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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.

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  • $\begingroup$ If your $\mathbb{P}^2$ is $\mathbb{P}(\mathbb{C}^3)$, you can identify the complement of the zero section in $L^{-1}$ with $\mathbb{C}^3\smallsetminus 0$. One possible way to extend your action is to have $t\in\mathbb{C}^*$ acts by $(x,y,z)\mapsto(x,ty,t^2z)$. Then $t$ acts trivially on $L_{e_0}$, as $t$ on $L_{e_1}$ and as $t^2$ on $L_{e_2}$, so that the weights on $L$ are $0,-1,-2$. Note however that you are free to add a fixed integer to these (the linearization is not unique). $\endgroup$
    – abx
    Commented Aug 10, 2020 at 15:53
  • $\begingroup$ Dear @abx, thanks a lot for the comment! I'm sorry, but I don't follow exactly your passages. You consider $\mathcal{O}(-1)$, and indeed $s_0(\mathbb{P}^2)^c=\mathbb{C}^3\setminus 0$. The you define an action on $\mathbb{C}^3\setminus 0$ with weights $(0,1,2)$, but I don't get the next passage, that is for example $t$ acts precisely as $t^2$ on $L_{e_2}$,I'm sorry The other points are fine, you swap sign since you're working with the dual bundle, and yes, I 'm aware of the last sentence. If you want to write it as an answer I'll surely accept it, thanks again for the patience! $\endgroup$
    – konoa
    Commented Aug 10, 2020 at 16:09
  • $\begingroup$ I have written my comment as an answer with some more details, let me know if this not completely clear. $\endgroup$
    – abx
    Commented Aug 10, 2020 at 16:21

1 Answer 1

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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).

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  • $\begingroup$ Dear @abx, thanks again for your patience, It's clearer now. What I was wondering is why you need to eliminate the zero section in your construction: intuitively I think you're removing the limit $\lim_{t\to 0}tp$, but maybe there's a deeper justification, Anyway thanks again, hope to receive other answer from you! Have a nice day $\endgroup$
    – konoa
    Commented Aug 10, 2020 at 19:48
  • $\begingroup$ Just because the fibration $V\smallsetminus 0\rightarrow \mathbb{P}(V)$, for $V$ a vector space, is not a line bundle but a $\mathbb{C}^*$-bundle, naturally isomorphic to the complement of the zero section in $L^{-1}$. You can also blow up $0$ in $V$, then you get a natural isomorphism with $L^{-1}$. $\endgroup$
    – abx
    Commented Aug 11, 2020 at 4:11

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