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

Could anybody help me with the following question ?

Assume we are given:

(1) a finite order (linear) automorphism $g$ of the complex projective space $\mathbb{P}^r$,

(2) a closed algebraic subvariety $Z \subset \mathbb{P}^r$ of codimension at least 2.

The problem is to find lines in $\mathbb{P}^r$ which are stable under $g$ and do not meet $Z$. This is not always possible. However, one can embed $\mathbb{P}^r$ in a bigger dimension projective space $\mathbb{P}^N$ by the Veronese embedding of degree $d$ and extend $g$ to a finite order automorphism of $\mathbb{P}^N$. Is it true, for sufficiently large $d$, that there exists a $g$-invariant line in $\mathbb{P}^N$ which does not meet the image of $Z$?

Thanks in advance!

share|improve this question
Yes, this is true. Since the order of $g$ is fixed, if $N$ is large there will be linear subspaces $L$ of large dimension preserved pointwise by $d$. Consider any line in such a subspace not meeting the image of $Z$. –  ulrich Apr 22 '12 at 7:19
Thanks ulrich! Could you please give more details? –  reference Apr 22 '12 at 7:26
Since $g$ is of finite order, you can assume it is given by a diagonal matrix in $SL_{r+1}$ whose entries are roots of unity. It is clear from the definition of the Veronese that $g$ will also act diagonally on $\mathbb{P}^N$, the space corresponding to the Veronese embedding. The diagonal entries are again roots of unity of the same (or smaller) order. Since there are only finitely many such, if $N$ is large there will be many repeated entries. The subspace corresponding to the repeated entries of any root of unity is fixed pointwise. –  ulrich Apr 23 '12 at 13:03

Your Answer


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

Browse other questions tagged or ask your own question.