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I suspect this is really obvious, but I'm not seeing it.

For an algebraic group $G$ acting on a variety $V$, and for a point $x \in \text{hom}(\text{Spec}(K),V)$, we define the orbit $G(x)$ to be $G(x)(R)=\{gx|g \in G(R)\}$ where $x$ is viewed as $x \in \text{hom}(\text{Spec}(R),V)$ via $\text{Spec}(R) \to \text{Spec}(K)$.

Is this representable?

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Could you please clarify your question? Is $x$ held fixed in the definition of this functor? Or are you defining a functor of orbits and then specifying that "$x$" gives an element of your functor on $V$. It is certainly not true that the functor of "orbits" is representable by a scheme, e.g., the standard action of $\mathbf{G}_m$ on $\mathbb{A}^n$. In fact, if I am understanding correctly your definition, this functor is not even a sheaf. –  Jason Starr Aug 16 '12 at 13:43
@Jason: It seems likely that $x$ is held fixed (with $K$ the ground field). @Qiuzhen Li: The reason your functor is "wrong" is that (for $G$ smooth) the locally closed orbit subscheme of $V$ through $x$ turns out (by proof, not by definition) to represent the quotient sheaf $G/G_x$ where $G_x$ is the scheme-theoretic stabilizer ($R$-points consist of those $g \in G(R)$ such that $gx = x$ in $V(R)$), using the fppf topology (or etale topology if $G_x$ is smooth). The difference between quotient sheaves and quotient presheaves is your error. –  user22479 Aug 16 '12 at 13:57
Maybe you are simply trying to ask the following: supposing that $G$ and $V$ are defined over an alg. closed field $K$, is the set $\{g(x)|g\in G(K)\}$ the set of closed points of a scheme over $V$ ? –  Damian Rössler Aug 16 '12 at 14:06
FYI, if you want to make braces show up you need to type two backslashes first. –  MTS Aug 16 '12 at 14:08
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1 Answer

Thanks to those who clarified the question. The answer by quasi-coherent is correct. Here is one explicit example: let $G$ be $\mathbf{G}_m$, and let $V$ be a second copy of $\mathbf{G}_m$. For an integer $n>1$ (prime to the characteristic), define the action $\sigma:G\times V \to V$ by $\sigma(s,t) = s^nt$. Define $x$ to be the element $1$ in $\mathbf{G}_m$. The stabilizer (of every orbit) is $\mathbf{\mu}_n$. The quotient by the stabilizer is $q:G\to V$, $q(s) = s^n$. This is a finite, étale morphism. The fiber product $G\times_V G$ as a closed subscheme of $G\times G$ is the image of the closed immersion $$i:\mathbf{\mu}_m\times \mathbf{G}_m\to \mathbf{G}_m \times \mathbf{G}_m, \ (r,s) \mapsto (rs,s).$$ The identity $\text{Id}_G:G\to G$ gives an element $[\text{Id}_G]\in G(x)(G)$. Also the two pullbacks to $G\times_V G$ are equal. So, were your functor representable and thus a sheaf for the étale topology, then $[\text{Id}_G]$ would be the pullback of some $[g]\in G(x)(V)$. In turn, this would be an equivalence class of morphisms $g:\mathbf{G}_m\to \mathbf{G}_m$ such that $q\circ g$ equals the identity. However, there is no such morphism $g$. Therefore your functor is not an étale sheaf. Therefore it is not representable.

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