# Two morphisms possess the same Viehweg's variation

Recall the definition of Viehweg's variation intimated from Weak Positivity and the Additivity of the Kodaira Dimension for Certain Fibre Spaces, E. Viehweg

Let $$f: V\rightarrow W$$ be a fiber space (surjective with connected geometric fiber $$V_w:=V\times_W \text{Spec }\overline{\mathbb C(W)}$$) between two non-singular projective varieties over the field of complex number $$\mathbb C$$. Then, the variation of $$f$$, denoted by $$\text{Var}(f)$$, is defined to be the minimal number $$k$$, such that there exists a subfield $$L$$ of $$\overline{\mathbb C(W)}$$ of transcendental degree $$k$$ over $$\mathbb C$$ and a variety $$F$$ over $$k$$ with $$F\times_{\text{Spec }(L)}\text{Spec }\overline{\mathbb C(W)}$$ is birationally equivalent to $$V_w$$.

Now Let $$f: X \rightarrow Y$$ be a surjective morphism between smooth projective varieties with connected fibers, and we have a commutative diagram:

such that

1. $$V$$ and $$W$$ are smooth projective varieties.
2. $$\alpha$$ and $$\beta$$ are birational.
3. All g-exceptional divisors are $$\alpha$$-exceptional.

Question: how to show $$\text{Var}(f)=\text{Var}(g)?$$

Since the definition only depends on the general fiber, and $$\beta$$ is birational, one may assume that $$\beta$$ is actually and isomorphism. So, then $$L$$ is defined as a subfield with minimal transcendence degree over $$k$$ such that $$F\times_{\text{Spec }(L)}\text{Spec }\overline{\mathbb C(W)}$$ is birationally equivalent to $$V_w$$ and $$X_w$$ respectively. But, since $$\alpha$$ is birational, $$V_w$$ and $$X_w$$ are birational, so if $$L$$ works for one, it works for the other as well.
An equivalent definition of $${\rm Var} f$$ is given in Def 2.8 in Subadditivity of the Kodaira Dimension: Fibers of General Type by János Kollár. This definition of $${\rm Var}f$$ is more convenient and has become the standard. With that version, the statement you are asking about is straightforward.
• Dear @Sándor Kovács, thanks for your reply. One little question, in your answer, do you want to say ''But, since $\alpha$ is birational...?'' Oct 7, 2022 at 9:39