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Dec
17 |
awarded | Yearling |
Dec
7 |
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Support of 0-dimensional sheaf and its dual
Note that the assumptions are too strong: the only relevant condition is that $E$ is Cohen-Macaulay (so that $Ext^i(E,\omega_R)=0$ for all but one value of $i$) and the same argument shows that the support of $E$ and the support of $E^D$ are equal as subschemes. (Here $E^D$ is the non-zero $Ext^i(E,\omega_R)$.) |
Nov
23 |
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Definition of étale (etc) for non-representable morphisms of algebraic stacks?
I would assume that an etale morphism of stacks $X\to Y$ is required to be `relative DM'; that is, that any base change from $Y$ to a scheme gives a DM stack. (Obviously, this is my opinion, but I don't see any other well-behaved definition.) |
Nov
23 |
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Is there a unique line bundle in the Kummer surface which pulls back to a totally symmetric line bundle?
First of all, do you want L' to pull back to L^2? If so then the answer to both of your questions (with L replaced by L^2 throughout) is yes, basically by a kind of descent (which works even though X->Y is not flat). |
Nov
10 |
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Vector bundles with symmetric perfect form
The `problem' with taking square roots is supposed to happen, because O(n) bundles are only locally trivial in etale topology, and taking square roots gives you an etale cover. It is too much to expect transition functions in the Zariski topology here. |
Oct
21 |
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Do all simple factors of jacobians of curves come from correspondences?
@Maarten Derickx: No, I don't, but it does not seem particularly hard. Fix base points $e\in E$, $c\in C$, we need to show that for any line bundle $L$ on $E\times C$, there exist $n_1$ and $n_2$ such that the line bundle $L(n_1(\{e\}\times C)+n_2(E\times \{c\}))$ is of the form $O(D)$ for smooth curve $D\subset E\times C$ (which then gives a correspondence between the two curves). However, $O((\{e\}\times C)+(E\times \{c\}))$ is ample, therefore, $L(n(\{e\}\times C)+n(E\times \{c\}))$ is very ample for $n\gg 0$, and the claim follows from Bertini's Theorem. |
Oct
20 |
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Do all simple factors of jacobians of curves come from correspondences?
P.S. A reference for the second fact: Theorem III.10.1 of Milne's jmilne.org/math/CourseNotes/AV.pdf (the section has the telling title `Abelian varieties are quotients of Jacobian varieties :) And the first fact should follow from Bertini's Theorem... |
Oct
20 |
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Do all simple factors of jacobians of curves come from correspondences?
It does seem pretty plausible. First, any morphism $J(E)\to J(C)$ comes from a line bundle on $C\times E$, which in turns comes from a divisor in $C\times E$... Which is kind of a curve (probably singular, but we can resolve this, also probably reducible, which is more annoying). Secondly, for any abelian variety $A$, we can find a curve $E\subset A$ that generates it (again, the obvious construction gives a reducible curve, so some argument would be required?), and then $A$ becomes a quotient of $J(E)$. Combining these two facts we'd get the required claim. |
Oct
20 |
answered | Invertible combinations of linear maps on infinite-dimensional vector spaces |
Oct
15 |
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Showing that closure of all lines through a projective variety $Y$ has degree strictly less than $Y$
Just take a generic L: avoid singularities of X and make sure L is transversal to smooth locus. (I don't have Hartshorne's book at hand, so I am not sure what would be the best reference.) |
Oct
15 |
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Showing that closure of all lines through a projective variety $Y$ has degree strictly less than $Y$
What I mean is, don't draw a plane through P. Draw a subspace L of dimension complementary to dimension of X which meets X at finitely many points (different from P). Now look at the subspace spanned by L and P; this will intersect X at a bunch of lines. How does it meet Y? |
Oct
15 |
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Showing that closure of all lines through a projective variety $Y$ has degree strictly less than $Y$
Have you tried going the other way? Choose a subspace L that is in generic position to X, and then see how the subspace spanned by L and P meets Y? |
Oct
14 |
reviewed | Approve Generalization for Leray Hirsch theorem for Principal $G$-bundle |
Oct
13 |
answered | Isotropic subspaces in a symplectic vectorspace over $GF(q)$ |
Oct
12 |
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Why should curves be two-dimensional?
Of course, one can argue that when you look at $\pi_1(X)$, this means you are already looking at etale topology :) |
Oct
12 |
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When does a “universal” quot scheme exist?
Generally speaking, points of $M$ parametrize S-equivalence classes of sheaves. If $E_1$ and $E_2$ are S-equivalent, there does not seem to be any reason for an isomorphism $Quot(E_1)\simeq Quot(E_2)$. So in this case, your $Q$ probably does not exist. On the other hand, if you restrict to stable locus (where no S-equivalence happens) the universal family exists locally in etale topology on $M$, so $Q(v)$ exists as an algebraic space. Whether it happens to be a scheme is another question. |
Oct
12 |
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Lie algebra and base change
On the other hand, if you want your Lie algebra to be a scheme over $\mathbb Z_p$, it is supposed to commute with the base change, but there is no contradiction: the Lie algebra of $\mathbb{G}m$ is the constant scheme with one-dimensional fiber, and it contains a non-flat subscheme sitting over the closed point. |
Oct
12 |
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Lie algebra and base change
Well, what happens depends on the kind of object you want the Lie algebra to be. Most of your question is written for `naive' object, where you expect the Lie algebra to be a $\mathbb{Z}_p$-module. In this setup, the formation of Lie algebra does not commute with base change (unless the group is smooth). |
Oct
11 |
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Twisting locally free sheaves in characteristic $p$
The point is, even if $p|rk(E)$, the map $L'\mapsto L'^{\otimes rk(E)}$ is surjective on $\ker(Pic(X)→Pic(\tilde X))$, because the kernel is isomorphic to $(\mathbb{G}m)^k$ (the curve is nodal), and the map is surjective on $\mathbb{G}m$. So you can find $L'$ such that $L'^{\otimes rk(E)}\simeq L\otimes(\det E)^{-1}$. |
Oct
11 |
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Twisting locally free sheaves in characteristic $p$
Sorry, didn't notice you want your curve nodal. Then of course the answer is positive: $ker(Pic(X)\to Pic(\tilde X))$ is a torus, and the power map is surjective. |