It is known that the tangent bundle $T$ on $\mathbb{P}^2$ is stable. Is the symmetric power $\operatorname{Sym} T^2$ stable?
I know it is a vector bundle of rank $3$ with Chern character $(3,9,21/2)$. One can also find a resolution of it using line bundles see here. But I do not know how to prove or disprove its stability.
Thank you in advance.
To check stability one can use Hoppe's criterion --- for a vector bundle $E$ of rank r it amounts to verifying that $$ H^0(\mathbb{P}^2, \wedge^p E(-a_p)) = 0, $$ where $1 \le p \le r-1$ and $a_p$ is the minimal integer such that the slope of $\wedge^p E(-a_p)$ is nonpositive.
In the case of $E = \operatorname{Sym}^2 T$ using the isomorphism $$ \wedge^2 \operatorname{Sym}^2 T \cong \operatorname{Sym}^2 \Omega(9) $$ the criterion reduces to the computation of $$ H^0(\mathbb{P}^2, \operatorname{Sym}^2 T(-3)) \quad\text{and}\quad H^0(\mathbb{P}^2, \operatorname{Sym}^2 \Omega(3)). $$ The first is a direct summand in $\operatorname{Hom}(T,T) = \Bbbk$, while the other summand is $$ H^0(\mathbb{P}^2, \wedge^2 T(-3)) = H^0(\mathbb{P}^2, \mathcal{O}) = \Bbbk, $$ hence it is zero. The same argument shows that the other space is zero as well. Therefore, $\operatorname{Sym}^2 T$ is stable.