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This question is about a result in the section 16.3 of the book Linear Algebraic Groups from Humphreys.

The follow can be deduced from a proposition in the section 16.3 of the book:

Corollary: Let $G$ be a diagonalizable algebraic group, $V$ a connected variety and $\varphi:V\times G\rightarrow G$ a morphism of varieties. Assume also the map $x\mapsto \varphi_x$, given by $\varphi_x(y)=\varphi(x,y)$, is a group homomorphism for each $x\in V$. Them the map $x\mapsto \varphi_x$ is constant.

The book says its implies a diagonalizable group is 'rigid', i.e., there are few automorphisms. My question is if the following application is correct.

Example: Let $G$ be a diagonalizable algebraic group and $\varphi:\mathbb P^1\times G\rightarrow G$ a morphism of varieties with $\varphi_x\in Aut(G)$ (automorphisms of algebraic groups) for each $x\in \mathbb P^1$. Them by the corollary $x\mapsto \varphi_x$ is constant and there is only one automorphism indeed. This shows not exist a 'one parameter family' of automorphisms of $G$.

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migrated from mathoverflow.net Dec 14 '13 at 15:35

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  • $\begingroup$ You corollary is obviously false (take $V=G$ and $\varphi$ the group law), do you forget to assume something more about $\varphi$? $\endgroup$ – YCor Dec 13 '13 at 14:17
  • $\begingroup$ @Yves Cornulier: I forget a important hypothesis: the maps $\varphi_x$ are group homomorphisms. Thank you. $\endgroup$ – Rick Rischter Dec 13 '13 at 21:48
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Yes, your application is correct: you just specify the corollary to $V=\mathbb{P}^1$.

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