Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $X$ be a nice algebraic variety (say smooth, projective) over a field of characteristic 0. Let $G$ be an abelian group acting on $X$. For each subgroup $H$ of $G$, denote by $X^H$ the closed subvariety of points fixed by $H$.

Proposition. $X^H$ is smooth

Of course, the intersection between different $X^H$ could be non-empty. That's why people look at the "inertia stratification"

$X_H=X^H- \bigcup_{H' \subsetneq H} X^{H'}$

My question is: assuming the above proposition, why is each $X_H$ smooth?

Thanks for you help

share|improve this question
1  
$X_H$ is an open subset of $X^H$. I voted to close as "too localized". –  Angelo Apr 7 '13 at 17:05
    
@Angelo, so the argument is just that an open of a smooth is smooth? –  inert89 Apr 7 '13 at 17:11

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.