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Let $X$ denote $\mathbb{P}^1$ and let $\textbf{Aut}(X)$ denote the functor sending an affine scheme $A$ over $\mathbb{C}$ to the group $Aut_A(X \times A)$ of automorphisms of $X \times A$ over $A$.

We know that $\textbf{Aut}(X)$ is representable by the Lie group $PGL(2)$.

Let us now consider $\textbf{Aut}(X)(D)$, where $D=\mathbb{C}[\epsilon]/(\epsilon)^2$. We have that $\textbf{Aut}(X)(D)= Aut_D(X \times D)=PGL(2)(D)$.

On the other hand, we know that the set of automorphisms of $Aut_D(X \times D)$ which restrict to the identity morphism on the fiber are given by $Lie(Aut(X))=Lie(PGL(2))$. These automorphisms locally look like $z \mapsto z + \epsilon(a_0 + a_1 z + a_2 z^2)$. (Somehow this corresponds to $sl_{2}$ but I don't know how exactly.)

But, these automorphism should also correspond to elements in $PGL(2)(D)$? Which is matrix with entries from $D$. I don't see how this is possible given the local form I have written down.

How do automorphisms of the type $z \mapsto z + \epsilon(a_0 + a_1 z + a_2 z^2)$ corresponds to $PGL(2)(D)$?

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1 Answer 1

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From the identification, $$ \frac{(1+a\epsilon)z+b\epsilon}{c\epsilon z + (1+d\epsilon)}=\big((1+a\epsilon)z+b\epsilon\big)\big(1-d\epsilon-c\epsilon z\big)=z+\epsilon\big( b+(a-d)z-cz^2 \big), $$ we have that, $$ z\mapsto \frac{(1+a\epsilon)z+b\epsilon}{c\epsilon z + (1+d\epsilon)}, $$

which corresponds to, $$ u \mapsto (1 + a \epsilon)v + b \epsilon u$$ $$ v \mapsto (1 + d \epsilon) v + d \epsilon u$$

which is clearly corresponds to an element in $PGL(2)(D)$.

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