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Let $G$ a group split over a local field $F$. We call a $z$-extension a group $G'$ such that $G'_{der}$ is simply connected, $G'$ is a central extension of $G$ by a central torus $Z$.

Can we find a $z$-extension of $G$, $G'$ such that the center of $G'$ is connected?

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1 Answer 1

No, if the center $Z(G)$ is not connected, we cannot construct $G'$ with connected $Z(G')$.

Indeed, let $\pi\colon G'\to G$ be an epimorphism of connected reductive $F$-groups with central kernel. Choose a maximal torus $T\subset G$. We have $$ Z(G)=\ker[T\to {\rm Aut}\ {\rm Lie}(G)]=\ker[T\to {\rm Aut}\ {\rm Lie}( G_{der})]. $$ Set $T'=\pi^{-1}(T)\subset G'$, then $$ Z(G')=\ker[T'\to {\rm Aut}\ {\rm Lie}( G'_{der})]. $$ It follows easily that $\pi(Z(G'))=Z(G)$.

Now, if $Z(G')$ is connected, then $Z(G)=\pi(Z(G'))$ must be connected. Therefore, if $Z(G)$ is not connected, then we cannot construct $G'$ with connected center $Z(G')$.

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The way that Lie algebras are handled in this argument seems to implicitly assume that the ground field is of characteristic 0...not that there's anything wrong with that. :) –  user30379 Feb 15 '13 at 2:58
    
One can formulate this argument so it works in any characteristic. Namely, if $H$ is a connected reductive group over a field $k$ with schematic center $Z_H$ and if $Z \subset Z_H$ is a closed $k$-subgroup scheme then we claim $H/Z$ has schematic center $Z_H/Z$ (so if $Z$ is connected then $H$ has connected center if and only if $H/Z$ does). This amounts to the $k$-group $(H/Z)/(Z_H/Z) = H/Z_H$ having trivial schematic center. The equality $T = Z_H(T)$ forces $Z_H \subset T$, and likewise $T/Z_H$ contains the schematic center of $H/Z_H$. But $\Phi(H/Z_H,T/Z_H)$ spans $X^{\ast}(T/Z_H)$. QED –  user30379 Feb 15 '13 at 3:08

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