rafaelm

Zagreb, Croatia

16h
revised Is highest weight of a module necessarilly dominant?
added 28 characters in body
16h
revised Is highest weight of a module necessarilly dominant?
deleted 17 characters in body
17h
asked Is highest weight of a module necessarilly dominant?
Jul
2
awarded Curious
Jun
30
comment Is it true if $[LL] = L$ then $L$ is a semisimple Lie algebra?
No. Same question has been asked on MathOverflow: mathoverflow.net/questions/60498/lie-algebra-semisimple
Apr
16
accepted Action of $SO_n$ on $\mathbb{S}^{n-1}$ induces fibre bundle.
Apr
15
comment Action of $SO_n$ on $\mathbb{S}^{n-1}$ induces fibre bundle.
Please don't delete your answer, because someday I might be able to fill all the details myself.
Apr
13
comment Action of $SO_n$ on $\mathbb{S}^{n-1}$ induces fibre bundle.
Could you please provide reference for the fact that compact Lie group and closed subgroup induce principal bundle?
Apr
13
comment Action of $SO_n$ on $\mathbb{S}^{n-1}$ induces fibre bundle.
And I couldn't figure out how to connect story from last paragraph with my desired trivializations. How can I write $\alpha^+(A) = (A e_n,$ some $SO_{n-1}$ matrix$)$?
Apr
13
comment Action of $SO_n$ on $\mathbb{S}^{n-1}$ induces fibre bundle.
Moreover, I don't see how is $p^{-1}(U^\pm)$ identified via $\Pi^\pm$, and how is $(\Sigma^\pm(x),e_1,\ldots,e_n)$ its orthonormal frame. How to think about $p^{-1}(U^+)$, besides matrices which don't have $e_n$ as fixed point?
Apr
13
comment Action of $SO_n$ on $\mathbb{S}^{n-1}$ induces fibre bundle.
Thank you. I don't understand the last part, which starts with "More explicitly" :). I see that conformality of stereographich projection implies that Jacobian $D\Sigma^\pm$ is orthogonal operator, but why does it have positive determinant?
Apr
13
asked Action of $SO_n$ on $\mathbb{S}^{n-1}$ induces fibre bundle.
Mar
23
awarded Nice Question
Mar
3
comment Cocompletion and functor categories
So, your claim $[A^{op},B^{op}] \cong [A,B]$ is false in general.
Mar
3
comment Cocompletion and functor categories
Just write down the definition of natural transformation and of opposite category. :)
Mar
3
comment Cocompletion and functor categories
If we have functor $F \colon A \to B$, then we have functor $F^{op} \colon A^{op} \to B^{op}$ given with the same action as $F$. But if we have natural transformation $\alpha \colon F \to G$ in $[A,B]$, then the corresponding natural transformation in $[A^{op},B^{op}]$ goes in the opposite direction: $\alpha^{op} \colon G^{op} \to F^{op}$.
Mar
2
comment Cocompletion and functor categories
In general, $[A,B]^{op} \cong [A^{op},B^{op}]$.
Feb
16
awarded Nice Answer
Jan
10
revised Commutativity of the square diagram coming from an adjoint triple
added 1 characters in body
Jan
10
accepted Commutativity of the square diagram coming from an adjoint triple
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