# Variety with nonalgebraic automorphism group

In Algebraic Geometry - A First Course by Joe Harris, an example in section 10 on algebraic groups reads that there exist varieties whose automorphism group is not an algebraic group.

To give an example the author proposes the following. Take two general cubic curves in $\mathbb{P}^2$ and consider the blow-up $X$ at the nine points of intersection. Then $X$ is a family of elliptic curves with eight sections, therefore $Aut(X)\cong \mathbb{Z}^8$.

Can someone explain me this last sentence - at least intuitively - because I can’t make sense of it. I’m familiar with the material in chapter 1 of Hartshorne.

As a side question: is there a nice criterion for a variety to have an algebraic automorphism group? Most basic examples like segre/veronese varieties or grassmannians do.

• By the way - note that affine counterexamples are not hard to come up with: $\mathbb{A}^2$ already works. I believe the author restricts to (quasi-)projective examples.
– ArtW
May 14, 2018 at 20:11

An intuitive idea goes as follows: if $C_1: g(x:y:z)=0$ and $C_2: h(x:y:z)=0$ are two of the cubics passing through the 9 given points, then for all scalars $\alpha,\beta$, the cubic of equation $\alpha.g(x:y:z) + \beta.h(x:y:z) = 0$ also goes through the 9 points. In fact, one can prove that all the cubics through these 9 points have such an equation and, thus, each of these cubics correspond to a point $[\alpha : \beta]\in\Bbb P^1$.
One can prove that these cubics cover all the projective plane so this should give a map $\Bbb P^2\to \Bbb P^1$ which maps a point to the $[\alpha:\beta]$ corresponding to the cubic containing this point. But, what value of $[\alpha:\beta]$ can we choose if this point is one of the intersection points of the 9 cubics (and hence of all our cubics)? We can't know! That's why we obtain only a rational map $\Bbb P^2\dashrightarrow\Bbb P^1$.
Now, blowing up these 9 points removes indeterminacies of this map, since in the surface $S$ obtained with this blow-up, each of these points is replaced by a $\Bbb P^1$. Pulling back all our cubics of $\Bbb P^2$ on $S$ gives us a family of elliptic curves which don't intersect anymore and we obtain an elliptic fibration $S\to \Bbb P^1$, i.e. a family of elliptic curves indexed by $\Bbb P^1$ with some singular fibers (like nodal or cuspidal cubics). But the smooth fibers are elliptic curves, thus one can obtain translations on them because of their group law.
The $\Bbb Z^8$ comes from this: the 8 sections can give translations on each smooth fiber (simply between the points of intersection between these sections and the fiber) and if our 9 points aren't in special positions, then these translations (which only give birational self-maps of $S$ in general, because translations aren't defined on singular fibers) can be extended on the whole surface $S$ to give automorphisms of $S$ which are "sufficiently independent" to give a subgroup of $\text{Aut }S$ isomorphic to $\Bbb Z^8$.