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I am looking for an example of a smooth Fano $3$-fold $X$ over $\mathbb{C}$, with a non-trival $\mathbb{C}^{*}$-action, which satisfies the following properties:

  1. There is a $\mathbb{C}^{*}$-action such that the fixed point set is finite.

  2. $Aut(X)$ contains no copy of $(\mathbb{C^{*}})^{2}$.

  3. The rank of the Picard group is at least $2$.

(Note that if we relax any of these three conditions, then there are examples. If we relax 1. then we can take $X = \mathbb{P}^{1} \times S$, where $S$ is a del Pezzo surface with discrete automorphism group. If we relax 2. then there are toric examples. If we relax $3.$, then the Mukai-Umemura Fano $3$-fold works. Hence, I believe that there is a reasonable chance that such an example should exist.)

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  • $\begingroup$ Do you mean all fixed points are isolated, or that there exist isolated fixed points? If the latter, you could take the blow up of P^3 along a twisted cubic: this has a Moebius group action, since P^3=Sym^3(P^1), and the twisted cubic is an orbit (triples of coincident points). A C^* inside the Moebius group has four isolated fixed points, two on the twisted cubic, so I guess these become fixed P^1s after blowing up, but the other two fixed points are still isolated. $\endgroup$ – Jonny Evans Jul 17 '18 at 19:30
  • $\begingroup$ I meant that the fixed point set it isolated. But thanks for the example! I did not know that one. $\endgroup$ – Nick L Jul 17 '18 at 21:11
  • $\begingroup$ I don't understand the first part of your comment, since condition 1) is that there is finite fixed points, so to relax this condition means that we don't require there to be finite fixed points; all I am saying is that $\mathbb{P}^{1} \times S$ satisfies 2. and 3 but not 1. $\endgroup$ – Nick L Jul 23 '18 at 6:10
  • $\begingroup$ @NickL You are absolutely right. Sorry for this. I removed my comment. $\endgroup$ – Jérémy Blanc Jul 23 '18 at 6:37
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Choose $X$ to be the blow-up of a smooth quadric $Q\subset \mathbb{P}^4$ at a curve $\Gamma\subset Q$ being a smooth normal rational quartic curve.

In coordinates, you can for instance choose $\Gamma$ to be the image of $$\mathbb{P}^1\to \mathbb{P}^4, [u:v]\mapsto [u^4:u^3v:u^2v^2:uv^3:v^3]$$ and $Q$ to be given by $$x_2^2+x_1x_3-2x_0x_4=0$$ so the action of $\mathbb{C}^*$ given by $[u:v]\mapsto [u:\xi v]$ extends to a unique action of $\mathbb{C}^*$ on $\mathbb{P}^4$, which has only $5$ fixed points, four of them being on $Q$ and $2$ of them on $\Gamma$, namely $[1:0:0:0:0]$ and $[0:0:0:0:1]$. You can check in coordinates that the action has finitely many fixed points on the blow-up.

Moreover, the group $\mathrm{Aut}^{\circ}X$ cannot contain $(\mathbb{C}^*)^2$ as this group would come from an action on $Q$ preserving $\Gamma$. (In fact, we can check that $\mathrm{Aut}^{\circ}X$ is $\mathrm{PGL}_2$ has index $2$ in $\mathrm{Aut} X$).

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