Lee Introduction to Smooth Manifolds - Why is Riemannian metric continuous?

Question

In Lemma 13.28, the book says $|v|_g$ (where $g$ is the Riemannian metric) is continuous. Why? (preferably just using stuff established in Lee's book up to that point)

More generally, is it true that smooth covariant tensor fields, viewed as maps on $T_p M \times \cdots \times T_p M$ (where $M$ is the manifold), are continuous? I know that smooth covariant tensor fields, when viewed as maps from the manifold $M$, are continuous from Prop 12.19 in Lee's book.

Additionally, it is not proved in the book up to that point that $|v|_g$ is in fact a norm. How can we prove that it is a norm?

Answer 1

More generally, is it true that smooth covariant tensor fields, viewed as maps on $T_p M \times \cdots \times T_p M$ (where $M$ is the manifold), are continuous?

Yes. Since $T_pM \cong \mathbb{R}^n$ topologically. And every $m$-linear map $T\colon \mathbb{R}^n \times \ldots \times \mathbb{R}^n \to \mathbb{R}$ is continuous (in fact smooth). The previous is easy by first proving the not-so trivial result that every norm induces the same topology in euclidean space. I would like to add that this implies that every section of the bundle $T^2(T^*M) \to M$ induces a smooth map $T_pM \times T_pM \to M$. However, the important smoothness property is the one $M \to T^2(T^*M)$ that implies that our metric depends smoothly on $M$.

Additionally, it is not proved in the book up to that point that || is in fact a norm. How can we prove that it is a norm?

This is immediate form the fact that $\langle \bullet, \bullet\rangle_g$ is an inner product.

Answer 2

However the map is defined abstractly, when the definitions with all the $T$'s and $p$'s and $g$'s and $|\bullet|$'s and $\langle\bullet,\bullet\rangle$'s and so on are unwound, what you get is a map whose domain is some open set of one Euclidean space, and whose range is another Euclidean space. Smoothness of the map therefore means smoothness in the ordinary "advanced calculus" sense for maps from open subsets of one Euclidean space to another Euclidean space, namely: all partial derivatives of all orders exist and are continuous. And for such maps, smoothness implies continuity, as you learn in advanced calculus; in fact, you really only need existence and continuity of first partials to deduce continuity of the map.