3,774 reputation
722
bio website
location
age
visits member for 2 years
seen yesterday
stats profile views 868

olivier dot begassat dot cours at gmail dot com


Jun
10
comment Motivation of stable homotopy theory
Wow, this answer is so brilliant, I had to favorite the question so that I would find it, and come back to your answer, later.
Jun
9
comment My son's Sum of Some is beautiful! But what is the proof or explanation?
This question and its answers have generated an absurd amount of votes.
Jun
4
comment Let A and B be $n \times n$ real matrices with same minimal polynomial.
Why downvote? The Question shows some work, and is otherwise perfectly fine.
May
31
comment Cech homology groups
@PeterTamaroff read the edit history, what happened?
May
30
comment The “set” of equivalence classes of things.
@AsafKaragila So in order to talk about the set of representatives for extensions of any pair of modules I need to invoke the axiom of choice, right?
May
30
comment The “set” of equivalence classes of things.
@ZhenLin So is this an example where things fail : two sets with fibers of cardinality $2$ over $Y$, well-orderable, that cannot be in bijection?
May
30
comment The “set” of equivalence classes of things.
Ok, so the existence of sections depends on some choice principle for families of given size whose members are of constant cardinality, but in absence of those, can we have two sets $X,X'$ that fiber over $Y$, with all fibers of equal cardinality, yet $X$ and $X'$ having different cardinalities? And if so, can we at least say that the cardinalities that occur are bounded from above?
May
30
comment The “set” of equivalence classes of things.
A free abelian group is always projective, but it's the converse that requires choice (and apparantly is equivalent to it).
May
30
comment The “set” of equivalence classes of things.
Thanks. There is a little typo when you talk about the projective and free abelian groups.
May
28
comment Geodesic and Euler - Lagrange equation
I think $L:TM\to M$ is defined as $X\mapsto |x|^2=g(X,X)$.
May
27
comment Simple non-closed geodesic.
I know almost nothing about the subject, but I think the answer is yes, and it is related to laminations.
May
26
comment Submit papers: arxiv or vixra?
I saw this one too ^^ you just can't make this stuff up!
May
25
comment Submit papers: arxiv or vixra?
@julien just type Goldbach in the search field on the top right of the page :D
May
24
comment Questions about epimorphisms and projectives in functor categories
@ZhenLin Thanks!
May
24
comment Questions about epimorphisms and projectives in functor categories
@ZhenLin what is this left adjoint please?
May
24
comment Questions about epimorphisms and projectives in functor categories
could you please give me some details? I don't understand what you mean by "Each representable presheaf $\mathcal I(i,−)$ is free because it occurs as the image of $1$ under the left adjoint of the evaluation-at-$i$ functor."
May
23
comment Questions about epimorphisms and projectives in functor categories
Thank you for your answer. There are several things I don't understand. I assume by representable presheaf you mean a presheaf of the form $\mathrm{Hom}_I(i,-):I\to\mathbf{Set}$ for some $i\in I$. I don't understand what it means for this to be free, or why they are projective. Also, what is that canonical construction as a quotient of a coproduct?
May
23
comment Why is the dimension of $SL(2,\mathbb{H})$ equal to $15$?
The dimesion statement in my comment is false.
May
23
comment Why is the dimension of $SL(2,\mathbb{H})$ equal to $15$?
There is a notion of a determinant for matrices with coefficients in a noncommutative ring due to Dieudonné. Also the quaternions can be seen as a subring of $M_{2\times 2}(\Bbb C)$, so that $M_{2\times 2}(\Bbb H)\subset M_{4\times 4}(\Bbb C)$ as a subring. That gives you at least one possible definition of a $\Bbb C$-valued determinant. You could search "quaternionic determinant" on google, and "Dieudonné determinant". As for dimension, I would expect that the above determinant is a submersion onto $\Bbb C^*$ so that $SL(2,\Bbb H)$ should have complex dimension 15.
May
22
comment Are the quotient groups in a composition sequence necessarily subgroups?
Why do you add the requirement $G/N$ simple?