# Counterexample for non-homotopic vector bundles

Let $E_1\to X$ and $E_2\to X$ be two vector bundles over a compact space $X$, such that there exists a homotopy between the two bundles: That is, they are both pullbacks of another bundle $F\to X\times I$ by maps $f:X\to X\times I$ and $g:X\to X\times I$ respectively which are homotopic: $[f]=[g]$ (see for example Husemoeller page 30 Corollary 4.6). Then it is well known that $$\underbrace{f^{\ast}(F)}_{=E_1}\cong \underbrace{g^*(F)}_{=E_2}$$ and we assume $\left.F\right|_{X\times\{0\}}=E_1$ and $\left.F\right|_{X\times\{1\}}=E_2$.

Question: I'm looking for a counter-example where two bundles are isomorpic yet NOT homotopic in the above sense.

There is a lot of theory about classifying vector bundles up to isomorphism. What about classifying vector bundles up to homotopy equivalence?

• To clarify, when you say homotopic, you mean that there's a vector bundle over $X \times I$ such that the restriction to each end gives you $E_1$ and $E_2$, respectively? – user98602 Dec 9 '15 at 19:43
• Isomorphism and homotopy equivalence are the same equivalence relation on vector bundles. – Qiaochu Yuan Dec 9 '15 at 19:48
• @MikeMiller, yes, exactly. – PPR Dec 9 '15 at 20:08

It does not make any sense to say $F|_{X\times \{0\}}=E_1$. Two vector bundles (or groups or manifolds or vector spaces) cannot be equal to each other unless they are the same set. These equalities need to be replaced by isomorphisms, and in that case the obvious extension of either bundle to $X\times [0,1]$ gives the desired homotopy. If $E_1$ and $E_2$ are say sub-bundles of some fixed bundle (e.g. the tangent bundle), then such an equality could be made meaningful. In fact, the classification of sub-bundles of $TX$ up to homotopy is not the same as sub-bundles of $TX$ up to isomorphism.
For instance consider the following two trivial line bundles in $T(S^2 \times \Bbb R)$. Let $e_0$ be the line bundle corresponding to the $\Bbb R$ direction, and let $e_1$ be the line bundle corresponding to the first coordinate in a trivialization of $T(S^2\times \Bbb R)$ (say by embedding $S^2\times \Bbb R$ into $\Bbb R^3$ smoothly). Fix a Riemannian metric on $S^2 \times \Bbb R$ and let $e_0^\perp$ and $e_1^\perp$ be the corresponding orthogonal complements of $e_0$ and $e_1$. If $e_0$ and $e_1$ were homotopic through line fields $e_t$, then $e_t^\perp$ would give a homotopy through plane fields from $e_0^\perp$ to $e_1^\perp$. Then by the result in the question we would have $e_0^\perp$ and $e_1^\perp$ were isomorphic as vector bundles. By the hairy ball theorem, this is not the case.
• Thanks for your help. So can you give a concrete example of when two sub-bundles of $TX$ are isomorphic but not homotopic? Can you give reference to the theory of homotopy classification? – PPR Dec 10 '15 at 6:16