable.</p> http://mathoverflow.net/questions/40945/compelling-evidence-that-two-basepoints-are-better-than-one/128257#128257 Answer by John Klein for Compelling evidence that two basepoints are better than one John Klein 2013-04-21T16:40:36Z 2013-04-22T01:10:16Z <p>I just wanted to add something to the discussion about the utility of adding additional basepoints. It turns out this is crucial for understanding certain aspects of embedding theory. See the bottom of this answer for an explanation.</p> <p>For a map of spaces $A \to B$, let $T(A\to B)$ be the category of spaces which factorize this map. This has objects given by factorizations $A \to X \to B$ and morphisms maps $X \to X'$ which are factorization compatible in the obvious sense. Let's consider the case of the constant map $S^0 \to \ast$. Clearly, an object of $T(S^0\to \ast)$ is just a <em>space with a preferred pair of basepoints.</em></p> <p>Then <strong>unreduced</strong> fiberwise suspension can be regarded as a functor $$S: \text{Top}(\emptyset \to \ast) \to \text{Top}(S^0 \to *) .$$ That is, the functor which assigns to an unbased space its unreduced suspension, considered as a space with two basepoints.</p> <p>Now a desuspension question in this context asks given an object $X \in \text{Top}(S^0 \to *)$, is there an object $Y \in \text{Top}(\emptyset \to \ast)$ and a weak equivalence $$SY \simeq X ?$$ More generally, I've gotten a lot of mileage out of the <strong>fiberwise version</strong> of this question. </p> <p>Given a space $B$ we can consider the unreduced fiberwise suspension of $\emptyset \to B$ as the projection map $B \times S^0 \to B$ (here unreduced fiberwise suspension of $Y\to B$ means the double mapping cylinder of the diagram $B \leftarrow Y \to B$, or concretely, it's $B \times 0 \cup Y \times [0,1]\cup B \times 1$. </p> <p><strong>Unreduced fiberwise suspension</strong> is then a functor $$S_B: \text{Top}(\emptyset \to B) \to \text{Top}(B\times S^0 \to B) ,$$ and one can consider the problem of whether an object $X \in \text{Top}(B\times S^0 \to B)$ can be written as $S_B Y$ up to weak equivalence.</p> <p><strong>Why I care about this problem</strong></p> <p>This problem naturally arises in embedding theory: if $P \to N \times [0,1]$ is an embedding, where $P$ and $N$ are closed manifolds and if $W$ is the complement of $P$ in $N \times [0,1]$ then $W$ is an object of the category $\text{Top}(N\times S^0 \to N$) and a necessary obstruction to compressing $P$ as an embedding into $N$ is that $W$ should fiberwise desuspend over $N$. Furthermore, in certain instances the existence of fiberwise unreduced desuspension suffices to finding the compression of the embedding. (This story is explained in detail in the paper: Poincaré duality embeddings and fiberwise homotopy theory, <em>Topology</em> <strong>38</strong>, 597–620 (1999).)</p> <p><strong>Postscript</strong></p> <p>In the fiberwise context there is a real difference between the <em>reduced</em> and <em>unreduced</em> cases of the desuspension problem. For example, in the case of the compression problem $P \to N \times I$ described above, the two inclusions $N \times i \to P$ for $i = 0,1$ might have distinct (fiberwise) homotopy classes. If this is the case, then there's no chance that the complement data $W$ can underly a reduced fiberwise suspension, for if it did, then the map $N \times S^0 \to P$ would factor through $S_N N \cong N \times D^1$, giving a homotopy of the two inclusions $N \times i \to P$.</p> <p>(For $Y \in \text{Top}(\text{id}:B \to B)$, the reduced fiberwise suspension $\Sigma_B Y$ is given by $$\Sigma_B Y = \text{colim}(B \leftarrow S_B B \to S_B Y) .$$ This is an endo-functor of $\text{Top}(\text{id}:B \to B)$.</p> <p>An even more mundane example is this: when $B = \ast$, we can consider $S^0$ with its two distinct basepoints. Clearly $S^0 = S\emptyset$, but $S^0$ is not, even up to weak equivalence, the reduced suspension of any based space.</p> http://mathoverflow.net/questions/40945/compelling-evidence-that-two-basepoints-are-better-than-one/128275#128275 Answer by Peter May for Compelling evidence that two basepoints are better than one Peter May 2013-04-21T18:52:04Z 2013-04-21T18:52:04Z <p>I'd like to expand on Dustin's point. There is simply no way to think sensibly about equivariant topology, whether algebraic or geometric, without taking account of multiple basepoints. Even taking account of them one runs into subtle difficulties invisible without them (see eg [65] on my web page). I'll give examples from algebraic topology, since that is what I know best, but examples from geometric topology must abound, as illustrated in other answers.</p> <p>Take a compact Lie group, or even just a finite group, and consider a smooth closed $G$-manifold $M$. What does it mean for $G$ to be orientable, and what is an orientation? These are seriously interesting questions, necessary to make sense of equivariant Poincar\'e duality, and they are difficult except in the boringly simple-minded case (treated in [53] on my website) when the tangent $G_x$-representation $T_x$ is isomorphic to the restriction to $G_x$ of an ambient $G$-representation $V$ for all $x\in M$. Usually there is no such $V$, and then I can't imagine answers that do not use functors defined on equivariant fundamental groupoids (which themselves are not altogether obvious to define.) Three references which give rather different answers to these questions are [93] and [100] on my web site, and <a href="http://front.math.ucdavis.edu/0310.5237" rel="nofollow">http://front.math.ucdavis.edu/0310.5237</a>, by Costenoble and Waner. I actually do not know how to compare these answers or to calculate with them. </p> <p>Again, while one can (twistedly) escape explicit use of fundamental groupoids when setting up the Serre spectral sequence with local coefficients nonequivariantly, one cannot do so equivariantly.</p> <p>Perhaps invoking equivariant theory is overkill, but the fundamental groupoid is such a natural thing, and so elementary, that it seems a little perverse to try to avoid it!</p>