Skip to main content

All Questions

Filter by
Sorted by
Tagged with
2 votes
0 answers
96 views

Is the Leray projection continuous with respect to the Frechet topology of smooth periodic vector fields in $3$ dimensions?

Let $\mathbb{T}^3:=(\mathbb{R}/\mathbb{Z})^3$ be the $3$-torus and $C^\infty(\mathbb{T}^3,\mathbb{R}^3)$ be the Frechet space of smooth periodic vector fields on $\mathbb{T}^3$. By Helmholtz ...
Isaac's user avatar
  • 3,477
4 votes
0 answers
334 views

Hodge decomposition on non-compact manifolds

Let $(\mathcal{M},g)$ be a compact Riemannian manifold without boundary. Then we have the well-known Hodge decomposition $$\Omega^{k}(\mathcal{M})\cong\mathcal{H}^{k}(\mathcal{M})\oplus\mathrm{ran}(\...
B.Hueber's user avatar
  • 1,171
3 votes
2 answers
956 views

Hodge decomposition on open manifold

For the open manifold like $X\times \mathbb R$ or $X\times \mathbb R^+$, where $X$ is a closed manifold. Is there any decomposition like (Hodge Decomposition) of the Differential forms on it.
DLIN's user avatar
  • 1,915
11 votes
0 answers
364 views

Carleson's Theorem on Manifolds

Let $M$ be an oriented, compact, differentiable manifold with some Riemmanian metric $g$, so that $(M,g)$ has a nice volume form and one can define $L^2(M,g)$ as the completion of $C^\infty(M)$ under ...
Greg Zitelli's user avatar
  • 1,104
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 ...
Tobias Diez's user avatar
  • 5,824
5 votes
0 answers
584 views

Constructing a Sobolev space containing the differential k-forms of a Riemannian manifold

I am currently writing a paper about the Hodge theorem for an algebraic topology course. The specific formulation I am proving can be stated thus. Let $M$ be a compact, orientable n-dimensional ...
user avatar
6 votes
3 answers
3k views

Can the adjoint of the exterior derivative in semi-Riemannian geometry be defined without the Hodge * operator?

The adjoint of the exterior derivarive is defined by $\delta:=(-1)^k\ast^{-1}d\ast$, but I need a way which avoids the Hodge $\ast$ operator. Is there another definition? For example, for ...
Cristi Stoica's user avatar