$(BU,f)$ structures on manifolds via stable normal bundles and stable tangent bundles

Question

Background

Consider $BU=colim \, BU_k$ where we take $BU_k$ to be the specific model of classifying space for the group $U(k)\subseteq O(2k)$ given by the quotient space of the infinite real Stiefel manifold $V_{2k}$ by the action of $U(k)$. The spaces $BU_k$ as described come with maps $f_k : BU_k \rightarrow BO_{2k}$ that are fibrations.

With the above setup, we can define a $(BU,f)$-structure on a stable vector bundle $\xi : X \rightarrow BO_{2k}$ as a particular lift $\tilde{\xi} : X \rightarrow BU_k$. We consider lifts $\tilde{\xi}_0, \tilde{\xi}_1$ equivalent if there is $k>>0$ and a fiberwise homotopy $H:X\times [0,1] \rightarrow BU_k$ between the two lifts. (Fiberwise homotopy means $f_k \circ H = \xi$).

There is a map $I: BO \rightarrow BO$ given by sending a subspace $A\subseteq \mathbb{R}^n$ to its orthogonal complement $A^{\perp} \subseteq \mathbb{R}^n$.

The question

In Stong's notes on cobordism theory, he shows that a $(BU,f)$-structure on the stable normal bundle is equivalent to an $(I^*BU,f^*)$-structure on the stable tangent bundle; this is OK. Is it possible, though, to construct a bijection between $(BU,f)$-structures on the stable normal bundle and $(BU,f)$-structures on the stable tangent bundle?

Some thoughts

For other kinds of $(B,f)$-structures it is doable, I believe. Certainly for $(BO,1)$-structures you can do it. Also, for $(BSO,f)$-structures you can do it. If $TX$ is the tangent bundle of $X$ and $N$ is the normal bundle to $X$ for some embedding in $\mathbb{R}^{n+k}$, $k>>0$, one has a canonical trivialization $TX \oplus N \cong \epsilon^{n+k}$. As the trivial bundle has a canonical choice of orientation, given an orientation of $TX$, we can get an orientation on $N$ by requiring the induced orientation on $\epsilon^{n+k}$ agrees with the canonical one. One can do the same in the other direction.

A note in Davis & Kirk claims you can do it for complex structures (Exercise 137), but I don't think the discussion is correct. It works for complex vector bundles, but that is weaker than having complex structures. E.g. the case of $X=pt$, with a trivial 2-dimensional bundle $\epsilon^2$. There are two possible (inequivalent) lifts of the bundle to $BU_1$ as defined above, but only one lift to $G_1(\mathbb{C})$.

Answer 1

This is a question about general vector bundles, not tangent/normal bundles. So let $X$ be a finite complex, and $V,W \to X$ be two real vector bundles such that $V \oplus W$ is trivialized and $n$-dimensional. If a stable complex structure on $V$ is given, pick an embedding $\iota:V \oplus \mathbb{R}^r \to X \times \mathbb{C}^N$ of complex vector bundles for some large $N$. The orthogonal complement $V^{\bot}$ of the image of $\iota$ is a complex vector bundle, and we have a specific isomorphism of real bundles

$$W\oplus \mathbb{C}^N \cong W \oplus V \oplus \mathbb{R}^r \oplus V^{\bot} \cong \mathbb{R}^{n+r} \oplus V^{\bot},$$

giving the bundle $W$ a stable complex structure. If $N$ is large enough, then two such embeddings are isotopic, and so the orthogonal complements are concordant, hence isomorphic (as complex vector bundles). If you compose $\iota$ with the embedding $\mathbb{C}^N \to \mathbb{C}^{N+1}$, you add a trivial line bundle to the orthogonal complement. This shows that the stable complex structure on $W$ is uniquely determined.

The whole construction is symmetric in $V$ and $W$, and therefore induces the desired bijection.