14
votes
Accepted
Why are ordinary spheres not strictly invertible?
An $E_{\infty}$ structure extending the $E_1$ structure on $R(n)$ in particular yields maps $(\mathbb{S}^{2n})^{\otimes p}_{hC_p} \to \mathbb{S}^{2pn}$ splitting the inclusion of the bottom cell. The ...
- 10.8k
12
votes
Accepted
Pic^0 and H^0(K,Pic^0)
By the long exact sequence of low degree terms for the Leray spectral sequence computing $H^r_{\text{et}}(C,\mathbb{G}_m)$ via $H^p_{\text{et}}(\text{Spec}\ K,R^q f_*\mathbb{G}_m)$, the cokernel of ...
12
votes
Accepted
Picard group of connected linear algebraic group
$\DeclareMathOperator\Pic{Pic}$The statement is false over most imperfect fields, even for smooth affine group schemes.
In particular, it is false over any separably closed imperfect field $k$. I will ...
- 2,923
12
votes
Accepted
Reference request: Generic k3 surface has Picard number 1
Welcome new contributor. I am just writing my comment as an answer, and expanding on the observation of Prof. Arapura. For a smooth, projective scheme $X$ over a field $k$, the space of first order ...
11
votes
Accepted
Picard group of a finite type $\mathbb{Z}$-algebra
This is false. A counterexample is given in [Kahn06, Rmq. 1 (6)]. The example uses the cuspidal cubic $B = A[x^2,x^3]$ over a finite type $\mathbb Z$-algebra $A$ that is not a finitely generated $\...
- 28.7k
11
votes
Accepted
p-torsion in the Picard group of a regular projective curve
Take $p=3$ and $C\subset \mathbb{P}^2$ with equation $y^2 z=x^3 - t z^3$ where $t\in K$ is not a cube. Then $C$ is regular but, putting $L:=K(t^{1/3})$, $C_L$ is isomorphic to the usual cuspidal cubic ...
- 19.6k
9
votes
Accepted
Extension of line bundle defined over an open subscheme
As explained here Extending vector bundles on a given open subscheme the only possible such extension is
$$
\tilde{L} = (i_*L)^{\vee\vee},
$$
where $i \colon U \to X$ is the embedding. The sheaf $\...
- 39.3k
9
votes
Accepted
Del Pezzo surfaces and Picard-Lefschetz theory
Indeed you can see it this way. This is my symplectic geometer's
perspective on it (I blame Paul Seidel's Lecture notes on four-dimensional Dehn twists).
Consider the $n$-point blow-up of $\mathbf{CP}...
- 7,005
9
votes
Why are ordinary spheres not strictly invertible?
The following solution to the question, which is very close to the one given by Achim, is basically what leads to the entire proof that $spic(\mathbb{S}) \simeq \widehat{\mathbb{Z}}$ (which is proven ...
- 4,189
9
votes
Accepted
A question on "Ample subvarieties of algebraic varieties"
I suspect that you are supposed to view the projective variety $X$ as being given with a chosen projective embedding $X\subset \mathbb P^n$, and therefore a distinguished ample divisor $\mathcal{O}_X(...
- 1,306
9
votes
Accepted
When $R $ is a cusp then $K_0(R) \ncong K_0(R[s])$
Presumably you are looking at the conductor square on the left below, where $\epsilon^2=0$:
$$\matrix{k[t^2,t^3]&\rightarrow& k[t]\cr
\downarrow&&\downarrow\cr
k&\rightarrow &k[...
- 23k
8
votes
What is the right definition of the Picard group of a commutative ring?
1) About the second definition:
$\alpha$) It is not true that for an arbitrary ring a) is equivalent to c):
Indeed Bourbaki in Algèbre commutative, Chapitre II, Exercices §5, 12) c) exhibits a ring ...
- 47.5k
8
votes
Accepted
Picard-surjectivity and Morita-equivalence
Yes, the basic algebra of $A$ will be Picard-surjective.
The basic algebra is the endomorphism algebra $\operatorname{End}_A(\bigoplus_{i=1}^{n}P_i)$ of the direct sum of indecomposable projective (...
- 35.2k
8
votes
Class numbers of functions fields and spanning trees
I think you remembered correctly except that it should be the modular curve $X_0(N)$. Note that $N$ is prime here.
The Eichler-Shimura relation says the eigenvalues of Frobenius on the Tate module of ...
- 148k
7
votes
Picard group of symplectic group modulo orthogonal group
With the suggested choice of the symplectic and orthogonal form, there is a direct sum decomposition of $\mathbb{C}^{2n}$ into the sum of two Lagrangian (with respect to the both forms) subspaces:
$$
...
- 39.3k
7
votes
Accepted
On a morphism from the Brauer group to the Picard group
It seems to me that the involution of $Q \otimes Q$ that exchanges $a \otimes b$ and $b\otimes a$ is inner, which means that the homomorphism you describe should always be trivial.
Here is the ...
- 27k
7
votes
Picard group and reduced schemes
Let $f : X \to Y$ be a universal homeomorphism of schemes. Then as you noted above, the pullback functor on (small) étale sites
$$\begin{eqnarray} Y_{\text{ét}} &\to& X_{\text{ét}}\\
U &\...
- 2,607
7
votes
Accepted
Picard group of a cubic hypersurface
It is cyclic, generated by $\mathscr{O}(1)$. Indeed this is true for $X$ by the Lefschetz theorem (SGA2, Exp. XII, Cor. 3.7), and the restriction map $\operatorname{Pic}(X)\rightarrow \operatorname{...
7
votes
Picard group of a cubic hypersurface
Another way to find $\mathrm{Pic}(X)$ is the following. Note that the cubic $X$ is the symmetric determinantal cubic and it has a resolution of singularities
$$
\tilde{X} = \mathbb{P}_{\mathbb{P}^2}(S^...
- 39.3k
6
votes
Accepted
Why is it useful for the (relative) Picard functor to be representable?
Let me give one cute example: the Theorem of the Cube (cf. e.g. Mumford's "Abelian varieties") can be proved quite easily if we have Picard schemes at our disposal. In contrast, the proof in Mumford's ...
- 16.1k
6
votes
Accepted
Cohomological interpretation of G-equivariant line bundles
See Theorem 4.2.2 in https://www-fourier.ujf-grenoble.fr/~mbrion/lin.pdf
In particular, in your example properness of $X=G/B$ simplifies the left part of the sequence, turning it into $$0\to \hat{G}\...
- 7,367
6
votes
Accepted
Picard group of toric varieties
Edit: I have elaborated on this approach to the Picard group in Section 2 of my preprint.
The question was answered in the comments above, but only for the case of torsion-free Picard group. However, ...
6
votes
Accepted
Picard group of derived category of sheaves
Thanks to Drew Heard's comment I was able to find answers to my questions. In his paper "Picard groups of derived categories" H. Fausk proves the following theorem (see Theorem 4.2).
Theorem: Let $(\...
- 9,870
6
votes
Accepted
Galois invariant Picard group elements
Updated. The example by @Lucifer is completely correct. Thanks to @Count Dracula who explained the example proposed by @Lucifer.
That example is fine. I am keeping the counterexamples below, since ...
6
votes
Examples of smooth projective varieties with "nice" Picard group
Here is a simple example. Let $X = P^1 \times Q^3$ and $H = O(1,1)$. The Hilbert polynomial of the line bundle $O(a,b)$ is
$$
P(t) = (t+a+1)(t+b+1)(t+b+2)(t+b+3/2)/3.
$$
It has three integral roots $-...
- 39.3k
6
votes
Accepted
Picard group modulo codimension 2
The group $G_X$ can be identified with the group of rank 1 reflexive sheaves on $X$ ($F$ is reflexive if the canonical morphism $F \to F^{\vee\vee}$ is an isomorphism) by taking a sheaf $F$ to the ...
- 39.3k
6
votes
Accepted
Are algebras with invertible linear duals always Frobenius?
For a finite dimensional algebra $A$, $A^{\ast}$ being an invertible
bimodule is equivalent to $A$ being self-injective (which is the same
as quasi-Frobenius for finite dimensional algebras).
One ...
- 35.2k
6
votes
Picard group of a cone over the elliptic curve
The restriction map $\operatorname{Pic}(\mathbb{P}^3) \rightarrow \operatorname{Pic}(X)$ is an isomorphism. This is actually true for any cone provided the depth at the vertex is $>1$: see ...
- 38k
6
votes
Accepted
Very general quartic hypersurface in $\mathbb{P}^3$ has Picard group $\mathbb{Z}$
That is the Noether-Lefschetz theorem. Searching online should find plenty of results in web pages and lecture notes. If you want a published source, how about: Mark Green, A new proof of the explicit ...
- 6,237
6
votes
Characterizing principal polarizations of abelian surfaces
The usual way of presenting elements of the Neron-Severi group of an abelian variety is as Hermitian forms on the tangent space at the identity (subject to the condition that the imaginary part of the ...
- 148k
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