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Suppose we have have a co-ordinate mapping $\phi: M \to \mathbb{R}^n$ for a smooth Riemannian manifold $M$. Then the differential (pushforward) of this mapping $d\phi : TM \to \mathbb{R}^n\times \mathbb{R}^n$, where $TM$ is the tangent bundle, is a smooth linear map. Consider the mapping $$\sigma : M \to \mathcal{L}(T_xM,\mathbb{R}^n)\text{ given by } x\to d\phi(x),$$ where $\mathcal{L}(T_xM,\mathbb{R}^n)$ is the space of linear operators between $(T_xM$ and $\mathbb{R}^n)$. Note that the operator norm $$\| d \phi(x)\| \leq C_{x}<\infty$$ at each point $x$ (being a linear map between finite dimensional spaces). Can we claim that if $K$ is a compact set (in the usual topology for the manifold), then $$\sup_{x \in K}\|d \phi(x)\| \leq C $$ for some constant $C$.

Thanks in advanc and any help would be appreciated

user143234
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  • Is a continuous function on a compact space always bounded? – Moishe Kohan Jul 05 '17 at 04:13
  • Yes, for a metric space a continuous function maps compact sets to compact sets (Note that a compact set is totally bounded and complete - Heine Borel for general metric space). For general topological spaces the concept of boundedness is not defined – user143234 Jul 05 '17 at 04:19
  • If you choose a fixed coordinate system once and also fix the norm then $||d\phi(x)||$ is a continuous function. The detour via $C_x$ is not helpful. – Thomas Jul 05 '17 at 04:55
  • Can we claim that $x \to d\phi(x)$ is continuous ? ($d\phi : TM \to \mathbb{R}^n \times \mathbb{R}^n $ obviously is ) – user143234 Jul 05 '17 at 05:56

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