30 votes
Accepted

Embedding abelian varieties into projective spaces of small dimension

Recall that any smooth projective variety of dimension $g$ embeds into $\mathbf{P}^{2g+1}$. Consider now an abelian variety $A$ of dimension $g$ which embeds into $\mathbf{P}^{2g}$. Van de Ven proves (...
ssx's user avatar
  • 2,729
23 votes
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Square root of the determinant line

There is no such isomorphism (at least for $g \geq 9$). In O. Randal-Williams, The Picard group of the moduli space of r-Spin Riemann surfaces. Advances in Mathematics 231 (1) (2012) 482-515. I ...
Oscar Randal-Williams's user avatar
13 votes

Classification of line bundles by second cohomology of a manifold

For what it is worth, here is another approach, in the algebraic geometry style. By taking out the zero section, complex line bundles correspond bijectively to $\mathbb{C}^*\!$-principal bundles on $M$...
abx's user avatar
  • 37.3k
12 votes

What is the Theorem of the Cube?

I'm a bit late to the party, but since these question are clearly still getting views, I'll answer the second question a bit. Given a nice category, one can form the pointed category $C$, and consider ...
Pax's user avatar
  • 831
12 votes

Square root of the determinant line

I'm not sure what you mean by 'canonical'. For example, when $\Sigma$ has genus $1$, there are 4 distinct spin structures, one representing the trivial line bundle, whose space of holomorphic ...
Robert Bryant's user avatar
11 votes
Accepted

One-point compactification of ample line bundle

This is EGA 2, Prop. 8.8.2. It basically says that if $L$ is ample then one can contract the zero section of the geometric realization $\mathbb V(L)$ of $L$ to a point. The result is called the affine ...
Friedrich Knop's user avatar
10 votes
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Classification of line bundles by second cohomology of a manifold

I think $H^2(M;\mathbb{Z})$ cannot mean the de Rham cohomology group. The coefficients are wrong. Anyway: $\mathbb{CP}^\infty$ is an amazing space. It is both a model for $K(\mathbb{Z},2)$ and a ...
Thomas Rot's user avatar
  • 7,373
9 votes

Is a torsion free sheave of rank one on a reducible curve the pushforward of a line bundle on a normalization?

This is (exactly as stated in your question) for example in Proposition 10.1 of Oda, Tadao; Seshadri, C. S. Compactifications of the generalized Jacobian variety. Trans. Amer. Math. Soc. 253 (1979), ...
Reimundo Heluani's user avatar
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(...
Tom Ducat's user avatar
  • 1,276
8 votes
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The existence of the extension of a non-trivial line bundle

This is a bordism problem, and as such can be answered using algebraic topology. I'll answer in the unoriented setting, then indicate how to modify things if $M$ and $W$ are required to be oriented. ...
Mark Grant's user avatar
8 votes
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Compact complex non-Kähler manifolds with nef canonical bundle

Let $X$ and $Y$ be compact complex manifolds. Note that $K_{X\times Y} \cong \pi_1^*K_X\otimes \pi_2^*K_Y$. If $Y$ has trivial canonical bundle, then $K_{X\times Y} \cong \pi_1^*K_X$. Now the pullback ...
Michael Albanese's user avatar
7 votes
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The kernel of a nef line bundle

Consider $V=\mathbb{P}^1\times \mathbb{P}^2$ with projections $p_1\colon V \rightarrow \mathbb{P}^1 \text{ and } p_2\colon V \rightarrow\mathbb{P}^2.$ Let $L = p_1^*(\mathcal{O}_{\mathbb{P}^1}(1))$, ...
Chiles's user avatar
  • 86
7 votes
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Pull-back of the canonical divisor via a rational map

There are several issues with this question. One issue is that if $f$ is a rational map and not a morphism, then you have to say what you mean by $f^*$. Another issue is that if $K_Y$ is not at least ...
Sándor Kovács's user avatar
7 votes
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Global choice of eigenvectors on an open surface

Not necessarily. To construct a counter-example, start from the other direction. Suppose that the tangent bundle of $M$ can be split as the direct sum $TM = L_1\oplus L_2$ where $L_1$ and $L_2$ are ...
Robert Bryant's user avatar
7 votes

Question regarding the definition of linearization of line bundles

I think that you should regard the first definition as an imprecise version of the second definition. For example, suppose that $ X $ is a point and so $ L $ is simply a 1-dimensional vector space. ...
Joel Kamnitzer's user avatar
6 votes
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Universal covering of symmetric product

In fact, the universal cover of $C^{(n)}$ will not be $\mathcal H^n$ once $n \gg 0$. Indeed, if $C$ is a compact Riemann surface of genus $g \geq 2$ (so the universal cover is $\mathcal H$) with a ...
R. van Dobben de Bruyn's user avatar
5 votes
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Lifting line bundles

This cannot always be done. If $X$ is a supersingular K3 surface then there are $22$ ample line bundles $L$ that give independent classes; if every $L^{\otimes p}$ lifted to char. zero, then you'd ...
inkspot's user avatar
  • 3,092
5 votes

Classification of line bundles by second cohomology of a manifold

Although Milnor and Stasheff is an excellent suggestion, I was also led to wonder where one would find this result in more recent textbooks. Most ingredients are in May's "Concise introduction to ...
Neil Strickland's user avatar
5 votes
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Existence of a connection $A$ on a holomorphic line bundle $L$, s.t $F(A)=(\deg L)\omega$

Let me expand a bit Henri's (hi Henri!) answer, even if this is completely standard. In general, given a compact Kähler manifold $X$ of any dimension, given a holomorphic line bundle $L\to X$, and ...
diverietti's user avatar
  • 7,862
5 votes
Accepted

Line bundles trivial outside of codimension 3

This will hold for CW structures on manifolds coming from handle decompositions, e.g. induced by a Morse function. Complex line bundles on $X$ are classified by the homotopy class of maps to $\mathbb{...
Ian Agol's user avatar
  • 66.9k
5 votes

What are meromorphic line bundles?

You can define "meromorphic vector bundle" as locally free sheaf of modules over a sheaf of meromorphic functions. This is a highly non-trivial object, because (in contrast with rational ...
Misha Verbitsky's user avatar
4 votes
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What is known about the cohomology of the relative tangent bundle on a conic bundle?

Such conic bundle is given by a rank three vector bundle, say $E$ on $X$, and a line subbundle $L \subset Sym^2E$ (just take $E$ to be the pushforward of $\omega^{-1}_\pi$, and $L$ corresponds to the ...
Sasha's user avatar
  • 37.3k
4 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, ...
Justus Springer's user avatar
4 votes

Existence of a connection $A$ on a holomorphic line bundle $L$, s.t $F(A)=(\deg L)\omega$

Let $h_0$ be any hermitian metric on $L$, with curvature $F(h_0)$. By Hodge theory, there exists a function $f\in \mathcal C^{\infty}(X)$ such that $$\Delta_{\omega} f =\Lambda_{\omega} F(h_0) - \...
Henri's user avatar
  • 2,697
4 votes
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Weights on the linearization

If your $\mathbb{P}^2$ is $\mathbb{P}(\mathbb{C}^3)$, you can identify the complement of the zero section in $L^{-1}$ with $\mathbb{C}^3\smallsetminus 0$, viewed as a bundle over $\mathbb{P}^2$ via ...
abx's user avatar
  • 37.3k
4 votes
Accepted

The set of isomorphism classes of Z/nZ-equivariant line bundles over a 2 dimensional Z/nZ-CW complex

You can replace $GL_1(\mathbb C)$ with its maximal compact subgroup, which is $S^1$. Since $S^1$ is an abelian compact Lie group, there is a natural $\mathbb Z/n$-equivariant equivalence $$B_{\mathbb ...
Gregory Arone's user avatar
4 votes

Differential refinement of homology

Differential cohomology groups are computed as homotopy groups $\hat{\rm H}^n(M)=π_0(F_n(M))$, where $F_n\colon{\sf Man}^{\rm op}\to{\sf Sp}$ is a sheaf of spectra on the site of smooth manifolds. (...
Dmitri Pavlov's user avatar
4 votes
Accepted

Galois invariant line bundle and base change

I am posting my comment as an answer. This result is discussed in many sources. I do not have Serre's "Galois cohomology" with me at this moment, but I am certain that it is discussed ...
4 votes
Accepted

Extension of first order deformations of a line bundle

Under some conditions on $X,V$, your line bundle can be extended to $X_{\varepsilon}$. Indeed, let $\imath_X:X\hookrightarrow X_{\varepsilon}$ and $\imath_V:V\hookrightarrow V_{\varepsilon}$ be two ...
nariri's user avatar
  • 350

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