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Question: Let $X$ be a scheme separated and of finite type over $\mathbb C$ with a $\mathbb C^*$-action. Suppose that

  1. The fixed point locus is proper.
  2. Each orbit has proper closure.

Does this imply that $X$ is proper? Proper means proper over $\mathbb C$, of course.

Some comments: It would be handy if we had this criterion for properness. I have been trying to find a counterexample but did not succeed. I tried to start with a proper scheme and then throw away some non-constant orbits. But it seems that this will always give us something like $\{(x,y)\in \mathbb A^2, y= 0 \text{ when }x=0\}$, which is not a scheme. I am not sure if this picture is correct in general, especially when $X$ is highly singular. Also there could (a priori) be examples which can not be compactified respecting the $\mathbb C^*$-action.

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  • $\begingroup$ By the usual Chevalley results, it suffices to prove that the underlying reduced scheme $X_{\text{red}}$ is proper, and then it suffices to prove properness of the normalization of $X_{\text{red}}$. Thus, assume that $X$ is integral and normal. By Chow's Lemma there exists a projective morphism $\nu:\widetilde{X}\to X$ that is an isomorphism over a dense open $X^o\subset X$, and there exists a dense open immersion $\iota:\widetilde{X}\to \overline{X}$ with $\overline{X}$ projective. The scheme $X$ is proper if and only if $\widetilde{X}$ is proper if and only if $\iota$ is surjective . . . $\endgroup$ – Jason Starr Jan 31 '17 at 16:48
  • $\begingroup$ . . . If $\iota$ is not surjective, then there exists a smooth, affine curve $C$, a closed point $p\in C$, and a $k$-morphism $u:C\to \overline{X}$ such that $u(p)\not\in \text{Image}(\iota)$, yet $\iota^{-1}(\widetilde{X}^o)$ equals $C^o = C\setminus\{p\}$. The action defines a morphism $v:\mathbb{G}_m\times C^o\to \widetilde{X}^o$. By the valuative criterion of properness, this extends to a regular morphism on $\mathbb{P}^1\times C$ away from finitely many points. There exists a blowing up $\lambda:S\to \mathbb{P}^1\times C$ such that $v\circ \lambda$ is everywhere regular . . . $\endgroup$ – Jason Starr Jan 31 '17 at 16:55
  • $\begingroup$ . . . By hypothesis, the strict transform of $\{0\}\times C$ maps into $\widetilde{X}$. In particular, $(0,p)\in \mathbb{P}^1\times C$ maps to a point of $X$. For any open affine neighborhood $U$ of the image point, $\lambda^{-1}(\nu^{-1})(U)$ is a dense open that contains a neighborhood of $(0,p)$. Up to replacing $C$ by a dense open that contains $p$, we may assume that there exists a section $\sigma:C\to \lambda^{-1}(\nu^{-1})(U)$ with $\sigma(p)=(0,p)$. So now start over with $w=\nu\circ v\circ \lambda\circ C$ as a morphism $C\to X$ . . . $\endgroup$ – Jason Starr Jan 31 '17 at 17:13
  • $\begingroup$ . . . Again via the action, $w$ gives a morphism $t:\mathbb{G}_m\times C \to X$. Up to removing finitely many closed points of $C$ that are different from $p$, this extends to a regular morphism $\mathbb{P}^1\times C \setminus\{(\infty,p)\} \to X$. By Zariski's Main Theorem, if this is not regular at $(\infty,p)$, then after blowing up $(\infty,p)$, the induced rational transformation from the exceptional curve $E$ is generically finite. But this rational transformation is $\mathbb{G}_m$-equivariant by construction. Thus, the indeterminacy locus for the new rational map is invariant . . . $\endgroup$ – Jason Starr Jan 31 '17 at 17:17
  • $\begingroup$ . . . So the indeterminacy is contained in $0$ and $\infty$ on $E$. Now iterate. At each stage, by the same argument, the indeterminacy locus has to be $\mathbb{G}_m$-invariant and finite, i.e., contained in $0$ and $\infty$. Also, by measuring the degrees of the strict transforms of these curves in $\overline{X}$ (with respect to an ample invertible sheaf) and comparing to the degree of the fibers $\mathbb{P}^1\times\{q\}$, $q\neq p$, there can be only finitely many blowing ups before the rational map becomes regular. $\endgroup$ – Jason Starr Jan 31 '17 at 17:21
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Edit. I am making the answer shorter by eliminating the repeated argument.

Let $k$ be a perfect field. Let $X$ be a finite type, separated $k$-scheme. Let $$m:\mathbb{G}_m \times_{\text{Spec}\ k} X \to X$$ be a $k$-action of $\mathbb{G}_m$ on $X$. Since $k$ is perfect, $m$ induces a $k$-action of $\mathbb{G}_m$ on $X_{\text{red}}$, the reduced scheme of $X$. By the universal property of the normalization, this in turn induces an action of $\mathbb{G}_m$ on the normalization, $X^{\text{nor}}_{\text{red}}$. The $k$-scheme $X$ is proper if and only if $X^{\text{nor}}_{\text{red}}$ is proper. Thus, without loss of generality, assume that $X$ is integral and normal.

For every $k$-scheme, $q:T\to \text{Spec}\ k$, and for every $k$-morphism to $X$, $f:T\to X$, as in Geometric Invariant Theory, denote by $$\psi_f:\mathbb{G}_m\times_{\text{Spec}\ k} T \to X\times_{\text{Spec}\ k} T$$ the $T$-morphism whose projection to $X$, $\text{pr}_X\circ f_m$, equals the composition, $$\mathbb{G}_m\times_{\text{Spec}\ k} T \xrightarrow{(\text{Id}_{\mathbb{G}_m},f)} \mathbb{G}_m\times_{\text{Spec}\ k} X \xrightarrow{m} X.$$ Let $\mathbb{G}_m\times_{\text{Spec}\ k} \mathbb{P}^1 \to \mathbb{P}^1$ be the standard extension to $\mathbb{P}^1$ of the regular action of $\mathbb{G}_m$ on the dense open $\mathbb{G}_m = \mathbb{P}^1\setminus\{0,\infty\}$. Then the morphism $\psi_f$ is a $T$-rational transformation, $$\psi_f:\mathbb{P}^1\times_{\text{Spec}\ k} T \dashrightarrow X\times_{\text{Spec}\ k} T.$$

Closed Orbit Hypothesis. For every morphism $f$ with $T = \text{Spec}\ K$ for $K$ a field extension of $k$, the morphism $\psi_f$ extends to a regular morphism $\mathbb{P}^1\times_{\text{Spec}\ k} \text{Spec}\ K \to X\times_{\text{Spec}\ k} \text{Spec}\ K$.

Assume the closed orbit hypothesis. Then when $T$ is Spec of a DVR, say $R$ with residue field $\text{Spec}\ \kappa$ and with fraction field $\text{Spec}\ K$, then the rational transformation $\psi_f$ is regular on the generic fiber $\mathbb{P}^1\times_{\text{Spec}\ k}\text{Spec}\ K$. In particular, restricting this regular morphism to $\{0\}\times_{\text{Spec}\ k} \text{Spec}\ K$ gives a $k$-morphism, $$\psi_{f,0}: \text{Spec}\ K \to X\times_{\text{Spec}\ k} \text{Spec}\ R.$$ Since $0\in \mathbb{P}^1$ is a $\mathbb{G}_m$-invariant point, also $\psi_{f,0}$ maps into the fixed locus $X^{\mathbb{G}_m}$.

Fixed Locus Hypothesis. For every morphism $f$ with $R$ a DVR containing $k$, the associated regular morphism $\psi_{f,0}:\text{Spec}\ K \to X\times_{\text{Spec}\ k} \text{Spec}\ R$ extends to a regular morphism on $\text{Spec}\ R$.

Via the valuative criterion of properness, this is equivalent to properness of the fixed locus $X^{\mathbb{G}_m}$. It is convenient to modify $\psi_f$ a bit. Denote by $\phi_f$ the unique morphism, $$(\text{pr}_{\mathbb{P}^1},\psi_f):\mathbb{P}^1 \times_{\text{Spec}\ k} \text{Spec}\ K \to \mathbb{P}^1\times_{\text{Spec}\ k} X\times_{\text{Spec}\ k} \text{Spec}\ K.$$ This is a locally closed immersion. Denote by $W_f$ the Zariski closure of the image of $\phi_f$. This is a finite type, separated, integral $R$-scheme, but it may be that $W_f$ is not normal. Thus, denote by $W_f^{\text{nor}}\to W_f$ the normalization. By the fixed locus hypothesis, $W_f$ contains the image of the regular morphism, '$$\phi_{f,0}:\text{Spec}\ R \to X\times_{\text{Spec}\ k} \text{Spec}\ R =\{0\} \times X\times_{\text{Spec}\ k} \text{Spec}\ R.$$ Thus the closed fiber of $W_f^{\text{nor}}$ is not empty.

Claim. After a finite sequence of blowings up of $\mathbb{P}^1\times_{\textbf{Spec}\ k}\text{Spec}\ R$ at $\mathbb{G}_m$-fixed closed points, the rational transformation $\phi_f$ extends to a regular $\mathbb{G}_m$-equivariant morphism to $W_f^{\text{nor}}$. In particular, the strict transform of $\{1\}\times \text{Spec}\ R$ gives an extension of $f$ to a regular morphism $\text{Spec}\ R \to X$.

Altogether, the projection, $$\text{pr}_{\mathbb{P}^1,R}:W_f^{\text{nor}} \to \mathbb{P}^1\times_{\text{Spec}\ k} \text{Spec}\ R,$$ is a finite type, separated, $\mathbb{G}_m$-equivariant $R$-morphism that is an isomorphism on $\text{Spec}\ K$-fibers. The remainder of the argument is an analysis of this morphism. However, to prove boundedness, it is convenient to use an ad hoc, auxiliary morphism. Since $W_f^{\text{nor}}\to \text{Spec}\ R$ is a flat, separated, finite type morphism of relative dimension $1$, it is a quasi-projective relative curve. Thus, there exists a dense open $R$-immersion, $\iota_f:W_f^{\text{nor}} \to \overline{W}_f$ where $\overline{W}_f\to \text{Spec}\ R$ is a projective curve. Denote by $\mathcal{L}$ a very ample invertible sheaf on $\overline{W}_f$. Moreover, let $s:R^{\oplus 3} \to H^0(\overline{W}_f,\mathcal{L})$ be global sections that generate $\mathcal{L}$. Denote by $d$ the degree of $\mathcal{L}$ on the fibers of $\overline{W}_f\to \text{Spec}\ R$.

By the valuative criterion of properness, the $R$-rational transformation, $$\phi_f:\mathbb{P}^1\times_{\text{Spec}\ k}\text{Spec}\ R \dashrightarrow \overline{W}_f,$$ extends to a regular morphism away from finitely many points. The pullback of $\mathcal{L}$ by this regular morphism extends to an invertible sheaf on all of $\mathbb{P}^1\times_{\text{Spec}\ k} \text{Spec}\ R$, and the global sections $s$ generate the extension away from the finitely many indeterminacy points. The degree of the extension of $\mathcal{L}$ equals $d$ since $\phi_f$ is an isomorphism of $\text{Spec}\ K$-fibers.

The restriction of $\phi_f$ to the $\text{Spec} \ \kappa$-fiber is an $R$-rational transformation $\mathbb{P}^1\times_{\text{Spec}\ k} \text{Spec}\ \kappa \to \overline{W}_f$. By the valuative criterion, this extends to a regular morphism. The pullback of $\mathcal{L}$ is a globally generated invertible sheaf. Thus the degree $d_0$ of this invertible sheaf satisfies $0\leq d_0\leq d$. The number of blowings up necessary to regularize $\phi_f$ is $\leq d - d_0$.

We will iteratively blow up the $\mathbb{G}_m$-invariant closed points in the indeterminacy locus of $\phi_f$ over $0$. Thus, let $$P_m \xrightarrow{\nu_m} P_{m-1} \xrightarrow{\nu_{m-1}} \dots \xrightarrow{\nu_1} \mathbb{P}^1 \times_{\text{Spec}\ k} \text{Spec}\ R$$ be a sequence of blowings up of a single $\mathbb{G}_m$-fixed point that maps to $\{0\}\times \text{Spec}\ \kappa$ and that is in the indeterminacy locus of $\phi_f$. By Zariski's Main Theorem, each exceptional divisor $E_i$ of $\nu_i$ is mapped quasi-finitely to $\overline{W}_f$. By the valuative criterion of properness, the restriction to $E_i$ of $\phi_f$ extends to a regular morphism, and this is finite to its image. Thus, the pullback of $\mathcal{L}$ on $E_i$ by this regular, finite morphism has degree $\geq 1$. Since the sum of $d_0$ and the degrees of the exceptional divisors is $\leq d$, the number $m$ of such blowings up is $\leq d-d_0$. So for some $m\leq d-d_0$, the associated rational transformation $$\phi_f:P_m \dashrightarrow \overline{W}_f$$ is regular at every $\mathbb{G}_m$-equivariant point lying over $0$.

Denote by $P_m^o\subset P_m$ the maximal open subscheme on which $\phi_f$ is regular and maps $P_m^o$ into the open subset $W_f^{\text{nor}} \subset \overline{W}_f$. This is a $\mathbb{G}_m$-invariant open subset of $P_m$. In particular, $\phi_f$ is regular on the strict transform of $\{0\}\times \text{Spec}\ R$. Since $\phi_{f,0}$ maps this into $W_f^{\text{nor}}$, $P_m^o$ contains the strict transform of $\{0\}\times \text{Spec}\ R$. Thus, $P_m^o$ also intersects the unique $\nu$-exceptional component $E_0$ of $P_m\times_{\text{Spec}\ R} \text{Spec}\ \kappa$ that intersects this strict transform. The complement of $P_m^o$ in this component consists of at most $1$ closed point, and this is $\mathbb{G}_m$-equivariant. By hypothesis, $\phi_f$ is regular at this closed point. And then, by the closed orbit hypothesis, this closed points is contained in $W_f^{\text{nor}}$. Now we can repeat this argument with every $\nu$-exceptional irreducible component $E_{i+1}$ of $P_m\times \text{Spec}\ \kappa$ that intersects an irreducible component $E_i$ on which it has already been proved that $\phi_f$ is regular and maps $E_i$ into $W_f^{\text{nor}}$. In particular, since the strict transform of the original closed fiber $\mathbb{P}^1 \times \text{Spec}\ \kappa$ intersects one of these exceptional components, $\phi_f$ is regular at the unique closed point of the strict transform that maps to $\{0\} \times \text{Spec}\ \kappa$, and this closed point maps into $W_f^{\text{nor}}$. So, by the same argument, the complement of $P_m^o$ in this strict transform consists of at most one point, the $\mathbb{G}_m$-invariant closed point $\infty$, and then $\phi_f$ is not regular at $\infty$.

Now we can repeat the argument above with $\infty$ in place of $0$. Instead of using the strict transform of $\{0\}\times \text{Spec}\ R$, use the strict transform of $\mathbb{P}^1\times \text{Spec}\ \kappa$. Thus, let $$P_n \xrightarrow{\nu_n} P_{n-1} \xrightarrow{\nu_{n-1}} \dots \xrightarrow{\nu_{m+1}} P_m$$ be a sequence of blowings up at $\mathbb{G}_m$-fixed points that map to $\{\infty\}\times \text{Spec}\ \kappa$ and each of which is in the indeterminacy locus of $\phi_f$. By Zariski's Main Theorem, each exceptional divisor $E_i$ of $\nu_i$ is mapped quasi-finitely to $\overline{W}_f$. Thus, the pullback of $\mathcal{L}$ on $E_i$ has degree $\geq 1$. Since the sum of $d_0$ and the degrees of the exceptional divisors is $\leq d$, the number $n$ of such blowings up is $\leq d-d_0$. So for $n\leq d-d_0$, the associated rational transformation $$\phi_f:P_n \dashrightarrow \overline{W}_f$$ is regular at every $\mathbb{G}_m$-equivariant point lying over $\infty$.

By the same argument as above, the open subset $P_n^o\subset P_n$ contains the entire closed fiber $P_n\times_{\text{Spec}\ R} \text{Spec}\ \kappa$. Since $P_n$ is proper over $\text{Spec}\ R$, $P_n^o$ equals all of $P_n$. This proves the claim, and this implies that $X$ is proper.

Comment. For a product of multiplicative groups, e.g., $G\times H \cong \mathbb{G}_m\times \mathbb{G}_m$, the same result holds. The point is that the closed orbit hypothesis for $G\times H$ implies the closed orbit hypothesis for both $G$ and $H$. Thus, for an action of $G\times H$ on $X$, the $G\times H$-closed orbit hypothesis for $X$ implies the $H$-closed orbit hypothesis for the $H$-invariant closed subset $X^G\subset X$. Of course the $H$-fixed locus inside $X^G$ equals the $G\times H$-fixed locus. Thus the $G\times H$-fixed locus hypothesis for $X$ implies the $H$-fixed locus hypothesis for $X^G$. By the argument above, $X^G$ is proper. Therefore $X$ satisfies both the $G$-closed orbit hypothesis and the $G$-fixed locus hypothesis. So by the argument above, $X$ is proper.

Please note, however, that the result does not hold for more general reductive groups. For $G$ a nontrivial semisimple group, for a parabolic subgroup $P$, for the natural action of $G$ on $G/P$, for any separated, finite type $k$-scheme $Y$ that is not proper, the induced action of $G$ on $X=(G/P)\times_{\text{Spec}\ k} Y$ satisfies the $G$-closed orbit hypothesis. Moreover, it vacuously satisfies the $G$-fixed locus hypothesis since the $G$-fixed locus is empty. Yet $X$ is not proper.

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  • $\begingroup$ Thanks so much for the detailed answer. I am still trying to digest it. I am a little bit confused about the last part. Could you explain why $P^0_m$ is $\mathbb G_m$-invariant? Are we assuming that the action on $W^{nor}_f$ extends to $\overline W_f$? $\endgroup$ – YZhou Feb 3 '17 at 8:06
  • $\begingroup$ Also, I hope you don't mind that I corrected some typos for you. Before the "Claim", should $\{\text{Spec }k\}$ be $\text{Spec }R$? $\endgroup$ – YZhou Feb 3 '17 at 8:07
  • $\begingroup$ @YZhou. "Could you explain why $P^o_m$ is $\mathbb{G}_m$-invariant?" The scheme $P_m$ has a natural action of $\mathbb{G}_m$, as an iterated blowing up of $\mathbb{G}_m$-equivariant schemes at $\mathbb{G}_m$-invariant closed points. The scheme $W_f^{\text{nor}}$ has a natural action of $\mathbb{G}_m$. The rational transformation $P_m\dashrightarrow W_f^{\text{nor}}$ is $\mathbb{G}_m$-equivariant. Thus, the maximal open subscheme on which this rational transformation is regular is a $\mathbb{G}_m$-invariant open subscheme. $\endgroup$ – Jason Starr Feb 3 '17 at 11:57
  • $\begingroup$ Great! Now I am happy. Thanks for the nice argument. I guess we only have to verify the Closed Orbit Hypothesis for $K$ an algebraic extension of $k$, am I right? $\endgroup$ – YZhou Feb 3 '17 at 23:22

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