Let $E\to M$ be a Riemannian vector bundle over an oriented Riemannian manifold $(M,g)$ with a connection $\nabla$. Let $\Gamma(E)$ denote the vector space of sections of $E\to M$. For $\sigma \in \Gamma(E)$, its Sobolev $k$-norm $(k=1,2,\dots)$ is defined by $$ |\sigma|_k^2:=\int_M ||\sigma||^2+||\nabla \sigma||^2+\cdots+||\nabla^k \sigma||^2 v_g$$
where $v_g$ is the volume form of $M$. The completion of the space $V^k(E):=\{\sigma \in \Gamma(E):|\sigma|_k<\infty\}$ is denoted by $W^k(E)$.
I want to compare three different topologies on $V^k(E)$.
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By definition of $W^k(E)$ we have $V^k(E)\subset W^k(E)$, and the norm $|\cdot |_k$ on $W^k(E)$ defines a topology on $W^k(E)$, and hence on $V^k(E)$.
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Second, since $\Gamma(E)$ is a vector space it has a topology: the weak topology determined by all finite-dimensional subspaces (cf. Infinite Dimensional Topological Vector Space). Since $V^k(E)\subset \Gamma(E)$, we have another topology on $V^k(E)$.
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Finally, there is a topology $V^k(E)$ induced by the $C^\infty$-topology (https://en.wikipedia.org/wiki/Whitney_topologies#Whitney_C%E2%88%9E-topology) of $C^\infty(M,E)$.
Are these three topologies the same? Is there a "natural" choice of a topology of $\Gamma(E)$ that are used commonly?
Best Answer
These topologies are all distinct, as can be seen by considering the trivial vector bundle over $\mathbb{R}^n$. In this case, the space $V^k(E)$ is the same as the space $V := C^\infty(\mathbb{R}^n) \cap W^{k, 2}(\mathbb{R}^n)$ of smooth functions on $\mathbb{R}^n$ with bounded $W^{k, 2}$ norm. Then we see that:
The norm $|\cdot|_k$ is the $W^{k, 2}$ norm on $V$.
To see that the topology (2) is not equal to the topology (1), it suffices to exhibit a subset $C \subseteq V$ which is closed in (2) but not in (1). Let $C = \{v_i\}_{i \in I}$ be an algebraic basis of $V$. Then for any finite-dimensional subspace $W \subseteq V$, the intersection $C \cap W$ is finite, hence closed in $W$. So $C$ is closed in (2). But it is not difficult to find a basis $C$ which is not closed in (1). (Hint: Find a sequence of linearly independent elements $v_i \in V$ which converge to $0$, and extend $v_i$ to a basis of $V$.)
The $C^\infty$-topology is the union of the $C^k$-topologies, where the $C^k$-topology is defined to be the topology generated by the basis $$ B(f, \delta) = \{g \in V \mid \text{$|\partial^\alpha f - \partial^\alpha g| < \delta$ for all $|\alpha| \leq k$}\}, $$ where $f \in V$ and $\delta : \mathbb{R}^n \to (0, \infty)$ is a continuous function.
A similar argument as before shows that (3) is not equal to (2). To show that (3) is not equal to (1), it is not difficult to exhibit a sequence which converges in (1) but not in (3). Alternatively, we can see that (3) is not first countable (and thus not metrizable) as follows. Let $\delta_k : \mathbb{R}^n \to (0, \infty)$ be a countable family of functions. Then choose $\delta : \mathbb{R}^n \to (0, \infty)$ such that for all $k$, there exists a point $x_k \in \mathbb{R}^n$ such that $\delta(x_k) < \delta_k(x_k)$. Then it is straightforward to check that $B(0, \delta)$ does not contain any of the sets $B(0, \delta_k)$, so in particular $0 \in V$ does not admit a countable neighborhood basis.
Note that similar arguments can be used to show that the three topologies are distinct for any vector bundle $E$ (except for trivial cases).
I don't have a great answer for what the most natural topology on $\Gamma(E)$ is. It seems there are several different topologies which are useful in different situations, analogous to the way that many different function spaces in analysis are useful.