1
vote
1answer
185 views

Are smooth functions dense in the space $\{u \in H^1(Q) \text{ with } \Delta_\Gamma u \in L^2(Q)\}$?

Define $$Q = \bigcup_{t \in (0,T)}\Gamma \times \{t\}$$ where $\Gamma$ is a compact (without boundary) hypersurface. Assume whatever smoothness is required. Define $L^2(Q) := L^2(0,T;L^2(\Gamma))$ ...
2
votes
0answers
104 views

How to show the identity $\int_0^T \int_{\Gamma(t)}f(s,t)\;dsdt = \int_S f(\sigma)(1+(\mathbf w \cdot \mathbf n)^2)^{-\frac{1}{2}}\;d\sigma$?

I am reading this paper. Let $\Gamma(t)$ be a smooth closed connected oriented hypersurface for each $t \in [0,T]$. Define the set $$S = \bigcup_{t \in (0,T)}\Gamma(t) \times \{t\}.$$ On page 5 of ...
1
vote
0answers
42 views

Rational homogenous functions

I'm interested in the set $\mathcal{S}$ of rational functions $F \colon \mathbb{R}^3 \to \mathbb{R}$ verifying: \begin{align} \Delta F=0 \quad \text{et} \quad F(\lambda x)= \lambda^d F(x) \quad d \in ...
10
votes
1answer
337 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 ...
7
votes
1answer
439 views

complete metric space

Hallo, I have the following question: Let $(X,d)$ be a complete metric space. Is then $(X,\operatorname{dist})$ also complete? Here by $\operatorname{dist}$ I mean the metric induced by $d$ by: ...
4
votes
3answers
397 views

Manifold-Valued Sobolev Spaces

I have the following basic question about Sobolev-spaces which take their values in a Riemannian manifold $(M,g)$, i.e. functions $u:\Omega \to M$, $\Omega \subset \mathbb{R}^n$ bounded, such that ...
3
votes
1answer
588 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 ...
17
votes
0answers
574 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 ...
3
votes
1answer
642 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 ...