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Results tagged with ag.algebraic-geometry
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user 5301
Algebraic varieties, stacks, sheaves, schemes, moduli spaces, complex geometry, quantum cohomology.
2
votes
What is the relation between vector bundles on a manifold and grassmanians?
It is fun to give an answer when there is no question :-)) LOL...
Anyway, pick your vector bundle $V$ on a manifold $X$ and consider its top exterior power $\Lambda^m V$. Now you have Plucker map fr …
- 12.1k
2
votes
Vector bundle and inverse image
For noetherian schemes the pull-back of coherent (quasicoherent, locally free) sheaf retains that property, cf. II.5.8 in Hartshorne. The notion of coherence becomes subtle in non-noetherian case but …
- 12.1k
1
vote
Obstruction for a real algebraic surface to be a complex algebraic curve
If it is orientable, you have a complex structure and the field of meromorphic functions.
Putting my ears into the firing line, I suggest that something should go wrong with the transcendence degree …
- 12.1k
7
votes
2
answers
621
views
$n$-path-connected components of a variety
This question is motivated by the question Path Connectedness of Varieties and some funny little theorem I was trying to prove. Let $X$ be a (quasiprojective smooth connected) algebraic variety over a …
- 12.1k
1
vote
line bundles on smooth affine variety
No, there is no reason for an an $O$-module, even locally free rank 1, to be a $D$-module.
- 12.1k
5
votes
1
answer
756
views
When is the image of a regular map an algebraic variety?
Let $f:X\rightarrow Y$ be a regular map of smooth connected algebraic varieties (say over an algebraically closed field). I know that the image $f(X)$ is only a constructible set, in general, but I am …
- 12.1k
3
votes
What finite group schemes can act freely on a rational function field in one variable?
I guess by looking at it algebraically one can at least rule out the forms of $\mu_p$. Let $H$ be the function algebra of the group scheme, $H^\ast$ its dual. $H^\ast$ is a cocommutative Hopf algebra. …
- 12.1k
2
votes
Accepted
What is a right-handed Dehn twist of a cut curve of a Riemann surface?
Cut the curve with a scalpel, going along the curve (it is oriented), rotate the right side 360 degrees, and glue it back in...
- 12.1k
4
votes
Algebraic geometry
There is no easy way into Algebraic Geometry as this is one of the most imressive mathematical machineries ever created. Make yourself a favour and start reading one of the two mainstream books: Harts …
2
votes
Quasi-coherent envelope of a module
Let us do the case of an affine scheme $X$ first. This is easy. If $M$ is an $O_X$-module, we define $\tilde{M}$ as the quasicoherent $O_X$-module defined by the global sections $M(X)$. Notice that th …
- 12.1k
11
votes
1
answer
316
views
The Locus of Complete Intersection Points
Let $X$ be an algebraic variety over an algebraically closed field. Consider the two subsets $X_0\subseteq X_1 \subseteq X$:
$$X_0 = \{a\in X| a \mbox{ is a scheme-theoretic complete intersection in } …
- 12.1k
3
votes
Prime ideals over the ring of power series over $\mathbb{Z}$
Your $R$ is not a ring. I replace it with $R^\prime$, the ring of all power series, whose convergence radius is at least 1.
Then one of the ideals is distinct. Indeed, take $f(T) = \sum a_k T^k \in I …
- 12.1k
15
votes
5
answers
6k
views
Sheaves without global sections
The line bundle $O(-1)$ on a projective space or $O(-\rho)$ on a flag variety has a property that all its cohomology vanish. Is there a story behind such sheaves?
Here are more precise questions. Let …
- 12.1k
2
votes
One point compactification of the tangent bundle
Did you mean $P^1({\mathbb C})$?
Then, no, because it would a smooth rational projective surface, and we know all of them: blow-ups of $P^2$ or $n$-th Hirzebruch surface.
- 12.1k
1
vote
Complex torus, C^n/Λ versus (C*)^n
The reason why ${\mathbb C}^*$ is called torus is clear by looking at it real points. It could be a split real torus ${\mathbb R}_+^*$ or a compact torus $U_1({\mathbb C}) \cong {\mathbb R}/{\mathbb Z …
- 12.1k
0
votes
Is there a free action on a given variety?
Wild-wild guess it is but hopefully it will send you on the right track: $G=Z/pZ$ is Cartier-self dual. So the action on $V$ is the same as grading on functions $A=C[V]$ for which I presume you know g …
- 12.1k
2
votes
Accepted
Quotient by p-th roots of unity in characteristic p
If you cannot generalize, you may try to simplify:-)) Representations of $\mu_p$ are still completely reducible: they are just graded vector spaces by the cyclic group of order $p$. The first part of …
- 12.1k
21
votes
group scheme neither affine, nor an abelian variety
Yes, take a product of an abelian variety with an affine group scheme...
- 12.1k
5
votes
1
answer
717
views
Are there enough meromorphic functions on a compact analytic manifold?
Let $X$ be a compact complex analytic manifold, $D\subset X$ an irreducible smooth divisor, given as zeroes of a global meromorphic function $f\in {\mathfrak M} (X)$. Are there enough other meromorphi …
- 12.1k
4
votes
0
answers
165
views
Components of variety of subalgebras
This question is motivated by the question Subalgebras of matrices and its answer by Mariano. We consider $X_{n,d}$, the variety of $d$-dimensional subalgebras not necessarily with 1 (with 1 makes sen …
- 12.1k
2
votes
Is the Jordan decomposition for reductive groups algebraic?
I upvoted the two answers in the comments but the fact that you are asking also follows easily from algebraicity of holomorphic representations. Four possible proofs are outlined in the link above. Th …
- 12.1k
1
vote
Replacing Spectrum with Valuations of a Field - An Alternative to Schemes?
You cannot define the variety as the set of valuations. Birational varieties will have the same set of valuations.
As far as I remember, logicians were looking at it for a while. If you have spare 2 …
- 12.1k
5
votes
Geometric (or intuitive) interpretation of additional derivatives in characteristic p > 0
In a nutshell, observe that $\partial_x^p(x^p)=0$ in characteristic $p$. This means that you can divide either $\partial_x^p$ or $x^p$ by $p!$ and get a sensible object, i.e., the second object will i …
- 12.1k
6
votes
0
answers
428
views
Ever seen a ringed group?
A locally ringed space is a common generalization of schemes and various manifolds. I am wondering about a locally ringed group which should be a common generalization of group schemes and various Lie …
- 12.1k
5
votes
Why should the tensor product of $\mathcal{D}_X$-modules over $\mathcal{O}_X$ be a $\mathcal...
There is no natural action of $D$ on $Hom_R(M,N)$ on my carrot patch. Try to act by $x \frac{d}{dx}-\frac{d}{dx}x +1$ on $f$ by your formula and see that it does not act by zero. For this fellow to …
- 12.1k
10
votes
Accepted
Why care about Grothendieck topology?
Etale topology, required to define etale cohomology, is not a topology in the usual sense. It is Grothendieck topology only.
In the category of topological manifolds, an etale cover of $X$ is a surje …
- 12.1k
3
votes
Accepted
The Lie algebra of the subgroup of $GL(n)$ preserving a given variety
Doc, you are right but only amorally. You need to replace the tangent vectors with jets to capture the behavior of your cone.
Let $I(Y)$ be the ideal of zeroes of your $Y$. Then
$$
Lie (G_Y) = \{ X …
- 12.1k
4
votes
1
answer
810
views
2d Weil conjecture
Does there exist a two variable analogue of the Weil conjecture?
What I mean is that the usual Weil involves a one-variable zeta-function which you get by using numbers $V_n = V ( GF(p^n))$ of points …
- 12.1k
3
votes
Accepted
Tannakian fundamental group of automorphic representations
It is not a tannakian category. The issue is the tensor product. Let $V$ and $W$ be automorphic representations. The algebraic tensor product $V\otimes W$ is no longer automorphic. You need some kind …
- 12.1k
5
votes
Lie Algebras and Simple Connectivity for general algebraic groups
No, it does not. The additive $G$ and the multiplicative $H$ groups have isomorphic Lie algebras but only trivial group homomorphisms between them.
The Lie algebra of $G$ has a restricted structure! …
- 12.1k
3
votes
Elementary reference for algebraic groups
If you have gone through Shafarevich, you don't really need it to be "elementary". Go for Jantzen's Representations of Algebraic Groups. Part 1 is groups and no representation theory and Part 2 has as …
- 12.1k
5
votes
Accepted
Faithfully flat modules over a group algebra
No way. Let $G={\mathbb Z}$ so that ${\mathbb Z}[G]={\mathbb Z}[x,x^{-1}]$. Then use the complex
$$\ldots \rightarrow 0 \rightarrow 0 \rightarrow {\mathbb Z}[x,x^{-1}] \xrightarrow{1-x+x^2} {\mathbb Z …
- 12.1k