28

No. If you blow up a smooth subvariety $X$ of $\mathbb{P}^n$, the Hodge conjecture for the resulting variety is equivalent to the Hodge conjecture for $X$. So the Hodge conjecture for rational varieties (= birational to $\Bbb{P}^n$) implies the Hodge conjecture in general.

25

In this generality, the answer is no. The projective curve $X$ given by $2y^2z^2 = x^4 - 17z^4$ over the rationals satisfies the HP, since it has local points everywhere (the affine part $z \neq 0$ is given by $2y'^2=x'^4-17$, which is the famous Reichardt-Lind equation which is known to be everywhere locally, but not globally, soluble) and it has the unique ...

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20

For the first question I am not that pessimistic. At least there are candidates as follows: Recall that $Z$ is stably rational if there is $n\ge0$ such that $Z\times\mathbf A^n$ is rational. Now suppose there is such a $Z$ such that the minimal $n$ is $\ge2$. I would be extremely surprised if that didn't exist. Then put $Y=Z\times\mathbf A^{n-2}$ and $X=Y\...

18

Of course, abx is completely correct in saying that the truth of the Hodge conjecture is not a birational invariant. That said, something slightly weaker is true: if $X$ and $Y$ are $K$-equivalent, then the Hodge conjecture is true for $X$ if and only if it is true for $Y$.
Here we say two smooth projective varieties $X, Y$ are $K$-equivalent if there ...

17

The secant variety $Sec_k(V^n_2)$ is the variety parametrizing $(n+1)\times (n+1)$ symmetric matrices modulo scalar of rank at most $k$ that is of corank at least $n+1-k$.
Then by Proposition 12(b) in
J. Harris; L. W. Tu, On symmetric and skew-symmetric determinantal varieties, Topology 23 (1984), no. 1, 71–84.
the degree of $Sec_k(V^n_2)$ is given by
$$\deg(...

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15

I assume you mean $H^0(X, K_X)^{\otimes m}$ rather than $\oplus_{i=1}^m H^0(X, K_X)$. If $X$ is a smooth projective connected complex curve of genus $g \geq 2$, then the map
$$H^0(X, K_X)^{\otimes m} \longrightarrow H^0(X, m K_X),$$
is surjective for any $m \geq 0$, as long as $X$ is not hyperelliptic. This is a theorem of M. Noether.

14

To complete the answer of Divierietti and the comment of Roy Smith, here is a statement which might interest you:
Theorem If $X,Y$ are varieties over a field $k$, assume $X$ is smooth and $Y$ proper containing no rational curves. Then any rational map $X\dashrightarrow Y$ is everywhere defined.
You can find that statement in Debarre's book Higher ...

13

You don't need your variety, say $X$, to be Fano, only $\mathrm{Pic}(X)=\mathbb{Z}$. A pseudo-automorphism $u$ of $X$ induces an automorphism of $\mathrm{Pic}(X)$, which must be the identity. Let $L$ be a very ample line bundle on $X$; since $u^*L\cong L$, $u$ induces an automorphism of $H^0(X,L)$ (here you use Hartogs theorem, as Jason pointed out).
Then ...

13

Any smooth projective toric variety is rational, in particular simply connected.
Then, by the Lefschetz hyperplane theorem for global complete intersections, if $\dim X \geq 3$ is a smooth complete intersection into a smooth toric variety then $\pi_1(X)=\{1\}$.
In particular, for instance, abelian varieties of dimension at least $3$ cannot be realized as ...

answered Mar 19 '18 at 9:02

Francesco Polizzi

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13

The result is true in all characteristics. See O. Zariski, Illinois J. Math. 2(1958), 303-315.

answered Mar 19 '19 at 9:14

Laurent Moret-Bailly

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12

I want to mention the positive direction. Let $X$ be a smooth, projective variety over $\mathbb{C}$, resp. over an algebraically closed field of arbitrary characteristic. Let $Z\subset X$ be a proper closed subset. If $X$ is (separably) rationally connected and if $Z$ has codimension $\geq 2$ in $X$, then $X\setminus Z$ is simply connected for the ...

12

EDIT: I've just realized that this holds under somewhat weaker assumptions. It is not necessary that the fibers of $g$ are connected.
EDIT#2: Apparently, in my previous edit I weakened the conditions too far... properness of $g$ is back as it is needed, but connectivity of fibers is not as it is not. I think it is OK now. :)
Cool question. Actually, I ...

11

It also holds for minimal surfaces of Kodaira dimension $\kappa\geq0$.

11

In general your polynomial $P(k)$ has degree $n$ as soon as there is a positive dimensional subscheme in the base locus of the linear system. Consider for instance the case $n=3$, $s = 2$, $m=2$, $d = 3$. Then the line through the two base points is in contained in the base locus of the linear system with multiplicity $1$.
The rational map induced by this ...

11

Here is some of the general philosophy of birational invariants, at least those coming from (co)homology (I don't think this approach quite works for homotopical invariants.)
Philosophy. If $f \colon X \dashrightarrow Y$ is a birational map, consider the cycles $\bar \Gamma_f \subseteq X \times Y$ and $\bar \Gamma_{f^{-1}} \subseteq Y \times X$. Then $\bar\...

answered Nov 5 '16 at 15:57

R. van Dobben de Bruyn

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11

Two such examples were given by Achinger and Zdanowicz [AZ17], both of which satisfy a whole bunch of other good properties (e.g. their classes in the Grothendieck ring of varieties are polynomials in the Lefschetz motive $\mathbb L = [\mathbb A^1]$). The easiest one to state is probably the following:
Example. (Achinger–Zdanowicz [AZ17, §4]) Let $k = \bar{\...

answered Mar 26 '18 at 22:29

R. van Dobben de Bruyn

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10

The group $\mathrm{Aut}(X)$ is finite because, as you point out, it preserves a very ample line bundle; hence it is an algebraic group (a closed subgroup of a projective group). Therefore it suffices to prove that its Lie algebra $H^0(X,T_X)$ is trivial. By Serre duality this is dual to $H^3(X,\Omega ^1_X)=H^{1,3}$, which is conjugate to $H^{3,1}=H^1(X,\...

10

Since you are talking about rational maps, I assume you mean "open dense" in the Zariski topology, so that $X$ and $Y$ are algebraic varieties. Therefore we have a particular case of the following well-known statement in algebraic geometry.
Proposition 1. Let $k$ be an an algebraically closed field of characteristic zero and $f \colon X \to Y$ a ...

answered Sep 29 '14 at 9:17

Francesco Polizzi

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10

A reference for birational equivalence of $CH_0$ is Fulton's Intersection Theory [1], Example 16.1.11. In the example, he makes the assumption that $k$ is algebraically closed, but he never uses it. Since the argument is fairly short, let me repeat it here.
Theorem. Let $k$ be a field, and let $X$ and $Y$ be smooth proper $k$-varieties. If $X$ and $Y$ are ...

answered Jun 11 '16 at 3:59

R. van Dobben de Bruyn

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10

Basically, any invariant $T$ which satisfies the following purity property will turn into a birational invariant for proper smooth varieties (aka complex compact manifolds): let $X$ be a smooth variety and $U$ be an open subspace of $X$ such that $\mathop{\rm codim}_X(X\setminus U)\geq 2$; then the injection $U\hookrightarrow$ induces an isomorphism $T(X)\...

10

As $\mathcal{O}_{\mathbb{P}^2}$ is trivial, then multiplicativity of Chern classes in exact sequences implies:
$$
c_*(\mathcal{E}) = c_*(\mathcal{I}_p(-1)).
$$
We can compute $c_*(\mathcal{I}_p(-1))$ by noting that $p\in \mathbb{P}^2$ is a complete intersection of 2 lines. Therefore, the $(-1)$-twist of the Koszul complex is exact:
$$
0\rightarrow \mathcal{O}...

10

EDIT. An alternative reference is
Husemoller, Dale H.: Finite automorphism groups of algebraic varieties, Finite groups, Santa Cruz Conf. 1979, Proc. Symp. Pure Math. 37, 611-619 (1980). ZBL0466.14018.
A lot of work has been done in order to bound $|\mathrm{Aut}(X)|$ in terms of the dimension and the volume of $X$. You can look at
Hacon, Christopher D.; ...

answered Sep 8 '18 at 8:59

Francesco Polizzi

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10

If I have understood the OP's definitions correctly, the answer is no for rational chain connectedness. Let $E$ be a smooth genus $1$ curve in $\mathbb{P}^2$ and let $X \subset \mathbb{P}^3$ be the cone on $E$, with vertex $x_0$. For any $x_1$ and $x_2$ in $X$, the lines joining $x_1$ to $x_0$ and $x_0$ to $x_2$ are in $X$, so $x_1$ and $x_2$ are connected ...

answered Sep 18 '18 at 12:52

David E Speyer

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10

For lc pair or slc pair, it is true. This is Gongyo’s result. See [J. ALGEBRAIC GEOMETRY 22 (2013) 549–564].
BTW, the relative version is also true, which is not a trivial generalization of the absolute case. It is proved by Hacon and Xu [On Finiteness of B-representation and Semi-log Canonical Abundance].

9

It looks like you are mixing up flips and flops. 1) looks more like a flip, though not really the usual definition. 2) is indeed a flop, but this is also not the usual definition. For certainty look up the definition in Kollár-Mori 98. (Also, flips go from $K$ being anti-ample to ample, not the other way around. This is actually important!!)
The essence of ...

9

This is also true for every smooth Fano variety $X$, with any Picard number. One can see it using Mori dream spaces: $X$ is a Mori dream space (by BCHM), and hence has (up to isomorphisms) only finitely many "small $\mathbb{Q}$-factorial modifications" (SQM) = $f\colon X$-->$Y$ birational, isomorphism in codimension one, with $Y$ projective, normal, and $\...

9

Let me start with a little nitpicking:
What on Earth do you mean by $\mu^*L-\epsilon E$ when you said that $L$ was a line bundle? You cannot add a line bundle and a divisor! So, let's assume that you said that $L$ was a Cartier divisor (or at least $\mathbb Q$-Cartier).
It looks like you are assuming that Y is smooth, or at least that it is smooth along Z, ...

9

For non-smooth $Z$ the answer is in general no.
Take $P=\mathbb{P}^4$ and let $Z \subset \mathbb P^4$ be a surface with a non-normal double point $p$ (i.e. a singularity locally analytically isomorphic to the one given by two planes intersecting in a single point, it is no difficult to construct irreducible examples, for instance taking general projections ...

answered Dec 16 '15 at 8:58

Francesco Polizzi

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