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Let $M$ be an oriented, compact, finite-dimensional Riemannian manifold and let $\Omega^k$ denote the space of differential $k$-forms on $M$. Endow $\Omega^k$ with its usual Fréchet topology (where differential forms are seen as sections of a exterior product bundle). Denote the Fréchet space of vector fields on $M$ by $\mathcal{X}$.

Are the following operations on differential forms continuous?

  1. Wedge product: $\wedge: \Omega^k \times \Omega^l \to \Omega^{k+l}$
  2. Contraction: $\iota: \mathcal{X} \times \Omega^k \to \Omega^{k-1}$
  3. Exterior derivative: $d: \Omega^k \to \Omega^{k+1}$
  4. Hodge star: $\star: \Omega^k \to \Omega^{n-k}$
  5. Integration: $\int_M: \Omega^n \to \mathbb{R}$

The wedge product, contraction and the Hodge star should be continuous because they arise from fibre-preserving operations on the underlying bundles (correct?). What about the derivative and integration?

Side-question/remark: If all these operations are continuous, then the map $\Omega^k \to \mathbb{R}$ defined by $\alpha \mapsto \int_M \alpha \wedge \star \alpha$ would be a continuous norm on $\Omega^k$, which I suspect is in contradiction to the fact that the topology on $\Omega^k$ is defined by a family of seminorms.

Tobias Diez
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    Your last sentence is not true, there is no such contradiction. Compare to the example $X=C_0^\infty(\Omega)$ for some open $\Omega\subseteq\mathbb{R}^n$ with one of its usual locally convex topologies. $|\cdot|_{L^2}$ is a continuous norm on $X$ even though $X$ is not normable. Being not normable only means that there is no norm that induces the topology, it does not mean that there aren't any continuous norms at all. They just do not induce the topology, that's all. – Johannes Hahn Nov 20 '13 at 12:47
  • Continuity is checked locally, so if you write down all the operations in local coordinates using a chosen local basis of differential forms, you can see that they are indeed continuous in the Fréchet topology (even smooth). Since your $M$ is compact, you need only a finite number of local charts to check the continuity in. For non-compact $M$ there might be slightly more trouble. –  Nov 20 '13 at 13:36

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