# Does every automorphism of a separably rationally connected variety have a fixed point?

Let $k$ be an algebraically closed field. Let $X$ be a smooth, projective variety over $k$ that is separably rationally connected, i.e., there exists a $k$-morphism $u:\mathbb{P}^1_k \to X$ such that $u^*T_X$ is isomorphic to $\mathcal{O}_{\mathbb{P}^1_k}(a_1)\oplus \dots \oplus \mathcal{O}_{\mathbb{P}^1_k}(a_n)$ for positive integers $a_1,\dots,a_n$. Let $f:X\to X$ be a $k$-automorphism.

Question. Does there exist a $k$-point of $X$ that is fixed by $f$?

This is true if $k$ is of characteristic $0$ by the Atiyah-Bott fixed point theorem. That theorem was extended to positive characteristic by many authors; one reference is Corollaire 6.12 (Appendice), Exposé III, SGA 5. This extension does not quite give the result above; it does give the result if $h^i(X,\mathcal{O}_X)$ vanishes for all $i>0$, but that is unknown for arbitrary separably rationally connected, smooth, projective varieties.

By a theorem of Kollár, this is also true in positive characteristic if $f$ has finite order. By the "spreading out" technique, this implies the result if there exists an ample invertible sheaf $\mathcal{L}$ such that $f^*\mathcal{L}$ is isomorphic to $\mathcal{L}$, i.e., if for some $n>0$, the iterate $f^n$ is in the identity component of the automorphism group scheme of $X$. In particular, the result is true if $X$ is separably rationally connected and Fano. However, there do exist pairs $(X,f)$ with $X$ a separably rationally connected variety and $f$ an automorphism that preserves no ample divisor class, e.g., translations on the elliptic surface obtained from $\mathbb{P}^2$ by blowing up the base locus of a pencil of plane cubics.

This question is related to the following questions: let $p$ be a prime integer, let $q$ be $p^r$, and let $Y/\mathbb{F}_q$ be a smooth, projective variety whose base change to $\overline{\mathbb{F}}_q$ is separably rationally connected. Let $g:Y \to Y$ be an $\mathbb{F}_q$-automorphism. What can we say about the induced permutation of the finite set $Y(\mathbb{F}_q)$ (whose cardinality is congruent to $1$ modulo $q$, by work of Esnault)? Does there exist an integer $N$ depending only on $p$ (not $q$) and geometric properties of $Y$ and $g$ such that there exists an orbit of size $\leq N$?

This is not a full answer, but a partial result. Recall that a rationally chain connected variety has rational decomposition of the diagonal: there exists $N \in \mathbb Z_{> 0}$ and a cycle $Z$ supported on $X \times D$ such that $$N \cdot \Delta_X = N \cdot (\{x\} \times X) + Z \in \operatorname{CH}^n(X \times X).$$ This always forces $H^0(X,\Omega_X^i) = 0$ for $i > 0$ (in characteristic $p > 0$ we need $(N,p) = 1$, cf. Totaro [Tot], Lemma 2.2). (But for separably rationally connected, we always have $H^0(X,\Omega_X^i) = 0$ for $i > 0$, even in positive characteristic.)

In characteristic $0$, this forces $H^i(X,\mathcal O_X) = 0$ for $i > 0$, by Hodge symmetry. This would imply the claim, as you note, using a Woods Hole trace formula.

In analogy with Totaro's theorem, we have the following:

Theorem. Suppose $(N,p) = 1$, i.e. $X$ has decomposition of the diagonal in $\operatorname{CH} \otimes \mathbb Z/p\mathbb Z$. Then $H^i(X,\mathcal O_X) = 0$ for all $i > 0$.

Proof. Working in $\operatorname{CH} \otimes \mathbb Z/p\mathbb Z$, we may assume $N = 1$ after multiplying by its inverse. We use the cycle class map to Hodge cohomology, cf. Chatzistamatiou–Rülling [CR]. By Proposition 3.2.2 (1) of [loc cit], the map $$\bigoplus_{i,j} H^i(X,\Omega_X^j) \to \bigoplus_{i,j} H^i(X,\Omega_X^j)$$ induced by $Z$ vanishes on the part where $j = 0$. Thus, on $H^i(X,\mathcal O_X)$, the map induced by $\{x\} \times X$ is the identity. But the map induced by $\{x\} \times X$ factors through $H^i(\{x\}, \mathcal O_{\{x\}})$, which is $0$ for $i > 0$. $\square$

Thus, any rationally connected variety with $H^i(X, \mathcal O_X) \neq 0$ needs to have $N$ (as above) divisible by $p$. We do not know if this is possible when $X$ is separably rationally connected (which is what you're interested in), but it certainly is possible for merely rationally connected:

Example. Let $X$ be a supersingular K3 surface. Then $X$ is unirational (at least if $p \geq 5$), by Liedtke [Lie]. However, since $X$ is a K3 surface, we have $h^2(X,\mathcal O_X) = h^0(X,\Omega_X^2) = 1$. But of course $X$ cannot be separably rationally connected.

To conclude, the difference between rational, mod $p$, and integral decomposition of the diagonal, as well as between rationally (chain) connected and separably rationally connected, remains very subtle. But maybe this gives some ideas for how to construct counterexamples: we need some $p$ stuff happening.

References.

[CR] Andre Chatzistamatiou and Kay Rülling. Higher direct images of the structure sheaf in positive characteristic. Algebra & Number Theory 5 (2011), no. 6, 693–775. MR2923726

[Lie] Christian Liedtke. Supersingular K3 surfaces are unirational. Invent. Math. 200 (2015), no. 3, 979--1014. MR3348142

[Tot] Burt Totaro. Hypersurfaces that are not stably rational. J. Amer. Math. Soc. 29 (2016), no. 3, 883–891. MR3486175.

• That is interesting. My own search for counterexamples focuses on separably rationally connected varieties arising together with an automorphism fixing no ample divisor. Nov 12, 2016 at 10:20