Let E be the colimit of $E_n-n$ where $E_n$ is the universal bundle over $BO(n)$. $E$ is a virtual bundle of dimension $0$ over $BO$.
Given an $O$-structure on a $d$-dimensional manifold $M$, that is, the principal $O$-bundle $P_O(TM)$=bundle of orthonormal frames of the tangent bundle $TM$. $P_O(TM)$ determines a map $f: M\to BO$.
Question: What is the pullback of $E$ along $f$?
I guess it is $TM-d$.
Can you help me to prove or disprove it? Thank you!
Choose an isometric embedding $\iota:M\hookrightarrow \mathbb{R}^D$ for some $D\geq d$. Now the tangent maps $\iota_*:T_pM\rightarrow T_{\iota(p)}\mathbb{R}^D\cong\mathbb{R}^D$ are linear isometries for each $p\in M$ and identify each tangent space with a $d$-dimensional subspace of $\mathbb{R}^d$. Since the tangent bundle of $\mathbb{R}^D$ is trivial, the embedding $\iota$ induces an isometry of bundles $T\iota:TM\hookrightarrow \mathbb{R}^D\times\mathbb{R}^D$ into the trivial bundle over $\mathbb{R}^D$.
Thus we get a map $f:M\rightarrow Gr_d(\mathbb{R}^D)$ from $M$ into the Grassmannian of $d$-planes in $\mathbb{R}^D$ that sends a point $p\in M$ to the subspace $\iota_*(T_pM)\subseteq\mathbb{R}^D$. Given its definition in the previous paragraph, and the definition of the smooth structure on $G_d(\mathbb{R}^D)$, you can check that $f$ is smooth.
Now sitting above $Gr_d(\mathbb{R}^D)$ is the Stiefel manifold $V_d(\mathbb{R}^D)$ of orthogonal $d$-frames in $\mathbb{R}^D$, which is the total space of a principal $O(d)$-bundle. Another thing for you to check is that the pullback $f^*V_d(\mathbb{R}^D)$ is exactly the orthogonal frame bundle $P_O(M)$.
Now the universal rank $d$ vector bundle over $Gr_d(\mathbb{R}^D)$ has total space $E_d(\mathbb{R}^D)=\{(V,v)\in Gr_d(\mathbb{R}^D)\times \mathbb{R}^D\mid v\in V\}$ and this is isomorphic to the Borel construction $V_d(\mathbb{R}^D)\times_{O(d)}\mathbb{R}^d$. Simply choose an abitrary basis $(e_1,\dots,e_d)$ for $V\in Gr_d(\mathbb{R}^D)$ and send the pair $(V,v)\in E_d(\mathbb{R}^D)$ to the pair to $((e_1,\dots,e_d),(v^i))\in V_d(\mathbb{R}^D)\times_{O(d)}\mathbb{R}^d$, where $(v^i)$ are the components of $v$ with respect to the chosen basis. You should check that this is well-defined and does indeed produce an isomorphism of vector bundles.
The point is that
$$f^*E_d(\mathbb{R}^D)\cong f^*(V_d(\mathbb{R}^D)\times_{O(d)}\mathbb{R}^d)\cong (f^*V_d(\mathbb{R}^D))\times_{O(d)}\mathbb{R}^d\cong P_O(M)\times_{O(d)}\mathbb{R}^d\cong TM.$$
The inclusions $\mathbb{R}^n\cong\mathbb{R}^n\oplus0\hookrightarrow\mathbb{R}^{n+1}$ induce embeddings $Gr_d(\mathbb{R}^d)\hookrightarrow Gr_d(\mathbb{R}^{d+1})$ and $V_d(\mathbb{R}^d)\hookrightarrow V_d(\mathbb{R}^{d+1})$. Letting $D$ go to infinity we get the Grassmannian of $d$-planes $Gr_d=Gr_d(\mathbb{R}^\infty)$ in $\mathbb{R}^\infty$, which is a model for the classifying space $BO(d)$. The naturality of these embeddings means that if we consider the universal bundle $E_d\rightarrow Gr_d$ then we get
$$f^*E_d\cong f^*E_d(\mathbb{R}^d)\cong TM.$$
On the other hand, if we stabilise as $j_d:Gr_d(\mathbb{R}^D)\hookrightarrow Gr_{d+1}(\mathbb{R}^{D+1})$, by sending a $d$-plane $V$ in $\mathbb{R}^\infty$ to the $(d+1)$-plane $V\oplus \langle e_{D+1}\rangle$, then
$$j_d^*E_{d+1}(\mathbb{R}^{D+1})\cong E_d(\mathbb{R}^{D+1})\oplus\epsilon^1$$
where $\epsilon^1$ is the trivial real line bundle. We conclude that for $n\geq 0$,
$$j_d^*E_{d+n}(\mathbb{R}^{D+n})\cong E_d(\mathbb{R}^{D+1})\oplus \epsilon^n.$$
Now the pullback of a trivial bundle by any map is again a trivial bundle. Therefore if we consider the virtual bundle $(E_{d+N}-\epsilon^{d+N})$ for large $N$ we get
$$j_d^*(E_{d+N}-\epsilon^{d+N})=(j_d^*E_{d+N}-j_d^*\epsilon^{d+N})=(E_{d}\oplus\epsilon^N-\epsilon^{d+N})=(E_{d}-\epsilon^d)$$
as virtual bundles. Now our map $f:M\rightarrow Gr_d(\mathbb{R}^D)$ gives us the composite
$$f'=j_d\circ f:M\rightarrow Gr_d(\mathbb{R}^D)\rightarrow Gr_\infty(\mathbb{R}^\infty)\simeq BO$$
and letting $E=colim (E_n-\epsilon^n)$ we get
$$f'^*E=f^*(j_d^*E)=f^*(E_d-\epsilon^d)=TM-\epsilon^d$$
as a virtual bundle.