User hans stricker - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T17:55:47Z http://mathoverflow.net/feeds/user/2672 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/35286/origins-of-names-of-algebraic-structures Origins of names of algebraic structures Hans Stricker 2010-08-11T22:30:53Z 2013-04-29T13:08:43Z <p>Consider the names of basic algebraic structures: 'group', 'ring', 'space', 'field', '<em>Körper</em>', even the name 'structure' itself - all of them time-honoured terms, deeply rooted in our history and culture. </p> <p>But what has an algebraic field to do with an acre? What has an algebraic group to do with a group of people?</p> <p>Even when it's known who coined these names (of algebraic structures), it's not obvious <em>why</em> they were choosen and what the connection is between the named structures and what was named originally (or later on). Only those who coined the names could tell.</p> <blockquote> <p>Are there etymological studies concerning these names - 'group', 'ring', 'space', 'field',... - which elucidate this connection?</p> </blockquote> http://mathoverflow.net/questions/119455/visualizing-polyhedra-from-their-1-skeletons Visualizing polyhedra from their 1-skeletons Hans Stricker 2013-01-21T10:58:43Z 2013-03-28T10:17:42Z <p>Except for a few simple cases (typically pyramids and prisms) I find it hard to visualize a polyhedron from its 1-skeleton <a href="http://en.wikipedia.org/wiki/Graph_embedding" rel="nofollow">embedded</a> in the plane, e.g. the hexahedral graph 5, as can be seen <a href="http://mathworld.wolfram.com/PolyhedralGraph.html" rel="nofollow">here</a>.</p> <p>Tools that are able to take an arbitrary polyhedral graph as input and draw the corresponding polyhedron perspectively will most surely rely on an abstract representation of the graph, e.g. by its adjacency matrix. From this abstract representation - presumably - they will also draw the embedded version of the graph (without edges crossing).</p> <p>I am interested in the underlying algorithms and/or heuristics of</p> <blockquote> <ol> <li><p>drawing the embedded graph from the adjacency matrix</p></li> <li><p>drawing the polyhedron from the adjacency matrix</p></li> <li><p>drawing the polyhedron from the embedded graph </p></li> <li><p>drawing the embedded graph from the polyhedron</p></li> </ol> </blockquote> <p>I am asking for references.</p> <p>Computer programs will most certainly deal with (1) and (2) while humans typically have to solve problems (3) and (4). </p> <p>I guess that experts have some mental techniques to visualize a polyhedron from looking at its 1-skeleton.</p> <blockquote> <p>Can these techniques be described, made explicit, and taught?</p> </blockquote> <p><em>[Side question: If anyone could give me a visualization of the hexahedral graph 5, I would be thankful.]</em></p> http://mathoverflow.net/questions/124542/probing-a-manifold-with-closed-curves Probing a manifold with closed curves Hans Stricker 2013-03-14T17:37:00Z 2013-03-14T21:28:47Z <p>Since fedja's excellent comment on Joseph's question on <a href="http://mathoverflow.net/questions/81622/probing-a-manifold-with-geodesics" rel="nofollow">probing a manifold with geodesics</a> remained uncommented (especially by topologists), I'd like to make a question out of it:</p> <blockquote> <p><strong>Conjecture:</strong> Given an orientable 2-dimensional manifold and two closed curves on it which <a href="http://mathworld.wolfram.com/TransversalIntersection.html" rel="nofollow">intersect transversally</a> in exactly one point. Then the two curves cannot be homotopic.</p> </blockquote> <p>(An immediate consequence of this would be that living on a surface with two such curves, one would know, that it is not homeomorphic to the sphere.)</p> <blockquote> <p>How to proof this conjecture (if it's true)? </p> </blockquote> http://mathoverflow.net/questions/123243/isometric-but-differently-shaped-closed-surfaces-in-mathbbr3 Isometric but differently shaped closed surfaces in $\mathbb{R}^3$ Hans Stricker 2013-02-28T17:25:31Z 2013-02-28T22:07:03Z <p>Starting from the following <a href="http://mathoverflow.net/questions/76955/determining-a-surface-in-mathbbr3-by-its-gaussian-curvature/76957#76957" rel="nofollow">inclusions for surfaces $M_1,M_2$ in $\mathbb{R}^3$</a>:</p> <blockquote> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;$M_1,M_2$ have the same <strong>shape</strong>, i.e. are related by an ambient isometry</p> <p>&rarr; $M_1,M_2$ have the same <strong>metric</strong></p> <p>&rarr; $M_1,M_2$ have the same Gaussian <strong>curvature</strong></p> </blockquote> <p>the only <a href="http://math.stackexchange.com/questions/315888/isometric-but-differently-shaped-surfaces-in-mathbbr3" rel="nofollow">examples of isometric but differently shaped surfaces</a> in $\mathbb{R}^3$ I have seen so far do have boundaries:</p> <ul> <li>cone and cylinder</li> <li>catenoid and helicoid </li> <li>other associate families of isometric minimal surfaces. </li> </ul> <p>I wonder if there <em>are</em> no (examples of) isometric but differently shaped <em>closed</em> surfaces, and why that could be. (I am particularly interested in <em>smooth</em> surfaces.)</p> <p>And I am still looking for (closed) surfaces with the same Gaussian curvature but different metrics.</p> <p><strong>A picture gallery would be highly welcome, because I really would like to</strong> <em>see</em> <strong>two such (non-)isometric surfaces.</strong></p> http://mathoverflow.net/questions/123170/some-questions-about-ideal-knots Some questions about ideal knots Hans Stricker 2013-02-27T23:49:37Z 2013-02-28T01:07:19Z <p>The <em>ropelength</em> of a knot curve $C$ is defined as the ratio $L(C) = \lambda(C)/ \tau(C)$, where $\lambda(C)$ is the length of $C$ and $\tau(C)$ is the <a href="http://en.wikipedia.org/wiki/Knot_thickness" rel="nofollow">thickness</a> of the knot defined by $C$ [from <a href="http://en.wikipedia.org/wiki/Ropelength" rel="nofollow">Wikipedia</a>]. Intuitively the thickness is the maximally possible thickness that $C$ as a rope can have (without intersecting itself).</p> <p>The infimum ropelength of the realizations of a knot type is a knot invariant. A knot that minimizes ropelength is called an <em>ideal knot</em>.</p> <blockquote> <p><strong>Question 1</strong>: How can it be proved that there are ideal knots at all (besides the plane circle which is the ideal unknot)?</p> </blockquote> <p>Another question concerns the number (of <a href="http://en.wikipedia.org/wiki/Shape#Rigid_shape_definition" rel="nofollow">shapes</a>) of ideal knots of a given knot type.</p> <blockquote> <p><strong>Question 2</strong>: Is the number of ideal knots known, at least for some knot typyes? Can it be calculated from other knot invariants, eventually?</p> </blockquote> <p>(Assuming that this number is finite, it would be another knot invariant &ndash; the more interesting the more it can vary.)</p> <blockquote> <p><strong>Question 3</strong>: Are there ideal knots of constant curvature (besides the plane circle)?</p> </blockquote> <p>Alternatively: Are there approximations of ideal knots that approximate constant curvature?</p> <p>(The last one is a refinement of <a href="http://math.stackexchange.com/questions/309531/closed-space-curves-of-constant-curvature" rel="nofollow">another question</a>.)</p> http://mathoverflow.net/questions/18475/rado-graph-containing-infinitely-many-isomorphic-subgraphs Rado graph containing infinitely many isomorphic subgraphs Hans Stricker 2010-03-17T10:26:41Z 2013-02-27T12:33:15Z <p>The Rado graph contains every finite graph as an induced subgraph. It surely contains <em>some</em> finite graphs infinitely often as an induced subgraph, e.g. $K_2$. Does it contain <em>all</em> finite graphs infinitely often as an induced subgraph? Or can an example of a graph be given that is <em>not</em> contained infinitely often?</p> http://mathoverflow.net/questions/122835/possible-curvatures-of-the-topological-sphere Possible curvatures of the topological sphere Hans Stricker 2013-02-24T22:13:22Z 2013-02-26T19:37:50Z <p>Consider the family $\mathbb{S}$ of compact oriented surfaces homeomorphic to the 2-sphere $\mathcal{S} = S^2$. Consider arbitrary continuous mappings $k: \mathcal{S} \rightarrow \mathbb{R}$ which obey the condition $\int_\mathcal{S} k = 4\pi$. The latter looks like the Gauss-Bonnet condition for 2-spheres, and that's why I want to call those mappings &ldquo;curvature-like&rdquo;.</p> <blockquote> <p>For which curvature-like mappings $k$ does exist a surface $S \in \mathbb{S}$ which $k$ is the Gaussian curvature of?</p> </blockquote> <p>In other words: For which curvature-like mappings $k$ does exist a surface $S \in \mathbb{S}$ with a homeomorphism $s: S \rightarrow \mathcal{S}$ such that $k \circ s$ equals the Gaussian curvature $\kappa: S \rightarrow \mathbb{R}$?</p> <p><strong>Background (and for comparison's sake)</strong>: When one considers (plane) Jordan curves - which are homeomorphic to the 1-sphere $S^1$ - and continuous &ldquo;curvature-like&rdquo; mappings $k: S^1 \rightarrow \mathbb{R}$ which obey the condition $\int_{S^1} k = 2\pi$ &ndash; in accordance to <a href="http://warwickmaths.org/files/Curvature.pdf#page=4" rel="nofollow">Hopf's Umlaufsatz</a> &ndash; then the additional condition for a curvature-like mapping to be a &ldquo;real&rdquo; curvature seems to be given by the <a href="http://math.stackexchange.com/a/280308/1792" rel="nofollow">four-vertex theorem</a>.</p> http://mathoverflow.net/questions/122863/possible-curvatures-of-the-topological-torus Possible curvatures of the topological torus Hans Stricker 2013-02-25T08:07:06Z 2013-02-25T14:11:03Z <p>Consider the family $\mathbb{T}$ of compact oriented surfaces homeomorphic to the torus $\mathcal{T} = S^1 \times S^1$. Consider arbitrary continuous mappings $k: \mathcal{T} \rightarrow \mathbb{R}$ which obey the condition $\int_\mathcal{T} k = 0$. </p> <blockquote> <p>For which curvature-like mappings $k$ does exist a surface $T \in \mathbb{T}$ which $k$ is the Gaussian curvature of?</p> </blockquote> <p>For the <a href="http://mathoverflow.net/questions/122835/possible-curvatures-of-the-topological-sphere" rel="nofollow">topological sphere</a> the answer seems to be: for all. Can the proof of the latter - which I do not know - be generalized for arbitrary compact oriented surfaces? Or is it already a general proof for arbitrary surfaces?</p> http://mathoverflow.net/questions/120814/geodesics-in-polyhedral-graphs Geodesics in polyhedral graphs Hans Stricker 2013-02-05T00:46:38Z 2013-02-05T22:43:02Z <p>Let $e = \lbrace u,v\rbrace$, $e' = \lbrace v,u'\rbrace$ be edges of an undirected graph $G$ and $ee'$ be the path from $u$ through $v$ to $u'$. The following defintions make sense for <em>every</em> graph and thus are purely combinatorial:</p> <blockquote> <p><strong>Definition 1</strong>: $ee'$ is a <strong>pre-geodesic</strong> when it is the unique shortest path between $u$ and $u'$</p> <p><strong>Definition 2</strong>: A pre-geodesic $ee'$ is a <strong>geodesic</strong> when there is no other edge $f$ through $v$ such that $ef$ or $e'f$ is a pre-geodesic.</p> </blockquote> <p>The first condition is motivated by the fact that geodesics in differential geometry are locally <em>shortest</em> paths and as such locally <em>unique</em>. It prevents the graph</p> <p><img src="http://i.stack.imgur.com/lsP2Q.png" alt="enter image description here"></p> <p>from having geodesics.</p> <p>The second condition reflects the fact, that geodesics usually do not &ldquo;split&rdquo;. It prevents the graph</p> <p><img src="http://i.stack.imgur.com/6ReZe.png" alt="enter image description here"></p> <p>from having geodesics.</p> <p>But now, let's restrict to <strong><a href="http://en.wikipedia.org/wiki/Polyhedral_graph" rel="nofollow">polyhedral graphs</a></strong>. I wonder if it's true for polyhedral graphs that only a small number of patterns of geodesics going through a given vertex $v$ are possible:</p> <blockquote> <ul> <li><p>Which patterns?</p></li> <li><p>How to prove this?</p></li> </ul> </blockquote> <p>Have a look at this gallery of vertices $v$ (grey, drawn with <em>all</em> their neighbours) with degree at most 6 and at most three geodesics (red, blue, green) passing through them (all the other vertices may or must have <em>further</em> neighbours):</p> <p><img src="http://i.stack.imgur.com/Xc7BM.png" alt="enter image description here"></p> <blockquote> <ul> <li><p>Is this gallery essentially complete? </p></li> <li><p>If so: how to prove it?</p></li> </ul> </blockquote> <p>Otherwise: </p> <blockquote> <ul> <li><p>Is there a polyhedral graph with a vertex of degree 7 with a geodesic passing through it?</p></li> <li><p>Is there a polyhedral graph with a vertex with four geodesics passing through it?</p></li> </ul> </blockquote> http://mathoverflow.net/questions/120597/ways-to-look-at-a-polyhedral-graph Ways to look at a polyhedral graph Hans Stricker 2013-02-02T15:48:04Z 2013-02-03T16:01:01Z <h3>Motivation</h3> <p>There are at least three <em>interpretations</em> of an abstract polyhedral (= planar 3-vertex-connected) graph:</p> <ul> <li><p>the <strong>1-skeleton of a convex polyhedron</strong> (when embedded into $\mathbb{R}^3$)</p></li> <li><p>a <strong>polygonization of the sphere</strong> (when embedded into the sphere $\mathbb{S}^2$)</p></li> <li><p>a <strong>polygonization of a polygon</strong> &ndash; for each of its faces (when embedded into the plane $\mathbb{R}^2$)</p></li> </ul> <p>In any case there are many geometric <em>realizations</em>: </p> <ul> <li><p>of a polyhedron</p></li> <li><p>of a polygonization of the sphere</p></li> <li><p>of a polygon and its polygonizations</p></li> </ul> <h3>Question</h3> <p>I'd like to understand in an abstract setting:</p> <blockquote> <p>What do these interpretations and realizations have to do with each other?</p> </blockquote> <h3>Example</h3> <p>For many (but not all) polyhedral graphs there is a realization as a convex polyhedron that is <a href="https://digital.lib.washington.edu/dspace/bitstream/handle/1773/2276/Graphs%20of%20polyhedra.pdf;jsessionid=722305F151F9B613CC1379D5C04FF5B7?sequence=1" rel="nofollow">inscribable</a> into the sphere. A central projection of this polyhedron back onto the sphere induces a polygonization of the sphere.</p> <p>Taken for granted is <a href="http://en.wikipedia.org/wiki/Steinitz%27s_theorem" rel="nofollow">Steinitz' theorem</a>. The question is <em>not</em> about this.</p> <hr> <p><strong>EDIT</strong>: For completeness' sake I should mention embeddings of a polyhedral graph into:</p> <ul> <li><p>the hyperbolic space $\mathbb{H}^2$</p></li> <li><p>the 3-dimensional sphere $\mathbb{S}^3$</p></li> <li><p>the hyperbolic space $\mathbb{H}^3$</p></li> </ul> http://mathoverflow.net/questions/119607/degree-of-freedom-restricted-by-inequalities Degree of freedom restricted by inequalities Hans Stricker 2013-01-22T23:01:20Z 2013-01-22T23:45:00Z <h3>Motivational example</h3> <p>Consider a <a href="http://en.wikipedia.org/wiki/Polyhedral_graph" rel="nofollow">polyhedral graph</a> $G$. A realization of $G$ is given by a convex polyhedron which is - essentially - characterized by the angles between the edges emanating from each vertex. There are $\sum_v d(v) = 2E$ such angles, so the degree of freedom $\mathsf{d}$ of a polyhedron is $\mathsf{d}\leq 2E$ (with $E$ the number of edges). But the angles cannot be choosen freely: for every (flat) face of the polyhedron the interior angle sum must obey a (strict) equality. Thus, $\mathsf{d}\leq 2E - F$ (with $F$ the number of faces). Because the polyhedron is supposed to be convex there is another restriction: for every vertex $v$ of the polyhedron the sum of angles between the edges emanating from $v$ has to be less than $2\pi$. Thus, we have $V$ additional restricting inequalities (with $V$ the number of vertices). But because inequalities are somehow "weaker" than equalities we cannot simply deduce that </p> <p>$$\mathsf{d}\leq 2E - F - V = E - \chi = E - 2$$</p> <p>(with $\chi$ the Euler characteristic of the graph/polyhedron). But by which amount $\mathsf{d}$ might be greater than $E-2$?</p> <h3>Question</h3> <p>The original question is (more abstractly) concerned with the question how <strong>inequalities</strong> enter into the calculation of a "degree of freedom":</p> <blockquote> <p>How can an (eventually non-integer) degree of freedom be defined when inequalities are involved?</p> </blockquote> http://mathoverflow.net/questions/119523/the-notion-of-logical-difference The notion of logical difference Hans Stricker 2013-01-22T00:08:24Z 2013-01-22T00:30:55Z <h3>Motivating example</h3> <p>All vertices $v$ in a 3-connected graph have degree $d(v) \geq$ 3 (because every two vertices are connected by three independent paths).</p> <p>What is the "logical difference" between (i) 3-connectedness and (ii) minimal degree $\delta(G) = 3$? That is: which properties $\phi(G)$ make the statement</p> <blockquote> <p>The graph $G$ is 3-connected if and only if $\delta(G) = 3$ and $\phi(G)$</p> </blockquote> <p>true? </p> <h3>Question</h3> <p>Consider a setting with a fixed language and theory. Let two properties $F(x)$ and $G(x)$ with $(\forall x) F(x) \rightarrow G(x)$ be given. Ask the question: Are there properties $H(x)$ such that $(\forall x) F(x) \leftrightarrow G(x) \wedge H(x)$? And which?</p> <blockquote> <p>For which languages and theories and/or in which cases might such a question have a definite and sensible answer? By which means could the answer be derived?</p> </blockquote> <p>It's - very loosely speaking - about resolving some kind of "logical equation": </p> <p>$F > G \Rightarrow (\exists H)\ F = G + H \Rightarrow \bf{H = F-G}$.</p> http://mathoverflow.net/questions/119047/unique-circular-ordering-of-edges-around-a-vertex Unique circular ordering of edges around a vertex Hans Stricker 2013-01-16T09:50:48Z 2013-01-18T01:53:30Z <p>Consider the property of a vertex $v$ of a planar graph $G$ that the <a href="http://en.wikipedia.org/wiki/Rotation_system" rel="nofollow">circular ordering</a> of its edges is the same (upto orientation) for every <a href="http://en.wikipedia.org/wiki/Graph_embedding" rel="nofollow">graph embedding</a> $\pi$ of $G$ into the plane $\mathbb{R}^2$.</p> <blockquote> <ol> <li><p>Does this property have an official name?</p></li> <li><p>(How) can it be defined purely combinatorially?</p></li> <li><p>(How) can planar graphs be characterized in which every vertex has this property?</p></li> </ol> </blockquote> http://mathoverflow.net/questions/119047/unique-circular-ordering-of-edges-around-a-vertex/119086#119086 Answer by Hans Stricker for Unique circular ordering of edges around a vertex Hans Stricker 2013-01-16T17:07:12Z 2013-01-17T06:52:13Z <p>It might be simpler than I believed:</p> <blockquote> <p>ad 2. The <em>circular</em> ordering of the edges (= neighbours) of a vertex $v$ of a planar graph $G$ is unique (upto orientation) when the neighbours of $v$ lie on a <em>cycle</em> that does not contain $v$.</p> </blockquote> <p>If they happen to lie on more than one cycle their circular ordering doesn't depend on which.</p> http://mathoverflow.net/questions/118877/jordan-like-cycles-in-graphs Jordan-like cycles in graphs Hans Stricker 2013-01-14T14:11:28Z 2013-01-16T09:46:09Z <p><em>[Added another complementary question below.]</em></p> <h2>Motivation</h2> <p>The 1-skeleton of the <a href="http://en.wikipedia.org/wiki/Triangular_bipyramid" rel="nofollow">triangular bipyramid</a> seems to be the smallest connected planar graph $G$ with the following </p> <blockquote> <p><strong>Property:</strong> There is a cycle $\gamma$ ($\color{red}{\mathsf{red}}$) in $G$ with exactly two connected components $G_1, G_2$ of $G - \gamma$ such that for every <a href="http://en.wikipedia.org/wiki/Graph_embedding" rel="nofollow">graph embedding</a> $\pi$ of $G$ into the plane $\mathbb{R}^2$ the component $G_1$ ($\color{blue}{\mathsf{blue}}$) is contained in the <strong>interior</strong> of $\pi(\gamma)$ and the other component $G_2$ ($\color{black}{\mathsf{black}}$) is contained in the <strong>exterior</strong> of $\pi(\gamma)$ &ndash; or vice versa.</p> </blockquote> <p><img src="http://i.stack.imgur.com/Xye1v.png" alt="enter image description here"></p> <blockquote> <p><strong>Definition</strong>: A cycle $\gamma$ in the graph $G$ is a <strong><em>Jordan cycle</em></strong> if $G - \gamma$ splits up into exactly two connected components $G_1, G_2$ such that for every graph embedding $\pi$ of $G$ into the plane $\mathbb{R}^2$ the component $G_1$ is contained in the interior of $\pi(\gamma)$ and the other component $G_2$ is contained in the exterior of $\pi(\gamma)$ &ndash; or vice versa.</p> </blockquote> <h2>Questions</h2> <blockquote> <ol> <li><p>(How) can the property of <strong>being a Jordan cycle</strong> $\gamma$ be defined purely combinatorial, without mentioning graph embeddings $\pi$ and Jordan curves $\pi(\gamma)$?</p></li> <li><p>(How) can planar graphs <strong>containing a Jordan cycle</strong> be characterized purely combinatorially?</p></li> <li><p><strong>ADDED: Can planar graphs be characterized in which <em>every</em> cycle is a Jordan cycle?</strong></p></li> </ol> </blockquote> http://mathoverflow.net/questions/9255/can-you-determine-whether-a-graph-is-the-1-skeleton-of-a-polytope Can you determine whether a graph is the 1-skeleton of a polytope? Hans Stricker 2009-12-18T09:57:57Z 2013-01-14T10:38:54Z <p>How do I test whether a given undirected graph is the <a href="http://en.wikipedia.org/wiki/Geometric_graph_theory" rel="nofollow">1-skeleton of a polytope</a>?</p> <p>How can I tell the dimension of a given 1-skeleton?</p> http://mathoverflow.net/questions/22184/almost-or-probably-complete-graph-invariants Almost or probably complete graph invariants? Hans Stricker 2010-04-22T14:06:55Z 2013-01-13T23:57:40Z <p>Assuming that there are no known complete graph invariants in the spirit of <a href="http://mathoverflow.net/questions/11631/complete-graph-invariants" rel="nofollow">Harrsion's question</a> that do not depend on any labelling (see <a href="http://en.wikipedia.org/wiki/Graph_invariant" rel="nofollow">graph property</a> at Wikipedia), I wonder if there are graph invariants that are</p> <ul> <li>almost complete, i.e. discriminating almost all finite graphs up to isomorphism</li> <li>almost complete in a weaker sense, i.e. discriminating all finite graphs except for a small, but finite fraction (the smaller the fraction the greater the invariant's discriminating power)</li> <li>probably complete, i.e. not proven to be incomplete yet (e.g. by counterexamples)</li> </ul> <p>Can anyone provide examples or references?</p> http://mathoverflow.net/questions/25215/reconstruction-conjecture-can-other-decks-do-the-job Reconstruction conjecture: Can other decks do the job? Hans Stricker 2010-05-19T09:06:23Z 2013-01-09T11:39:09Z <p>The standard reconstruction conjecture states that a graph is determined by its <a href="http://en.wikipedia.org/wiki/Reconstruction_conjecture#Formal_statements" rel="nofollow"><strong>deck of vertex-deleted subgraphs</strong></a>.</p> <blockquote> <p><strong>Question</strong>: Have other decks been investigated, finding out that only vertex-deleted subgraphs can do the job? If so: Which property of vertex-deleted subgraphs makes them exceptional?</p> </blockquote> <p>I have three candidates in mind, others are conceivable. (For the sake of simplicity I consider only simple connected graphs $G$.)</p> <ol> <li><p>the <strong>deck of sub-maximal neighbourhoods</strong>: Let the <em>sub-maximal neighbourhood of $v$</em> be the $v$-rooted graph constructed from $G$ by deleting all vertices with maximal distance from $v$.</p></li> <li><p>the <strong>deck of distinguishing neighbourhoods</strong>: Let the <em>distinguishing neighbourhood of $v$</em> be the smallest $n$-neighbourhood of $v$ which distinguishes it from all vertices not conjugate to it ($n$-neighbourhood = the $v$-rooted induced subgraph containing all vertices $w$ with distance $d(v,w) \leq n$).</p></li> <li><p>the <strong>deck of crossref-deleted subgraphs</strong>: Let the <em>crossref-deleted subgraph with respect to $v$</em> be the $v$-rooted graph constructed from $G$ by deleting all edges between vertices that have the same distance from $v$.</p></li> </ol> <p>Note that the <em>vertex-deleted subgraph with respect to $v$</em> is nothing but the $v$-rooted graph constructed from $G$ by deleting all edges between $v$ and its neighbours.</p> <p>I am not good in systematically constructing counterexamples, and I do not have very much intuition about general graphs. So, any counterexample to one of the candidates above would be very welcome.</p> <p>What I <em>do</em> know is that a) trees are trivially reconstructible from their deck of crossref-deleted subgraphs, that b) graphs with one node of which the distinguishing neighbourhood is the whole graph are trivially reconstructible from their deck of distinguishing neighbourhoods, and that c) reconstructing (very) small graphs from one of the decks above is fun.</p> http://mathoverflow.net/questions/118162/approximating-jordan-curves Approximating Jordan curves Hans Stricker 2013-01-05T23:50:30Z 2013-01-06T14:00:06Z <p>I'd like to capture the intuitive notion that a Jordan curve $\gamma_2$ &ldquo;follows&rdquo; or &ldquo;approximates&rdquo; another Jordan curve $\gamma_1$, i.e. goes somehow &ldquo;parallel&rdquo; to it or &ldquo;oscillates&rdquo; around it.</p> <p>Consider a differentiable Jordan curve $\gamma_1: [0,1] \rightarrow \mathbb{R}^2$ and its normals, seen as straight lines crossing the curve perpendicularly. Consider another Jordan curve $\gamma_2$ with the following properties:</p> <ol> <li><p>Each normal of $\gamma_1$ crosses $\gamma_2$ at least once. I.e., for each normal $n(s_1)$ of $\gamma_1$ there is an $s_2$ such that the point $\gamma_2(s_2)$ lies on $n(s_1)$.</p></li> <li><p>Furthermore when $s_1 &lt; s_1'$ then there are unique $s_2 \le s_2'$ such that $\gamma_2(s_2)$ lies on $n(s_1)$ and $\gamma_2(s_2')$ lies on $n(s_1')$.</p></li> </ol> <blockquote> <p>Do these conditions suffice to capture the notion described above? For which &ldquo;pathological &rdquo; cases do they eventually fail? How then would they have to be adjusted to capture the notion?</p> </blockquote> <p>Further questions:</p> <ol> <li><p>Under which extra conditions does &ldquo;$\gamma_2$ follows $\gamma_1$&rdquo; imply that $\gamma_1$ follows $\gamma_2$? </p></li> <li><p>When $\gamma_2$ follows $\gamma_1$, (how) can the area enclosed by $\gamma_1$ and $\gamma_2$ be calculated via the distance function $d(s_1) = |\gamma_1(s_1) - \gamma_2(s_2)|$ <br/>($s_2$ the unique parameter according to condition 2 above)?</p></li> </ol> <p>(Let the area enclosed by $\gamma_1$ and $\gamma_2$ be the symmetric difference $X_1 \triangle X_2 = (X_1 \cup X_2) \setminus (X_1 \cap X_2)$ of the areas $X_1, X_2$ enclosed by $\gamma_1$ and $\gamma_2$.)</p> http://mathoverflow.net/questions/117657/generalization-of-the-isoperimetric-inequality Generalization of the isoperimetric inequality Hans Stricker 2012-12-30T17:48:02Z 2013-01-01T15:27:53Z <h3>Preliminaries</h3> <p>Suppose that $\Gamma$ is a simple (sufficiently smooth) closed curve in $\mathbb{R}^2$ with length $L$ and enclosing an area $A$. Suppose furthermore that $\gamma:[0,L]\rightarrow\mathbb{R}^2$ is a parametrization-by-arclength for $\Gamma$ so that $|\gamma '(s)|=1$ for every s, i.e. </p> <p>$$L = \int_0^{L} |\gamma'(s)|$$</p> <p>The <a href="http://www2.mat.dtu.dk/people/S.Markvorsen/DISPLAY/TEMP/PROOPG/KnottedCurves/Milnor.pdf" rel="nofollow">total (absolute) curvature</a> can be defined by</p> <p>$$C = \int_0^{L} |\gamma''(s)| = \int_0^{L} \kappa(s)$$</p> <p>It can be shown that $C\ge 2\pi$. Equality holds iff $\Gamma$ is convex. </p> <p>The <a href="http://cornellmath.wordpress.com/2008/05/16/two-cute-proofs-of-the-isoperimetric-inequality/" rel="nofollow">isoperimetry inequality</a></p> <p>$$4\pi A\le L^2$$</p> <p>(equality holds iff $\Gamma$ is a circle) can be stated using another derivative of $\gamma$ </p> <p>$$\Omega = \int_0^{L} ||\gamma''(s)|'| = \int_0^{L} |\kappa'(s)|$$</p> <p>A closed curve $\Gamma$ is a circle $\mathsf{O}$ iff $\Omega = 0$. (<strong>Edit</strong>: Pietro correctly pointed out that the initial formula $\int_0^{L} |\gamma'''(s)|$ would not make sense.) This implies</p> <p>$$\left(( L_{\Gamma} = L_{\mathsf{O}}) \wedge (C_{\Gamma} = C_{\mathsf{O}}) \wedge (\Omega_{\Gamma} \ge \Omega_{\mathsf{O}})\right) \rightarrow (A_{\Gamma} \le A_{\mathsf{O}})$$</p> <p>for all closed curves $\Gamma$ and circles $\mathsf{O}$.</p> <h3>Question</h3> <p>I wonder if and how it could be shown that a generalization of the last proposition holds:</p> <blockquote> <p>$$\left((L_{\Gamma} = L_{\Gamma'}) \wedge (C_{\Gamma} = C_{\Gamma'}) \wedge (\Omega_{\Gamma} \ge \Omega_{\Gamma'})\right) \rightarrow (A_{\Gamma} \le A_{\Gamma'})$$</p> </blockquote> <p>for all closed curves $\Gamma, \Gamma'$, i.e. of two curves of the same length and the same absolute curvature the one with greater change rate of curvature has the smaller area.</p> <p>How would this &ndash; probably <a href="http://mathoverflow.net/questions/27164/why-is-fourier-analysis-so-handy-for-proving-the-isoperimetric-inequality" rel="nofollow">using Fourier analysis</a> &ndash; be shown for ellipses with the same perimeter?</p> http://mathoverflow.net/questions/115884/combinatorial-geodesics Combinatorial geodesics Hans Stricker 2012-12-09T12:50:14Z 2012-12-12T17:56:32Z <p>[There has been a flaw in my definition - as Sergei and Andreas pointed out. I hope I could fix it.]</p> <p>I want to understand how the concepts of directions, straight (or shortest) lines, and geodesics &ndash; all from differential geometry &ndash; can be understood in abstract graphs, i.e. what their combinatorial counterparts might be.</p> <p>Specifically, I'd like to know whether the following relation among adjacent edges in a <a href="http://en.wikipedia.org/wiki/Graph_%28mathematics%29#Simple_graph" rel="nofollow">simple graph</a> has been explored, under which name, and in which context:</p> <blockquote> <p><strong>Definition</strong>: Two adjacent edges <em>e, e' have the same direction</em> iff</p> <ol> <li><p>the path <em>ee'</em> is the unique shortest path between its endpoints</p></li> <li><p>neither <em>e</em> nor <em>e'</em> is part of another unique shortest 2-edge-path <strong>through their common vertex</strong></p></li> </ol> </blockquote> <p>The first condition is motivated by the fact that <a href="http://en.wikipedia.org/wiki/Geodesic#Existence_and_uniqueness" rel="nofollow">geodesics in differential geometry</a> are locally <em>shortest</em> paths and as such locally <em>unique</em>. It prevents the graph</p> <p><img src="http://i.stack.imgur.com/lsP2Q.png" alt="enter image description here"></p> <p>from having edges with the same direction.</p> <p>The second condition reflects the fact, that geodesics usually do not &ldquo;split&rdquo;. It prevents the graph</p> <p><img src="http://i.stack.imgur.com/6ReZe.png" alt="enter image description here"></p> <p>from having edges with the same direction.</p> <blockquote> <p><strong>Definition</strong>: Paths of length at least 2 in which adjacent edges have the same direction are called <em>combinatorial geodesics</em>.</p> </blockquote> <p>Note, that even a single pair of edges having the same direction is a (minimal) combinatorial geodesic. <em>But no single edge is a combinatorial geodesic.</em></p> <p>It's an easy excercise to find combinatorial geodesics in some planar graphs and/or 3-vertex-connected graphs (seen as polygonizations/tilings of surfaces).</p> <ul> <li>In trees (with minimal non-leaf degree 3) there are <em>no</em> geodesics at all.</li> <li>In finite, infinite, and &ldquo;torified&rdquo; <a href="http://en.wikipedia.org/wiki/Triangular_grid" rel="nofollow">triangular</a> and <a href="http://en.wikipedia.org/wiki/Square_tiling" rel="nofollow">rectangular</a> grid graphs there <em>are</em> geodesics.</li> <li>In infinite and in &ldquo;torified&rdquo; <a href="http://en.wikipedia.org/wiki/Hexagonal_grid" rel="nofollow">hexagonal</a> grid graphs there are <em>no</em> geodesics. </li> <li>In <a href="http://en.wikipedia.org/wiki/Polyhedral_graph" rel="nofollow">polyhedral graphs</a> there are &ndash; in general &ndash; <em>no</em> geodesics.</li> </ul> <p>Things get really interesting, I believe, when one considers e.g. <em>non-regular tilings of the plane</em> &ndash; from &ldquo;slightly non-regular&rdquo; (= &ldquo;regular with a few perturbations&rdquo;) to &ldquo;completely random&rdquo;. One can investigate the &ldquo;geodesic structure&rdquo; of such graphs: which geodesics are there, and how are they interconnected? [geodesic structure &rlarr; <a href="http://en.wikipedia.org/wiki/Cycle_space" rel="nofollow">cycle structure</a>]</p> <p>[The notion of &ldquo;<em>true</em> shortest paths&rdquo; between two vertices with several non-unique shortest paths between them &ndash; like in the rectangular grid &ndash; shall be subject of a follow-up question. It can be easily defined based on combinatorial geodesics.]</p> <p>To repeat my questions from above:</p> <blockquote> <p>Have <em>this</em> relation of &ldquo;having the same direction&rdquo; and <em>this</em> concept of &ldquo;combinatorial geodesics&rdquo; been explored, under which name, and in which context?</p> </blockquote> http://mathoverflow.net/questions/114911/abstract-characterization-of-polygonizations Abstract characterization of polygonizations Hans Stricker 2012-11-29T18:44:42Z 2012-11-30T08:57:39Z <p>Consider a <a href="http://en.wikipedia.org/wiki/Triangulation_%2528geometry%2529" rel="nofollow">polygonization</a> of the plane by convex polygons of a given minimal size that meet edge-to-edge and vertex-to-vertex.</p> <p><img src="http://i.stack.imgur.com/cYI09.png" alt="enter image description here"> <em><sup>What's the &ldquo;official&rdquo; name of such a polygonization?</sup></em></p> <p>Such polygonizations of the plane induce infinite graphs.</p> <blockquote> <p>How can such abstract graphs be characterized?</p> </blockquote> <p>Somehow like this: &ldquo;A graph is induced by a polygonization of the plane iff it is infinite, planar, 3-vertex-connected, and <em>P</em>.&rdquo; (The question asks for property <em>P</em>, since infinite, planar and 3-vertex-connected those graphs obviously are.)</p> <blockquote> <p>Is it true, that the graphs that are induced by a polygonization of the <em>sphere</em> are exactly the <a href="http://en.wikipedia.org/wiki/Polyhedral_graph" rel="nofollow">polyhedral graphs</a> which in turn are exactly the finite planar <a href="http://en.wikipedia.org/wiki/Steinitz%27s_theorem" rel="nofollow">3-vertex-connected graphs</a>?</p> </blockquote> <p>Finally I want to know:</p> <blockquote> <p>Can the graphs be characterized that are induced by a polygonization of <em>any</em> surface?</p> </blockquote> <p><sup>For the record: I asked this question at <a href="http://math.stackexchange.com/questions/246929/graphs-that-polygonize-a-manifold" rel="nofollow">MSE</a> before but it didn't earn a lot of interest.</sup></p> http://mathoverflow.net/questions/114117/sets-structured-sets-without-structure Sets = structured sets without structure Hans Stricker 2012-11-22T00:18:51Z 2012-11-22T01:51:25Z <h2>Motivation</h2> <p>There is presumably no single and widely accepted formal definition of <strong>structured sets = sets <em>plus</em> structure</strong> based on sets as primitive objects, but several approaches are around. See e.g. <a href="http://en.wikipedia.org/wiki/Structure_%28mathematical_logic%29" rel="nofollow">structures</a> (model theory), <a href="http://books.google.de/books?id=IL-SI67hjI4C&amp;pg=PA289&amp;lpg=PA289&amp;dq=echelon%20bourbaki&amp;source=bl&amp;ots=nbgnz98aNL&amp;sig=qsJnl4NV_bRxRn-buNgRUuXfSCs&amp;hl=de&amp;sa=X&amp;ei=NkerUMbbOIWHswahkIH4Cw&amp;ved=0CDgQ6AEwAg#v=onepage&amp;q=echelon%20bourbaki&amp;f=false" rel="nofollow">echelons</a> (Bourbaki), <a href="ftp://210.45.114.81/math/2007_07_06/UTM/Y.Moschovakis%20Notes%20on%20Set%20Theory.pdf#page=56" rel="nofollow">frames</a> (Moschovakis). I don't want to discuss these approaches, but for simplicity's and specifity's sake I want to pick out one especially simple definition of rather generic structured sets &mdash; <strong>graphs</strong> (which among other things are able to <a href="http://books.google.de/books?id=Rf6GWut4D30C&amp;pg=PA228&amp;dq=%22classes+which+interpret+any+structure%22&amp;hl=de#v=onepage&amp;q=%22classes%20which%20interpret%20any%20structure%22&amp;f=false" rel="nofollow">interpret <em>any</em> structured set</a>).</p> <p>When we start with the &bdquo;graph&ldquo; of <strong>sets</strong> $U = \langle V,\in\rangle$, sets as primitive objects are &bdquo;defined&ldquo; just by</p> <p>$$\text{Set}(X) :\equiv X\ \epsilon\ V$$</p> <p>with $\epsilon$ indicating class membership. Graphs (structured sets), then, are - rather sophisticatedly - defined by</p> <p>$$\text{Graph}(X) :\equiv (\exists S,R\ \epsilon\ V)\ R \subseteq S^2 \wedge X = \langle S,R\rangle$$</p> <p>To complete the picture we define</p> <p>$$\text{Relation}(X) :\equiv (\exists S\ \epsilon\ V)\ X \subseteq S^2$$</p> <p>Thus, a graph is a set <em>plus</em> a relation over it, usually written as $G = \langle V,E\rangle$.</p> <h2>Interlude</h2> <p>Oppose this set-based picture to the category $\mathsf{Graph} = \langle \mathcal{O},\mathcal{M},\dots\rangle$ with $\mathcal{O}$ the class of all graphs as primitive objects, related by <a href="http://en.wikipedia.org/wiki/Graph_homomorphism" rel="nofollow">graph homomorphisms</a> $\mathcal{M}$, etc. Thus graphs as primitive objects are defined by</p> <p>$$\text{Graph}(X) :\equiv X\ \epsilon\ \mathcal{O}$$</p> <p>To define &bdquo;set&ldquo; as a now <em>derived</em> concept one might try to resolve the above &bdquo;equation&ldquo; informally to obtain <strong>sets = structured sets <em>minus</em> structure</strong>. Thus, sets don't have any structure (anymore), so morphisms as structure-<em>preserving</em> functions don't have to preserve any structure (anymore), so <em>every</em> function from a set to any other graph (= structured set) is a morphism in $\mathcal{M}$. According to the standard definition of <a href="http://en.wikipedia.org/wiki/Graph_homomorphism" rel="nofollow">graph homomorphism</a>, the graphs being sets are exactly the edgeless graphs (which complies with intuition). </p> <p>Translating this into categorical terms we obtain:</p> <p>$$\text{Set}(X) :\equiv (\forall\ Y\ \epsilon\ \mathcal{O})(\exists\ f\ \epsilon\ \mathcal{M})\ f: X \rightarrow Y$$</p> <p>Making use of categorical terminology we can equivalently say: Let $\mathcal{C}/$ be the <a href="http://en.wikipedia.org/wiki/quotient_category" rel="nofollow">quotient category</a> of $\mathcal{C}$ which identifies all morphisms (if present) from $A$ to $B$ as one. Then:</p> <p>$$\text{Set}(X) :\equiv X\ \text{is an initial object in } \mathsf{Graph}/$$</p> <h2>Question(s)</h2> <p>I am aware that I didn't even mention the category $\mathsf{Set}$ of sets and the notion of a concrete category (even though $\mathsf{Graph}$ <em>is</em> a concrete category).</p> <p>What I'd like to learn is - among other things - how the above definition of being a <em>set</em> relates to the usual notions:</p> <ul> <li><p>Which concrete categories don't have sets?</p></li> <li><p>Which non-concretizable categories <em>do</em> have sets?</p></li> </ul> http://mathoverflow.net/questions/112122/a-weaker-concept-of-graph-homomorphism A weaker concept of graph homomorphism Hans Stricker 2012-11-11T22:38:51Z 2012-11-12T17:21:41Z <p>In the category $\mathsf{Graph}$ of <a href="http://mathworld.wolfram.com/SimpleGraph.html" rel="nofollow">simple graphs</a> with <a href="http://en.wikipedia.org/wiki/Graph_homomorphism" rel="nofollow">graph homomorphisms</a> we'll find the following situation (the big circles indicating objects, labelled by the graphs they enclose, arrows indicating the existence of a homomorphism):</p> <p><img src="http://i.stack.imgur.com/3YIpO.png" alt="enter image description here"></p> <p>Speaking informally, the "obvious" structural relatedness between the two circle graphs $C_3$ and $C_4$ reduces to its two common subgraphs $P_3$ and $P_4$, with $P_3$ being a subgraph of $P_4$.</p> <p>But even though there is an "undeniable" structural relatedness between $C_3$ and $C_4$, there is no single graph homomorphism between the two. This in complete contrast to the category <a href="http://en.wikipedia.org/wiki/Category_of_topological_spaces" rel="nofollow">$\mathsf{Top}$</a>, where they are even <a href="http://en.wikipedia.org/wiki/Topological_graph_theory#Graphs_as_topological_spaces" rel="nofollow">isomorphic</a>. (Instead of this: no arrows between $P_i$ and $C_j$!)</p> <p>But why is there no graph homomorphism between $C_3$ and $C_4$? There are two interrelated reasons:</p> <ol> <li><p>If all vertices of a graph were forced to have a self-loop, there <em>would</em> be a homomorphism from $C_4$ to $C_3$, since two adjacent vertices $x,y$ were allowed to be mapped onto the same vertex $f(x)=f(y)$. Furthermore, there would also be a homomorphism from $P_4$ to $P_3$.</p></li> <li><p>If one insists on graphs to be loop-less (as in topological graph theory?), one might instead weaken the definition of a graph homomorphism. Instead of defining $f:G\rightarrow G'$ to be a homomorphism when $(x,y) \in E(G)$ implies $(f(x),f(y)) \in E(G')$, one might define it like this:<br/></p> <blockquote> <blockquote> <p>$f:G\rightarrow G'$ is a (weak) homomorphism when $(x,y) \in E(G)$ implies $f(x) = f(y) \vee (f(x),f(y)) \in E(G')$.</p> </blockquote> </blockquote></li> </ol> <p>My questions are:</p> <ol> <li><p>In which contexts does this definition of a (weak) homomorphism between simple graphs have drawbacks other than not being standard, e.g. technical ones?</p></li> <li><p>Is there a known or imaginable "problem" that might be easier to handle with weak homomorphisms (other than the missing homomorphism between $C_4$ and $C_3$ which isn't really a problem)?</p></li> <li><p>Might weak homomorphisms be conceptually more appropriate, i.e. catch the "meaning" of structure preserving better?</p></li> <li><p>Where can I find weak homomorphisms in the literature, maybe under another name?</p></li> </ol> http://mathoverflow.net/questions/110484/geodesics-on-a-twisted-torus Geodesics on a twisted torus Hans Stricker 2012-10-23T22:18:55Z 2012-10-27T18:41:57Z <p><sup>This is a repost of a question I posted at <a href="http://math.stackexchange.com/questions/219499/geodesics-on-the-torus" rel="nofollow">MSE</a>.</sup></p> <p>Mark L. Irons' paper <a href="http://www2.rdrop.com/~half/math/torus/torus.geodesics.pdf" rel="nofollow">The Curvature and Geodesics of the Torus</a> gives a concise overview of the <a href="http://en.wikipedia.org/wiki/Geodesic" rel="nofollow">geodesics</a> on the <a href="http://en.wikipedia.org/wiki/Torus" rel="nofollow">torus</a>: </p> <ul> <li>There are five clear-cut families of geodesics.</li> <li>Most of the geodesics are "ergodic": aperiodic and covering either the entire surface - by spiraling endlessly around - or substantial parts of it.</li> <li>Some of the geodesics are "boring": the meridians, the inner and the outer equator</li> <li>A few of them are "æsthetically pleasing": returning to their starting point after just a few circuits. </li> </ul> <blockquote> <p>Does the structure of geodesics change when twisting the "hose" before gluing its ends?</p> </blockquote> <p><img src="http://upload.wikimedia.org/wikipedia/commons/6/60/Torus_from_rectangle.gif" alt="alt text"></p> <p>E.g., there might be no equators anymore because after twisting the (two) equators lost their "ends".</p> http://mathoverflow.net/questions/24978/spectral-graph-theory-interpretability-of-eigenvalues-and-vectors Spectral graph theory: Interpretability of eigenvalues and -vectors Hans Stricker 2010-05-17T08:09:00Z 2012-10-27T00:36:53Z <p>I thought "Wow!" when I learned that the eigenvector of the adjacency matrix of a cycle graph $C_n$ corresponding to the second largest eigenvalue gives <strong>the coordinates of the vertices</strong> when equally distributed on the unit cycle: the $n$-th roots of unity (from complex analysis) come up in a completely discrete context! What's more: The coordinates of the eigenvector can be "interpreted" straight forwardly when assigned properly to the vertices.</p> <p>There's another straight-forward interpretation of an adjacency eigenvector: the eigenvector corresponding to the largest eigenvalue gives <strong>the relative importances of the vertices</strong>, being proportional to the sum of the relative importances of its neighbors.</p> <blockquote> <p><strong>Question:</strong> Can more - or eventually all - adjacency eigenvectors be sensibly "interpreted"?</p> </blockquote> <p>Or does it in general depend on the type of graph, whether and how the eigenvectors can be interpreted, and the first example above is just a curio? </p> <p>What about the interpretation of the eigen-<em>values</em>? Do at least some of them "mean" something conceivable?</p> http://mathoverflow.net/questions/109498/the-abc-of-categories-abstract-vs-concrete The ABC of categories: ABstract vs Concrete Hans Stricker 2012-10-12T22:37:48Z 2012-10-13T01:08:52Z <p>From <a href="http://en.wikipedia.org/wiki/Concrete_category" rel="nofollow">Wikipedia</a>:</p> <blockquote> <p>A <strong>concrete category</strong> is a category that is equipped with a faithful functor to the category of sets. </p> <p>This functor makes it possible to think of the objects of the category as sets with additional structure, and of its morphisms as structure-preserving functions.</p> </blockquote> <p>This definition gives rise to the well-established dichotomy between concrete and abstract categories.</p> <p>The examples of abstract (= non-concrete) categories I've heard of come in two flavours:</p> <ol> <li><p>categories with structured sets as objects, and <strong>some <em>other</em> sort</strong> of structure-preserving maps than standard homomorphisms, e.g. <a href="http://en.wikipedia.org/wiki/Interpretation_%28model_theory%29" rel="nofollow">interpretations</a>. (I understand that - basically - interpretations <em>are</em> structure-preserving maps, just involving re-definitions of individuals (as <em>n</em>-tupels) and relations.)</p></li> <li><p>categories with structured set as objects, and <strong>equivalence classes</strong> of structure-preserving maps as morphisms, e.g. the <a href="http://en.wikipedia.org/wiki/Homotopy_category_of_topological_spaces" rel="nofollow">homotopy category of topological spaces</a> or much simpler: the category $C'$ which collapses all arrows $X \rightarrow Y$ of a (concrete) category $C$ into one (what's its name?)</p></li> </ol> <p>[<em>Side remark:</em> The trivial equivalence relation on morphisms: $f \sim g :\equiv f = g$ leaves a given category $C$ unchanged.]</p> <p>Given such a vast variety of possible definitions of structure-preserving maps and equivalence relations between them, I wonder why only classical homomorphisms and the trivial equivalence relation give rise to so-called <strong>concrete</strong> categories?</p> <p>The other way around: </p> <blockquote> <p>What is an example of an abstract category (in the standard sense) with structured sets as objects, such that we <strong>cannot</strong> think of its morphisms as <em>some</em> equivalence class of <em>some</em> sort of structure-preserving maps?</p> </blockquote> http://mathoverflow.net/questions/11631/complete-graph-invariants/107550#107550 Answer by Hans Stricker for Complete graph invariants? Hans Stricker 2012-09-19T10:56:58Z 2012-09-19T10:56:58Z <p>The sequence of homomorphism numbers $|Hom(F_i,G)|$ for all (isomorphism types of) graphs $F_i$ is an invariant of $G$ (see Lovász, <em><a href="http://bolyai.math.elte.hu/~lovasz/scans/opstruct.pdf" rel="nofollow">Operations with structures</a></em>). </p> <p>(Does this fit your bill? Or do you want finite invariants only?)</p> http://mathoverflow.net/questions/107266/is-homotopy-definable-by-categorical-means Is homotopy definable by categorical means? Hans Stricker 2012-09-15T16:00:17Z 2012-09-15T16:00:17Z <p>Is being homotopic - as a relation between two continuous functions, i.e. morphisms in <strong>Top</strong> - definable by categorical means? Can one detect from the context of dots and arrows, whether two parallel arrows in <strong>Top</strong> are homotopic to each other? Or from another ("higher") categorical point of view?</p> <p>(If this were the case the relationship between <strong>Top</strong> and its quotient category <a href="http://en.wikipedia.org/wiki/Homotopy_category" rel="nofollow"><strong>hTop</strong></a> was purely categorical and not grounded on extra-categorical properties.)</p> http://mathoverflow.net/questions/11845/theory-mainly-concerned-with-lambda-calculus Theory mainly concerned with $\lambda$-calculus? Hans Stricker 2010-01-15T11:08:24Z 2012-09-11T18:54:23Z <p>Automata theory is mainly concerned with Turing machines and all its relatives-in-spirit. $\lambda$-calculus is rather rarely mentioned in textbooks on automata theory.</p> <p>What's the common name of the theory mainly concerned with $\lambda$-calculus and its relatives? (I think, "mathematical logic", "computability theory", "programming language theory" and "recursion theory" are too general, compared to "automata theory". But there should be an "$\lambda$-theory", shouldn't it?)</p> http://mathoverflow.net/questions/124542/probing-a-manifold-with-closed-curves/124543#124543 Comment by Hans Stricker Hans Stricker 2013-03-16T18:25:59Z 2013-03-16T18:25:59Z @Ryan: Thanks again for the hint to Guillemin &amp; Pollack. I received the book today - and it's love of first sight. http://mathoverflow.net/questions/8846/proofs-without-words Comment by Hans Stricker Hans Stricker 2013-03-15T17:53:02Z 2013-03-15T17:53:02Z Why has this question been closed? How can it be &quot;no longer relevant&quot;? (I mean: the longer people contribute examples, the better.) (BTW: I <i>do</i> have an astonishing example.) http://mathoverflow.net/questions/124542/probing-a-manifold-with-closed-curves/124543#124543 Comment by Hans Stricker Hans Stricker 2013-03-14T20:21:29Z 2013-03-14T20:21:29Z @Mariano &amp; Ryan: Sorry for the dumb question - of course I should have tried googling. (I have unwantedly switched into &quot;chat mode&quot;.) http://mathoverflow.net/questions/124542/probing-a-manifold-with-closed-curves/124543#124543 Comment by Hans Stricker Hans Stricker 2013-03-14T19:16:42Z 2013-03-14T19:16:42Z What does &quot;injectivity radius&quot; mean? Why not just &quot;radius&quot;? http://mathoverflow.net/questions/124542/probing-a-manifold-with-closed-curves/124554#124554 Comment by Hans Stricker Hans Stricker 2013-03-14T19:07:30Z 2013-03-14T19:07:30Z @Mariano: I grinned over your &quot;few seconds&quot; - for others it takes a few years! http://mathoverflow.net/questions/124542/probing-a-manifold-with-closed-curves/124554#124554 Comment by Hans Stricker Hans Stricker 2013-03-14T19:05:25Z 2013-03-14T19:05:25Z I only begin to learn that and why orientability is crucial for my question. So excuse me for not having mentioned orientability in my original question.) http://mathoverflow.net/questions/124542/probing-a-manifold-with-closed-curves/124543#124543 Comment by Hans Stricker Hans Stricker 2013-03-14T18:58:25Z 2013-03-14T18:58:25Z Another side question: Do &quot;tubular neighbourhoods&quot; in your understanding have to do with the thickness/ropelength of knots? http://mathoverflow.net/questions/124542/probing-a-manifold-with-closed-curves/124543#124543 Comment by Hans Stricker Hans Stricker 2013-03-14T18:46:53Z 2013-03-14T18:46:53Z @Ryan: Great! (Didn't expect to get such a concise answer.) http://mathoverflow.net/questions/124542/probing-a-manifold-with-closed-curves/124543#124543 Comment by Hans Stricker Hans Stricker 2013-03-14T18:13:07Z 2013-03-14T18:13:07Z @Ryan: Can you tell - by the way - how to aquire a power of imagination like yours (which I envy you for). How can one come to <i>see</i> at once that &quot;the central circle of a Moebius band self-intersects once with many transverse perturbations of itself&quot;? (I assume this is something you <i>see</i>, but for ordinary people like me it requires a very lot of <i>thinking about it</i>.) The same question - by the way - goes to Mariano's first comment to my original question. So you don't have to take it too serious. http://mathoverflow.net/questions/124542/probing-a-manifold-with-closed-curves/124543#124543 Comment by Hans Stricker Hans Stricker 2013-03-14T17:54:14Z 2013-03-14T17:54:14Z Sorry, I just added &quot;orientable&quot; (thanks to maproom's hint). Is your additional assumption equivalent to require the manifold to be orientable? http://mathoverflow.net/questions/124542/probing-a-manifold-with-closed-curves Comment by Hans Stricker Hans Stricker 2013-03-14T17:51:34Z 2013-03-14T17:51:34Z @maproom: Indeed, I had only orientable surface in mind. So I added it. Thanks for the correction! http://mathoverflow.net/questions/81622/probing-a-manifold-with-geodesics Comment by Hans Stricker Hans Stricker 2013-03-12T09:34:37Z 2013-03-12T09:34:37Z @fedja: Is it, that one just has to show that two such loops - with just one intersection which is not a &quot;kissing&quot; - cannot be homotopic? Such that the fundamental group is not trivial? http://mathoverflow.net/questions/81622/probing-a-manifold-with-geodesics Comment by Hans Stricker Hans Stricker 2013-03-10T16:34:41Z 2013-03-10T16:34:41Z Is there someting new concerning fedja's interesting comment, in the meanwhile? http://mathoverflow.net/questions/1714/best-online-mathematics-videos/1766#1766 Comment by Hans Stricker Hans Stricker 2013-03-04T23:23:46Z 2013-03-04T23:23:46Z It's not only nice: it's great! http://mathoverflow.net/questions/123243/isometric-but-differently-shaped-closed-surfaces-in-mathbbr3/123266#123266 Comment by Hans Stricker Hans Stricker 2013-03-04T09:46:05Z 2013-03-04T09:46:05Z Robert Connelly's flexible polyhedron at IHES: <a href="http://www.youtube.com/watch?v=reO7Qx3HiIg&amp;feature=youtube_gdata_player" rel="nofollow">youtube.com/&hellip;</a>