Dtjytyk

Let $mathcal{E}$ be a rank three vector bundle on $mathbb{P}^1$. It splits as $mathcal{E}congmathcal{O}(a_1)oplusmathcal{O}(a_2)oplusmathcal{O}(a_3)$. It's not hard to see that any automorphism of $mathbb{P}(mathcal{E})$ acts trivially on the Picard group, which is generated by the class $f$ of a fiber and the class $h$ of $mathcal{O}_{mathbb{P}(mathcal{E})}(1)$. In particular, because any automorphism takes fibers to fibers, we obtain a surjective map $text{Aut}(mathbb{P}(mathcal{E}))rightarrow text{Aut}(mathbb{P}^1)$. We get an exact sequence
$$
1rightarrow Grightarrow text{Aut}(mathbb{P}(mathcal{E}))rightarrow text{Aut}(mathbb{P}^1)rightarrow 1
$$
We think of $G$ as automorphisms that fix each fiber as a set and leave $mathbb{P}^1$ fixed. Algebraically, we can see $G=mathbb{P}text{Aut}(mathcal{E})$. We have $text{Aut}(mathcal{E})subset text{End}(mathcal{E})=H^0(mathbb{P}^1,mathcal{E}otimes mathcal{E}^{vee})$. We can view an element of $H^0(mathbb{P}^1,mathcal{E}otimes mathcal{E}^{vee})$ as a $3times 3$ matrix with entries $a_{ij}in H^0(mathbb{P}^1,mathcal{O}(a_i-a_j))$. I'd like to get my hands on what these automorphisms do in a geometric sense. It may be instructive to consider a specific example. Let $mathcal{E}=mathcal{O}(3)oplusmathcal{O}(3)oplusmathcal{O}(4)$. Our matrix is
$begin{pmatrix}
a & b & 0 \
c & d & 0 \
l_1 & l_2 & e
end{pmatrix} $ where $a,b,c,d,e$ are scalars and $l_1, l_2in H^0(mathbb{P}^1,mathcal{O}(1))$. The condition that this matrix defines an automorphism is that the determinant doesn't vanish, which just means that $(ad-bc)eneq 0$. After projectivizing, we can set $e=1$. Intuitively, this matrix breaks up in to a few parts. The top left $2times 2$ block seems to be a $GL_2$. The bottom left part is a $H^0(mathcal{O}(1))^{oplus 2}=mathbb{A}^4$. How can I interpret these parts of the matrix geometrically? What types of automorphisms do they correspond to? One idea I have is the following. There is a natural divisor $mathbb{P}(mathcal{O}(3)oplusmathcal{O}(4))$ inside $mathbb{P}(mathcal{E})$ given by one of the obvious inclusions. Using the well-known Chow ring of $mathbb{P}(mathcal{E})$, I was able to compute that this divisor lies in class $h-3f$. Using the projection formula, we see that $h^0(mathbb{P}(mathcal{E}),mathcal{O}(h-3f))=4$. From here, it seems like the bottom left part of the matrix moves around divisors in the class $h-3f$. Is this a reasonable interpretation? What about the $GL_2$ block? Perhaps it corresponds in some way to the divisor $mathbb{P}(mathcal{O}(3)oplusmathcal{O}(3))$?