Etale Slice Theorem

Sketched proof. This is simply because $\widehat{\mathfrak{m}_{x} /\mathfrak{m}_{x}^{2}}\cong \mathfrak{m}_{x}/\mathfrak{m}_{x}^{2}$.

We omit the other direction. ◻

Sketched proof. Any flat morphism between varieties is open. This is a highly nontrivial commutative algebra result.

Note that for any $x\in X$, $\dim \mathcal{O}_{X,x}=\dim \widehat{\mathcal{O}_{X,x}}$. Therefore the generic point of $X$ is mapped to a point in $Y$ with codimension 0.

This follows from the fact that flat and $\Omega_{X /Y}=0$ is stable under base change. ◻

Proof. See ⁵ for the proof. ◻

Proof. This is a direct corollary of Remark 5.

We shall recall the fibre $\pi_{X}^{-1}(u)$ is actually given by the fibre product:

$$\pi_{X}^{-1}(u)=\mathop{\mathrm{Spec}}k(u)\times_{X //G}X=\mathop{\mathrm{Spec}}k(u)\times_{X //G}(Y\times_{Y //G}X //G)=\mathop{\mathrm{Spec}}k(u)\times_{Y //G}Y,$$

and $\mathop{\mathrm{Spec}}k(\bar{f}(u))\times_{Y //G}Y$ is the fibre $\pi_{Y}^{-1}(\bar{f}(u))$. By commutative algebra, the residue field of $\mathcal{O}_{X,u}$ and the residue field of $\widehat{\mathcal{O}_{X,u}}$ are the same, since $\widehat{\mathcal{O}_{X,u}}=\widehat{\mathcal{O}_{Y,\bar{f}(u)}}$, we conclude $k(u)\cong k(\bar{f}(u))$. Thus $\pi_{X}^{-1}(u)\cong \pi_{Y}^{-1}(\bar{f}(u))$.

For the second assertion, $V\subseteq \pi_{Y}^{-1}\pi_{Y}(V)$ is clear. For any $y\in \pi_{Y}^{-1}\pi_{Y}(V)$, choose $u$ such that $\bar{f}(u)=\pi_{Y}(y)$. Then $\pi_{Y}^{-1}(\pi_{Y}(y))\cong \pi_{X}^{-1}(u)$ and there exists $v\in \pi_{X}^{-1}(u)$ such that $f(v)=y$.

Let $u$ be the point $\pi_{X}(Gx)$, and then $Gx$ is in the fibre $\pi_{X}^{-1}(u)$. Therefore, $f|_{Gx}$ factor through $\pi_{X}^{-1}(u)\xrightarrow{\simeq} \pi_{Y}^{-1}(\bar{f}(u))$, which is clearly injective.

Note $\pi_{X}$ maps $\overline{Gx}$ to the point $u$, so $Gx$ is closed if and only if $Gx=\pi_{X}^{-1}(u)$. Similarly, $Gf(x)\subseteq \pi_{Y}^{-1}(\bar{f}(u))$. We also have $f(Gx) \subseteq Gf(x)$, for any point $x'\in \partial Gx$, since $f$ preserves dimension, $f(x')$ must lies in $\partial Gf(x)$. Thus $Gf(x)$ is closed implies $Gx$ is closed. Therefore, by the isomorphism on fibres, we know $\pi_{X}^{-1}(u)=Gx$ if and only if $\pi_{Y}^{-1}(\bar{f}(u))=Gf(x)$. We conclude that $x$ has closed orbit if and onl if $f(x)$ has closed orbit. ◻

Proof. There’s no easy approach to this theorem. See ⁴ for an approach and ¹ Theorem 6.5.21 for a modern approach. ◻

Proof. Since our base field $k$ is algebraically closed, the residue field $k(x)$ is just the base field. So $G\cong G /G_{x}$. If $X$ is affine, all the closed subvariety is also affine, and the corollary follows. ◻

Proof. $G /G_{x}$ is principal $G_{x}$-bundle, consider an étale morphism $U\to G /G_{x}$, by the definition of principal bundle, we have $G\times_{G //G_{x}}U\cong G_{x}\times U$. The morphism $G\to G\times S$, $g\mapsto (g,s)$ for any $s$ give rise to to a morphism $G\times^{G_{x}}S\to G /G_{x}$.

We first show that $U\times_{G /G_{x}}(G\times^{G_{x}}S)\cong U\times S$.

Since $H$ act on $U$ trivially, the GIT quotient pulls back, i.e.

$$(G\times S) //G_{x}\times_{G /G_{x}} U\cong ((G\times_{G /G_{x}}U)\times S) //G_{x}.$$

Then

$$((G\times_{G /G_{x}}U)\times S) //G_{x}\cong (G_{x}\times U\times S) //G_{x}.$$

The action of $G_{x}$ on $G_{x}\times U\times S$ is given by $h\cdot (g,u,s)=(gh^{-1},u,hs)$, so $(G_{x}\times U\times S) //G_{x}\cong U\times S$.

Therefore $U\times_{G /G_{x}}(G\times^{G_{x}}S)\cong U\times S$, $U\to G /G_{x}$ is étale so $U\times S\to G\times^{G_{x}}S$ is étale.

The trivialization is also clear:

$$\begin{aligned} U\times S\times_{G\times^{G_{x}S}}(G\times S)&\cong (U\times_{G /G_{x}}(G\times^{G_{x}}S))\times_{G\times^{G_{x}}S}(G\times S)\\&\cong U\times_{G /G_{x}}G\times S\\&\cong U\times S\times G_{x}. \end{aligned}$$

So we conclude that $G\times S\to G\times^{G_{x}}S$ defines a principal $G_{x}$-bundle.

Consider the morphism $f:S\to G\times S$ and the projection $g:G\times S\to S$. We first note that the $g\circ f$ gives the identity on $S$. Consider the diagram:

the morphism composed from the upper paths defines an $G$-invariant morphism $S\to (G\times^{G_{x}}S)//G$ since $G\times^{G_{x}}S\to (G\times^{G_{x}}S)//G$ is $G$-invariant. So by the universal property of GIT, it factor through $S //G_{x}$. We get

$$\bar{f}:S //G_{x}\to (G\times^{G_{x}}S)//G.$$

Similarly, we can factor the projection morphism $G\times S\to S$ through $G\times^{G_{x}}S\to S //G_{x}$. Since $S$ and $S //G_{x}$ has trivial $G$ action, we can factor it through

$$\bar{g}: (G\times^{G_{x}}S)//G\to S //G_{x}.$$

By the universal property, clearly $\bar{g}\circ \bar{f}$ is the identity morphism. Now we check $\bar{f}\circ \bar{g}$ is also identity. Note that $\bar{f}\circ\bar{g}$ is induced by the morphism $G\times S\to G\times S$, $(g,s)\mapsto (g,s)$, which maps the orbit of $G$-orbit of $G\times^{G_{x}}S$ to the same $G$-orbit. So $f\circ g$ induce the identity morphism. Thus $S //G_{x}\cong (G\times^{G_{x}}S)//G$.

Pick an étale morphism $U\to G\times^{G_{x}}S$ whose image contains $y$. By the definition of principal $G_{x}$ bundle, $U\times G_{x}\cong G\times S$, so $T_{y}U\oplus T_{e}G_{x}\cong T_{e}G\oplus T_{s}S$, and $T_{y}(G\times^{G_{x}} S)\cong (T_{e}G\oplus T_{s}S)/T_{e}G_{x}$. ◻

Proof. Very complicated. See ² Theorem 4.18 for a classical approach. See ¹ Theorem 6.5.28 for a modern approach. ◻

Proof. We may embed $X\to \mathbb{A}^{n}$ then we can regard $\mathfrak{m}_{x}\subseteq k[x_{1},\dots, x_{n}]$ a maximal ideal. Define the natural homomorphism

$$d:\mathfrak{m}_{x}\to \mathfrak{m}_{x} /\mathfrak{m}_{x}^{2}=T_{x}X^{*},\quad t\mapsto t+\mathfrak{m}_{x}^{2}.$$

The linearly reductive group $G_{x}$ has an action on $\mathfrak{m}_{x}$(i.e. $\mathfrak{m}_{x}^{G_{x}}=\mathfrak{m}_{x}$) since $x$ is invariant under the action of $G_{x}$, and so as $\mathfrak{m}_{x} /\mathfrak{m}_{x}^{2}$. Since $G_{x}$ is linearly reductive, there’s a invariant $k$-vector subspace $W\subseteq \mathfrak{m}_{x}$ such that

$$d|_{W}:W\to \mathfrak{m}_{x} /\mathfrak{m}_{x}^{2}$$

is an isomorphism. So we could define

$$i=(d|_{W})^{-1}:\mathfrak{m}_{x} /\mathfrak{m}_{x}^{2}\to W$$

such that $d\circ i$ is the identity map on $\mathfrak{m}_{x}/\mathfrak{m}_{x}^{2}$. Let $S^{*}i:S^{*}(\mathfrak{m}_{x}/\mathfrak{m}_{x}^{2})\to S^{*}W$ be the induced homomorphism on the symmetric algebra. The inclusion $j:W\hookrightarrow \mathcal{O}(X)$ also gives an symmetric algebra homomorphism

$$S^{*}j:S^{*}W\to S^{*}\mathcal{O}(X)=\mathcal{O}(X).$$

$S^{*}j\circ S^{*}i$ is $G_{x}$-equivariant since $\mathfrak{m}_{x}\hookrightarrow \mathcal{O}(X)$ is $G_{x}$-equivariant, and $S^{*}i$ thus induces an $G_{x}$-equivariant morphism $\phi:X=\mathop{\mathrm{Spec}}\mathcal{O}(X)\to \mathop{\mathrm{Spec}}S^{*}(\mathfrak{m}_{x} /\mathfrak{m}_{x}^{2})$.

Now we check $\phi$ is the desired morphism. Let $f_{1},\dots, f_{n}$ generates $W\cong \mathfrak{m}_{x} /\mathfrak{m}_{x}^{2}$, the number of generator is exactly $n$ since $X$ is smooth at $x$. Then $S^{*}W=k[f_{1},\dots, f_{n}]$. $\phi$ is given by

$$\phi:X\to \mathop{\mathrm{Spec}}k[f_{1},\dots, f_{n}],\quad x\mapsto (f_{1}(x),\dots,f_{n}(x)).$$

Since $f_{1},\dots, f_{n}\in \mathfrak{m}_{x}$, $f_{1}(x)= \cdots =f_{n}(x)=0$, $\phi$ maps $x$ to $0$.

The remaining part is to show $\phi$ is étale at $x$. $T_{x}X=(\mathfrak{m}_{x} /\mathfrak{m}_{x}^{2})^{*}$ and $T_{\phi(x)}\mathop{\mathrm{Spec}}k[f_{1},\dots, f_{n}]= ((f_{1},\dots,f_{n}) /(f_{1},\dots, f_{n})^{2})$, the morphism $\phi$ maps $\mathfrak{m}_{x}$ to $(f_{1},\dots, f_{n})$. So

$$d\phi:(\mathfrak{m}_{x}/\mathfrak{m}_{x}^{2})^{2}\to ((f_{1},\dots,f_{n}) /(f_{1},\dots, f_{n})^{2})$$

is an isomorphism. Hence $\phi$ is étale at $x$. ◻

Proof. We only prove for the case $X$ is smooth at $x$.

Note that $\mathop{\mathrm{Spec}}S^{*}(\mathfrak{m}_{x}/\mathfrak{m}_{x}^{2})=\mathcal{T}_{x}X$ is the tangent bundle of $X$ at $x$, we have the and exact sequence

$$0\to \mathcal{T}_{x}Gx\to \mathcal{T}_{x}X\to \mathcal{N}_{Gx /X}\to 0,$$

where $\mathcal{N}_{Gx/X}$ is the normal bundle.

Let $I$ be the ideal corresponding to the closed subvariety $Gx$, then $I^{G_{x}}=I$ since $Gx$ is invariant under $G_{x}$-action. So $(\mathfrak{m}_{x}/I) /(\mathfrak{m}_{x}/I)^{2}$ is invariant under $G_{x}$-action. There’s a $G_{x}$-action on the bundle $\mathcal{T}_{x}Gx$ and the morphism $\mathcal{T}_{x}Gx\to \mathcal{T}_{x}X$ is $G_{x}$-equivariant. Thus the exact sequence splits and $\mathcal{T}_{x}X\cong \mathcal{T}_{x}Gx\oplus \mathcal{N}_{Gx /X}$ and $\mathcal{N}_{Gx /X}$ also has a $G_{x}$-action.

Let $\phi:X\to \mathcal{T}_{x}X$ be the map in in Lemma 17. Let $U=\phi^{-1}(\mathcal{N}_{Gx /X})$. $\mathcal{N}_{Gx /X}$ is invariant under $G_{x}$-action and $\phi$ is $G_{x}$-equivariant, so $U$ is invariant under $G_{x}$-action. Also, clearly $U$ is smooth at $x$. We may construct the morphism $\varphi:G\times^{G_{x}}U\to X$ as in part 2.

Let $y$ be the image of $(e,x)$ under the quotient morphism $\pi_{G\times U}:G\times U\to G\times^{G_{x}}U$. $d\varphi: T_{y}(G\times^{G_{x}}U)\to T_{x}X$ defines a homomorphism of tangent spaces. Let’s examine this differential in detail: $d\varphi$ is induced from the differential $d\tau: T_{e}G\oplus T_{x}U\to T_{x}X$, where $\tau:G\times U\to X$ is the action. Define

$$\delta:G\to Gx,\quad g\mapsto gx$$

then

$$d\tau: (v,u)\mapsto d\delta(v)+u.$$

From our construction of $U$, $T_{x}U\oplus T_{x}Gx=T_{x}X$, so $(v,u)\in \mathop{\mathrm{ker}}d\tau$ implies $u=0$ and $d\delta(v)=0$. On the other hand, the inclusion $T_{e}G_{x}\to T_{e}G\oplus T_{x}U$ is given by

$$i:G_{x}\to G\times X, \quad h\mapsto (h^{-1},h\cdot x),$$

so

$$d i:T_{e}G_{x}\to T_{e}G\oplus T_{x}U,\quad v\mapsto (-v,0).$$

Therefore, $\mathop{\mathrm{ker}}d\tau \subseteq T_{e}G_{x}$ and $d\varphi:(T_{e}G\oplus T_{x}U) /T_{e}G_{x}\to T_{x}X$ is injective.
Computing the dimension:

$$\begin{aligned} \dim T_{e}G-\dim T_{e}G_{x}&=\dim G-\dim G_{x}\\ &=\dim Gx= \dim T_{x}Gx=\dim \mathcal{T}_{x}Gx\\ &=\dim \mathcal{T}_{x}X-\dim \mathcal{N}_{Gx /X}\\ &=\dim X-\dim U\\ &=\dim T_{x}X-\dim T_{x}U,\end{aligned}$$

so $d\phi$ defines an isomorphism. Thus $\phi$ is étale at $x$.

Now we show that $\varphi|_{Gy}$ is injective. If $\varphi(g\cdot y)= \varphi(g'\cdot y)$, then by lifting to $G\times U$, $g\cdot x=g'\cdot x$. Then $g^{-1}g'x=x$ and thus $g^{-1}g'\in G_{x}$. Therefore $g\cdot y=g'\cdot y$ in $(G\times U) //G_{x}$.

By Lemma 16, we can find locally closed subvariety $S\hookrightarrow X$ that $S$ is an étale slice. ◻

Proof. First $Gx$ is closed, otherwise there’s a point $y\in \overline{Gx} \backslash Gx$. such that $Gy$ is closed and $\dim Gy<\dim Gx$. Since $\dim Gx+\dim G_{x}=\dim Gy+\dim G_{y}=\dim G$, we have $\dim G_{y}>\dim G_{x}=0$, that contradicts with the assumption.

Now for any $u\in X //G$, take $x \in X$ such that $\pi_{X}(x)=u$. Let $S$ be the étale slice through $x$. Since $G_{x}=\{e\}$, $G\times^{G_{x}}S= G\times S$. We have the strongly étale morphism $G\times S\to X$ and all horizontal morphisms in the fibre product diagram

are étale. So $S$ gives the local étale trivialization and $\pi_{X}:X\to X //G$ is principal $G$ bundle. ◻

Proof. By Theorem 19, we can take étale slice $S$ smooth at $x$. Since $G_{x}=\{e\}$, $G\times^{G_{x}}S= G\times S$. So we also have a fibre product diagram:

with horizontal étale morphisms. Since $S$ is smooth,

$$\dim S=\dim T_{x}S=\dim T_{\pi_{X}(x)}X //G=\dim X //G.$$

So $X //G$ is smooth at $\pi_{X}(x)$. ◻