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.
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asked Apr 1, 2019 at 16:00
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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).
answered Apr 1, 2019 at 16:12
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$\begingroup$ 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? $\endgroup$ Apr 1, 2019 at 16:16
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$\begingroup$ 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). $\endgroup$ Apr 1, 2019 at 16:20
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1$\begingroup$ Every line bundle $\mathcal{O}(a)$ can be realized as a subbundle of the trivial rank two bundle if $a\leq 0$. $\endgroup$– MohanApr 1, 2019 at 18:20
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1$\begingroup$ 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. $\endgroup$– MohanApr 1, 2019 at 20:10
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1$\begingroup$ 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 $\mathcal O(1)\oplus\mathcal{O}(-1) \ncong \mathcal O \oplus \mathcal O$ $\endgroup$ Apr 1, 2019 at 20:38