User gabe cunningham - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T01:14:42Z http://mathoverflow.net/feeds/user/913 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/1886/infinite-matrices Infinite matrices Gabe Cunningham 2009-10-22T15:40:21Z 2013-02-06T12:30:25Z <p>Suppose we have an infinite matrix A = (a<sub>ij</sub>) (i, j positive integers). What is the "right" definition of determinant of such a matrix? (Or does such a notion even exist?) Of course, I don't necessarily expect every such matrix to have a determinant -- presumably there are questions of convergence -- but what should the quantity be? The problem I have is that there are several ways of looking at the determinant of a finite square matrix, and it's not clear to me what the "essence" of the determinant is.</p> http://mathoverflow.net/questions/1890/describe-a-topic-in-one-sentence Describe a topic in one sentence. Gabe Cunningham 2009-10-22T16:28:18Z 2012-10-07T22:16:28Z <p>When you study a topic for the first time, it can be difficult to pick up the motivations and to understand where everything is going. Once you have some experience, however, you get that good high-level view (sometimes!) What I'm looking for are good one-sentence descriptions about a topic that deliver the (or one of the) main punchlines for that topic.</p> <p>For example, when I look back at linear algebra, the punchline I take away is "Any nice function you can come up with is linear." After all, multilinear functions, symmetric functions, and alternating functions are essentially just linear functions on a different vector space. Another big punchline is "Avoid bases whenever possible."</p> <p>What other punchlines can you deliver for various topics/fields?</p> http://mathoverflow.net/questions/1785/how-do-you-keep-your-research-notes-organized How do you keep your research notes organized? Gabe Cunningham 2009-10-22T02:38:40Z 2012-05-18T12:46:44Z <p>One of the things I struggle with most in doing research is keeping my notes organized. Since research tends to do a lot of branching, keeping notes in a linear fashion seems useless to me. On the other hand, this means that I end up with several notebooks that have ideas and dead-ends everywhere. Then if I want to piece parts together, or if I eventually want to go back and re-investigate what looked like a dead-end at the time -- perhaps because I have learned some new tool -- it takes me a long time to find what I am looking for. How do other people surmount these obstacles?</p> http://mathoverflow.net/questions/26297/which-finite-groups-are-generated-by-n-involutions Which finite groups are generated by n involutions? Gabe Cunningham 2010-05-28T18:35:14Z 2011-04-05T08:19:39Z <p>One of the interesting problems in abstract polytope theory is to determine, for a given finite group, when that group is the automorphism group of a regular abstract polytope. This is equivalent to the following question: Given a finite group G, when is G generated by involutions $\rho_0, \ldots, \rho_n$ such that $(\rho_i \rho_j)^2 = 1$ if $|i - j| \geq 2$ and such that for all $I, J \subset \{0, \ldots, n\}$ we have $\langle \rho_i \mid i \in I \rangle \cap \langle \rho_i \mid i \in J \rangle = \langle \rho_i \mid i \in I \cap J \rangle$?</p> <p>The last property can be difficult to check, so let's relax that requirement for now. If a finite group G is generated by n involutions such that non-adjacent generators commute, what can we say about the structure or size of G? Of particular interest: what if G is simple?</p> <p>Here are a few simple observations:</p> <ol> <li>For each n, the smallest (abstract) n-polytope has an automorphism group that is isomorphic to the direct product of n copies of $C_2$, corresponding to the trivial Coxeter diagram on n nodes. So a finite group G cannot be the automorphism group of an abstract regular n-polytope for $n > \log_2(|G|)$. </li> <li>A (nontrivial) group generated by involutions has even order.</li> <li>The abelianization of a group generated by involutions such that nonadjacent generators commute is a quotient of the group in (1), the direct product of n copies of $C_2$.</li> </ol> http://mathoverflow.net/questions/45951/sexy-vacuity/46150#46150 Answer by Gabe Cunningham for Sexy vacuity .... Gabe Cunningham 2010-11-15T20:22:29Z 2010-11-15T20:22:29Z <p>Over the reals, $\sup \emptyset = -\infty$ and $\inf \emptyset = \infty$.</p> http://mathoverflow.net/questions/44165/reference-request-lattice-operations-on-the-class-of-finitely-presented-groups Reference request: lattice operations on the class of finitely presented groups Gabe Cunningham 2010-10-29T18:28:18Z 2010-11-04T23:15:06Z <p>In my research, I work with certain finitely presented quotients of Coxeter groups. These are the automorphism groups of abstract polytopes, which are combinatorial generalizations of "usual" polytopes. (Essentially, an abstract polytope is an incidence complex.) Now, in this context, there is a useful combinatorial operation that has a nice effect on the automorphism groups. In fact, it's easy to generalize the operation on groups, so I'm curious whether any work has been done with this.</p> <p>Let $G = \langle X \mid R \rangle$ and $H = \langle X \mid S \rangle$ be finitely presented groups. (I'm not sure that the finiteness of the presentation is essential, but let's assume it for now.) In other words, we have that $G = F(X) / \overline{R}$ and $H = F(X) / \overline{S}$, where $F(X)$ is the free group on $X$ and $\overline{R}$ is the normal closure of $R$ in $F(X)$. Then if $K$ naturally covers $G$ and $H$ (that is, if the identity map on $X$ extends to (surjective) homomorphisms from $K$ to $G$ and $K$ to $H$), we have that $K$ covers the group $F(X) / (\overline{R} \cap \overline{S})$. Similarly, if $G$ and $H$ naturally cover $K$, then the group $F(X) / (\overline{R} \overline{S})$ with presentation $\langle X \mid R \cup S \rangle$ naturally covers $K$ as well.</p> <p>Therefore, the group $F(X) / (\overline{R} \cap \overline{S})$ is the minimal natural cover of $G$ and $H$, and $F(X) / (\overline{R} \overline{S})$ is the maximal natural quotient of $G$ and $H$. The first group is the fibre product of $G$ and $H$ over the second group. </p> <p>These seem like such natural operations that I would guess they have been studied before, but I am having trouble finding anything. Any references would be greatly appreciated.</p> <p>EDIT: Let me expand a little bit. As Mark Sapir points out, I'm essentially looking at the lattice of normal subgroups of F(X). I've looked a little at the general theory of subgroup lattices, but I'm really only interested in normal subgroups of F(X) or other finitely presented groups. Also, I find it difficult to work with the normal subgroups of F(X) directly, whereas it's not too hard to work with the quotients by these normal subgroups via presentations. So I'm hoping to find something that's somewhat more specific than just subgroup lattices; ideally, something that works with presentations.</p> <p>Here are some examples of the type of questions I'm interested in:</p> <ol> <li>Is there a simple way to write down a presentation for $F(X) / (\overline{R} \cap \overline{S})$ without changing the generators?</li> <li>Given G and H, what are some conditions on the relations under which $\overline{R} \overline{S} = F(X)$? (This obviously won't be all-inclusive, but some instructive examples would be nice.)</li> </ol> http://mathoverflow.net/questions/32179/common-quotients-of-direct-products Common quotients of direct products Gabe Cunningham 2010-07-16T14:34:24Z 2010-10-01T02:51:00Z <p>In a paper I read recently, the authors use the fact that if two groups G and H have no nontrivial common quotient, then neither do GxG and HxH. It's unclear from the context whether this is true for all groups, or just groups of the type that are important for this paper, and they don't prove the claim.</p> <p>I've been trying to prove the statement for general groups without any success. Is it true?</p> <p>I've also tried restricting to the case I really care about, as follows. Let K, K', M, and M' be normal subgroups of U, with $U/K \simeq U/K'$ and $U/M \simeq U/M'$. Suppose that KK' = KM = KM' = K'M = K'M' = MM' = U. Is it also the case that $(K \cap K')(M \cap M') = U$? The reason this is connected to the above is that $U/(K \cap K')(M \cap M')$ is a common quotient of $U/(K \cap K') \simeq U/K \times U/K' \simeq U/K \times U/K$ and $U/(M \cap M') \simeq U/M \times U/M' \simeq U/M \times U/M$. So if U/K and U/M have no nontrivial common quotients, is the same true of $U/(K \cap K')$ and $U/(M \cap M')$?</p> <p>Finally, if the result is not true in general for either of these cases, what about if we restrict to finite groups G and H?</p> http://mathoverflow.net/questions/26439/minimal-basis-of-set-of-positive-integers/26480#26480 Answer by Gabe Cunningham for Minimal basis of set of positive integers Gabe Cunningham 2010-05-30T18:19:02Z 2010-05-30T18:19:02Z <p>Consider a set that starts with 0, 1, 2, ..., n, 2n+1, 3n+2, .... For each k from 0 to n, the only way to get 2n+1+k is by summing 2n+1 with k, so we can't remove any of the terms 0, 1, ..., n. Then this set contains a minimal basis that begins with 0, 1, ..., n, and there are arbitrarily "small" minimal bases.</p> http://mathoverflow.net/questions/24674/negative-vs-minus/24679#24679 Answer by Gabe Cunningham for "negative" vs "minus" Gabe Cunningham 2010-05-15T01:22:15Z 2010-05-15T01:22:15Z <p>To me, "minus x" sounds awkward because operators are grammatically close to verbs. "Negative", on the other hand, is clearly an adjective, so even with the given defects, I think "negative x" sounds much better than "minus x".</p> <p>That said, I echo Steven Gubkin's comment above; if I'm worried that "negative x" will be misinterpreted, I use "the opposite of x". (I don't know how other people were taught, but I learned the the numerator for the quadratic formula began with "the opposite of b...")</p> http://mathoverflow.net/questions/23743/what-are-some-good-group-theory-references What are some good group theory references? Gabe Cunningham 2010-05-06T16:39:13Z 2010-05-07T05:53:40Z <p>I'm curious about what books people use for a group theory reference. I don't currently own a dedicated group theory book, and I think I'd find such a book very helpful in my research. What is your favorite book on group theory? Please tell us why you like it -- and what sort of groups it focuses on (finite? discrete? finitely generated? etc.)</p> <p>(For my part, I'm interested mainly in discrete, finitely generated groups, but I enjoy the "flavor" of general group theory books more than combinatorial group theory books.)</p> http://mathoverflow.net/questions/23692/what-are-the-normal-subgroups-of-a-direct-product What are the normal subgroups of a direct product? Gabe Cunningham 2010-05-06T11:39:32Z 2010-05-06T14:37:33Z <p>Let $N$ be a normal subgroup of $G \times H$, and let $\pi_1: G \times H \to G$ and $\pi_2: G \times H \to H$ be the canonical projections. Then $\pi_1(N)$ is normal in $G$ and $\pi_2(N)$ is normal in $H$. What else can we say? I know that it is not true, in general, that $N \simeq \pi_1(N) \times \pi_2(N)$.</p> <p>I'm particularly interested in the case where $G$ and $H$ are simple. In that case, $N \simeq \pi_1(N) \times \pi_2(N)$ except possibly in the case where $\pi_1(N) = G$ and $\pi_2(N) = H$. In that case, what do we know?</p> http://mathoverflow.net/questions/7721/what-methods-exist-to-prove-that-a-finitely-presented-group-is-finite What methods exist to prove that a finitely presented group is finite? Gabe Cunningham 2009-12-03T22:24:31Z 2010-02-11T18:42:06Z <p>Suppose I have a finitely presented group (or a family of finitely presented groups with some integer parameters), and I'd like to know if the group is finite. What methods exist to find this out? I know that there isn't a general algorithm to determine this, but I'm interested in what plans of attack do exist. </p> <p>One method that I've used with limited success is trying to identify quotients of the group I start with, hoping to find one that is known to be infinite. Sometimes, though, your finitely presented group doesn't have many normal subgroups; in that case, when you add a relation to get a quotient, you may collapse the group down to something finite.</p> <p>In fact, there are two big questions here:</p> <ol> <li>How do we recognize large finite simple groups? (By "large" I mean that the Todd-Coxeter algorithm takes unreasonably long on this group.) What about large groups that are the extension of finite simple groups by a small number of factors?</li> <li>How do we recognize infinite groups? In particular, how do we recognize infinite simple groups?</li> </ol> <p>(For those who are interested, the groups I am interested in are the symmetry groups of abstract polytopes; these groups are certain nice quotients of string Coxeter groups or their rotation subgroups.)</p> http://mathoverflow.net/questions/6583/which-changes-of-metric-fix-all-open-balls-of-a-metric-space Which changes of metric fix all open balls of a metric space? Gabe Cunningham 2009-11-23T16:22:16Z 2010-01-21T03:26:31Z <p>In an earlier question, I was interested in counting the number of metric spaces on N points, where I considered two metric spaces to be the same if they had the same collection of open balls. Two questions:</p> <ol> <li>What are the usual notions of metric space equivalence? Are any of them nontrivial for finite metric spaces? (For instance, the obvious one that two metric spaces are equivalent if their topology is the same is trivial for finite vector spaces).</li> <li>If we say that two (labeled) metric spaces are equivalent if they have the same collection of open balls, in what ways can we operate on the metric d such that we get the same collection of open balls?</li> </ol> <p>For example, any two metric spaces on 2 points are equivalent in this way, so any allowable operation on metrics yields an equivalent metric space. Clearly we can always scale d by any positive real without changing the equivalence class. Now consider the two metric spaces</p> <pre><code>x ---3--- y x ---3--- y \ / \ / 3 4 4 5 \ / \ / z z </code></pre> <p>In both cases, the nontrivial open balls are {x, z} and {x, y}, so these metric spaces are equivalent. How can I describe the operation I'm performing on d in general terms? What other operations on d will yield metric spaces that are equivalent in this sense?</p> http://mathoverflow.net/questions/6925/how-can-we-formalize-the-naturality-of-certain-characteristic-subgroups How can we formalize the naturality of certain characteristic subgroups? Gabe Cunningham 2009-11-27T00:59:35Z 2010-01-08T14:59:14Z <p>I'm trying to get a better handle on characteristic subgroups, and many nice examples are given with some sort of "natural" definition. For example, it's clear that the center, torsion subgroup, and commutator subgroup of a given group are all characteristic, just because of the way they are defined. How can we formalize this "naturality"? The latter two have functors associated to them, but I'm not entirely certain if that's the reason for the subgroups being characteristic (and there isn't a similar functor for the center of a group).</p> <p>EDIT: Let me explain why I'm asking this question. It's useful to know how various characteristic subgroups interact with direct products, quotients, and other group constructions. Henry's construction below is perhaps the right formalism, but it isn't clear to me what extra that buys us. On the other hand, knowing that a certain subgroup arises as the image of a functor means that we ought to be able to use categorical considerations to determine some properties of this subgroup. So here are a few related questions:</p> <ol> <li>Is it possible to use definable subsets to see how characteristic subgroups act with respect to direct product, quotients, or other group constructions?</li> <li>What properties does the Comm functor on Grp have? What about the Tors functor on Ab? Furthermore, which properties does a functor have to have in order to define a characteristic subgroup? (Other than the obvious property that F(G) is a subgroup of G for all G!)</li> </ol> http://mathoverflow.net/questions/1677/number-of-metric-spaces-on-n-points Number of metric spaces on N points Gabe Cunningham 2009-10-21T15:27:57Z 2009-12-22T18:40:16Z <p>Given X = {x1, ..., xn}, how many collections C of subsets of X are there such that C is the listing of all open balls of some metric space?</p> <p>The first nontrivial example is n=3; let's call the points x, y, and z. Also, let a = d(x, y), b = d(y, z), and c = d(z, x). For any collection C to be a listing of all the open balls, it must contain all the singleton sets and the whole set X. Let C0 = {{x},{y},{z},X}. If x, y, and z are equidistant, these are the only open balls. If, say, a &lt; b &lt; c, then we get C = C0 \union {{x,y},{y,z}}. Through careful case enumeration, we can answer this for small n, but the process quickly becomes unwieldy. Has anyone ever looked at this before, and is there a recursive formula or a generating function for this? What about if the points are unlabeled? For n=3, I count 7 possibilities for C if the points are labeled, and 3 if the points are unlabeled. It's already somewhat time-consuming to count when n=4.</p> http://mathoverflow.net/questions/9478/searching-the-symmetric-group Searching the symmetric group Gabe Cunningham 2009-12-21T15:24:43Z 2009-12-21T17:25:55Z <p>You want to design a set of yes/no questions for quickly searching the symmetric group. The questions have to be of the form "Does your permutation move $a_1$ to $b_1$ or $a_2$ to $b_2$ or ... or $a_k$ to $b_k$?" Given a random permutation, you will ask all of your questions about that permutation, and your goal is to know what the permutation is once you know the answers to your questions.</p> <p>In particular, note that you aren't allowed to have later questions depend on the answers to earlier questions. You always ask the same questions of whichever random permutation you get.</p> <p>Here's a slightly different way to phrase the type of question you're allowed to ask. If we represent the elements of the symmetric group by permutation matrices, then the questions you can ask involve picking a (0,1)-matrix and asking if the random permutation matrix has any 1 where your chosen matrix has a 1 (i.e., if I apply the componentwise AND function to the random matrix and your chosen matrix, do I get 0 or 1?)</p> <p>Two questions:</p> <ol> <li>How many such questions are needed in order to search $S_n$? How much bigger than $log_2 n!$ is this?</li> <li>Are there "good" strategies for designing sets of questions that can be used for any n? Can we achieve the minimum number of questions for each n with a single strategy?</li> </ol> http://mathoverflow.net/questions/2851/algorithmic-combinatorics-resources Algorithmic Combinatorics resources? Gabe Cunningham 2009-10-27T17:41:11Z 2009-11-30T19:47:00Z <p>Some branches of combinatorics lend themselves naturally to algorithms; graph theory is a natural example. However, straight-up enumerative combinatorics relies much more on analytic and algebraic methods. As a result, you often don't have any idea of how you could systematically generate the objects you want. In order to do so with a computer, you'd need to pick a "nice" encoding, impose a total order on the coded objects, and then generate them one-by-one, checking against previous ones for whatever equivalence you're concerned with. Burnside's or Polya's Theorem will tell you when you've found them all, but in the meanwhile, you want an encoding and an order that's easy to work with, and that generates successive objects fairly quickly.</p> <p>Are there any good resources for this sort of algorithmic combinatorics? I'd like something that's not tailored to just a specific problem, unless that problem is representative of a large class of problems. Essentially, I'd like to know some particularly useful encodings and some useful generation algorithms. (For instance, for some problems, it might be more efficient to generate the objects randomly instead of using a total order).</p> http://mathoverflow.net/questions/6081/ways-to-synthesize-topics-in-linear-algebra/6122#6122 Answer by Gabe Cunningham for Ways to Synthesize Topics in Linear Algebra Gabe Cunningham 2009-11-19T16:07:06Z 2009-11-19T18:10:38Z <p>When you first start learning about vector spaces, you tend to think of vectors as just being some ordered collection of numbers. This is very natural, since the first vector spaces you deal with are $\mathbb R^n$ and $\mathbb C^n$.</p> <p>Then, at some point, someone suggests to you that your vectors don't have to be collections of numbers -- they could be continuous functions! It makes sense to add two functions in $n$ real variables or to scale such a function by a real number, and each function has an additive inverse, and so forth, so they form a real vector space. All the real (finite-dimensional) vector spaces you had encountered before were isomorphic to $\mathbb R^n$ for some $n$. Now, have we discovered a "genuinely new" vector space, or is this just $\mathbb R^n$ in another guise?</p> <p>Well, things are a little wild if we let our functions be anything at all; let's look at the vector space formed just by the linear maps from $\mathbb R^n$ to $\mathbb R$. Now you can fix a basis for $\mathbb R^n$ and look at the dual basis, and convince yourself that the dual basis really is a basis of the vector space of linear maps. Finally, you just have to convince yourself that the map taking a basis vector $b_i$ of $\mathbb R^n$ to its dual <code>$b_i^*$</code> is actually an isomorphism. So any time we want an $n$-dimensional real vector space, we can use <code>$(\mathbb R^n)^*$</code> instead of $\mathbb R^n$, which is often convenient.</p> <p>Chances are, the dual space wasn't motivated this way to you -- people often start with a basic vector space like $\mathbb R^n$ and dualize it for no apparent reason, except to show you that you get a vector space. I think it's more informative to start with the space of linear maps, since that's a very natural space to discover.</p> <p>(Of course, you may have learned already that every finite-dimensional real vector space is isomorphic to $\mathbb R^n$ for some $n$. But this approach really lets you get your hands on the spaces.)</p> http://mathoverflow.net/questions/6083/how-to-sufficiently-motivate-organization-of-proofs-in-math-books/6113#6113 Answer by Gabe Cunningham for How to sufficiently motivate organization of proofs in math books Gabe Cunningham 2009-11-19T15:32:09Z 2009-11-19T15:32:09Z <p>I think the advantages and disadvantages depend on the purpose of the book and on the specific subject area. Some books are reference books, packed with results, and it's okay to be light on the motivation there. However, if you want or expect people to follow your proofs line-by-line, a little motivation goes a long way. Personally, when I'm reading (or listening to) a proof, I cringe whenever I see a magic number or function pulled out of thin air that "just happens" to work. Tell me what you were looking for and how you found it. You don't have to go into excruciating detail, but a little context makes it much clearer, and it helps make sure the reader knows how to find these "magic" functions in his or her own research.</p> <p>I think almost any field of math would benefit from a conversational introductory book. Why are certain theorems important? Which ones are routine but tedious, and which ones rely on very novel ideas? What are some good heuristics in the field, and when do they break down? Which theorems generalize nicely, and for which is it not known? Reading such a book would be like discussing the field with a knowledgeable colleague, and I think it would encourage mathematicians to learn more math outside their field.</p> <p>Come to think of it, the Princeton Companion achieved this somewhat. Now let's expand each article to a whole book!</p> http://mathoverflow.net/questions/4442/is-there-a-theorem-that-says-that-there-is-always-more-than-one-way-to-continue/4456#4456 Answer by Gabe Cunningham for Is there a theorem that says that there is always more than one way to "continue a finite sequence"? Gabe Cunningham 2009-11-06T22:32:53Z 2009-11-06T22:32:53Z <p>Here's how I think about this. Suppose you're given n terms <img src="http://latex.mathoverflow.net/png?a%5F1%2C%20%2E%2E%2E%2C%20a%5Fn%2E" alt="a\sb 1, ..., a\sb n." title="" /> Define <img src="http://latex.mathoverflow.net/png?b%5Fi%20%3D%20a%5Fi%20%2F%20%28i%2D1%29%5Ccdots%28i%2Dn%29" alt="b\sb i = a\sb i / (i-1)\cdots(i-n)" title="" />, where the denominator skips the factor <img src="http://latex.mathoverflow.net/png?%28i%2Di%29" alt="(i-i)" title="" />. Consider the function f(x) = <img src="http://latex.mathoverflow.net/png?b%5F1%28x%2D2%29%28x%2D3%29%5Ccdots%28x%2Dn%29%20%2B%20%28x%2D1%29b%5F2%28x%2D3%29%5Ccdots%28x%2Dn%29%20%2B%20%5Ccdots%20%2B%20%28x%2D1%29%5Ccdots%28x%2Dn%2B1%29b%5Fn" alt="b\sb 1(x-2)(x-3)\cdots(x-n) + (x-1)b\sb 2(x-3)\cdots(x-n) + \cdots + (x-1)\cdots(x-n+1)b\sb n" title="" /> <img src="http://latex.mathoverflow.net/png?%2B%20%28x%2D1%29%5Ccdots%28x%2Dn%29c" alt="+ (x-1)\cdots(x-n)c" title="" />. Then for i an integer between 1 and n, <img src="http://latex.mathoverflow.net/png?f%28i%29%20%3D%20b%5Fi%20%2A%20%28i%2D1%29%20%2A%20%5Ccdots%20%2A%20%28i%2Dn%29" alt="f(i) = b\sb i \ast (i-1) \ast \cdots \ast (i-n)" title="" />, except that you skip the factor (i-i). Thus, f(i) = a<sub>i</sub>. But by changing c, you can make the next term f(n+1) whatever you want.</p> http://mathoverflow.net/questions/4086/does-every-finitely-generated-group-have-a-maximal-normal-subgroup/4090#4090 Answer by Gabe Cunningham for Does every finitely generated group have a maximal normal subgroup? Gabe Cunningham 2009-11-04T14:39:47Z 2009-11-04T14:39:47Z <p>Assuming you mean "does a maximal normal subgroup always exist?" (and that you don't care about computing it), here is a way to restate the problem. Notice that if G has no maximal normal subgroups, that means that every proper normal subgroup H of G is contained in a larger proper normal subgroup K of G. In particular, this means that the group G/H must not be finite; if it were, we could only find a finite chain of normal subgroups between H and G. So the question "does a maximal normal subgroup always exist" is the same as "must a finitely generated group have any finite nontrivial quotients?" I'm not sure what the answer to that is, but it seems like a useful restatement.</p> http://mathoverflow.net/questions/2380/number-of-subdivisions-of-an-n-gon Number of subdivisions of an n-gon Gabe Cunningham 2009-10-24T22:07:59Z 2009-10-29T22:25:16Z <p>Suppose I have a regular n-gon. I want to draw some noncrossing diagonals to subdivide it into smaller polygons. In how many ways can I do this? The vertices are unlabeled, so I don't distinguish between rotations or reflections of a given subdivision.</p> <p>A triangle has 1 subdivision (do nothing!); a square has 2, a pentagon has 3, and a hexagon has at least 9 -- I'm not certain that I haven't missed any.</p> <p>In fact, what I would really like is not just a count, but an algorithm for generating such subdivisions. There are obvious algorithms that generate some subdivisions multiple times, but what I'd really like is an algorithm that only generates distinct subdivisions, and that generates all of them.</p> http://mathoverflow.net/questions/3079/most-helpful-heuristic/3174#3174 Answer by Gabe Cunningham for Most helpful heuristic? Gabe Cunningham 2009-10-29T00:10:54Z 2009-10-29T00:10:54Z <p>Any identity of power series that holds for sufficiently small values of the variable(s) also holds for formal power series.</p> http://mathoverflow.net/questions/2981/has-anyone-studied-the-applications-which-map-open-sets-to-either-open-or-closed/3056#3056 Answer by Gabe Cunningham for Has anyone studied the applications which map open sets to either open or closed sets? Gabe Cunningham 2009-10-28T14:04:50Z 2009-10-28T14:04:50Z <p>Properties 1 and 2 seem difficult to work with, because the class of functions satisfying one of those properties isn't closed under composition. Similarly for properties 1' and 2'. However, the class of functions satisfying property 3 or 3' is closed under composition.</p> <p>It's hard to say what would be useful about these functions -- the question is, are there any useful properties of a topological space that are preserved by these functions? It seems to me that any such properties will have to be similarly wishy-washy; that is, that they will also have to depend on something being open or closed, but not caring which.</p> <p>Do you have a property in mind that makes these functions natural candidates?</p> http://mathoverflow.net/questions/2650/conjugacy-classes-in-finite-groups-that-remain-conjugacy-classes-when-restricted/2654#2654 Answer by Gabe Cunningham for Conjugacy classes in finite groups that remain conjugacy classes when restricted to proper subgroups Gabe Cunningham 2009-10-26T18:01:46Z 2009-10-26T18:01:46Z <p>Of course, if G is abelian, then the conjugacy classes of G are just the elements, and any pair (G, c) is nonsplitting. More generally, if x is in the center of G and c is the class of x, then (G, c) is nonsplitting. So the answer to Question 1 is yes. But I imagine that you're looking for more interesting examples to questions 1 and 2!</p> http://mathoverflow.net/questions/2358/most-harmful-heuristic/2371#2371 Answer by Gabe Cunningham for Most harmful heuristic? Gabe Cunningham 2009-10-24T21:39:53Z 2009-10-24T21:39:53Z <p>Not sure if this qualifies exactly, but I can never remember which theorems of group theory apply to finite groups, and which ones apply to groups in general. Anytime I remember a result, I have this sinking feeling that it appears in a textbook preceded by "for the remainder of this section, let G be a finite group." I'm not sure how well-founded this fear is (other than the theorems that obviously don't make sense for infinite groups, like the Sylow theorems).</p> http://mathoverflow.net/questions/1677/number-of-metric-spaces-on-n-points/1884#1884 Answer by Gabe Cunningham for Number of metric spaces on N points Gabe Cunningham 2009-10-22T15:21:00Z 2009-10-23T13:17:12Z <p>Well, here's a start. Suppose we have n points, and let k = n(n-1)/2. Thus there are k distances we have to pick. Let's take all of our distances to lie in the set {k+1, k+2, ..., 2k}, so that we don't have to worry about the triangle inequality. Now, the collection of open balls depends on how many times each distance is repeated. However, translating the set by an integer doesn't affect the collection of open balls. That is, if D is the multiset of distances, and C<sub>D</sub> is the collection of open balls induced by D, then C<sub>D+1</sub> = C<sub>D</sub>, where by D+1 I mean add 1 to each element of D. This is true because of what javier says: to find the collection of open balls at a point, we just start at that point and increase the radius of the ball by 1 at each step, writing down each open ball we get and stopping when we get the whole set.</p> <p>The upshot of this is that if k+1 is not the smallest element of D, we can translate D such that k+1 is the smallest element, without affecting the open ball structure. In fact, I'm pretty sure a stronger statement is true: if D has a gap, we can slide down the upper part of D to close that gap without affecting the open ball structure. That is, if r and s are elements of D such that r &lt; s-1 and there are no elements of D strictly between r and s, then we can translate s and everything above it down by 1. If D has r distinct values, then by doing such translations, we can get D to be a multiset with values in {k+1, ..., k+r}.</p> <p>Thus, if we have a particular open ball structure C on n points, we can find a multiset D with the above properties such that C = C<sub>D</sub>. So the number of such multisets provides an upper bound on the number of (unlabeled) metric spaces. I have no idea how good this bound is, but let's calculate it.</p> <p>Let f<sub>r</sub>(k) = # of multisets with k elements taking values from {k+1, ..., k+r}, and taking each value at least once. So we essentially have k-r free slots in D, and r different values, so the number of such multisets is the binomial coefficient B((k-r)-(r-1), r-1) = B(k-1, r-1). Then the total number of multisets is f<sub>1</sub>(k) + ... + f<sub>k</sub>(k) = 2^{k-1}.</p> <p>Now, that looks pretty huge: 2^{(n+1)(n-2)/2}. On the other hand, there are 2^{2^n} collections of subsets of X, and even when you account for the fact that you have to include all the singletons and the whole set, 2^{2^n - (n+1)} is hardly an improvement. For n=3, our new upper bound is 4, and the true value is 3, since the multisets {3, 4, 5} and {3, 3, 4} give the same open ball structure.</p> <p>Can anyone expand on this?</p> <p>Edit: On further reflection, there are multiple different open ball structures induced by a multiset. For example, if n=4 and D = {7, 7, 7, 7, 8, 8}, then we get different structures if we assign the two distances of 8 to the same vertex or to different vertices. So it seems we should look at ordered k-tuples instead of multisets, which makes our upper bound much larger (bigger than k factorial). So maybe this is less useful than I thought. But at least after thinking about it like this, I might have simplified the problem enough that I can write a program to calculate the next few values of the sequence.</p> http://mathoverflow.net/questions/1977/why-is-the-gradient-normal/1995#1995 Answer by Gabe Cunningham for Why is the gradient normal? Gabe Cunningham 2009-10-23T00:44:15Z 2009-10-23T00:44:15Z <p>If you are standing on a level set and want to walk some small distance d and get as far as possible from the level set, you want to walk along the normal. Otherwise, if the path you take has a tangent component, it will tend to keep you closer to the level set if d is small enough compared to the size of the level set. Furthermore, getting as far as possible from your level set is approximately the same as walking to the highest/lowest level curve in range, with the approximation improving as d shrinks.</p> http://mathoverflow.net/questions/1083/do-good-math-jokes-exist/1751#1751 Answer by Gabe Cunningham for Do good math jokes exist? Gabe Cunningham 2009-10-21T22:34:04Z 2009-10-21T22:34:04Z <p>There are 10 types of people in the world: those who understand binary, and 9 others.</p> http://mathoverflow.net/questions/44165/reference-request-lattice-operations-on-the-class-of-finitely-presented-groups/44189#44189 Comment by Gabe Cunningham Gabe Cunningham 2010-11-05T14:55:20Z 2010-11-05T14:55:20Z Thanks for your updated answer. Regarding your last paragraph: I know there's no general algorithm to determine whether $\overline{R} \overline{S} = F_k$. What I'm interested in is finding simple conditions on R and S that are sufficient to guarantee that $\overline{R} \overline{S} = F_k$. But maybe that question is too broad to have a good answer. Mostly, I was just hoping to find a treatment of the normal subgroup lattice of F(X) from the point of view of presentations of the quotient groups. http://mathoverflow.net/questions/39760/is-the-infinite-infinite-or-even-infiniter Comment by Gabe Cunningham Gabe Cunningham 2010-09-23T16:21:54Z 2010-09-23T16:21:54Z Because 'infinite' means 'not finite'. When we care to distinguish between the size or order-type of infinite sets (which is fairly rare), then we use the language of the infinite cardinals or ordinals. There's no need to try to shoehorn all this information into the word 'infinite'. http://mathoverflow.net/questions/26236/the-harmonic-series-beetle-live-illustrations-of-mathematical-theorems/33276#33276 Comment by Gabe Cunningham Gabe Cunningham 2010-07-25T13:58:08Z 2010-07-25T13:58:08Z All this shows is that 1 + (1/2)(2) + (1/3)(3) + ... grows without bound. http://mathoverflow.net/questions/32862/settling-a-circular-argument-room-for-one-more Comment by Gabe Cunningham Gabe Cunningham 2010-07-22T14:12:18Z 2010-07-22T14:12:18Z I'd suggest retitling the question since &quot;circular argument&quot; usually doesn't refer to an argument about circles! http://mathoverflow.net/questions/32179/common-quotients-of-direct-products/32185#32185 Comment by Gabe Cunningham Gabe Cunningham 2010-07-16T15:29:52Z 2010-07-16T15:29:52Z Thanks, Tom. I had gotten as far as AB = Q but hadn't seen how to finish the argument. http://mathoverflow.net/questions/31528/game-avoid-the-gaussian-primes Comment by Gabe Cunningham Gabe Cunningham 2010-07-12T13:12:33Z 2010-07-12T13:12:33Z Sorry, I should say not the closest, but the closest Gaussian integer to the lower-left; i.e., take the floor of each component. http://mathoverflow.net/questions/31528/game-avoid-the-gaussian-primes Comment by Gabe Cunningham Gabe Cunningham 2010-07-12T13:10:43Z 2010-07-12T13:10:43Z It says above that the midpoint of z+w is [(z+w)2], which is supposed to mean the closest Gaussian integer to (z+w)/2. http://mathoverflow.net/questions/26297/which-finite-groups-are-generated-by-n-involutions/26301#26301 Comment by Gabe Cunningham Gabe Cunningham 2010-05-31T15:48:05Z 2010-05-31T15:48:05Z Thank you for the detailed answer, Jack. I'll look into these to see if the satisfy the intersection property as well. http://mathoverflow.net/questions/26297/which-finite-groups-are-generated-by-n-involutions Comment by Gabe Cunningham Gabe Cunningham 2010-05-31T03:23:39Z 2010-05-31T03:23:39Z In 1), I meant the smallest abstract polytope, not the smallest convex polytope. In 2), I meant nontrivial. I'll update to clarify. http://mathoverflow.net/questions/26297/which-finite-groups-are-generated-by-n-involutions/26298#26298 Comment by Gabe Cunningham Gabe Cunningham 2010-05-29T01:44:46Z 2010-05-29T01:44:46Z It is only with the last condition that they classify the automorphism groups of regular abstract polytopes. In fact, you can build a regular abstract polytope out of such a group out of the cosets of certain subgroups. The precise construction is given in chapter 2 of Schulte and McMullen's book Abstract Regular Polytopes. http://mathoverflow.net/questions/26297/which-finite-groups-are-generated-by-n-involutions/26298#26298 Comment by Gabe Cunningham Gabe Cunningham 2010-05-28T19:26:06Z 2010-05-28T19:26:06Z The groups I'm dealing with are not just Coxeter groups but rather quotients of Coxeter groups. Also, the question here isn't so much &quot;when are these groups finite&quot; but &quot;which finite groups arise this way&quot;. For example: is there a presentation for the Monster group satisfying the above criteria? http://mathoverflow.net/questions/3973/what-should-be-offered-in-undergraduate-mathematics-thats-currently-not-or-isn/25302#25302 Comment by Gabe Cunningham Gabe Cunningham 2010-05-20T01:17:07Z 2010-05-20T01:17:07Z +1 for &quot;Computer science isn't programming&quot; and the great Dijkstra quote. http://mathoverflow.net/questions/24101/natural-generating-sets-for-symmetric-groups Comment by Gabe Cunningham Gabe Cunningham 2010-05-10T15:20:09Z 2010-05-10T15:20:09Z The definition I'm familiar with is that a permutation p is a shuffle if the sequence p(1), p(2), ..., p(n) consists of two increasing subsequences that have been interleaved. Informally speaking, shuffles are exactly those permutations that are obtained by performing a single riffle shuffle on a deck of cards. http://mathoverflow.net/questions/23692/what-are-the-normal-subgroups-of-a-direct-product/23693#23693 Comment by Gabe Cunningham Gabe Cunningham 2010-05-06T13:00:33Z 2010-05-06T13:00:33Z Perfect! So in particular, if I understand correctly, if $G$ and $H$ are distinct simple groups, then it is true that for a normal subgroup $N$, $N \simeq \pi_1(N) \times \pi_2(N)$. http://mathoverflow.net/questions/86/a-name-for-star-graph-with-long-laces/10693#10693 Comment by Gabe Cunningham Gabe Cunningham 2010-01-04T22:01:50Z 2010-01-04T22:01:50Z The family described above is more specific, since many subdivisions of star graphs don't have all arms the same length.