Can every line bundle $\mathcal{O}(a)$ on $ \mathbb{P}^1$ be realized as a subbundle of the trivial bundle $\mathcal{O}\oplus \mathcal{O}$ on $\mathbb{P}^1$? I know that this phenomena can happen in general: for example Moebius band gives an example of a nontrivial real subbundle of the rank-two real trivial bundle on $S^1$, however I can't think of any meaningful obstruction here.
Subbundles of a rank-two trivial bundle on $\mathbb{P}^1$.
algebraic-geometryalgebraic-topologyvector-bundles
Related Solutions
The difference between a rank $2k$ real bundle and a rank $k$ complex bundle is that in a complex bundle one is using $\mathbb C$ as the scalar field, whereas in a real bundle one is only using $\mathbb R$ as the scalar field.
Now let's suppose we are given a complex $n$-dimensonal bundle $E \mapsto X$.
It is true that if this bundle is trivial over $\mathbb C$, i.e. if there is a $\mathbb C$-linear isomorphism of bundles $E \mapsto X \times \mathbb C^n$ over $X$, then one immediately obtains an $\mathbb R$ linear isomorphism of bundles $E \mapsto X \times \mathbb R^{2n}$ over $X$: just restrict the scalar field from $\mathbb C$ to $\mathbb R$.
But the converse does not hold: if there is an $\mathbb R$ linear isomorphism of bundles $E \mapsto X \times \mathbb R^{2n}$ over $X$, you don't know how to extend this to a $\mathbb C$-linear isomorphism; roughly speaking, you don't know how to define multiplication by $i$ in $X \times \mathbb R^{2n}$ in such a way that makes the given map of bundles into a $\mathbb C$-linear isomorphism.
Connor Malin's comment pointed out that there is not likely to be a representability theorem of subbundles of a trivial bundle as in the infinite-dimensional Grassmannian case. This resolved most of my initial concerns.
Aside from that, I think I found a direct counter-example of 1&2&3 in the case $B=S^1\times S^1\times S^1$.
For simplicity, I will take $F=\mathbb{C}$ as the base field. The following construction is used in condensed matter physics to build a model of Hopf Insulator.
$$ f: B \longrightarrow Gr_1(\mathbb{C}^2)=\mathbb{C}P^1, \\ (k_1,k_2,k_3) \mapsto [\sin{k_1}+i\sin{k_2}:\sin{k_3}+i(\cos{k_1}+\cos{k_2}+\cos{k_3}-3/2)],$$
where $k_i$'s are the periodic coordinates on $B=T^3$.
$f$ defines the pullback line bundle $f^{*}E$ of the canonical bundle $E$ over $\mathbb{C}P^1$, which is trivial since it has a nonvanishing section
$$ \sigma:B\longrightarrow f^{*}E \subseteq B\times \mathbb{C}^2, \\ \mathbf{k} \mapsto \bigg(\mathbf{k},\big(\sin{k_1}+i\sin{k_2},\sin{k_3}+i(\cos{k_1}+\cos{k_2}+\cos{k_3}-3/2) \big)\bigg). $$
Hence hypotheses 1&2&3 in the question cannot be true simultaneously in the complex case unless $f$ is nulhomotopic.
In fact, $f$ restricts to the nulhomotopic maps on the surfaces ${1}\times S^1\times S^1, S^1\times {1}\times S^1, S^1\times S^1\times {1}$ and homotopy extension property of CW pairs enables us to see $f$ as $$ \widetilde{f}: S^3 \longrightarrow \mathbb{C}P^1\approx S^2. $$
$\widetilde{f}$ has nonzero Hopf invariant,
$$ \chi=-\frac{1}{4\pi^2}\int_{T^3} d\mathbf{k}\;\mathbf{F}\cdot\mathbf{A}=1, \quad\text{where} \\ \mathbf{A}(\mathbf{k})=i \overline{f(\mathbf{k})} \cdot \nabla_{\mathbf{k}} f(\mathbf{k}) \;(\text{here $f$ maps into $\mathbb{C}^2$}),\; \mathbf{F}(\mathbf{k})= \nabla_{\mathbf{k}} \times \mathbf{A}(\mathbf{k}), $$ which implies that $\widetilde{f}$ cannot be homotoped to a constant map.
Best Answer
Yes. For every $k > 0$, we have $\mathcal{O}(k)\oplus\mathcal{O}(-k) \cong \mathcal{O}\oplus\mathcal{O}$ as smooth bundles over $\mathbb{CP}^1$. To see this, note that
$$\operatorname{rank}_{\mathbb{R}}(\mathcal{O}(k)\oplus\mathcal{O}(-k)) = 4 > 2 = \dim_{\mathbb{R}}\mathbb{CP}^1,$$
so $\mathcal{O}(k)\oplus\mathcal{O}(-k)$ admits a nowhere-zero section, and hence $\mathcal{O}(k)\oplus\mathcal{O}(-k) \cong L\oplus\mathcal{O}$ for some complex line bundle $L$. As
$$c_1(L) = c_1(L\oplus\mathcal{O}) = c_1(\mathcal{O}(k)\oplus\mathcal{O}(-k)) = c_1(\mathcal{O}(k)) + c_1(\mathcal{O}(-k)) = 0,$$
and smooth complex line bundles are classified by their first Chern class, we have $L \cong \mathcal{O}$, and hence $\mathcal{O}(k)\oplus\mathcal{O}(-k) \cong L\oplus\mathcal{O} \cong \mathcal{O}\oplus\mathcal{O}$.
Note however that $\mathcal{O}(k)\oplus\mathcal{O}(-k)$ and $\mathcal{O}\oplus\mathcal{O}$ are not isomorphic as holomorphic vector bundles. To see this, note that
$$\Gamma(\mathbb{CP}^1, \mathcal{O}(k)\oplus\mathcal{O}(-k)) = \Gamma(\mathbb{CP}^1, \mathcal{O}(k))\oplus\Gamma(\mathbb{CP}^1, \mathcal{O}(-k)) = \Gamma(\mathbb{CP}^1, \mathcal{O}(k))\oplus\{0\}.$$
As $\Gamma(\mathbb{CP}^1, \mathcal{O}(k))$ can be identified with the set of degree $k$ homogeneous polynomials in two variables, it has dimension $k + 1$.
On the other hand,
$$\Gamma(\mathbb{CP}^1, \mathcal{O}\oplus\mathcal{O}) = \Gamma(\mathbb{CP}^1, \mathcal{O})\oplus\Gamma(\mathbb{CP}^1, \mathcal{O}) = \mathcal{O}(\mathbb{CP}^1)\oplus\mathcal{O}(\mathbb{CP}^1) = \mathbb{C}\oplus\mathbb{C}$$
which has dimension $2$. So for $k \neq 1$, we see that $\mathcal{O}(k)\oplus\mathcal{O}(-k)$ and $\mathcal{O}\oplus\mathcal{O}$ are not isomorphic as holomorphic vector bundles. I think there should be an easier way of demonstrating this fact which would also include the case $k = 1$, but I can't think of it right now (other than appealing to semistability directly).