Question

Suppose G is an algebraic group with an action G×X→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 Fix(T)={t∈X(T)|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.)

Answer 1

The question gives the "wrong" definition of 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) | G_T-action on X_T fixes x}
               = {x in X(T) | 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) | G(k) fixes x}
is the "wrong" notion of (X^G)(k), whereas
{x in X(k) | 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 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.

Answer 2

Let F be the field with two elements, and let G = G_m,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.