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No. First of all note that your line bundle $\mathcal{N}$ on $Y$ is trivial, so your assertion is $\mathcal{L}\cong\mathcal{M}$. Take a smooth projective curve $C$ of genus $\geq 1$ (say over $\math …

Not in general. Perhaps the simplest example is given by the Hilbert scheme of curves $C$ of degree 3 in $\mathbb{P}^3$ with $\chi (\mathscr{O}_C)=1$. This has 2 components, one (of dimension 12) corr …

1): The manifolds considered have $h^{1,0}=0$, so two line bundles $L$ and $L'$ on $\mathcal{X}_b$ are isomorphic if and only if $c_1(L)=c_1(L')$ in $H^2(\mathcal{X}_b,\mathbb{Z})$. If this holds at o …

This is true if you assume moreover that $(\mathcal{X}_K)_{red}$ is geometrically reduced -- in particular, in characteristic $0$. First of all, note that set-theoretically $\mathcal{X}'_b=\mathcal{X} …

The answer to the first question is no. In the moduli space $\mathcal{M}_{10}$ of curves of genus 10, the complete intersections $(3,3)$ in $\Bbb{P}^3$ form a strict subvariety $\mathcal{CI}$. Pick fo …

No, even if you assume $C_1$ and $C_2$ smooth irreducible. A rational equivalence class of curves in $\Bbb{P}^3$ is determined by its degree. Take for $C_1$ a complete intersection of 2 quadrics, you …

The space $H^{0}(X,f_{0}^{*}T_{Y})$ is the tangent space at $f_0$ to the variety $\mathrm{Hom}(X,Y)$, see J. Kollár, Rational curves on algebraic varieties. Thus if it is zero, $f_0$ is an isolated po …

It is an inclusion of analytic spaces — $\Delta $ is not a scheme. If $\mathfrak{A}=(f_1,\ldots ,f_p)$, $\Delta _{\mathfrak{A}}$ is the subspace of $\Delta $ defined by $f_1=\ldots =f_p=0$. I think $ …

This is actually true for all Chern classes, but you must first say how you identify $H^*(X,\mathbb{Z})$ and $H^*(X_t,\mathbb{Z})$. There is no problem for small deformations, that is if $B$ is a ball …

To develop what Jason says: if your curve deforms in a family of rational curves, it means that you can find a dominant rational map from a ruled surface onto your K3. This is forbidden (over $\mathbb …

Let $f:X\rightarrow \Delta $ be your family. $\ (*)$ implies that $f_*K_{X/\Delta }^{N}$ is a vector bundle on $\Delta $, with fiber $H^0(X_t, K_{X_t}^N)$ at $t\in\Delta$. The canonical homomorphism $ …

No. Take $X=\mathbb{P}^1\times \mathbb{P}^1$, $Y=\mathbb{P}^1$, $f$ the first projection, $\mathcal{L}=\mathcal{O}_{\mathbb{P}^1}(1)\boxtimes \mathcal{O}_{\mathbb{P}^1}(1)$, $s=X\otimes X'$, where $(X …

This map is injective. There is a Hochschild-Serre spectral sequence with $E^{pq}_2=H^p(\mathrm{Gal}(\bar{K}/K), H^q(X_{\bar{K}},\mathbb{G}_m))$ converging towards $H^{p+q}(X_{K},\mathbb{G}_m)$. This …

No for the first question. A counter-example is given by the Fano variety $X$ of lines in a cubic fourfold $V\subset \mathbb{P}^5$: for each hyperplane $H$ of $\mathbb{P}^5$, the lines contained in $H …

First of all, the image of your homomorphism is invariant under the Galois group $G:=\mathrm{Gal}(\bar{K}/K)$. So the right question is to ask whether the induced homomorphism $\mathrm{Pic}(X_{K})\rig …