Let $k$ be an algebraically closed field of characteristic zero.

Let $A$ be the $\mathbf{Z}$-subalgebra of the Grothendieck ring of $k$-varieties $K_0(\text{Var}_k)$ generated by classes of semi-abelian varieties.

Question 1. What can we say about the ring homomorphism $A\to K_0(\text{Var}_k)$?

It is obviously very far from being surjective. Mainly, is it integral/finite/flat/essentially of finite type?

Question 2. The class of smooth projective varieties over $k$ generates $K_0(\text{Var}_k)$ in the obvious sense. Is there a more manageable generating class of $k$-varieties, possibly containing semi-abelian varieties at the very least?

Question 3. In light of Question 2, does the class $\mathcal{P}$ of varieties over $k$, constructed below, have any hope to be a generating class for $K_0(\text{Var}_k)$, and if not, upon calling $B$ the $\mathbf{Z}$-subalgebra of $K_0(\text{Var}_k)$ it generates, same question as in Question 1, for the ring homomorphism $B\to K_0(\text{Var}_k)$.

We call $\mathcal{P}_0$ the class of $k$-varieties whose objects are:

  • abelian varieties
  • semi-abelian varieties (extension of abelian varieties by a torus)
  • smooth hypersurfaces in large projective spaces over $k$
  • smooth projective varieties of dimension $\le 3$
  • smooth projective toric varieties
  • Calabi-Yau varieties
  • products of the above

Let $\mathcal{P}_1$ be the class of $k$-varieties whose objects are obtained by blowing up an element of $\mathcal{P}_0$ along a closed $k$-subvariety that is again an element of $\mathcal{P}_0$.

Finally, let $\mathcal{P}$ be the class of all those $k$-varieties such that their class in the Grothendieck ring of varieties $K_0(\text{Var}_k)$ belongs to the subring $B\subset K_0(\text{Var}_k)$ generated by the classes of the elements of $\mathcal{P}_1$.

Question 4. Does there at least exist a morphism $f: Y\to X$ over $k$, with $Y$ in $\mathcal{P}$, and such that $f$ is surjective/flat/such that $f^*\omega_X\simeq\omega_Y$?

(This last question is expected to have negative answer for $f$ required to be surjective, and a good strategy to see this is suggested in the comments.)

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    $\begingroup$ A toric variety is rational. If (2) holds with $Y$ toric, $X$ is unirational. Similarly, most of your questions imply strong restrictions on $X$. $\endgroup$ – abx Nov 22 '17 at 20:28
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    $\begingroup$ 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?) $\endgroup$ – R. van Dobben de Bruyn Nov 23 '17 at 0:33
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    $\begingroup$ 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. $\endgroup$ – user97068 Nov 23 '17 at 1:35
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    $\begingroup$ 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.) $\endgroup$ – dhy Nov 25 '17 at 8:40
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    $\begingroup$ 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 $1$. $\endgroup$ – dhy Nov 25 '17 at 8:48

(Expanding my comments into an answer for more visibility.)

By Larsen-Lunts $K_0(\operatorname{Var}_k)/[\mathbb{A}^1]$ is the free abelian group on stable birational equivalence classes. It thus suffices to find a variety $X$ that is not stably birational to any variety in $\mathcal{P}_0$.

Take $X$ to be a smooth general type hypersurface in $\operatorname{Gr}(2,n),$ $n$ large. If two varieties are stably birational, then the bases of their MRC (maximally rationally connected) fibrations must also be birational. Now note that the operation of taking base of MRC fibration preserves the property of having a birational representative in $\mathcal{P}_0$. As $X$ is general type and thus its own MRC base, this reduces the problem to showing $X$ is not birational to any variety in $\mathcal{P}$.

Next we reduce birational equivalence to actual isomorphism. The operation of taking canonical model preserves the class of general type varieties in $\mathcal{P}_0$, and $X$ is its own canonical model. Thus we just need to rule out $X$ actually lying in $\mathcal{P}_0$.

As $\operatorname{Pic}X\cong\mathbb{Z}$ (by Lefschetz), $X$ can't be a product of two varieties. It remains to show $X$ is not a hypersurface in a projective space. Another application of Lefschetz shows that $H^4(X)\cong\mathbb{Z}\oplus\mathbb{Z},$ while a (high dimensional) hypersurface has $H^4\cong\mathbb{Z}$, giving us the final contradiction. QED.

Note that in the definition of $\mathcal{P}_1$, one can actually allow all (smooth) blow-ups! This is because blow-ups only change the class in $K_0(\operatorname{Var}_k)$ by a multiple of $[\mathbb{A}^1]$. This demonstrates that the Grothendieck ring of varieties is much more rigid than true motives - it really remembers the birational geometry of your variety.

Also, here is a sketch of an argument that for a very general high degree hypersurface $X$ in $\mathbb{P}^n\times\mathbb{P}^n$ so that it admits no dominant rational morphisms from varieties in $\mathcal{P}.$ The above arguments reduce this to showing $X$ admits no dominant rational morphisms from varieties in $\mathcal{P}_0$. My understanding is that Schoen proves in "Varieties dominated by product varieties" that a very general sufficiently ample hypersurface in any variety does not admit a dominant map from a product of varieties of smaller dimension, so it suffices to show that $X$ cannot be mapped to rationally by a smooth hypersurface in projective space.

Take $X$ to contain no rational curves. (To see this is possible, embed $\mathbb{P}^n\times\mathbb{P}^n$ in a projective space and intersect with a very general high degree hypersurface. It is known that this hypersurface will not contain any rational curves; I believe this statement was first proven by Clemens, in "Curves on generic hypersurfaces".) Under this assumption, any rational morphism to $X$ must be an actual morphism (see Kollar-Mori, Corollary 1.3.)

Now $X$ has an effective divisor class whose square is zero. But this stops us from being able to map a hypersurface in projective space dominantly to $X$, as the pullback of this divisor must also be effective of square zero, impossible on a high dimensional hypersurface (again, by Lefschetz).


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