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Timeline for $K_0$-equivalence of varieties

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S Nov 29, 2017 at 11:32 history bounty ended CommunityBot
S Nov 29, 2017 at 11:32 history notice removed user95222
Nov 26, 2017 at 1:55 vote accept CommunityBot
Nov 25, 2017 at 23:42 answer added dhy timeline score: 6
Nov 25, 2017 at 8:48 comment added dhy Call it $X$. By the Lefschetz hyperplane theorem $X$ will have Picard rank $1$ and $4$th Betti number equal to $2$ (assuming large degree in a large enough Grassmannian.) Under that assumption, X will be its own canonical model. Therefore $X$ can't be birational to a product of two varieties (or, taking canonical models of both, $X$ would need to have Picard rank at least $2$.) And it can't be birational to one of your specified varieties - the only general type (of large dimension) possibility is a hypersurface in projective space, also its own canonical model, but with $H^4$ rank $1$.
Nov 25, 2017 at 8:40 comment added dhy Does the following argument work? By Larsen-Lunts, $K_0(Var)/[\mathbb{A}^1]$ is the free abelian group on stable birational classes. It thus suffices to show that there is some variety which is not stably birational to a variety in class $P$. In fact, as the MRC base of any variety in $P$ is in $P$ (up to birational equivalence), it suffices to show that there is a general type variety not birational to a variety in P. To do this, consider a high degree hypersurface in a Grassmannian. (cont.)
S Nov 25, 2017 at 8:10 history bounty started CommunityBot
S Nov 25, 2017 at 8:10 history notice added user95222 Draw attention
S Nov 23, 2017 at 2:00 history suggested user97068 CC BY-SA 3.0
Edit to make question clearer.
Nov 23, 2017 at 1:59 review Suggested edits
S Nov 23, 2017 at 2:00
Nov 23, 2017 at 1:54 history edited user95222 CC BY-SA 3.0
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S Nov 23, 2017 at 1:51 history suggested user97068 CC BY-SA 3.0
I edited the question so as to make it more understandable.
Nov 23, 2017 at 1:51 review Suggested edits
S Nov 23, 2017 at 1:51
Nov 23, 2017 at 1:40 comment added user95222 My question was meant to ask if one can find a more handy generating class for the Grothendieck ring of $k$-varieties, instead of just all smooth projective varieties over $k$. Smooth projective varieties are known to be generators of $K_0(\text{Var}_k)$. I was merely asking if there is any serious reason why the class $\mathcal{P}$ cannot be a class of generators, why, and what is it that is missing to it in order to become one
Nov 23, 2017 at 1:35 comment added user97068 Side comment. I came to this question earlier and can't avoid noting it has been down voted. If you're going to down vote a question, at least make sure you explain why, out of courtesy. It really looks to me the OP is asking a meaningful question, though somewhat confusingly phrased. Unless you have a full argument as to why those classes of varieties, and the transformations allowed to construct the class $\mathcal{P}$ out of them, are not sufficient to generate the full $K_0(\text{Var}_k)$, then you should probably refrain from just down voting, until somebody provides a good answer.
Nov 23, 2017 at 0:33 comment added R. van Dobben de Bruyn The answer to (2) should be no, but it might not be so easy to write down an actual counterexample. (Maybe it's possible to write down a variety with a big irreducible sub-Hodge structure that cannot come from any of these varieties?)
Nov 22, 2017 at 20:41 review Close votes
Nov 23, 2017 at 12:12
Nov 22, 2017 at 20:28 comment added abx A toric variety is rational. If (2) holds with $Y$ toric, $X$ is unirational. Similarly, most of your questions imply strong restrictions on $X$.
Nov 22, 2017 at 20:16 history edited user95222 CC BY-SA 3.0
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Nov 22, 2017 at 19:31 history asked user95222 CC BY-SA 3.0