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Sep 4, 2018 at 19:08 history edited wonderich CC BY-SA 4.0
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Sep 4, 2018 at 19:03 history edited wonderich CC BY-SA 4.0
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Sep 4, 2018 at 18:56 history edited wonderich CC BY-SA 4.0
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Sep 1, 2018 at 3:48 comment added wonderich Thanks everyone. Maybe we need to know the statement how $\pi_0(H)=\mathbb{Z}_N \times \mathbb{Z}_2$ acts on $\pi_1(G)=\mathbb{Z}$? (if this is correct)
Sep 1, 2018 at 3:44 history edited wonderich CC BY-SA 4.0
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Sep 1, 2018 at 3:39 comment added annie marie cœur Dont we have $π1(G)=\mathbb{Z}$ and $\pi_0(H)=\mathbb{Z}_N \times \mathbb{Z}_2$?
Sep 1, 2018 at 3:38 comment added mme I clearly misread some of the notation. I apologize and have deleted my comment.
Sep 1, 2018 at 3:26 comment added annie marie cœur The $H$ looks complicated, it is not $G= (G/H) \times H$ for sure.
Sep 1, 2018 at 3:26 comment added annie marie cœur @Mike Miller, "Note here that H acts trivially on π1G" why is that true?
Sep 1, 2018 at 3:20 comment added annie marie cœur Sorry, isn't $$\pi_0(H)=\pi_0(\frac{SU(N)_{A,B}}{\mathbb{Z}_N}\times \mathbb{Z}_2)={\mathbb{Z}_N}\times \mathbb{Z}_2?$$ Should some of your answer be corrected?
Aug 31, 2018 at 21:59 history edited ThiKu CC BY-SA 4.0
Discrete group has trivial homotopy groups except in degree 0.
Aug 31, 2018 at 21:40 history asked wonderich CC BY-SA 4.0