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bio website mate.dm.uba.ar/~aldoc9
location Buenos Aires
age 40
visits member for 4 years, 11 months
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13h
comment 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.
13h
comment 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?
13h
comment 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.
2d
comment 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.
2d
comment 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
comment 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
comment 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
comment 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
comment Representations of the two dimensional non-abelian Lie algebra
@VladimirDotsenko, Ah, that's it.
Oct
19
comment 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
comment 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
comment 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
comment 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
comment 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
comment 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
comment 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
comment 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
12
comment exponential columns and rows in Linear programming
Edit the question, trying not to make it longer (if it gets too long, people are less likely to read it) and explain what you want to know there.