## 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
$X^{G} \neq \emptyset$.

smoothprojective varieties: a translation on an abelian variety has no fixed points. $\endgroup$ – Sándor Kovács Apr 16 '19 at 22:40