bio | website | mathnet.ru/php/… |
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location | ||
age | ||
visits | member for | 2 years, 5 months |
seen | 2 hours ago | |
stats | profile views | 1,360 |
Jul 21 |
answered | Hahn-Banach restricted to a pre-dual |
Jul 21 |
accepted | “Uniformly (co)well-powered” categories? |
Jul 21 |
comment |
“Uniformly (co)well-powered” categories?
Andreas, which reading do you recommend? |
Jul 20 |
revised |
“Uniformly (co)well-powered” categories?
deleted 466 characters in body |
Jul 20 |
comment |
“Uniformly (co)well-powered” categories?
Yes, I found this result about well-ordering of the class of all sets in A.Levy's book. Andreas, I think you should put what you wrote in form of answer, and I will close this question. I'll make corrections in the question so that your answer will be natural. Or I do not know what to do with this. Eric, I still do not understand what you wrote. That is interesting, please, give me a reference. |
Jul 20 |
comment |
“Uniformly (co)well-powered” categories?
@Eric and Andreas, I need a reference. This is the first time I hear this. The class of all sets can be well-ordered? And this trick with sending X to the set of all skeleta... Is there a text to look how people do this? |
Jul 20 |
comment |
“Uniformly (co)well-powered” categories?
To apply the axiom of choice (i.e. to choose this map $X\mapsto S_X$) we must already have a map which to every $X$ assigns a set which contains some $S$ as element. But we do not have such a map. Or you mean something else? |
Jul 20 |
comment |
“Uniformly (co)well-powered” categories?
@Eric Wolfsey: this must be a misunderstanding... I don't see how this follows from global choice. |
Jul 20 |
asked | “Uniformly (co)well-powered” categories? |
Jul 19 |
comment |
Tangent vectors on the algebra of trigonometric polynomials
@Qiaochu Yuan: Yes, I do not see the difference. To moderators: I flagged the previous comment by chance, excuse me, that was not intentionally! |
Jul 19 |
comment |
Tangent vectors on the algebra of trigonometric polynomials
For me it is not obvious that ${\sf Trig}(G)$ coincides with the algebra of pokynomials on the complexification of $G$. Why is this so? |
Jul 18 |
comment |
Tangent vectors on the algebra of trigonometric polynomials
Yes, this is simple... However you should have some experience in algebraic groups to feel free in this field. Can you recommend a textbook for reading and references? |
Jul 18 |
accepted | Tangent vectors on the algebra of trigonometric polynomials |
Jul 18 |
comment |
Tangent vectors on the algebra of trigonometric polynomials
Ah, OK! Thank you anyway! |
Jul 18 |
comment |
Tangent vectors on the algebra of trigonometric polynomials
Aakumadula, could you give a reference, please? |
Jul 18 |
comment |
Tangent vectors on the algebra of trigonometric polynomials
Peter, I am not sure I understood. Are you saying that the problem appears only for non-compact $G$, or for all $G$? |
Jul 18 |
revised |
Tangent vectors on the algebra of trigonometric polynomials
added 32 characters in body |
Jul 18 |
comment |
Tangent vectors on the algebra of trigonometric polynomials
Mariano, I have made the corrections, now ${\sf Trig}(G)$ is defined. |
Jul 18 |
revised |
Tangent vectors on the algebra of trigonometric polynomials
added 335 characters in body |
Jul 18 |
asked | Tangent vectors on the algebra of trigonometric polynomials |