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I am trying to prove the following fact about the holomorphic vector bundles over the complex projective line:

  • Stable vector bundles are the line bundles, i.e. $\mathcal{O}(a)$, the holomorphic line bundle of degree $a\in\mathbb{Z}$.
  • Semistable vector bundles are the sum of line bundles of the same degree: $\mathcal{O}(a)^k$, $k\in\mathbb{N}_0$.

I take the differential-geometric approach of slope-stability: if $E\rightarrow \mathbb{P}_C^1$ is a holomorphic vector bundle over its slope is defined by $\mu(E) = \frac{\deg E}{\text{rank}E}$, where the degree is usually computed via the first Chern class, $\deg E=\int_{\mathbb{P}_C^1} c_1(E)$.

It is fairly easy to see$^\dagger$ that line bundles are stable: any holomorphic subbundle of a line bundle is either trivial or the same bundle; stability is satisfied vacuously.

It is also easy to see using Grothendieck's decomposition theorem (any holomorphic bundle over $\mathbb{P}_C^1$ is equivalent to some $\bigoplus_i \mathcal{O}(a_i)^{k_i}$ with all $a_i$ distinct. Should we have more than one term in that sum, taking the maximal $a_M$ involved here would yield a destabilizing subbundle of the form $$ \mathcal{O}(a_M)^{k_M}\hookrightarrow \bigoplus_i \mathcal{O}(a_i)^{k_i} $$ and we have tha $\mu(\mathcal{O}(a_M)^{k_M}) = a_M > \frac{\sum_i a_i k_i}{\sum_j k_j} = \mu\left( \bigoplus_i \mathcal{O}(a_i)^{k_i} \right)$.

Now I'd only need to prove that all $\mathcal{O}(a)^k$ are semistable, meaning that any holomorphic subbundle $E\hookrightarrow \mathcal{O}(a)^k$ has at most slope $a$. I can't find any good argument for this. I tried considering the quotient bundle $Q$ this induces $$ 0 \rightarrow E \hookrightarrow \mathcal{O}(a)^k \twoheadrightarrow Q \rightarrow 0, $$ but I am unable to proceed any further. How could I show that $E$ does not destabilize $\mathcal{O}(a)^k$?

$\dagger$ Remark: I'd need to prove that stability can be tested with subbundles rather than subsheaves. I take this for granted over Riemann surfaces, but as a side note I would like an explanation of this fact or a reference.

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    $\begingroup$ Hint: What is $Hom(\mathcal{O}(b),\mathcal{O}(a))$ if $b>a$? $\endgroup$
    – Notone
    Commented Nov 10, 2021 at 9:31
  • $\begingroup$ In that case $Hom(\mathcal{O}(b),\mathcal{O}(a))\simeq \mathcal{O}(a-b)$ which has no global sections. However, even if I know $E$ must be some sum of line bundles, I cannot grasp how that structure allows me to assume something regarding an arbitrary injection $E\hookrightarrow \mathcal{O}(a)^k$, or how to apply the hint above. $\endgroup$ Commented Nov 10, 2021 at 9:57
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    $\begingroup$ Now just use the isomorphism $Hom(E,L^k)\cong Hom(E,L)^k$ $\endgroup$
    – Notone
    Commented Nov 10, 2021 at 11:03
  • $\begingroup$ I am sorry for again asking for further hints: using that isomorphism I can see the injection as a direct sum of its components $f = \bigoplus f_i$. However, the injectivity of $f$ does not imply the injectivity of its components. Should I have the injectivity of at least one component, I'd deduce that $E$ is necessarily of rank 1, and isomorphic to $\mathcal{O}(a)$, and I'd be done. $\endgroup$ Commented Nov 10, 2021 at 15:40

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Let's prove this by contradiction and assume that the slope of $E$ is bigger than that of $\mathcal{O}(a)^k$.

Let $E\hookrightarrow \mathcal{O}(a)^k$ be an injective morphism. Decompose $$ E\cong \bigoplus_{i=1}^n E_i$$ into line bundles. Since the slope of $E$ is greater than of $\mathcal{O}(a)^k$, we must have that one of the $E_i$ is isomorphic to $\mathcal{O}(b)$ where $b>a$. We can compute $$ Hom(\mathcal{O}(b),\mathcal{O}(a)^k)\cong Hom(\mathcal{O}(b),\mathcal{O}(a))^k=0$$ meaning that $$\mathcal{O}(b)\hookrightarrow E\hookrightarrow \mathcal{O}(a)^k $$ has to be the zero map. This is a contradiction.

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