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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 (essentially by applying the self-intersection formula to the normal bundle of $A$ in $\mathbf{P}^{2g}$) that the degree of $A$ in $\mathbf{P}^{2g}$ is given by ...

answered Jan 21 '19 at 10:10

The Sand Reckoner

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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 computed the Picard groups of moduli spaces of Spin Riemann surfaces (for $g \geq 9$). Grothendieck--Riemann--Roch shows that, in the notation of that paper, the ...

answered Oct 10 '17 at 11:23

Oscar Randal-Williams

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Let me expand my comment in a short answer.
One of the most common way to build a rational map $f \colon X \dashrightarrow \mathbb{P}(a_1, \ldots, a_n)$ is to consider a line bundle $\mathscr{L}$ on $X$ together with sections $$\sigma_1 \in H^0(X, \, \mathscr{L}^{a_1}), \quad \sigma_2 \in H^0(X, \, \mathscr{L}^{a_2}), \ldots, \sigma_n \in H^0(X, \, \...

answered Jul 12 '14 at 13:49

Francesco Polizzi

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12

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 sections has dimension $1$, and three nontrivial, whose spaces of holomorphic sections have dimension $0$. If the above isomorphism were 'canonical' in the three ...

answered Oct 10 '17 at 11:27

Robert Bryant

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11

It has been proved by Thomsen [Thomsen J. F.. “Frobenius direct images of line bundles on toric varieties” Journal of Algebra 226, no. 2 (2000)] that such a push-forward is a direct sum of line bundles. To be precise, his theorem applies to the Frobenius map on a smooth toric variety in characteristic $p$, but the same result holds for the "toric ...

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 ...

10

The answers are the following.
(1) It is well known that the singularity at the vertex of the cone over the Veronese surface is isomorphic to a quotient singularity of type $\frac{1}{2}(1, \, 1, \,1)$, that is the isolated singularity given by the quotient of $\mathbb{C}^3$ by the action of $\mathbb{Z}/2 \mathbb{Z}$ of the form $(x, \,y, \,z) \mapsto (-x, \,...

answered Dec 30 '14 at 15:38

Francesco Polizzi

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By Corollary 5.6.1 here
https://arxiv.org/pdf/1712.05564.pdf
if $\text{deg}(\Delta)\leq 4$ then $X$ is rational.
If $\text{deg}(\Delta) = 5$ then $X$ could be rational or not depending on whether the double cover $\widetilde{\Delta}\rightarrow\Delta$ is defined by an even or an odd theta characteristic. For instace, by blowing-up a line in a smooth cubic $3$-...

9

Complex line bundles over a manifold are classified by their first Chern class; we have a bijection
$$\{\text{isomorphism classes of complex line bundles on $X$}\} \leftrightarrow H^2(X; \Bbb Z),$$
$$L \mapsto c_1(L).$$
The first Chern class is additive with respect to tensor products, so we see that
$$c_1(L^{\otimes 2}) = 2c_1(L). \tag{$\ast$}$$
Now if $K$ ...

ag.algebraic-geometry at.algebraic-topology dg.differential-geometry mp.mathematical-physics line-bundles

answered Nov 19 '13 at 18:48

Henry T. Horton

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9

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 functors $F:C\to Ab$. There are cannonical maps $\beta:F(X_0\times\dots \times X_n)\to \prod_i F(X_0\times \dots \times X_{i-1}\times X_{i+1}\times \dots \...

9

Let me start being a little nitpicking with the formulation of the question. The fact that $X$ is $\mathbb Q$-factorial does not in itself imply that such $a$ and $b$ exists. One also needs the fact that the Picard number of $X$ is $1$. This is indeed true, but perhaps should be mentioned. In fact, the Picard group of $X$ is $\mathbb Z$ which was silently ...

9

There are holomorphic line bundles over a compact Riemann surface $X$ that are topologically trivial, yet not holomorphically trivial. To see this, note that smooth complex line bundles are classified by a complete invariant, called the degree. By contrast, we have the Picard group $Pic(X)$ of isomorphism classes of holomorphic line bundles on $X$.
One ...

9

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), 1–90.
Although as the idea (by Mumford) of compactifying the Jacobian by torsion free sheaves is older, this can surely be found explicitly in older works, ...

9

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 model of $BU(1)$. Homotopy classes into $K(\mathbb Z,2)$ is in bijection with $H^2(M;\mathbb{Z})$ (By pulling back the fundamental class) and homotopy classes into ...

9

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 $M$. Such bundles are classified by the cohomology group $H^1(M, \mathcal{O}_M^*)$, where $\mathcal{O}_M$ is the sheaf of $\mathcal{C}^{\infty}$ complex-valued ...

8

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 $\mathbb Q$-Cartier, then $f^*K_Y$ doesn't make sense even if $f$ is a morphism. (More precisely, there is no good way to define $f^*K_Y$. In other words, $f^*$...

answered Oct 25 '16 at 19:03

Sándor Kovács

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7

Here is a proof of $d = \frac{1}{2}$. Then you can argue as Francesco did in point two of his answer.
Consider the action:
$$
\begin{array}{ccc}
\mu_{2}\times\mathbb{A}^{3} & \longrightarrow & \mathbb{A}^{3}\\
(\epsilon,x_{1}, x_{2}, x_3) & \longmapsto & (\epsilon x_{1},\epsilon x_{2}, \epsilon x_3)
\end{array}
$$
The ring of invariants is ...

7

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 $L = p_1^*(\mathcal{O}_{\mathbb{P}^1}(1))$, let $X = p_2^{-1}(\ell_1)$ and $Y=p_2^{-1}(\ell_2)$ for two distinct lines $\ell_1$ and $\ell_2\subset \mathbb{P}^2$. Then $Z = X\cap Y \cong \mathbb{P}^1$ is a ...

7

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.
Complex line bundles $\mathcal{L}$ over $W$ are classified by maps $f:W\to BU(1)\simeq \mathbb{C}P^{\infty}$. We want to decide if there is a $4$-manifold $M$ ...

5

1) Take $X=\mathbb{P}^1\times \mathbb{P}^1$, which satisfies $\mathrm{Pic}(X)=\mathbb{Z}f_1+\mathbb{Z} f_2$, where $f_1,f_2$ are fibres of the two projections. Choose $D=2f_1$. Then $\lvert D\rvert$ is movable but any element corresponds to the union of two fibres (since $D\cdot f_1=0$, it is contained in fibres).
EDIT: I do not know if this example is what ...

5

No in general, for instance if $K$ is a number field : assuming that $C$ has a rational point, the group $\mathrm{Pic}(C)$ is naturally isomorphic to $JC(K)$, the group of rational points of the Jacobian of $C$, which is finitely generated by the Mordell-Weil theorem. So $\mathrm{Pic}(C)$ may very well contain points which are non-torsion and non-divisible ...

5

The answer is 'yes'. One way to argue this is to first find a Cartier divisor $D$ on $U$ whose associated line bundle is $\mathcal{L}$ (the existence of such a divisor is ensured, for instance, by [EGA IV$_4$, 21.3.4 a)]), extend $D$ to a Cartier divisor $\widetilde{D}$ on the whole $X$ (e.g., by applying [EGA IV$_4$, 21.9.4]), and then let $\widetilde{\...

answered Apr 30 '15 at 21:16

Kestutis Cesnavicius

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5

A more direct approach is the following. Let $U=U_0\cup \dots \cup U_r$ be an open cover of $U$ such that $\mathscr L\left|_{U_i}\right.\simeq \mathscr O_{U_i}$ for all $i=0,\dots,r$. Define $X_0:=U_0\cup Z$, $X_i=U_i$ for $i>0$ and let $\overline{\mathscr L_i}:= \mathscr O_{X_i}$. Now glue $\overline{\mathscr L_i}:= \mathscr O_{X_i}$ together by the ...

5

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 have a K3 surface in char. zero with $22$ independent line bundles, which is impossible. (Or $X$ could be any liftable unirational surface with $p_g(X)>0$.)

5

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 algebraic topology". Specifically, May proves on page 177 that $H^2(X;\mathbb{Z})\simeq [X,K(\mathbb{Z},2)]$. Here $K(\mathbb{Z},2)$ is officially defined as $B^...

answered Aug 17 '18 at 11:04

Neil Strickland

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5

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 $\mathbb{CP}^{\infty}$, by pulling back the canonical line bundle $\gamma(\mathbb{CP}^\infty)$. Also, $\mathbb{CP}^\infty = K(\mathbb{Z},2)$, so $\pi_i(\mathbb{CP}^\...

5

That is true. Just note that
$$deg(x_1^{<m,u_1>+a_1}\cdots x_k^{<m,u_k>+a_k})$$
is the class in $Cl(X_{\Sigma})$ of the divisor
$$\sum_{\rho} (<m,u_\rho>+a_\rho)D_\rho = \sum_{\rho}<m,u_\rho>D_\rho + \sum_{\rho}a_{\rho} D_\rho \sim div(\chi^m)+D\sim D$$

4

Just so I can use this later, let me start by saying that obviously $X$ has to be irreducible for this to be interesting.
EDIT: In an earlier version of this answer I ruminated on what a movable divisor might mean as it seems to conflict with the definition of a movable curve. Following Mark's advice in the comments (thanks!) I unearthed a definition in a ...

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