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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 $G_1(\mathbb{C})$.

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up vote 3 down vote accepted

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.

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Is there just an extra $\oplus$ in the displayed equation? When you pick the embedding $\iota$, are you assuming the almost complex structure on $V\oplus \mathbb{R}^r$ is induced from the standard almost complex structure on $\mathbb{C}^n$? I don't think this can always be done. Consider the example in the question, $X=pt$, with $X\times\mathbb{C}$ and the almost complex structure given by $x+iy\to y-ix$; it is not the restriction of the standard complex structure on $\mathbb{C}^n$ for any embedding $\mathbb{C}\rightarrow\mathbb{C}$. I agree that this works for stable complex vector bundles. – Glen M Wilson Sep 5 '13 at 14:16
I've also never heard the term "concordant" used. Do you mean it in the sense here: ? – Glen M Wilson Sep 5 '13 at 14:19
@Glen: 1.) the opposite structure \it{is} induced from an embedding (namely $z \mapsto \bar{z}$). 2.) The fact that each bundle has such an embedding is a standard bundle lemma (I assumed $X$ to be compact). 3.) A ''concordance'' of vector bundles $V_i\to X$ $i=0,1$, is a vector bundle $V \to X \times [0,1]$ that restricts to $V_i$ on $X \times \{i\}$. By a known theorem (homotopy invariance of vector bundles), the concordance relation is the same as the isomorphism relation. The notion on nlab seems to be ultimately the same, but buried under tons of higher category stuff. – Johannes Ebert Sep 6 '13 at 7:32
Johannes, thank you so much. You are of course correct; I see it now. Thanks for the help! – Glen M Wilson Sep 6 '13 at 14:04

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