# Grothendieck group of constructible sets

Let $$K_0$$ be the Grothendieck group of complex algebraic varieties. This is the group generated by all complex algebraic varieties, subject to the relations:

(i) $$[X]=[Y]$$ if $$X,Y$$ are isomorphic,

(ii) $$[X]= [Z]+[X-Z]$$ if $$Z \subset X$$ is a closed subset.

(See the following notes: https://sites.math.northwestern.edu/~mpopa/571/chapter6.pdf)

Let $$\chi_c(.)$$ denote the Euler characteristic with compact support of a complex algebraic variety. It can be shown that it has the property that $$\chi_c(X)= \chi_c(Z)+ \chi_c(X-Z)$$. Hence, $$\chi_c(.)$$ gives a well-defined map $$\phi: K_0 \to \mathbb{Z}$$ taking a variety $$V \mapsto \phi(V)= \chi_c(V)$$.

Suppose that $$C \subset X$$ is a constructible set. In general, $$C$$ is not a variety. However an element $$[C]\in K_0$$ can still be defined as follows:

By definition of $$C$$ being constructible, we can write $$C= V_1 \cup\dots\cup V_n$$ where the $$V_i$$ are disjoint locally closed subvarieties of $$X$$. Let us now write $$V_i = Y_i - Z_i$$, with $$Y_i, Z_i \subset X$$ closed. Then $$[C]$$ can be viewed as an element in $$K_0$$ by setting $$[C]:= \sum_i [Y_i]-[Z_i]$$.

Question: is $$\chi_c(C)= \phi([C])= \sum_i \chi_c(V_i)$$?

Remark: if $$C$$ were a variety, then this would be true by definition the map $$\phi$$. However, $$C$$ is merely assumed here to be a constructible set.

• It's obviously wrong as written, because your decomposition $C = V_1 \cup \ldots \cup V_n$ does not see points that are included multiple times. For example, you can add $V_{n+1} = \{p\}$ for any point $p \in C$, which doesn't change $C$ but it does change $[C]$. You should revise your definition. – R. van Dobben de Bruyn Dec 20 '18 at 22:38
• @R.vanDobbendeBruyn Thank you for your comment. I added the additional assumption that the $V_i$ are assumed to be disjoint. – user142700 Dec 20 '18 at 22:47