Cotangent bundle of the projective space

 Here is a sketch of an argument why 

Ω:=ΩPnk$\mathrm{\Omega }:={\mathrm{\Omega }}_{{\mathbb{P}}_k^n}$  is not decomposable into direct sum of OPnk${\mathcal{O}}_{{\mathbb{P}}_k^n}$-modules (this answers one of the questions asked in meeting today):

If we show Hom(Ω,Ω)≅k$Hom(\mathrm{\Omega },\mathrm{\Omega })\cong k$  then we are done, because suppose Ω=F0⊕F1$\mathrm{\Omega }=F_0\oplus F_1$  for some O$\mathcal{O}$ -modules F0,F1$F_0,F_1$ . Then dimHom(Ω,Ω)≥dimHom(F0,F0)+dimHom(F1,F1)≥2$\dim Hom(\mathrm{\Omega },\mathrm{\Omega })\ge \dim Hom(F_0,F_0)+\dim Hom(F_1,F_1)\ge 2$ , which is impossible. 

To prove the claim, the long exact sequence of cohomology of the Euler's sequence0→Ω→O(−1)n+1→O→0$0\to \mathrm{\Omega }\to \mathcal{O}(-1{)}^{n+1}\to \mathcal{O}\to 0$ gives the exact sequence 0→H0(Ω)→H0(O(−1)n+1)→H0(O)→H1(Ω)→H1(O(−1))n+1→⋯$$\begin{gathered} 0\to H^0(\mathrm{\Omega })\to H^0(\mathcal{O}(-1{)}^{n+1})\to H^0(\mathcal{O})\to \\ {H}^1(\mathrm{\Omega })\to H^1(\mathcal{O}(-1){)}^{n+1}\to \cdots \end{gathered}$$ 
But since H0(O(−1))=H1(O(−1))=0$H^0(\mathcal{O}(-1))=H^1(\mathcal{O}(-1))=0$  we find that H0(Ω)=0,H1(Ω)≅H0(O)≅k$H^0(\mathrm{\Omega })=0,H^1(\mathrm{\Omega })\cong H^0(\mathcal{O})\cong k$  . (This shows that when n=1$n=1$ we have  Ω≅O(−2)$\mathrm{\Omega }\cong \mathcal{O}(-2)$, because this is the only invertible sheaf on P1k${\mathbb{P}}_k^1$  with these cohomology groups.)

Next, tensor the Euler's sequence by O(1)$\mathcal{O}(1)$  and then take cohomology. This gives an exact sequence
0→H0(Ω(1))→H0(O)n+1→H0(O(1))→ H1(Ω(1))→H1(O)n+1→⋯$$\begin{gathered} 0\to H^0(\mathrm{\Omega }(1))\to H^0(\mathcal{O}{)}^{n+1}\to H^0(\mathcal{O}(1))\to \\ {H}^1(\mathrm{\Omega }(1))\to H^1(\mathcal{O}{)}^{n+1}\to \cdots \end{gathered}$$ 
By the construction of the Euler's sequence the third arrow from the left is an isomorphism and H1(O)=0$H^1(\mathcal{O})=0$  therefore, H0(Ω(1))=H1(Ω(1))=0$H^0(\mathrm{\Omega }(1))=H^1(\mathrm{\Omega }(1))=0$ .

Finally, apply Hom(−,Ω)$Hom(-,\mathrm{\Omega })$  to the Euler's sequence. We get the exact sequence  0→H0(Ω)→H0(Ω(1))n+1→Hom(Ω,Ω)→H1(Ω)→H1(Ω(1))n+1→⋯$$\begin{gathered} 0\to H^0(\mathrm{\Omega })\to H^0(\mathrm{\Omega }(1){)}^{n+1}\to Hom(\mathrm{\Omega },\mathrm{\Omega })\to \\ {H}^1(\mathrm{\Omega })\to H^1(\mathrm{\Omega }(1){)}^{n+1}\to \cdots \end{gathered}$$   
So by what we proved above, Hom(Ω,Ω)≅H1(Ω)≅k$Hom(\mathrm{\Omega },\mathrm{\Omega })\cong H^1(\mathrm{\Omega })\cong k$  .