bio | website | mate.dm.uba.ar/~aldoc9 |
---|---|---|
location | Buenos Aires | |
age | 41 | |
visits | member for | 5 years, 9 months |
seen | 14 hours ago | |
stats | profile views | 20,361 |
Jul 26 |
awarded | Good Answer |
Jul 18 |
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Commutative algebra books representing the edge of research
You should subscribe to arXiv announcements on that subject and, if you have access, browse MathSciNet. |
Jul 16 |
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Limits of complex projective varieties in the Hausdorff topology on closed subsets of CP^n
Ah. I am sorry. I thought you might be trying to be helpful. My mistake. |
Jul 16 |
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Limits of complex projective varieties in the Hausdorff topology on closed subsets of CP^n
But homogeneous polynomials do not "take values" on points of projective space. |
Jul 10 |
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Lie algebra cohomology with values in the ring of smooth functions of a $G$-manifold
The original version looked intriguing :-) |
Jul 10 |
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Lie algebra cohomology with values in the ring of smooth functions of a $G$-manifold
Your cocycles are just smooth maps? They are usually linear. |
Jul 6 |
awarded | Guru |
Jun 23 |
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Do representations of the universal enveloping algebra $\mathrm{U}\mathfrak{su}_2$ retain the Hopf algebra structure?
In what way is your $U_J\mathfrak{su}_2$ an algebra? |
Jun 23 |
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Do representations of the universal enveloping algebra $\mathrm{U}\mathfrak{su}_2$ retain the Hopf algebra structure?
There are no Hopf algebras which are simple as algebras (except the ground field, of course), precisely because of the existence of the counit. On the other hand, there is no finite dimensional Hopf algebra over a field of characteristic zero with a non-zero primitive element. |
May 28 |
awarded | Necromancer |
May 27 |
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Example of a $G$-sphere that is not a $G$-representation sphere
@QiaochuYuan, can a finite group not fix a smooth structure? |
May 27 |
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The construction of the 257gon
@FranzLemmermeyer, thanks! Do you know what the tables appearing in pictures 8 to 4 (counting from the last) are? |
May 26 |
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What is the exterior derivative intuitively?
@FallenApart, ah. No , I do not mean that $\Omega^1(M)$ is the module of Kähler differentials of $C^\infty(M)$ (mostly, because it isn't! :) ) The operator $d:C^\infty(M)\to\Omega^1(M)$ can be characterized in terms of its functorial properties. This is surely done in detail in the book Natural Operations in Differential Geometry by Kolar, Michor and Slovak. |
May 26 |
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What is the exterior derivative intuitively?
@FallenApart, I don't understand exactly what statement you mean. |
May 14 |
awarded | Good Answer |
May 13 |
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Homotopy type of embeddings of circle in the plane
My guess is because the group of self-homeos of the circle has contractible identity component (a homeo lifts to a map $\mathbb R\to\mathbb R$, and the latter is a strictly increasing or decreasing function which you can deform to a linear one) |
May 13 |
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Homotopy type of embeddings of circle in the plane
What do you mean by embedding,exxactly? If the maps are injective then you have only two contractible components, no? |
May 3 |
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Singular projective variety where the Cartan homomorphism is not an isomorphism?
The Hilbert series in that situation converges to a rational function which can be evaluated at everything which is not a pole. |
May 3 |
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Singular projective variety where the Cartan homomorphism is not an isomorphism?
You talk about a linked question in your question but there is no link? |
Apr 23 |
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When does Vopěnka's principle hold?
What is $0^\sharp$? :-| |