Timeline for General Linear Group as a Direct Product?

7 events

when toggle format	what		by	license	comment
Oct 31, 2015 at 17:20	history	edited	Drew Armstrong	CC BY-SA 3.0	deleted 6 characters in body
Oct 30, 2015 at 23:55	comment	added	YCor		@JulianRosen thanks, you're right. My claim in the comment is not correct (only one implication holds); this is fixed in the answer below.
Oct 30, 2015 at 23:54	answer	added	YCor		timeline score: 10
Oct 30, 2015 at 22:54	comment	added	Julian Rosen		@YCor The square roots of unity are a direct summand of $\mathbb{R}^\times$: $\mathbb{R}^\times = \{\pm 1\}\oplus \mathbb{R}^\times_{>0}$, but there can be no retraction $\varphi:GL_2(\mathbb{R})\to SL_2(\mathbb{R})$ because $A:=[-1,0;0,1]$ and $B:=[0,1;1,0]$ both square to $1$, every element of $SL_2(\mathbb{R})$ which squares to $1$ is central, so we would need $1=[\varphi(A),\varphi(B)]=\varphi([A,B])=\varphi([-1,0;0,-1])=[-1,0;0,-1]$. This same argument seems to work for any $K$ of characteristic not $2$.
Oct 30, 2015 at 17:58	comment	added	Jay Taylor		Just an observation that if $K$ has characteristic $p> 0$ and $n$ is a power of $p$ then $GL_n(K) = SL_n(K) \times Z(GL_n(K))$.
Oct 30, 2015 at 17:53	comment	added	YCor		It's if and only if the set of $n$-th roots of unity in $K^$ has a direct summand in $K^$.
Oct 30, 2015 at 16:50	history	asked	Drew Armstrong	CC BY-SA 3.0