Fixed points under a finite group action on projective variety

Setup

Let $G$ be a finite group acting on a smooth projective variety $X$, and let $$ \rho: G \times X \to X $$ be the action morphism. For any $g \in G$ let $\rho_g$ denote the composition $$ X \simeq \{g\} \times X \subset G \times X \xrightarrow{\rho} X $$

The fixed point locus $X^G \subset X$ is a closed subscheme

(possibly empty):

Edit incorporating Peter McNamara's comment: $X^G$ is always closed when $G$ is an affine algebraic group (link in comments).

When $G$ is finite there's an elementary argument: note that for any $g \in G$, the $g$-fixed points $X^g$ fit into the cartesian diagram $\require{AMScd}$ \begin{CD} X^g @>>> X \\ @VVV @V \mathrm{id} \times \rho_g VV \\ X @> \Delta >> X \times X \end{CD}

Since $\Delta$ (and $\mathrm{id} \times \rho_{g}$ for that matter) are closed immersions and closed immersions are compatible with base change, $X^{g} \to X$ is a closed immersion. Since $X^{G} = \bigcap_{g \in G} X^{g}$, $X^{G}$ is also a closed subscheme of $X$.

Remark: This heavily relies on the fact that $G$ is a finite (or at least discrete) group, which is fine since that was the question, but I'd be interested in whether/when/how-to-show $X^{G}$ is closed in the case where $G$ is, say, a positive dimensional linearly reductive affine group scheme acting on $X$.

where $\Delta$ is the diagonal. As Sándor pointed out in a comment it definitely can happen that $X^{G}$ is empty, e.g. if $X$ is an abelian variety and $G \subset X$ is a finite subgroup acting by translations.

The fixed point locus is smooth

This is more difficult to prove -- an analytic argument due to Cartan appears in Algebraic geometry and topology, and an alternative approach is Luna's étale slice theorem.

The idea behind both approaches is to get local enough that the local geometry at a fixed point $x \in X^G$ is modeled by the tangent space $T_{X,x}$ with it's linear $G$--action (if $\rho_{g}: X \to X$ is the action of $g \in G$ on $X$, then $g$ acts on $T_{X,x}$ via $d \rho_{g}: T_{X, x} \to T_{X, \rho_{g}(x)} = T_{X, x}$). One then proves that $T_{X^{G}, x} = T_{X,x}^{G}$, the $G$-invariant subspace.

I won't attempt to go into further detail.

The fixed locus can have arbitrary dimension:

Let $\rho: G \times V \to V$ be a linear representation of a finite group $G$. Then $\mathbb{P}(V)$ is a smooth projective variety with an induced $G$-action $$ \bar{\rho}: G \times \mathbb{P}(V) \to \mathbb{P}(V), \, \, \text{ where } \bar{\rho}[v] = [\rho(v)] $$ Observe that for a non-0 vector $v \in V$, $[v] \in \mathbb{P}(V)^{G}$ if and only if for each $g \in G$ there's a scalar $\lambda_{g} \in \mathbb{C}^{\times}$ so that $\rho_{g}(v) = \lambda_{g} v$. Evidently this occurs if and only if $v$ lies in the isotypical summand of a character (1-dimensional representation) $L$ of $V$.

This observation together with some representation theory of finite groups yields examples where $\dim X^{G}$ takes arbitrary values in $0, \dots, \dim X$.

Criteria for $X^{G}$ to be non-empty

There are various ``fixed point theorems'' which guarantee the existence of, well, fixed points. See for instance the Lefschetz fixed point theorem and the holomorphic Lefschetz fixed point theorem. The later is especially powerful and shows for instance that if $$ H^{i}(X, \mathscr{O}_{X}) = 0 \,\, \text{ for } i > 0 $$ (for example, if $X$ is Fano or even just rationally connected) then $X^{G} \neq \emptyset$.

The answer to the edited question is no. Take a double covering $\pi :X\rightarrow \mathbb{P}^2$ branched along a smooth quartic curve $C$ (so $X$ is a Del Pezzo surface), and $G=\langle \sigma \rangle$, where $\sigma $ is the involution such that $\pi \circ\sigma =\pi $. Then $\operatorname{rk} H^*(X)=10$, but $\operatorname{rk} H^*(X^G)=\operatorname{rk} H^*(C)=8$.