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.)

  • $\begingroup$ As Scott points out below, Anton's definition does not give a functor. Perhaps the following is a better definition. Let Stab be the stabilizer group scheme over X (the fiber product of the action map GxX --> XxX with the diagonal). Let Fix be the subsheaf of points of X where Stab --> G is an isomorphism. (In Scott's example of Gm acting on A<sup>n</sup> this gives the origin, as it should.) $\endgroup$ Nov 1, 2009 at 2:50

2 Answers 2


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.

  • $\begingroup$ Thanks a lot! I hope you don't mind I've edited your question to include a link to a pdf of the book. $\endgroup$ Feb 10, 2010 at 17:42
  • $\begingroup$ By the way, you can type math between dollar signs to get nice LaTeX-like output. For example, typing \$x\in X^G(T)\$ will produce $x\in X^G(T)$. $\endgroup$ 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.

  • $\begingroup$ Yikes! I thought for sure the answer would be yes. It looks like the problem is basically that Fix is not a sheaf. Do you know if it becomes representable if you take the (fppf) sheafification? $\endgroup$ Oct 29, 2009 at 3:05
  • 2
    $\begingroup$ I think the problem is that Fix isn't even a functor. It doesn't seem to respect morphisms. $\endgroup$
    – S. Carnahan
    Oct 29, 2009 at 3:25
  • 1
    $\begingroup$ Double Yikes ! $\endgroup$ Oct 29, 2009 at 18:48

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