Projective semilinear group, fractional linear mapping and Frobenius automorphisms

finite-fieldsfinite-groups

For a finite field $\mathbb{F}$, the projective general linear group $ \text{PGL}_2(\mathbb{F}) $ is a group of linear transforms $\mathbb{F}\cup \{\infty \}\to \mathbb{F}\cup \{\infty \}$

$$ u\to \frac{\text{$\alpha $u}+\beta }{\text{$\gamma $u}+\delta } $$

Projective semilinear group $\text{P$\Gamma $L}_2(\mathbb{F}) $ is an extension $ \text{PGL}_2(\mathbb{F}) $ by the Frobenius endomorphisms $\rho :\mathbb{F}\to \mathbb{F} $

However I dont know whether the extension contains all of the transformations

$$ u\to \frac{\alpha \rho _1 (u) + \beta }{\gamma \rho _2 (u)+ \delta } $$

or them with $ \rho _1 (u) = \rho _2 (u) $ only.

Or both definitions are equivalent?

Best Answer

A curious question that I'm a bit ashamed I had to think about for a while:-)

I am convinced that you can only include the transformations with $\rho_1=\rho_2$. At least two things will go wrong otherwise. In my examples $F$ stands for the Frobenius automorphism $F(x)=x^p$.

  1. The set of mappings will not be closed under composition. Consider $\sigma(u)=u+1$ and $\tau(u)=(u+1)/F(u)$. Then $$\sigma(\tau(u))=\frac{u+F(u)+1}{F(u)},$$ which is not of the prescribed form.
  2. The transformations won't necessarily be bijections of $\Bbb{F}\cup\{\infty\}$ otherwise. Let $\Bbb{F}=\Bbb{F}_{p^p}$. A standard exercise shows that $x^p-x=1$ has $p$ solutions in $\Bbb{F}$. If $\alpha$ is one zero of that irreducible polynomial the other zeros are $\alpha+j, j=0,1,2,\ldots,p-1$. Implying that the "transformation" $$\delta(u)=\frac{u-1}{F(u)}$$ takes the value $1$ for $p$ distinct choices of $u=\alpha+j$.
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