Suppose $G$ is an algebraic group with an action $G\times X\to X$ on a scheme. Does the fixed locus (the set of points x∈X fixed by all of $G$) have a scheme structure? You can obviously define the functor $\operatorname{Fix}(T)=\{t\in X(T)\mid \text{$t$ is fixed by every element of $G(T)$}\}$. Is this functor always representable?
(This question was "broken off" of a compound question of mine after Scott Carnahan answered the other part so wonderfully that I had to accept his answer.)
The question gives the "wrong" definition of $\operatorname{Fix}(T)$, hence the resulting confusion.
A more natural definition of the subfunctor $X^G$ of "$G$-fixed points in $X$" is
$$
X^G(T) = \{x \in X(T) \mid\text{$G_T$-action on $X_T$ fixes $x$}\}
= \{x \in X(T) \mid\text{$G(T')$-action on $X(T') $ fixes $x$ for all $T$-schemes $T'$}\}.
$$
(Of course, can just as well restriction to affine $T$ and $T'$ for "practical" purposes.)
By way of analogy with more classical situations, if the base is a field k then a moment's reflection with the case of finite k shows that
$\{x \in X(k) \mid\text{$ G(k)$ fixes $x$}\} $
is the "wrong" notion of $X^G(k)$, whereas $
\{x \in X(k) \mid\text{$G$-action on $X$ fixes $x$}\}$ is a "better" notion, and is what the above definition of $X^G(k)$ says.
From this point of view, if (for simplicity of notation) the base scheme is an affine $\operatorname{Spec}(k)$ for a commutative ring $k$ then the "scheme of $G$-fixed points" exists whenever $G$ is affine and $X$ is separated provided that $k[G]$ is $k$-free (or becomes so after faithfully flat extension on $k$). So this works when $k$ is a field, or any $k$ if $G$ is a $k$-torus (or "of multiplicative type"). See Proposition A.8.10(1) in the book Pseudo-reductive groups.
\$x\in X^G(T)\$ will produce $x\in X^G(T)$.
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Commented
Feb 10, 2010 at 17:44
Let F be the field with two elements, and let G = Gm,F. Let X = An, affine space of dimension n (at least 1), with G acting by dilations. Then G(F) is trivial, so Fix(F) = X(F), which has elements other than the origin. Fix(F4) is the origin, but it should contain all the F4-points that factor through the canonical map to Spec F. Fix is therefore not representable.
