rafaelm

Zagreb, Croatia

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
Jan
10
comment Commutativity of the square diagram coming from an adjoint triple
Thanks, I corrected it.
Jan
10
revised Commutativity of the square diagram coming from an adjoint triple
edited body
Jan
10
asked Commutativity of the square diagram coming from an adjoint triple
Dec
4
comment The twisted cubic is an affine variety.
You are welcome.
Dec
4
comment The twisted cubic is an affine variety.
Here is a reference: books.google.hr/… (page 173).
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