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Feb 24, 2019 at 21:06 vote accept Ali Taghavi
Feb 22, 2019 at 12:29 answer added emiliocba timeline score: 3
Feb 22, 2019 at 10:09 comment added YCor @AndySanders Of course, bi-invariant is sufficient but not necessary. For instance it's obviously true when $N$ has dimension $1$, or when $N$ is normal. In particular, for $G$ non-abelian of dimension 2, it's true for all $N$.
Feb 21, 2019 at 23:30 comment added Francois Ziegler @AndySanders Not necessarily, as they can be isometric via some map other than $\operatorname{Ad}_g:n\mapsto gng^{-1}$. (E.g. the identity, when $N=G$.)
Feb 21, 2019 at 21:38 comment added Andy Sanders It seems you would want a bi-invariant metric for this to be true...
Feb 21, 2019 at 20:52 answer added Francois Ziegler timeline score: 7
Feb 21, 2019 at 10:25 history edited Ali Taghavi CC BY-SA 4.0
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Feb 21, 2019 at 9:52 history asked Ali Taghavi CC BY-SA 4.0