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

Suppose that $X\subset\mathbb P^n$ is a $d$-dimentional smooth projective variety (not a linear subspace) over an algebraically closed field. If $\gamma\colon X\to\mathrm{Gr}(d,\mathbb P^n)$ is Gauss mapping that attaches to each point $x\in X$ the embedded Zariski tangent space to $X$ at $x$, then it is known that $\gamma$ is finite.

If characteristic is zero, it is known that $\gamma$ is not just finite but birational onto its image. My question is whether $\gamma$ is generically one to one in finite characteristic.

Edit: removed the question about birationality in finite characteristic, thanks to the example given by Felipe Voloch.

Thanks in advance,
Serge

share|improve this question
2  
You can start reading here: link.springer.com/article/10.1007%2Fs10711-008-9334-1#page-1 –  M P Apr 25 '13 at 10:05
    
@MP: Great! Thanks for the reference. –  Serge Lvovski Apr 25 '13 at 15:33
    
There are papers by Kleiman-Piene that discuss this question. My best recollection is that they tend to be inseparable, but finite. –  aginensky Apr 25 '13 at 15:34
1  
MP's reference provides counter-examples that are smooth space curves. On the other hand, all smooth plane curves have generically one-to-one Gauss map, by [Hajime Kaji : On the Gauss maps of space curves in characteristic p, Corollary 4.5]. –  Olivier Benoist May 12 '13 at 14:06
add comment

1 Answer

No. The plane curve $x^{p+1}+y^{p+1}=1$ has an inseparable Gauss map.

share|improve this answer
    
@Felipe: Thank yuo for the example (I will edit the question accordingly). However, this Gauss mapping is 1-1. Does there exist an example where it is not generically one to one? –  Serge Lvovski Apr 25 '13 at 15:26
    
See MP's reference. –  Felipe Voloch Apr 25 '13 at 18:12
add comment

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.