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Algebraic varieties, stacks, sheaves, schemes, moduli spaces, complex geometry, quantum cohomology.
4
votes
Accepted
normalization of a bijection
We should be able to construct a counterexample as follows:
Let $Y$ be an affine curve which is smooth away from a single node. We obtain $X$ from the normalization of $Y$ by removing one of the poin …
1
vote
0
answers
292
views
Algebraic fiber bundles for various topologies
Let's fix a scheme $S$ and consider some Grothendieck topology $\tau\in\{Zariski,\acute{e}tale,fppf\}$ on the category of schemes over $S$. Define a $\tau$ fiber bundle with fiber type $F$ to be a map …
9
votes
Accepted
Is there a "free abelian group of rank 1" in the category of affine group schemes?
The affine group scheme $G$ you describe is not finite type. It is possible to describe $G$ explicitly.
For $R$ a $k$-algebra, the $R$-points of $G$ are the tensor functorial automorphisms of $\mathr …
2
votes
change of the residue of meromorphic differential by a covering map.
Another way to see this: if $\gamma$ is a small loop in $C_1$ winding once around $x$, then $f(\gamma)$ is a small loop in $C_2$ winding $e_x$ times around $f(x)$, so
$$
\mathrm{Res}_x f^*\omega=\frac …
6
votes
Accepted
commutative algebra, diagonal morphism
Let $I'\subset B\otimes_A B$ be the ideal generated by the elements $b_i\otimes 1-1\otimes b_i$, and define
$$
R=\{b\in B:b\otimes 1-1\otimes b\in I'\}.
$$
It’s not hard to check that $R$ is an $A$-su …
6
votes
Accepted
Reals with integer powers bounded away from integers?
We should be able to construct such a real number $x$. Let $\epsilon\in(0,1/2)$ be fixed. If $S\subset \mathbb{R}_{>0}$ and $r>0$, we write $S^r$ for the set of positive $r$-th powers of elements of $ …
13
votes
Accepted
Motives associated to a Number Field
Write $X=\mathrm{Spec}\, k$, which is a $0$-dimensional variety. Motives of $0$-dimensional varieties are called Artin motives, and they are pure. The Betti realization is the Betti cohomology of $X$, …
17
votes
The underlying space of a scheme remembers its affineness?
Here is a counterexample. Fix a field $k$, and let $Y$ be built from two copies of the affine nodal curve $y^2=x^3+x^2$, glued together on the complement of the singular point. In other words $Y$ is a …
33
votes
Accepted
Classical algebraic varieties VS $k$-schemes VS schemes
Let's say $k=\mathbb{C}$ (although something like this should work over any algebraically closed field). Let $V_1=\mathbb{P}^1-\{0,1,\infty,\pi\}$ and $V_2 = \mathbb{P}^1-\{0,1,\infty,e\}$. One can se …
19
votes
3
answers
2k
views
$p$-adic periods
For a variety $X$ defined over $\mathbb{Q}$, there's a (functorial) comparison isomorphism
$$
H^i_{dR}(X)\otimes\mathbb{C}\to H^i_B(X,\mathbb{Q})\otimes\mathbb{C}.
$$
If we pick $\mathbb{Q}$-bases for …
5
votes
$p$-adic periods
There is a notion of $p$-adic period coming from the Frobenius action on $p$-adic cohomology. Suppose $X$ is a smooth variety over $\mathbb{Q}$, and let $p$ be a prime. If there is a smooth model $\ti …
4
votes
Is torsion submodule of a $p$-adically complete and separated $\mathbb{Z}_{p}$-module closed?
Take
$$
R=\prod_{n=1}^\infty \mathbb{Z}/p^n\mathbb{Z},
$$
with the product topology. Every non-empty open set in $R$ contains an element that is $0$ in all but finitely many factors, and this element …
5
votes
0
answers
198
views
Algebraic fundamental group without regularity at infinity
Suppose $X$ is a smooth (connected) variety over $\mathbb{C}$. Let $\mathscr{C}$ be the category of finite rank vector bundles on $X$ equipped with an integrable connection, and let $\mathscr{C}'$ be …
6
votes
Accepted
Lie algebra preserving ideal of functions
A counterexample is $X=\mathbb{A}^2\backslash\{y=0\}$, $A=\mathbb{C}[x,y,y^{-1}]$, and $\mathfrak{g}$ the span of the derivation $D(x)=1$, $D(y)=y$. Now $\mathfrak{g}$ is $1$-dimensional and $X$ is $2 …
4
votes
How to cook up an Artin motive from a positive-dimensional variety
Tate twists do indeed shift the Hodge filtration. Since the de Rham realization of the Tate motive $\mathbb{Q}(1)$ is concentrated in the degree $-1$ part of the Hodge filtration, we have $F^n M(m)_{d …