Algebraic Group Actions, Orbits, Stabilizers, and Controlled Quotients — Examples

The algebra $A=k[x]\mathrm{/}(x^2-a)$ is a field $K=k[x]\mathrm{/}(x^2-a)$: the polynomial $x^2-a$ has no root in $k$ because $a$ is a nonsquare, hence is irreducible of degree two, and it is separable because its derivative $2x$ is nonzero as $\mathrm{char}k\mathrm{\ne }2$ and $a\mathrm{\ne }0$. Thus $A$ is free of rank two over $k$ and finitely presented, and ${\mathrm{\Omega }}_{A\mathrm{/}k}=Adx\mathrm{/}(2xdx)=0$ because $x$ is a unit with $x^2=a\in k^{\times }$ and $2$ is a unit; by [F3], $X\to \mathrm{Spec}k$ is finite étale of degree two. A $k$-algebra map $A\to k$ would send $x$ to an element $q\in k$ with $q^2=a$, so $X(k)=\mathrm{\varnothing }$ and the orbit set $X(k)\mathrm{/}G(k)$ is empty.

Put $G=\mathrm{Spec}k[t]\mathrm{/}(t^2-1)$ with comultiplication $t\mapsto t\otimes t$, counit $t\mapsto 1$ and antipode $t\mapsto t$, the last well defined because $t^2=1$; for every $k$-algebra $R$ the set $G(R)=\{r\in R:r^2=1\}$ is a group under multiplication, naturally in $R$, so by [F1] these maps make $G$ a finite-type $k$-group scheme with these groups of points. The formula $g\cdot z=gz$ defines a natural action of $G(R)$ on $X(R)=\{z\in R:z^2=a\}$: it is associative, unital, and stays in $X(R)$ because $(gz{)}^2=g^2{z}^2=a$, so by [F2] there is a morphism $\alpha :G{\times }_{k}X\to X$ with $\alpha (g,z)=gz$.

The morphism $\phi :G{\times }_{k}X\to X{\times }_{k}X$, $(g,z)\mapsto (gz,z)$, is an isomorphism. On coordinate rings it is the $k$-algebra map $k[x,y]\mathrm{/}(x^2-a,y^2-a)\to k[t,x]\mathrm{/}(t^2-1,x^2-a)$ with $x\mapsto tx$ and $y\mapsto x$; in the source ring $x$ and $y$ are units with $x^2=y^2=a\in k^{\times }$, and the assignment $t\mapsto x\mathrm{/}y$, $x\mapsto y$ defines an inverse: it respects the relations since $(x\mathrm{/}y{)}^2=x^2\mathrm{/}{y}^2=1$ and $y^2=a$, and the two composites fix each generator.

The relation $R=G{\times }_{k}X\rightrightarrows X$ has $s(g,z)=z$ and $t(g,z)=gz$, so $j=(t,s)$ is the isomorphism $\phi$ of step 2.1, in particular an equivalence relation. The ring $B=A[t]\mathrm{/}(t^2-1)$ is free over $A$ via $s$ with basis $1,t$, so $s$ is finite locally free. The involution $(g,z)\mapsto (g,gz)$ carries $s$ to $t$, so $t$ is finite locally free as well. The invariant ring is $C=k$: for $f=\alpha +\beta x$ one computes $s^{\mathrm{♯}}(f)=\alpha +\beta x$ and $t^{\mathrm{♯}}(f)=\alpha +\beta tx$ in $B\cong K\times K$ (evaluation at $t=1,-1$). Equality is equivalent to $\beta x(t-1)=0$; since $x$ is a unit and $t-1$ has components $(0,-2)$, this forces $\beta =0$. By [F4] the quotient is represented by $\mathrm{Spec}C=\mathrm{Spec}k$, so $(X\mathrm{/}G)(k)=\{\ast \}$.

Consider the identity section ${\mathrm{i}\mathrm{d}}_X\in X(X)$ and its class $[{\mathrm{i}\mathrm{d}}_X]$ in the naive quotient presheaf $P(X)=X(X)\mathrm{/}G(X)$ of [F5]. Its two pullbacks along the projections $X{\times }_{k}X\rightrightarrows X$ are the first and second projections, which differ by the element $u=y\mathrm{/}x\in G(X{\times }_{k}X)$: indeed $u^2=y^2\mathrm{/}{x}^2=a\mathrm{/}a=1$, so $u$ is a ${\mu }_2$-valued point, and $u\cdot {\mathrm{p}\mathrm{r}}_1={\mathrm{p}\mathrm{r}}_2$ because $u\cdot x=y$. Hence the two pullbacks of $[{\mathrm{i}\mathrm{d}}_X]$ in $P(X{\times }_{k}X)$ agree, so the class of the identity section is a compatible family of sections of the naive presheaf over the fppf covering $X\to \mathrm{Spec}k$.

That compatible family does not descend: $P(k)=X(k)\mathrm{/}G(k)$ is empty by step 1.1, so there is no class in $P(k)$ whose pullback along $X\to \mathrm{Spec}k$ could be the class of the identity section. Therefore the naive quotient presheaf is not an fppf sheaf and is not the quotient sheaf $X\mathrm{/}G$; since $X\mathrm{/}G$ is represented by $\mathrm{Spec}k$ with $(X\mathrm{/}G)(k)=\{\ast \}$ by step 3.1, while $X(k)\mathrm{/}G(k)=\mathrm{\varnothing }$, the orbit set does not compute the $k$-points of the quotient sheaf. Sheafification is strictly necessary, and the quotient sheaf here is representable: the counterexample separates the orbit set from the quotient sheaf, not representability from the quotient sheaf.

The closed subscheme of $G$ cut out by the two off-diagonal coordinates has $R$-points the invertible diagonal matrices, which form a subgroup of ${\mathrm{GL}}_2(R)$ for every $R$; by [F2] it is a closed subgroup scheme, and we call it $T$. As a scheme $T$ is the open subscheme $ad\mathrm{\ne }0$ of the affine plane with coordinates $a,d$, and by [F1] both $G$ and $T$ are smooth and affine of finite type over $k$.

By [F1] the standard representation identifies ${\mathrm{GL}}_V$ with $G$, and by [F3] the two factors $\mathbb{P}(V^{\vee })\cong {\mathbb{P}}^1$ carry the actions induced by it, whose product is the stated action of $G$ on $Y$; the point $o=(⟨e_1⟩,⟨e_2⟩)$ is a $k$-point of $Y$. A matrix $g=(ab;cd)$ preserves the line $⟨e_1⟩$ exactly when $c=0$ and preserves $⟨e_2⟩$ exactly when $b=0$, and the two stabilizer functors are closed; hence the scheme-theoretic stabilizer $G_o$ has $G_o(R)=T(R)$ for every $k$-algebra $R$ and equals $T$ by [F2].

Let $\mathrm{\Delta }\subseteq Y$ be the diagonal and define $q:G\to Y\setminus \mathrm{\Delta }$ by $q(g)=(g⟨e_1⟩,g⟨e_2⟩)$; it is well defined because $g$ is invertible and $⟨e_1⟩\mathrm{\ne }⟨e_2⟩$. For a field $K$, every $K$-point $(L_1,L_2)$ of $Y\setminus \mathrm{\Delta }$ has linearly independent generators $u\in L_1$, $v\in L_2$, and the matrix with columns $u,v$ is an invertible element of $G(K)$ mapping $o$ to $(L_1,L_2)$; hence $q$ is surjective on $K$-points and its image set is $Y\setminus \mathrm{\Delta }$.

Put $M=Y\setminus \mathrm{\Delta }$ and apply [F3] to its identity point, obtaining the two universal line subbundles $L_1,L_2\subseteq V_M$. On an open $U\subseteq M$ where both have frames $u,v$, the determinant $u\wedge v$ is nonzero in every residue field: two lines in a two-dimensional vector space are dependent exactly when they coincide, and the diagonal has been removed. Thus the determinant lies in no maximal ideal of any affine chart of $U$ and is a unit; the column matrix $a=(u,v)$ is invertible. It defines a section of $q$ over $U$ and proves $L_1\oplus L_2=V_U$. The isomorphism $U{\times }_{k}T\to q^{-1}(U)$, $(m,h)\mapsto a(m)h$, has inverse $g\mapsto (q(g),a(q(g){)}^{-1}g)$: the second component preserves the two coordinate lines and hence belongs to the diagonal torus by step 1.2. These opens cover $M$, so $q$ is locally a projection with fibre the torus $T$, flat and locally of finite presentation by [F1] and [F5]. It is surjective by step 1.3, hence faithfully flat.

The morphism of the criterion $G{\times }_{k}T\to G{\times }_{Y\setminus \mathrm{\Delta }}G$, $(g,h)\mapsto (g,gh)$, is an isomorphism: on $R$-points for every $k$-algebra $R$ it is a bijection onto the pairs $(g,g^{\mathrm{\prime }})\in G(R{)}^2$ with $q(g)=q(g^{\mathrm{\prime }})$, with inverse $(g,g^{\mathrm{\prime }})\mapsto (g,g^{-1}{g}^{\mathrm{\prime }})$, and $g^{-1}{g}^{\mathrm{\prime }}\in T(R)$ exactly when $q(g)=q(g^{\mathrm{\prime }})$ by the stabilizer computation of step 1.2.

Applying the criterion [F4] to $U=G$, $R=G{\times }_{k}T$ with $s(g,h)=g$, $t(g,h)=gh$, $M=Y\setminus \mathrm{\Delta }$ and $q$ from steps 2.1 and 2.2 shows that $Y\setminus \mathrm{\Delta }$ represents the fppf quotient sheaf $G\mathrm{/}T$ and that the quotient morphism is $q$. Since the image of $q$ is $Y\setminus \mathrm{\Delta }$ by step 1.3, the orbit subscheme $O_o$ and $Y\setminus \mathrm{\Delta }$ agree; so $O_o$ represents $G\mathrm{/}T$, the quotient morphism is ${\varrho }_o=q$, and the conclusions of (a) and (b) follow, including the representability of $G\mathrm{/}T$.

For (c): the quotient $G\mathrm{/}T$ is isomorphic to $O_o$ by step 3.1; the orbit $O_o$ is smooth over $k$ and of finite type by Smooth orbits are locally closed and their orbit maps are faithfully flat over every field, and it is separated over $k$ because it is a locally closed subscheme of the separated finite-type $k$-scheme $Y$ by [F5], an immersion being separated (Open and closed immersions are separated) and separatedness being stable under composition (Separated morphisms compose). Base changing to an algebraic closure $\bar{k}$, the orbit-stabilizer identity [F6] applied to the connected smooth group $G_{\bar{k}}$ acting on $Y_{\bar{k}}$ and the orbit $O_{o,\bar{k}}$ gives $\dim O_{o,\bar{k}}=\dim G_{\bar{k}}-\dim T_{\bar{k}}=4-2=2$, and $(G\mathrm{/}T{)}_{\bar{k}}\cong O_{o,\bar{k}}$, which is the stated dimension. Here ${\mathrm{GL}}_2$ is the nonempty open subset $d\mathrm{\ne }0$ of ${\mathbf{A}}^4$, of dimension $4$ by [F5] and Affine and projective n-space have dimension n, and $T$ is the nonempty open subset $ad\mathrm{\ne }0$ of ${\mathbf{A}}^2$, of dimension $2$ by the same two results; $G_{\bar{k}}$ is connected because it is a nonempty open subscheme of the irreducible ${\mathbf{A}}_{\bar{k}}^4$, whose coordinate ring $\bar{k}[x_1,\dots ,x_4]$ is a domain (A polynomial ring over an integral domain is an integral domain), so that the zero ideal corresponds to ${\mathbf{A}}^4$ under the Nullstellensatz correspondence (Affine algebraic sets correspond to radical ideals, and irreducible ones to prime ideals) and nonempty open subschemes are irreducible by Irreducibility via nonempty open subsets, connectedness and open subspaces.

The first projection $Y\setminus \mathrm{\Delta }\to {\mathbb{P}}^1$ is the composite of the isomorphism $G\mathrm{/}T\cong Y\setminus \mathrm{\Delta }$ with ${\mathrm{p}\mathrm{r}}_1$; over a point $y$ put $K=\kappa (y)$ and let $L_1$ be its canonical $K$-point; the fibre is $\{L_2\in {\mathbb{P}}_K^1:L_2\mathrm{\ne }{L}_1\}$, which is ${\mathbb{P}}_K^1$ with one closed point removed. Finally the natural map $G(K)\mathrm{/}T(K)\to (G\mathrm{/}T)(K)$ is surjective because every $K$-point of $Y\setminus \mathrm{\Delta }$ is $q(g)$ for some $g\in G(K)$ by step 1.3, and injective because $q(g)=q(g^{\mathrm{\prime }})$ forces $g^{-1}{g}^{\mathrm{\prime }}\in T(K)$ by step 2.2; hence it is a bijection for every extension field $K\mathrm{/}k$. This completes the verification of (c).