User danny calegari - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T16:24:46Z http://mathoverflow.net/feeds/user/1672 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/5277/simplicial-volume Simplicial volume Danny Calegari 2009-11-12T23:01:31Z 2012-05-22T10:06:39Z <p>Is there a finite dimensional closed manifold $M$ which is a $K(\pi,1)$, whose fundamental group is not word-hyperbolic, but which has a positive simplicial volume (ie "Gromov norm")?</p> <p>(Added:) The answers of Jim and Richard are both excellent; another example is any closed, irreducible locally symmetric manifold (of non-positive curvature). But these examples are all CAT(0); I wonder if there is an example which is not CAT(0)? (of course then it is hard to see the example is a $K(\pi,1)$ . . .)</p> http://mathoverflow.net/questions/84522/is-there-a-simplicial-volume-definition-of-chern-simons-invariants/84590#84590 Answer by Danny Calegari for Is there a simplicial volume definition of Chern Simons invariants? Danny Calegari 2011-12-30T14:31:59Z 2011-12-30T14:31:59Z <p>If you use eta invariant in place of Chern-Simons invariant, there is almost such a definition, at least in a closely related context. If we restrict to surface bundles over the circle with fiber of fixed genus, then the eta invariant of a fibered 3-manifold can be thought of as a certain kind of class function on the mapping class group. Such eta invariants exist for many different kinds of (unitary) representations (of subgroups of mapping class groups), and under suitable circumstances (see e.g. <a href="http://arxiv.org/abs/1003.4977" rel="nofollow">http://arxiv.org/abs/1003.4977</a>) the functions they define on (subgroups of) mapping class groups are examples of what are known as homogeneous quasimorphisms. </p> <p>Quasimorphisms arise in the theory of bounded cohomology; there is a duality theory relating them to a (relative) 2-dimensional Gromov norm, called stable commutator length. Gromov norm here is just a synonym for simplicial volume, as in your question.</p> http://mathoverflow.net/questions/76890/how-to-estimate-the-pressure How to estimate the pressure? Danny Calegari 2011-09-30T22:05:22Z 2011-10-01T01:34:47Z <p>I have a finite collection of diffeomorphisms $g_1,\cdots,g_n$ taking the unit interval $I$ to disjoint subintervals $I_1, I_2,\cdots,I_n$. If $G$ is the semigroup they generate, the limit set of $G$ (also called the attractor of the IFS) is a Cantor set, and under suitable hypotheses, Bowen (and under slightly weaker hypotheses, Urbanski) showed that the Hausdorff dimension of this Cantor set is the smallest zero of the pressure function $P$, defined by $$P(t) = \lim_{n \to \infty} \frac 1 n \log \sum_{w \in G_n} \|w'\|^t$$ where $G_n$ is the set of elements in $G$ of word length $n$, and $\|\cdot\|$ is the sup norm (the hypotheses for Bowen's theorem is that the $g_i$ are uniformly contracting; Urbanski proves the same theorem when the $g_i$ are allowed to have neutral fixed points; my examples have such points).</p> <p>This is all well and good, but how do I actually estimate the least zero of the pressure function for an explicit example? (yes, I mean numerically) My $g_i$ are all given by the restrictions of explicit polynomial functions of low degree, but the computational bottleneck seems to be the large number of elements in $G_n$. </p> <p>Or is there a better method to estimate the Hausdorff dimension in practice? Note that although I just want to estimate the dimension, I would like to be able to (computer-assisted if necessary) give rigorous bounds on the error. </p> http://mathoverflow.net/questions/5140/monotone-invariants-of-braid-forcing Monotone invariants of braid forcing Danny Calegari 2009-11-12T01:46:17Z 2011-09-30T21:47:10Z <p>Let $\phi$ be a diffeomorphism of the unit disk $D^2$, fixed on the boundary, and suppose that $Q$ is a finite subset of the interior permuted by $\phi$. The isotopy class of $\phi$ relative to $Q$ and relative to the boundary determines a conjugacy class in a braid group $B_n$ where $n$ is the cardinality of $Q$.</p> <p>One says that a conjugacy class of braid $b\in B_n$ forces another conjugacy class $b' \in B_m$ (where possibly $n$ is not equal to $m$) if every diffeomorphism $\phi$ representing $b$ permutes some finite subset $P$ in the interior of $D^2$ in such a way that the isotopy class of $\phi$ relative to $P$ represents $b'$.</p> <p>A function from conjugacy classes in (all) braid groups to the non-negative reals is monotone if it can only go down under braid forcing; i.e. if the value of the function on $b'$ is less than or equal to its value on $b$ as above.</p> <p>One way to define such a monotone invariant is to define some dynamical invariant of the conjugacy class of a diffeomorphism, and to take the infimum over all representatives. Since monotone braid classes give rise to inclusion of representative diffeomorphisms, such functions are necessarily monotone. One well-known (nontrivial) example is the (topological) entropy. Are there any other dynamically defined monotone invariants? What if one "stiffens" the structure, eg. by restricting to area-preserving diffeomorphisms?</p> http://mathoverflow.net/questions/51343/generalization-of-rogers-dilogarithm generalization of (Rogers) dilogarithm Danny Calegari 2011-01-06T21:25:59Z 2011-05-13T12:22:19Z <p>Let $C$ and $S$ be abbreviations for $\cosh$ and $\sinh$, and consider the following function:</p> <p>$$f(x,y) = \int_{-y\le r+l \le y} \frac{ (C(x)S(l)C(r) - C(l)S(r))(C(x)C(l)S(r)-S(l)C(r)) } {(C(x)C(l)C(r) - S(l)S(r))^2-1} dl dr$$</p> <p>If $y=\infty$, this specializes (I think!) to $4\mathcal{L}(1/C^2(x/2))$ where $\mathcal{L}$ is the Rogers dilogarithm (maybe some constants and factors are missing). The question is whether the function $f$ is studied anywhere. References would be appreciated.</p> <p>Note: This function arises as the volume of a certain region in the unit tangent bundle of a hyperbolic surface; therefore I am not looking for an answer which just translates it back into its geometric origin.</p> http://mathoverflow.net/questions/8023/when-is-a-riemannian-metric-equivalent-to-the-flat-metric-on-mathbb-rn/8033#8033 Answer by Danny Calegari for When is a Riemannian metric equivalent to the flat metric on $\mathbb R^n$? Danny Calegari 2009-12-06T21:14:18Z 2009-12-06T21:14:18Z <p>Greg's comment on Deane's answer is sort of correct (given suitable hypotheses), but maybe a bit misleading in the context of this discussion. Since the character count doesn't allow it, I'm adding this comment as an "answer" (though it is not an answer to the original question).</p> <p>There are non-isometric 2-spheres $S_1,S_2$ for which there is a diffeomorphism $f$ from $S_1$ to $S_2$ so that the curvature at each point $p \in S_1$ is equal to the curvature of $f(p)$ in $S_2$. For example, let $O$ be a curve in the plane with dihedral $D_2$ symmetry whose curvature has 4 critical points (is this called an "oval"? I forget). If $S$ is a surface of revolution of $O$, then $S$ is foliated (in the complement of two "poles") by "latitude" circles of constant curvature, and the value of the curvature moves monotonically between two extreme values as one moves from the "poles" to the "equator". One can easily produce nonisometric surfaces with "the same" curvature function. The length of the circle with a given curvature value is an invariant of the isometry type which is not captured by the curvature itself (thought of as smooth function on $S$). </p> <p>I think this example (and more discussion) is in Berger's book "A panoramic view of Riemannian geometry".</p> http://mathoverflow.net/questions/6950/why-is-the-standard-definition-of-cocycle-the-one-that-always-comes-up/7619#7619 Answer by Danny Calegari for Why is the standard definition of cocycle the one that _always_ comes up?? Danny Calegari 2009-12-02T23:23:22Z 2009-12-02T23:23:22Z <p>Hey Kevin - </p> <p>You write: "The standard decision was the rather clunky $$c(g_1,g_2, \cdots g_i)=f(1,g_1,g_1g_2,g_1g_2g_3,\cdots, g_1g_2g_3\cdots g_i)$$</p> <p>A topologist thinks of group cocycles as operating on "simplices labeled by elements of a group". Since orderings matter, we have "the" standard $n$-simplex, which comes with a labeling of its vertices by the numbers $0,1,2,\cdots,n$. The way the group comes in is that each <em>edge</em> of the simplex from vertex $i-1$ to vertex $i$ is labeled by an element of the group, i.e. $g_i$ is the label going from vertex $i-1$ to vertex $i$. This is because the simplex is "really" part of a $K(G,1)$ with one vertex, and one edge for each element of the group, one triangle for each element of the multiplication table of the group (i.e. for each expression $g \times h = gh$) and so on. These $g_i$ are the terms on the LHS of your equation.</p> <p>The universal cover of this $K(G,1)$ is a contractible simplicial complex whose <em>vertices</em> are now labeled by elements of the group (because the group "is" the deck group of the cover, and acts simply transitively on vertices). Your original simplex has a unique lift to the cover taking the vertex $0$ to some specific vertex, which might as well be the one labeled by the identity. The vertex $i$ then lifts to the <em>composition</em> $g_1g_2\cdots g_i$. So the terms on the RHS of your equation are the labels on the <em>vertices</em>.</p> <p>I am really only repeating what Mariano wrote in very slightly more topological language; you didn't seem happy with his answer, so maybe you won't be happy with mine either.</p> <p>Best,</p> <p>DC</p> http://mathoverflow.net/questions/4971/stable-w-length/7479#7479 Answer by Danny Calegari for Stable w-length Danny Calegari 2009-12-01T18:50:51Z 2009-12-01T18:50:51Z <p>In case any one is still thinking about this question, it turns out one can say a lot. For example, $sl(w|w)=1/2$ whenever $w$ is a word of the form $[a,b^n]$ (and several other examples), one has $2/3 \le sl(w|w) \le 4/5$ when $w=[a,b]^2$, and if $\gamma_n$ is the iterated commutator $[x_1,[x_2,\cdots[x_{n-1},x_n]\cdots]$ one has $sl(\gamma_n|\gamma_n)\le 1-2^{1-n}$. In fact, I would explicitly conjecture that $sl(w|w)&lt;1$ (i.e. strict inequality) for every $w$.</p> <p>Getting systematic lower bounds on $sl(w|w)$, other than $scl(w)/(scl(w)+1/2)$ seems difficult; one imagines that the (currently nonexistent) theory of nonabelian Bavard duality might do the trick.</p> http://mathoverflow.net/questions/4971/stable-w-length Stable w-length Danny Calegari 2009-11-11T02:58:54Z 2009-12-01T18:50:51Z <p>Let $F$ be a free group, and $w$ an element of $F$. In any group $G$, a $w$-word is the image of $w$ or $w^{-1}$ under a homomorphism from $F$ to $G$. The subgroup of $G$ generated by $w$-words is denoted $G(w)$.</p> <p>For any $g \in G(w)$, the $w$-length of $g$, denoted $l(g|w)$, is the minimum number of $w$-words in $G$ whose product is $g$, and the stable $w$-length of $g$, denoted $sl(g|w)$, is the limit $sl(g|w) = lim_{n \to \infty} l(g^n|w)/n$.</p> <p>If $w$ is not in the commutator subgroup of $F$, the stable $w$-length of every element in any group is trivial. Otherwise, one has a universal inequality $$1/2 \le sl_F(w|w) \le 1$$ (where the subscript $F$ indicates that stable $w$-length is being calculated in the free group $F$ containing $w$ itself.)</p> <p>The lower bound of $1/2$ is realized e.g. by the word $w=xyx^{-1}y^{-1}$ (i.e. a standard commutator) in $F_2$ but I don't know how to compute (or even approximate!) $sl(w|w)$ in (essentially) any other case.</p> <p>What values are achieved by $sl(w|w)$? Are they all rational? Are they dense? Is $1$ ever achieved? Is $1/2$ ever achieved for a word other than $xyx^{-1}y^{-1}$? </p> <p>(Added:) After reading FC's answer, it is probably worth pointing out that the lower bound $1/2 \le sl_F(w|w)$ comes from the inequality $scl_G(g) \le sl_G(g|w)(scl_F(w)+1/2)$ for any $g$ in any $G$, so one gets a better lower bound on $sl_F(w|w)$ if one knows $scl_F(w)>1/2$ (the estimate $scl_F(w)\ge 1/2$ is always true). Upper bounds can be established by exhibiting identities (like FC's identity below). Does a blind computer search yield any interesting examples?</p> http://mathoverflow.net/questions/5161/groupoid-of-moves-on-trivalent-fatgraph Groupoid of moves on trivalent fatgraph Danny Calegari 2009-11-12T05:56:00Z 2009-12-01T18:10:32Z <p>Let $T$ be a finite trivalent fatgraph - i.e. a graph with a cyclic order of the edges at each vertex. Then there are certain basic "moves" we can perform on $T$: an embedded edge can be collapsed and then uncollapsed in a different way (a "rotation", or "2-2 move"), or the circular order of the three edges incident at a vertex can be reversed (a "flip"). </p> <p>Define a set $\mathcal{T}$ whose elements are trivalent fatgraphs $T'$ homotopic to $T$ with a labeling of the edges from $1$ to $n$ and a labeling of the vertices from $1$ to $m$ (note $m = 2n/3$). A "move" is a pair $(T,c)$ where $T \in \mathcal{T}$, and $c$ is an element of the set $e1, e2, \cdots, en, v1, v2, \cdots, vm$. The move acts on the labeled fatgraph $T$, and turns it into a new labeled fatgraph $T'$ obtained from $T$ by performing a rotation on edge $ei$ if $c=ei$ or a flip on vertex $vi$ if $c=vi$. It is clear how a flip affects the labels (it doesn't). A rotation destroys one edge labeled by $ei$ and creates a new edge, so label this new edge $ei$.</p> <p>Now define a <em>marked</em> fatgraph to be a labeled fatgraph (i.e. an element of $\mathcal{T}$) together with a homotopy class of homotopy equivalence to some fixed <img src="http://latex.mathoverflow.net/png?%24K%28%5Cpi%5F1%28T%29%2C1%29%24" alt="$K(\pi\sb 1(T),1)$" title="" />. The moves defined above generate a new groupoid on marked fatgraphs, by acting on the labeled fatgraph part. This groupoid - the groupoid acting on marked fatgraphs - I will denote by $V(T)$ (the notation $V(T)$ is suggested by the similarity to Thompson's group $V$).</p> <p>This groupoid - or something like it - turns up in many different contexts, so as a preliminary question, it would be nice to know how it is referred to. (Or: does this construction even make sense?)</p> <p>More substantially: what is known about the algebraic structure of $V(T)$? What can be said about the cohomology of its classifying space? What is the relation to the group $Out(F)$, where $F$ is the (free) fundamental group of $T$? (note that $Out(F)$ acts on marked fatgraphs in a way that commutes with $V(T)$, by acting by homotopy equivalences of the <img src="http://latex.mathoverflow.net/png?%24K%28%5Cpi%5F1%28T%29%2C1%29%24" alt="$K(\pi\sb 1(T),1)$" title="" /> and thereby changing the marking). Is there a good reference?</p> <p><hr /></p> <p>Since the answers I am getting are not really what I am after, I think I need to make the question more pointed. A marked fatgraph determines a certain amount of algebraic structure on a free group (i.e. the fundamental group of $T$), namely a pair $(l,e)$ where $l$ is a length function, and $e$ is a bounded 2-cocycle. The first part of the data comes from the "thin" underlying graph, and is just the translation length of each element on its axis. The second part of the data comes from the fattening, and is an explicit cocycle representing the Euler class of the thickened surface. The first kind of move affects $l$, the second kind affects $e$. Crucially, both $l$ and $e$ are <em>integer valued</em> (this is the point of discussing discrete combinatorial objects, namely fatgraphs, instead of eg. discrete faithful representations of $F$ into $PSL(2,R))$. </p> <p>Many, many papers discuss length functions, and many, many papers discuss Euler classes, but I would like to have a (presumably homological) <em>algebraic</em> framework which treats the two components as a single object with, presumably, more structure. The question is: what is this structure? Is it something that is already well-studied? Is there a reference?</p> http://mathoverflow.net/questions/7454/what-are-cr-manifolds-like/7455#7455 Answer by Danny Calegari for What are CR manifolds like? Danny Calegari 2009-12-01T16:29:51Z 2009-12-01T16:42:23Z <p>CR does stand for Cauchy-Riemann.</p> <p>CR structures on 3 dimensional manifolds arise as the boundaries of complex (or almost-complex) 4 manifolds; if these boundaries are strictly <em>pseudo-convex</em> (i.e. convex in "holomorphic directions") the CR structure on the 3-manifold is a contact structure (if the boundary is only (pseudo-)convex or (Levi) flat, the CR structure integrates to a confoliation or a foliation respectively). There can be infinite dimensional families of foliations on a 3-manifold; more generally, whenever the CR structure is "non-generic" or integrable, one has continuous moduli, otherwise (eg in the contact structure case) one has discrete moduli (to be explicit: what has discrete moduli is the contact structure, not the "CR+contact structure".)</p> http://mathoverflow.net/questions/6345/hermite-normal-form-in-families Hermite normal form in families Danny Calegari 2009-11-21T03:06:55Z 2009-11-29T07:10:09Z <p>How does Hermite normal form (over $Z$) vary in families? I.e. if I have an $n\times m$ matrix $M$ whose entries are integral polynomials in some integral variable $x$, how does the Hermite normal form of the integral matrix $M(p)$ (obtained by setting $x$ equal to $p$) vary as a function of $p$? What about the special case that the entries are (at most) linear in $x$?</p> <p>The question is a bit open ended so answers could be of several kinds, eg: (i) how certain integer programming problems associated to $M(p)$ depend on $p$; (ii) by explaining how the answer can be expressed in a way that generalizes Ehrhart theory; (iii) by specializing to an important case that is well-understood; (iv) in some other interesting way.</p> <p>I would also really appreciate a pointer to any relevant literature.</p> http://mathoverflow.net/questions/6142/circle-bundles-over-rp2/6145#6145 Answer by Danny Calegari for Circle bundles over $RP^2$ Danny Calegari 2009-11-19T18:13:53Z 2009-11-19T18:13:53Z <p>Such manifolds are examples of Seifert fibered spaces, which have, indeed, been classified. A good reference is Montesinos "Classical Tessellations and Three-Manifolds". Basically, such manifolds (over any nonorientable surface base) are classified by their Euler class, which measures the obstruction to the existence of a section. </p> http://mathoverflow.net/questions/4966/fibered-nots-non-geometric-hnn-extensions-of-free-groups-normally-generated-by-t Fibered nots (non-geometric HNN extensions of free groups normally generated by the monodromy) Danny Calegari 2009-11-11T02:48:43Z 2009-11-15T00:26:11Z <p>A fibered knot is a knot $K$ in the $3$-sphere whose complement is a surface bundle over a circle. If $S$ is the fiber, the fundamental group of $S$ is free (of even rank), and the fundamental group of the complement is an HNN extension <img src="http://latex.mathoverflow.net/png?%24%24%5Cpi%5F1%28S%29%20%5Cto%20%5Cpi%5F1%28S%5E3%2DK%29%20%5Cto%20Z%24%24" alt="$$\pi\sb 1(S) \to \pi\sb 1(S^3-K) \to Z$$" title="" /> where the $Z$ is generated by the meridian of the knot.</p> <p>Since surgery on $K$ recovers the $3$-sphere, the group $\pi_1(S^3-K)$ has the interesting property that it is normally generated by (the conjugacy class of) the meridian.</p> <p>What I didn't realize until recently is that there are many examples of "non-geometric" automorphisms $\phi$ of free groups $F$ for which the associated HNN extension $F \to G \to Z$ is normally generated by the conjugacy class of the monodromy. One simple example is the case $F = \langle a,b,c \rangle$ and $\phi$ is the automorphism $a \to c^{-1}abac, b \to bac, c\to bc$. </p> <p>Is there any systematic way of generating such examples? Is there a classification? One reason to be interested is that such examples can be used to construct smooth $4$-manifolds which are topologically $S^4$ but not obviously diffeomorphically $S^4$.</p> <p>Edit: a link to the construction is <a href="http://lamington.wordpress.com/2009/11/09/4-spheres-from-fibered-knots/" rel="nofollow">http://lamington.wordpress.com/2009/11/09/4-spheres-from-fibered-knots/</a> (this explains the construction in the case of a fibered knot, but the group-theoretic condition is the only important ingredient).</p> http://mathoverflow.net/questions/5351/whats-an-example-of-a-space-that-needs-the-hahn-banach-theorem/5370#5370 Answer by Danny Calegari for What's an example of a space that needs the Hahn-Banach Theorem? Danny Calegari 2009-11-13T15:25:56Z 2009-11-13T16:16:34Z <p>If $G$ is a group, Bavard showed that the stable commutator length vanishes on $[G,G]$ if and only if $G$ admits no nontrivial "homogeneous quasimorphisms". These functions (on the space of group $1$-boundaries) are constructed using the Hahn-Banach theorem, but are usually very hard (or impossible) to write down explicitly.</p> <p>Another example: let $M$ be a triangulated manifold, and suppose we orient every edge of every simplex in such a way that the orientations come from a "total order" on each triangle. We would like to assign positive "lengths" to each edge in such a way that on each triangle, the sum of the values on the "short edges" is equal to the value on the "long edge" (where "short" and "long" are defined according to the orientations). The (finite-dimensional!) Hahn-Banach theorem tells us we can do this if and only if every oriented loop in the 1-skeleton is homologically essential; i.e. "homological positivity" can be improved to "chain positivity". Of course the finite-dimensional Hahn-Banach theorem is just a psychological crutch, but versions of this construction in other categories need the "real" Hahn-Banach theorem (applied to certain spaces of de Rham currents).</p> http://mathoverflow.net/questions/5262/lie-groups-and-manifolds/5272#5272 Answer by Danny Calegari for Lie Groups and Manifolds Danny Calegari 2009-11-12T22:45:12Z 2009-11-12T23:19:05Z <p>$SO(3)$ is homeomorphic to $RP^3$, not to $S^2$. The relationship between $SO(3)$ and $S^2$ is that $SO(3)$ is the group of (orientation-preserving) isometries of $S^2$ in its round metric. If $M$ is any Riemannian manifold, the group of isometries of $M$ is a Lie group (this is an old theorem of Kobayashi (edit: I mean Myers-Steenrod; see comments). </p> http://mathoverflow.net/questions/5137/monopole-classes-on-hyperbolic-3-manifolds Monopole classes on hyperbolic 3-manifolds Danny Calegari 2009-11-12T01:37:51Z 2009-11-12T20:46:21Z <p>Let $M$ be a closed hyperbolic $3$-manifold, and $e \in H^2(M)$ an integral cohomology class which is the first Chern class of a $Spin^c$ structure on $M$. Suppose there is a solution to the monopole equations on $M$ in this $Spin^c$ class with respect to the hyperbolic metric on $M$. Does it follow that $e$ is a monopole class? More generally, is it true that as a metric on a $3$-manifold evolves by (nonsingular) Ricci flow, pairs of solutions to the monopole equations can disappear, but new ones never appear?</p> <p>(Added:) One reason to be interested is that if the answer is "yes", then one can deduce that a given class is a monopole class directly from geometry. One can then hope to use this information to show that certain families of classes (eg. on families of manifolds obtained by hyperbolic Dehn surgery on some fixed cusped manifold) are all monopole classes. More abstractly, if one has a "natural" PDE on a manifold which depends on the metric, then one should try to understand how/whether solutions to that PDE evolve under "natural" flows on the space of metrics. A similar (more geometric) question might be: are minimal surfaces destroyed by Ricci flow but never created? Eg. if a hyperbolic $3$-manifold contains an embedded minimal surface, is there an isotopic minimal surface in every other metric on the manifold? The reason to speculate about a connection between the monopole equations and Ricci flow is the key role of scalar curvature in both cases: via the Weitzenbock formula for the Dirac operator on the one hand; and via Hamilton's monotonicity formula for the infimum of scalar curvature under (rescaled) Ricci flow on the other.</p> http://mathoverflow.net/questions/3973/what-should-be-offered-in-undergraduate-mathematics-thats-currently-not-or-isn/5233#5233 Answer by Danny Calegari for What should be offered in undergraduate mathematics that's currently not (or isn't usually)? Danny Calegari 2009-11-12T19:14:24Z 2009-11-12T19:14:24Z <p>Following Greg's lead, I wish that undergrads who want to become math majors didn't "skip out" of differential equations classes (in their eagerness to get to the good stuff like Drinfeld chtoukas, or whatever). I can't think of a more important foundational subject that tends to be systematically avoided by the "best" undergraduate mathematics students.</p> http://mathoverflow.net/questions/453/what-is-an-example-of-a-function-on-m-g/5225#5225 Answer by Danny Calegari for What is an example of a function on M_g? Danny Calegari 2009-11-12T18:54:35Z 2009-11-12T18:54:35Z <p>The period mapping lets you embed $M_g$ in the quotient of the Siegel upper half-space by the integral symplectic group; so any function on this locally symmetric space restricts to a function on $M_g$.</p> http://mathoverflow.net/questions/5117/independence-of-the-continuum-hypothesis-on-zfc/5125#5125 Answer by Danny Calegari for Independence of the continuum hypothesis on ZFC Danny Calegari 2009-11-11T23:37:17Z 2009-11-11T23:37:17Z <p>Raymond Smullyan and Melvin Fitting wrote a long (but very readable) monograph, called "Set theory and the continuum problem" (Oxford Logic Guides, 34. Oxford Science Publications. The Clarendon Press, Oxford University Press, New York, 1996. xiv+288 pp. ISBN: 0-19-852395-5) which starts from the very beginning, introducing the von Neumann-Bernays-Godel "class-set" formalism, establishing all the basic properties, etc. culminating in a complete, self-contained exposition of Cohen's result and the basics of forcing.</p> <p>There are three parts. The first part is foundational. The second is an exposition of Godel's relative consistency result, which says that the continuum hypothesis is consistent with NBG (or ZFC if you prefer). The third is about forcing, and Cohen's result.</p> <p>In very, very broad outline, the point of Godel's argument is to exhibit a "model" of class-set theory with as few sets as possible: the only sets are those that are absolutely required to be there by some application of the axioms. Each set therefore carries with it a formula which requires it to exist. There are so few objects in this model that the continuum hypothesis is seen to be true because every element of the continuum that is required to exist is required by some explicit "reason", and the reasons can be enumerated.</p> <p>The point of forcing is to show that one can build a new model in which there are many more objects, by adding new objects when there is no explicit reason why they can't exist (i.e. they are "forced" to exist) and keeping careful track of how many such objects you can add without reaching a contradiction. Certain tools (eg "compactness") are required to be able to add infinitely many new objects in this way.</p> http://mathoverflow.net/questions/5068/dolbeault-cohomology/5072#5072 Answer by Danny Calegari for Dolbeault cohomology Danny Calegari 2009-11-11T17:11:59Z 2009-11-11T17:11:59Z <p>Chern's book "Complex manifolds without potential theory" is a good book, and it does explain Dolbeault Cohomology. But it's a short book, and it explains it concisely. If you need more details, you could also try Griffiths-Harris (but I greatly prefer Chern's book).</p> <p>Kodaira's book "Complex manifolds and deformations of complex structures" is much more leisurely, and with great attention paid to exposition and detail (it doesn't appeal to some people, but I enjoyed it a lot).</p> http://mathoverflow.net/questions/5057/rational-group-cohomology/5064#5064 Answer by Danny Calegari for Rational Group Cohomology Danny Calegari 2009-11-11T15:58:00Z 2009-11-11T15:58:00Z <p>Stallings showed that if $f:\Gamma \to \Delta$ is a homomorphism between finitely presented groups where <img src="http://latex.mathoverflow.net/png?%24f%5Fi%3AH%5Fi%28%5CGamma%2CQ%29%20%5Cto%20H%5Fi%28%5CDelta%2CQ%29%24" alt="$f\sb i:H\sb i(\Gamma,Q) \to H\sb i(\Delta,Q)$" title="" /> is bijective for $i\le 1$ and surjective for $i=2$ then $f\otimes Q: \Gamma \otimes Q \to \Delta \otimes Q$ is an isomorphism between the $Q$-unipotent completions. So "taking the $Q$-unipotent completion" is a functor with some interesting properties with respect to rational group (co)homology. I'm not sure if this is the kind of thing you're after . . .</p> <p>Stallings' paper is </p> <p>MR0175956 (31 #232) Stallings, John Homology and central series of groups. J. Algebra 2 1965 170--181. </p> http://mathoverflow.net/questions/4934/symmetrical-presentation-of-4-dimensional-rotation-matrix/4936#4936 Answer by Danny Calegari for Symmetrical Presentation of 4-Dimensional Rotation Matrix Danny Calegari 2009-11-10T22:15:47Z 2009-11-10T22:15:47Z <p>A 4-d rotation does not have to fix a 2-d axis. For example, (complex) multiplication by $e^{i\theta}$ rotates every vector of unit length in $C^2$ the same way''.</p> http://mathoverflow.net/questions/4918/can-you-fool-snappea/4921#4921 Answer by Danny Calegari for can you fool SnapPea? Danny Calegari 2009-11-10T20:51:56Z 2009-11-10T20:51:56Z <p>If you are asking whether SnapPea rigorously certifies the existence of the hyperbolic structures that it "finds", then I think the answer is that it does not. Andrew Casson wrote a couple of programs to hyperbolize closed and cusped 3-manifolds, called "geo" and "cusp" respectively, and rigorously proved that they do not give "false positives" (this is, of course, relative to the correct functioning of the compiler, the computer hardware etc.)</p> <p>If you are asking whether SnapPea in practice thinks that it has found a hyperbolic structure where none exists (or the structure it finds corresponds to a non-faithful or indiscrete representation), I think it can sometimes be fooled by analytically continuing a family of Dehn fillings, and where what is really being hyperbolized is the image of the manifold under a degree 1 map. Note that this information is from my experience playing with SnapPea ~15 years ago, so it may easily be out of date.</p> http://mathoverflow.net/questions/3965/minimal-surface-in-a-ball/4911#4911 Answer by Danny Calegari for Minimal surface in a ball Danny Calegari 2009-11-10T19:38:50Z 2009-11-10T19:38:50Z <p>One obvious observation (of which you are probably already aware) is that if the boundary of the surface is connected, it must have length at least $2\pi\sqrt{1-r^2}$, or else it is contained in a lune whose convex hull does not contain a point at distance $r$ from the center. In the very special case that your surface is a topological disk transverse to a foliation of the ball by concentric spheres, the coarea formula (obtained by integrating the lengths of the intersection of your surface with concentric spheres, and using this observation) gives an estimate for the area, but a quick calculation shows that it is not good enough to prove what you want.</p> http://mathoverflow.net/questions/4769/algorithm-generalizing-continued-fractions-for-non-quadratic-algebraic-numbers/4910#4910 Answer by Danny Calegari for Algorithm generalizing continued fractions for non-quadratic algebraic numbers Danny Calegari 2009-11-10T19:32:05Z 2009-11-10T19:32:05Z <p>One generalization is to the theory of sails. If $A$ is an $n\times n$ integer matrix whose eigenvalues are all real, positive, irrational and distinct, a collection of $n$ suitable eigenvectors spans a polyhedral cone which is invariant under $A$. The convex hull of the set of integer lattice points in this cone is a polyhedron, and the vertices of this polyhedron are the best'' integral approximations to the eigenvectors. Also see Arnold, e.g. </p> <p>MR1704965 (2000h:11012) Arnold, V. I.(RS-AOS) Higher-dimensional continued fractions. (English, Russian summary) J. Moser at 70 (Russian). Regul. Chaotic Dyn. 3 (1998), no. 3, 10--17. </p> http://mathoverflow.net/questions/3405/is-a-subgroup-of-a-free-abelian-group-free-abelian/4908#4908 Answer by Danny Calegari for Is a subgroup of a free abelian group free abelian? Danny Calegari 2009-11-10T19:24:32Z 2009-11-10T19:24:32Z <p>A variety of groups $V$ is said to have the Schreier property if every subgroup of a free group in the variety is free. It is a classical theorem of Peter Neumann and James Wiegold that the only varieties of groups with the Schreier property are: the (absolutely) free groups, the free abelian groups, and the free exponent $p$ abelian groups for $p$ prime. </p> http://mathoverflow.net/questions/1010/what-is-the-right-way-to-think-about-represent-general-tilings/4889#4889 Answer by Danny Calegari for What is the right way to think about / represent general tilings? Danny Calegari 2009-11-10T17:01:17Z 2009-11-10T17:01:17Z <p>Aperiodic tilings can be thought of (in a sometimes useful way) as leaves of laminations; the groupoid in question (as in Emily's answer) is then the holonomy groupoid of the lamination.</p> <p>There is a standard description of the Penrose tiles in this way; think of an irrational plane (i.e. an $R^2$) in $R^n$ for some $n>2$, and consider the set of 2-dimensional faces of the $Z^n$ lattice in $R^n$ that intersect a (uniform) thickened tubular neighborhood of your plane. Project each such 2-dimensional face perpendicularly down to your plane; the result is an aperiodic tiling. If the irrational plane happens to be chosen with extra symmetries (eg it could be an eigenspace of a finite order element in $GL(n,Z)$) one gets quite a tile set with extra "partial symmetries". The Penrose tiling is of this kind: think of $Z/5Z$ permuting the coordinate axes in $R^5$. This fixes the vector $(1,1,1,1,1)$ and has two perpendicular irrational eigenspaces on which it acts as an order 5 rotation; translates of these eigenspaces give rise to the "standard" Penrose tilings.</p> <p>The lamination in this case is the "irrational foliation" of the torus $R^5/Z^5$ by planes with slope equal to the slope of the $R^2$ (and one can easily imagine generalizations).</p> http://mathoverflow.net/questions/4011/whats-a-non-abelian-totally-ordered-group/4886#4886 Answer by Danny Calegari for What's a non-abelian totally ordered group? Danny Calegari 2009-11-10T16:48:05Z 2009-11-10T16:48:05Z <p>Thurston's construction of a left-order on braid groups uses hyperbolic geometry, but it doesn't have to. For the sake of argument, let S be a closed, oriented surface of non-positive Euler characteristic. The universal cover of S is homeomorphic to the plane. An essential embedded loop in S lifts to a properly embedded line in the plane. The mapping class group of S permutes the set of isotopy classes of essential embedded loops in S, and a certain extension of this group acts on the universal cover, permuting the set of embedded lines covering properly embedded curves on the surface. Any collection of proper embedded rays in the plane whose pairwise intersections are compact inherits a natural circular order from the topology of the plane; it is this circular order that gives rise to circular orders on (certain extensions of) mapping class groups, and left orders on braid groups.</p> http://mathoverflow.net/questions/35514/pair-of-curves-joining-opposite-corners-of-a-square-must-intersect-proof/54283#54283 Comment by Danny Calegari Danny Calegari 2011-10-01T12:15:51Z 2011-10-01T12:15:51Z +1 for retaining something from one of our conversations . . . http://mathoverflow.net/questions/5140/monotone-invariants-of-braid-forcing Comment by Danny Calegari Danny Calegari 2011-10-01T06:33:57Z 2011-10-01T06:33:57Z @Agol: essentially all the invariants discussed in Ghys' paper (eg helicity, Ruelle invariant, Calabi quasimorphism etc.) as well as many variations (Polterovich, Py, etc.) are obtained by taking some local topological invariant of the dynamics on finitely many points, and integrating it over the degrees of freedom of the choice of the finitely many points wrt an invariant measure. If you take any braid type and &quot;shrink&quot; the dynamics down to be concentrated in a very small disk, the value of these invariants typically goes to zero. Did you have any specific part of the paper in mind? http://mathoverflow.net/questions/76890/how-to-estimate-the-pressure/76904#76904 Comment by Danny Calegari Danny Calegari 2011-10-01T06:20:20Z 2011-10-01T06:20:20Z Just out of curiosity, what did you have in mind when you say &quot;the upper bound may be accessible by other means&quot;? http://mathoverflow.net/questions/76890/how-to-estimate-the-pressure/76904#76904 Comment by Danny Calegari Danny Calegari 2011-10-01T06:16:26Z 2011-10-01T06:16:26Z Thanks for the references and the suggestions. I am in fact interested in the non-uniformly contracting (OK, expanding) case, so I'll take a look at the Climenhaga-Thompson paper. http://mathoverflow.net/questions/51343/generalization-of-rogers-dilogarithm/51355#51355 Comment by Danny Calegari Danny Calegari 2011-01-07T04:02:50Z 2011-01-07T04:02:50Z I know that paper, and I don't know precisely which integrals you mean. Can you be more specific? http://mathoverflow.net/questions/5161/groupoid-of-moves-on-trivalent-fatgraph/7457#7457 Comment by Danny Calegari Danny Calegari 2009-12-01T18:35:44Z 2009-12-01T18:35:44Z Hmm, I'll think about it. Thanks. http://mathoverflow.net/questions/5161/groupoid-of-moves-on-trivalent-fatgraph/7457#7457 Comment by Danny Calegari Danny Calegari 2009-12-01T17:32:32Z 2009-12-01T17:32:32Z Greg, I know you know that I know about Outer space, the Hatcher-Thurston complex, etc. I think it is my fault for not making it clear what I am really interested in - apologies. I have had a stab at rewriting the question. Maybe there is still no good answer; c'est la vie. http://mathoverflow.net/questions/7114/why-are-planar-graphs-so-exceptional/7116#7116 Comment by Danny Calegari Danny Calegari 2009-12-01T16:38:05Z 2009-12-01T16:38:05Z The (Koebe-Andreev)-Thurston theorem is true (and was actually proved first in this generality) for graphs on arbitrary surfaces (the question explicitly asks about planar graphs <i>as compared to</i> graphs embedded in other surfaces, so I thought I'd mention it). This was one of the steps in Thurston's original proof of geometrization of Haken manifolds (a later simplification, due to Otal I think, showed that it sufficed to prove the theorem for planar graphs). http://mathoverflow.net/questions/6345/hermite-normal-form-in-families/6938#6938 Comment by Danny Calegari Danny Calegari 2009-12-01T16:31:20Z 2009-12-01T16:31:20Z OK, I will take away your &quot;tick&quot; and then put it back again. Thanks. http://mathoverflow.net/questions/6498/what-is-the-right-definition-of-the-free-abelian-group-on-a-set/6510#6510 Comment by Danny Calegari Danny Calegari 2009-11-23T17:37:15Z 2009-11-23T17:37:15Z I have absolutely no argument with this answer (which is excellent and to the point). But I think that saying &quot;anything you want to know about this object will follow most transparently from the universal mapping property&quot;, while perhaps true of free abelian groups, does not generalize well. For example, free (nonabelian) groups are universal objects in the category of nonabelian groups, and from this universality many things follow, but I would hardly say that &quot;anything&quot; you want to know about free groups is most easily seen in this categorical way. http://mathoverflow.net/questions/6345/hermite-normal-form-in-families Comment by Danny Calegari Danny Calegari 2009-11-21T07:22:13Z 2009-11-21T07:22:13Z Unfortunately, I don't think it is that simple. For example, a family of 2x2 upper triangular matrices with 2s on the diagonal and x in the upper right has &quot;periodic&quot; Hermite normal form over Z (the upper right entry alternates between 0 and 1). This is why I speculated about Ehrhart theory (eg. maybe the entries are eventually quasipolynomials) but this is just a guess too. I do think your guess is probably true for some &quot;generic&quot; families, in a sense that needs a bit more clarification . . . http://mathoverflow.net/questions/4971/stable-w-length/5312#5312 Comment by Danny Calegari Danny Calegari 2009-11-19T18:54:25Z 2009-11-19T18:54:25Z I don't consider this question &quot;answered&quot;, but FC's answer is evidently the best answer that is likely to come along. http://mathoverflow.net/questions/6142/circle-bundles-over-rp2/6146#6146 Comment by Danny Calegari Danny Calegari 2009-11-19T18:50:37Z 2009-11-19T18:50:37Z Hi F.G: the bundles with nontrivial Euler class all have S^3 geometry (i.e. are quotients of S^3). http://mathoverflow.net/questions/4966/fibered-nots-non-geometric-hnn-extensions-of-free-groups-normally-generated-by-t Comment by Danny Calegari Danny Calegari 2009-11-15T00:29:30Z 2009-11-15T00:29:30Z Oops - I guess the edit was not due to Greg (sorry!) Anyway - please leave it as it is. http://mathoverflow.net/questions/4966/fibered-nots-non-geometric-hnn-extensions-of-free-groups-normally-generated-by-t Comment by Danny Calegari Danny Calegari 2009-11-15T00:27:43Z 2009-11-15T00:27:43Z What Sam said - homology implies that the fundamental group of S has even rank. But it is a good question, and it would be nice to have some simple criteria to show that an even-rank example is not geometric. The title <i>was</i> a deliberate joke, for the reason Henry gives, and I have changed it back. Greg, what gives?!