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The origin of the hypersurface defined by $$x_0^2 + x_1^4 + x_2^4 + \cdots + x_n^4 = 0$$ is a canonical singularity for $n = 3$, and a terminal singularity for $n \ge 4$ (see e.g. Theorem 2 in this pa …

Are there complex projective elliptic threefolds $f: X \to S$ with infinitely many rational sections satisfying the following properties? $S$ is a smooth surface of Picard rank $1$. $X$ is smooth wit …

This is an answer to your last question. In general we can't represent the class of a Kähler current by a semi-positive smooth form $\alpha$. Consider the blowup $\tilde{S} \to S$ of a compact Kähler …

If $X'$ is a Mukai flop of a compact hyper-Kähler manifold $X$, then $X'$ is in Fujiki class $\mathcal{C}$ and carries a holomorphic symplectic form $\sigma$. Taking the real part or the imaginary par …

Edit notice: As Evgeny Shinder pointed out in the comment, it is unclear why we have $S = J^{-1}(\{0,1,\infty\})$ where $J : C \to \mathbf{P}^1$ is the $j$-invariant map. The problem is that $X \to S$ …

This is related to Mori's theorem through Grauert's ampleness criterion in Hartshorne's "Ample vector bundles" (Proposition 3.5). Let's assume that $M$ is projective and $\dim M \ge 2$. Let $\alpha : …

For a reference in English, you could take a look at this paper of Varouchas, which contains a definition of Kähler spaces and relatied concepts (e.g. Kähler morphisms) and some of their fundamental …

First of all, Deligne-Illusie worked with relative Frobenius $F = F_{X/S}: X \to X'$, so the target of the composition should be $\Omega_{X'/S}^p$ instead of $\Omega_{X/S}^p$. Your question about why …

Edit notice: The answer is completely rewritten due to user2520938's comment. My original answer was that the linear operator $P$ depends on $t$, and we have $(1)$ as long as $P(t) = 0$. But as user25 …

Let $C$ be a smooth projective curve (say over $\mathbf{C}$ for simplicity). Given $L,M \in \mathrm{Pic}^0(C)[n]$, recall that the Weil pairing $e_n(L,M)$ is defined as follows. First of all, for an …

This is not an answer but rather a lengthy comment. A necessary condition for the pencil to be isotrivial is that a smooth member in that pencil has a non-trivial automorphism: By blowing-up the base …

In an answer of this MO question, Frank Loray constructed an example of analytic singularity which is not algebraic. On the other hand, as I learned from one of Joël's comments in that question, Arti …

For complex manifolds, simultaneous contractions in a family has been studied by Riemenschneider in the paper "Deformations of Rational Singularities and their Resolutions". Theorem 1 in that article …

Apart from the reference given in the comment, you can also find a proof of the existence of the fiber product in Fischer's "Complex Analytic Geometry", Corollary 0.32. For direct products, a more st …

If $k$ is an algebraically closed field of characteristic $0$, then the map $$X \mapsto N(X) := \#(\text{connected components of }X)$$ defined for smooth projective varieties extends to a ring homom …