User viktor bundle - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T07:19:17Z http://mathoverflow.net/feeds/user/15856 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/116501/upper-bound-on-first-non-zero-eigenvalue-of-the-laplacian upper bound on first non-zero eigenvalue of the Laplacian Viktor Bundle 2012-12-16T01:09:40Z 2012-12-19T15:41:46Z <p>I'm looking for an upper bound on the first non-zero eigenvalue of the Laplace-Beltrami operator on compact manifolds of dimension greater than four that have constant negative curvature. In particular, I would like to know whether or not there is an upper bound that is less than $\frac{1}{2}n^2 - 2$, where $n$ is the dimension, for the case when the sectional curvature is $-1$. I can find quite a few papers on lower bounds, but the few I could find on upper bounds required information that is hard to obtain. Any help would be greatly appreciated.</p> http://mathoverflow.net/questions/70952/smooth-real-analytic-interpolation-of-monotonic-sequence smooth/real analytic interpolation of monotonic sequence Viktor Bundle 2011-07-22T02:45:28Z 2012-10-02T09:16:58Z <p>Let $x_n$ be a increasing sequence of negative real numbers that converge to $0$. Let $f$ be a function defined on $x_n$ such that $f(x_n)$ is a increasing sequence of negative numbers that converge to zero. Can I find a $C^\infty$ or real-analytic function $g$ defined on a ball around zero such that $g(x_n)=f(x_n)$ for all $n$ larger than some positive integer? </p> http://mathoverflow.net/questions/68810/vanishing-on-bad-sets Vanishing on Bad Sets Viktor Bundle 2011-06-25T20:16:14Z 2012-05-29T20:01:15Z <p>Let $f: \Bbb{R}^n \rightarrow \Bbb{R}$ be a non-negative function that vanishes on a set $\Omega$ that is compact and has positive measure. What is the minimial amount of regularity required of $f$ to guarantee that $\Omega$ contains an open set? I'm interested in classes of the form $C^k$ or $C^{k,\alpha}$ ($k$-times continuous differentiability and Hölder spaces, respectively).</p> http://mathoverflow.net/questions/91436/unique-continuation-principle unique continuation principle Viktor Bundle 2012-03-17T00:38:51Z 2012-03-19T01:56:53Z <p>I recently encountered a paper by Protter ("Unique continuation of elliptic equations") that starts out by saying "any solution of an elliptic equation that is defined on a domain $D$ must vanish on all of $D$ if it vanishes on an open set in $D$" and calls this the unique continuation principle. The problem is that this statement is so general that I'm afraid to use it. Does anyone know the rigorous statement of the unique continuation principle? In particular I wonder what are the conditions on the elliptic equation and the solution involved. </p> http://mathoverflow.net/questions/90752/ellipticity-independent-of-metric ellipticity independent of metric? Viktor Bundle 2012-03-09T19:17:36Z 2012-03-09T21:00:23Z <p>I am new to the theory of pseudo-differential operators on compact manifolds, but I need to use a result related to this theory in a proof I'm working on. The problem is as follows: Let $(M,g)$ be a closed, compact Riemannian manifold. It is clear that $L_g:=(\Delta_g + 1)^k$, where $\Delta_g$ is the Laplace-Beltrami operator for the metric $g$, is an elliptic pseudo-differential operator for all positive real numbers $k$. Does it follow that $L_g$ is elliptic on $(M,h)$, where $h$ is a metric conformal to $g$? </p> <p>This result is easy to prove (in the affirmative) for integral values of $k$, but I am uncomfortable with the sketch of a proof I have for non-integral $k$. Any thoughts on the matter would be appreciated. The main idea that I have to work with is my unproven notion that a symbol for a pseudo-differential operator is independent of the metric, but I having problems proving this idea in the manifold setting. </p> http://mathoverflow.net/questions/90598/estimating-norms-of-derivatives Estimating norms of derivatives Viktor Bundle 2012-03-08T17:24:22Z 2012-03-08T20:33:51Z <p>Let $(M,g)$ be a closed, compact Riemannian manifold. Let $u$ be a smooth function. Let $H^{-k}(M)$,, $k$ is a positive integer, be the dual Hilbert space of $H^{k}(M)$. Does it follow that $|| |\nabla^l u| ||^{-k}$, where $||\circ||^{-k}$ is the norm on $H^{-k}(M)$ and $l \in \Bbb{Z}_+$, can be bounded by $|| u ||^{l-k}$, where $||\circ||^{l-k}$ is the norm on $H^{l-k}(M)$? </p> http://mathoverflow.net/questions/88941/ellipticity-and-invertible-differential-operators ellipticity and invertible differential operators Viktor Bundle 2012-02-19T15:59:10Z 2012-02-19T16:59:32Z <p>Let $(M,g)$ be a closed, compact Riemannian manifold. Let $P$ be a $2r$th order pseudo-differential operator, where $r \in \Bbb{R}_+$. Suppose that the differential equation $Pu=f$ has a unique $H^r(M)$ solution for all $f$ in the dual space of $H^r(M)$. Does it follow that $P$ is elliptic?</p> http://mathoverflow.net/questions/69427/obstructions-to-einstein-metrics-in-high-dimensions Obstructions to Einstein metrics in high dimensions Viktor Bundle 2011-07-03T23:40:20Z 2012-01-29T19:32:19Z <p>It is well known that there exists three and four manifolds that do not admit an Einstein metric, but I wonder if this question is still open for manifolds of dimension higher than four. That is, does anyone know of a compact n-manifold, $n>4$, that does not admit an Einstein metric?</p> http://mathoverflow.net/questions/80789/linear-maps-on-incomplete-inner-product-spaces Linear maps on incomplete inner-product spaces Viktor Bundle 2011-11-13T01:11:09Z 2011-12-25T21:13:26Z <p>Let $(M,g)$ be a compact Riemannian manifold without boundary. Let $V$ be the vector space $C^4(M) \times C^4(M)$, where addition and scalar multiplication are defined in the obvious way. Suppose we turn $V$ into an incomplete inner-product space with a inner product given by the second variation of a $C^2$ functional $E$ on $V$, $E=E(w,v)$, with respect to the second argument $v$. (It is symmetric and positive definite in this case.) Now let $X$ be a nowhere dense subset of $V$. Given this, can there exist a linear map $B$ from $X$ to $V$ that is one-to-one and onto? </p> http://mathoverflow.net/questions/82675/a-proposed-version-of-a-implicit-function-theorem-on-banach-spaces A proposed version of a implicit function theorem on Banach spaces Viktor Bundle 2011-12-05T05:29:25Z 2011-12-05T05:29:25Z <p>Let $X$ be a banach space, and let $E: X \times X \rightarrow \Bbb{R}$ be a $C^2$ function. Suppose that at the point $(0,0)$ we have it that $E(0,0)=0$, and that $dE = 0$ at that point as well. Suppose as well that $E_{vv}$, the second variation of $E$ with respect to $v$, is positive definite at $(0,0)$. Does it follow that we can express the level manifold for $E$ at the value $0$ with an equation of the form $v=f(w)$ near $(0,0)$? Note that I don't need any type of smoothness, other than continuity, for the function $f$.</p> http://mathoverflow.net/questions/81138/kernel-of-the-adjoint-operator-and-the-cokernel-of-the-operator kernel of the adjoint operator and the cokernel of the operator Viktor Bundle 2011-11-17T03:44:34Z 2011-11-17T03:44:34Z <p>Let $L$ be a closed linear operator from Banach space $X$ to Banach space $Y$. Under the heading of Fredholm operators in the "Encyclopedic Dictionary of Mathematics" it says that if the range of $L$ is closed and the domain dense, then the dimension of the cokernel is equal to the dimension of the kernel of the adjoint of $L$. Unfortunately there is some ambiguity as to whether or not they are already supposing that the operator is Fredholm. So, my question is whether or not one needs to assume that $L$ is Fredholm in addition to the hypotheses stated above to conclude that the dimension of the cokernel is equal to the dimension of the kernel of the adjoint. </p> http://mathoverflow.net/questions/80850/non-negativity-to-positivity non-negativity to positivity Viktor Bundle 2011-11-13T22:47:15Z 2011-11-13T22:47:15Z <p>Let $(M,g)$ be a closed Riemannian manifold of dimension $n>7$. In this setting I have been able to prove that the Green's function of a positive Paneitz-Branson operator is non-negative. Furthermore, we can also conclude that the Green's function can't vanish on open sets. I need the Green's function to be positive, though, so I was wondering if anyone had an idea as to how to establish positivity of the Green's function under the hypotheses used above.</p> <p>The definition of the Paneitz-Branson operator $P_g$ of the metric is as follows:</p> <p>$P_g:= (\Delta_g)^2 - div_g(a_n R_g g + b_n Ric_g)d + Q_g.$</p> <p>Here $\Delta_g$ is the Laplace-Beltrami operator, $a_n = \frac{(n-2)^2 +4}{2(n-1)(n-2)}$, $b_n = \frac{-4}{n-2}$, and $Q$ is the $Q$-curvature of the metric $g$. Note that it is a fourth-order operator and hence classical maximum principles can't be used in general. Also, if the hypothesis that $g$ is locally conformally flat is used, that would be ok too. Of course, if a counter-example is found that would be of interest as well.</p> http://mathoverflow.net/questions/79972/vanishing-on-sets-with-non-zero-measure Vanishing on sets with non-zero measure Viktor Bundle 2011-11-03T20:22:03Z 2011-11-03T20:22:03Z <p>Let $(M,g)$ be a smooth, compact, Riemannian manifold without boundary. Let $P$ be a positive, formally self-adjoint elliptic differential operator. Let $G$ be the Green's function of $P$ and let $f$ be a non-negative smooth function with small enough support so that $u$, the solution of $Pu = fu$, is positive on the support of $f$. Suppose that $u$ is negative somewhere. Can we conclude that $u$ vanishes on a set of measure zero? By unique continuation properties of elliptic operators we know that it can't vanish on open sets, but it seems that there is still the possibility of vanishing on a compact set of non-zero measure if we don't know whether or not the gradient vanishes on the level submanifold $u=0$.</p> <p>What I want to conclude is that $fu^s$ is a $L^1$ function, where $s$ is a negative real number. The only problem is that there will be places where $f=0$ and $u^s = \infty$, so I want to conclude that that is on a set of measure zero.</p> http://mathoverflow.net/questions/69432/looking-for-ideas-concerning-the-teaching-of-lower-division-differential-equation Looking for ideas concerning the teaching of lower-division differential equation courses... Viktor Bundle 2011-07-04T01:02:02Z 2011-11-02T23:29:34Z <p>I'm looking for problems/lessons plans that could be used in a lower-division differential equations course that involve discerning properties of solutions of an equation, IVP, or BVP, without looking for an explicit/implicit solution (general or particular, given the context). Flow lines are an example of this, but I'm looking for something more advanced. One idea I've used is using first-order autonomous equations to figure out the dynamical properties of solutions for different initial conditions. I'm looking for similar ideas. Also: recommendations on how to present existence/uniqueness issues, besides showing a lot of examples, would be appreciated. (Boyce and DiPrima try to give a sketch of a proof of the basic existence/uniqueness result for first order IVPs. I wonder if this can be done without a course on analysis under your belt.)</p> <p>Bonus: What about introducing group theoretic concepts at an early level? There are textbooks that claim to do this, but I wonder if this is as untenable as trying to teach measure theory in a calculus course. </p> http://mathoverflow.net/questions/68438/einstein-metrics-and-conformal-geometry Einstein metrics and conformal geometry Viktor Bundle 2011-06-21T21:25:09Z 2011-11-02T04:05:10Z <p>I recall reading somewhere that if a conformal class contains an Einstein metric then that metric is the unique metric with constant scalar curvature in its conformal class, with the exception of the case of the round sphere. Does this sound right? If it is true: where can I find the proof of this result? </p> http://mathoverflow.net/questions/78614/meromorphic-family-of-pseudo-differential-operators meromorphic family of pseudo-differential operators Viktor Bundle 2011-10-19T21:10:51Z 2011-10-20T03:22:55Z <p>Let $(M,g)$ be a closed, Riemannian manifold. Let $S(z)$ be a holomorphic family of pseudo-differential operators, with $z \in \Bbb{C}$. Let $u$ be a smooth function. Does it follow that $\lim_{y \rightarrow z} ||S(y)u - S(z)u||_\infty = 0$?</p> http://mathoverflow.net/questions/73264/asymptotics-of-a-bubble Asymptotics of a bubble Viktor Bundle 2011-08-20T06:15:32Z 2011-08-21T03:04:23Z <p>Let $n$ be a positive integer greater than seven. Let $u_a = (\frac{a}{a^2 + r^2})^\frac{n-4}{2}$, where $a$ is a positive real number. Let $\Delta u_a$ be the Laplacian of $u_a$. What is the dominant term of the small $a$ asymptotic expansion of the integral $\int_{r=b}^\infty (\Delta u_a)^2 r^{n-1} dr$, where $b$ is a small positive real number? </p> <p>This problem arises naturally when considering the convergence of Palais-Smale sequences of the Paneitz-Branson functional. I've tried using Maple to solve this problem, but it can't handle the problem due to the fact that $n$ is left undetermined. If $n$ is odd, I believe one can use the method of contour integrals in the complex plane to obtain a solution, so I'm primarily interested in the case where $n$ is even.</p> http://mathoverflow.net/questions/71039/hilbert-space-inner-products-with-converging-metrics Hilbert space inner products with converging metrics Viktor Bundle 2011-07-23T02:36:45Z 2011-07-23T02:47:12Z <p>Let $(M,g)$ be a closed, smooth, Riemannian manifold of dimension $n>4$. Let $P^g$ be a positive, self-adjoint elliptic differential operator on $(M,g)$. Then we have it that $(u,P^g v)=({P^g}^{1/2}u,{P^g}^{1/2}v)=\int_M {P^g}^{1/2}u {P^g}^{1/2}v dv_g$ is an inner-product on $H^2(M)$. Now let $g^n$ be a sequence of Riemannian metrics on $M$ that converges to $g$ in $C^4$ norm. Let $u^n$ and $v^n$ be a sequence of smooth functions that converge in $H^2$ norm to $u$ and $v$, respectively. Suppose $(u^n,P^{g^n} v^n) \rightarrow 0$, with $(u^n,P^{g^n} v^n) \neq 0$, as $n \rightarrow \infty$. Suppose as well that $(u^n,P^g v^n) \rightarrow 0$ as $n \rightarrow \infty$. Does it follow that $(u^n,P^g v^n)$ over $(u^n,P^{g^n} v^n)$ goes to 1 as $n \rightarrow \infty$?</p> http://mathoverflow.net/questions/70590/kernel-of-the-conformal-laplacian kernel of the conformal Laplacian Viktor Bundle 2011-07-18T03:53:54Z 2011-07-18T03:53:54Z <p>Let $M$ be a smooth, closed manifold of dimension $n>2$. Let $L_g$ be the conformal Laplacian of the metric $g$. That is, $L_g=-\Delta_g + \frac{n-2}{4(n-1)}R_g$, where $R_g$ is the scalar curvature of the metric $g$. Is the set of $C^2$ Riemannian metrics on $M$ such that conformal Laplacian has a trivial kernel dense with respect to the $C^2$ norm? </p> http://mathoverflow.net/questions/70390/bound-on-q-implies-bound-on-r Bound on $Q$ implies bound on $R$? Viktor Bundle 2011-07-15T01:38:07Z 2011-07-15T01:38:07Z <p>Let $(M,g)$ be a smooth, closed Riemannian manifold with dimension $n>4$. Define the $Q$-curvature through the formula</p> <p>$Q = \Delta R + \frac{n^3-4n^2+16n-16}{(n-1)(n-2)^2} R^2 - \frac{8(n-1)}{(n-2)^2}|Ric|^2,$</p> <p>where $\Delta = -div\nabla$, $R$ is the scalar curvature, and $|Ric|$ is the norm of the Ricci tensor. This quantity arises in many problems in conformal differential geometry. The question is whether or not a uniform sup-norm bound on a sequence of $Q$ curvatures implies there exists a uniform sup-norm bound on the corresponding scalar curvatures, provided that all of the scalar curvatures change sign?</p> http://mathoverflow.net/questions/70279/estimate-on-the-size-of-flat-balls-where-the-weyl-tensor-vanishes Estimate on the size of flat balls where the Weyl tensor vanishes Viktor Bundle 2011-07-14T00:35:23Z 2011-07-14T03:05:17Z <p>Let $(M,g)$ be a Riemannian manifold and suppose that the Weyl tensor of $g$ vanishes at a point $p \in M$. Can one estimate the size of the largest geodesic ball around $p$ that we can make $g$ flat on through conformal deformation in terms of geometric data, i.e., $g$ and the usual curvature invariants? Is this just a function of the injectivity radius? </p> <p>Also: suppose $g$ is locally conformally Einstein. Can we produce similiar estimates on the largest geodesic ball around a given point that we can make $g$ Einstein on? </p> http://mathoverflow.net/questions/70279/estimate-on-the-size-of-flat-balls-where-the-weyl-tensor-vanishes/70284#70284 Answer by Viktor Bundle for Estimate on the size of flat balls where the Weyl tensor vanishes Viktor Bundle 2011-07-14T01:49:14Z 2011-07-14T01:49:14Z <p>I believe that the answer to the first question is the injectivity radius if the manifold is locally conformally flat. The injectivity radius provides us with an estimate on the size of a chart that contains the point $p$. In this chart the metric has to take the form $h dx^i dx^j$, where $h$ is a smooth, positive function, if the Weyl tensor vanishes everywhere. It follows that we can make the metric flat through conformal deformation on the whole of the interior of the chart. </p> http://mathoverflow.net/questions/69371/geometric-interpretation-of-q-curvature Geometric Interpretation of $Q$-curvature Viktor Bundle 2011-07-02T23:06:38Z 2011-07-12T23:03:25Z <p>Let $(M,g)$ be a Riemannian manifold of dimension $n>2$. Thanks to the late T.Branson we have the following definition of the so-called $Q$-curvature:</p> <p>$Q= \Delta R + \frac{n^3-4n^2+16n-16}{4(n-1)(n-2)^2} R^2 - \frac{8(n-1)}{(n-2)^2}|Ric|^2.$</p> <p>Here $\Delta = -div\nabla$, $R$ is the scalar curvature, and $|Ric|$ is the norm of the Ricci tensor. There has been much research on $Q$ curvature since its discovery in the eighties, motivated in large part by the conformal transformation properties that it possesses. A question that appears to be open, though, is whether or not there is a (relatively) concise geometric interpretation of this scalar curvature invariant. For $R$ we have the nice interpretation that it determines the rate at which the growth of a ball around a point differs from the flat case. Similarily the Ricci tensor measures the deviation of a solid angle from the Euclidean case. Can you think of a geometric interpretation of $Q$-curvature that is similarily elegant?</p> http://mathoverflow.net/questions/68893/higher-order-sobolev-inequality Higher order Sobolev inequality Viktor Bundle 2011-06-27T00:44:58Z 2011-06-28T02:29:03Z <p>Let $(M,g)$ be a closed, Riemannian manifold of dimension $n>4$. Let $K$ be the best constant for the Sobolev inequality</p> <p>$||u||^2_p \leq K \int_{{\Bbb{R}}^n} (\Delta u)^2 dx,$</p> <p>where $p=\frac{2n}{n-4}$. Here we are assuming that $u \in H^2(\Bbb{R}^n)$, so constants are excluded. Then by work in Djadli et al. we have it that there exists a constant $B$ such that for any $\epsilon$ greater than zero, we have it that</p> <p>$||u||_p^2 \leq (K+\epsilon) \int_M (\Delta u)^2 + B(|\nabla u|^2 + u^2) dv_g,$</p> <p>for all $u \in H^2(M)$. My question is whether or not there is a $H^3(M)$ generalization of this result -- something like the following: for every $\epsilon > 0$ there exists a constant $B$ such that</p> <p>$||u||_q^2 \leq (M + \epsilon) \int_M |\nabla \Delta u|^2 + B((\Delta u)^2 + |\nabla u|^2 + u^2) dv_g,$</p> <p>for all $u \in H^3(M)$, where $q = \frac{2n}{n-6}$, $n>6$, and $M$ is the best constant for the Sobolev inequality on $\Bbb{R}^n$,</p> <p>$||u||^2_q \leq M \int_{{\Bbb{R}}^n} |\nabla \Delta u|^2 dx.$</p> <p>I tried to modify the proof given by Djadli et al., but it breaks down because $\nabla \Delta u$ will involve second order derivatives of the metric tensor. </p> http://mathoverflow.net/questions/68893/higher-order-sobolev-inequality/68983#68983 Answer by Viktor Bundle for Higher order Sobolev inequality Viktor Bundle 2011-06-27T23:27:10Z 2011-06-28T02:29:03Z <p>The desired inequality is correct. It is easily generalized from Aubin's proof (which is given in Lee&amp;Parker's "Yamabe Problem") of the classical Sobolev inequality for Riemannian manifolds. One simply use a partition of unity argument to transfer the $\Bbb{R}^n$ result to the compact manifold. This requires numerous applications of the Cauchy-Schwarz inequality and the Cauchy inequality after one expands the derivatives of the product of the function with the partition functions. The key idea is that you can estimate all of the second, first, and zero order terms in a crude fashion, because the third order term is the only one that needs to be estimated carefully. This process seems like it could be generalized easily to get similiar Sobolev inequalities for $H^k(M)$ where $k>3$.</p> http://mathoverflow.net/questions/33614/does-elliptic-regularity-guarantee-analytic-solutions/68441#68441 Answer by Viktor Bundle for Does elliptic regularity guarantee analytic solutions? Viktor Bundle 2011-06-21T21:52:54Z 2011-06-24T21:51:21Z <p>Also: there is a classical result due to Charles Morrey, "Analyticity of the solutions of analytic non-linear elliptic systems of partial differential equations", that says that if $F(x,u,\nabla u,\nabla^2 u,...)$ is analytic in its arguments and elliptic then the solution of $F(x,u,\nabla u, \nabla^2 u,...)=0$ will be as well. (It actually goes one step further to deal with systems, but the notion of ellipticity is complicated to explain.) This result generalizes work done since the early 1900's; references can be found in Fritz John's (and two other author's I can't recall) pde book. </p> http://mathoverflow.net/questions/68329/constant-scalar-curvatureconstant-sigma-2c-g-curvature Constant scalar curvature+Constant $\sigma_2(C_g)$ curvature = ? Viktor Bundle 2011-06-20T23:49:47Z 2011-06-22T01:00:38Z <p>Let $(M,g)$ be a closed, smooth, Riemannian manifold of dimension $n>4$. Suppose both the scalar curvature and norm of the Ricci tensor are constant. In addition suppose that $g$ satisfies the following condition: $(\frac{1}{2} \Delta Ric_{ij} + Ric^{ml}W_{milj} - \frac{3n}{4(n-1)^2}Ric_{ij} -\frac{2}{n-2}Ric^l_j Ric_{il}) = \lambda g_{ij}.$</p> <p>Here $r$ is the scalar curvature, $Ric$ is the Ricci tensor, $W$ is the Weyl tensor, $\Delta = g^{-1}\nabla\nabla$, and $\lambda$ is a real number. This condition is the Euler-Lagrange equation of a quadratic Riemannian functional. Notice that Einstein metrics satisfy it. Can one conclude that $g$ is Einstein?</p> http://mathoverflow.net/questions/30035/some-questions-about-scalar-curvature/68435#68435 Answer by Viktor Bundle for Some questions about scalar curvature Viktor Bundle 2011-06-21T21:12:53Z 2011-06-21T21:12:53Z <p>Also: when one is talking about positive scalar curvature, the Yamabe invariant is important. The Yamabe invariant is the supremum over all conformal classes of the Yamabe constants of a manifold. The Yamabe invariant is positive if and only if the manifold supports a metric with positive scalar curvature. For a comprehensive guide to the somewhat recent state of affairs on the classification of such manifolds look for Jonathan Rosenberg's preprint on his website. Also, Bray and Neves have computed the Yamabe invariant for some three manifolds. This can be found on the wikipedia website for the Yamabe invariant. </p> http://mathoverflow.net/questions/68325/positivity-of-second-order-elliptic-differential-operators Positivity of Second-Order Elliptic Differential Operators Viktor Bundle 2011-06-20T22:45:19Z 2011-06-21T17:05:37Z <p>Let $(M,g)$ be a closed, smooth Riemannian manifold. Let $\Delta = -div\nabla$ be the Laplace-Beltrami operator. Let $h$ be a smooth function on $M$. Is there a condition on $h$ weaker than non-negativity such that $\Delta + h$ is a positive or non-negative operator? </p> <p>I'm thinking of something akin to the following: For the conformal Laplacian, non-negativity of the Yamabe constant is sufficient for any $h$ that is a scalar curvature of a metric in the conformal class.</p> http://mathoverflow.net/questions/68325/positivity-of-second-order-elliptic-differential-operators/68328#68328 Answer by Viktor Bundle for Positivity of Second-Order Elliptic Differential Operators Viktor Bundle 2011-06-20T23:07:25Z 2011-06-20T23:07:25Z <p>I believe the answer is no if $n>2$. Let $g$ be a metric with a negative Yamabe constant. There will be a metric $h$ in the conformal class of $g$ such that $\int_M R_h dv_g> 0$. Let $L_h$ be the conformal Laplacian of the metric $h$. It will possess a negative eigenvalue due to the negativity of the Yamabe constant. </p> http://mathoverflow.net/questions/88941/ellipticity-and-invertible-differential-operators/88942#88942 Comment by Viktor Bundle Viktor Bundle 2012-02-20T06:40:42Z 2012-02-20T06:40:42Z Great! Thanks for the reference. It's exactly what I needed. http://mathoverflow.net/questions/80789/linear-maps-on-incomplete-inner-product-spaces Comment by Viktor Bundle Viktor Bundle 2011-11-13T22:51:01Z 2011-11-13T22:51:01Z @Nate Eldredge: Thanks. When I wrote this I didn't believe that $B$ was continuous, but I have been able to establish this now. This, of course, completely alters the situation. http://mathoverflow.net/questions/73264/asymptotics-of-a-bubble Comment by Viktor Bundle Viktor Bundle 2011-08-20T16:12:59Z 2011-08-20T16:12:59Z @Fedor Petrov: You are correct. @fedja: We are actually working in $R^n$. http://mathoverflow.net/questions/70390/bound-on-q-implies-bound-on-r Comment by Viktor Bundle Viktor Bundle 2011-07-17T17:16:00Z 2011-07-17T17:16:00Z Deane: an example of this type of sequence would be one where $g_n = u^\frac{4}{n-4}g$, where $u$ is a bubble in the sense of Hebey and Robert. Here we are also assuming that the Yamabe constant of $g$ is negative. I believe there is hope for such a bound, because it might be impossible for $\Delta R$ to blow-up at the same rate as $R^2$ and $|Ric|^2$. http://mathoverflow.net/questions/70390/bound-on-q-implies-bound-on-r Comment by Viktor Bundle Viktor Bundle 2011-07-15T22:38:02Z 2011-07-15T22:38:02Z Deane: In higher dimensions you could be able to use the maximum principle to get a bound in terms of $Q$ and $E$ (traceless Ricci tensor), but the aim is to get it to depend on $Q$ only. http://mathoverflow.net/questions/70390/bound-on-q-implies-bound-on-r Comment by Viktor Bundle Viktor Bundle 2011-07-15T22:19:32Z 2011-07-15T22:19:32Z Deane: In four dimensions the formula is special because it contains the conformal Laplacian of the scalar curvature, so, yes, higher dimensions are different. By &quot;translate to infinity&quot; I mean to rule out $R$ becoming unbounded like the sequence $\{R+n\}$. The idea is that if the Laplacian of $R$ is controlled it my prevent blow-up. http://mathoverflow.net/questions/70390/bound-on-q-implies-bound-on-r Comment by Viktor Bundle Viktor Bundle 2011-07-15T13:38:30Z 2011-07-15T13:38:30Z Deane: The mode situation is where there is a sequence of metrics with constant $Q$ curvature, where the constant is bounded uniformly. The changing signs hypothesis is there to prevent situations where the scalar curvature simply &quot;translates its way to infinity&quot;. http://mathoverflow.net/questions/69432/looking-for-ideas-concerning-the-teaching-of-lower-division-differential-equation/69435#69435 Comment by Viktor Bundle Viktor Bundle 2011-07-04T02:38:20Z 2011-07-04T02:38:20Z Autonomous equations seem like a good source for this type of problem. You could ask to show that a solution of $u''=|u|$ will be non-negative if the initial position and velocity are non-negative. http://mathoverflow.net/questions/69432/looking-for-ideas-concerning-the-teaching-of-lower-division-differential-equation Comment by Viktor Bundle Viktor Bundle 2011-07-04T01:36:08Z 2011-07-04T01:36:08Z I'm actually not familiar with it. Is it suitable for undergraduates? I'm interested in students early exposure to differential equations. I can see the argument for focusing on finding explicit solutions, but it seems to make the subject seem tedious. http://mathoverflow.net/questions/69371/geometric-interpretation-of-q-curvature Comment by Viktor Bundle Viktor Bundle 2011-07-02T23:42:22Z 2011-07-02T23:42:22Z In four dimensions the total $Q$ curvature integral makes up the non-conformally invariant part of the geometric side of Chern-Gauss-Bonnett formula. This gives a global interpretation. I still haven't seen a local interpretation. http://mathoverflow.net/questions/68893/higher-order-sobolev-inequality Comment by Viktor Bundle Viktor Bundle 2011-06-27T20:28:50Z 2011-06-27T20:28:50Z Deane: Right. There is another hypothesis needed. http://mathoverflow.net/questions/68810/vanishing-on-bad-sets Comment by Viktor Bundle Viktor Bundle 2011-06-25T20:57:21Z 2011-06-25T20:57:21Z @Jagy: point taken. http://mathoverflow.net/questions/25630/major-mathematical-advances-past-age-fifty/25772#25772 Comment by Viktor Bundle Viktor Bundle 2011-06-25T02:59:12Z 2011-06-25T02:59:12Z 50 is pretty high related to the old saw from physics where &quot;if you haven't done great work by the age of 28, you might as well give up.&quot; http://mathoverflow.net/questions/68329/constant-scalar-curvatureconstant-sigma-2c-g-curvature Comment by Viktor Bundle Viktor Bundle 2011-06-24T21:07:59Z 2011-06-24T21:07:59Z Deane: I like your argument. But why does the second term vanish? http://mathoverflow.net/questions/68329/constant-scalar-curvatureconstant-sigma-2c-g-curvature Comment by Viktor Bundle Viktor Bundle 2011-06-21T19:59:42Z 2011-06-21T19:59:42Z Yang: That does the job. I'm going to add another hypothesis to it, to see if it forces Einstein-ness