# Direct sum of two stable bundles of same slope

How to prove that the direct sum of two stable vector bundles of the same slope over a smooth curve is a semistable bundle?

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One possible answer could be: by opening any book on vector bundles, and looking at the proposition right after the definition of stable vector bundle. :)

Another possible answer is as follows.

What you ask is valid in much more generality on any compact Kähler manifold.

Proposition. Let $\mathcal F$ and $\mathcal G$ two torsion free coherent sheaves over a compact Kähler manifold $(X,\omega)$. Then, $\mathcal F\oplus\mathcal G$ is $\omega$-semistable if and only if $\mathcal F$ and $\mathcal G$ are both $\omega$-semistable with $\mu(\mathcal F)=\mu(\mathcal G)$.

Here, $\mu(\mathcal E):=\deg_\omega\mathcal E/\operatorname{rank}\mathcal E$, where $$\deg_\omega\mathcal E:=\int_X c_1(\mathcal E)\wedge\omega^{n-1},\quad\dim X=n,$$ for any torsion free coherent sheaf $\mathcal E$.

Proof. Suppose first that $\mathcal F$ and $\mathcal G$ are both $\omega$-semistable with the same slope $\mu$. Then, $\mu(\mathcal F\oplus\mathcal G)=\mu$. Given a subsheaf $\mathcal E$ of $\mathcal F\oplus\mathcal G$, set $\mathcal E_1=\mathcal E\cap(\mathcal F\oplus 0)$ and $\mathcal E_2$ to be the image of $\mathcal E$ under the projection $\mathcal F\oplus\mathcal G\to\mathcal G$. By the semistability of $\mathcal F$ and $\mathcal G$ you have $$\deg_\omega(\mathcal E_i)\le\mu\cdot\operatorname{rank}\mathcal E_i,\quad i=1,2.$$ Therefore, $$\mu(\mathcal E)=\frac{\deg_\omega(\mathcal E_1)+\deg_\omega(\mathcal E_2)}{\operatorname{rank}\mathcal E_1+\operatorname{rank}\mathcal E_2}\le\mu,$$ and $\mathcal F\oplus\mathcal G$ is $\omega$-semistable.

Conversely, since both $\mathcal F$ and $\mathcal G$ are at the same time quotient sheaves and subsheaves of $\mathcal F\oplus\mathcal G$, you have $$\mu(\mathcal F\oplus\mathcal G)=\mu(\mathcal F)=\mu(\mathcal G).$$
But any subsheaf $\mathcal E$ of $\mathcal F$ or $\mathcal G$ is a subsheaf of their direct sum, as well. Hence, $$\mu(\mathcal E)\le\mu(\mathcal F\oplus\mathcal G)=\mu(\mathcal F)=\mu(\mathcal G).\quad\square$$

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