Norm topology and compact open topology coincide for vector bundles?

Question

Let $B$ be a compact Hausdorff base space, $p:E \rightarrow B$ a vector bundle.

1. As outlined in Hatcher's, pg 45 we may associate a norm to bundle endomoprhisms, denoted $End(E)$, a norm. Given by $$ |\alpha| = \sup_{z \in B} ||\alpha_z||$$ where $||\alpha_z||$ is the norm of the linear operator $\alpha$ when restricted to a fiber of $z$. $|| \cdot ||$ being a norm induced by a choice of inner product on $E$.

2. $End(E) \subseteq E^E$, hence, it inherits as a subspace, of the compact open topology of mappings.

Are these topologies the same? Is one coarser than the other?

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Though I do not really understand the proof given, I have written down a proof that hopefully explicates user10354138's proof.

Answer 1

They are equivalent.

A linear map $\mathbb{R}^n\to\mathbb{R}^n$ is determined by its $S^{n-1}\to\mathbb{R}^n$ (similarly for $\mathbb{C}$). So the compact-open topology of $\operatorname{End}E$ can be taken as coming from $E^\Sigma$, where $\Sigma$ is the (fiber, not vector) subbundle of $E$ consisting of unit vectors. Then $\Sigma$ is compact. And $\{v\in E\mid\|v\|<r\}$ is open, so $\{|\alpha|<r\}$ is an open set in the compact-open topology. Obviously we can also translate, so $\{\alpha\mid |\alpha-\beta|<r\}$ is also open. This proves one inclusion, and the other direction is obvious.

Answer 2

I am really slow at making sense of all this. But I think there are some subtleties.

On the inner product: In each local trivialization $$ p^{-1}(U)\rightarrow U \times \Bbb R^n \quad \quad (*)$$

we take an inner product of $\Bbb R^n$, $b_U(-,-):\Bbb R^n \times \Bbb R^n \rightarrow \Bbb R^n$. Then we take a partition of unity, and define $$b(-,-):=\sum \varphi_Ub_U(-,-)$$ as our inner product. This is one way to define an inner product on $E$ and will be assumed to be.

Locally, only finitely many supports are nonzero, hence restricting $U$ in $(*)$ above, shows that the set $U \times B(\epsilon)$, $B(\epsilon):=\{x \in \Bbb R^n, b(x,x) < \epsilon \}$ is an open set.

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Norm topology is coarser: Let $b \in B$, and $b \in U$ a local trivialization, such that $b(-,-)$ as an inner product on $E$ allows us to assume it restricts to an inner product on in
$$h: p^{-1}(U) \rightarrow U \times \Bbb R^n. $$ $b$ is contained in a precompact neighborhood $V \subseteq \bar{V} \subseteq U$ as $B$ is compact Hausdorff. Thus, the preimage of $h^{-1}(\bar{V} \times S^1)$, is a comapct subset of $p^{-1}(U)$, hence that of $E$.

A subbasis element is of the form $W(K,Z)$, $Z$ being an open subset of $E$. For example, we can let $$K= h^{-1}(\bar{V} \times \Bbb R^n), Z= h^{-1}(U \times B(\epsilon))$$

Then a subbasis element is

$$ \{\alpha \in End(E) \, : \, ||\alpha_x|| \le \epsilon \text{ for all } x \in \bar{V} \}. $$

Thus, as $B$ is compact, and basis elements are finite intersections subbasis ones, $$ \{ \alpha \in End(E) \, : \, ||\alpha_x || \le \epsilon \text{ for all } x \in B \}. $$ is thus an open set. Hence, the topology of $End(E)$ as subspace is coarser.

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Compact open topology is coarser: Note that if $K_1 \supseteq K_2, Z_1 \subseteq Z_2$, then $W(K_2 , Z_1) \subseteq W(K_1, Z_1)$. Thus, we maximize $K$ and minimize $Z$, to see that $W(B,U_\epsilon)$ forms a basis, where $$ U_\epsilon:= h^{-1}(U \times B_n(\epsilon))$$ But this is precisely the norm topology.