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

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.

• Hint: What is $Hom(\mathcal{O}(b),\mathcal{O}(a))$ if $b>a$? Nov 10, 2021 at 9:31
• 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. Nov 10, 2021 at 9:57
• Now just use the isomorphism $Hom(E,L^k)\cong Hom(E,L)^k$ Nov 10, 2021 at 11:03
• 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. Nov 10, 2021 at 15:40

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.