User ari - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T18:36:49Z http://mathoverflow.net/feeds/user/673 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/95182/construction-of-finite-element-differential-forms-based-on-derham-sequences/95219#95219 Answer by Ari for Construction of finite element differential forms based on deRham sequences Ari 2012-04-26T02:10:20Z 2012-04-26T02:10:20Z <p>As robot suggested, a good place to start would be the work of Arnold, Falk, and Winther -- particularly their 2006 paper in <em>Acta Numerica</em> and 2010 paper in <em>Bulletin of the AMS</em>. (Both available at <a href="http://ima.umn.edu/~arnold/publications.html" rel="nofollow">http://ima.umn.edu/~arnold/publications.html</a>.)</p> <p>One key insight of their work is that, to get a stable discretization, we need two properties. First, the discrete complex needs to be a <em>subcomplex</em>, in the sense that the subspace inclusions commute with the differentials. Second, there needs to be a bounded projection mapping the continuous complex to the discrete complex, and this projection must also commute with the differentials. In $\mathbb{R}^n$, there are two main families of piecewise-polynomial differential forms, which Arnold et al. call $\mathcal{P}_r$ and $\mathcal{P}_r^-$, where $r$ denotes the degree of the polynomials. (The lowest-order Whitney forms are $\mathcal{P}_1^-$.) They show that these spaces can be used, systematically, to construct subcomplexes that satisfy the two stability conditions.</p> <p>This defines the key relationship between the choice of basis functions and the de Rham complex: the discrete inclusion and projection functions must "respect the structure" of the de Rham complex, in the sense of commuting with the differentials, otherwise one is not guaranteed stability. As you suggest, there are of course many other ways of constructing interpolants, and as long as they satisfy the conditions above, you are guaranteed a stable discretization. However, piecewise-polynomials are especially useful for computation, and their approximation properties are well understood (which is part of the reason they are used in classical finite element methods, not to mention FEEC).</p> http://mathoverflow.net/questions/84852/reduced-symplectic-form/85969#85969 Answer by Ari for reduced symplectic form Ari 2012-01-18T08:48:41Z 2012-01-18T08:48:41Z <p>Your question is a bit vague, but it seems like you're asking whether it's possible to perform reduction several times, in sequence, instead of all at once. If that's the case, you may be interested in so-called <em>reduction by stages</em>, cf. <a href="http://www.cds.caltech.edu/~marsden/books/Hamiltonian_Reduction_Stage.html" rel="nofollow">http://www.cds.caltech.edu/~marsden/books/Hamiltonian_Reduction_Stage.html</a></p> http://mathoverflow.net/questions/71571/example-of-symplectic-and-hamiltonian-diffeomorphism-on-s2-and-t2/71618#71618 Answer by Ari for Example of symplectic and hamiltonian diffeomorphism on $S^2$ and $T^2$ Ari 2011-07-29T20:53:43Z 2011-07-29T20:53:43Z <p>On an orientable 2-manifold (such as $S^2$ or $T^2$), the symplectic 2-form $\omega$ can be given by the signed area. In this case, symplectic diffeomorphisms are just those which preserve area and orientation.</p> <p>Now, let's set aside the difference between diffeomorphisms and flows of vector fields (which is a complicated enough issue by itself), and just focus on the difference between symplectic and Hamiltonian vector fields.</p> <p>A vector field $X$ is Hamiltonian if $i_X \omega = dH$ for some Hamiltonian function $H$ (where $i_X$ is the contraction by $X$ and $d$ is the exterior derivative). On the other hand, $X$ is symplectic if the Lie derivative $L_X \omega$ vanishes. Applying Cartan's "magic formula" $L_X \omega = di_X \omega + i_X d \omega = d i_X \omega$ (since $\omega$ is closed), this means that $X$ is symplectic when $d i_X \omega = 0$. In summary, $X$ is Hamiltonian when $i_X \omega$ is exact, and symplectic when $i_X \omega$ is closed. The difference between these is given by the 1st de Rham cohomology of the manifold in question.</p> <p>So, as Weiwei said, "symplectic" and "Hamiltonian" are identical on $S^2$, since $S^2$ is simply connected. On the other hand, they're <em>not</em> the same on $T^2$, since the torus has nontrivial 1st cohomology. Putting coordinates $(\theta,\phi)$ on $T^2$, the vector fields $\partial/\partial \theta$ and $\partial/\partial \phi$ are both symplectic but not Hamiltonian.</p> http://mathoverflow.net/questions/60960/integration-of-a-laplacian/61004#61004 Answer by Ari for integration of a laplacian Ari 2011-04-07T23:08:10Z 2011-04-07T23:08:10Z <p>Denis has this exactly right, if your goal is really to calculate these integrals. However, if your <em>real</em> goal (as you say) is to calculate the residual, then this isn't what you want to do at all.</p> <p>In a weak sense, the Laplacian is a map $\Delta \colon H^1 (\Omega) \to H^{-1} (\Omega)$, so the PDE $\Delta u = f$ makes sense when $u \in H^1 (\Omega)$ and $f \in H^{-1} (\Omega)$. Denoting the approximate FEM solution by $u_h$, the residual is $f - \Delta u_h \in H^{-1} (\Omega)$, so it really makes sense to measure the residual in the $H^{-1} (\Omega)$ norm, not the $L^1 (\Omega)$ or $L^2 (\Omega)$ norm. That is, $$\lVert f - \Delta u_h \rVert _{H^{-1}(\Omega)} = \sup _{ \lVert v \rVert _{H^1 (\Omega)}= 1} \langle f - \Delta u_h , v \rangle _{H^{-1} (\Omega) \times H^1 (\Omega) }.$$</p> <p>On the other hand, maybe you don't really want to measure the residual itself; you want to estimate the <em>a posteriori</em> error $e _h = u - u_h$. In this case, $e _h \in H^1 (\Omega)$ solves the residual equation $$\Delta e _h = \Delta (u - u_h) = f - \Delta u_h .$$ You can measure $e _h$ a number of ways, e.g., using the energy norm. Typically, of course, you can't <em>actually</em> solve for $e_h$ (since that would mean solving the original PDE exactly!), but you can estimate it by using a more accurate finite-element method for the residual equation (e.g., finer mesh and/or higher-order elements) than you used for $u_h$.</p> <p>To learn more about these sorts of things, you should look up <em>residual-based a posteriori error estimation</em> (Google returns lots of hits for this phrase).</p> http://mathoverflow.net/questions/60200/fem-on-a-laplacian/60774#60774 Answer by Ari for FEM on a Laplacian Ari 2011-04-06T06:38:40Z 2011-04-06T06:38:40Z <p>Step (3) is, essentially, a way of defining the weak version of the Laplacian. Given $u \in H^1$, the classical Laplacian $\Delta u$ is generally not defined. However, for any test function $v \in H^1$, one can define $(\Delta u, v ) = -(\nabla u, \nabla v)$. In other words, we have $\Delta \colon H^1 \to H^{-1}$, so if $f \in H^{-1}$, then the weak problem is precisely equivalent to the operator equation $\Delta u = f$.</p> http://mathoverflow.net/questions/50156/ellp-version-of-singular-values/50208#50208 Answer by Ari for $\ell^p$ version of singular values Ari 2010-12-22T23:21:39Z 2010-12-23T03:33:32Z <p>It might be more useful to pose the problem as follows. Let $X = (\mathbb{R}^n, \| \cdot \| _p )$ and $X^\ast = (\mathbb{R}^n, \| \cdot \| _{p^\ast})$, where $p^\ast$ is the conjugate exponent to $p$. Rather than considering $M$ as a map from $X \to X$, it may be more useful to treat it as $M \colon X \to X^\ast$. (Of course, when $p=p^{*}=2$, these are the same.) <strike>In that case, one can make sense of the compositions $M^\ast M \colon X \to X$ and $M M^\ast \colon X ^\ast \to X^\ast$, and take the singular values as the square root of the eigenvalues of these maps.</strike></p> <p>EDIT: This is equivalent to looking at $\| Mx \| _{p^\ast} / \| x \|_p$ instead of $\| Mx \| _p / \| x \|_p$, so it ties into the work that Suvrit mentioned in his response.</p> <p>EDIT 2: Sorry, I made a stupid mistake in the struck-out sentence above. Of course, if $M \colon X \to X^\ast$, then we again have $M^\ast \colon X \to X^\ast$ -- <em>not</em> $X^\ast \to X$ as I had written above. Ultimately, you may have to resort to the fact that $\ell^p$ is isomorphic to $\ell^2$ (since $n$ is finite), so one can map between $X$ and $X^\ast$ -- but this has gotten sufficiently far from my original answer that I'll just stop at that.</p> http://mathoverflow.net/questions/50201/question-about-banachs-matchbox-problem/50207#50207 Answer by Ari for Question about Banach's matchbox problem. Ari 2010-12-22T23:04:16Z 2010-12-22T23:31:54Z <p>Your mistake has to do with the definition of the problem. The man does not stop taking matches as soon as one of the boxes is empty (as in your code). He stops taking matches after he <em>picks a box and finds it empty</em>. This means that when Box A is empty, he doesn't immediately stop -- he continues until he picks Box A again (or until he empties Box B as well).</p> <p>EDIT: In fact, it's trivial to fix your code -- you don't even have to rewrite your function! Just add the following:</p> <pre><code>def bar (n=3): return foo(n+1) - 1 </code></pre> <p>This just exploits the fact that picking the empty matchbox in the $n$-match problem is equivalent to emptying the box in the $(n+1)$-match problem.</p> http://mathoverflow.net/questions/28265/proving-hodge-decomposition-without-using-the-theory-of-elliptic-operators/28692#28692 Answer by Ari for Proving Hodge decomposition without using the theory of elliptic operators? Ari 2010-06-18T23:39:39Z 2010-06-18T23:39:39Z <p>The Hodge decomposition can be proved, in a very nice, abstract-functional-analysis setting, on so-called <em>Hilbert complexes</em>. Brüning and Lesch wrote an <a href="http://dx.doi.org/10.1016/0022-1236%2892%2990147-B" rel="nofollow">excellent paper</a> on the topic in J. Funct. Anal., first developing the theory on arbitrary Hilbert complexes, and then discussing the application to elliptic complexes.</p> http://mathoverflow.net/questions/26018/what-are-variational-crimes-and-who-coined-the-term/26066#26066 Answer by Ari for What are "variational crimes" and who coined the term? Ari 2010-05-26T21:00:53Z 2010-05-26T21:00:53Z <p>Thanks for your interest in the paper! (It's also nice to see something on Math Overflow that I know something about.) Your summary of variational crimes is actually pretty close to the mark: it refers to certain "abuses" of the Galerkin method, where some of the assumptions are violated, and thus the standard error estimates (e.g., <a href="http://en.wikipedia.org/wiki/C%C3%A9a%27s_lemma" rel="nofollow">Céa's lemma</a>) are no longer valid.</p> <p>Since you have the basic idea right, let me try to provide some context and motivation. As a simple example, consider Poisson's equation on some domain $U \subset \mathbb{R}^n$ with Dirichlet boundary conditions,</p> <p>$$- \Delta u = f \text{ on } U, \quad u \rvert _{\partial U} = 0 .$$ </p> <p>This can be written as a variational problem on $V = \{ v \in H^1(U) : v \rvert_{\partial U} = 0 \}$: Find $u \in V$ such that</p> <p>$$\int _U \nabla u \cdot \nabla v \ dx = \int _U f \thinspace v \ dx , \quad \forall v \in V .$$</p> <p>(You can see, using integration by parts, that any classical solution solves this variational problem.) If we define the bilinear form $B(u,v) = \int _U \nabla u \cdot \nabla v \ dx$ and functional $F(v) = \int _U \thinspace f \thinspace v \ dx$, then this problem can be written in the usual abstract form: Find $u \in V$ such that</p> <p>$$B(u,v) = F(v) ,\quad \forall v \in V .$$</p> <p>To apply the Galerkin method, we need to take a subspace $V _h \subset V$ (e.g., the span of some finite element basis) and solve the Galerkin variational problem: Find $u _h \in V _h$ such that </p> <p>$$B(u_h, v ) = F(v), \quad \forall v \in V_h .$$</p> <p>The problem is that, for many practical purposes, this is impossible to compute. First, the bilinear form $B(\cdot, \cdot)$ requires us to calculate an integral exactly. In practice, this is usually not possible, so people instead approximate the integral using <em>numerical quadrature</em>. However, this is a variational crime, since using numerical quadrature replaces $B (\cdot, \cdot)$ by some $B_h (\cdot, \cdot) \approx B(\cdot, \cdot)$ in the Galerkin variational principle; likewise, numerical quadrature also replaces $F(\cdot)$ by some $F_h(\cdot) \approx F(\cdot)$. (This is aside from the fact that computers only use finite-precision arithmetic, so even if we had an closed formula for these integrals, there would always be some floating-point error involved.)</p> <p>Moreover, if $U \subset \mathbb{R}^n$ is polyhedral, then it can be triangulated exactly, so we can get $V_h \subset V$ to be some finite element space supported on this piecewise-linear mesh. However, if $U$ has a curved boundary, then a piecewise-linear (or piecewise-polynomial, in the case of isoparametric elements) mesh only approximates the actual domain. Since the functions in $V_h$ are defined on a slightly different domain than those in $V$, in this case $V_h \not\subset V$.</p> <p>Now, in the real world of numerical computation (engineering, etc.), people didn't worry too much about using these approximations instead of the exact Galerkin variational problem; it was a practical necessity, and the approximations seemed to converge just fine. However, these "variational crimes" meant that the abstract Galerkin error analysis was no longer valid for the modified methods. Strang pointed this out, and his lemmas quantify the additional errors introduced by these "crimes."</p> <p>As far as the history/terminology: to the best of my knowledge, Strang himself coined the term "variational crime." The earliest reference I know is [Strang, G. (1972), Variational crimes in the finite element method. In The mathematical foundations of the finite element method with applications to partial differential equations (Proc. Sympos., Univ. Maryland, Baltimore, Md., 1972), pages 689–710. Academic Press, New York.], although I haven't been able to find an electronic copy. He followed this up with a more easily-located article for a wider audience: <a href="http://projecteuclid.org/euclid.bams/1183535131" rel="nofollow">Strang, G. (1973), Piecewise polynomials and the finite element method. Bull. Amer. Math. Soc., 79, 1128–1137.</a> Finally, an excellent account is given in the book <a href="http://dx.doi.org/10.1007/978-0-387-75934-0" rel="nofollow">Brenner, S. C., and L. R. Scott (2008), The mathematical theory of finite element methods, volume 15 of Texts in Applied Mathematics. Springer, New York, third edition.</a>; Chapter 10 is entirely about variational crimes.</p> http://mathoverflow.net/questions/3476/is-there-a-categorification-of-the-integers-in-terms-of-graded-sets/3484#3484 Answer by Ari for Is there a categorification of the integers in terms of "graded sets"? Ari 2009-10-30T23:15:34Z 2009-10-30T23:15:34Z <p>Not sure if you've seen this already, but it looks like Baez talks about this in <a href="http://math.ucr.edu/home/baez/twf%5Fascii/week102" rel="nofollow">one of his "This Week's Finds" columns</a>, where he shows that the morphisms are given by tangles.</p> http://mathoverflow.net/questions/2556/real-world-applications-of-mathematics-by-arxiv-subject-area/2730#2730 Answer by Ari for Real-world applications of mathematics, by arxiv subject area? Ari 2009-10-27T00:02:27Z 2009-10-27T00:02:27Z <p><strong>math.SG Symplectic Geometry</strong></p> <p>Symplectic integrators are used for numerical simulation of Hamiltonian mechanics. Prominent applications include molecular dynamics, solar system dynamics, computer animation, and a wide variety of problems in mechanical engineering.</p> http://mathoverflow.net/questions/1470/what-are-some-conserved-quantities-of-poisson-brackets/1520#1520 Answer by Ari for What are some conserved quantities of Poisson brackets? Ari 2009-10-20T23:07:56Z 2009-10-20T23:07:56Z <p>Not sure if this is what you're asking for, but conserved quantities correspond to symmetries of the Hamiltonian (Noether's theorem). In the example you just gave, I think the conserved quantity p_i - p_j corresponds to the action sending x_i to x_i + y while sending x_j to x_j - y, which leaves the Hamiltonian invariant. In fact, you can take x_i to x_i + y_i for i=1,...,n, where y_1 + ... + y_n = 0, since this leaves the sum x_1 + ... +x_n (and hence the Hamiltonian) unchanged.</p> http://mathoverflow.net/questions/1443/algorithm-to-find-all-the-cycle-bases-in-a-graph/1471#1471 Answer by Ari for Algorithm to Find all the Cycle Bases in a Graph Ari 2009-10-20T18:58:16Z 2009-10-20T18:58:16Z <p>It seems like the OP is looking for a list of faces and their boundaries for a planar graph. However, without coordinates or an embedding in the plane, this is definitely ill-posed. As a simple counterexample, consider the complete graph K4. This has 4 possible faces (123,124,134,234), but any embedding in the plane has only 3 of them. This leads to 4 different possible answers, for the same graph, depending on which vertex is placed in the center of the other three. This means that, without more information, the problem doesn't have a unique answer.</p> http://mathoverflow.net/questions/1114/whats-a-groupoid-whats-a-good-example-of-a-groupoid/1456#1456 Answer by Ari for What's a groupoid? What's a good example of a groupoid? Ari 2009-10-20T17:22:08Z 2009-10-20T17:22:08Z <p>While the categorical definition of groupoid is the most concise, you can also think of a groupoid as being like a group, except where multiplication is only partially defined, rather than being defined for any pair of elements. Here are a few of my favorite examples:</p> <ul> <li><p>Given a vector bundle E, the general linear groupoid GL(E) is the groupoid of linear isomorphisms between fibers. Given a map from Ex -> Ey, and another from Ey' -> Ez, we can only compose them if y=y'. When E is just a vector space over a single point, then this is the usual general linear group. In differential geometry, this gives a very natural way to think about frames and G-structures on a differentiable manifold M: just look at the general linear groupoid GL(TM). G-structures can be understood as subgroupoids: for example, a Riemannian structure corresponds to the orthogonal subgroupoid O(TM), consisting of elements of GL(TM) which are also isometries.</p></li> <li><p>More generally, given a principal G-bundle, the gauge groupoid consists of G-equivariant maps between fibers. This is useful for talking about connections, holonomy, etc., without having to fix a particular gauge.</p></li> <li><p>Given a directed graph, one can construct the free groupoid generated by the edges. As a special case, the free group on n elements is generated by the graph with one vertex and n self-loops. (There is also a forgetful functor from groupoids to graphs, which is adjoint to the free functor.)</p></li> </ul> <p>The first two examples happen to be <em>Lie groupoids</em>, and they have corresponding <em>Lie algebroids</em>, which generalizes the relationship between Lie groups and Lie algebras. Whereas a Lie algebra is a vector space with a bracket between elements, a Lie algebroid is a vector bundle with a bracket between sections (as well as an additional structure called the anchor map). For example, if Q is a principal G-bundle, then the gauge groupoid is (Q x Q)/G, while the corresponding gauge algebroid is TQ/G. This comes in handy in geometric mechanics, particularly in reduction theory. If we have a Lagrangian L: TQ -> R, which is invariant with respect to the action of a Lie group G, then is useful to look at the reduced Lagrangian \ell: TQ/G -> R. There are some subtleties arising from the fact that TQ/G is not a tangent bundle, but it <em>is</em> still a Lie algebroid, so this has motivated the study of mechanics on Lie algebroids.</p> http://mathoverflow.net/questions/1053/what-m-minimizes-em-x3-for-a-random-variable-x/1054#1054 Answer by Ari for What m minimizes E(|m-X|^3) for a random variable X? Ari 2009-10-18T17:42:34Z 2009-10-18T17:57:49Z <p>I assume you mean |m-X| as opposed to |m-EX|? Otherwise, |m-EX| is not a random variable, so E(|m-EX|^k) = |m-EX|^k is always zero (and hence minimized) when m = EX -- i.e., the mean -- and that's probably not what you're asking.</p> <p>After a bit of Googling around, it looks like you might be talking about the third absolute central moment E(|X-EX|^3), which is related to something called the Barry-Esseen inequality ... <a href="http://books.google.com/books?id=AEIhcqLvJo4C&amp;lpg=PA47&amp;ots=FX9f7qgT1R&amp;dq=%22absolute%20central%20moment%22&amp;pg=PA152#v=onepage&amp;q=%22absolute%20central%20moment%22&amp;f=false" rel="nofollow">see here</a>.</p> http://mathoverflow.net/questions/60890/weakest-condition-for-an-integrable-almost-symplectic-manifold Comment by Ari Ari 2011-04-07T22:29:15Z 2011-04-07T22:29:15Z Do you mean Liouville's theorem (the Hamiltonian vector field preserves the symplectic form) or Liouville integrability (maximal set of first integrals in involution)? http://mathoverflow.net/questions/60200/fem-on-a-laplacian/60774#60774 Comment by Ari Ari 2011-04-06T19:10:59Z 2011-04-06T19:10:59Z $H^{-1}$ is the dual space of $H^1$. This is actually even weaker than being in $L^2$ (a.k.a. $H^0$), since it's only a continuous functional when applied to weakly differentiable test functions. For example, consider the 1-D Dirac $\delta$-function, $\delta = \mathrm{H} '$, where $\mathrm{H}$ is the Heaviside step function. This is obviously not in $L^2$; however, it <i>is</i> in $H^{-1}$, since $(\delta, v) = (\mathrm{H}', v) = -(\mathrm{H}, v') \leq \lVert \mathrm{H} \rvert _{L^2} \lVert v \rVert _{H^1}$. This means that you can even make sense of the PDE $u'' = \delta$. http://mathoverflow.net/questions/6043/splitting-book-into-chapters/6049#6049 Comment by Ari Ari 2009-11-19T02:27:40Z 2009-11-19T02:27:40Z I agree that \include and \includeonly are the best way to go. This creates a separate .aux file for each included chapter -- so it maintains cross-references, page numbering, etc. -- but only outputs the chapter(s) selected with \includeonly. I used this recently when I had to output my list of references as a separate PDF file, and it worked like a charm. http://mathoverflow.net/questions/5450/cocktail-party-math/5551#5551 Comment by Ari Ari 2009-11-15T22:38:14Z 2009-11-15T22:38:14Z The idea is that, if you pick up a slice of pizza by the crust, gravity makes the tip of the slice droop downwards. This makes it difficult to eat, and the toppings can slide off. However, if you use your hand to fold it lengthwise a bit (i.e., perpendicular to the crust), then the pizza stays straight and no longer droops. (See <a href="http://slice.seriouseats.com/images/FoldHoldHalf-thumb.jpg" rel="nofollow">slice.seriouseats.com/images/&hellip;</a> for a picture of this.) Essentially, you make one of the principal curvatures positive -- but since the Gaussian curvature is zero, this forces the other principal curvature to be zero. http://mathoverflow.net/questions/3476/is-there-a-categorification-of-the-integers-in-terms-of-graded-sets/3484#3484 Comment by Ari Ari 2009-10-31T15:09:26Z 2009-10-31T15:09:26Z Do you mean that it's boring because every morphism (tangle) ends up being an isomorphism? If so, I wonder if tangles can be generalized so that this is not the case. (For example, one might relax the requirement that each point pairs with exactly one other point.) It would be nice to get the usual FinSet category in the cases where S1 is empty. http://mathoverflow.net/questions/2250/thematic-programs-for-2010-2011 Comment by Ari Ari 2009-10-27T02:10:00Z 2009-10-27T02:10:00Z A master list of thematic programs might be a good idea for a page on the Notable Math Wiki (<a href="http://notable.math.ucdavis.edu/wiki/Main_Page" rel="nofollow">notable.math.ucdavis.edu/wiki/Main_Page</a>), alongside the existing Math Jobs Wiki that they host.