Grothendieck class and Normalization

Question

Does the Grothendieck class behave well under normalization? Here by Grothendieck class, I mean the class of a variety in $K_0(\mathcal{var/\mathbb{C}})$

Answer 1

If $\tilde{X}$ is the normalization of $X$, then $\tilde{X}$ and $X$ are birational (see here for example).

So there exists open sets $U\subset X$ and $\tilde{U}\subset \tilde{X}$ that are isomorphic via the normalization map. Let $D=X-U$ and $\tilde{D}=\tilde{X}-\tilde{U}$. Then $[\tilde{X}]=[X]+([\tilde{D}]-[D])$.

This generalizes the example given by Allen Knutson in the comments since in the case of a nodal cubic curve, normalization is desingularization (as with any such curve) and so over the node, the normalization has two points; thus, $[\tilde{D}]-[D]=[pt]$.

Of course this observation is true for any birational map, and really says nothing about the normalization itself (aside from that property).

Here is another example. The cusped cubic curve has both $[\tilde{D}]=[pt]$ and $[D]=[pt]$, so $[\tilde{D}]-[D]=0$. In that case the normalization map is injective.

More generally, now suppose that the normalization map is injective.

Lemma: $X \to Y$ is a dominant morphism of irreducible algebraic varieties. If there exists a dense subset $Z \subset Y$ such that for any $y \in Z$ the fiber consists of a single point, then the map is birational.

So the restriction of the normalization map to $\tilde{D}$ is bijective and finite onto $D$ and hence is birational (by the lemma). So on each component of $\tilde{D}$ we get an isomorphism on open sets to open sets in components of $D$. By induction on dimension, we eventually get down to an isomorphism. This shows that $[\tilde{D}]-[D]=0$ whenever the normalization map is injective.

More generally, radical surjective maps $f:X\to Y$ give equality $[X]=[Y]$ in characteristic $0$, and if the field is algebraically closed then $f$ is radicial if it is injective. So over $\mathbb{C}$ bijections $f:X\to Y$ imply $[X]=[Y]$.

Now, if one considers instead of $\tilde{X}$, the seminormalization $X^{sn}$, then there is $\tilde{X}\to X^{sn}\to X$ where the map $X^{sn}\to X$ is finite and bijective (hence birational by the above lemma). So the same argument as above shows that $[X]=[X^{sn}]$, as commented by Allen Knutson.

In general, $[\tilde{D}]-[D]$ has to be handled case-by-case however, but one can WLOG assume that $X$ is seminormal.