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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

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    $\begingroup$ You can start reading here: link.springer.com/article/10.1007%2Fs10711-008-9334-1#page-1 $\endgroup$
    – M P
    Commented Apr 25, 2013 at 10:05
  • $\begingroup$ @MP: Great! Thanks for the reference. $\endgroup$ Commented Apr 25, 2013 at 15:33
  • $\begingroup$ There are papers by Kleiman-Piene that discuss this question. My best recollection is that they tend to be inseparable, but finite. $\endgroup$
    – meh
    Commented Apr 25, 2013 at 15:34
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    $\begingroup$ 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]. $\endgroup$ Commented May 12, 2013 at 14:06

1 Answer 1

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

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  • $\begingroup$ @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? $\endgroup$ Commented Apr 25, 2013 at 15:26
  • $\begingroup$ See MP's reference. $\endgroup$ Commented Apr 25, 2013 at 18:12

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