MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Fix an algebraically closed field $k$. Why is the general curve over $k$ of genus $g \ge 3$ automorphism-free?

I am particularly interested in seeing an argument that does not go by induction and specialization to a singular genus $g$ curve.

Let's say a curve is a smooth, projective, connected $1$-dimensional $k$-scheme.

share|cite|improve this question
Hartshorne has an exercise proposing to show this for curves of genus $3$ by counting, for each $n$, the dimension of the family of curves with an automorphism of order $n$. – Mariano Suárez-Alvarez Feb 10 '11 at 6:52
up vote 12 down vote accepted

One way to do it is through deformation theory, provided we only consider automorphism groups $G$ of order not divisible by the characteristic (one may of course assume that it is cyclic of prime order). Then the the moduli space (or just a miniversal deformation) of all curves of genus $g>1$ is smooth with tangent space at the curve $C$ equal to $H^1(C,T^1_C)$. The tangent space of the sublocus where the action of $G$ extends is equal to $H^1(C,T^1_C)^G$ and hence all curves in a neigbourhood of $C$ has an action of $G$ only when $G$ acts trivially on $H^1(C,T^1_C)$. The (Brauer) character of the action can be computed by the holomorphic Lefschetz trace formula (resp. of a lifting of $(C,G))$ and is seen to be non-trivial. (To this I guess one has to add that there is a finite stratification of the moduli space where the automorphism group is fixed on each stratum.)

Addendum: In principle this method could also handle automorphisms of order equal to the characteristic (say), what needs to be shown is that they act non-trivially on $H^1(C,T^1_C)$. I haven't thought about that though.

As suggested by Mariano one can also count parameters: Assume $G$ is cyclic of prime order. If the order is not equal to the characteristic one can use the Hurwitz formula to get bounds on the genus for the quotient curve and the number of critical values of the quotient map. Counting parameters for the quotient curve and the critical values always gives a value which is smaller than $3g-3$ (not a difficult calculation but also not altogether pleasant).

Addendum: The calculation turns out not to be that difficult. Indeed, if we are dealing with an automorphism of prime order $\ell$, if $C\to C'$ is the quotient map and we have $r$ critical values, then the Hurwitz formula gives $$ 3(g-1) = 3\ell(g'-1) + \frac32(\ell-1)r, $$ where $g'=g(C')$, and on the other hand we want to show (when $g'>1$) that $3(g-1)>3(g'-1)+r$ which follows immediately as $\frac32(\ell-1)>1$. When $g'=1$ we want to show that $3(g-1)>1+r-1=r$ ($1$ parameter for varying the elliptic curve and by automorphisms we may fix one critical value), i.e., $\frac32(\ell-1)r>r$, i.e., $(3\ell-5)r>0$ which is always OK. Finally, with $g'=0$ we are OK if $3(g-1)>\max(r-3,0)$. We may assume $r>3$ and then $r\geq 6$ for reasons of divisibility which easily gives that we are OK unless $r=6$ and $\ell=2$ which gives $g=2$.

The case when the order is equal to the characteristic is even messier, one has to look at the local contribution at a critical value to the genus of $C$ which is of Artin-Schreier type and then bound the number of local parameters in such covers (we here have parameters even when the critical values and the quotient curve stay constant). One could also however use the result of Oort saying that a curve with such an automorphism lifts equivariantly.

share|cite|improve this answer
@Ekedahl: Thanks! – jlk Feb 11 '11 at 5:30
There's one delicate point when the (algebraically closed) ground field is countable. The counting argument only shows that the set of curves of a fixed genus g>2 lies in a countable union of closed subsets of dimension <3g-3. So one also needs to bound the number of automorphisms independently of the curve. This is done using Weierstrass gaps; in characteristic p, the argument, which is a bit tricky, was carried out by H.L. Schmid in the 1930's. – paul Monsky Feb 11 '11 at 14:50
You are perfectly right. Another proof which works in all characteristics is to use that the representation of the automorphism group on points of order dividing $\ell$, a prime different from $p$ and at least $3$, is faithful. – Torsten Ekedahl Feb 11 '11 at 15:24

All you need to do is write down one curve without automorphisms for every genus $g$. For $g$ of the form $(d-1)(d-2)/2$ use smooth plane curves. The automorphisms of such curves are induced by linear automorphisms of the plane, so it's easy to show that the general smooth plane curve has no automorphisms. For the general case, you might try plane curves with few singularities but I am not sure whether this goes through.

A slightly different approach is to look at the general hyperelliptic curve of genus $g$ and show that it only has the hyperelliptic involution as automorphism. Any other automorphism comes from a linear automorphism of the line (quotient of the curve by the hyperelliptic involution). Now you are reduced to exhibit a curve of genus $g$ with no automorphism of order two. Maybe $y^3=f(x)$ will work similar to the hyperelliptic case (as long as $\deg f > 3$, ha!).

share|cite|improve this answer
Why is it enough to exhibit one curve? A semicontinuity argument? – Mariano Suárez-Alvarez Feb 10 '11 at 6:53
@Mariano: I am assuming he meant general in the algebraic geometry sense, i.e., in a non-empty Zariski open set of moduli. But it is not hard to show that curves with automorphisms form a Zariski closed subset of moduli (using the finiteness of the automorphism group), so once you know there is one, you conclude that most don't have automorphisms. – Felipe Voloch Feb 10 '11 at 7:28
@Felipe Voloch: I did mean "general" in the sense of algebraic geometry. – jlk Feb 10 '11 at 7:42
It seems to me that this can be done very painlessly if one constructs a curve without automorphisms instead in the Deligne-Mumford boundary of $M_g$. To find an automorphism-free curve of genus three you could take a general curve of genus two and glue together any two points of it which are not conjugate under the hyperelliptic involution and not Weierstrass points. Then you could just iterate: by gluing together two arbitrary points of an automorphism-free curve of genus g you get an automorphism-free curve of genus g+1. – Dan Petersen Feb 10 '11 at 16:56
Here is a paper of Cornalba which enumerates all subloci in M(g) of curves with automorphisms. – roy smith Feb 11 '11 at 16:27

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.