User s.a.a - MathOverflow

Variational problems whose lagrangian density depends on derivatives higher than 1.

2012-10-16T04:00:35Z

The usual theory of calculus of variations, as far as I know, is concerned with lagrangian densities which depend on the function and its gradient, namely we try to minimise $\int L(Dw,w,x) dx$. Sometimes in differential geometry, however, one runs into problems in which some function of second derivative is involved. One can, of course, introduce new variables and reduce the order of system and demand that a certain compatibility be satisfied. Namely, we can substitute $D^2 w$ by n vectors, each representing a row of the hessian matrix, and add a constraint, and possibly use the method of Lagrange multipliers.

My question is whether there is a more `intrinsic' approach to this kind of variational problem.

Addendum: I am mainly concerned with the analytical aspects, such as existence and regularity of minimisers. For example, in the usual theory, wherein the lagrangian density is a function of gradient, we know that under certain, say, convexity conditions on $F$, that is, ellipticity of the Euler-Lagrange equation, we have solutions in some appropriate space. Nevertheless, when we include $D^2u$ in the lagrangian, the equation we obtain is a 4-th order one and it won't be as straightforward as the second order case.

Elliptic Differential Equations with rough boundary data

2012-10-13T03:17:54Z

Question stated roughly: Consider the Dirichlet problem for an elliptic equation on a ball. How much can we say about regularity at the boundary of non-linear elliptic equations? Further, how can one approach solving a quasi-linear or non-linear elliptic Dirichlet problem with rough data?

Evidence: Existence of Poisson kernels guarantee existence of solutions to the Laplace equation even with rough boundary. In the interior nevertheless, we have infinite smoothness. It's true that laplacian has all the nice properties, nevertheless, one might hope to be able to obtain similar, but possibly weaker, results in more general classes of equations.

The case of linear equations: Let us consider the case of a ball, $B$. In case we have a linear equation in the `divergence' form, the usual Hilbert space approach theory we get existence of solutions so long as the boundary data lies in the image of the trace operator, $\operatorname{Tr}: H^1(B)\rightarrow L^2(\partial B)$. Equivalently, if the boundary data can be extended to an $H^1$ function in the interior, the standard theory yields a generalised solution in $H^1$.

Nevertheless, if the equation is not in divergence form, this approach will not work. Therefore, one might restrict oneself to the case where the standard Schauder theory works: divergence equations with co-efficients in a H\"older space. The problem with the conventional Schauder theory is that it only works when the boundary data is $C^{2,\alpha}$. What I have been able to find has been an article by Gilbarg and H\"ormander which extends the regularity and existence results to less regular boundary data (actually with less restriction even on the co-efficients).

Question: Are there major approaches other than weighted H\"older spaces to this problem? In case the equation is fully non-linear, say $F(x,D^2 u)=0$, is it reasonable to expect, under certain structural conditions on the operator, to be able to prove existence and regularity in appropriate weighted H\"older spaces? Are there any major results in these directions?

Green's function for a certain elliptic equations with rough coefficients

2012-09-26T00:04:36Z

We know the laplacean operator has a Green function which is smooth away from the boundary. Now, consider a linear operator of the form $\partial_i(a^{ij} \partial_j u)$.We can prove that this operator has a Green function as well. Now, if we just assume that $a^{ij} \in L^\infty$, and that $a^{ij}$ is uniformly elliptic, how much smoothness can be expected for the Green function?

Addendum: Well, we already know that away from the point on which the Dirac mass is supported, the solution is regular at least up to $C^{\alpha}$ by virtue of the theorem of De Giorgi-Nash-Moser.

Rellich-Kondrachov compactness theorem in arbitrary smooth metric measure spaces

2012-06-22T01:13:58Z

Consider a smooth metric measure space in which the integral of a gradient is meaningful. For example in the sense of upper gradients of Heinonen, or on a riemannian manifold with the associated measure. Therefore, one can define Sobolev spaces $W^{1,p}$. In the case of a compact riemannian metric, and in particular a domain in the euclidean space, one has the Rellich's theorem which guarantees compactness of the inclusion $W^{1,p} \hookrightarrow L^q$ for $q< p^*$ , where $p^{*}$ is the exponent in the Sobolev embedding theorem.

My question is whether sufficient conditions are known which may guarantee the same compactness property for Sobolev spaces defined in more abstract measure spaces. As a particular example, I'd like to know if anything is known about the conditions on the weight function $w(x)$ when the measure $d \nu$ is defined as $d \nu= w (x) d \mu$, wherein $d \mu$ is the Lebesgue measure on $\mathbb{R}^n$. Another example is the case of riemannian manifolds with possibly degenerate metrics.

Degeneration of riemannian metrics with curvature bounds

2012-09-26T00:23:08Z

In short, I'm curious to know what modes of degeneration of metric might still keep the curvature bounded. More precisely, assume we are keeping the total volume of the manifold fixed and deform the metric, e.g by a conformal factor that preserves the volume. If we allow the metric to degenerate, namely allow it to become semi-definite, $g_{ij} \geq 0$ on some set, is it possible to keep the Ricci curvature bounded below and yet let the metric degenerate? If not, is it possible to still keep some $L^p$ bound on the sectional, Ricci, or scalar curvature?

Edit: It seems that the post requires some clarification: I am aware that there is whole field of studying degeneration of metrics, convergence etc. I was mostly curious to see examples of degeneration of metrics while curvature is bounded. Examples that could, for example, illustrate when you can keep the scalar curvature bounded but Ricci, or any $L^p$ norm of Ricci, might blow up. More than the conformal deformation I was curious to see an example of the case when metric in a K\"ahler class degenerates as one varies the potential, but one curvature functional, say $\Vert Ric \Vert_p$ remains bounded.

Geometric conditions for isoperimetric, Sobolev, Poincar\'e inequalities on a riemannian manifold

2012-06-22T21:50:43Z

By a theorem of Lichnerowicz, on a riemannian manifold $M^{(m)}$ with positive Ricci curvature, the reciprocal of Sobolev constant(ie. the first eigenvalue of laplacian) can be bounded from below by the quantity $inf |Ric|$. It has been generalised to bounding the Sobolev constant via $K=||Ric||_{L^p}$ by S.Gallot, wherein $p > \frac{m}{2}$. I assume that this holds even if the metric degenerates to a semi-positive metric on a subset of $M$ while keeping $K$ bounded. Theorem of Lichnerowicz does not, obviously, hold for degenerate metrics. My question is whether other geometric -sufficient- conditions are known that can be used to bound the Sobolev constant on manifolds with -possibly- degenerate metrics.

Edit: The Poincar\'e constant I consider is $C$ in the following: $||u - \bar{u}||{2} $ $ \leq C ||\nabla u||{2}$.

Answer by S.A.A for Rank of ABA where B is positive definite

2012-08-30T00:02:33Z

This is rather tangential, but I can't help mentioning that this is related to the fact that in a hilbert space, for a (compact) linear operator one has: $Ker A^* \perp Im A$.

Answer by S.A.A for Some questions about scalar curvature

2012-08-27T00:24:14Z

In the setting of K\"ahler geometry, a necessary condition for the existence of a k\"ahlerian metric of constant scalar curvature is vanishing of the so-called Futaki invariant. (cf. Futaki On compact Kähler manifolds of constant scalar curvatures).

Answer by S.A.A for Hermitian Christoffel Symbols

2012-07-04T23:37:58Z

Just another reference: the book of Bochner and Yano, curvature and Betti numbers, in the chapter where they deal with k\"ahlerian metrics, they do some calculation in co\"ordinates which you might find helpful.

Answer by S.A.A for Question on Hilbert Manifolds

2012-06-21T23:45:05Z

It probably is NOT a smooth manifold. I think finding a chart around the point corresponding to constants, namely, the fixed points of the action of the group of rotation, is problematic.More precisely, at a fixed point, there is not a well-defined tangent space.

Answer by S.A.A for Measures on Riemannian manifolds which are not induced by the volume form of some Riemannian metric

2012-06-22T21:04:29Z

I think the answer to your question is given at least within the class of k\"ahlerian manifolds. I would refer you to standard texts of complex geometry for the preliminaries. But in the case of k\"hlerian manifolds, any volume form, that is, a measure, whose Radon-Nikodym density with respect to a fixed reference metric $g$ is smooth enough, and has the same total volume as $d Vol_g$ can be realised as the volume form of a k\"ahlerian metric $g^{\phi}$ in the same cohomology class, ie. $g_{i \bar{j}}^{\phi} = g_{i \bar{j}} + \partial_{i} \partial_{\bar{j}}$. This was Calabi's volume conjecture.

Also, you can realise an -reasonably- arbitrary volume forms by conformal change of the metric. However, in the conformal case the total volume can change.

Answer by S.A.A for German mathematical terms like "Nullstellensatz"

2011-10-03T22:22:45Z

If you think of the symbols, you can also see Gothic, alternatively called German, letters. Also, in algebraic topology, it is common to show the cycles by $Z$, which is the first letter of Zykel.

Also, many words that are Latin or Greek, in terms of the ingredients, were first coined and used in German, like Topologie which used to be called Analyse Situs.

It was common to show curvature by $K$, which stands for Krummung. Also, it was common to show a domain by B, for Bereiche. Or in riemannian geometry, the metric tensor is represented by $g$, which stands for Gravit\"at Also, Faltung used to be common in English before the word convolution took over.

I can also add Umlaufssatz in the differential geometry of surfaces.

There are so many more...

Embedding of riemannian manifolds into space forms

2011-09-29T23:56:11Z

Hi;

By a celebrated theorem of J.Nash, we know that any riemannian manifold with smooth enough metric tensor can be realised as an embedded submanifold of $\mathbb{E}^N$ for some $N$.

Can one hope to be able to embed (compact) manifolds into some space with constant curvature? If the answer is yes and we have the isometric embedding $(M,g) \hookrightarrow Q(k)$, where $Q(k)$ is a space form of constant curvature $k$, do we have any estimates on the largest possible $|k|$? One possible bound comes from boundedness of diameter of positive space forms in terms of their curvature, although, the embedded submanifold might have longer geodesics than the ambient space.

Also, if we let the embedding be merely conformal, that is the induced metric belong to the same conformal class as that of $(M,g)$, what are the possible obstructions?

Space of metrics with positive sectional curvature

2011-09-28T23:54:02Z

Hello;

We know that the space of riemannian metrics on a compact manifold is an open cone in the space of symmetric 2-tensors. Is it reasonable to think that metrics with positive sectional curvature (even positive at a specific point $x \in M$) also form a convex cone?

This is a question about the local behaviour of metrics, so I am not imposing the condition that the sectional curvature be positive everywhere.

Also, due to certain corollary of this statement for a class of metrics, I am quite certain that this cannot hold for metrics of negative curvature.

Comment by S.A.A

2012-11-10T05:51:34Z

Along the same lines, you can take convolution of the fundamental solution or Green function and a distribution, that is $f$, and that will give the solution.

Comment by S.A.A

2012-11-10T04:00:23Z

@ Semyon: Yes, before Evans-Krylov theory (like the case of Aubin and Yau's original proof) they needed $C^3$. Having $C^{2,\alpha}$ you can differentiate the equation and prove higher order estimates.

Comment by S.A.A

2012-10-21T19:45:37Z

By 'that' I meant problems like extremal metrics.

Comment by S.A.A

2012-10-21T19:45:06Z

That was the original motivation, but I was curious as to how this problem is approached by analysts, probably in a more general setting.

Comment by S.A.A

2012-10-13T18:15:18Z

Than you for your complete answer. In case of the counter-example you mentioned for equation $\det (D^2) =f$, is there a controlled rate at which the second normal derivative blows up? If so, there seems to still be some hope for applying continuity method in an appropriate Banach space. Isn't that so?

Comment by S.A.A

2012-10-13T17:56:33Z

@Deane: Do you mean even in the non-linear case?

Comment by S.A.A

2012-09-28T23:58:48Z

Thanks. I'll have a look.

Comment by S.A.A

2012-09-26T02:45:09Z

@ Deane, I'm convinced. I could have asked a more precise question. Nevertheless, making it too precise, I feared, could have dismissed some possibilities that I had not thought about.

Comment by S.A.A

2012-09-26T00:47:02Z

@Deane Yang: Do you mean the fact that it requires the $L^p$ bound of the negative part of the Ricci curvature?

Comment by S.A.A

2012-06-27T20:44:04Z

By a degenerate metric, I mean a metric that might be semi-definite at some points. (Let's assume that this set is a no wehere dense subset). What probably is important is the `rate' of degeneration. One concrete example might be the following: let $g$ be a positive definite metric and consider $u g$, obtained by a conformal change, wherein $u$ is merely non-negative. Under what conditions on $u$, or geometric properties of the metric $u g$ can we guarantee a bound on Poincar\'e constant? By the way, I think the definition of Sobolev constant in your example is the reciprocal of mine.

Comment by S.A.A

2012-06-27T18:57:55Z

Thank you, edited accordingly.

Comment by S.A.A

2012-06-27T02:56:38Z

Thank you Dr. Rajala. One further question: do you know of an analogue of ricci bound from below in terms of bounding some integral of the quantity $-inf_{|x|=1} Ric(x,x)$?

Comment by S.A.A

2012-06-22T03:40:51Z

I would like to add a comment to Mr. Haslhofer's answer: This was conjectured by H.Hopf that any immersed surface of constant mean curvature is the round sphere, however, a counterexample by H.Wente proved it incorrect.

Comment by S.A.A

2011-09-29T00:35:17Z

I added the word convex to the question. Thanks.

Comment by S.A.A

2011-09-29T00:34:34Z

Yes Reimundo, the question is whether it is convex or not.