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Tyler Lawson
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EDIT: This answer is incorrect, for the reason indicated in the other answer; it should be consulted for a correct curve.


Here is a recipe for constructing some examples.

Suppose that $f$ is a polynomial of degree 5 over $\Bbb C$ without repeated roots. Suppose that $f$ has the following two properties:

  • The only linear fractional transformations in $x$ fixing the union of the zeros of $f(x)$ with $\infty$ is the identity.

  • We have $-\overline{f(-\overline x)} = f(x)$.

(For example, you could take $f(x) = (x^2-1)(x^2-2)(x-i)$.)

Consider the resulting hyperelliptic curve $C$ of genus two of the form $y^2 = f(x)$. The projection $C \to \Bbb P^1$ determined by the canonical bundle on $C$ is projection onto the $x$-coordinate, and the first condition we've imposed on $f$ guarantees that any automorphism of $C$ must fix the $x$-coordinate (as it must fix the six ramification points). As a result, the only possible such automorphisms are the identity and $y \mapsto -y$.

Thus we have a hyperelliptic curve whose automorphism group is exactly $\Bbb Z/2$.

Now we also find that the second condition on $f$ means that there is an isomorphism of $C$ with its complex conjugate $\overline{C}$, given by $y \mapsto iy$, $x \mapsto -x$. This means that the field of moduli has to be strictly smaller than $\Bbb C$, and so it must be $\Bbb R$.

However, this curve cannot be defined over $\Bbb R$. The obstruction, as is typical when the automorphism group is abelian, shows up as an element in $H^2(Gal(\Bbb C/\Bbb R), Aut(C))$. In terms of group cohomology, this is determined by an extension $$ 1 \to Aut(C) \to G \to Gal(\Bbb C/\Bbb R) \to 1. $$ Here $G$ is the collection of automorphisms of $C$ which are either $\Bbb C$-linear or conjugate-linear.

In this case, both groups on the outside are $\Bbb Z/2$, and so we want to show that the extension is $\Bbb Z/4$. To calculate the extension, we take the nontrivial automorphism of $\Bbb C$, lift it to an isomorphism $g: C \to C$ which is conjugate-linear (which we already calculated above), and then form the composite $g^2: C \to C$. This automorphism This automorphism is the hyperelliptic involution, and so the extension is nontrivial. (Here is where the hyperelliptic involutionmistake was, and so the extension is nontrivialI was not careful enough.)

I believe that the same method works over $\Bbb Q(i)$, but you need a sturdier Galois cohomology computation.

Here is a recipe for constructing some examples.

Suppose that $f$ is a polynomial of degree 5 over $\Bbb C$ without repeated roots. Suppose that $f$ has the following two properties:

  • The only linear fractional transformations in $x$ fixing the union of the zeros of $f(x)$ with $\infty$ is the identity.

  • We have $-\overline{f(-\overline x)} = f(x)$.

(For example, you could take $f(x) = (x^2-1)(x^2-2)(x-i)$.)

Consider the resulting hyperelliptic curve $C$ of genus two of the form $y^2 = f(x)$. The projection $C \to \Bbb P^1$ determined by the canonical bundle on $C$ is projection onto the $x$-coordinate, and the first condition we've imposed on $f$ guarantees that any automorphism of $C$ must fix the $x$-coordinate (as it must fix the six ramification points). As a result, the only possible such automorphisms are the identity and $y \mapsto -y$.

Thus we have a hyperelliptic curve whose automorphism group is exactly $\Bbb Z/2$.

Now we also find that the second condition on $f$ means that there is an isomorphism of $C$ with its complex conjugate $\overline{C}$, given by $y \mapsto iy$, $x \mapsto -x$. This means that the field of moduli has to be strictly smaller than $\Bbb C$, and so it must be $\Bbb R$.

However, this curve cannot be defined over $\Bbb R$. The obstruction, as is typical when the automorphism group is abelian, shows up as an element in $H^2(Gal(\Bbb C/\Bbb R), Aut(C))$. In terms of group cohomology, this is determined by an extension $$ 1 \to Aut(C) \to G \to Gal(\Bbb C/\Bbb R) \to 1. $$ Here $G$ is the collection of automorphisms of $C$ which are either $\Bbb C$-linear or conjugate-linear.

In this case, both groups on the outside are $\Bbb Z/2$, and so we want to show that the extension is $\Bbb Z/4$. To calculate the extension, we take the nontrivial automorphism of $\Bbb C$, lift it to an isomorphism $g: C \to C$ which is conjugate-linear (which we already calculated above), and then form the composite $g^2: C \to C$. This automorphism is the hyperelliptic involution, and so the extension is nontrivial.

I believe that the same method works over $\Bbb Q(i)$, but you need a sturdier Galois cohomology computation.

EDIT: This answer is incorrect, for the reason indicated in the other answer; it should be consulted for a correct curve.


Here is a recipe for constructing some examples.

Suppose that $f$ is a polynomial of degree 5 over $\Bbb C$ without repeated roots. Suppose that $f$ has the following two properties:

  • The only linear fractional transformations in $x$ fixing the union of the zeros of $f(x)$ with $\infty$ is the identity.

  • We have $-\overline{f(-\overline x)} = f(x)$.

(For example, you could take $f(x) = (x^2-1)(x^2-2)(x-i)$.)

Consider the resulting hyperelliptic curve $C$ of genus two of the form $y^2 = f(x)$. The projection $C \to \Bbb P^1$ determined by the canonical bundle on $C$ is projection onto the $x$-coordinate, and the first condition we've imposed on $f$ guarantees that any automorphism of $C$ must fix the $x$-coordinate (as it must fix the six ramification points). As a result, the only possible such automorphisms are the identity and $y \mapsto -y$.

Thus we have a hyperelliptic curve whose automorphism group is exactly $\Bbb Z/2$.

Now we also find that the second condition on $f$ means that there is an isomorphism of $C$ with its complex conjugate $\overline{C}$, given by $y \mapsto iy$, $x \mapsto -x$. This means that the field of moduli has to be strictly smaller than $\Bbb C$, and so it must be $\Bbb R$.

However, this curve cannot be defined over $\Bbb R$. The obstruction, as is typical when the automorphism group is abelian, shows up as an element in $H^2(Gal(\Bbb C/\Bbb R), Aut(C))$. In terms of group cohomology, this is determined by an extension $$ 1 \to Aut(C) \to G \to Gal(\Bbb C/\Bbb R) \to 1. $$ Here $G$ is the collection of automorphisms of $C$ which are either $\Bbb C$-linear or conjugate-linear.

In this case, both groups on the outside are $\Bbb Z/2$, and so we want to show that the extension is $\Bbb Z/4$. To calculate the extension, we take the nontrivial automorphism of $\Bbb C$, lift it to an isomorphism $g: C \to C$ which is conjugate-linear (which we already calculated above), and then form the composite $g^2: C \to C$. This automorphism is the hyperelliptic involution, and so the extension is nontrivial. (Here is where the mistake was, I was not careful enough.)

I believe that the same method works over $\Bbb Q(i)$, but you need a sturdier Galois cohomology computation.

Source Link
Tyler Lawson
  • 52.6k
  • 9
  • 187
  • 251

Here is a recipe for constructing some examples.

Suppose that $f$ is a polynomial of degree 5 over $\Bbb C$ without repeated roots. Suppose that $f$ has the following two properties:

  • The only linear fractional transformations in $x$ fixing the union of the zeros of $f(x)$ with $\infty$ is the identity.

  • We have $-\overline{f(-\overline x)} = f(x)$.

(For example, you could take $f(x) = (x^2-1)(x^2-2)(x-i)$.)

Consider the resulting hyperelliptic curve $C$ of genus two of the form $y^2 = f(x)$. The projection $C \to \Bbb P^1$ determined by the canonical bundle on $C$ is projection onto the $x$-coordinate, and the first condition we've imposed on $f$ guarantees that any automorphism of $C$ must fix the $x$-coordinate (as it must fix the six ramification points). As a result, the only possible such automorphisms are the identity and $y \mapsto -y$.

Thus we have a hyperelliptic curve whose automorphism group is exactly $\Bbb Z/2$.

Now we also find that the second condition on $f$ means that there is an isomorphism of $C$ with its complex conjugate $\overline{C}$, given by $y \mapsto iy$, $x \mapsto -x$. This means that the field of moduli has to be strictly smaller than $\Bbb C$, and so it must be $\Bbb R$.

However, this curve cannot be defined over $\Bbb R$. The obstruction, as is typical when the automorphism group is abelian, shows up as an element in $H^2(Gal(\Bbb C/\Bbb R), Aut(C))$. In terms of group cohomology, this is determined by an extension $$ 1 \to Aut(C) \to G \to Gal(\Bbb C/\Bbb R) \to 1. $$ Here $G$ is the collection of automorphisms of $C$ which are either $\Bbb C$-linear or conjugate-linear.

In this case, both groups on the outside are $\Bbb Z/2$, and so we want to show that the extension is $\Bbb Z/4$. To calculate the extension, we take the nontrivial automorphism of $\Bbb C$, lift it to an isomorphism $g: C \to C$ which is conjugate-linear (which we already calculated above), and then form the composite $g^2: C \to C$. This automorphism is the hyperelliptic involution, and so the extension is nontrivial.

I believe that the same method works over $\Bbb Q(i)$, but you need a sturdier Galois cohomology computation.