Semistable bundles on $\mathbb{P}_C^1$

Question

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.

Answer 1

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.