bio | website | mate.dm.uba.ar/~aldoc9 |
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location | Buenos Aires | |
age | 40 | |
visits | member for | 4 years, 11 months |
seen | 20 hours ago | |
stats | profile views | 18,826 |
Oct 22 |
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Spectral sequence and HOM functor
Not really. If M has finite projective dimension, then convergence of the spectral sequence coming from the filtration by columns is immediate; the first oage, which you get by taking homology wrt the differential of Q is easy to get, because each P_i is projective, and the next one is just the complex you usebto compute the ext from P. |
Oct 22 |
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Does Schur's Lemma hold in this case? Regular representations of $S_n$ over $\mathbb R$
Is there a way to prove absolute irreducibility without having to find all irreps? |
Oct 22 |
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Spectral sequence and HOM functor
Provided it converges (for example, if M has finite projective dimension, but it may converge always), it converges to Ext_A(M,N) in two steps. |
Oct 20 |
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Is every polynomial ring over any field regular?
@Alex, the Koszul complex for a regular sequence is always exact ---whether the ring is local or not. |
Oct 20 |
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Representations of the two dimensional non-abelian Lie algebra
An alternative argument for Vladimir's observation is that Lie's theorem implies that the only simple finite dimensional modules are one dimensional, together with the immediate fact that $y$ acts by zero on any one-dimensional module. |
Oct 20 |
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Is every polynomial ring over any field regular?
There is no need to consider B, in that it is difficult to provr that B has finite global dimension without proving atmthe same time that A has finite global dimension :-). A itself has finite global dimension: this follows at once from the fact that its projective dimension as a bimodule over itself is finite, and this last can be seen by looking at a Koszul complex. |
Oct 19 |
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Does a classification of simultaneous conjugacy classes in a product of symmetric groups exist?
When $d$ is $1$ the number of conjugacy classes grows like $n^{-1}\exp(n^{1/2})$ (with various scalars I am not writing) so you cannot enumerate them in polynomial time in $n$. |
Oct 19 |
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A metric on $S^{2}$
Isn't the equivalence you ask about an easy exercise? Also, the map $p$ is equivariant with the respect to the action of unit quaternions, so your metric on the sphere is invariant under the orthogonal group. That answers your questions, I think. |
Oct 19 |
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Representations of the two dimensional non-abelian Lie algebra
@VladimirDotsenko, Ah, that's it. |
Oct 19 |
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Representations of the two dimensional non-abelian Lie algebra
(I heard a talk by an expert saying that very recently this has been done for sl2; I'll ask for a reference to check if I remember correctly) |
Oct 18 |
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A functorial isomorphism in derived category
If you contruct Rf(A) by constructing functorially a Cartan-Eilenberg resolution and using it to compute, one of the hypercohomology spectral sequences gives you those isomorphisms naturally, no? |
Oct 17 |
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Can I find a resolution of singularities that is both smooth and projective?
Don't you need $X$ to be proper over $\operatorname{Spec}k$ for that? |
Oct 17 |
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Can I find a resolution of singularities that is both smooth and projective?
If X is not projective, then finding a surjective map to it from a projective varierty is not easy. |
Oct 16 |
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Poincare duality for (co)homology of Lie algebras?
Have tou tried computing the traces in your two examples? There is no difficulty in doing that, really. |
Oct 16 |
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Poincare duality for (co)homology of Lie algebras?
A Lie algebra $g$ is unimodular if $\operatorname{ad}_X$ has zero trace for all $X\in g$. |
Oct 16 |
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Poincare duality for (co)homology of Lie algebras?
For a Lie algebra free over a commutative base ring, I'd guess exactly the same thing holds provided one is talking about Lie algebra cohomology relative to the base ring. |
Oct 16 |
answered | Poincare duality for (co)homology of Lie algebras? |
Oct 16 |
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Poincare duality for (co)homology of Lie algebras?
@VladimirDotsenko, the OP does consider only Lie algebras free over the base ring. |
Oct 15 |
awarded | Sheriff |
Oct 11 |
awarded | Necromancer |