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: enter image description here

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 $\text{Var}(f)=\text{Var}(g)?$


1 Answer 1


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.

  • $\begingroup$ Dear @Sándor Kovács, thanks for your reply. One little question, in your answer, do you want to say ''But, since $\alpha$ is birational...?'' $\endgroup$
    – Invariance
    Oct 7, 2022 at 9:39
  • $\begingroup$ Yes, that's what I meant. Thanks for catching this. :) $\endgroup$ Oct 8, 2022 at 19:37

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