User deane yang - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T06:57:21Z http://mathoverflow.net/feeds/user/613 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/130904/can-one-block-diagonalize-the-curvature-matrix-of-2-forms-on-a-riemannian-manif/130913#130913 Answer by Deane Yang for Can one (block) diagonalize the curvature matrix of 2 forms on a Riemannian manifold? Deane Yang 2013-05-17T06:01:33Z 2013-05-17T06:01:33Z <p>If you are asking whether there always exists an orthonormal basis where $R_{ijkl} = 0$ unless ${i,j} = {k,l}$, the answer is yes in dimension 2 or 3 and no in higher dimensions. In dimension 3 it's the same as diagonalizing the Ricci tensor. As for higher dimensions, in a paper with Dennis DeTurck on diagonal coordinates we show that those conditions hold if and only if the Weyl tensor vanishes. </p> http://mathoverflow.net/questions/130883/is-there-any-proof-that-you-feel-you-do-not-understand/130898#130898 Answer by Deane Yang for Is there any proof that you feel you do not "understand"? Deane Yang 2013-05-17T03:07:03Z 2013-05-17T03:07:03Z <p>I hate to sound dumb, but I still don't really understand the Pythagorean theorem as well as I like. I've seen lots of proofs but they all feel too clever for me. To me, understanding a theorem or its proof means being able to see why it's true without having to work out the details of the proof.</p> http://mathoverflow.net/questions/120875/ring-with-three-binary-operations Ring with three binary operations Deane Yang 2013-02-05T16:34:01Z 2013-05-06T06:05:54Z <p>A rather precocious student studying abstract algebra with me asked the following question: Are there interesting rings where there are not just two but three binary operations along with some appropriate distributivity properties?</p> http://mathoverflow.net/questions/127061/is-it-possible-to-get-another-math-phd/127093#127093 Answer by Deane Yang for Is it possible to get another math PhD Deane Yang 2013-04-10T13:25:25Z 2013-04-10T13:25:25Z <p>I agree with item 2 of Alexandre Eremenko's answer. Lower tier departments who care more about how they appear to the dean rather than the actual quality of the research do sometimes go after Ph.D.'s from top schools because it looks better. But departments that have actual aspirations of having strong research will look at anyone who has a strong track record, regardless of where that person got a Ph.D.</p> <p>Here are some thoughts:</p> <p>1) Get to know as many strong senior mathematicians as possible, especially those who work in your field or closely related ones. Try to make sure they are familiar with your work by giving seminars at their schools, attending and giving talks at conferences, posting your papers on arxiv, and sending a PDF of your paper to people you think might be interested (but don't overdo this).</p> <p>These people are invaluable in at least two ways: 1) Putting in a good word for you, whether in a formal letter or informally by email or phone. 2) Providing useful guidance to you on which departments are hiring and where you stand the best chance.</p> <p>Who writes strong letters for you matters at least as much as where you got a Ph.D.</p> <p>2) When visiting a department that you think you have a shot at getting a job at, don't be afraid to ask them what the prospects are and what they are looking for. Getting a better sense of how each department chooses how to hire is invaluable.</p> <p>3) Maintain good and friendly relations with absolutely everyone in your field. For example, when you write a paper, cite people generously, including anyone who has done anything related to your work, whether you used the work or not in your paper. Showing your appreciation for others' work by citing them is a good way to build a lot of goodwill.</p> http://mathoverflow.net/questions/126128/on-closed-simple-curve-with-curvature-at-most-1/126153#126153 Answer by Deane Yang for On closed simple curve with curvature at most 1 Deane Yang 2013-04-01T04:57:31Z 2013-04-01T04:57:31Z <p>This is not an answer, since I don't know a reference. But I now remember the proof when the curve is known to be convex. Here's a sketch:</p> <p>The trick is to put the disk of radius 1 so that it osculates the curve at a point where the curvature is largest. Now parameterize the curve as a function of the angle $\theta$ that the outer unit normal makes with the positive $x$-axis with the origin placed at the center of the disk. Define the support function $h$ as a function of theta to be the dot product of the corresponding point on the curve with the outer unit normal. First, note that the disk is inside the curve if and only if $h \ge 1$ for all $\theta$. It can be shown that $h$ satisfies the ODE $$h'' + h = \rho$$ where $\rho$ is the reciprocal of curvature. The result now follows by the Sturm comparison theorem. This proof works nicely in the sense that it can be generalized to higher dimensions using the second fundamental form.</p> <p>This all uses standard stuff from convex geometry, so I think it's likely it was known to Blaschke if not even earlier than that. I imagine that the proof above could be adapted to nonconvex curves, but I haven't tried. If so, I would imagine that this was also known a long time ago.</p> <p>Finally, the Harvard graduate student who figured out the proof somehow knew about the support function and that it satisfies the ODE above. The rest of us knew only the more usual differential geometric definitions of curvature, which were useless for this question.</p> http://mathoverflow.net/questions/123866/support-functions-of-3d-convex-bodies-in-spherical-polar-coordinates/123872#123872 Answer by Deane Yang for Support Functions Of 3D Convex Bodies In Spherical Polar Coordinates Deane Yang 2013-03-07T13:40:31Z 2013-03-07T13:40:31Z <p>For me, the simplest way to figure out whether a function, which is often defined as a function of the unit sphere, is a support function or not is to extend it to all of $\mathbb{R}^n$ as a function homogeneous of degree $1$. Then the function is a support function if and only if it is convex.</p> <p>You can figure out whether a smooth function of $\theta$ and $\phi$ is a support function or not by finding the formula for this function in terms of the extended function and differentiating twice.</p> http://mathoverflow.net/questions/123310/construct-embedding-given-metric/123325#123325 Answer by Deane Yang for Construct embedding given metric Deane Yang 2013-03-01T13:34:20Z 2013-03-01T16:27:00Z <p>Although, as jc says, this question was addressed in earlier questions, the answer is scattered among the accepted answer as well as the comments. So here is a summary:</p> <ol> <li><p>By the Nash-Kuiper theorem (which I believe is one of the original uses of the so-called h-principle), there always exists a global $C^1$ isometric embedding of a surface with a $C^1$ or better Riemannian metric. This holds whether the surface is closed, open, or has boundary and whether the metric is complete or not. This is both amazing and somewhat unsatisfactory. If the embedding is only $C^1$, there is no second fundamental form, which is in some sense the most important geometric invariant of a surface in $R^3$</p></li> <li><p>If the surface is closed and has a metric with strictly positive curvature, then the question is known as the Weyl problem. The existence of a global isometric embedding, if the metric is sufficiently smooth, was proved by Nirenberg in a 1953 CPAM paper. Its uniqueness was proved much earlier by Cohn-Vossen. Nirenberg's paper is a landmark paper, because it was one of the first to develop and use a priori estimates for nonlinear elliptic PDE's to prove a theorem in global differential geometry. This approach to differential geometry led eventually to the important and exciting work by, among many others, Yau, Schoen, Uhlenbeck, Taubes, Donaldson, Hamilton. But the solution of the Weyl problem is also credited to Pogorelov and maybe also Alexandrov. Unfortunately, their work was not well understood back in the 50's and 60's and did not get the attention it deserved until much later.</p></li> <li><p>If the curvature is strictly negative, it's easy to prove that no global isometric embedding can exist.</p></li> <li><p>If the curvature changes sign, then no global theorem is known. ADDED: This is not quite right. I just found a paper by Lin and Han that gives sufficient conditions for a 2-torus to be isometrically embedded in $R^3$. But the conditions are quite restrictive.</p></li> <li><p>With suitable assumptions on curvature, the existence of local isometric embeddings in the neighborhood of a point is known. But that's another long story.</p></li> </ol> <p>As Alexandre Eremenko says, you can find all the details in the book by Han and Hong. (Names of authors corrected)</p> http://mathoverflow.net/questions/122879/understanding-gibbss-inequality/122912#122912 Answer by Deane Yang for Understanding Gibbs's inequality Deane Yang 2013-02-25T18:44:55Z 2013-02-25T18:44:55Z <p>The trick is to use appropriate units or scaling for the different edges of the rectangular parallelopiped when computing its volume. More specifically, apply the uniform probability argument (i.e., the isoperimetric inequality) to the probabilities $$\tilde{p}_i = \frac{1}{n}\text{ and } \tilde{q_i} = \frac{q_i}{p_i}\left(\sum \frac{q_i}{p_i}\right)^{-1}.$$</p> http://mathoverflow.net/questions/122879/understanding-gibbss-inequality/122905#122905 Answer by Deane Yang for Understanding Gibbs's inequality Deane Yang 2013-02-25T17:01:42Z 2013-02-25T17:29:37Z <p>I don't have a geometric interpretation of this inequality, but I do like to think of it as a limiting case of the Holder inequality: Given $0 &lt; \alpha &lt; 1$,</p> <p>$$\frac{1}{1-\alpha}\log \sum_i p_i\left(\frac{q_i}{p_i}\right)^\alpha \le \frac{1}{1-\alpha}\log\left(\sum_i q_i\right)^\alpha \left(\sum_i p_i\right)^{1-\alpha} = 0.$$</p> <p>If you now take the limit $\alpha \rightarrow 1$, you get the Gibbs inequality. The inequality above can be viewed as the analogue of the Gibbs inequality but for Renyi entropy instead of Gibbs or Shannon entropy.</p> http://mathoverflow.net/questions/122729/why-is-it-important-that-partial-derivatives-commute/122730#122730 Answer by Deane Yang for Why is it important that partial derivatives commute? Deane Yang 2013-02-23T15:51:03Z 2013-02-23T15:51:03Z <p>To me, a Riemannian metric and the Levi-Civita connection associated with the metric represent the intrinsic geometric properties of a submanifold in Euclidean space induced by the inner product and natural flat connection on Euclidean space. Since they are intrinsic, their definitions can be extended from submanifolds of Euclidean space to abstract manifolds.</p> <p>If you don't assume the connection is torsion-free, then there are an infinite number of connections that are compatible with the metric (instead of exactly one), so the link between the geometric properties of the metric and that of the connection is much weaker.</p> http://mathoverflow.net/questions/121565/mathematicians-whose-works-were-criticized-by-contemporaries-but-became-widely-ac/121658#121658 Answer by Deane Yang for Mathematicians whose works were criticized by contemporaries but became widely accepted later Deane Yang 2013-02-12T23:45:56Z 2013-02-12T23:45:56Z <p>Louis de Branges and his proof of the Bieberbach conjecture. </p> http://mathoverflow.net/questions/121293/determining-a-convex-set-by-mixed-volumes/121365#121365 Answer by Deane Yang for determining a convex set by mixed volumes Deane Yang 2013-02-09T22:58:57Z 2013-02-09T22:58:57Z <p>As Yoav says in his comment, this is essentially the Minkowski problem, which asks whether, given a measure on the unit sphere, it is the surface area measure of a convex body and whether the convex body is unique. This was originally solved by Alexandrov and independently by Fenchel and Jessen. The convex body is unique only up to translation. If you add the assumption that the center of mass of the body is at the origin, then you do get uniqueness.</p> http://mathoverflow.net/questions/119940/tensor-contraction-and-covariant-derivative/119959#119959 Answer by Deane Yang for Tensor contraction and Covariant Derivative Deane Yang 2013-01-26T20:00:43Z 2013-01-26T20:00:43Z <p>I like the question. Below is a somewhat sketchy version of how I see this.</p> <p>I think the importance of tensors and contraction of tensors originates from trying to do basic differential geometry or vector calculus from a co-ordinate-free point of view. The most basic objects are curves and velocity vectors of curves. The key observation is that the set of all possible velocity vectors at a single point in space is naturally an abstract vector space. Given any set of co-ordinates, the $n$ velocity vectors you get by holding all but one co-ordinates fixed form a basis of this vector space. If you change co-ordinates, then chain rule leads to the appropriate change of basis formula for velocity vectors. This space of all velocity vectors at a point in space is also known as the tangent space at that point. The construction described above defines the tangent space at each point in a smooth manifold.</p> <p>But once you have a naturally defined abstract vector space, all of linear and multilinear algebra becomes available for use. For example, the dual space to the space of velocity vectors is the cotangent space, and at each point there is a natural contraction between a tangent vector and a cotangent vector. It is also natural to consider tensor products of copies of the tangent and cotangent spaces, leading to higher order tensors. Whether this is useful or not is not immediately obvious and becomes clear only after you go on further and discover how to construct tensors that have important geometric, physical, or topological meaning. But the upshot is that I like to think of a smooth manifold as being, among other things, a parameterized family of abstract vector spaces. Of course, that <em>is</em> exactly what a vector bundle is.</p> <p>As for the covariant derivative, this can be introduced on its own. It arises, because if you try to find a natural co-ordinate-free way to differentiate vector fields, the best you can do is the Lie derivative or Lie bracket of two vector fields and a closer study of this shows that it is limited in its properties and usefulness. In particular, you want to, say, be able to define the directional derivative of one vector field $V$ in the direction of another vector field $W$. Moreover, just as the directional derivative at a point of a function in the direction $W$ depends only on the value of $W$ at that point (and not a neighborhood), you want the same property for the directional derivative of $V$. Moreover, the Leibniz rule for a scalar multiple of a vector field is a natural property to want. But you find that these properties don't determine a unique connection, so a connection is an additional geometric assumption. This is somehow unsatisfying.</p> <p>The covariant derivative becomes more compelling, when it is introduced in the context of a Riemannian metric, which given the discussion about tangent vectors above, is a natural extension of the concept of Euclidean space and the inner product. Note that the Riemannian metric, as defined, is naturally a tensor. Next, there is something that, as far as I know is simply a miracle, namely that there is a unique naturally compatible torsion-free connection known as the Levi-Civita connection. After this, the value of tensors and the Levi-Civita connection is justified both because they correspond to naturally defined and interesting geometric concepts for submanifolds of Euclidean space and because they lead to so much insight and deep results in geometry, analysis, and topology.</p> http://mathoverflow.net/questions/119118/totally-geodesic-submanifolds/119120#119120 Answer by Deane Yang for Totally Geodesic Submanifolds Deane Yang 2013-01-16T23:27:45Z 2013-01-16T23:27:45Z <p>As for an example where $N$ is complete: Slice a 2-sphere just above and below a great circle. Keep the piece containing the great circle. Glue flat disks along the resulting boundaries and smooth the surface near the boundaries.</p> http://mathoverflow.net/questions/118626/real-symmetric-matrix-has-real-eigenvalues-elementary-proof/118856#118856 Answer by Deane Yang for real symmetric matrix has real eigenvalues - elementary proof Deane Yang 2013-01-14T03:32:30Z 2013-01-14T03:32:30Z <p>This is just the details of the first step of Alexander Eremenko's answer (so upvote his answer if you like mine), which I think is by far the most elementary. You only need two facts: A continuous function on a compact set in $R^n$ achieves its maximum (or minimum), and the derivative of a smooth function vanishes at a local maximum. And there's no need for Lagrange multipliers at all.</p> <p>Let $C$ be any closed annulus centered at $0$. The function $$R(x) = \frac{x\cdot Ax}{x\cdot x},$$ is continuous on $R^n\backslash{0}$ and therefore achieves a maximum on $C$. Since $R$ is homogeneous of degree $0$, any maximum point $x \in C$ is a maximum point on all of $R^n\backslash{0}$. Therefore, for any $v \in R^n$, $t = 0$ is a local maximum for the function $$f(t) = R(x + tv).$$ Differentiating this, we get $$0 = f'(0) = \frac{2}{x\cdot x}[Ax - R(x) x]\cdot v$$ This holds for any $v$ and therefore $x$ is an eigenvector of $A$ with eigenvalue $R(x)$.</p> http://mathoverflow.net/questions/117391/how-to-understand-the-diffeomorphism-in-the-cheeger-gromov-compactness/117474#117474 Answer by Deane Yang for How to understand the diffeomorphism in the Cheeger-Gromov compactness Deane Yang 2012-12-29T03:51:13Z 2012-12-29T03:51:13Z <p>I also don't understand the question at all. The diffeomorphisms are constructed to "normalize" the Riemannian metrics, so that the metrics differ as little as possible. So by construction they are as far fron being wild as possible. Cheeger-Gromov compactness is the statement that such "tame" diffeomorphisms exist and, since they behave so nicely, subsequences of both the diffeomorphisms and metrics converge on any compact domain.</p> <p>If you know something more about the metrics, then you can often use the additional information to construct particularly nice diffeomorphisms.</p> <p>If all of the metrics are rotationally symmetric on $\mathbb{R}^n$, then there is no need to construct diffeomorphisms at all. Cheeger-Gromov compactness, for example assuming bounded sectional curvature, in this situation is easily verified using the metrics written with respect to polar co-ordinates and the Jacobi equation.</p> http://mathoverflow.net/questions/116947/center-of-mass-from-the-abstract-point-of-view-or-could-the-ancient-greeks-inven/117003#117003 Answer by Deane Yang for Center of mass from the abstract point of view, or could the ancient Greeks invent modern analysis? Deane Yang 2012-12-22T02:33:58Z 2012-12-24T00:22:54Z <p>If I'm not mistaken (but I often am), the physicists already have a rather simple way of defining the center of mass. But I don't think you can do it with just sets. You have to associate a mass with each set. The critical axiom is simply the one we all know:</p> <p>If $A$ and $B$ are disjoint sets with masses $m(A)$ and $m(B)$ and center of masses $c(A)$ and $c(B)$ respectively, then the mass of $C$ is $m(C) = m(A) + m(B)$ and the center of mass of the set $C = A \cup B$ is given by $$c(C) = \frac{m(A)}{m(C)}c(A) + \frac{m(B)}{m(C)}c(B).$$</p> <p>You do need one more axiom to get started somehow. I believe physicists like to start with point masses (where the definition of the center of mass is easy) and then view a body as a limit of point masses. That's more or less what Liviu has proposed. But it also suffices to say that the center of mass of a square or cube is its geometric center. Or, more generally, the center of mass of any set with sufficient symmetry is its center.</p> <p>Of course, if you really want arbitrary shapes, then you do need a countable version of the first axiom. But I think that's all you need. Note that this approach allows for bodies with different and even non-constant mass densities.</p> <p>ADDED (in response to fedja's edit): It's worth noting explicitly that my answer above requires no notion of volume or choice of measure (such as Lebesgue measure) on the ambient space. It works on any length space.</p> <p>But I don't see any way reduce this to just geometry (and not physics) without a notion of volume. In essence, you do it by just assuming all objects have the same constant mass density, so the mass is essentially equal to volume.</p> <p>More generally, there has to be a way to measure the relative size of two sets, in order to determine the center of mass of the union of the two sets. If I understand correctly, axiom (4) in the question is an attempt to set this up.</p> <p>[PREVIOUS DISCUSSION REPLACED BY THE FOLLOWING]</p> <p>But for me it seems simpler to define a notion of size first and define the center of mass. And, as alvarezpaiva points out in a comment to Liviu's answer, valuations provide the appropriate setting, especially if we restrict to convex polytopes, which are objects that I believe the Greeks understood pretty well. This also allows us to avoid any issues of having to work with infinite sums or unions.</p> <p>Here, a valuation $f$ is a finitely additive function on the space of convex polytopes. In other words, given polytopes $A$ and $B$, $$f(A \cup B) + f(A \cap B) = f(A) + f(B).$$ The first observation is that the "critical axiom" stated above is equivalent to saying that $C \mapsto m(C)c(C)$ is a valuation. However, Monika Ludwig showed in her paper <em>Moment vectors of polytopes</em> that the only $R^n$-valued measurable valuation on convex polytopes that behaves appropriately under affine transformations is the volume of the polytope times the standard center of mass.</p> <p>Ludwig also showed in her Advances article <em>Valuations on polytopes containing the origin in their interiors</em> that any real-valued measurable $SL(n)$-invariant valuation homogeneous of positive degree must be a constant times volume. So it is reasonable to assume $m$ is volume. This therefore implies that $c$ must be the standard center of mass.</p> <p>Moreover, if you examine Ludwig's proofs, you will see that although they are quite nontrivial, the technology used was arguably within the grasp of the Greeks.</p> http://mathoverflow.net/questions/115427/pde-with-the-jacobian-determinant/115544#115544 Answer by Deane Yang for PDE with the Jacobian Determinant Deane Yang 2012-12-05T20:38:03Z 2012-12-05T20:38:03Z <p>This is not really an answer, but I prefer the luxury of the answer box instead of the rather spartan comment box.</p> <p>You have only one equation for $n$ unknown functions (the components of the vector-valued function $u$), so the equation is underdetermined. This gives you a lot of flexibility on what to do. Roughly speaking, you get to impose $n-1$ additional conditions in order to get a well-posed boundary value problem for a system of PDE's.</p> <p>There might be a clever choice of the $n-1$ conditions that makes your question very easy to answer, but I don't know and haven't been able to think of any.</p> <p>The most common way to do this is to set $u$ equal to the gradient of a scalar function $\phi$, which turns your equation into a real Monge-Ampere equation. If the function $f$ were always positive, then you can try to solve for $\phi$ convex. There is a well-developed theory of this situation, because the PDE becomes second order elliptic and many techniques are available.</p> <p>However, if $f$ changes sign, then much, much less is known. But here are some partial answers:</p> <p>1) If you restrict to $n=2$, assume $\Omega$ is a nice smooth domain, and assume that $f$ changes sign "cleanly" (i.e., $f$ vanishes along a smooth curve in $\Omega$ and its gradient is everywhere nonzero along this curve), then the equation becomes what is known as a nonlinear Tricomi equation, which has been studied. You can probably set up some appropriate boundary value problem to solve your equation on $\Omega$ this way.</p> <p>2) The ideas in 1) probably could be generalized to higher dimensions, but I'm unaware of any prior work on this. The idea is to solve for $\phi$ to have positive definite Hessian where $f$ is positive and signature $(n-1,1)$ where $f$ is negative. Then the PDE becomes elliptic where $f$ is positive and hyperbolic where $f$ is negative.</p> <p>3) All of this might be able to be recast in the form of a nonlinear symmetric positive system as defined and studied by K. O. Friederichs.</p> <p>4) Maybe the most promising but still rather difficult direction is to <em>not</em> use the Monge-Ampere equation but try to find a different set of $n-1$ additional first order equations to impose on $u$, (so you get an $n$-by-$n$ system of first order PDE's) such that the resulting system is symmetric positive. Then you try to identify the right boundary conditions that both leads to the existence of a solution and implies that $u$ vanishes on the boundary of $\Omega$.</p> http://mathoverflow.net/questions/114762/what-is-the-real-osculating-space-of-a-minimal-immersion/114784#114784 Answer by Deane Yang for What is the "real osculating space" of a (minimal) immersion? Deane Yang 2012-11-28T16:22:13Z 2012-11-28T16:22:13Z <p>View the immersion $x$ as an immersion into $R^{n+1}$. Then for each $p \in S^2$, there is a unique polynomial map $O_k: R^2 \rightarrow R^{n+1}$ of degree $k$ such that $O_k(0) = x(p)$ and the partials of $O_k$ of order $k$ or less at $0$ are equal to the corresponding covariant derivatives of $x$ at $p$.</p> http://mathoverflow.net/questions/114531/divergence-form-elliptic-pde-removable-singularity-regularity-question/114774#114774 Answer by Deane Yang for Divergence form Elliptic PDE Removable Singularity/Regularity Question Deane Yang 2012-11-28T14:33:24Z 2012-11-28T14:33:24Z <p>With a linear elliptic PDE, there's no way to bootstrap. What you see is what you get. The regularity of $A^{ij}\partial_j u$ cannot be made any better than the regularity of $g^i$. So if all you assume about $g^i$ is that it is $L^\infty$, then that's all you get for $A^{ij}\partial_j u$. Once this observation is made, it's easy to find the simplest possible example (where $u$ is Lipschitz, $\partial u$ is $L^\infty$, and you just set $A^{ij} = \delta^{ij}$ and $g^i = \partial_iu$). And that's exactly the one found by Daniel, which works in all dimensions.</p> http://mathoverflow.net/questions/114555/does-physics-need-non-analytic-smooth-functions/114575#114575 Answer by Deane Yang for Does Physics need non-analytic smooth functions? Deane Yang 2012-11-26T19:45:55Z 2012-11-27T12:30:30Z <p>It is worth noting that it is impossible to solve the initial value problem for the standard heat equation in the real analytic category. Here, there are asymptotic expansions available but no Taylor series.</p> <p>ADDED: It should also be noted that the directionality of time, as exhibited in the heat and diffusion processes, is a phenomenon lives outside the real analytic category. I believe that any PDE in the real analytic category that is well-posed as an initial value problem can be solved in both positive and time directions. That's not true for the heat equation in the smooth category. So the need for going outside the real analytic category appears already in fundamental classical physics.</p> <p>This can be avoided, I suppose, by working purely with discrete models, but that for some of us is a cure worse than the original "problem".</p> http://mathoverflow.net/questions/114201/reference-for-cohomology-theories/114443#114443 Answer by Deane Yang for reference for (co)homology theories Deane Yang 2012-11-25T20:02:00Z 2012-11-25T20:02:00Z <p>Chapter 5 of Frank Warner's book, Foundations of Differentiable Manifolds, (http://www.amazon.com/Foundations-Differentiable-Manifolds-Graduate-Mathematics/dp/0387908943) presents four different cohomology theories and why they are all isomorphic.</p> http://mathoverflow.net/questions/111915/a-question-about-first-order-hyperbolic-equations/111931#111931 Answer by Deane Yang for a question about first-order hyperbolic equations Deane Yang 2012-11-09T19:33:47Z 2012-11-09T19:33:47Z <p>If I understand correctly, you have a PDE of the form $$(D_t - \lambda_1(t, x, D_x) + \lambda_0(t, x, D_t, D_x))u = f,$$ where $\lambda_1$ is a first order pseudodifferential operator and $\lambda_0$ is a zero-th order pseudodifferential operator. It seems to me that the proofs of many if not all estimates, including energy integral estimates, and theorems about regularity, uniqueness, and existence for the equation $$(D_t - a(t,x)D_x + b(t,x))u = f,$$ as presented in books like Taylor can be extended to your PDE.</p> http://mathoverflow.net/questions/110184/refined-kato-inequality/110242#110242 Answer by Deane Yang for refined Kato inequality Deane Yang 2012-10-21T15:54:56Z 2012-10-21T15:54:56Z <p>As Matt says, the improved Kato inequality I know about involves the trace of the Hessian squared and not the determinant. The oldest proof I'm aware of for this inequality is in Stein's book, "Singular Integral Operators and Differentiability Properties of Functions", where he uses it to get improved estimates for functions in a Hardy space. But I believe that he does not call it a "Kato inequality".</p> <p>This inequality was later put to extremely good use by Uhlenbeck, Taubes, Yau, and others and plays a critical role in many major theorems on minimal hypersurfaces, self-dual Yang-Mils fields, and special Riemannian metrics, especially where the PDE involved is "critical" or "scale-invariant". You can, for example, find a rather messy proof of the inequality (Stein's is cleaner) in the paper of Schoen-Simon-Yau, "Curvature estimates for minimal hypersurfaces".</p> <p>Anyway, here's my take on it. The improved Kato inequality is a pointwise tensor space inequality that holds in many different situations. The desired inequality is, given a section $T$ of a tensor bundle $$|\nabla |\nabla T|| \le c |\nabla^2T|.$$ With $c = 1$, this follows easily from the identity $$|\nabla |\nabla T|| = \frac{|\nabla T \cdot \nabla^2T|}{|\nabla T|}$$ and the Cauchy-Schwarz inequality $$|\langle v, t\rangle| \le c|v||t|$$ with $c = 1$, where $t \in V\otimes W$, $v \in V$, and $V$ and $W$ are vector spaces with inner products, and $\langle\cdot,\cdot\rangle$ is the inner product in $V$.</p> <p>So here's the key point: If you happen to know that $t$ always lies in a known subspace of $V\otimes W$, then it is sometimes possible that the last inequality holds with a constant less than 1. To identify the optimal constant, it is easiest to differentiate $$\frac{|\langle v, t\rangle|^2}{|v|^2|t|^2}$$ with respect to $t$ and solve for the critical points of $t$ restricted to the subspace.</p> <p>In the specific example cited in the question, $W = V$, so $V\otimes W$ is the space of all matrices but $t$ is restricted to the space of traceless matrices. The same argument gives for example a more general improved Kato inequality for divergence free tensor fields. This was used, for example, in the work of Taubes and Uhlenbeck on self-dual Yang-Mills.</p> http://mathoverflow.net/questions/109778/variational-problems-whose-lagrangian-density-depends-on-derivatives-higher-than/109902#109902 Answer by Deane Yang for Variational problems whose lagrangian density depends on derivatives higher than 1. Deane Yang 2012-10-17T12:26:49Z 2012-10-17T12:26:49Z <p>You might find it helpful to study previous work on specific functionals like this. I suggest searching for journal articles that talk about extremal Kahler metrics (which usually means studying the integral of scalar curvature squared on a closed Kahler manifold) or Q-curvature (which is a curvature integral defined for a conformal structure).</p> http://mathoverflow.net/questions/102321/have-derivatives-of-determinants-along-1-psgs-ever-been-coherently-computed-vi/109809#109809 Answer by Deane Yang for Have derivatives of determinants along 1-psg's ever been 'coherently' computed via Jacobi's formula? Deane Yang 2012-10-16T12:21:26Z 2012-10-16T12:21:26Z <p>The formula $d(\det A) = (\det A)\operatorname{tr} A^{-1}dA$ or equivalently $d(\log\det A) = \operatorname{tr}A^{-1}dA$ is <em>extremely</em> useful. I'm surprised there aren't more answers to this question. A simple but very important example is the Bishop-Gromov inequality in Riemannian geometry.</p> http://mathoverflow.net/questions/109474/bounded-curvature-derivatives-and-shis-estimates/109476#109476 Answer by Deane Yang for Bounded curvature (derivatives) and Shi's estimates Deane Yang 2012-10-12T18:10:24Z 2012-10-13T15:36:52Z <ol> <li><p>The norm defined by the Riemannian metric $g$ on the tangent and cotangent bundles naturally induces a norm on each tensor bundle. This follows from the fact that given vector spaces $V$ and $W$ with inner products, there is a naturally induced inner product on the vector space $V\otimes W$.</p></li> <li><p>The metric at $t = 0$ is initially prescribed data and therefore has nothing to do with the Ricci flow itself. So your question is equivalen to: Given a Riemannian metric $g$, when do the covariant derivatives of order $k$ of Riemann curvature have pointwise bounded norm? A sufficient condition is that $g$ can be written in local co-ordinates as $g_ij\,dx^i\,dx^j$, where the function $g_{ij}$ are $C^{k+2}$ functions of the co-ordinates $x^1, \dots x^n$.</p></li> </ol> <p>EDIT: My answer to #2 is incomplete, since we're working on a noncompact manifold. You also need uniform pointwise upper and lower bounds on the eigenvalues of $g_{ij}$, as well as its derivatives up to order $k+2$ with respect to local co-ordinates.</p> http://mathoverflow.net/questions/109042/reference-request-parabolic-pde/109104#109104 Answer by Deane Yang for Reference request: parabolic PDE Deane Yang 2012-10-07T22:47:28Z 2012-10-07T22:47:28Z <p>If it's the Ricci flow you're really interested in, I recommend checking out the books on the Ricci flow written by Bennett Chow and his co-authors.</p> http://mathoverflow.net/questions/108519/quasi-linear-system-of-first-order-p-d-e-s-of-mixed-type/108677#108677 Answer by Deane Yang for Quasi-linear System of First Order P.D.E.s of "Mixed" type Deane Yang 2012-10-03T00:33:08Z 2012-10-03T00:44:26Z <p>I gave an answer to your earlier question, but although it was technically correct, I see now that since you have only two independent variables my answer wasn't really the right one. Let me try one more time (but I can't guarantee I'm right).</p> <p>I'm going to address only case 4, since that's the one you care about. Given your assumptions, there exist invertible-matrix-valued functions $P(x,u)$ and $Q(x, u)$ such that the system $$P(A\partial_1 u + B\partial_2 u) = Pf$$ can be written as $$\partial_1 v + C(x,v)\partial_2v = g(x,v),$$ where $v = (v_1, v_2, v_3, v_4) = Q(x,u)u$. $$C(x,v) = \begin{bmatrix} a(x,b)&amp; 0 &amp; 0 &amp; 0 \newline 0 &amp; b(x,v) &amp; 0 &amp; 0 \newline 0 &amp; 0 &amp; 0 &amp; -1 \newline 0 &amp; 0 &amp; 1 &amp; 0 \end{bmatrix}$$ If you fix a rectangular domain, say, $D = [0,T]\times[0,S]$, then this system is well-behaved from the point of view of uniqueness and, with the right additional assumptions, existence if you specify initial data consisting of $v_1$ and $v_2$ on ${0} \times [0,S]$ and $v_3$ on the boundary of $D$.</p> <p>This is a coupled hyperbolic-elliptic system that is hyperbolic in $v_1$ and $v_2$ and elliptic in $v_3$ and $v_4$. My guess is that linear systems like this been analyzed and solved before, but unfortunately I don't know any specific reference. You would handle the quasilinear version using the linear theory using the usual techniques (inverse function theorem, fixed point theory, etc.). If you know the theory of linear first order elliptic systems of PDE's and of linear first order hyperbolic systems of PDE's in two independent variables really well, it's reasonably straightforward to adapt it to a coupled system like this.</p> http://mathoverflow.net/questions/108123/degeneration-of-riemannian-metrics-with-curvature-bounds/108159#108159 Answer by Deane Yang for Degeneration of riemannian metrics with curvature bounds Deane Yang 2012-09-26T12:38:38Z 2012-09-26T12:38:38Z <p>The seminal work on how a Riemannian manifold can degenerate with pointwise bounds on curvature are Cheeger and Gromov's JDG papers, "Collapsing Riemannian manifolds while keeping their curvature bounded" as well as papers by Fukaya, which can be found in the references of the Cheeger-Fukaya-Gromov JAMS paper, "Nilpotent structures and invariant metrics on collapsed manifolds".</p> <p>You can then find other papers on the subject by looking up on Mathscinet which papers cite these papers. You've mentioned degenerations under conformal deformations and degeneration of Kahler metrics. You should be able to find any existing work on this by doing the above. I haven't looked at this for a long time, so unfortunately I can't help more.</p> <p>As for degenerations with only an integral bound on curvature, searching on "integral bound and curvature" in the Anywhere box on Mathscinet seems to produce lots of hits but I suspect the vast majority of papers study situations where degeneration does not occur. I have a Duke paper, "Riemannian manifolds with small integral norm of curvature" that proves an $L_p$ analogue of the Cheeger-Gromov collapse theorem and also describes an example of what can happen with only an integral bound but doesn't happen with pointwise bounded curvature. I don't know of any other work in this direction, but I'm hoping that someone else does.</p> http://mathoverflow.net/questions/131051/the-relations-between-the-perelmans-entropy-functional-and-notions-of-entropy-fr Comment by Deane Yang Deane Yang 2013-05-18T15:39:00Z 2013-05-18T15:39:00Z Although it might seem like nothing more than a formal correspondence, the power of using entropy-type functionals for certain types of elliptic and parabolic PDE's indicates strongly to me that there is a deeper connection to the physical and information theoretic definitions of entropy than what we currently understand. http://mathoverflow.net/questions/130883/is-there-any-proof-that-you-feel-you-do-not-understand/130898#130898 Comment by Deane Yang Deane Yang 2013-05-18T15:29:52Z 2013-05-18T15:29:52Z alvarezpaiva, sorry about that. I did realize you were joking but couldn't help responding to it seriously. http://mathoverflow.net/questions/130883/is-there-any-proof-that-you-feel-you-do-not-understand/130898#130898 Comment by Deane Yang Deane Yang 2013-05-18T15:29:12Z 2013-05-18T15:29:12Z Dustin, I have the impression that to identify the geometric meaning of the dot product and write down the formulas for the rotation group requires the Pythagorean theorem itself. http://mathoverflow.net/questions/131053/differentiable-manifolds-by-serge-lang-question Comment by Deane Yang Deane Yang 2013-05-18T15:27:18Z 2013-05-18T15:27:18Z I second Roy's comment. Lang's book is suitable (but not necessarily the best) only if you are already familiar with finite-dimensional manifolds and have a specific need for learning about infinite-dimensional manifolds. Even then, it might be easier to focus first only on the specific infinite-dimensional spaces that you want to work with, rather than learning the general theory first. http://mathoverflow.net/questions/131053/differentiable-manifolds-by-serge-lang-question Comment by Deane Yang Deane Yang 2013-05-18T14:58:54Z 2013-05-18T14:58:54Z It depends on what you mean by &quot;traditional treatment&quot;. There are in fact many different &quot;traditional treatments&quot; of tensors on manifolds, and some of them are indeed quite messy. http://mathoverflow.net/questions/131051/the-relations-between-the-perelmans-entropy-functional-and-notions-of-entropy-fr Comment by Deane Yang Deane Yang 2013-05-18T14:43:22Z 2013-05-18T14:43:22Z The more standard notions of entropy, notably Boltzmann and Shannon are roughly of the form $\int u\log u\,d\mu$ and if in Perelman's definiton you set $u = e^{-f}$ you get one term like this. The gradient term looks to me more like Fisher information, which can be viewed as the derivative of entropy with respect to time under Brownian motion. I suppose that the scalar curvature arises because everything is on a curved instead of flat space. The constant term arises from normalization. But this is all just my speculation. http://mathoverflow.net/questions/130883/is-there-any-proof-that-you-feel-you-do-not-understand/130898#130898 Comment by Deane Yang Deane Yang 2013-05-18T03:51:28Z 2013-05-18T03:51:28Z alvarezpaiva, isn't it a little unfair to compare the Pythagorean theorem to its converse? The concept of scaling and similar triangles is, as Palais and Tao show, a rather powerful tool. http://mathoverflow.net/questions/130883/is-there-any-proof-that-you-feel-you-do-not-understand/130898#130898 Comment by Deane Yang Deane Yang 2013-05-18T03:49:38Z 2013-05-18T03:49:38Z Francois, Tao does indeed give a nice version of the scaling proof. http://mathoverflow.net/questions/130883/is-there-any-proof-that-you-feel-you-do-not-understand/130898#130898 Comment by Deane Yang Deane Yang 2013-05-17T05:46:32Z 2013-05-17T05:46:32Z Dick, that is indeed the most natural proof I've ever seen. Thanks. I never knew that one before. http://mathoverflow.net/questions/130883/is-there-any-proof-that-you-feel-you-do-not-understand/130905#130905 Comment by Deane Yang Deane Yang 2013-05-17T04:25:02Z 2013-05-17T04:25:02Z Or vice versa, using the difference quotient to estimate the derivative on each subinterval and observe that if you multiply each difference quotient by the size of the subinterval and add them all up, you get the difference between the values of the function at the endpoints of the full interval. http://mathoverflow.net/questions/130883/is-there-any-proof-that-you-feel-you-do-not-understand/130905#130905 Comment by Deane Yang Deane Yang 2013-05-17T04:22:43Z 2013-05-17T04:22:43Z I think of the fundamental theorem of calculus as breaking up the interval into smaller subintervals and applying the tangent line approximation on each successive subinterval. http://mathoverflow.net/questions/130883/is-there-any-proof-that-you-feel-you-do-not-understand/130898#130898 Comment by Deane Yang Deane Yang 2013-05-17T04:18:45Z 2013-05-17T04:18:45Z I must say that I'm gratified that at least 8 other people are also willing to admit (if anonymously) that they're as dumb as me. http://mathoverflow.net/questions/130883/is-there-any-proof-that-you-feel-you-do-not-understand/130898#130898 Comment by Deane Yang Deane Yang 2013-05-17T04:17:52Z 2013-05-17T04:17:52Z You lost me at &quot;4 congruent triangles to play with&quot;. I do know this proof. I agree that once you're given the 4 congruent triangles, you have a good chance of figuring out this proof. But where does the idea of playing with 4 congruent triangles come from? http://mathoverflow.net/questions/130881/constructing-a-sobolev-space-containing-the-differential-k-forms-of-a-riemannian Comment by Deane Yang Deane Yang 2013-05-17T02:14:05Z 2013-05-17T02:14:05Z And you shouldn't be shy about interrupting these busy professors. No matter how much research they're doing, they have a responsibility to teach their students and answer questions. You can tell them I said so. http://mathoverflow.net/questions/130881/constructing-a-sobolev-space-containing-the-differential-k-forms-of-a-riemannian Comment by Deane Yang Deane Yang 2013-05-17T02:12:11Z 2013-05-17T02:12:11Z Actually, the only detail that you didn't get right in the summary is that the Laplacian is not an operator from $H$ to $H$, where $H$ is a fixed Sobolev space. The Sobolev space you want to work with is $W^{2,k}$, where the norm of $\omega$ is the sum of the $L^2$ norms of the derivatives of $\omega$ up to order $k$. Then the Laplacian is a bounded linear map from $W^{2,k}$ to $W^{2,k-2}$. You construct the Sobolev space as the completion of smooth forms on $M$.