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Let $\mathcal{M}_{d}(G)$ be the moduli space of $G$-Higgs bundles. $\mathcal{M}_{d}(G)$ have a non-trivial $\mathbb{C}^{*}$-holomorphic action by multiplication of the Higgs field, $$ z\cdot (E, \varphi)=(E, z\varphi). $$ I'm interested in finding the fixed points of this $\mathbb{C}^*$-action.

It is known that the restriction of this action to $S^1\subset \mathbb{C}^{*}$ is Hamiltonian with a momentum map $\mu$. So, we know how to find the fixed points of the action of $S^{1}$ once they will be exactly the critical points of $\mu$.

In Florentino, Gothen, and Nozad - Homotopy type of moduli spaces of $G$-Higgs bundles and reducibility of the nilpotent cone, on page 5, the authors state that the sets of fixed points of the actions of $S^1$ and $\mathbb{C}^{*}$ coincide. One of the inclusions is clear. However, I can't see why the other one should also hold.

Does anyone have any suggestions of how can I prove this?

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    $\begingroup$ You are working over a fixed compact Riemann surface, presumably. You can get the desired conclusion by putting the following facts together: (1) the moduli space $\mathcal{M}_{d}(G)$ is an algebraic variety, (2) action of $\mathbb{C}^*$ is on it is algebraic, (3) $S^1\subset \mathbb{C}^*$ is a Zariski dense subgroup. $\endgroup$ Oct 8, 2019 at 14:06
  • $\begingroup$ Hi, @DonuArapura. Thanks for your comment! Is it a general result of the algebraic group's actions? If yes, could you provide me some reference where I can find more about this subject? $\endgroup$ Oct 8, 2019 at 16:04

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Let me expand my comment "You can get the desired conclusion by putting the following facts together: (1) the moduli space $\mathcal{M}_𝑑(𝐺)$ is an algebraic variety, (2) the action of $\mathbb{C}^*$ on it is algebraic, (3) $S^1\subset \mathbb{C}^*$ is a Zariski dense subgroup."

(1) and (2) follows from the construction. See for example, Simpson, Moduli of representations of the fundamental group of a smooth projective variety I, II for details of that. I don't have a reference for (3), but it's elementary:

Lemma. Suppose that $G$ is a (say) complex algebraic group acting algebraically on a variety $X$, and let $K\subset G$ be Zariski dense subgroup. Then a point of $X$ fixed by $K$ is fixed by $G$.

Proof: Suppose that $x\in X$ is fixed by $K$. Its stabilizer is a Zariski closed subgroup $H\subseteq G$. But $K\subseteq H$, so $H=G$.

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