2
$\begingroup$

As anyone who follows the algebraic geometry tag on arXiv will probably know, there has been a lot of papers recently showing various varieties are non-stably rational. What I am interested in however is ways of showing a variety is stably rational.

Obviously, showing a variety is rational would do the trick. There are examples of stably rational varieties which are non-rational, the first as far I'm aware was in "Variétés stablement rationnelles non rationnelles" by A. Beauville, J.-L. Colliot-Thélène, J.-J. Sansuc, and P. Swinnerton-Dyer. Another proof came later in "Stably rational irrational varieties" by N. Shepherd-Barron. The variety and it's proof arises out of group actions and geometric invariant theory, an area which I am not very knowledgeable on, so I can't really give a better description.

The second potential method I've found is from the result of Larsen and Lunts on the Grothendieck ring of varieties. I.e. If $X$ is a smooth and complete variety, then it is stably birational if and only if $[X] = 1 \text{ mod } \mathbb{L} $, where $[X]$ is it's class in the Grothendieck ring and $\mathbb{L} = [\mathbb{A}^1] $. This seems to be quite a nice criterion, but calculating classes explicitly in the Grothendieck ring seems to be in general rather hard. I have yet to see this be used to give an example of a non-rational, stably rational variety.

Are there any other potential methods I have not mentioned?

$\endgroup$
  • $\begingroup$ The title would seem to make more sense if the last word was "rational". $\endgroup$ – potentially dense May 19 '16 at 8:36
  • $\begingroup$ Yes, it would. Sorry, it was quite late when made this post! $\endgroup$ – Gabriel_s_syme May 19 '16 at 9:13
3
$\begingroup$

Though it applies to quite particular varieties, one might mention the so-called "no-name method": let $G$ be a reductive group which admits an almost free representation $V$ such that the quotient $V/G$ is rational ("almost free" means that there is a $G$-stable Zariski open subset of $V$ on which $G$ acts freely). Then for all almost free representations $W$ of $G$, the quotient $W/G$ is stably rational. This applies to many groups, e.g. $GL(n)$ (take for $V$ the space of $n\times n$ matrices, with $GL(n)$ acting by multiplication), $SL(n)$, $SO(n)$ etc. A good reference is Dolgachev's survey Rationality of fields of invariants in the Proceedings of the Bowdoin AMS conference.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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