All Questions
59 questions
10
votes
3
answers
834
views
Rigorous justification that overdetermined systems do not have a solution
There is the following well known and very useful heuristic principle: Assume one has a natural map from the space of $k$-tuples of functions in $n$ variables into the space of $K$-tuples of functions ...
- 21.8k
3
votes
1
answer
326
views
Spectral multipliers vis-a-vis Differential geometry
Let us mention two papers for examples: this one by Seeger and Sogge and this by Cheeger, Gromov and Taylor. One can also mention papers by Stein, for example, this one. There are also many others of ...
- 141
2
votes
0
answers
149
views
Variational inequality on Manifold
Let $(M,g)$ be a Riemannian manifold. Consider $A : W^{1,r}(M,\mathbb{R}) \rightarrow W^{-1,r'}(M,\mathbb{R}), k \mapsto Ak$, where $Ak$ is defined by $(Ak)(\varphi) = \int_{M}g(\nabla k, \nabla \...
- 21
1
vote
1
answer
547
views
de Rahm Laplace operator on forms bounded
Let $M$ be a closed differentiable manifold. Let $E^{p}(M)$ be the vector space of $p$-forms on $M$ equipped with the $L^{2}$-inner product $(\alpha, \beta) = \int_{M}\alpha \wedge \star \beta$. The ...
- 11
3
votes
0
answers
189
views
Can we obtain topology results using analysis in metric measures spaces?
Let $M$ be a smooth compact manifold. It is known that a lower bound on the Ricci curvature is equivalent to the convexity of the entropy on $\mathcal{P}^2(M)$ (Von Rennesse and Sturm '05), but I don'...
- 215
18
votes
3
answers
4k
views
Formal adjoint of the covariant derivative
Let $E \to M$ be a vector bundle over some Riemannian metric $(M, g)$ and endow it with some fibre metric. Assume that covariant derivative $\nabla$ is compatible with the metric.
It is essentially ...
- 5,824
7
votes
1
answer
1k
views
Helmholtz-Decomposition on compact Riemannian manifolds
For smooth domains $\Omega$ in $\mathbb{R}^n$ it is known that one can decompose vector fields in $L^p(\Omega)^n$, $1 < p <\infty $ into a "gradient"- and a "divergence-free"-part such that
$L^...
- 73
21
votes
0
answers
876
views
Are the eigenvalues of the Laplacian of a generic Kähler metric simple?
It is a theorem of Uhlenbeck that for a generic Riemannian metric, the Laplacian acting on functions has simple eigenvalues, i.e., all the eigenspaces are 1-dimensional. (Here "generic" means the set ...
- 6,247
5
votes
1
answer
1k
views
Orthogonal complements in Hilbert bundles
It's a standard fact that for a finite-dimensional vector bundle with an inner product, the othogonal complement of any subbundle is itself a locally trivial vector bundle.
What is known about the ...
- 8,803