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Suppose I have two Euclidean metrics $\mu_1, \mu_2$ on a given vector bundle $\xi$. Does anyone know of necessary and/or sufficient conditions to ensure that there is a homeomorphism $\phi: E(\xi) \to E(\xi)$ between fibers such that $\mu_1 \circ \phi = \mu_2$?

I vaguely recall results relating to this question, but it's been a while since I studied algebraic topology.

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1 Answer 1

It turns out that there is no additional requirement: a Euclidean structure on a vector bundle $\xi$ is unique up to isometric isomorphism of Euclidean vector bundles. This is Exercise 2-E of Milnor and Stasheff's Characteristic Classes.

In fact, we can say more.

Theorem. Let $\xi_1 = (E_1, p_1, B)$ and $\xi_2 = (E_2, p_2, B)$ be two vector bundles over a common base space $B$, and let $\mu_1$ be a Euclidean metric on $\xi_1$ and $\mu_2$ a Euclidean metric on $\xi_2$. If $\phi: \xi_1 \longrightarrow \xi_2$ is a vector bundle isomorphism, then $\phi$ is homotopic through vector bundle isomorphisms to an isometric isomorphism $\tilde{\phi}: (\xi_1,\mu_1) \longrightarrow (\xi_2,\mu_2)$.

Proof. First, suppose $\xi_1$ and $\xi_2$ are trivial bundles, so that the isomorphism $$\phi: B \times \mathbb{R}^n \longrightarrow B \times \mathbb{R}^n$$ is given by $$\phi(x,v) = (x,g(x)v)$$ for some map $$g: B \longrightarrow \mathrm{GL}(n).$$ Now $\mathrm{O}(n)$ is a deformation retract of $\mathrm{GL}(n)$, so $g$ is homotopic to some map $$h: B \longrightarrow \mathrm{O}(n).$$ Write $g_t$ for this homotopy, so that $g_0 = g$ and $g_1 = h$. Define $$\phi_t: B \times \mathbb{R}^n \longrightarrow B \times \mathbb{R}^n$$ by $$\phi_t(x,v) = (x,g_t(x)v).$$ Then $\phi_t$ is a homotopy of vector bundle isomorphisms from $\phi$ to $\tilde{\phi} = \phi_1$, and clearly $\tilde{\phi}$ is an isometric isomorphism of vector bundles.

For the general case where $\xi_1$ and $\xi_2$ are not trivial bundles, apply the above to a locally finite cover of $B$ that simultaneously trivializes $\xi_1$ and $\xi_2$. $~\Box$

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I don't see the last part of your argument. If you choose a locally finite cover $\{U_\alpha\}$ and apply the above you get that every $\phi_\alpha:=\phi|_{U_\alpha}$ is homotopic through isomorphisms to an isometry $\tilde \phi_\alpha$ on $U_\alpha$. But for $U_\alpha \cap U_\beta \neq \emptyset$ why would the isometries $\tilde \phi_\alpha$ and $\tilde \phi_\beta$ agree on the overlap? Of course you could paste the various $\tilde \phi_\alpha$ together with a partition of unity. This would yield a well-defined map, but one might lose the isometry property on the overlaps. –  Meneldur May 6 '12 at 10:36

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