bio | website | math.sunysb.edu/~jstarr/… |
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location | Stony Brook University, Stony Brook, NY USA | |
age | ||
visits | member for | 4 years, 11 months |
seen | 2 hours ago | |
stats | profile views | 3,578 |
Sep 14 |
comment |
What are the exact holomorphic Lagrangians in complex 2-space?
To add to Robert Bryant's point: are you asking that the one-form be exact in the category of holomorphic functions, or in the category of polynomial functions? If you use the category of holomorphic functions, there are tons of examples because of the Poincar'e lemma, e.g., all algebraic embeddings of $\mathbb{C}$ in $\mathbb{C}^2$. |
Sep 10 |
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Does there exist a Fano variety with torsion in $H^3$?
Oops, I spoke too soon, and I definitely should have known better! If the complete intersection in $\mathbb{P}V//G$ is disjoint from the set with nontrivial stabilizier, i.e., the singular set, then its inverse image in $\mathbb{P}V$ would be a finite, 'etale cover. But (by Campana or by Debarre-Koll'ar), the fundamental group of every Fano manifold (or rationally connected variety) is trivial. So the construction I suggest does not work. |
Sep 10 |
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Does there exist a Fano variety with torsion in $H^3$?
Actually, I believe Daniel Loughran's point could lead to examples. First of all, I believe that there are examples of a finite group $G$ such that $H^3(BG)$ has nonvanishing torsion. Thus, for a "sufficiently large", "sufficiently faithful" linear representation of $G$ on $\mathbb{P}V$, the GIT quotient $\mathbb{P}V//G$ will have nonvanishing torsion in $H^3$. Of course this quotient is singular. But for an appropriate complete intersection, assuming $V$ is "very large", the complete intersection might be Fano and disjoint from the singular set. |
Sep 10 |
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Does there exist a Fano variety with torsion in $H^3$?
@DanielLoughran: Yes, but I believe the group in that case is disconnected. In Corti's description, the group is a product of copies of $\textbf{GL}_n$, for various $n$. |
Sep 10 |
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Does there exist a Fano variety with torsion in $H^3$?
This is not really an explanation, but Alessio Corti shows that all Fano 3folds can be obtained as the zero schemes of invariant sections of an equivariant vector bundle on a certain GIT quotient of affine space. It may be that all varieties constructed in this way have vanishing torsion in $H^3$. |
Sep 9 |
answered | are K3 surfaces complete intersections in their polarization? |
Sep 9 |
comment |
Is the cotangent complexes of groupoids bounded above by degree $1$?
The standard references for cotangent complexes of stacks are Laumon - Moret-Bailly and Olsson. |
Sep 8 |
comment |
Understanding a proof of a lemma in elliptic surfaces
What paper? What page? |
Sep 8 |
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The locus of rational/elliptic curves on a special surface in $\mathbb{P}^3$
"So if we assume this conjecture it is enough to prove ..." Actually, you are only using the direction that is known (and trivial). Every nonconstant rational map from an Abelian variety induces (many) entire curves whose images sweep out the image of the Abelian variety. So the union of (nonconstant) images of Abelian varieties is contained in the union of (nonconstant) images of entire maps. |
Sep 6 |
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Flat Connections on the Cotangent Complex
I recommend you look up "Atiyah extension" in Illusie's first volume. There is a definition using the cotangent complex in place of the sheaf of relative differentials (as in the "usual" definition). |
Sep 1 |
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excess intersection theory
Well, because it is what I know (not because I think it is a likely approach), I would specialize to the normal cone of the locus parameterizing "large intersections". For every point $p=(p_1,\dots,p_k)$ in $X = (\mathbb{P}^1)^k$, there is a codimension $k+1$ linear section $X\cap L_p$ that is a union $\cup_{i<j} Z_{i,j}$, where $Z_{i,j}$ parameterizes points $(q_1,\dots,q_n)$ such that $(q_i,q_j)$ equals $(p_i,p_j)$. Inside the Grassmannian parameterizing linear subspaces, I would consider the deformation to the copy of $X$ parameterizing these subspaces $L_p$. |
Aug 31 |
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excess intersection theory
For your variety, the number of points contained in $\Sigma$ can vary from $1$ up to, at least, $2^k-k-1$. So I do not see how an excess intersection theory computation will lead to so many different answers. Do you have some extra information about the hyperplane sections in your case of interest? |
Aug 31 |
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excess intersection theory
I doubt it. What can you tell us about your variety $X$? Is it linearly nondegenerate? |
Aug 29 |
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Complex conjugation of positive roots
Is your torus compact? Is that what is going on? |
Aug 29 |
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Complex conjugation of positive roots
What if $G$ is split and $T$ is a split torus? |
Aug 27 |
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linear section of codimension $k+1$ of a variety of dimension $k$
You should at least add the hypothesis that the span of $X$ is not a linear space of codimension $k-1$. Otherwise your bound is wrong. |
Aug 25 |
revised |
Weil restriction
added 20 characters in body |
Aug 25 |
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Weil restriction
It is still false, for instance, if $k$ equals $\mathbb{Q}$, if $X$ equals $\text{Spec}(\mathbb{Q})$, if $Y$ equals $\text{Spec}(\mathbb{Q}[i])$, and if $G$ equals $\mathbb{G}_m$. Since you can get the previous counterexample from this one by basechange, there cannot be an isomorphism. |
Aug 25 |
answered | Weil restriction |
Aug 24 |
comment |
Characteristic polynomials of reductive subgroup over C
For one direction, presumably it would suffice to know that every maximal torus of $H$ extends to a maximal torus of all of $\textrm{SL}(n,\mathbb{C})$, since you already know the morphism is finite on maximal tori of $\textrm{SL}(n,\mathbb{C})$. |