User lee mosher - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T06:11:31Z http://mathoverflow.net/feeds/user/20787 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/131577/hyperbolic-pair-of-pants/131587#131587 Answer by Lee Mosher for Hyperbolic pair of pants. Lee Mosher 2013-05-23T12:00:32Z 2013-05-23T12:29:12Z <p>No it does not. Suppose that the three boundary components have equal and very long length $R$. Then the pair of pants is almost isometric to a graph having two vertices $V,W$ and three edges of length $R/2$ each connecting $V$ to $W$; by "almost isometric" I mean that there is a function from the pants to the graph which distorts length of closed geodesics and of geodesics hitting the boundary at right angles by an additive amount which is bounded for $R$ near $+\infty$. Each of the three boundary curves has its length preserved in this graph. However, the arc you ask about maps to a single edge, of length $R/2$, so back in the pants this arc has length as close to $R/2$ as you like.</p> <p>This can be made rigorous using formulas of hyperbolic trigonometry found, for example, in Thurston's book <a href="http://library.msri.org/books/gt3m/PDF/" rel="nofollow">The geometry and topology of 3-manifolds</a>.</p> http://mathoverflow.net/questions/131506/p1-minus-k-points/131521#131521 Answer by Lee Mosher for $P^1$ minus k points Lee Mosher 2013-05-22T22:24:19Z 2013-05-22T22:24:19Z <p>One can always find a fundamental domain for $G$ which is an ideal polygon $P \subset \mathbb{H}$ having $2k-2$ vertices at infinity and $2k-2$ sides, so that the sides are written in cyclic order around $P$ as $a_1 a_2 \ldots a_{k-1} \bar a_{k-1} \ldots \bar a_2 \bar a_1$, and so that the group $G$ is generated by $k-1$ side pairings which are isometries identifying $a_i$ to $\bar a_i$ for $i=1,\ldots,k-1$. This gives a fairly explicit description of $G$, expressed in the language of the Poincare polygon theorem.</p> <p>One can indeed use this idea, or closely related ideas, to parameterize $\cal M_k$: see the paper of Epstein and Bowditch, "Natural triangulations associated to a surface" Topology 27 (1988); and the paper of Penner, "The decorated Teichmüller space of punctured surfaces" Comm. Math. Phys. 113 (1987). </p> http://mathoverflow.net/questions/130431/mapping-class-group-of-once-punctured-torus/130494#130494 Answer by Lee Mosher for Mapping class group of once-punctured torus Lee Mosher 2013-05-13T15:08:42Z 2013-05-13T16:16:14Z <p>I think the correct answer is the book "Thurston's work on surfaces", or under its original French title "Travaux de Thurston sur les surfaces", by Fathi, Laudenbach, and Poenaru; english translation by Kim and Margalit.</p> <p>The explicit statement you want is in the beginning of section 1.5.</p> http://mathoverflow.net/questions/130306/the-role-of-the-automatic-groups-in-the-history-of-geometric-group-theory/130499#130499 Answer by Lee Mosher for The role of the Automatic Groups in the history of Geometric Group Theory Lee Mosher 2013-05-13T16:08:35Z 2013-05-13T16:14:06Z <p>I think there are a few different ways in which Automatic Groups affected the history of Geometric Group Theory.</p> <p>One was mentioned by Derek Holt, which I will spin in a slightly different way: if you really want to know the group, and if it has an automatic structure, you had better know that structure. For an individual group, the way to find out is to plug its presentation into the programs that Derek mentions. For classes of groups things can be trickier, but the effort is rewarding and sometimes even quite beautiful, for instance the proofs on automaticity of braid groups and Euclidean groups in "Word processing in groups", and the proof by Niblo and Reeves on biautomaticity of cubulated groups; and I retain a lot of affection for my paper proving automaticity of mapping class groups (and no offense taken, Misha).</p> <p>Another thread of influence is the theory of biautomatic groups. To me, this theory is not yet played out, although as others say perhaps the open questions are hard. However, there are various beautiful pieces of this theory which had some interesting effects, and I think there is still some gold to mine here. The theory starts with the papers of Gersten and Short. Some applications of that theory are: the proof by Bridson and Vogtmann that $Out(F_n)$ is not biautomatic when $n \ge 3$; and my proof that direct factors of biautomatic groups are biautomatic. I also like the beautiful papers of Neumann and Shapiro which describe completely all possible automatic and biautomatic structures on $Z^n$, and the paper of Neumann and Reeves and its followup by Neumann and Shapiro which describe how to construct biautomatic structures on central extensions. I like to think that one of the effects of the Gersten/Short theory is that it gives a hint to a "hierarchical" structure on a biautomatic group. Farb's thesis follows this idea up with his notion of relatively automatic and bi-automatic groups, and the thesis of my student Donovan Rebbechi pins some of this down by giving rigorous proofs of some statements in Farb's thesis regarding bi-automatic structures on relatively hyperbolic groups. The theory of biautomatic groups is definitely still alive; poking around on MathSciNet just this very moment I find a paper that slipped my notice and that I want to go read right now, by Bridson and Reeves studying the isomorphism problem for biautomatic groups.</p> <p>A separate and very important thread of influence is how the theory of automatic groups stoked interest in certain classes of quasi-isometry invariants. Gromov had already proved that hyperbolicity of a group is equivalent to linearity of the Dehn function (the isoperimetric function). One of the big applications of an automatic structure is Thurston's theorem that an automatic group has a quadratic (or better) Dehn function, which combined with the theorem that every subquadratic Dehn function is actually linear proves that nonhyperbolic automatic groups have quadratic Dehn functions on the nose. Thurston's proof of the quadratic upper bound introduced the concept of a combing of a group, and this led to a whole industry of studying different classes of combings, and the upper bounds they impose on the Dehn function. I particularly like the proof by Hatcher and Vogtmann finding an exponential upper bound to the Dehn function of $Out(F_n)$, which proceeds by finding a quite broadly stretched (pun intended) combing for $Out(F_n)$. Finding bounds on Dehn functions, and pinning down exact Dehn functions can be far from obvious, e.g. Robert Young's proof that $SL(n,Z)$ has quadratic Dehn function for $n \ge 5$, and the proof by Handel and myself of the exponential lower bound for the Dehn function of $Out(F_n)$. The issue of lower bounds is not particularly connected to automatic groups in any mathematical sense, but the historical connection is what I am trying to emphasize. Combings and Dehn functions have continued to grow from these and various seeds, and although I don't want to overstate the particular influence of Thurston's theorem in this context, nonetheless it is (besides Gromov's) one of the earliest concrete computations of Dehn functions. </p> http://mathoverflow.net/questions/130081/converse-to-milnors-theorem-on-manifolds-with-nonnegative-ricci-curvature/130094#130094 Answer by Lee Mosher for Converse to Milnor's theorem on manifolds with nonnegative Ricci curvature. Lee Mosher 2013-05-08T16:23:05Z 2013-05-08T18:17:03Z <p>For a compact counterexample, take any nilmanifold $N/H$ modulo the action of a freely acting cocompact lattice $\Lambda$, assuming $N/H$ is not just Euclidean space and $\Lambda$ is not just virtually abelian. For instance, $N=N/H$ is the $3 \times 3$ real Heisenberg group and $\Lambda$ is the $3 \times 3$ integer Heisenberg group. The proof that this is a counterexample is to apply Wilking's theorem in Anton's answer, and the theorem that a finitely generated nilpotent group has polynomial growth.</p> http://mathoverflow.net/questions/126995/classification-of-geometric-outer-automorphisms-of-free-groups/129906#129906 Answer by Lee Mosher for Classification of geometric outer automorphisms of free groups Lee Mosher 2013-05-06T22:20:31Z 2013-05-06T22:20:31Z <p>I think the short answer is "No", you cannot deduce <em>from this result</em> a classification of all geometric outer automorphisms. </p> <p>I think it might eventually be possible to obtain a classification of geometric outer automorphisms from the more powerful "relative train track / lamination" machinery developed in the works of Bestvina, Feighn, and Handel, although that has not been done. Nonetheless one can deduce bits and pieces of such a classification. </p> <p>For instance, suppose that $\phi \in Out(F_n)$ has an attracting lamination $\Lambda$ with the property that the smallest free factor of $F_n$ that supports $\Lambda$ is the whole free group. In this case one can prove that $\phi$ is geometric if and only if there exists a finite $\phi$-invariant set of conjugacy classes $c_1,...,c_k$ such that the smallest free factor of $F_n$ that supports $c_1,...,c_k$ is also the whole free group. The outer automorphisms of this kind are exactly those that are represented by pseudo-Anosov homeomorphisms on compact surfaces with no restriction on the number of boundary components. This statement can be found in a slightly different form in Proposition 2.38 of the paper <a href="http://front.math.ucdavis.edu/0908.1255%20%22Subgroup%20classification%20in%20%24Out%28F_n%29%24%22" rel="nofollow">Subgroup classification in $Out(F_n)$</a> by Handel and myself, and in this exact form in the soon-to-appear Part III of the expanded version "Subgroup decomposition in $Out(F_n)$".</p> http://mathoverflow.net/questions/129883/periodic-automorphisms-of-free-groups-and-surface-homeomorphisms/129899#129899 Answer by Lee Mosher for Periodic automorphisms of free groups and surface homeomorphisms Lee Mosher 2013-05-06T22:03:21Z 2013-05-06T22:03:21Z <p>Assuming that you mean outer automorphisms of $F_n$ (as per my comment), the answer is that if $n \ge 3$ then there is a nongeometric finite order element of $Out(F_n)$, for example the outer automorphism class of $a_1 \mapsto a_1^{-1}$ and $a_i \mapsto a_i$ for $2 \le i \le n$. </p> <p>For the proof, given $h : M \to M$ that induces a nonidentity, finite order, outer automorphism $\phi$ of $\pi_1 M$, I'll show that there does not exist a free factor of $\pi_1 M$ of rank $\ge 2$ whose conjugacy class is fixed by $\phi$ and whose elements all have conjugacy classes fixed by $\phi$, in contrast with the above example having such a free factor of rank $n-1$.</p> <p>By Nielsen realization, after an isotopy of $h$ we may assume that $h$ is a finite order isometry with respect to some hyperbolic structure on $M$ with totally geodesic boundary. If there is a free factor $G &lt; \pi_1 M$ of rank $\ge 2$ with conjugacy class fixed by $h$ then $h$ lifts to the $G$-cover $\tilde M$, and that lift restricts to an isometry of the convex hull $\tilde h : \mathcal{H}(\tilde M) \to \mathcal{H}(\tilde M)$. Since $rank(G) \ge 2$, the convex hull $\mathcal{H}(\tilde M)$ is itself a compact hyperbolic surface with totally geodesic boundary (if $rank(G)$ were equal to $1$ then the convex hull would just be a closed geodesic, and no contradiction would arise). Since $h$ fixes the conjugacy classes of all the elements of $G$, and since $G$ is a free factor (malnormality of $G$ suffices), it follows that $h$ induces the identity outer automorphism of $\pi_1(G)$. From this it follows in turn that $\tilde h$ is the identity isometry on $\mathcal{H}(\tilde M)$. It follows by projection that $h$ is locally the identity at some points of $M$, which by analytic continuation implies that $h$ is globally the identity. So $h$ induces the identity outer automorphism.</p> http://mathoverflow.net/questions/129507/hyperbolic-riemann-surface/129537#129537 Answer by Lee Mosher for Hyperbolic Riemann Surface Lee Mosher 2013-05-03T13:57:51Z 2013-05-03T13:57:51Z <p>Here's an answer from a different point of view then Henri's. It may happen that $X'$ is disconnected but in that case I'll just argue one component at a time. The Riemann surface $X'$ is noncompact, and if it is nonhyperbolic then by the Riemann mapping theorem it must be conformally equivalent (biholomorphic) to either $\mathbb{C}$ or $\mathbb{C}-$(point). Each of its one or two ends must therefore be a removable singularity. But that's impossible if $r_x > 0$ (justifying this impossibility probably involves a case analysis not unlike what Henri proposes).</p> http://mathoverflow.net/questions/127889/is-rigour-just-a-ritual-that-most-mathematicians-wish-to-get-rid-of-if-they-could/128744#128744 Answer by Lee Mosher for Is rigour just a ritual that most mathematicians wish to get rid of if they could? Lee Mosher 2013-04-25T18:17:57Z 2013-04-25T18:17:57Z <p>I disagree a little with Greg Martin's answer. In my mind, the correct analogy with physics is the question: is physical reality just a ritual that most physicists wish to get rid of if they could? Physics is at its root a laboratory science where good results must accord with physical reality. Of course there is a lot of creative activity in physics which does not take place in the laboratory, during which physical theories are developed outside the setting of the laboratory; sometimes people who practice this are called "theoretical physicists". But in the end, physical theories that contradict physical reality either die or undergo changes that put them back in accord with physical reality. Physical theories that survive are ones which are actually verified to be in accord with physical reality; the people who do this part of physics are sometimes called "laboratory physicists". Even a theoretical physicist worth his or her salt needs to have a good "physical intuition" in order not to spin out physical nonsense.</p> <p>I like to think that mathematics at its root is a laboratory science where good results must accord with logic. Of course there is a lot of creative activity in mathematics where logic is set aside, where one instead uses intuition or analogy or common sense or beauty or naturality or one of many other "illogical" activities in order to discover a solution to a mathematical problem. But in the end, once a potential solution has been discovered, it must be tested by logic, that is to say, it must be proved correct. </p> <p>Extending what Milnor is quoted by the OP as saying, regarding "Some mathematicians... while some...", I can imagine a world where mathematical activity is divided into "laboratory mathematics" and "theoretical mathematics": the theoretical mathematicians just do the creative part, developing solutions of problems; the laboratory mathematicians do the grunt work to provide the actual proofs. My wording is chosen as a way to play devil's advocate, I'm not sure I see any actual value in such a division. At the very least, a mathematician worth his or her salt needs to have a good "logical intuition" in order not to spin out mathematical nonsense. </p> http://mathoverflow.net/questions/128711/what-does-it-mean-that-homotopy-is-generic/128714#128714 Answer by Lee Mosher for What does it mean that homotopy is generic? Lee Mosher 2013-04-25T13:38:51Z 2013-04-25T13:38:51Z <p>Intuitively, for the homotopy to be generic has the same meaning as for the Morse function itself to be generic: the set of points where it fails to be a submersion is as simple as possible. Practically, this means that for all but a finite number of times (values of the $t$ parameter) the function $H(x,t)$ is a Morse function of the variable $x$ and the Morse singularities trace out smooth curves transverse to the $x$ direction; and for each of those exceptional times $t$ the function $H(x,t)$ is Morse at all values of $x$ except for a single value $x$ (I'm assuming compactness here) where it undergoes one of a certain class of very special singularities obtained by collapsing two Morse singularities, called "birth-death" singularities. Just as with Morse singularities themselves, which are described locally by specific functions of $x$ in some coordinate system, birth-death singularities are described locally by specific functions of $x,t$ in some coordinate system. The function $H(x,t) = x^2 - t$ at $t=0$ is an example of a birth-death singularity in one dimension.</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/127867/when-does-one-obtain-different-3-manifolds-by-pasting-two-tori/127880#127880 Answer by Lee Mosher for When does one obtain different 3-manifolds by pasting two tori? Lee Mosher 2013-04-17T19:06:57Z 2013-04-17T19:06:57Z <p>Let me address your third question, about the Thurston geometries. What I have to say is covered in standard references such as Peter Scott's article "The geometries of 3-manifolds".</p> <p>Thurston's result about geometries of compact 3-manifolds requires one to first decompose the 3-manifold canonically into pieces; only after doing that will the pieces have one of eight geometries. Your manifold is already prime, so only the decomposition along incompressible tori is needed. Every prime manifold can be cut along a unique minimal collection of incompressible tori (unique up to isotopy, that is) so that the complementary pieces are either Seifert fibered or atoroidal. </p> <p>In your case, the starting manifold $M$ is indeed Seifert fibered, in fact it is locally trivially fibered over the 3-punctured sphere, with circle fibers. One can work out from your description of $M$ how the circle fibers restrict to the two boundary tori, for example the circle fibers on the outer boundary torus form a foliation by circles of slope $(1,2)$ running parallel to the inner torus. The geometry of $M$ is $\mathbb{H}^2 \times \mathbb{R}$, because the 3-punctured sphere embeds in $M$ transverse to the fibers so that the monodromy map is (up to isotopy) of finite order $2$. </p> <p>After you glue the two boundary tori to obtain the manifold $M_f$, it may or may not be Seifert fibered, depending on whether the gluing map $f$ matches up the circle fibers of the inner boundary with the circle fibers of the outer boundary (up to isotopy). </p> <p>If $f$ does not match up the inner and outer boundary fibers then $M_f$ is not Seifert fibered, and it is not atoroidal, so it has no geometry itself; you will have to recut first along the glued torus, to get the Seifert fibered manifold $M$. In this case $M_f$ is a "graph manifold", which simply means a prime manifold such that the pieces of its torus decomposition are all Seifert fibered. </p> <p>If $f$ does match up the inner and outer boundary fibers then $M_f$ is Seifert fibered and its geometry has one of the two Seifert fibered geometries over $\mathbb{H}^2$, either $\mathbb{H}^2 \times \mathbb{R}$ or $PSL(2,\mathbb{R})$: if $f$ matches up the outer meridian curve to the inner meridian curve then $M_f$ has $\mathbb{H}^2 \times \mathbb{R}$ geometry; otherwise $M_f$ has $PSL(2,\mathbb{R})$ geometry.</p> http://mathoverflow.net/questions/126615/noncompact-manifold-with-two-ends-splits/126623#126623 Answer by Lee Mosher for noncompact manifold with two ends splits? Lee Mosher 2013-04-05T14:15:00Z 2013-04-05T14:57:48Z <p>Take the infinite ladder surface, a 2-manifold with 2 ends and infinitely generated $H_1$. One description, an embedding in $\mathbb{R}^3$, is the boundary of an $\epsilon$-neighborhood of the ladder graph $$(\{0,1\} \times 0 \times \mathbb{R}) \cup ([0,1] \times 0 \times \mathbb{Z})$$ In this description, the geodesic $\beta$ will just be a vertical line, and the Busemann function is just projection to the $z$-axis.</p> <p>Another description is to start with the closed surface of genus 2 whose fundamental group has the presentation $\langle a,b,c,d \,\, | \,\, [a,b][c,d]=1 \rangle$ and then take the infinite cyclic regular covering space associated to the homomorphism to $\mathbb{Z}$ defined by $a \mapsto 1$, $b,c,d, \mapsto 0$. In this description, the geodesic $\beta$ will be the lift of a closed geodesically embedded circle representing the conjugacy class of $a$ in the fundamental group.</p> <p>Assuming that by a "regular function" you mean one for which every value is a regular value, the error in your proof is simply the statement that the Busemann function is regular. For any complete $\mathbb{Z}$-invariant Riemannian metric on the infinite ladder surface, even if the Busemann function is Morse there will be infinitely many singular points of index 1.</p> http://mathoverflow.net/questions/126044/whats-the-name-of-twisted-semidirect-products/126051#126051 Answer by Lee Mosher for What's the name of "twisted semidirect products"? Lee Mosher 2013-03-30T21:05:12Z 2013-03-30T21:05:12Z <p>This group fits into a short exact sequence $1 \to \Lambda \to G \to K \to 1$. In this situation there is a plain vanilla terminology that $G$ is an "extension group". If you want to bring in mention of the kernel $\Lambda$ and the quotient $K$, then there is an unfortunately ambiguous terminology saying either that $G$ is an "extension of $K$ by $\Lambda$", or that $G$ is an "extension of $\Lambda$ by $K$". Another notation/terminology that is still evolving is to read the short exact sequence from left to right, leave the word "extension" out of it, and say that $G$ is a "$\Lambda$-by-$K$ group". </p> <p>Once upon a time I was in the lucky situation of entitling a paper with complete lack of ambiguity as "A hyperbolic-by-hyperbolic hyperbolic group".</p> http://mathoverflow.net/questions/125840/a-direct-proof-of-the-harer-zagier-recursion-enumerating-the-ways-to-paste-a-2n-g/125845#125845 Answer by Lee Mosher for A direct proof of the Harer-Zagier recursion enumerating the ways to paste a 2n-gon to get a genus g surface? Lee Mosher 2013-03-28T18:20:17Z 2013-03-28T18:20:17Z <p>How about </p> <blockquote> <p>Goulden, I. P.; Nica, A. A direct bijection for the Harer-Zagier formula. J. Combin. Theory Ser. A 111 (2005), no. 2, 224–238.</p> </blockquote> <p>or one of the references therein?</p> http://mathoverflow.net/questions/125737/chain-recurrent-set-of-a-isometry/125752#125752 Answer by Lee Mosher for Chain Recurrent Set of a Isometry Lee Mosher 2013-03-27T18:01:17Z 2013-03-27T18:01:17Z <p>Given $x \in X$, consider the bi-infinite sequence $T^n(x)$. Given $\epsilon>0$, the space $X$ is covered by a finite collection of sets of diameter $&lt;\epsilon$, and by the pigeonhole principle one of those sets contains infinitely many entries of the sequence $T^n(x)$. It follows that for every $M > 0$ there exists $i &lt; j$ such that $j-i \ge M$ and $d(T^i(x),T^j(x)) &lt; \epsilon$, so $d(x,T^{j-i}(x)) &lt; \epsilon$ and so $$x, T(x), T^2(x), \ldots, T^{j-i}(x)$$ is an $\epsilon$-chain from $x$ to itself.</p> http://mathoverflow.net/questions/125527/does-the-poset-of-free-factors-of-a-free-group-form-a-lattice/125542#125542 Answer by Lee Mosher for Does the poset of free factors of a free group form a lattice? Lee Mosher 2013-03-25T16:33:41Z 2013-03-25T22:28:20Z <p>ORIGINAL ANSWER, ADDRESSING A SLIGHTLY DIFFERENT QUESTION: There is a closely related poset for which greatest lower bounds and least upper bounds indeed exist. Instead of an individual free factor $A$, first consider its conjugacy class $[A]$. Then, instead of individual conjugacy classes of free factors $[A]$, consider a "free factor system" (in the language of Bestvina-Feighn-Handel): a finite set $\mathcal{F} = \{[A_1],\ldots,[A_k]\}$ such that there exists a free factorization of the form $F_n = A_1 * \cdots * A_k * B$ where $B$ may or may not be trivial but all the $A_i$'s are nontrivial. The partial ordering $\mathcal{F}\sqsubset \mathcal{F}'$ is defined by requiring that for each $[A] \in \mathcal{F}$ there exists $[A'] \in \mathcal{F}'$ such that $A$ is conjugate to a subgroup of $A'$. </p> <p>The Kurosh Subgroup Theorem can be translated into the statement that this poset has greatest lower bounds. The greatest lower bound of $\mathcal{F}$ and $\mathcal{F}'$ is $$\mathcal{F} "meet" \mathcal{F}' = \{[A \cap A'] \, | \, [A] \in \mathcal{F}, [A'] \in \mathcal{F}', A \cap A' \ne 1\}$$ (I don't know how to get the "meet" operator in this version of TeX).</p> <p>CORRECTION: You have to allow the "trivial" free factor system in order for this meet to be well-defined, because it is possible that the definition above produces the emptyset. In which case I should have left out the condition "$A \cap A' \ne 1$" in my definition of the meet.</p> <p>The meet operator can then be extended to an operator on arbitrary sets of free factor systems, and using this one gets least upper bounds too: the least upper bound of $\mathcal{F}$ and $\mathcal{F}'$ is the meet of all free factor systems $\mathcal{F}''$ (including the improper free factor system $\{[F]\}$) such that $\mathcal{F} \sqsubset \mathcal{F}''$ and $\mathcal{F}' \sqsubset \mathcal{F}''$. This $\mathcal{F}''$ is called the free factor support'' of $\mathcal{F}$ and $\mathcal{F}'$.</p> <p>ADDITION, ADDRESSING THE ORIGINAL QUESTION: Now that I've had a chance to think more about the Kurosh subgroup theorem myself, I realize that the answer to the original question is just "yes".</p> <p>First, for greatest lower bound: the intersection of any collection of free factors of the finite rank free group $F_n$ is a free factor. For two free factors $A,B$ this is a consequence of the Kurosh subgroup theorem which says that the set of nontrivial intersections of $A$ with conjugates of $B$ is a finite set $\{C_1,\ldots,C_K\}$ with the property that there exists a free factorization $A = C_1 * \cdots * C_k * D$ where $D$ may be trivial; so $A \cap B$, if nonempty, is one of the $C$'s, and is therefore a free factor of $A$, and a free factor of a free factor of $F_n$ is a free factor of $F_n$. In general, for any collection of free factors, write them in a sequence, intersect each initial segment of the sequence, and use the fact that in a finite rank free group, a free factor strictly included in another has smaller rank (proved by abelianizing).</p> <p>Then, for least upper bound: given any collection $\mathcal A$ of free factors, the intersection of all free factors (including the improper free factor $F_n$) containing each element of $\mathcal A$ is the least a free factor containing each element of $\mathcal A$.</p> http://mathoverflow.net/questions/125544/conditions-for-a-graph-to-be-the-1-skeleton-of-a-surface/125547#125547 Answer by Lee Mosher for Conditions for a graph to be the 1- skeleton of a Surface Lee Mosher 2013-03-25T17:50:47Z 2013-03-25T17:50:47Z <p>It suffices to consider a connected graph. Start from a point, which is the 1-skeleton of a sphere. By induction, consider a connected graph $G$ and an edge $E$, and let $S$ be the surface in which the complementary graph $G \setminus E$ is embedded as the 1-skeleton of a CW structure.</p> <p>If $E$ disconnects $G$ into two connected subgraphs, $S$ has two components and you can embed $G$ in their connected sum. If $E$ is a loop edge with base vertex $v$, you can extend the embedding $G \setminus E \to S$ to an embedding $G \to S$ by including $E$ into a corner of some 2-cell near $v$. If $E$ is a nonloop edge with end vertices $v,w$, but $E$ does not disconnect, there are two cases. If there is just one 2-cell touching $v,w$ then you can embed $E$ into that two cell, connecting one corner to another and cutting that 2-cell into two 2-cells. If there are two or more 2-cells touching $v,w$, then there exist 2-cells $C \ne D$ touching $v,w$ respectively, and you can take the "self-connected sum" of $S$ by taking the connected sum of $C$ and $D$ to form an annulus, and then extend the embedding of $G$ by embedding $E$ in that annulus, cutting from one boundary of the annulus to the opposite and subdividing that annulus into a 2-cell.</p> http://mathoverflow.net/questions/30156/demystifying-complex-numbers/125293#125293 Answer by Lee Mosher for Demystifying complex numbers Lee Mosher 2013-03-22T16:43:34Z 2013-03-22T16:43:34Z <p>This answer is an expansion of the answer of Yuri Bakhtin. </p> <p>Here is a kind of mime show.</p> <p>Silently write the formulas for $cos(2x)$ and $sin(2x)$ lined up on the board, something like this: $$cos(2x) = cos^2(x) \hphantom{+ 2 cos(x) sin(x)} - sin^2(x)$$ $$sin(2x) = \hphantom{cos^2(x)} + 2 cos(x) sin(x) \hphantom{- sin^2(x)}$$</p> <p>Do the same for the formulas for $cos(3x)$ and $sin(3x)$, and however far you want to go: $$cos(3x) = cos^3(x) \hphantom{+ 3 cos^2(x) sin(x)} - 3 cos(x) sin^2(x) \hphantom{- sin^3(x)}$$ $$sin(3x) = \hphantom{cos^3(x)} + 3 cos^2(x) sin(x) \hphantom{- 3 cos(x) sin^2(x)} - sin^3(x)$$</p> <p>Maybe then let out a loud noise like "hmmmmmmmmm... I recognize those numbers..."</p> <p>Then, on a parallel board, write out Pascal's triangle, and parallel to that write the application of Pascal's triangle to the binomial expansions $(x+y)^n$. Make some more puzzling sounds regarding those pesky plus and minus signs.</p> <p>Then maybe it's time to actually say something: "Eureka! We can tie this all together by use of an imaginary number $i = \sqrt{-1}$". Then write out the binomial expansion of $$(cos(x) + i \, sin(x))^n$$ break it into its real and imaginary parts, and demonstrate equality with $$cos(nx) + i \, sin(nx)$$</p> http://mathoverflow.net/questions/124998/was-the-early-calculus-inconsistent/125077#125077 Answer by Lee Mosher for Was the early calculus inconsistent? Lee Mosher 2013-03-20T14:45:22Z 2013-03-20T14:53:52Z <p>I found a copy of the relevant passage from Berkeley's works at <a href="http://www.maths.tcd.ie/pub/HistMath/People/Berkeley/Analyst/Analyst.html" rel="nofollow">this web site</a>. I have cut and pasted from that site, and I have reformatted the mathematics; apologies to the good Bishop for any alterations in meaning.</p> <blockquote> <p>XIV. To make this Point plainer, I shall unfold the reasoning, and propose it in a fuller light to your View. It amounts therefore to this, or may in other Words be thus expressed. I suppose that the Quantity $x$ flows, and by flowing is increased, and its Increment I call $o$, so that by flowing it becomes $x + o$. And as $x$ increaseth, it follows that every Power of $x$ is likewise increased in a due Proportion. Therefore as $x$ becomes $x + o$, $x^n$ will become $(x + o)^n$: that is, according to the Method of infinite Series, $$x^n + nox^{n-1} + \frac{n^2-n}{2} o^2 x^{n-2} + \text{ etc.}$$ And if from the two augmented Quantities we subduct the Root and the Power respectively, we shall have remaining the two Increments, to wit, $$o \text{ and } nox^{n-1} + \frac{n^2-n}{2} o^2 x^{n-2} + \text{ etc.}$$ which Increments, being both divided by the common Divisor o, yield the Quotients $$1 \text{ and } nx^{n-1} + \frac{n^2-n}{2} ox^{n-2} + \text{ etc.}$$ which are therefore Exponents of the Ratio of the Increments. Hitherto I have supposed that $x$ flows, that $x$ hath a real Increment, that $o$ is something. And I have proceeded all along on that Supposition, without which I should not have been able to have made so much as one single Step. From that Supposition it is that I get at the Increment of $x^n$, that I am able to compare it with the Increment of $x$, and that I find the Proportion between the two Increments. I now beg leave to make a new Supposition contrary to the first, i. e. I will suppose that there is no Increment of $x$, or that $o$ is nothing; which second Supposition destroys my first, and is inconsistent with it, and therefore with every thing that supposeth it. I do nevertheless beg leave to retain $nx^{n - 1}$, which is an Expression obtained in virtue of my first Supposition, which necessarily presupposeth such Supposition, and which could not be obtained without it: All which seems a most inconsistent way of arguing, and such as would not be allowed of in Divinity. </p> </blockquote> <p>It looks to me that Berkeley's argument amounts to an argument raised by every discerning student in a nonrigorous first semester calculus course: Is the increment zero? or not zero? How can it be both? That's inconsistent!'' In which case I would invite the good Bishop to come to my office hours where I would introduce him to $\epsilon$, $\delta$ proofs. </p> <p>I bet I could even convince the Bishop that Divinity would allow it: "Suppose the Devil gives you any $\epsilon > 0$. This $\epsilon$, although positive, might be very, very, very small, as small as the Devil likes...".</p> http://mathoverflow.net/questions/123772/fundamental-domain-for-subgroup-of-fuchsian-schottky-group/123790#123790 Answer by Lee Mosher for Fundamental domain for subgroup of fuchsian Schottky group. Lee Mosher 2013-03-06T17:16:50Z 2013-03-06T17:16:50Z <p>Here is a concrete construction of a fundamental domain for $H$ that works regardless of normality and finite generation. The translates of the fundamental domain $F$ by the action of $G$ form a tiling of $\mathbb{H}^2$. Let $T$ denote the dual tree of this tiling, with one vertex for each translate of $F$, and with two vertices connected by an edge if the corresponding translates of $F$ intersect along a line. Note that each oriented edge $E$ of $T$ corresponds to a transversely oriented line $L_E$ of the tiling which subdivides $\mathbb{H}^2$ into two halfplanes, and the transverse orientation on $L_E$ points into one of those half-planes which I'll denote $C_E$.</p> <p>The action of $G$ on $\mathbb{H}^2$ induces a properly discontinuous action of $G$ on $T$ by simplicial isomorphisms. Consider the restricted action of $H$ on $T$, also properly discontinuous. The quotient graph $T/H$ has fundamental group identified with $H$. Let $\tau$ be a maximal tree in $T/H$. Let $\tilde\tau \subset T$ be a homeomorphic lift of $\tau$. As $E \subset T$ varies over all oriented edges not contained in $\tilde\tau$ but with initial endpoint in $\tau$, the collection of half-planes $C_E$ demonstrates that $H$ is a Schottky group, and the complement of their union is a fundamental domain for $H$.</p> http://mathoverflow.net/questions/123372/discrete-subgroups-of-lie-groups-and-actions-on-homogeneous-spaces/123387#123387 Answer by Lee Mosher for discrete subgroups of Lie groups and actions on homogeneous spaces Lee Mosher 2013-03-02T01:24:43Z 2013-03-03T04:35:40Z <p>Here is a simple counter-example. Let $p : S_3 \to S_2$ be a degree 2 covering map from the closed genus 3 surface to the closed genus 2 surface, inducing an index 2 injection $p_* : \pi_1(S_3) \to \pi_1(S_2)$. Pulling back any hyperbolic structure from $S_2$ to $S_3$ via $p$ defines an embedding of Teichmuller spaces $p^* : T(S_2) \to T(S_3)$, the domain having dimension 6 and the range having dimension 12. Pick a point in $T(S_3)$ which is not in the image of this embedding nor in any translate of the image under the action of the mapping class group $MCG(S_3)$ on $T(S_3)$. This point represents a discrete subgroup $\Gamma \approx \pi_1(S_3)$ of $Isom(H^2) = PSL(2,R)$ such that, under the index 2 inclusion of $\Gamma \hookrightarrow \tilde\Gamma \approx \pi_1(S_2)$, the action of $\Gamma$ does not extend to an action of $\tilde\Gamma$.</p> <p>ADDED AFTER INITIAL COMMENTS: The OP has clarified that he intends only to ask about the existence of topological actions of $\tilde\Gamma$, in which setting this is not a counter-example as pointed out by Yves.</p> <p>However, in the other direction, rigidity theorems apply to give many interesting positive examples. For instance, suppose $\Gamma$ is a uniform lattice in $Isom(H^n) = SO(n,1)$, $n \ge 3$, meaning a discrete cocompact subgroup. If $\tilde\Gamma$ is a finite index supergroup of $\Gamma$, and if $\tilde\Gamma$ has no finite normal subgroups, then the given isometric action of $\Gamma$ extends to a faithful isometric action of $\tilde\Gamma$.</p> <p>To prove it, since $\tilde\Gamma$ is quasi-isometric to $\Gamma$ which is quasi-isometric to $H^n$, applying the Sullivan-Tukia quasi-isometric rigidity theorem for $H^n$ it follows that there is a cocompact discrete action of $\tilde\Gamma$ on $H^n$ with finite kernel, but we have hypothesized the kernel away so this action is faithful. We now have two discrete cocompact actions of $\Gamma$: the given one, and the restriction of the one on $\tilde\Gamma$ that we got by applying Sullivan-Tukia. We can therefore apply Mostow rigidity and conjugate the $\tilde\Gamma$ action by an isometry of $H^n$, so that its restriction to $\Gamma$ agrees with the given $\Gamma$ action, and we are done.</p> <p>Basically the same argument will work whenever the symmetric space satisfies Mostow rigidity and some reasonably strong form of quasi-isometric rigidity, which is true of just about every symmetric space associated to an irreducible semi-simple Lie group (except for $SL(2,R)$ of course).</p> <p>And if it is OK for the action to have a finite kernel, then just drop the hypothesis that $\Gamma$ has no finite index normal subgroup, and the same argument still works.</p> <p>ADDED AFTER FURTHER COMMENTS: See Misha's comment for an improvement on this argument that avoids QI-rigidity altogether, using solely Mostow rigidity.</p> http://mathoverflow.net/questions/123224/is-there-any-relation-between-automorphism-group-of-a-cayley-graph-over-a-group-a/123269#123269 Answer by Lee Mosher for Is there any relation between automorphism group of a Cayley graph over a group and over its subgroup? Lee Mosher 2013-02-28T21:54:41Z 2013-02-28T22:01:36Z <p>I don't know how interested you are in the kinds of things that happen in infinite groups, but this question does have some general interest in that setting. </p> <p>It seems to be quite common, for example, that $G$ has finite index in the automorphism group of its Cayley graph. This is true, for example, for many examples of lattices in Lie groups, as Alex Furman has noted. For a very specific example, this is true for the fundamental group of a closed surface of genus $g$ with the standard presentation $$\langle a_1,b_1,\ldots,a_g b_g \quad | \quad [a_1,b_1] \ldots [a_g,b_g] \rangle$$</p> <p>At a different extreme, for free groups with their standard generating set, the automorphism group of the Cayley graph is a locally compact topological group locally homeomorphic to the Cantor set, and hence the free group has uncountable index. </p> http://mathoverflow.net/questions/123243/isometric-but-differently-shaped-closed-surfaces-in-mathbbr3/123266#123266 Answer by Lee Mosher for Isometric but differently shaped closed surfaces in $\mathbb{R}^3$ Lee Mosher 2013-02-28T20:59:15Z 2013-02-28T21:58:23Z <p><a href="https://www.google.com/search?q=flexihedra&amp;ie=utf-8&amp;oe=utf-8&amp;aq=t&amp;rls=org.mozilla%3aen-US%3aofficial&amp;client=firefox-a" rel="nofollow">Flexihedra</a></p> http://mathoverflow.net/questions/122630/topology-of-mathbb-rn/122640#122640 Answer by Lee Mosher for Topology of ${\mathbb R}^n$ Lee Mosher 2013-02-22T15:40:00Z 2013-02-22T15:40:00Z <p>You can avoid the Kunneth formula, "all" you need is the existence of a single nontrivial homology group of $M$ in positive dimensions, namely $H_{\text{dim} \, M}(M;\mathbb{Z}/2\mathbb{Z}) = \mathbb{Z}/2\mathbb{Z}$. Once you know that, it follows that the identity map on $M$ is not homotopic to a constant, and so the inclusion map $M \to M \times N$ is not homotopic to a constant.</p> http://mathoverflow.net/questions/121571/connected-sum-of-topological-manifolds/121635#121635 Answer by Lee Mosher for Connected sum of topological manifolds Lee Mosher 2013-02-12T19:17:12Z 2013-02-13T21:06:57Z <p>In the topological category the proof that connected sum is well-defined depends on the Annulus Theorem, first proved by Kirby; the necessity of the Annulus Theorem is seen from Bruno Martelli's answer. So you are not likely to find a proof before Kirby's paper. Perhaps someone jotted a proof down, maybe someone who thought about the Annulus Theorem when it was still a conjecture, and realized that well-definedness of connected sum was a good application. But, I do not know. </p> <p>Anyway, the proof is straightforward once you have the Annulus Theorem. Here's a sketch.</p> <p>There's a couple of missing hypotheses. One must assume $M,M'$ are connected. One must also assume $M,M'$ are orientable. And one must assume the balls $B,B'$ are "nicely embedded"; at the minimum, assume that the boundary spheres $S,S'$ are locally bicollared, which implies globally bicollared by Brown's theorem. This rules out nastiness like an Alexander horned ball.</p> <p>Now one shows that the connected sum is independent of the choice of gluing map $S \to S'$. This follows from the fact that any two homeomorphisms $S \to S'$ which agree on orientations are isotopic: once that is known, one absorbs the isotopy into the collar neighborhoods. Proving this fact may already require the Annulus Theorem.</p> <p>For the rest, it suffices to prove that for any two nicely embedded balls $B_1,B_2 \subset M$ there exists an orientation preserving homemeorphism of $M$ taking $B_1$ to $B_2$, in fact an ambient isotopy. Using the boundary bicollaring, we may assume $B_1,B_2$ are contained respectively in open balls $U_1,U_2$, which are centered on points $p_1,p_2$ in some coordinate chart. We can also assume that $p_1=p_2$, because there is an ambient isotopy of $M$ taking $p_1$ to $p_2$: connect $p_1$ to $p_2$ by a path, cover the path by finitely many charts, and concatenate a sequence of ambient isotopies supported in these finitely many charts, moving $p_1$ along the path step by step to $p_2$. We can also replace $B_1$ by an arbitrarily small subball in $U_1$ centered at $p_1$, and similarly for $B_2$; this is straightforward to check using an ambient isotopy supported in the coordinate charts for $U_1$ and $U_2$. In particular, we can assume $B_1$ is contained in the interior of $B_2$.</p> <p>Now apply the annulus theorem: the difference $B_2 \setminus B_1$ is homeomorphic to a sphere crossed with an interval. Using this, one can then ambiently isotope $B_2$ to $B_1$.</p> http://mathoverflow.net/questions/121193/when-is-the-teichmuller-space-a-group/121198#121198 Answer by Lee Mosher for when is the Teichmuller space a group? Lee Mosher 2013-02-08T13:57:19Z 2013-02-08T13:57:19Z <p>Since you do not say what group operation you have in mind, your question is rather difficult to answer. But what you seem to be proposing in your "Or, basically..." sentence does not work.</p> <p>For $f : \mathbb{H} \to \mathbb{H}$ to be compatible with $G$ means that the Fuchsian groups $G$ and $f G f^{-1}$ are conjugate under some automorphism of $G$, which means that the points of $T(G)$ represented by those two Fuchsian groups are in the same orbit under the action on $T(G)$ of the mapping class group of the quotient surface $\mathbb{H} / G$ (I am assuming implicitly that $G$ has no torsion and so the quotient is indeed a surface as opposed to an orbifold). The mapping class group of $\mathbb{H} / G$, aka the Teichmuller modular group, is a finitely generated group acting properly discontinuously on $T(G)$, in particular the mapping class group orbit of any point of $T(G)$ is a discrete set. You might wish to say that the effect is to identify the orbit of a point of $T(G)$ with the mapping class group itself, and so you might wish to conclude that this puts a group structure on the orbit (this is itself problematical because orbits of group actions need not correspond bijectively to the group; but that is beside the point of your question). The real point is that this identification misses every point of $T(G)$ which is not on that orbit. </p> http://mathoverflow.net/questions/120612/trichotomies-in-mathematics/121003#121003 Answer by Lee Mosher for Trichotomies in mathematics Lee Mosher 2013-02-06T18:46:18Z 2013-02-06T18:46:18Z <p>Every finitely generated group is either of polynomial growth, intermediate growth, or exponential growth. </p> <p>As a statement, there is not much to this, the only mathematical content is that the growth function of every finitely generated group has an exponential upper bound. </p> <p>But as a method of classifying finitely generated groups, it has been very fruitful: Gromov's theorem on groups of polynomial growth; the incredibly rich theory that arose from Grigorchuk's original construction of an intermediate growth group; and the emergence of rich classes of exponential growth groups such as word hyperbolic groups.</p> http://mathoverflow.net/questions/120612/trichotomies-in-mathematics/120673#120673 Answer by Lee Mosher for Trichotomies in mathematics Lee Mosher 2013-02-03T15:12:25Z 2013-02-03T15:12:25Z <p>Every isometry of a proper $CAT(-1)$ space is either elliptic, parabolic, or hyperbolic: elliptic means fixes a finite point; hyperbolic means fixes two infinite points connected by a translation axis, equivalently translation distance bounded away from zero; parabolic means translation distance limiting to zero but no fixed point. There are versions for proper Gromov hyperbolic spaces, and even for the nonproper case, if you are willing to "quasify" the statements of the cases, and if you are willing to let the trichotomy degenerate to a dichotomy.</p> <p>Every isometry of Teichmuller space is either elliptic, parabolic, or hyperbolic. This is Bers' form of Thurston's trichotomy for mapping classes: finite order, reducible, pseudo-Anosov. This trichotomy also has an interpretation in terms of the action of the mapping class group on the curve complex which is a nonproper Gromov hyperbolic space by a theorem of Masur and Minsky. </p> <p>For elements of $Out(F_n)$, the outer automorphism group of a rank $n$ free group, there are related trichotomies and other -otomies coming from the work of Bestvina, Feighn, and Handel on relative train track theory. The simplest one is that every element of $Out(F_n)$ is either of finite order, or of polynomial growth, or of exponential growth. </p> http://mathoverflow.net/questions/119845/continuity-of-length-function-l-tx-times-mf-to-mathbb-r/119873#119873 Answer by Lee Mosher for continuity of length function $l: T(X) \times MF \to \mathbb R$ Lee Mosher 2013-01-25T18:25:33Z 2013-01-25T18:30:44Z <p>I suppose that to "see" these continuities, you are not asking about a proof, but about some intuition. </p> <p>For the first question, I like the intuition one gets from Bonahon's paper MR0931208, under which $MF$ and $T(X)$ are both embedded in a common space on which an obviously continuous intersection number is defined. This is the space of $\pi_1(X)$-invariant Borel measures defined on the "Mobius band beyond infinity" $M = ((\partial \pi_1(X) \times \partial \pi_1(x)) - \Delta) / (\mathbb{Z}/2)$, using the weak topology on the space of Borel measures on $M$. $T(X)$ embeds via the Liouville measure of a hyperbolic metric, and $MF$ embeds in an obvious fashion. Given two such measures $\mu,\nu$, using a local Fubini product description Bonahon defines a product measure on $M \times M$, mods out by $\pi_1(M)$, and the total mass of the quotient; the result is "obviously" continuous, by simple facts about weak topologies on measure spaces. </p> <p>Another way to say almost the same thing is to think of $T(X)$ as a continuously varying family of hyperbolic structures, and of $MF$ as a continuously varying family of measured geodesic laminations, then take the local Fubini product of the transverse measure on the lamination and the length measure along leaves from the hyperbolic structure, integrate, and you get the intersecton number; the whole picture varies continuous in both the $T(X)$ variable and the $MF$ variable, the Fubini product measure varies continuously in the weak topology on Borel measures, and its total measure varies continuously.</p> <p>For the second question, one can see continuity in a similar manner. Think of $Q(X)$ as a continuously varying family of singular Euclidean structures on $X$. For elements of $MF$, think of straightening them in each singular Euclidean structure to become "singular Euclidean measured geodesic laminations". Convince yourself that this whole picture varies continuously. Again, take Fubini product measures locally, integrate, and you get the intersection number. The one twist here is that when you straighten an element of $MF$ in a singular Euclidean structure, leaves need not stay disjoint, but that's ok because all you really need to do is to pull back the Euclidean length measure along leaves to some abstract model of the lamination.</p> http://mathoverflow.net/questions/131506/p1-minus-k-points/131521#131521 Comment by Lee Mosher Lee Mosher 2013-05-23T16:45:35Z 2013-05-23T16:45:35Z The derivation of the matrices should be covered in any textbook on hyperbolic geometry. In outline, if you fix the polygon $P$ in the upper half plane model, each side pairing $a_i \mapsto \bar a_i$ takes the endpoints of $a_i$ to the endpoints of $\bar a_i$. Once the image of a third point at infinity is determined, the matrix is determined. There is also a completeness condition for each cusp: the monodromy around that cusp must be parabolic. With that, you get a very explicit set of formulas parameterizing the matrices. http://mathoverflow.net/questions/131426/does-this-qualify-as-self-plagiarism-or-something Comment by Lee Mosher Lee Mosher 2013-05-22T12:58:00Z 2013-05-22T12:58:00Z Here is a general rule which I tell my students when they need to write background material which they learned from another source, and which perhaps applies to material from ones own earlier writings. Don't be lazy: learn the old references in your heart of hearts, and then rewrite it anew the way you need it for your current paper. http://mathoverflow.net/questions/131356/what-does-a-singular-simplex-with-real-coefficient-mean/131358#131358 Comment by Lee Mosher Lee Mosher 2013-05-21T18:12:43Z 2013-05-21T18:12:43Z Perhaps you meant &quot;references on singular homology&quot;. I would recommend Hatcher's book, which explains the distinctions between simplicial homology and singular homology. It is available at <a href="http://www.math.cornell.edu/~hatcher/AT/ATpage.html" rel="nofollow">math.cornell.edu/~hatcher/AT/ATpage.html</a> http://mathoverflow.net/questions/131343/antiderivative-of-an-absolute-function Comment by Lee Mosher Lee Mosher 2013-05-21T14:15:19Z 2013-05-21T14:15:19Z <a href="http://mathoverflow.net/howtoask#specific" rel="nofollow">mathoverflow.net/howtoask#specific</a> http://mathoverflow.net/questions/131200/what-does-vertex-solution-mean Comment by Lee Mosher Lee Mosher 2013-05-20T13:09:42Z 2013-05-20T13:09:42Z A citation of the paper or book in which you found this phrase, for example. http://mathoverflow.net/questions/131080/what-is-an-interpretation-of-the-relation-in-the-cohomology-of-the-pure-braid-gro Comment by Lee Mosher Lee Mosher 2013-05-18T19:03:33Z 2013-05-18T19:03:33Z At least you will have a lot of time to draw pictures of braids in the sand. http://mathoverflow.net/questions/131069/who-first-computed-the-integral-cohomology-ring-of-a-weighted-projective-space-w Comment by Lee Mosher Lee Mosher 2013-05-18T18:01:52Z 2013-05-18T18:01:52Z This is a duplicate of <a href="http://mathoverflow.net/questions/131067/who-first-computed-the-integral-cohomology-ring-of-a-weighted-projective-space-w" rel="nofollow" title="who first computed the integral cohomology ring of a weighted projective space w">mathoverflow.net/questions/131067/&hellip;</a> http://mathoverflow.net/questions/130972/use-mathematical-induction-to-prove-the-equality-thank-you Comment by Lee Mosher Lee Mosher 2013-05-17T15:16:36Z 2013-05-17T15:16:36Z MathOverflow is a site for questions of interest to research mathematicians; read the FAQ. You might try your question on another site such as math.stackexchange.com. http://mathoverflow.net/questions/130828/two-series-questions Comment by Lee Mosher Lee Mosher 2013-05-16T12:44:42Z 2013-05-16T12:44:42Z @Edwardo: you are on the wrong web site, math overflow is a site for research mathematics. http://mathoverflow.net/questions/130797/what-are-these-compact-sets-called Comment by Lee Mosher Lee Mosher 2013-05-16T11:58:44Z 2013-05-16T11:58:44Z How about just a &quot;piecewise smooth set&quot;? http://mathoverflow.net/questions/130431/mapping-class-group-of-once-punctured-torus/130494#130494 Comment by Lee Mosher Lee Mosher 2013-05-13T17:27:18Z 2013-05-13T17:27:18Z I guess what I mean is that what I would write in this situation is: &quot;The proof for $T$ is given in [FLP], and the proof for $S$ follows immediately from the definitions.&quot; http://mathoverflow.net/questions/130493/fundamental-class-and-simplicial-volume Comment by Lee Mosher Lee Mosher 2013-05-13T15:13:54Z 2013-05-13T15:13:54Z To say the &quot;the $l^1$ norm of the fundamental class is majorized by the number of top dimensional simplices to triangulate $V$&quot; means &quot;the $l^1$ norm is less than or equal to the number of top dimensional simplices needed to triangulate $V$&quot;. This is obvious, because the fundamental class is represented by a chain with coefficient $\pm 1$ on every simplex in the triangulation and coefficient $0$ on every other singular simplex. http://mathoverflow.net/questions/130296/show-that-is-continuous-xyxy-etc Comment by Lee Mosher Lee Mosher 2013-05-10T20:25:40Z 2013-05-10T20:25:40Z This is not a question, nor is it related to mathematical research. Math Overflow is a site for questions related to mathematical research. Please read the FAQ, formulate a question, and try your question at another site such as math.stackexchange.com. http://mathoverflow.net/questions/130196/jordan-curve-theorem-can-every-point-on-the-curve-be-reached-from-each-region Comment by Lee Mosher Lee Mosher 2013-05-09T22:48:35Z 2013-05-09T22:48:35Z Actually, what you state is not the full &quot;Jordan Schonflies theorem&quot;. The full statement of the Schonflies theorem is that there is a homeomorphism of the plane taking $C$ to $S^1$, from which your property follows easily. http://mathoverflow.net/questions/130167/is-simple-extension-qcos-9q-algebraic-and-normal Comment by Lee Mosher Lee Mosher 2013-05-09T13:42:19Z 2013-05-09T13:42:19Z Koam: MathOverflow is a site for research mathematical questions. Your question looks like homework, and is more suitable for other sites such as math.stackexchange.com; read the faq. If you do post elsewhere, you'll get better answers if you explain where the question came from (homework?), what you have tried, etc.