Consider, just as an example, an action of $\mathbb{C}^\*$ on $\mathbb{P}^2$ of the form


$$t\cdot p=[p\_0:tp\_1:t^2p\_2]$$


There are $3$ fixed points, namely $e\_1,e\_2,e\_3$. If I consider a $\mathbb{C}^\*$-linearizable line bundle -like $L=\mathcal{O}(1)$-, then I have an induced action
$$\phi:\mathbb{C}^\*\times L\to L,$$
which is linear along the fibers and equivariant with respect to the previous action. If we consider for example $e\_1=[1:0:0]$, I have a linear action $$\phi:\mathbb{C}^\*\times L\_{e\_1}\to L\_{e\_1}, \text{ i.e. } \mathbb{C}^\*\times\mathbb{C}\to\mathbb{C}$$
and I would like to understand what is the weight of the $\mathbb{C}^\*$-action here. I'm pretty confident there must be a way to recover the weight of the action from the action on $\mathbb{P}^2$, but I've no idea how to do it and I'm curious (I considered a specific example just to for better understanding).


Any hint, help or reference would be much appreciate, thanks in advance.



Weights on the linearization
If your $\mathbb{P}^2$ is $\mathbb{P}(\mathbb{C}^3)$, you can identify the complement of the zero section in $L^{-1}$ with $\mathbb{C}^3\smallsetminus 0$, viewed as a bundle over $\mathbb{P}^2$ via the projection $p:\mathbb{C}^3\smallsetminus 0\rightarrow \mathbb{P}^2$. One possible way to extend your action is to have $t\in \mathbb{C}^\*$ acts on $\mathbb{C}^3\smallsetminus 0$ by $(x,y,z)\mapsto (x,ty,t^2z)$. Then $t$ acts trivially on $L\_{e\_0}\smallsetminus 0=p^{-1}(e\_0)= \{(x,0,0)\} $, as $t$ on $L\_{e\_1}\smallsetminus 0= \{(0,y,0)\}$, and as $t^2$ on $L\_{e\_2}\smallsetminus 0= \{(0,0,z)\}$, so that the weights on $L$ are $0,-1,-2$. Note however that you are free to add a fixed integer to these weights (the linearization is not unique).


