sphere bundles isomorphic to vector bundle.

Question

Bredon claims that sphere bundles in certain cases are isomorphic to vector bundles. For example he says just replace $S^{n-1}$ with $R^n$. But for example the circle and the plane are not even isomorphic. How can we talk about bundle isomorphism when even the fibers are not isomorphic?

"A disk or sphere bundle gives rise to a vector bundle with the orthogonal group as the structure group just by replacing the fiber $D^n$ or $S^{n-1}$ with $R^n$ and using the same change of coordinate function".

We cannot just "replace" the fibers with something not isomorphic can we?

Answer 1

Let $F$ be a topological space and $G$ be a topological group acting on $G$. Very generally, you can construct an $F$-bundle on a space $X$ with structure group $G$ from the following data:

• An open cover $(U_\alpha)$ of $X$
• For each $\alpha$ and $\beta$, a map $f_{\alpha\beta}:U_\alpha\cap U_\beta\to G$ such whenever $x\in U_\alpha\cap U_\beta \cap U_\gamma$, $$f_{\beta\gamma}(x)f_{\alpha\beta}(x)=f_{\alpha\gamma}(x).$$

Namely, take the trivial bundles $F\times U_\alpha$ over each $U_\alpha$, and glue them together to a bundle on $X$ by identifying $(s,x)\in F\times U_\alpha$ with $(f_{\alpha\beta}(x)s,x)\in F\times U_\beta$ when $x\in U_\alpha\cap U_\beta$. Conversely, any $F$-bundle on $X$ with structure group $G$ can be obtained (up to isomorphism) in this way, by taking an open cover $(U_\alpha)$ over which the bundle is trivialized and taking the $f_{\alpha\beta}$ to be the transition functions between the trivializations.

Furthermore, any isomorphism between the bundle obtained from an open cover $(U_\alpha)$ and functions $f_{\alpha\beta}$ and the bundle obtained from an open cover $(V_\gamma)$ and functions $g_{\gamma\delta}$ can be described as follows. For each $\alpha$ and $\gamma$, we must give a map $h_{\alpha\gamma}:U_\alpha\cap V_\gamma\to G$ such that whenever $x\in U_\alpha\cap U_\beta\cap V_\gamma\cap V_\delta$, $$h_{\beta\delta}(x)f_{\alpha\beta}(x)=g_{\gamma\delta}(x)h_{\alpha\gamma}(x).$$ The isomorphism is then defined by sending $(s,x)\in F\times U_\alpha$ to $(h_{\alpha\gamma}(x)s,x)\in F\times V_\gamma$ for any $x\in U_\alpha\cap V_\gamma$.

Now note that none of this data depends on $F$: it only depends on $G$! So you can describe $F$-bundles with structure group $G$ (and isomorphisms between them) in terms of just the group $G$. In your case, $G=O(n)$, and $G$ acts on several different spaces: $\mathbb{R}^n$, $D^n$, or $S^{n-1}$. So, for instance, you can take a sphere bundle with structure group $O(n)$, extract the data above (which involves only the group $O(n)$) from it, and then use that same data to construct a vector bundle with structure group $O(n)$.

Answer 2

Apparently, in this context you have a sphere bundle say $M$ with the orthogonal group as structure group. This is a very special case of a sphere bundle. This means that you have a family of trivialization $\phi _i$ of your bundel $F$ over open sets $U_i$, say $\phi _i : U_i\times S \to M$ such that the change of chart over $U_i\cap U_j$ is of the form $\phi _{i,j} (u, v)= O_{i,j}(u).v$ where $O_{i,j}: U_i\cap U_j \to O(n)$ is a certain (continuous or smooth) function. Bredon says that you can use the same functions to define an $\bf R^n$ bundle $E$. First defined its trivialization over $U_i$, say $E_i$ by the formula $\Phi _i : U_i\times {\bf R}^n = E _i$. Then $E$ is the union of the $E_i=U_i\times {\bf R}^n$ glued over $U_i \cap U_j$ by the map $\phi _{i,j} (u, v)= O_{i,j}(u).v$.

Another approach is through the theory of principal bundles. To a sphere bundle with $O(n)$ as structure group you can associate its frame bundle $P$, a principal $O(n)$ bundle, which is the bundle whose fiber at a point $p$ is the set of isometries between the fiber over the point and the standard $n$ sphere. Once this frame bundle is defined, you can construct its associate euclidan bundle by the formula $E= P\times _{O(n)} {\bf R}^n$