Representability and smoothness of concentrator subschemes

Suppose first that $X=\mathrm{Spec}A$ is affine, with gradation $A={⨁}_{n\in \mathbb{Z}}{A}_n$ induced by the action, and let $\mathfrak{b}$ be the ideal generated by $\mathfrak{a}\cap A_0+{\sum }_{n<0}{A}_n$, so that $X(Z)=\mathrm{Spec}(A\mathrm{/}\mathfrak{b})$ in the sense of [F1]. For a $k$-algebra $R$ and a point $x\in X(R)$, identified with a homomorphism $x:A\to R$ by [F2], the orbit map corresponds to $A\to R[T,T^{-1}]$, ${\sum }_n{f}_n\mapsto {\sum }_{n}x(f_n)T^n$; the limit exists exactly when $x(A_n)=0$ for all $n<0$, and then the limit point is the composite $f\mapsto x(f_0)$ together with the point of $Z$. The limit lies in $Z(R)$ exactly when this composite kills $\mathfrak{a}$, that is, when $x(\mathfrak{a}\cap A_0)=0$; combining, $x$ satisfies the defining property of the concentrator functor exactly when $x$ kills $\mathfrak{b}$, i.e. exactly when $x$ factors through $A\mathrm{/}\mathfrak{b}$. Hence the affine concentrator represents the functor and the inclusion is a closed immersion.

Keep $X$ affine and suppose $X$ and $Z$ smooth. Work over an algebraic closure and let $x$ be a geometric point of $X(Z)$, with fixed limit $y$. Localize $A$ by $A_0\setminus {\mathfrak{m}}_0$, where ${\mathfrak{m}}_0$ is the ideal of $y$ in $A_0$. Evaluation at the fixed point $y$ kills every nonzero-degree piece, so $\mathfrak{m}={\mathfrak{m}}_0\oplus {⨁}_{n\mathrm{\ne }0}{A}_n$ is now a homogeneous maximal ideal. The local rings of $X$ and $Z$ at $y$ are regular, so [F3] shows that the local ring of $X(Z)$ at $y$ is regular. The quotient $A\mathrm{/}\mathfrak{b}$ has only nonnegative degrees, giving an ${\mathbf{A}}^1$-action that contracts $x$ to $y$. The nonregular locus of this finite-type scheme is closed and ${\mathbf{G}}_m$-stable; if it contained $x$, its closedness would also put $y$ in it, a contradiction. Hence every geometric point $x$ is regular and $X(Z)$ is smooth. In the affine case uniqueness as a smooth closed subscheme follows from reducedness and equality of geometric point sets.

Choose a finite cover by invariant affine opens $X_i$ and put $Z_i=Z\cap X_i$. The limit functor is local on the test scheme, because orbit extensions glue uniquely by separatedness. Its subfunctor consisting of extensions landing in $X_i$ is open: it is the condition that the limit morphism to $Z$ lands in $Z_i$. Indeed, an invariant closed complement $X\setminus X_i$ contains the limit of any orbit starting in it; hence a limit in $X_i$ forces the entire orbit extension into $X_i$, checked on geometric fibers. The affine schemes $X_i(Z_i)$ of step 1.1 therefore glue along these open subfunctors and cover the representing scheme $X(Z)$. On each such open the realization is a closed immersion into $X_i$, making it a local immersion. Also $p^{-1}(Z_i)=X_i(Z_i)$ is affine, so $p$ is affine. Smoothness follows locally from step 2.1, and representability supplies uniqueness of the scheme and both realization morphisms.

Base change: for a field extension $k^{\mathrm{\prime }}\mathrm{/}k$ the graded description of step 1.1 is stable under $A\mapsto A{\otimes }_k{k}^{\mathrm{\prime }}$ and limits of $R{\otimes }_k{k}^{\mathrm{\prime }}$-points are computed after this base change, so $X(Z{)}_{k^{\mathrm{\prime }}}=X_{k^{\mathrm{\prime }}}(Z_{k^{\mathrm{\prime }}})$ compatibly with $i$ and $p$; for a ${\mathbf{G}}_m$-equivariant morphism $f:X\to X^{\mathrm{\prime }}$ carrying $Z$ into $Z^{\mathrm{\prime }}$ the same functorial description shows $f(X(Z))\subseteq X^{\mathrm{\prime }}(Z^{\mathrm{\prime }})$, giving the morphism of pairs. For the conjugation action of ${\mathbf{G}}_m$ on a smooth affine group $G$ with cocharacter $\lambda$, let $H=G(\{g:{\lim }_{t\to 0}t\cdot g=e\})$. The defining condition is closed and its points form a subgroup of $G(R)$ for every $R$, because conjugation is a group homomorphism, $(gh{)}_t=tg{t}^{-1}th{t}^{-1}$, and the limit of a product is the product of the limits; hence $H$ is an algebraic subgroup. It is normal in $P_G(\lambda )$: for $g\in P_G(\lambda )(R)$ the limit $q={\lim }_{t\to 0}t\cdot g$ is defined, and for $h\in H(R)$ the conjugates $gh{g}^{-1}$ have limit $qe{q}^{-1}=e$, so $gH{g}^{-1}\subseteq H$, and symmetry gives $gH{g}^{-1}=H$. AC is used only through the geometric suppliers named in the deps.