Let $X^n$ be a compact Kähler manifold with $K_X$ semi-ample, i.e., a sufficiently high power of $K_X$ is basepoint free. The associated pluricanonical system $| K_X^{\ell} |$ furnishes a birational map $$f : X \dashrightarrow \mathbb{P}^{\dim H^0(X, K_X^{\ell})-1}$$ onto some normal projective variety $Y \subset \mathbb{P}^{\dim H^0(X, K_X^{\ell})-1}$ of dimension $\kappa(X)$. Here, $\kappa(X)$ denotes the Kodaira dimension of $X$. The semi-ampleness of $K_X$ further implies that $K_X^{\ell} \simeq f^{\ast} \mathcal{O}(1)$. In particular, for every $y \in Y$ that is not contained in the discriminant locus of $f$, $K_X^{\ell} \vert_{f^{-1}(y)} \simeq \mathcal{O}_{X_y}$. Since $f$ is a submersion near $f^{-1}(y)$, we have the adjunction-type relation $K_{f^{-1}(y)} \simeq K_X \vert_{f^{-1}(y)}$ and therefore the fibres of $f$ are Calabi--Yau manifolds of dimension $n- \kappa(X)$.

In the Kähler geometry literature, it is common to refer to this map $f$ as a Calabi--Yau fibration. My question may be extremely obvious, but nevertheless:

Question: Is this a fibration in the sense of homotopy theory, i.e., does this map satisfy the homotopy lifting property?


1 Answer 1


I would say that the answer is in general no.

Think of an elliptic surface $X$ with Kodaira dimension $1$ and whose elliptic fibration contains a cuspidal curve. Then the general fibre is not homotopically equivalent to the special one (the former is homeomorphic to $S^1 \times S^1$, the latter to $S^1$, in particular their fundamental groups are different), whereas all the fibres of a Hurewicz fibration have the same homotopy type.

Edit. Actually, any elliptic surface $X$ with Kodaira dimension $1$ and whose elliptic fibration contains a nodal curve also provides a counterexample. In fact, a nodal cubic is homeomorphic to a torus "with one cycle shrunk away”; in particular, it has the homotopy type of $S^1 \vee S^2$ and the previous argument applies.

  • $\begingroup$ Thankyou! Do you have a reference for these homeomorphisms? Admittedly, I have never thought about homotopy types of the cuspoidal/nodal curves. $\endgroup$
    – AmorFati
    Commented Aug 14, 2020 at 20:55
  • 1
    $\begingroup$ It is rather elementary. The cuspidal cubic has affine equation $y^2=x^3$, and the projection $(x, \, y) \mapsto y$ induces a homeomorphism onto $\mathbb{P}^1$, which is topologically $S^2$. Regarding the nodal cubic, in can be seen as $\mathbb{P}^1$ with two points identified. Hence, topologically, we are identifying two points on $S^2$; this is clearly the same thing as shrinking a cycle on a torus, and the corresponding quotient space is homotopically equivalent to a sphere with a circle attached at a point. $\endgroup$ Commented Aug 14, 2020 at 21:24

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