Rank 2 bundles on $\mathbb P^2$ Let $\mathcal E$ be an $SL(2)$-bundle on $\mathbb P^2$.
Assume that the restriction of $\mathcal E$ to any $\mathbb P^1$ is non-trivial.
Is it true that $\mathcal E$ is a direct sum of two line bundles?
 A: No. A simple counterexample is a nontrivial extension
$$
0 \to O(1) \to E \to I_x(-1) \to 0,
$$
where $x$ is a point and $I_x$ is its ideal sheaf.
EDIT: basics of Serre's construction.
First, note that
$$
Ext^1(I_x(−1),O(1)) \cong
Ext^2(O_x,O(1)) \cong 
H^0(\mathbb{P}^2,\mathcal{E}\mathit{xt}^2(O_x,O(1))) \cong H^0(\mathbb{P}^2,O_x).
$$
Next, let us rewrite the defining sequence of $E$ as
$$
0 \to O(1) \to E \to O(-1) \to  O_x \to 0.
$$
Applying $\mathcal{H}\mathit{om}(-,O(1))$ to it, we get 
$$
0 \to O(2) \to  E^\vee(1) \to O \to \mathcal{E}\mathit{xt}^2(O_x,O(1)) \to \mathcal{E}\mathit{xt}^1(E,O(1)) \to 0.
$$
So, to check that $\mathcal{E}\mathit{xt}^1(E,O(1)) = 0$ and hence $E$ is locally free, it is enough to show that the map
$$
O \to \mathcal{E}\mathit{xt}^2(O_x,O(1)) \cong O_x
$$
is surjective. But by the above chain of isomorphisms, this map corresponds to the original extension class, hence is non-zero. And any non-zero map to $O_x$ is surjective.
A: Let's look at $T_{\mathbb{P}^2}$ as Jason Starr suggests. We have the Euler exact sequence:
$$ 0 \rightarrow \mathcal{O}_{\mathbb{P}^2} \rightarrow \mathbb{C}^3 \otimes \mathcal{O}_{\mathbb{P}^2}(1) \rightarrow T_{\mathbb{P}^2} \rightarrow 0$$
Note that $\mathrm{Ext}^1(L_1,L_2) = 0$ for any two line bundles on $\mathbb{P}^2$. Hence, if $T_{\mathbb{P}^2}$ splits as direct sum of line bundles (say $L_1 \oplus L_2$), we have:
$$ \mathbb{C}^3 \otimes \mathcal{O}_{\mathbb{P}^2}(1) = \mathcal{O}_{\mathbb{P}^2} \oplus L_1 \oplus L_2,$$
which is clearly impossible. On the other hand, $T_{\mathbb{P}^2}$ is obviously $\mathrm{SL}_3$-homogeneous and by the normal exact sequence, the restriction of $T_{\mathbb{P}^2}$ to any $\mathbb{P}^1$ is $\mathcal{O}(1) \oplus \mathcal{O}(2)$, which is non-trivial.
You might be interested in the following result:
Theorem (Grauert-Mülich) : Let $E$ be a rank $2$, $\mu$-semi-stable vector bundle on $\mathbb{P}^2$ with $c_1(E) = 0.$ Then, for a generic line $l \subset \mathbb{P}^2$, the restriction $E|_{l}$ is trivial.
You could also have a look at this paper https://link.springer.com/chapter/10.1007/BFb0099972 by Maruyama. In there he studies the singularities of the curve of lines $l \in \mathbb{P}^2$ such $E|_{l}$ is non-trivial. 
EDIT I misunderstood what was meant by $\mathrm{SL}_2$-bundle. By $\mathrm{SL}_2$-bundle, you mean a rank-$2$ vector bundle with trivial first Chern class. So if your vector bundle does not restrict to the trivial bundle on the generic $\mathbb{P}^1$, then the Grauert-Mülich Theorem implies that it is not $\mu$-semi-stable. In particular, it has a destabilizing sub-line bundle.
It is not obvious to me that all unstable rank $2$ vector bundle on $\mathbb{P}^2$ do split, but that is perhaps true...
EDIT(bis) By proposition 4 of this paper : http://www.seminariomatematico.unito.it/rendiconti/67-1/15.pdf, a rank $2$-vector bundle (say $E$) on $\mathbb{P}^2$ with $c_1 = 0$ splits if and only if $h^1(E(-1)) = 0$.
