# Subbundles of a rank-two trivial bundle on $\mathbb{P}^1$.

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.

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).

• Great! I have a following-up question then. Let's take the subbundle $F = \mathcal{O}(k)$ of $E = \mathcal{O} \oplus \mathcal{O}$ from your construction. When is $E / F$ torsion free? Commented Apr 1, 2019 at 16:16
• The quotient of a vector bundle by a subbundle is always a vector bundle, so it is torsion-free (or rather, it's sheaf of sections is). Commented Apr 1, 2019 at 16:20
• Every line bundle $\mathcal{O}(a)$ can be realized as a subbundle of the trivial rank two bundle if $a\leq 0$. Commented Apr 1, 2019 at 18:20
• Do you know the tautological line bundle $\mathcal{O}(1)$ and that there is a natural surjection from the trivial bundle onto this? Then, the kernel is a line bundle and easy to see that it is $\mathcal{O}(-1)$. It is now easy to get all $a<0$, by pulling back to a morphism of degree $a$ from the projective line to itself. $a=0$ is of course trivial. Commented Apr 1, 2019 at 20:10
• Dear Michael, for the case $k=1$ just notice that $\mathcal{O}\oplus\mathcal{O}(-2) \ncong\mathcal{O}(-1)\oplus\mathcal{O}(-1)$ (because the first bundle has non-zero holomorphic functions while the second hasn't) so that twisting by $\mathcal O(1)$ you obtain the desired non-isomorphism $\mathcal O(1)\oplus\mathcal{O}(-1) \ncong \mathcal O \oplus \mathcal O$ Commented Apr 1, 2019 at 20:38