User ryan thorngren - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T18:16:24Z http://mathoverflow.net/feeds/user/5312 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109749/trace-of-a-functor-or-dimension-of-a-category-in-extended-2d-tqfts Trace of a functor (or dimension of a category) in extended 2d TQFTs Ryan Thorngren 2012-10-15T19:14:19Z 2012-10-22T06:14:57Z <p>In an extended 2d TQFT $Z$, a point (with orientation + or -) is assigned a category $Z(+)$ or $Z(-)$. This category should be as close to a vector space as possible: $\mathbb{C}$-linear, monoidal, etc. $Z( + \cup +)$ should be something like $Z( + ) \otimes Z( + )$, the empty set of points should get the unit category for this tensor operation, Vect$_\mathbb{C}$, and $Z(+)$ and $Z(-)$ should be dual.</p> <p>If we consider a circle as broken up into two opposite U shapes, these properties tell us that $Z(S^1)$ (a monoidal $\mathbb{C}$-linear functor $Z(empty)\rightarrow Z(empty)$, ie. $V\otimes -$ for some vector space $V$) is something like the dimension of $Z(+)$ or the trace of the identity functor.</p> <p>Can we make sense of this enough to compute it for some simple categories? Eg. the category of $\mathbb{C}$-representations of a finite group?</p> <p>I'm sure that this wouldn't be hard to answer if I knew more about what the tensor product should be when I write $Z(+)\otimes Z(+)$. All I know about this operation is that the unit should be the category of $\mathbb{C}$-vector spaces.</p> <p>How about for higher dimensional TQFTs? Does someone know a good reference?</p> <p>Thanks.</p> http://mathoverflow.net/questions/107820/homotopy-pullbacks-of-tensor-product-chain-complexes-towards-kunneth-formula-in homotopy pullbacks of tensor product chain complexes (towards Kunneth formula in diff cohomology) Ryan Thorngren 2012-09-22T07:09:31Z 2012-09-24T20:04:38Z <p>I have editted this question from the previous version which did not obtain much attention.</p> <p>Suppose I have two diagrams of chain complexes:</p> <p>$A^* \rightarrow C^* \leftarrow B^*$</p> <p>$\tilde{A}^* \rightarrow \tilde{C}^* \leftarrow \tilde{B}^*$</p> <p>We can form the tensor product diagram</p> <p>$(A\otimes \tilde{A})^* \rightarrow(C\otimes \tilde{C})^* \leftarrow(B\otimes \tilde{B})^*$.</p> <blockquote> <p>How can we express the homotopy pullback of the tensor product diagram (more importantly, its homology) in terms of the homotopy pullbacks of the first two diagrams?</p> </blockquote> <p>Now I'll give some motivation, because maybe someone knows the answer to the question the above seeks to find. I am interested in whether there is a Kunneth formula for (mainly ordinary) differential cohomology. The model of ordinary differential cohomology I am currently working with is as the homotopy pullback of</p> <p>$\Omega_{\mathbb{Z}}^k(M)\rightarrow H^k(M,\mathbb{R})\leftarrow H^k(M,\mathbb{Z})$,</p> <p>where the subscript $\mathbb{Z}$ denotes forms with integral periods. <strike>Taking homology commutes with taking homotopy pullbacks since we can replace anything by something quasi-isomorphic without affecting the quasi-isomorphism type (so I assume...), so</strike> Somehow we can from the above description produce a chain complex whose homology is ordinary differential cohomology. That is, the above diagram comes from</p> <p>$(...\rightarrow\Omega ^k_\mathbb{Z}\stackrel{0}{\rightarrow}\Omega^{k+1}_\mathbb{Z}\rightarrow...)\rightarrow(\mathrm{de Rham complex})\leftarrow(\mathrm{singular complex})$,</p> <p>which has as homotopy pullback the chain complex $\Omega^k_\mathbb{Z}\times C^{k-1}_\mathrm{dR}\times C^k$ with differential</p> <p>$(F,\mu,c)\rightarrow (0,F-c+d\mu,dc)$.</p> <p>We have Eilenberg-Zilber q-isoms $\Omega_\mathbb{Z}(M\times N)\rightarrow\Omega_\mathbb{Z}(M)\otimes \Omega_\mathbb{Z}(N)$ etc for each of the three chain complexes above such that the diagram for $M\times N$ becomes the tensor product of the diagrams for $M$ and $N$. I expect that this implies something nice about the pullback, but I am not sure what.</p> <p>I would also appreciate any answers using another model of differential cohomology (eg. Deligne cohomology), being ultimately interested in geometrical application, but I would like to understand this piece of homological algebra.</p> <p>Thanks!</p> http://mathoverflow.net/questions/105601/characterizing-flat-2-connections-by-their-holonomy Characterizing flat 2-connections by their holonomy Ryan Thorngren 2012-08-27T05:56:44Z 2012-08-27T15:53:00Z <p>Hello,</p> <p>A flat principal $G$-bundle over $X$ is determined by its holonomies, which are (after picking a trivialization) group homomorphisms $\pi_1(X)\rightarrow G$. The fiber of the bundle is not canonically identified with $G$, so these maps are only determined up conjugation by $G$. Equivalently these are gauge transformations at the basepoint where $\pi_1$ is evaluated.</p> <p>Now let $H \overset{t}\rightarrow G$ present a 2-group. With respect to some trivialization, a flat 2-connection on a principal 2-bundle assigns an element of $G$ to a closed curve, and an element $h$ of $H$ to a surface bounding a curve $\gamma$ such that $\gamma$ gets $t(h)$. Flat means that $h$ only depends on the homotopy class of this surface (homotopies fixing $\gamma$). </p> <p>Thus, I expect flat 2-connections are determined by a functor from the path 2-group of $X$ into $(H\rightarrow G)$. My question is : what is the degeneracy of this presentation? In other words, what is analogous to "up to conjugation in $G$" for flat 1-bundles?</p> <p>Thanks.</p> http://mathoverflow.net/questions/43538/wonderful-applications-of-the-vandermonde-determinant/100269#100269 Answer by Ryan Thorngren for Wonderful applications of the Vandermonde determinant Ryan Thorngren 2012-06-21T18:55:22Z 2012-06-21T18:55:22Z <p>Probably not an application for just any audience, but I thought I'd share... </p> <p>The Vandermonde determinant shows up in matrix models of quantum field theories. Roughly speaking, in these we consider integrals of the form</p> <p>$\int dM f(M)$,</p> <p>over the space of Hermitian matrices, where $f$ is invariant under conjugation by unitary matrices, and $dM$ is the (also conjugation invariant) measure</p> <p>$dM = (\prod_i dM_{ii})(\prod_{i\lt j} dM_{ij})$.</p> <p>We want to calculate the integral by gauge fixing. In other words, to integrate over a judiciously chosen set of representatives of each $U(N)$ orbit. Usually the best set of representatives to take are the diagonal matrices. The procedure for evaluating the integral in that case is exactly that of the Weyl integration formula. In doing so we get what physicists call the Faddeev-Popov determinant, which turns out to be the Vandermonde determinant! In other words, we get an equivalent integral</p> <p>$\int d\lambda_1 ... d\lambda_N \Delta(\lambda_1, ...,\lambda_N) f(diag(\lambda_1,...,\lambda_N))$,</p> <p>where $\Delta$ is the Vandermonde determinant.</p> http://mathoverflow.net/questions/69558/information-theory-of-decision-machines Information Theory of "decision machines" Ryan Thorngren 2011-07-05T18:10:14Z 2011-07-20T19:22:11Z <p>Hello, everyone. I am considering the following type of situation.</p> <p>Suppose I have a decision machine (DM) that I can ask any yes/no question and I want to use this to measure an n-ary random variable. Measuring a binary random variable using the DM with prior probability distribution ${p,1-p}$ gives an average change in uncertainty $S(p)$. The information measures giving the average change in uncertainty for measuring n-ary random variables with the DM will be built up from $S(p)$ depending on how the measurement is done. </p> <p>For example, when measuring a ternary random variable $X \in {x_1,x_2,x_3}$, with prior probability distribution $p_1,p_2,p_3$ I can first ask "is $X=x_1$?", and if the answer is no, "is $X=x_2$", after which I will certainly know the value of $X$. This will give an information measure $S(p_1)+(1-p_1)S(p_2/(1-p_1))$. Similarly I can first ask "is $X=x_3$?", followed by "is $X=x_1$"?, giving an average change in uncertainty $S(p_3)+(1-p_3)S(p_1/(1-p_3))$.</p> <p>My goal is to relate this type of information measure to a particular nonassociative structure which I am studying. This "semiring" is constructed given an information measure, and the associativity of addition in the semiring is equivalent to</p> <p>$S(p_3)+(1-p_3)S(p_1/(1-p_3))=S(p_1)+(1-p_1)S(p_2/(1-p_1))$,</p> <p>a sort of associativity for binary information measures. Along with $S(p)=S(1-p)$, the only information measure satisfying this is the Shannon entropy.</p> <p>I would like to relate features of this structure to features of other information measures, to better understand the role information theory plays in this construction, which is a sort of Witt ring in characteristic one. However, all of the measures I have found are defined for arbitrary n-ary random variables in a way that $S(p_1,...,p_n)$ is not built up by asking yes/no questions as above.</p> <p>I was hoping one of you out there had some references to similar things that have been studied, because my own searches have largely come up empty-handed.</p> <p>Thanks.</p> http://mathoverflow.net/questions/61513/critical-radius-for-infinite-dimensional-sphere-packing Critical Radius for Infinite Dimensional Sphere Packing Ryan Thorngren 2011-04-13T07:30:10Z 2011-04-13T21:19:20Z <p>Hello. I'd like to consider the open unit ball in an infinite dimensional Hilbert space and ask when can we fit infinitely many open balls of radius $r&lt;1$ inside.</p> <p>For example, when $r=1/(1+\sqrt2)$, we can pick an orthonormal basis $(x_1,...)$ for our Hilbert space and put the centers of the balls at $(1-r)x_i = \sqrt2/(1+\sqrt2)x_i$ for each $i$. The distance between any two centers is thus $\sqrt2/(1+\sqrt2)\sqrt2 = 2r$, so indeed the balls just kiss each other.</p> <p>Can we fit any larger balls? What is the critical radius $r_\infty$ such that for $r>r_\infty$ we may only fit finitely many balls of radius $r$, but for smaller $r$ we may fit infinitely many?</p> http://mathoverflow.net/questions/33125/what-happens-to-the-boundary-conditions-as-a-pde-is-approximated-by-a-lesser-orde What happens to the boundary conditions as a PDE is approximated by a lesser order PDE? Ryan Thorngren 2010-07-23T19:00:37Z 2010-08-21T07:54:51Z <p>Consider a fourth order linear (biharmonic) PDE in two variables of the form</p> <p>$\nabla^4u + c\nabla^2u-\lambda u = F(x,y)$; $(x,y) \in \Lambda$</p> <p>To have uniqueness, we must specify two equations per point on $\partial \Lambda$. Now consider the limit where $c\rightarrow \infty$. The solution, $u$, is approximated by $\tilde u$, given by</p> <p>$c\nabla^2\tilde u-\lambda \tilde u = F(x,y)$; $(x,y) \in \Lambda$</p> <p>for which we must only specify one equation on the boundary.</p> <p>The question is: what boundary conditions do we pick for $\tilde u$?</p> <p>If I use a "clamped boundary" $u=u'=0$, the solution is approximated by using the boundary condition $\tilde u = 0$. If I use a "free boundary" $u''=u'''=0$, the appropriate approximation is $\tilde u'=0$.</p> http://mathoverflow.net/questions/34591/probability-of-generating-the-symmetric-group Probability of generating the symmetric group Ryan Thorngren 2010-08-05T08:22:10Z 2010-08-05T10:07:12Z <p>The statement is simple:</p> <p>What is the probability that a set of n-1 transpositions generates the symmetric group, $S_n$?</p> <p>The motivation is that I remembered reading that this was an open problem somewhere on the internet, and then I solved it. I'm curious to see other people's solutions, because I think it's a nice problem, and don't quite believe that it is hard enough to be open.</p> http://mathoverflow.net/questions/29958/eigenvalue-problems-with-linear-constraints Eigenvalue Problems with Linear Constraints Ryan Thorngren 2010-06-29T19:09:14Z 2010-06-30T23:35:16Z <p>The motivation for this problem comes from trying to develop a simple way to decompose domains into non-overlapping subdomains to solve for the eigenvalues of some differential operator. The idea is to construct the matrices for each subdomain, letting the points on the boundaries be redundant for each matrix.</p> <p>In the simplest one dimensional case with 2 subintervals, using the normal ordering, the last row of the left matrix and the first row of the right matrix will both describe the operator at the point shared between the two subintervals. We average the rows with some sort of weighting that maximizes the accuracy and add the two columns both corresponding to the redundant point.</p> <p>The problem is, the only boundary conditions this specifies between the two subintervals is continuity. For an operator of order 2, for example, we would also need to specify the continuity of the first derivative to maintain uniqueness.</p> <p>My question is how to specify the continuity of the first derivative in solving the eigenvalue problem.</p> <p>My first idea was to add another row and a column of zeros, the row being the difference in the derivatives from the left and right at the boundary point. Then to prescribe that this difference is zero, I would solve the generalized eigenvalue problem with mass matrix identity except for the difference in derivatives, which is given weight zero.</p> <p>Using the simplest numerics with a 2nd order centered difference and a first order backward (resp. forward) difference at the left (resp. right) of the boundary point, the eigenvalues I calculate reflect the eigenvalues of the subintervals considered separately.</p> <p>If someone has any ideas or a good reference to some literature that isn't completely geared towards finite element methods (I'm working on a different method altogether, and identities involving the weak form are not helpful), that would be much appreciated.</p> <p>UPDATE:</p> <p>A couple recent thoughts. Any linear scalar function on a vector space is a member of its dual, ie. is a row vector (using the general convention that vectors are columns, covectors are rows). For a given constraint, call this vector $u$. Define the operator $B = v u^T$, where v is one in the first index and zero elsewhere. If we wish the solve the eigenvalues of $A$ that are subject to this constraint, we need only find the simultaneous eigenvalues</p> <p>$A x = \lambda x$</p> <p>$(A - B) x = \lambda x$</p> <p>The problem with this is that it does not "mimic appropriately" the continuous case I wish to approximate. In an underdetermined Sturm-Liouville problem, any real number is an eigenvalue, so it will simply be luck that the above system has a solution for finite dimensional matrices.</p> <p>UPDATE:</p> <p>Another thought is, since I know the projection onto the null space is $I - v v^T$, perhaps there is some way to restrict my eigenvalue search for members of that subspace.</p> http://mathoverflow.net/questions/21409/the-delta-system-lemma-outside-set-theory the delta system lemma outside set theory Ryan Thorngren 2010-04-15T01:41:43Z 2010-04-21T15:51:33Z <p>The lemma:</p> <p>Any uncountable set $S$ of finite sets has an uncountable subset $\Delta \subseteq S$ and an $x$ such that $\forall a,b \in \Delta$, if $a \neq b$ then $a \cap b = x$. $\Delta$ is called a $\Delta$-system.</p> <p>I've seen this lemma used in independence proofs, such as the famous result that the negation of the continuum hypothesis is consistent with ZFC, but it seems like it would be useful in other fields as well. Does anyone have any examples with this lemma outside pure set theory?</p> <p>See also the finite and generalized infinite versions posted below.</p> http://mathoverflow.net/questions/21782/generalized-chinese-remainder-theorem/21801#21801 Answer by Ryan Thorngren for Generalized Chinese Remainder Theorem Ryan Thorngren 2010-04-19T01:23:45Z 2010-04-19T01:23:45Z <p>If we consider submodules generated by applying ideals to $M$, ie. $I \ M$, where $I \subseteq R$ is an ideal, we can generalize the two submodule case to any finite number of modules by induction from n = 2. All we need to check is that if $A_{i} \ M = U_{i}$ and the $A_{i}$'s are pairwise comaximal, then $A_{1} ... A_{n-1}$ and $A_{n}$ are comaximal.</p> http://mathoverflow.net/questions/21310/minimal-non-planar-extensions-of-a-graph Minimal Non-planar Extensions of a Graph Ryan Thorngren 2010-04-14T05:40:40Z 2010-04-14T19:39:39Z <p>Given a planar graph $G=(V,E)$ with vertices $V$ and edges $E$, call $\bar G = (V,\bar E)$ a non-planar extension of $G$ if $\bar G$ is non-planar and $E \subset \bar E$.</p> <p>I'm interested in minimal non-planar extensions in the sense that if $\bar G$ is a non-planar extension of $G$, there is no non-planar extension of $G$ that is a subgraph of $\bar G$.</p> <p>I first wondered whether these minimal extensions could be unique, but this is quickly disproved by the existence of maximally planar (also called triangulated) graphs. I refine this question slightly:</p> <p>(1) Is the minimal non-planar extension of an arbitrary planar graph unique up to isomorphism?</p> <p>(2) What if we define minimality as having the smallest number of edges for one of these extensions?</p> <p>and also:</p> <p>(3) Do these extensions mean anything interesting for representations of algebraic objects (eg. groups) on them?</p> http://mathoverflow.net/questions/21214/particular-problem-solved-by-solving-a-more-general-problem/21226#21226 Answer by Ryan Thorngren for Particular problem solved by solving a more general problem. Ryan Thorngren 2010-04-13T17:01:05Z 2010-04-13T17:07:55Z <p>Occasionally when trying to prove a certain type of object exists, it is easier to show that the set of those objects is very large.</p> <p>For instance, it's difficult to give an example of a transcendental number over the rationals. However, it is quite easy to show that the set of algebraic numbers is only countably infinite, so almost every real number is transcendental.</p> http://mathoverflow.net/questions/115948/what-is-the-transpose-variety Comment by Ryan Thorngren Ryan Thorngren 2012-12-10T05:51:20Z 2012-12-10T05:51:20Z Brilliant observation! Thank you. The order of the product hadn't even occurred to me. http://mathoverflow.net/questions/109749/trace-of-a-functor-or-dimension-of-a-category-in-extended-2d-tqfts/109759#109759 Comment by Ryan Thorngren Ryan Thorngren 2012-10-16T18:01:27Z 2012-10-16T18:01:27Z This is a nice answer, thanks again. And thanks Qiaochu and Dmitri for the references. http://mathoverflow.net/questions/109749/trace-of-a-functor-or-dimension-of-a-category-in-extended-2d-tqfts Comment by Ryan Thorngren Ryan Thorngren 2012-10-16T02:09:54Z 2012-10-16T02:09:54Z It strikes me now that the monoidal structure I've had in mind is non-canonical! It just comes from the action you mention being free and transitive. http://mathoverflow.net/questions/109749/trace-of-a-functor-or-dimension-of-a-category-in-extended-2d-tqfts Comment by Ryan Thorngren Ryan Thorngren 2012-10-16T01:30:35Z 2012-10-16T01:30:35Z Thank you, David. I do often enjoy your lecture notes. http://mathoverflow.net/questions/109749/trace-of-a-functor-or-dimension-of-a-category-in-extended-2d-tqfts Comment by Ryan Thorngren Ryan Thorngren 2012-10-16T01:21:55Z 2012-10-16T01:21:55Z I agree in that morphisms should be local boundary-changing operators and fusion of those gives composition. What I meant is that for boundary conditions that arise from some codimension 1 bulk operators, these can be fused as bulk operators onto the boundary. Sometimes all boundary conditions arise this way (eg. Wilson lines for 2d BF theory), but I'm not sure when one can say this for sure. http://mathoverflow.net/questions/109749/trace-of-a-functor-or-dimension-of-a-category-in-extended-2d-tqfts Comment by Ryan Thorngren Ryan Thorngren 2012-10-15T21:13:05Z 2012-10-15T21:13:05Z @Theo I believe that it should be the category of boundary conditions for a 2d tqft, and so fusion should give a product. There's no reason to assume this for very general 2d tqfts I suppose. Anyway it's not that pertinent to the question. And thanks for your answer! http://mathoverflow.net/questions/107820/homotopy-pullbacks-of-tensor-product-chain-complexes-towards-kunneth-formula-in Comment by Ryan Thorngren Ryan Thorngren 2012-09-24T19:57:18Z 2012-09-24T19:57:18Z @Fernando I see. Thank you. I am certain that the chain complex I describe has the homology I want, so it must be something nice in this case about how the three complexes fit together. http://mathoverflow.net/questions/105601/characterizing-flat-2-connections-by-their-holonomy/105607#105607 Comment by Ryan Thorngren Ryan Thorngren 2012-08-27T07:13:22Z 2012-08-27T07:13:22Z Thanks, Konrad. This is precisely what I was looking for. And yes, I see that I did not need to just talk about flat 2-connections and holonomy. I was just thinking towards my own problem :-). http://mathoverflow.net/questions/59186/regge-calculus-questions-of-consistency-resolved Comment by Ryan Thorngren Ryan Thorngren 2012-08-24T00:53:32Z 2012-08-24T00:53:32Z @timur: what Bob Marley gets from not brushing his teeth. http://mathoverflow.net/questions/97074/is-the-susy-algebra-isomorphic-for-all-kahler-manifolds Comment by Ryan Thorngren Ryan Thorngren 2012-05-16T02:42:39Z 2012-05-16T02:42:39Z Actually, having thought about it some, I now think that if all we want is a representation of the lie algebra, then this representation will be faithful. All we need to show is that there are nonzero smooth functions which remain nonzero when we apply these operators. I think that a bump function does the trick in each case. http://mathoverflow.net/questions/97074/is-the-susy-algebra-isomorphic-for-all-kahler-manifolds Comment by Ryan Thorngren Ryan Thorngren 2012-05-16T01:35:19Z 2012-05-16T01:35:19Z In some sense, yes. The Kahler identities specify all the a priori nontrivial commutation relations. This gives us some abstract lie algebra referred to as the (2,2) SUSY algebra. The algebra of differential forms on the Kahler manifold need not be a faithful representation of this algebra. http://mathoverflow.net/questions/69558/information-theory-of-decision-machines/69586#69586 Comment by Ryan Thorngren Ryan Thorngren 2011-07-06T19:14:51Z 2011-07-06T19:14:51Z I've edited the question hopefully to address this question. http://mathoverflow.net/questions/69558/information-theory-of-decision-machines/69586#69586 Comment by Ryan Thorngren Ryan Thorngren 2011-07-06T19:04:54Z 2011-07-06T19:04:54Z Indeed. The semiring I am looking at is constructed given an information measure, which only needs to be concave and symmetric. Its associativity is equivalent to this information measure having an associativity property that says the two schemes above for measuring a ternary random variable are equivalent. This is equivalent to the information measure being the Shannon entropy. I would like to look at other information measures, such as the Reyni entropy, but ones I have found are not of the form I describe above for n-ary random variables. http://mathoverflow.net/questions/34591/probability-of-generating-the-symmetric-group Comment by Ryan Thorngren Ryan Thorngren 2010-08-05T16:29:04Z 2010-08-05T16:29:04Z So I thought as well. I wonder what the actual statement was, or whether I had dreamt it. http://mathoverflow.net/questions/33158/use-of-conjectures-to-prove-a-theorem Comment by Ryan Thorngren Ryan Thorngren 2010-07-24T01:15:06Z 2010-07-24T01:15:06Z Shouldn't any such two proofs be able to be rewritten as cases?