Let $G=\mathrm{Gl}_n\mathbb C$ and let $X$ be an affine $G$variety. Let $\phi:\tilde X\to X$ be the normalization of $X$, i.e. the spectrum of the integral closure of $\mathbb C[X]$ in its fraction field. Can $\tilde X$ be given the structure of a $\tilde G$variety such that $\phi$ is equivariant?

2$\begingroup$ It is worth noting that this can fail for other choices of affine algebraic group $G$. $\endgroup$ – Jason Starr Oct 15 '13 at 12:21

2$\begingroup$ .. over a positive characteristic field $k$ (not over $\mathbb{C}$). $\endgroup$ – Jason Starr Oct 15 '13 at 15:51

8$\begingroup$ If you want to frighten yourself, you should check out Brian Conrad's work on pathologies for group schemes in positive characteristic (wait till Halloween, tune to a Vincent Price film, read by flashlight with the other lights turned off). $\endgroup$ – Jason Starr Oct 15 '13 at 15:54

$\begingroup$ Just about two more weeks. That'll be the best Halloween ever. Also, +1 and thanks for the comforting clarification ;). $\endgroup$ – Jesko Hüttenhain Oct 15 '13 at 16:21
Here is the sort of example I think Jason Starr was raising. (I looked at Brian's webpage, but it wasn't obvious which paper to read.) Take $k$ to be a perfect field of characteristic $p$, with $p \neq 0$, $2$. Let $A = k[x,y]/(y^2x^p)$. The normalization of $A$ is $\tilde{A} = k[t]$, with $y=t^p$ and $x=t^2$.
Let $G$ be the group scheme with underlying space $k[\epsilon]/\epsilon^p$ and multiplication given by the map $\epsilon \to \epsilon \otimes 1 + 1 \otimes \epsilon$. If, like me, you prefer to think in terms of functors of points, $G(R) = \{ \epsilon \in R : \epsilon^p =0 \}$ and the multiplication map $G(R) \times G(R) \to G(R)$ is $(\epsilon_1, \epsilon_2) \mapsto \epsilon_1 + \epsilon_2$.
Let $G(R)$ act on $A(R)$ by $(\epsilon, (x,y)) \mapsto (x+2 \epsilon, y)$. If you prefer maps of algebras, $x \mapsto 1 \otimes x + 2 \epsilon \otimes 1$, $y \mapsto 1 \otimes y$. I claim that this action does not lift to $\tilde{A}$. Suppose, to the contrary, that $t \mapsto 1 \otimes t_0 + \epsilon \otimes t_1 + \cdots + \epsilon^{p1} \otimes t_{p1}$, with the $t_i \in k[t]$. Writing out that the action must preserve the relation $t x^{(p1)/2} = y$ gives $$ \left( 1 \otimes t_0 + \epsilon \otimes t_1 + \cdots \right) (1 \otimes t^2 + 2 \epsilon \otimes 1)^{(p1)/2} = 1 \otimes t^p $$
Equating the coefficients of $1$ and $\epsilon$ gives $t_0 t^{p1} = t^p$ and $t_1 t^{p1} + (p1) t_0 t^{p3} = 0$. So $t_0=t$ and $t_1 = 1/t$. But $1/t$ isn't in $k[t]$.
Morally, the action wants to be $(\epsilon, t) \mapsto t \sqrt{1+2\epsilon t^{2}}=1+\epsilon t^{1}  (1/2) \epsilon^2 t^{3} + \cdots $. The trouble is that $\left( t \sqrt{1+2\epsilon t^{2}} \right)^k$ is in $k[t, \epsilon]/\epsilon^p$ when $k$ is even or is $\geq p$, but not in general.
This can never occur when $G$ is normal. Over a field of characteristic zero, all group schemes are regular, and regular implies normal, so there are no examples over a field of characteristic zero.
Recall the universal property of normalization: For any normal variety $Y$, the induced map $\mathrm{DomHom}(Y, \tilde{X}) \to \mathrm{DomHom}(Y,X)$ is bijective, where $\mathrm{DomHom}$ is the dominant homomorphisms.
Proof: If $G$ is normal, then $G \times \tilde{X}$ is normal. We have a map $G \times \tilde{X} \to G \times X \to X$, where the first map is $\mathrm{Id} \times \phi$ and the second map is the group action. By the universal property of the normalization, there is a map $G \times \tilde{X} \to \tilde{X}$ making the obvious diagram commute. We claim that this map gives an action of $G$ on $\tilde{X}$.
Consider the two maps $G \times G \times \tilde{X} \to \tilde{X}$. We must show they are equal. Again, by the universal property of normalization, it is enough to show that the two compositions $G \times G \times \tilde{X} \to X$ are equal. But these are the same as $G \times G \times \tilde{X} \to G \times G \times X$, follwed by the two maps $G \times G \times X \to X$, and these are equal because $G$ acts on $X$.

2$\begingroup$ I think you were supposed to wait until Halloween :) But I'm glad you didn't, since this is a nice answer. $\endgroup$ – Peter Samuelson Oct 16 '13 at 23:06

$\begingroup$ I thoroughly believe you, but I don't understand why group varieties in positive characteristic can be singular. Take a $g\in G$ which is a regular point of $G$, at least one of those exists. For any $h\in G$, the action of $gh^{1}$ on $G$ by left multiplication is an automorphism, and $h$ is the preimage of $g$. Hence, $h$ should also be smooth. Where did I use $\mathrm{char}(\Bbbk)=0$ ? $\endgroup$ – Jesko Hüttenhain Oct 18 '13 at 10:08

1$\begingroup$ Two issues: (1) A nonreduced scheme need not have a regular point. This is what it going on in the example: $\mathbb{k}[\epsilon]/\epsilon^p$ has one point, and that point is not regular. (2) The way you are talking about closed points and translation suggests you are working over an algebraically closed ground field. If your points $g$ or $h$ are defined over extensions of the ground field, particularly if they are defined over inseparable extensions, there are more issues. $\endgroup$ – David E Speyer Oct 18 '13 at 13:59

$\begingroup$ Oh, nice. That settles everything. Indeed the field is algebraically closed and I assume the group scheme to be a variety, in particular reduced. So I guess, then it will work all the time. Thanks even more! $\endgroup$ – Jesko Hüttenhain Oct 18 '13 at 20:34
I believe the answer is yes. Since $X$ is an affine $G$variety, $G$ acts on $\mathbb{C}[X]$ by $\mathbb{C}$algebra automorphisms. This yields an action of $G$ on the fraction field of $\mathbb{C}[X]$ by algebra automorphisms. This restricts to an action on the integral closure of $\mathbb{C}[X]$, so that inclusion into the integral closure is $G$equivariant. We can now apply Spec to this inclusion to obtain the desired equivariant map.

1$\begingroup$ That sounds nice, however I am worried about @JasonStarr's comment (see above)  where in your argument will one use the fact that $G=\mathrm{Gl}_n\mathbb C$ (or that $G$ is reductive, which I suppose will be the most relevant property)? $\endgroup$ – Jesko Hüttenhain Oct 15 '13 at 12:39

$\begingroup$ You are correct that I imposed no conditions on $G$ other than that it be affine algebraic. Technically, we want an algebraic action $G\times\tilde{X}\rightarrow\tilde{X}$. Such a thing presumably comes from a $\mathbb{C}$algebra morphism $\mathbb{C}[\tilde{X}]\rightarrow\mathbb{C}[\tilde{X}]\otimes\mathbb{C}[G]$ satisfying the properties of a rightaction. I think this should just be what I constructed, but I could be wrong. $\endgroup$ – Peter Crooks Oct 15 '13 at 14:02

2$\begingroup$ No you're right, I don't see why we would need $G$ to be reductive, or equal to $\mathrm{Gl}_n$. Defining $g.(a/b):=g.a/g.b$ is certainly a valid group action on the fraction field. If $a/b$ is integral over the coordinate ring, applying a group element $g$ to some integral relation it satisfied should yield an integral relation for the translate, because the coefficients remain in $\mathbb C[X]$. However, I might be missing something and my question was directed more @JasonStarr. $\endgroup$ – Jesko Hüttenhain Oct 15 '13 at 14:07

2$\begingroup$ The counterexamples are over a positive characteristic field $k$ (say, algebraically closed), where the group scheme $G$ is not uniquely determined by the group $G(k)$. This should never be an issue in characteristic $0$. $\endgroup$ – Jason Starr Oct 15 '13 at 15:50