I have a field $L$ and two field automorphisms $f, g: L \to L$ such that $f \circ g = g \circ f$. Under what conditions can they be extended to automorphisms of an extension $K \mid L$ that still commute? In particular they can certainly be extended to automorphisms of the algebraic clousure $K = \overline L$, but can they be extended to a pair of commuting automorphisms?



Extending commuting automorphisms of a field
**Reformulation in terms of group cohomology of Galois groups.** Assume that $f$ and $g$ both have finite order dividing the positive integer $n.$ The the fixed subfield $E=L^{f,g}$ of $f$ and $g$ gives a finite, Galois extension $L/E.$ The two commuting elements $f$ and $g$ define a surjective group homomorphism, $$\rho\_{f,g,n}:\mathbb{Z}/n\mathbb{Z}\times \mathbb{Z}/n\mathbb{Z} \twoheadrightarrow \text{Gal}(L/E).$$ 


For every finite, separable extension $K/L$, since also $K/E$ is finite and separable, there exists a minimal Galois extension $K'/E$ that contains $K/E$ as a subextension. By the Fundamental Theorem of Galois Theory, $K'/L$ is a Galois extension, the subgroup $\text{Gal}(K'/K)$ is a normal subgroup, $\text{Gal}(K'/L)\trianglelefteq \text{Gal}(K'/E),$ and the quotient by this subgroup is canonically isomorphic to $\text{Gal}(L/E).$ To summarize, there is a normal extension of groups, $$\Sigma\_{K'/L/E}: \ \ \ 1 \to \text{Gal}(K'/L)\to \text{Gal}(K'/E) \to \text{Gal}(L/E) \to 1.$$ In particular, there are lifts $\widetilde{f},\widetilde{g}\in \text{Gal}(K'/E)$ mapping to $f,g\in \text{Gal}(L/E).$ 


Up to increasing $n$, assume that the order of $\text{Gal}(K'/E)$ divides $n$. Then the pullback of $\Sigma\_{K'/L/E}$ by $\rho\_{f,g,n}$ is a normal extension of groups, $$\Sigma\_{K'/L,f,g,n}: \ \ \ \ 1 \to \text{Gal}(K'/L) \to G \to \mathbb{Z}/n\mathbb{Z}\times \mathbb{Z}/n\mathbb{Z} \to 0.$$ 


**Proposition.** There exist lifts $\widetilde{f}$ and $\widetilde{g}$ of $f$ and $g$ in $\text{Gal}(K'/E)$ that commute if and only if the normal extension $\Sigma\_{K'/L,f,g,n}$ is a semidirect product, i.e., if and only if the extension is split. If $\text{Gal}(K'/L)$ is Abelian, the extension $\Sigma\_{K'/L,f,g,n}$ is an element in the group cohomology $H^2(\mathbb{Z}/n\mathbb{Z}\times \mathbb{Z}/n\mathbb{Z},\text{Gal}(K'/L)).$


**Proof.** If there are lifts that commute, then the subgroup generated by $\widetilde{f}$ and $\widetilde{g}$ gives a section of $\Sigma\_{K'/L,f,g,n}.$ Conversely, if there is a section, then the images in $\text{Gal}(K'/E)$ of the standard generators of $\mathbb{Z}/n\mathbb{Z}\times \mathbb{Z}/n\mathbb{Z}$ commute. **QED**


**The "elementary obstruction" to existence of simultaneous fixed points.** The extension $\Sigma\_{K'/L,f,g,n}$ need not **always** split. The simplest such example is any Galois extension $K'/E$ with Galois group equal to the order $8$ quaternion group $Q$. Take $L$ to be the fixed field of the center of $Q$. The corresponding normal extension of groups is not a semidirect product. 


There are also many examples arising in algebraic geometry from the problem of trying to find simultaneous fixed points of two commuting automorphisms of a $k$-scheme. 


**Setup.** Let $k$ be a field of characteristic prime to $n$ that contains a primitive $n^{\text{th}}$ root of unity, $\zeta.$ Let $P$ be a projective, integerl $k$-scheme $P$ whose Picard group is cyclic generated by an ample invertible sheaf $\mathcal{O}(1).$ Let $f,g:P\to P$ be $k$-automorphisms of finite order dividing $n$ that commute with each other.


**Problem.** Does there exists a $k$-point of $P$ that is simultaneously fixed by $f$ and by $g?$


**N.B.** For either $f$ or $g$ individually, there are many fixed point theorems that insure the existence of a fixed point. For instance, if $k$ is algebraically closed and $P$ is smooth and separably rationally connected, then there exists a fixed point of $f$, resp. there exists a fixed point of $g$. However, there may exist no **simultaneous** fixed point, and one obstruction arises from failure of commutativity of lifts of $f$ and $g$. 


**Heisenberg group examples.** Let $P$ be $\mathbb{P}^{n-1}\_k = \text{Proj}\ k[t\_0,\dots,t\_{n-1}].$ Consider the following $k$-automorphisms, $$f:\mathbb{P}^{n-1}\_k \to \mathbb{P}^{n-1}\_k, \ \ [t\_0,t\_1,\dots,t\_{n-2},t\_{n-1}] = [t\_{1},t\_2,\dots,t\_{n-1},t\_0],$$
$$g:\mathbb{P}^{n-1}\_k \to \mathbb{P}^{n-1}\_k, \ \ 
[t\_0,t\_1,\dots,t\_{n-1}] \mapsto [\zeta^0 t\_0, \zeta^1t\_1,\dots,\zeta^{n-1}t\_{n-1}].$$ The fixed points of $g$ are the points $p\_i = [0,\dots,0,1,0,\dots,0],$ $i=0,\dots,n-1,$ with precisely one nonzero coordinate. The fixed points of $f$ are the points $[\zeta^{0\cdot a},\zeta^{1\cdot a}, \dots, \zeta^{(n-1)\cdot a}]$ for $\overline{a}\in \mathbb{Z}/n\mathbb{Z}.$ There is no simultaneous fixed point of $f$ and $g$.


**Cyclic Galois extensions.**
For each integer $e$, the automorphisms $f$ and $g$ "linearize" to a $k$-linear action of $\mathbb{Z}/n\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z}$ on the $k$-vector space $$V\_e=H^0(P,\mathcal{O}(n\cdot e)).$$ Since the characteristic is prime to $n,$ every such $k$-representation is a direct sum of $1$-dimensional $k$-subrepresentations. Let $L=\text{span}(\sigma)$ be a $1$-dimensional $k$-subrepresentation. Let $q:Y\_\sigma\to P$ be the cyclic $n$-sheeted cover on which there is a universal section $s$ of $f^\*\mathcal{O}(e)$ with $s^n$ equal to the pullback of $\sigma$. Over the dense Zariski open $D(\sigma)\subset P$, the morphism $q$ is a cyclic Galois cover with Galois group $\mu\_n(k)$ acting by scaling $s$. Thus, every lift $\widetilde{f},$ resp. $\widetilde{g},$ of $f,$ resp. $g,$ is unique up to scaling by an element $\phi,$ resp. $\gamma,$ in $\mu\_n(k).$


**Definition.** The **elementary obstruction** to the existence of a simultaneous fixed point is the commutator $o\_{f,g,\sigma}=\widetilde{f}\circ \widetilde{g}\circ \widetilde{f}^{-1}\circ \widetilde{g}^{-1}$ in $\mu\_n(k).$ Since $\phi\cdot \gamma\cdot \phi^{-1}\cdot \gamma^{-1}$ equals $1$ in the Abelian group $\mu\_n(k)$, the elementary obstruction is independent of the choice of lifts.


**Lemma.** If $P$ is smooth and if there exists a $k$-point of $P$ that is simultaneously fixed by $g$ and by $g$, then the elementary obstruction equals the identity element $1$ in $\mu\_n(k).$


**Proof.** By Lang-Nishimura, existence of a fixed points is a birational invariant. Thus, up to blowing up and taking strict transforms, assume that there exists a simultaneous fixed point $x$ that is contained in the open $D(\sigma)$. The fiber of $q$ over $x$ is a torsor for $\mu\_n(k)$. Thus, the lifts $\widetilde{f}$ and $\widetilde{g}$ restrict on this fiber to scaling by elements of $\mu\_n(k)$. Up to scaling by appropriate elements $\phi$ and $\gamma$, assume that these are the identity. Then the commutator equals $1$ on this fiber. Since the commutator is scaling by a constant element $\mu\_n(k),$ it follows that the commutator equals $1$. **QED**


As above, this obstruction is naturally an element in the group cohomology $H^2(\mathbb{Z}/n\mathbb{Z}\times \mathbb{Z}/n\mathbb{Z},\mu\_n(k)).$ In fact, this obstruction depends only on the integer $e$, not on the nonzero element $\sigma,$ i.e., $o\_{f,g,\sigma} = o\_{f,g,e}.$ Moreover, $o\_{f,g,e}$ equals $(o\_{f,g})^e$, where $o\_{f,g}$ equals $o\_{f,g,1}.$


**Computation of the elementary obstruction in the Heisenberg examples.**
For $P=\mathbb{P}^{n-1}\_k,$ for the invariant section $\sigma = t\_0^n + \dots + t\_{n-1}^n,$ over each basic open $D\_+(t\_i)$, the inverse image of $Y\_\sigma$ is equal to the smooth, integral, affine scheme, $$\text{Spec}\ k[s,(t\_j/t\_i)]\_{j\neq i} /\langle s^n - (t\_0^n+\dots+t\_{n-1}^n)/t\_i^n \rangle.$$ The automorphism $f$ lifts to an automorphism, $$(s,(t\_j/t\_i))\_{j\neq i} \mapsto (s(t\_{i+1}/t\_i)^{-1}, (t\_{j+1}/t\_i) (t\_{i+1}/t\_i)^{-1})\_{j\neq i},$$ and this is unique up to scaling $s$ by a scalar $\phi.$ The automorphism $g$ lifts to an automorphism, $$(s,(t\_j/t\_i))\mapsto (s,\zeta^{j-i}(t\_j/t\_i)),$$ and this is unique up to scaling $s$ by a scalar $\gamma.$ For the lifts just given, the commutator $o\_{f,g,\sigma} = \widetilde{f}\circ \widetilde{g}\circ \widetilde{f}^{-1}\circ \widetilde{g}^{-1}$ equals $\zeta^{-1}.$ Thus, this nontrivial obstruction "explains" the failure of simultaneous fixed points of $f$ and $g.$


Having said this much, I feel compelled to explain one family of **positive** examples.


**Theorem**[A. J. de Jong-Xuhua He-S.] Let $k$ be algebraically closed, and let $P$ be a smooth, projective $k$-scheme that is a specialization of a "rationally simply connected" variety of Picard rank $1,$ e.g., a projective homogeneous space of Picard rank $1$ (like a Grassmannian) or a complete intersection in projective space with both $\text{c}\_1(T\_P)$ and $\text{c}\_1(T\_P)^2 - 2c\_2(T\_P)$ positive ($2$-Fano complete intersection). If the elementary obstruction $o\_{f,g}$ vanishes, then there exists a simultaneous fixed point.


